Tilted Convexity and an Exact Offline-Online Separation

Tilted Convexity and an Exact Offline–Online Separation Mohammad Pedramfar Mila – Quebec AI Institute / McGill University Rajiv Khanna Purdue University

arXiv:2611.03472v1 [cs.LG] 2 Nov 2026

Christopher John Quinn Iowa State University Vaneet Aggarwal Purdue University

Abstract Convexity has three equivalent descriptions: support by affine functions, monotonicity of the subdifferential, and increasing secant slopes. We weaken all three by one distortion with parameter γ ∈ (0, 1], which multiplies positive changes by γ and divides negative ones by γ. The resulting classes are the convex functions at γ = 1 and, for γ < 1, are distinct and nested, with the parameter dropping by at most a square along the chain. For proper lower semicontinuous functions in these classes, the effective domain is convex, local minimizers are global, and the function is locally Lipschitz on the relative interior of its domain. The classes separate offline from online optimization over compact convex sets with efficient projection, with access to function values and bounded support vectors. A smooth member of any class can be minimized to additive accuracy ε with poly(1/ε) first-order queries and projections. Online, with nonpositive losses and α–regret measured √ against α times the best fixed point, a polynomial-time algorithm achieves γ–regret O( T ) on the largest class. For every fixed, effectively specified γ ∈ (0, 1) and every α ∈ (γ, 1], no randomized polynomial-time algorithm has sublinear expected α–regret unless NP = RP, already for smooth concave polynomial losses in the smallest class, each minimized by one linear optimization. Thus, unless NP = RP, γ is the exact online coefficient of each class, while the offline coefficient is one. Keywords: tilted convexity; tilted monotonicity; generalized convexity; abstract convexity; online optimization; regret minimization; computational hardness

i

Contents 1 Introduction 1.1 Main Result 1: The Three Classes: Relations and Structure . . . . . . . . . . . . . . 1.2 Main Result 2: Offline–Online Separation . . . . . . . . . . . . . . . . . . . . . . . .

1 2 5

2 Related Work

7

3 Background and notation

9

4 The Function Classes 4.1 The gradient inequality: γ–tilted convexity . . . . . . . . . . . . . . . . . . . . . . . 4.2 Monotone generalized gradients: γ–tilted monotonicity . . . . . . . . . . . . . . . . . 4.3 Secant slopes: γ–tilted secant convexity . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Envelope description of the secant class . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Endpoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 14 14 15 15

5 Structural Properties 5.1 Properties of γ–tilted convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Properties of γ–tilted monotone subdifferentials . . . . . . . . . . . . . . . . . . . . . 5.3 Properties of γ–tilted secant convex functions . . . . . . . . . . . . . . . . . . . . . . 5.4 Envelopes of tilted affine functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Separations and negative results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Relations between the classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18 18 22 25 29 29 33

6 Examples 6.1 Examples from monotone generalized gradients . . . . . . . . . . . . . . . . . . . . . 6.2 Examples from function values and envelopes . . . . . . . . . . . . . . . . . . . . . . 6.3 Increasing transformations of convex costs . . . . . . . . . . . . . . . . . . . . . . . .

34 35 35 37

7 Polynomial-Time Offline Optimization

37

8 γ-Approximate Online Optimization 8.1 Online model and approximate regret . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Lower-linearizability and the reduction to linear losses . . . . . . . . . . . . . . . . . 8.3 A comparison vector with coefficient γ . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Computing the comparison vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Regret bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 40 40 41 44 46

9 Hardness Beyond γ: A Common Framework 9.1 The gap property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 A scalar-profile criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Profiles with a sharp monotonicity parameter . . . . . . . . . . . . . . . . . . . . . .

47 47 48 49

10 Conclusion

51

A Useful lemmas

52

B Approximate γ–tilted convexity B.1 Local boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Local Lipschitz continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.3 Dense approximate γ–tilted convexity implies exact γ–tilted convexity over the relative interior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.4 Continuity along lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

54 55 56 58 59

C Proofs for Section 5 61 C.1 Proof of Theorem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

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C.2 Proof of Theorem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 C.3 Proofs of Theorems 6 and 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 C.4 Proof of Theorem 8: Alternative First-order Characterization . . . . . . . . . . . . . 63 C.5 Proof of Theorem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 C.6 Proof of Theorem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 C.7 Proof of Theorem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 C.8 Proof of Theorem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 C.9 Proof of Theorem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 C.10 Proof of Theorem 14: Alternative First-order Characterization . . . . . . . . . . . . 80 C.11 Proof of Theorem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 C.12 Proof of Theorem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 C.13 Proof of Theorem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 C.14 Proof of Theorem 18: Characterizations of γ–tilted secant convexity . . . . . . . . . 91 C.15 Proof of Theorem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 C.16 Proof of Theorem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 C.17 Proof of Theorem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 C.18 Proofs of the separating examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 C.19 Proof of Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 C.20 Proof of Theorem 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 C.21 Proof of Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 D Proofs for Section 6 115 D.1 Proof of Lemma 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 D.2 Proof of Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 D.3 Proof of Example 9 and Examples 10 and 17 . . . . . . . . . . . . . . . . . . . . . . 116 D.4 Proofs of Example 11 and Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . 119 D.5 Proofs of Example 13 and Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 120 D.6 Proof of the exactness claim in Example 15 . . . . . . . . . . . . . . . . . . . . . . . 121 D.7 Proof of Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 E Proofs for Section 8 122 E.1 Proof of Lemma 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 E.2 Proof of Theorem 25(i) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 E.3 Proof of Theorem 25(ii): inexact oracle . . . . . . . . . . . . . . . . . . . . . . . . . . 125 E.4 Proof of Theorem 25(iii): bit model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 E.5 Proof of Corollary 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 F Proofs for Section 9 130 F.1 Proof of Theorem 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 F.2 Proof of Theorem 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 F.3 Proofs of Lemmas 5 and 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 F.4 Proofs of Theorem 28 and Corollary 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 136 References

140

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1. Introduction Convexity makes offline and online optimization efficient for many problems, but many objectives in learning, statistics, and control are not convex [GL13, ZSS19, BRAK22]. Since convexity has numerous equivalent characterizations, it can be relaxed in several ways, and extensive work has sought relaxations that preserve properties enabling efficient optimization, such as a unique minimizer or monotone slopes [CM09, HKS06]. A convex function can be described in three equivalent ways: by the gradient inequality f (y) − f (x) ≥ ⟨v, y − x⟩ for a gradient v at x; by monotonicity of the gradient, ⟨u − v, y − x⟩ ≥ 0 for gradients v at x and u at y, which uses only gradients; and by Jensen’s inequality f (λy+(1−λ)z) ≤ λf (y) + (1 − λ)f (z), or equivalently by the increase of the slopes of secants, which uses only function values and is equivalent to f being a pointwise supremum of affine functions. The classical relaxations of convexity keep the sign in these inequalities but not their size. Quasiconvexity replaces the right side of Jensen’s inequality by max{f (y), f (z)}, so that every sublevel set is convex, and pseudoconvexity requires only that ⟨∇f (x), y − x⟩ ≥ 0 implies f (y) ≥ f (x) [Man65]; each of the two also has an equivalent form in terms of gradients alone [HKS06]. Both relaxations suffice for offline tractability: a quasiconvex function can be minimized by bisection on its value [BV04] and a pseudoconvex function by cutting planes [BTN01]. We study a common multiplicative distortion of the three convexity conditions.1 For γ ∈ (0, 1], put ψγ (s) = γs when s ≥ 0 and ψγ (s) = s/γ when s < 0. At points where the values are finite, the three conditions are
γ–tilted convexity: f (y) − f (x) ≥ ψγ ⟨v, y − x⟩ , (1)
γ–tilted monotonicity: ⟨u, y − x⟩ ≥ ψγ ⟨v, y − x⟩ , (2)   f (y) − f (x) f (x) − f (z) γ–tilted secant convexity: ≥ ψγ , (3) ∥y − x∥ ∥x − z∥ where in (1) and (2) ∂f is the Clarke subdifferential, v ∈ ∂f (x), and u ∈ ∂f (y), and in (3) y ̸= z lie in dom(f ), λ ∈ (0, 1), and x = λy + (1 − λ)z is required to lie in dom(f ) as well, so that the two sides compare the slopes of the secants on the two sides of
x. Condition (3) is equivalent to the Jensen-type inequality f (λy
+ (1 − λ)z)
≤ Mγ,λ f (y), f (z) (Proposition 2), where the mean Mγ,λ (a, b) = λa + γ(1 − λ)b / λ + γ(1 − λ) for a ≥ b, and symmetrically for a < b, reduces the weight of the smaller value by a factor γ; it is the upper expectile of the two-point distribution at level 1/(1 + γ) [NP87, BKMRG14]. For differentiable functions, (1) is exactly the equal-parameter case γn = γp = γ of tilted γ γ convexity introduced by Martı́nez-Rubio [MR22, Eq. (3)]. We write F γ , Fmono , and Fsec for tilted convex functions, functions with tilted monotone subdifferentials, and γ–tilted secant convex γ γ functions, respectively (Definitions 3, 4, and 6). To the best of our knowledge, Fmono and Fsec have not been previously proposed. Our contribution is the proper lower semicontinuous theory, the sharp relations among these formulations, and the exact coefficient that polynomial-time online algorithms can achieve on these classes. At γ = 1 all three classes coincide with the proper lower semicontinuous convex functions, and their unions over positive parameters agree and form a strict subclass of the pseudoconvex functions. The support and operator formulations have distinct 0+ and 0 endpoints, corresponding to pseudoconvexity and quasiconvexity for radially continuous functions, that is, functions whose restriction to every closed segment between points of their domain is continuous. The secant 1 While we view all three as tilted forms of convexity, we call (1) as “γ-tilted convexity” to be consistent with [MR22].

1

−2π

π

−π

t 2π

Figure 1: The function f (t) = 12 t2 − 2 cos t (solid) and the convex function 21 t2 − 2 (dashed). On the shaded intervals f ′′ (t) = 1 + 2 cos t < 0, so f is not convex. Its only stationary point is the minimizer t√= 0, and its√subdifferential is γ–tilted monotone, so it lies in all three classes, with γ = (4π − 3 3)/(2π + 3 3) (Example 10). formulation reaches quasiconvexity directly at γ = 0. Theorem A and Figure 2 give the precise inclusions. In (1), a predicted increase ⟨v, y − x⟩ > 0 must be realized up to a factor γ, and an actual decrease may exceed the predicted one by at most a factor 1/γ. Since ψγ preserves signs, a point at which 0 is a generalized gradient is still a global minimizer. Additive relaxations lose this property: −t2 on (−1, 1) becomes convex after t2 is added, which makes it weakly convex in the sense of Davis and Drusvyatskiy [DD19], yet its stationary point 0 is a maximizer. The multiplicative condition is also unchanged when f is multiplied by a positive constant. Each class contains objectives from applications (Section 6). Concave polynomials of one coordinate with a large enough linear term, radial functions, robust regression losses such as the Cauchy and Geman–McClure losses [BR96, Bar19], and distorted convex functions, which include worst-margin costs on a norm ball [CV95], satisfy the monotonicity condition (2) and therefore lie in all three classes. Ratios of a cost to a positive output, as in fractional programming [CC62, Din67, Sch76], including risk per unit of expected excess return in portfolio selection [Sha66] and energy per transmitted bit in wireless networks [ZJ15], satisfy (3) with a parameter at which they need not satisfy (2), and so do concave quadratics such as −t2 − ct on [0, 1] with c > 0. For example, f (t) = 12 t2 − 2 cos t on R (Figure 1) is not convex, since f ′′ (t) = 1 + 2 cos t < 0 on (2π/3, 4π/3), but its √only stationary point is its minimizer. It is radial and satisfies (2) with √ γ = (4π − 3 3)/(2π + 3 3), so it lies in all three classes with this parameter (Example 10). Overview of Main Results: Our first main result is that the three classes are nested within a γ squared parameter of each other, with Fsec equal to the class of proper pointwise suprema of the √ functions y 7→ c + ψ γ (⟨z, y − w⟩), and that they keep the main structural properties of convex functions, without any differentiability assumption (Theorems A and B). Our second main result separates the (offline) minimization of one function from the (online) minimization of a sequence of functions revealed one at a time: one smooth function from any of the classes can be minimized to any accuracy with polynomially many queries, while the largest factor α for which a polynomialtime online algorithm can guarantee sublinear regret against α times the best fixed point is exactly γ, unless NP = RP (Theorem C). 1.1 Main Result 1: The Three Classes: Relations and Structure How the classes relate. The three characterizations are equivalent for convex functions (γ = 1) γ γ γ but not after relaxation (γ < 1); there Fmono ⊆ Fsec ⊆ F γ , and passing from F γ to Fmono costs 2

γ at most a square of the parameter. The class Fsec also has a description by envelopes: it consists exactly of the proper pointwise suprema of the functions y 7→ c + ψ√γ (⟨z, y − w⟩), which we call tilted affine functions, just as lower semicontinuous convex functions are the suprema of affine ones; it is therefore a class of abstract convex functions in the sense of Rubinov [Rub13]. The positiveparameter classes form a strict hierarchy below the pseudoconvex class; the separately defined 0+ and 0 endpoints give pseudoconvexity and quasiconvexity under radial continuity. Pointwise γ γ suprema stay in F γ and in Fsec , while the maximum of two members of Fmono can leave that class. Sums leave all three classes, and sums of their members plus one convex function reach every smooth function whose curvature is bounded below, so minimizing each function individually, as offline methods do, does not by itself solve the online comparator problem (Section 1.2).

Theorem A (Relations and representation; informal). Let 0 < γ ′ < γ < 1. (i) The classes are nested, F

√

γ

γ γ ⊆ Fmono ⊆ Fsec ⊆ Fγ,

each inclusion is strict, and some functions lose the full square of the parameter between γ F γ and Fmono (Corollary 8, Examples 1–3, and Figure 2). The concave quadratic −t2 − ct γ γ and F γ but only c/(c + 2) in Fmono on [0, 1], with c > 0, has parameter c/(c + 1) in Fsec (Example 4). (ii) A proper lower semicontinuous function is γ–tilted secant convex exactly when it is a nonempty pointwise supremum of tilted affine functions y 7→ c + ψ√γ (⟨z, y − w⟩) and exactly when its values satisfy an inequality at every finite set of points (Theorems 21 and 18). +

(iii) Let F 0 be defined by (1) with ψγ replaced by its pointwise limit as γ ↓ 0, and let F 0 be defined in the same way except that the inequality is not required when ⟨v, y − x⟩ = 0 (Section 4.5). θ , or F θ for every θ, then θ , Fsec If C θ stands for Fmono [ [ ′ + C1 ⊊ Cγ ⊊ Cγ ⊊ Cθ = F θ ⊊ F 0 ⊊ F 0, 0<θ≤1

0<θ≤1 +

where C 1 consists of the convex functions and, among radially continuous functions, F 0 and F 0 consist of the pseudoconvex and the quasiconvex functions (Theorem 23). (iv) Sums of γ–tilted convex functions need not be γ–tilted convex (Example 5), and every twice continuously differentiable function on [0, 1]n whose Hessian is bounded below is the sum of a convex function and n concave quadratics that lie in all three classes with parameter γ (Theorem 22). γ (v) A proper pointwise supremum of members of F γ lies in F γ , and the same holds for Fsec . A γ γ γ2 maximum of two members of Fmono can lie outside Fmono , but it always lies in Fmono , and the square is sharp (Theorems 12 and 17, Example 7, and Section 5.5.4).

A convex-like theory. Convex functions have convex sublevel sets and local minima that are global, they are locally Lipschitz on the relative interiors of their effective domains, and convexity can be checked along lines and is preserved under pointwise limits. Theorem B establishes the corresponding properties for all three classes, including nonsmooth extended-valued functions; maxima of tilted affine functions and distorted worst-margin costs are nonsmooth examples (Section 6). γ γ Theorem B (Structure; informal). Let γ ∈ (0, 1] and let f belong to Fsec , F γ , or Fmono .

3

Quasiconvex = F 0 +

Pseudoconvex = F 0

S

0<θ≤1 F

θ

2

γ γ F γ Fsec Fmono

γ Fmono F

√

γ

Convex = F 1

Figure 2: Nested classes for 0 < γ < 1; each class contains the classes drawn inside it. The √ γ γ γ2 three families interleave as F γ ⊆ Fmono ⊆ Fsec ⊆ F γ ⊆ Fmono (Corollary 8), and the pattern repeats at each squared parameter. Every inclusion is strict (Examples 1–3 and Theorem 23). The identifications with the pseudoconvex and the quasiconvex functions hold for radially continuous functions. (i) The set where f is finite is convex, and f is quasiconvex. Every local minimizer is a global minimizer, and x is a minimizer exactly when 0 is a generalized gradient at x (Theorems 4 and 5, Corollary 7). (ii) The function f is locally Lipschitz on the relative interior of its effective domain, and its values there determine f on the whole domain (Theorems 6 and 10). (iii) Membership in each class can be checked on the restrictions of f to lines and is preserved under restriction to closed convex sets, and if members of a class converge pointwise to finite values on a dense subset of a convex set, the lower semicontinuous envelope of the limit lies in the class (Theorems 9, 11, 15, 16, 17, and 20, and Corollaries 3 and 5). Membership in F γ does not depend on which standard notion of generalized gradient is used in (1) (Theorem 8). We prove (i) and (ii) for F γ only, and they hold for the other two classes by inclusion (Theorem A). Techniques for the structural results. The proofs rest on a mean-value inequality for nonsmooth functions (Appendix A), which turns the pointwise condition (1) into statements about whole segments, as Jensen’s inequality does for convex functions. The weights in the resulting segment inequality depend on the sign of a directional derivative, and the argument along segments absorbs this dependence. This argument yields the convexity of the domain and both conversions between (1) and (2): (2) implies (1) with the same parameter, and (1) implies (2) with the sharp loss of γ 2 . Together with Rademacher’s theorem, which gives gradients on a dense subset of the relative interior, and a bound of the horizon subdifferential at boundary points by the normal cone of the domain (Appendix C.4), it also shows that all standard notions of generalized gradient define the same class F γ . Regularity and stability under limits come from a weaker notion, approximate γ–tilted convexity (Appendix B), in which the vector in (1) may depend on the desired accuracy and is required only on a dense set of points; such approximate supports force local Lipschitz continuity and become exact on the relative interior. Because these Lipschitz bounds are uniform along a locally bounded sequence, members that converge on a dense set converge uniformly on compact subsets of the relative interior, and the lower semicontinuous envelope of the limit is again in the √ γ class. For Fsec , an equivalent form of (3) applies a distortion with parameter γ to differences of values and extends to any finite number of points, which gives a membership test that uses function values alone (Theorem 18). 4

1.2 Main Result 2: Offline–Online Separation One function versus a sequence. Convexity makes optimization tasks tractable: offline, gradient or cutting-plane methods minimize a single convex function to any accuracy; online, where a learner picks xt in a convex set K in each round √ t = 1, . . . , T and an adversary then reveals a convex loss ft , gradient descent stays within O( T ) regret with respect to the best fixed point in hindsight [Zin03, PA26b]. For P nonconvex losses the two tasks can differ, because the online benchmark minimizes the sum t ft and by Theorem A (iv) such sums can encode general smooth nonconvex minimization even when each ft is easy to minimize. After ft is revealed, our learner may evaluate ft and its generalized gradients at points of its choice, and we measure its performance by approximate regret. Losses are nonpositive, so multiplying the benchmark by α ≤ 1 weakens it, and for α ∈ (0, 1] the α–regret is Regα (T ) =

T X

ft (xt ) − α min u∈K

t=1

T X

ft (u),

t=1

which is the usual regret when α = 1. A coefficient α is achievable for a class of losses if a possibly randomized algorithm that is given the horizon T and runs in time polynomial in the input size and T keeps the expected Regα (T ) below P T 1−ρ for every horizon and every loss sequence from the class, with ρ > 0 a constant and P polynomial in the input size, which bounds the descriptions of K and of a single loss but not the horizon (Section 8.1). The online coefficient is the supremum of the achievable coefficients, with value zero if none is positive; we show that it is attained at γ. The offline coefficient of a class is 1 when a single member can be minimized to additive accuracy ε with a number of oracle and projection calls polynomial in 1/ε and the numerical parameters. We write RP for the class of decision problems solved by randomized polynomial-time algorithms with one-sided error [AB09]. For α < 1 the α–regret PTis not invariant under adding constants to the losses: replacing ft by ft − Ut subtracts (1 − α) t=1 Ut from it, so the sign convention is part of the model. Losses with known upper bounds Ut ≥ supK ft can be shifted to nonpositive losses in the same class, and the choice Ut = supK ft gives the most demanding version. Both the algorithm (Section 8.1) and the lower bound (Section 9) use nonpositive losses, and the hard losses take values in [−1, 0]. Offline minimization to additive accuracy does not depend on this convention. Because the classes are nested, an algorithm for the largest class F γ runs on all three, a lower γ holds for all three, and Theorem C provides both. bound for the smallest class Fmono Theorem C (Offline–online separation; informal). Fix an effectively specified 0 < γ < 1, and let γ γ C be one of Fmono , Fsec , and F γ . Use the value-and-support and projection model of Section 7 on a nonempty compact convex set of diameter at most D, with G–Lipschitz functions and, at each query point x, a selected support vector, that is, a vector v satisfying (1) for every feasible y, of norm at most G. (i) Offline. An L–smooth member of C can be minimized to additive accuracy ε with O(1 + 2 LD2 /(γ p ε)) gradient evaluations and projections by projected gradient descent, and with −1 e e hides logarithmic facO(γ LD2 /ε) first-order queries by an accelerated method, where O tors (Corollary 9, based on [MR26]). (ii) Online, achievability. For a continuous nonpositive loss ft in the largest class F γ and a point xt ∈ K, some vector gt with ∥gt ∥ ≤ G satisfies ft (xt ) − γft (y) ≤ ⟨gt , xt − y⟩ 5

(y ∈ K)

(Theorem 24). A polynomial-time procedure computes such a vector up to an additive error in this inequality that can be made inverse-polynomially small, and passing the computed vectors √ to a linear online learner gives O(DG T ) static γ–regret. The algorithm makes polynomially many value, support, and projection queries to each revealed loss, and its running time is polynomial under the conventions of Section 8.1 (Theorem 25, Corollary 10). It applies to all three classes. (iii) Online, hardness. For every fixed α ∈ (γ, 1], a randomized polynomial-time learner with expected static α–regret at most P (N )T 1−ρ , for a fixed polynomial P and constant ρ > 0, on γ all sequences from Fmono would imply NP = RP. This holds for explicit [−1, 0]–valued, 1– Lipschitz, 1–smooth concave polynomials on products of simplices. Their degree and number of relevant coordinates depend only on (γ, α); their Hessians have rank at most one, and one linear optimization minimizes each loss exactly. Independent uniform samples from a polynomial-size family suffice. For F γ alone, concave quadratics suffice (Theorem 28). Each class therefore has online coefficient exactly γ, unless NP = RP, under the stated model and normalization, and this already holds for its smooth members, whose offline coefficient is one (Corollary 11, Table 1). For the hard losses of Theorem C(iii), minimizing a single loss is also polynomial in running time, since one linear optimization minimizes each loss exactly, so the lower bound concerns their combination. Convex optimization has a (trivial) approximation coefficient α = 1 in both settings; matching α < 1 offline and online coefficients are also known in several submodular settings (Section 2). We determine the exact nontrivial gap for three nested classes. To the best of our knowledge, the classes studied here are the first with an online coefficient α that is determined exactly (α = γ) and lies strictly between 0 and the offline coefficient (i.e. 1), and this coefficient can be any γ ∈ (0, 1). Techniques for the separation. Offline, smooth members are tilted convex and therefore quasar-convex [MR22, MR26], so known first-order methods apply (Section 7). Online, the comparison must hold against every point, because the best fixed point in hindsight rarely minimizes any single loss, and the vector of Theorem C(ii) provides it. Losses with such vectors are called lower-linearizable, and the reduction of Pedramfar and Aggarwal [PA24] passes the vectors to any online algorithm for linear losses. We obtain the vector by rescaling a generalized gradient v at x to tv with t ∈ [γ, 1]. Each comparator y forces t above or below a threshold according to the sign of ⟨v, y − x⟩. The thresholds are compatible because two comparators of opposite signs can be paired through the point between them at which ⟨v, · − x⟩ vanishes, and an interpolation inequality for −f applies on their segment (Lemma 3). Since the vector is defined by infinitely many constraints, we find t by bisection and test each candidate with a projected method that searches for a violated comparator, which takes polynomially many value, support, and projection queries (Theorem 25). The number of calls has no inverse-power dependence on γ. For the lower bound, a class has the online gap property if SAT reduces to finite families of losses whose average equals −Q up to scale, for a score Q that reaches 1 on satisfiable instances and stays below some τ < α on unsatisfiable ones (Definition 14). Running a learner with coefficient α on random members of such a family decides SAT with one-sided error, so no class with this property admits a polynomial-time algorithm with coefficient α unless NP = RP; this step uses no form of convexity (Theorem 26). We build the families from Label Cover, a constraint satisfaction problem for which the PCP theorem makes it hard to tell instances in which all constraints can be satisfied from instances in which only a small fraction can [MR10]. Each right vertex v and label b give a loss −H(zv,b ), where zv,b is the mass that a fractional labeling of the left vertices sends to b along the edges at v, and H is an increasing convex function with H(0) = 0 and H(1) = 1. The 6

linear part of H pays for labels whether or not they agree across edges, so on unsatisfiable instances the score is at most H ′ (0) plus a term that Label Cover makes small (Theorem 27). Membership γ of −H in Fmono forces H ′ (0) ≥ γH ′ (1) ≥ γ, and polynomial choices of H in the class bring H ′ (0) arbitrarily close to γ (Lemma 6), so the threshold of the reduction equals the parameter of the class. Contributions. • Three classes from one distortion. We apply one multiplicative distortion to the three standard characterizations of convexity (the support inequality; monotonicity of the subdifferential; and increasing secant slopes or, equivalently, Jensen’s inequality) and study the resulting classes for proper lower semicontinuous functions. The support version extends tilted convexity [MR22] from differentiable to nonsmooth extended-valued functions; the monotonicity and secant versions are new. • Convex-like structure without smoothness. For proper lower semicontinuous functions in each class, the domain is convex, local minimizers are global, the function is locally Lipschitz on the relative interior of its domain, and membership can be checked on lines and passes to limits (Theorem B). √

γ γ • Exact relations among the classes. The classes are nested, F γ ⊆ Fmono ⊆ Fγ, ⊆ Fsec every inclusion is strict, and some functions lose the full square of the parameter. The γ secant class Fsec consists exactly of the proper suprema of tilted affine functions, as the lower semicontinuous convex functions are the suprema of affine ones. None of the classes is closed under addition (Theorem A).

• An exact offline–online separation. A smooth function from any of the classes can be minimized with polynomially many queries by known methods for quasar-convex functions. Online, a polynomial-time algorithm achieves coefficient γ on the largest class, and unless NP = RP no polynomial-time algorithm achieves a larger coefficient on the smallest class (Theorem C and Table 1).

2. Related Work Generalized convexity and generalized monotonicity. Quasiconvexity, pseudoconvexity, and invexity each keep part of the global geometry of convex functions [Man65, Man75, Cra81, CM09]. Their nonsmooth versions are characterized through generalized monotonicity of subdifferentials, with quasiconvexity corresponding to quasimonotonicity [ACL94] and pseudoconvexity to pseudomonotonicity [PQ97]; see also [HKS06, HS24, SD07, RW98]. These conditions record the sign of a change in value but not its size. Our three classes give a quantitative, sign-dependent version of each standard characterization of convexity, they reduce to convexity at γ = 1, and F γ reaches pseudoconvexity and quasiconvexity at its endpoints 0+ and 0. The size bound is what yields local Lipschitz continuity without differentiability, a hierarchy that is strict at every parameter, and sharp conversions between the classes with a loss of at most a square in the parameter. Convexity with respect to means, and abstract convexity. Replacing the arithmetic mean in Jensen’s inequality by another mean is a classical way to define convexity classes: f is convex with respect to means P and Q if f (P (x, y)) ≤ Q(f (x), f (y)) [Nic03, AVV07, MP15]. In this language, the Jensen form of γ–tilted secant convexity uses the weighted arithmetic mean on the arguments and the mean Mγ,λ on the values, where Mγ,λ is an upper expectile [NP87, BKMRG14]. 7

Table 1: Offline–online separation. Offline coefficient 1 means that a single objective can be minimized to arbitrary additive accuracy with polynomially many queries. Online coefficients refer to α–regret bounded by a polynomial times T 1−ρ , achieved in polynomial time, for losses with values in [−B, 0] and B, D, and G polynomial in the input size. The convex and positive-parameter rows use Section 7. The endpoint offline rows assume the stronger feasibility or cut access specified in their cells, with the usual geometric and encoding conditions. Online lower bounds use explicit smooth individual losses and hold unless NP = RP. Class

Offline coefficient and individual minimization

Convex

1; classical cutting-plane or projected methods [BTN01] 1 for L–smooth members; projected gradient descent uses O(1 + LD2 /(γ 2 ε)) calls, and acceleration uses (36) queries (Section 7). 1 under an efficient valid-cut oracle and the usual cutting-plane assumptions [BTN01] 1 under efficient weak feasibility for convex sublevel sets and the usual geometric and encoding assumptions [BV04]

γ γ Fmono , Fsec , Fγ, 0<γ<1

Pseudoconvex

Quasiconvex

Online coefficient √ 1, with O(DG T ) regret [Zin03] γ is achievable on F γ ; every fixed α > γ implies NP = RP already on γ Fmono (Corollary 10, Theorem 28) Every fixed α > 0 implies NP = RP (Corollary 11) Every fixed α > 0 implies NP = RP (Corollary 11)

We characterize the resulting function class, including its finite Jensen inequalities and its envelope representation. Abstract convexity represents a function as the upper envelope of a chosen family of elementary functions, generalizing the description of a proper lower semicontinuous convex function as a supremum of affine functions, and closure of the resulting classes under suprema is γ standard [Rub13, DR90, Pen13]. The two viewpoints meet in Fsec : the upper envelopes of the 2 tilted affine functions y 7→ c + ψγ (⟨z, y − w⟩) are exactly the γ –tilted secant convex functions and are characterized by an inequality on finitely many function values, just as the proper upper envelopes of affine functions are exactly the proper lower semicontinuous functions that satisfy Jensen’s inequality. Additive and multiplicative relaxations. Weak convexity and semiconvexity bound negative curvature by adding a quadratic; prox-regularity gives a local quadratic support property [HUL13, DD19, PN15]. Our distortion rescales slopes instead, so the classes are unchanged when f is multiplied by a positive constant and stationary points remain minimizers, whereas weak convexity with a large modulus contains every twice differentiable function whose Hessian is bounded below. The tilted condition is distinct from multiplicative, or geometric, convexity, which compares positive function values through geometric means [Nic00]. Tilted convexity and quasar-convexity. For differentiable functions, the support condition (1) is tilted convexity with γn = γp = γ, introduced by Martı́nez-Rubio [MR22] in the analysis of Riemannian acceleration: a geodesically convex function on a subset of hyperbolic or spherical space, mapped to Euclidean space by a geodesic map, satisfies it with constants that depend on the radius of the subset. That work notes that the condition is stronger than quasar-convexity and uses 8

it for accelerated first-order methods on convex sets. We develop the nonsmooth extended-valued theory, sharp relationships with the secant and operator formulations, and an exact polynomialtime online coefficient. Tilted-convex functions are γ–quasar-convex [MR22, MR26], a condition introduced by Hardt, Ma, and Recht and by Hinder, Sidford, and Sohoni [HMR18, HSS20] and studied online by Pun et al. [PS24]. Since quasar-convex functions need not be quasiconvex, they form a strictly larger class; the rates for its constrained form in [MR26] give our offline bounds. Offline and online approximation coefficients. Several established settings have matching offline and online coefficients under their respective access and constraint models. Convex losses have coefficient 1 in both settings [Zin03]. For monotone submodular maximization, the greedy coefficient 1−1/e is attained online [SG08], and offline greedy algorithms of a general form convert into online algorithms with the same coefficient [NGW+ 23]. For monotone continuous DR-submodular maximization, the coefficient is 1 − 1/e both offline and online [CHHK18, ZCHK19, PNQA24], and Mualem and Feldman [MF23] showed that the two coefficients also coincide for non-monotone maximization over general convex sets. Linearizable formulations transfer offline guarantees of this kind to the online setting across a broad range of up-concave objectives [PA24, PA26a]. For the three classes of this paper the coefficients differ, 1 offline and exactly γ online for every fixed, effectively specified γ ∈ (0, 1), unless NP = RP. Online optimization beyond convexity. For convex losses, projected online gradient descent √ achieves O( T ) static regret with first-order full-information feedback [Zin03, Haz16]. The classical one-point bandit method of Flaxman, Kalai, and McMahan has a T 3/4 rate, whereas two-point feedback permits square-root-type regret [FKM05, ADX10]. Separate guarantees concern moving comparators or every time interval [ZLZ18, HS09, DGSS15]. For nonconvex losses, one line of work replaces the global benchmark by a local stationarity criterion [HSZ17] or by per-round minimizers [GLZ18]. Another keeps the global benchmark but assumes an oracle that minimizes the sum of past losses plus a linear perturbation [AGH19, SN20]. We keep the global benchmark and allow access only to individual losses, through their values and generalized gradients. Under this access, Theorem C shows that the benchmark can be approached up to the factor γ in polynomial time and, unless NP = RP, not beyond it, for each of the three classes. Linearizable optimization. Pedramfar and Aggarwal [PA24] defined linearizable and quadratizable optimization, in which a comparison inequality that holds for all comparators is passed to an online linear learner. Later work added uniform wrappers, decentralized and projection-free variants, and weakly up-concave classes [PQA25, LPA25, PA26a]. We use this reduction without change; what is new is the existence, for every continuous nonpositive γ–tilted convex loss, of a comparison vector with coefficient exactly γ, and its computation with polynomially many queries. Computational limits of online learning. Hazan and Koren [HK16] established computational limits on online learning with access to an offline optimization oracle. Our lower bound has a different source: each hard loss is minimized by one linear optimization, and the obstruction lies in their sum, which encodes a Label Cover instance whose gap follows from the PCP theorem γ [MR10, Mos15]. The hard losses lie in the smallest class Fmono , and the threshold of the reduction equals its parameter.

3. Background and notation We use R to denote the set of extended real numbers R∪{−∞, +∞}. For f : Rd → R, we define the (effective) domain of f as the set dom(f ) := {x ∈ Rd | f (x) ∈ R}. We say f is proper if dom(f ) ̸= ∅

9

and f > −∞. 2 Given a non-empty set X ⊆ Rd and a function f : X → R, its proper extension fe : Rd → R equals f on X and +∞ on Rd \ X , and it is proper if and only if f is proper. With the indicator δX , which is 0 on X and +∞ elsewhere, we also write fe = f + δX . Subdifferentials of f , and membership of f in the function classes of Section 4, always refer to fe. For a convex set K, we write aff(K) for its affine hull and ri(K) for its relative interior, the interior of K relative to aff(K). If f is differentiable at x ∈ ri(K) as a function on aff(K), its gradient with respect to aff(K) is the vector in the linear subspace parallel to aff(K) that represents this derivative. We say f : X → R is lower semi-continuous (lsc) at x ∈ X if f (x) ≤ lim inf f (y). y∈X ,y→x

The function f is locally lsc at x if it is lsc on an open neighborhood of x. The function f is lsc at x ∈ X if and only if its proper extension is lsc at x. 3 If f is locally lsc at x, its proper extension need not be locally lsc at x. 4 Given f : X → R, let fe equal f on X and +∞ outside X . Its lower semicontinuous envelope is lsc(f )(x) := lim inf fe(y) = sup inf{fe(y) : ∥y − x∥ < r}, y→x

r>0

where the lower limit is nonpunctured and inf ∅ = +∞. It is the largest lower semicontinuous function below fe, or equivalently the largest lower semicontinuous function g satisfying g ≤ f on X , so lsc(f ) = fe whenever fe is lower semicontinuous. For a proper f , the envelope equals +∞ outside dom(f ), but it need not be proper. For example, f (x) = 1/x for x ̸= 0 and f (0) = 0 is proper, whereas lsc(f )(0) = −∞. Definition 1. A function f : Rd → R is radially continuous, or continuous on lines, if, for all x, y ∈ dom(f ), the restriction of f to the closed line segment [x, y] is continuous. Given a sequence xk and a point y in X , we say xk converges f -attentively to y, denoted by f

xk − → y, when xk → y and f (xk ) → f (y). For sets A and B, a multi-valued operator F : A ⇒ B is a function from A to the power set of B, so that F (a) ⊆ B for each a ∈ A. The domain of F is dom(F ) := {a ∈ A | F (a) ̸= ∅}. We use the following generalized derivatives [RW98]. Definition 2. Let f : Rd → R. The Fréchet subdifferential of f at x ∈ dom(f ) is       f (y) − f (x) − ⟨v, y − x⟩ ˆ (x) := v ∈ Rd ∂f lim inf ≥ 0 . y→x   ∥y − x∥   y∈dom(f ),y̸=x The limiting (Mordukhovich) subdifferential of f at x ∈ dom(f ) is   f ˆ (xk ) with vk → v , ∂f (x) := v ∈ Rd ∃ xk ∈ dom(f ), xk − → x, ∃ vk ∈ ∂f 2

This convention suits minimization; for maximization one instead requires that −f be proper. The extension may fail to be lsc outside X . For example, the function f : (0, ∞) → R defined by f (x) = 0 has fe(x) = 0 for x > 0 and fe(x) = ∞ for x ≤ 0, so both f and fe are lsc on (0, ∞), but fe is not lsc at 0. 4 For example, f : {(0, 0)} ∪ (0, 1] × [0, 1] → R defined identically zero on its domain is locally lsc on its domain, but its proper extension is not locally lsc at (0, 0). 3

10

and the horizon limiting (Mordukhovich) subdifferential of f at x ∈ dom(f ) is   f   ∃ xk ∈ dom f, ∃ λk > 0, xk − → x, λk → 0, ∞ d := . ∂ f (x) v∈R   b (xk ) with λk vk → v ∃ vk ∈ ∂f Finally, if f is proper and locally lsc at x ∈ dom(f ), then the Clarke–Rockafellar subdifferential (or simply Clarke subdifferential) of f at x is 5 ∂f (x) := conv(∂f (x)) + conv(∂ ∞ f (x)). Proposition 1. Let f, g : Rd → R be proper and x ∈ dom(f ) ∩ dom(g). Then the following are true. (i) If f is locally lsc at x, then ˆ (x). ∂f (x) ⊇ ∂f (x) ⊇ ∂f f ˆ (xi ) is non-empty. Moreover, there is a sequence xi − → x where ∂f

(ii) If f is locally Lipschitz at x, then ∂f (x) is non-empty. Specifically, if n o ∇f (x) := v ∈ Rd ∃ xk ∈ dom(f ), xk → x, ∇f (xk ) exists , ∇f (xk ) → v , then ∇f (x) is a non-empty compact set and ∇f (x) ⊆ ∂f (x) ⊆ ∂f (x) = conv(∇f (x)). ˆ (x) = {∇f (x)}. (iii) If f is differentiable at x, then ∂f (x) ⊇ ∂f ˆ (x) = (iv) If f is continuously differentiable on a neighborhood of x, then ∂f (x) = ∂f (x) = ∂f {∇f (x)}. (v) If g is continuously differentiable on a neighborhood of x, then ˆ + g)(x) = ∂f ˆ (x) + ∇g(x). ∂(f The same identity holds for the limiting subdifferential. If f is also locally lsc at x, it holds for the Clarke–Rockafellar subdifferential as well. In particular, let K ⊆ Rd be nonempty, closed, and convex, let x ∈ K, and let NK (x) = {n ∈ Rd : ⟨n, y − x⟩ ≤ 0 for all y ∈ K} be the normal cone of K at x. If f is the indicator δK , then f + g is the proper extension ge of g|K , and ∂e g (x) = ∇g(x) + NK (x). Since 0 ∈ NK (x), this gives ∇g(x) ∈ ∂e g (x). ˆ (x), the constant sequences xk := x The inclusions in (i) follow from the definitions: for v ∈ ∂f f ˆ (xk ) with and vk := v show that v ∈ ∂f (x). If ∂f (x) ̸= ∅, then there are xk − → x and vk ∈ ∂f ∞ vk → v for some v, and λk := 1/k gives λk vk → 0, so 0 ∈ ∂ f (x) and ∂f (x) ⊆ ∂f (x)+0 ⊆ ∂f (x). The second claim in (i) is Corollary 8.10 in [RW98], and parts (ii)–(v) are Theorems 9.13 and 9.61 and Exercise 8.8 in [RW98]. For the indicator in (v), the limiting subdifferential ∂δK (x) and the horizon subdifferential ∂ ∞ δK (x) both equal NK (x) [RW98, Chapters 6 and 8]. Since NK (x) is a closed convex cone, the definition of the Clarke subdifferential gives ∂δK (x) = NK (x) + NK (x) = NK (x), and the sum rule in (v) for ∂ gives the formula for ∂e g (x). 5

When f is proper and locally lsc at x ∈ dom(f ), this formula agrees with the classical definition of the Clarke– Rockafellar subdifferential; see Theorem 3.57 of [Mor06].

11

The tilt function ψγ .

Classical convexity is characterized by the affine support inequality f (y) − f (x) ≥ ⟨y − x, v⟩,

v ∈ ∂f (x).

Scaling the left side of a scalar inequality a ≤ b by a factor γ relaxes it for only one sign of a: for γ ∈ (0, 1] and a ≥ 0, the inequality γa ≤ b is weaker than a ≤ b, but for a < 0 it is stronger. To relax the inequality for both signs, for γ ∈ (0, 1] we define the functions ψγ and ψ γ over extended real numbers with ψγ (z) := γ sign(z) z = min{γz, γ −1 z},

ψ γ (z) := γ − sign(z) z = max{γz, γ −1 z}.

For γ = 0 we set    0 ψ0 (z) := undefined   −∞

   ∞ ψ 0 (z) := undefined   0

if z > 0 if z = 0 , if z < 0

if z > 0 if z = 0 if z < 0

Lemma 1. Let a, b ∈ R, γ, γ ′ ∈ [0, 1]. Then the following are true. 1. If γ > 0, then both ψγ and ψ γ are continuous, monotone increasing, bijective, and positively homogeneous. 2. Whenever ψγ (a) and ψ γ (a) are defined, ψγ (a) ≤ a ≤ ψ γ (a). 3. ψ γ (a) = −ψγ (−a) whenever both sides are defined. 4. ψγ (ψγ ′ (a)) = ψγγ ′ (a) whenever both sides are defined. 5. If γ > 0, then ψγ (a) + ψγ (b) ≤ ψγ (a + b) and ψ γ (a) + ψ γ (b) ≥ ψ γ (a + b). 6. If γ > 0, then ψ γ (ψγ (a)) = ψγ (ψ γ (a)) = a. 7. If γ ′ < γ and ψγ ′ (a) is defined, then ψγ ′ (a) ≤ ψγ (a). Proof. For γ ∈ (0, 1] the definitions read ( γz, z ≥ 0, ψγ (z) = z/γ, z < 0,

( z/γ, ψ γ (z) = γz,

z ≥ 0, z < 0,

(4)

and ψγ (±∞) = ψ γ (±∞) = ±∞. In particular, for γ > 0 the numbers ψγ (z), ψ γ (z), and z have the same sign. For γ = 0, both scalar expressions are defined exactly when their argument is nonzero. The calculations below use finite arguments; the same identities extend to infinite arguments whenever all expressions and extended-real sums are defined. (1) By (4), ψγ consists of two linear pieces with positive slopes γ and 1/γ that both vanish at 0. Hence ψγ is continuous and strictly increasing, and it maps [0, ∞) onto [0, ∞) and (−∞, 0) onto (−∞, 0), so it is bijective. For t > 0 we have sign(tz) = sign(z), so ψγ (tz) = γ sign(z) tz = t ψγ (z). The same arguments apply to ψ γ . (2) Let γ > 0. If a ≥ 0, then (4) gives ψγ (a) = γa ≤ a ≤ a/γ = ψ γ (a). If a < 0, then it gives ψγ (a) = a/γ ≤ a ≤ γa = ψ γ (a). Let γ = 0, so that a ̸= 0. If a > 0, then ψ0 (a) = 0 < a < ∞ = ψ 0 (a), and if a < 0, then ψ0 (a) = −∞ < a < 0 = ψ 0 (a). 12

(3) Let γ > 0. Since sign(−a) = − sign(a), we get −ψγ (−a) = −γ − sign(a) (−a) = γ − sign(a) a = ψ γ (a). Let γ = 0. If a > 0, then −ψ0 (−a) = −(−∞) = ∞ = ψ 0 (a), and if a < 0, then −ψ0 (−a) = 0 = ψ 0 (a). (4) Let γ, γ ′ > 0. Since ψγ ′ (a) has the sign of a, we get ψγ (ψγ ′ (a)) = γ sign(a) γ ′ sign(a) a = ′ (γγ )sign(a) a = ψγγ ′ (a). Now let γγ ′ = 0. Then ψγγ ′ = ψ0 , and both sides being defined forces a ̸= 0. Let a < 0, so that ψ0 (a) = −∞. If γ ′ = 0, then ψγ ′ (a) = −∞ and ψγ (−∞) = −∞. If γ = 0, then ψγ ′ (a) < 0 and hence ψ0 (ψγ ′ (a)) = −∞. If a > 0 and γ ′ = 0, then ψγ ′ (a) = 0, at which ψγ is defined only when γ > 0, and then ψγ (0) = 0 = ψ0 (a). If a > 0 and γ = 0 < γ ′ , then ψ0 (γ ′ a) = 0 = ψ0 (a). (5) Let γ > 0. By definition ψγ (z) = min{γz, z/γ}. For real numbers p, q, r, s we have min{p, q} + min{r, s} ≤ min{p + r, q + s}, since the left-hand side is at most p + r and at most q + s. Applying this with p = γa, q = a/γ, r = γb, s = b/γ gives ψγ (a) + ψγ (b) ≤ min{γ(a + b), (a + b)/γ} = ψγ (a + b). The inequality for ψ γ (z) = max{γz, z/γ} follows in the same way from max{p, q} + max{r, s} ≥ max{p + r, q + s}. (6) Let γ > 0. Since ψγ (a) has the sign of a, we get ψ γ (ψγ (a)) = γ − sign(a) γ sign(a) a = a. The identity ψγ (ψ γ (a)) = a follows in the same way. (7) Let γ ′ < γ, so that γ > 0. If γ ′ > 0 and a ≥ 0, then ψγ ′ (a) = γ ′ a ≤ γa = ψγ (a). If γ ′ > 0 and a < 0, then 1/γ ′ > 1/γ gives ψγ ′ (a) = a/γ ′ < a/γ = ψγ (a). If γ ′ = 0, then a ̸= 0; for a > 0 we have ψ0 (a) = 0 < γa = ψγ (a), and for a < 0 we have ψ0 (a) = −∞ < ψγ (a).

4. The Function Classes This section applies the distortion ψγ of Section 3 to the gradient inequality, to monotonicity of the subdifferential, and to the increase of secant slopes, which is equivalent to Jensen’s inequality (Sections 4.1–4.3). Section 4.4 describes the secant class as the envelopes of tilted affine functions, and Section 4.5 describes the endpoints γ = 1, 0+ , and 0. Throughout, K ⊆ Rd denotes a convex set. 4.1 The gradient inequality: γ–tilted convexity Definition 3 (γ–tilted convexity). Let f : Rd → R be a proper lsc function, and let F : Rd ⇒ Rd be a multi-valued operator. For γ ∈ [0, 1], we say f is γ–tilted convex with respect to F if, for all x, y ∈ dom(f ) and v ∈ F (x), f (y) − f (x) ≥ ψγ (⟨y − x, v⟩)

(5)

whenever the right-hand side is well-defined. We say f is γ–tilted convex if it is γ–tilted convex with respect to ∂f . 6 For any such function we define γ ∗ (f ) := sup{γ ∈ [0, 1] : f is γ–tilted convex}, with sup ∅ := 0. 6

Similarly, we say a usc function f : K → R is γ–tilted concave with respect to F if −f is γ–tilted convex with respect to −F . Equivalently, for all x, y ∈ dom(f ) and v ∈ F (x), f (y) − f (x) ≤ ψ γ (⟨y − x, v⟩), whenever the right-hand side is well-defined. We say f is γ–tilted concave if −f is γ–tilted convex, that is, if f is γ–tilted concave with respect to −∂(−f ), which in general differs from ∂f .

13

For a differentiable function and v = ∇f (x), the support inequality is tilted convexity with constants γn = γp = γ [MR22, Eq. (3)]. On a closed convex feasible set, the Clarke subdifferential of a continuously differentiable ambient function adds normal vectors, whose inner products with feasible directions are nonpositive, so the identification persists. The definition above extends the condition to proper lsc extended-valued functions; convexity of the effective domain and equivalence of the subdifferential formulations are proved in Section 5. If F (x) ⊆ G(x) for every x and f is γ–tilted convex with respect to G, then f is γ–tilted convex with respect to F , so by Proposition 1(i) a γ–tilted convex function is also γ–tilted convex with ˆ . respect to ∂f and ∂f For γ ∈ [0, 1] we write F γ for the class of proper lsc γ–tilted convex functions f : Rd → R. By the convention of Section 3, a function on a set K lies in F γ when its proper extension does. 4.2 Monotone generalized gradients: γ–tilted monotonicity Definition 4 (γ–tilted monotone operator). For γ ∈ [0, 1], we say the multi-valued operator F : Rd ⇒ Rd is γ–tilted monotone if for all x, y ∈ dom F , x∗ ∈ F (x), and y∗ ∈ F (y), ∗

⟨y∗ , y − x⟩ ≥ ψγ (⟨x∗ , y − x⟩) = γ sign(⟨x ,y−x⟩) ⟨x∗ , y − x⟩,

(6)

whenever both sides are well-defined. γ Definition 5. For γ ∈ [0, 1], the class Fmono consists of the proper lsc functions f : Rd → R ∗ (f ) := sup{γ ∈ [0, 1] : such that ∂f is γ–tilted monotone. For a proper lsc function f , γmono ∂f is γ–tilted monotone}, with sup ∅ := 0.

4.3 Secant slopes: γ–tilted secant convexity A function is convex exactly when the slopes of its secants increase: if x lies strictly between z and y on a line, the slope of the secant over [x, y] is at least the slope of the secant over [z, x], and we apply ψγ to this latter slope. Definition 6 (γ–tilted secant convexity). Let γ ∈ [0, 1] and f : K → R ∪ {+∞}. We say that f is γ–tilted secant convex if, for all y ̸= z in dom(f ) and λ ∈ (0, 1), the point x := λy + (1 − λ)z lies in dom(f ) and the secant slopes s− :=

f (x) − f (z) , ∥x − z∥

s+ :=

f (y) − f (x) ∥y − x∥

(7)

satisfy s+ ≥ ψγ (s− )

(8)

whenever ψγ (s− ) is defined. For any such function we define ∗ γsec (f ) := sup{γ ∈ [0, 1] : f is γ–tilted secant convex},

sup ∅ := 0.

For γ = 1, (8) is the increase of secant slopes. In general, (8) applies the distortion ψγ to secant slopes, as (5) applies it to supports and (6) to subdifferentials. Requiring ψγ (s− ) to be defined excludes only the case γ = 0 and s− = 0. An equivalent form in terms of function values, which weakens Jensen’s inequality, is given in Section 5.3 (Proposition 2). ∗ ∗ are attained whenever they are positive. Each of the The three suprema γ ∗ , γmono , and γsec three conditions (5), (6), and (8) with parameter γ implies the same condition with every smaller parameter, by Lemma 1(7). For fixed z, the value ψγ (z) = γ sign(z) z depends continuously on 14

γ ∈ (0, 1], so a condition that holds for every γ < γ0 with γ0 > 0 also holds at γ0 . Hence, if ∗ γ ∗ (f ) > 0, then f is γ–tilted convex exactly for γ ∈ [0, γ ∗ (f )], and the same holds for γmono and ∗ γsec . γ For γ ∈ [0, 1] we write Fsec for the class of proper lsc γ–tilted secant convex functions f : Rd → R. 4.4 Envelope description of the secant class Proper lower semicontinuous convex functions have a fourth description: they are the pointwise suprema of affine functions. Applying the same distortion to the generators gives the tilted affine functions y 7→ c + ψγ (⟨y − w, z⟩) and their suprema, the Hγ –convex functions defined below. This produces no new class: Theorem 21 shows that the proper Hγ –convex functions are exactly the γ2 members of Fsec . The envelope form makes closure under suprema immediate, and it produces the nonsmooth examples of Section 6.2. We use the following standard notion of abstract convexity. Definition 7. Let X be a set and let H be a family of functions h : X → R. A function f : X → R is called H–convex if there exists a non-empty set U ⊆ H such that f (x) = sup h(x),

∀x ∈ X.

h∈U

Among proper functions on Rd , H–convexity for the family of affine functions is exactly lower semicontinuous convexity. Definition 8. Fix γ ∈ (0, 1]. Define Hγ := {y 7→ c + ψγ (⟨y − w, z⟩) : Rd → R | c ∈ R, z, w ∈ Rd }. A function f : Rd → R is thus Hγ –convex if f (x) = suph∈U h(x) for all x ∈ Rd and some non-empty U ⊆ Hγ . Each element of Hγ is piecewise affine, with gradient γz on one side of the hyperplane {y : ⟨y − w, z⟩ = 0} and gradient z/γ on the other. The family H1 consists of the affine functions, so its proper suprema are exactly the proper lower semicontinuous convex functions. The parameter is squared because the two slopes of a member of Hγ along z differ by the factor γ 2 , and Example 15 γ shows that γ 2 is then the exact secant parameter. Equivalently, Fsec consists of the proper suprema of members of H√γ . 4.5 Endpoints At γ = 1, the support, monotonicity, and secant conditions of Sections 4.1–4.3 all characterize convexity, and H1 is the family of affine functions; at γ = 0, the secant condition characterizes quasiconvexity. For the support and monotonicity conditions, retaining or omitting the inequality at a zero inner product gives the distinct endpoints 0+ and 0, which under radial continuity characterize pseudoconvexity and quasiconvexity, respectively. The union of the positive-parameter classes is a strict subset of the pseudoconvex class (Theorem 23). Theorems 1, 2, and 3 restate the corresponding classical characterizations in terms of ψγ . For the secant condition, both endpoints follow from Definition 6. At γ = 1, ψ1 is the identity, so (8) is the increase of secant slopes, which characterizes convexity. At γ = 0, ψ0 (s− ) equals 0 for s− > 0 and −∞ for s− < 0, so (8) says that f (x) > f (z) implies f (y) ≥ f (x), that is, f (x) ≤ max{f (y), f (z)}. Together with the requirement x ∈ dom(f ), this is quasiconvexity. Theorem 1. Let f : Rd → R be a proper lsc function. Then the following are equivalent. 15

(i) For all x, y ∈ dom(f ) and v ∈ ∂f (x), we have f (y) − f (x) ≥ ⟨y − x, v⟩. (ii) The function f is 1–tilted convex. (iii) The function f is convex. (iv) The operator ∂f is 1–tilted monotone. (v) The operator ∂f is monotone. Proof. For every z ∈ R we have ψ1 (z) = 1sign(z) z = z, and this value is always defined. Hence (5) with γ = 1 and F = ∂f reads f (y) − f (x) ≥ ⟨y − x, v⟩,

x, y ∈ dom(f ), v ∈ ∂f (x),

which is condition (i), so (i) and (ii) are equivalent. Likewise, (6) with γ = 1 and F = ∂f reads ⟨y∗ , y − x⟩ ≥ ⟨x∗ , y − x⟩, that is, ⟨y∗ − x∗ , y − x⟩ ≥ 0,

x, y ∈ dom(∂f ), x∗ ∈ ∂f (x), y∗ ∈ ∂f (y),

which is monotonicity of ∂f , so (iv) and (v) are equivalent. For a proper lsc convex function, the Clarke subdifferential is the usual convex subdifferential, so (iii) implies (i). Adding the inequalities in (i) with the two points interchanged shows that (i) implies (v). Finally, (v) implies (iii) by the classical monotonicity characterization of convexity; see [DH99, Theorem 2.1(i)]. Definition 9 (Quasimonotonicity). A set-valued operator F : Rd ⇒ Rd is quasimonotone if, for all x, y ∈ dom(F ), x∗ ∈ F (x), and y∗ ∈ F (y), ⟨x∗ , y − x⟩ > 0

=⇒

⟨y∗ , y − x⟩ ≥ 0.

Theorem 2. Let f : Rd → R be a proper, lsc, and radially continuous function. Then the following are equivalent. (i) For all x, y ∈ dom(f ) and v ∈ ∂f (x) such that ⟨y − x, v⟩ > 0, we have f (y) ≥ f (x). (ii) The function f is 0–tilted convex. (iii) The function f is quasiconvex. (iv) The operator ∂f is 0–tilted monotone. (v) The operator ∂f is quasimonotone. Proof. By definition, ψ0 (z) = 0 for z > 0, ψ0 (z) = −∞ for z < 0, while ψ0 (0) is undefined. Fix x, y ∈ dom(f ) and v ∈ ∂f (x), and put z := ⟨y − x, v⟩. Condition (ii) imposes f (y) − f (x) ≥ ψ0 (z) only when z ̸= 0. If z < 0, this reads f (y)−f (x) ≥ −∞, which always holds because f (x), f (y) ∈ R. If z > 0, it reads f (y) − f (x) ≥ 0. Consequently, condition (ii) is equivalent to requiring that f (y) ≥ f (x) whenever ⟨y − x, v⟩ > 0, which is condition (i), so (i) and (ii) are equivalent. Similarly, fix x, y ∈ dom(∂f ), x∗ ∈ ∂f (x), and y∗ ∈ ∂f (y). With γ = 0, inequality (6) is imposed only when ⟨x∗ , y − x⟩ ̸= 0. It always holds when ⟨x∗ , y − x⟩ < 0, since its right-hand side is then −∞; when ⟨x∗ , y − x⟩ > 0, its right-hand side is 0. So Definition 4 with γ = 0 reduces to the implication ⟨x∗ , y − x⟩ > 0 =⇒ ⟨y∗ , y − x⟩ ≥ 0, 16

which is exactly quasimonotonicity, so (iv) and (v) are equivalent. Under radial continuity, (i) and (iii) are equivalent by [Aus98, Theorem 2.1 and Proposition 2.1]. The equivalence of (iii) and (v) is the classical quasimonotonicity characterization of quasiconvexity; see [DH99, Theorem 2.1(ii)]. Remark 1. Under stronger regularity, similar results hold for ∂f : if f is locally Lipschitz, then [SD07] shows that (i) ⇔ (ii) ⇔ (iii) ⇒ (iv) ⇔ (v). Definition 10 (Pseudoconvexity). A differentiable function f : K → R is pseudoconvex if, for all x, y ∈ K, ⟨y − x, ∇f (x)⟩ ≥ 0 =⇒ f (y) ≥ f (x). More generally [HKS06], a proper, lsc, and radially continuous function f : Rd → R is pseudoconvex if, for all x, y ∈ dom(f ) and v ∈ ∂f (x), we have ⟨y − x, v⟩ ≥ 0

=⇒

f (y) ≥ f (x).

If radial continuity is omitted, pseudoconvexity has several inequivalent definitions [HKS06]. Definition 11 (Pseudomonotonicity). A set-valued operator F : Rd ⇒ Rd is pseudomonotone if, for all x, y ∈ dom(F ), x∗ ∈ F (x), and y∗ ∈ F (y), ⟨x∗ , y − x⟩ ≥ 0

=⇒

⟨y∗ , y − x⟩ ≥ 0.

Define ( 0 ψ0+ (z) := −∞

if z ≥ 0; if z < 0.

The functions ψ0 and ψ0+ differ only at z = 0, where ψ0 is undefined. Definition 12. Let f : Rd → R be a proper lsc function. We say f is 0+ –tilted convex if, for all x, y ∈ dom(f ) and v ∈ ∂f (x), we have f (y) − f (x) ≥ ψ0+ (⟨y − x, v⟩). A set-valued operator F : Rd ⇒ Rd is 0+ –tilted monotone if, for all x, y ∈ dom(F ), x∗ ∈ F (x), and y∗ ∈ F (y), ⟨y∗ , y − x⟩ ≥ ψ0+ (⟨x∗ , y − x⟩). +

We write F 0 for the class of proper lsc 0+ –tilted convex functions. Theorem 3. Let f : Rd → R be a proper, lsc, and radially continuous function. Then the following are equivalent. (i) For all x, y ∈ dom(f ) and v ∈ ∂f (x) such that ⟨y − x, v⟩ ≥ 0, we have f (y) ≥ f (x). (ii) The function f is 0+ –tilted convex. (iii) The function f is pseudoconvex. (iv) The operator ∂f is 0+ –tilted monotone. 17

(v) The operator ∂f is pseudomonotone. Proof. Fix x, y ∈ dom(f ) and v ∈ ∂f (x), and put z := ⟨y −x, v⟩. If z < 0, then ψ0+ (z) = −∞, and the inequality f (y) − f (x) ≥ ψ0+ (z) always holds. If z ≥ 0, then ψ0+ (z) = 0, and the inequality reads f (y) − f (x) ≥ 0. Hence condition (ii) is equivalent to requiring that f (y) ≥ f (x) whenever ⟨y − x, v⟩ ≥ 0, which is condition (i), and condition (i) is the definition of pseudoconvexity of a proper, lsc, and radially continuous function, that is, condition (iii). Therefore, (i), (ii), and (iii) are equivalent. Similarly, fix x, y ∈ dom(∂f ), x∗ ∈ ∂f (x), and y∗ ∈ ∂f (y). The inequality ⟨y∗ , y − x⟩ ≥ ψ0+ (⟨x∗ , y−x⟩) always holds when ⟨x∗ , y−x⟩ < 0, and it reads ⟨y∗ , y−x⟩ ≥ 0 when ⟨x∗ , y−x⟩ ≥ 0. So 0+ –tilted monotonicity in Definition 12 is equivalent to the implication ⟨x∗ , y − x⟩ ≥ 0

=⇒

⟨y∗ , y − x⟩ ≥ 0,

which is exactly pseudomonotonicity, so (iv) and (v) are equivalent. The equivalence of (iii) and (v) under radial continuity is the classical pseudomonotonicity characterization of pseudoconvexity; see [DH99, Theorem 2.1(iii)]. Smaller parameters give weaker conditions: for all γ ′ , γ ∈ (0, 1] with γ ′ < γ, ψ0 (z) ≤ ψ0+ (z) ≤ ψγ ′ (z) ≤ ψγ (z)

(9)

for all z ∈ R at which both sides of each inequality are defined. The last inequality is Lemma 1(7). The middle one holds because ψ0+ (z) = 0 ≤ γ ′ z = ψγ ′ (z) for z ≥ 0 and ψ0+ (z) = −∞ for z < 0, and the first holds with equality because ψ0 and ψ0+ agree wherever ψ0 is defined. Hence an inequality f (y) − f (x) ≥ ψθ (⟨y − x, v⟩) implies the same inequality for every smaller parameter in (9), and ′

+

F 1 ⊆ F γ ⊆ F γ ⊆ F 0 ⊆ F 0,

0 < γ ′ < γ ≤ 1.

(10)

By Theorem 1, F 1 is the class of proper lsc convex functions. By Theorems 3 and 2, a proper, lsc, + and radially continuous function lies in F 0 exactly when it is pseudoconvex, and in F 0 exactly when it is quasiconvex. Theorem 23 in Section 5.6 shows that every inclusion in (10) is strict when γ < 1.

5. Structural Properties Section 5.1 shows that, like convex functions, members of F γ have convex domains, are locally Lipschitz on the relative interior of their domains, are radially continuous, and their local minimizers γ γ are global. Sections 5.2 and 5.3 compare Fmono and Fsec with F γ and transfer these properties γ2 to them, and Section 5.4 identifies the Hγ –convex functions with Fsec . Section 5.5 separates the classes and collects properties of convex functions that they lack, and Section 5.6 combines the comparisons into one chain of inclusions and orders each class strictly in its parameter. 5.1 Properties of γ–tilted convex functions 5.1.1 Convex domains and global minimizers Convexity of the domain is a consequence of γ–tilted convexity rather than an assumption. Theorem 4. Let f : Rd → R be proper, lsc, and γ–tilted convex for some γ ∈ (0, 1]. Then dom(f ) is convex and f is quasiconvex. 18

The proof is in Appendix C.1. For convex functions every local minimizer is a global minimizer; Theorem 5, proved in Appendix C.2, extends this to γ–tilted convex functions and characterizes ˆ (x). their minimizers by 0 ∈ ∂f Theorem 5. Let f : Rd → R be proper, lsc, and γ–tilted convex for some γ ∈ (0, 1]. Then, for every x ∈ dom(f ), the following are equivalent. (i) x is a global minimizer of f . (ii) x is a local minimizer of f . ˆ (x). (iii) 0 ∈ ∂f (iv) 0 ∈ ∂f (x). 5.1.2 Regularity The regularity and limit results of this section rest on one tool, proved in Appendix B. A proper lsc function f is approximately γ–tilted convex with respect to an operator F (Definition 15) if, for every ε > 0, each x ∈ dom(f )∩dom(F ) has a vector v ∈ F (x) with f (y)−f (x) ≥ ψγ (⟨v, y −x⟩)−ε for all y ∈ dom(f ); the vector may depend on ε. Every member of F γ has this property with ε = 0 and F the Fréchet subdifferential, whose domain is dense in dom(f ). When K := dom(f ) is convex and dom(F ) is dense in K, approximate supports already force local Lipschitz continuity on ri(K), with a constant that depends only on a local bound for f , and continuity along every line through ri(K). This gives Theorems 6 and 7 below and the characterizations of Theorem 8; since the constant is uniform along a locally bounded sequence, it also gives the line and limit results of Section 5.1.4. Convex functions are locally Lipschitz on the relative interior of their domains, and so are γ–tilted convex functions. Theorem 6. Let f : Rd → R be proper, lsc, and γ–tilted convex for some γ ∈ (0, 1], and set K := dom(f ). Then f |ri(K) is locally Lipschitz. The proof is in Appendix C.3. Such a function need not be locally Lipschitz at the boundary, √ √ even when it is continuous: the convex function f (x) = − x on [0, 1] has f ′ (x) = −1/(2 x) → −∞ as x → 0+ , so it is not locally Lipschitz at 0. A proper lsc convex function need not even be continuous on the boundary of its domain. For example, take f : R2 → R defined by  2   x /y if y > 0; f (x, y) := 0 if x = y = 0;   ∞ otherwise. This function is convex and lsc but not continuous at (0, 0). It is, however, radially continuous (Definition 1), as is every proper lsc convex function. Theorem 7 proves this property for γ–tilted convex functions on R, and Section 5.1.4 extends it to Rd . Theorem 7. Let f : R → R be proper, lsc, and γ–tilted convex for some γ ∈ (0, 1], and set K := dom(f ). Then f |K is continuous. The proof is in Appendix C.3. 19

5.1.3 Equivalent characterizations For γ ∈ (0, 1], the Clarke subdifferential in the definition of γ–tilted convexity can be replaced by the limiting or the Fréchet subdifferential, or by gradients at the points of the relative interior where they exist. Theorem 8. Let γ ∈ (0, 1] and let f : Rd → R be a proper lsc function with effective domain K := dom(f ). Then the following are equivalent. (i) f is γ–tilted convex. (ii) f is γ–tilted convex with respect to ∂f . ˆ . (iii) f is γ–tilted convex with respect to ∂f ˆ |ri(K) , i.e., we have (iv) K is convex and f is γ–tilted convex with respect to ∂f f (y) − f (x) ≥ ψγ (⟨v, y − x⟩), ˆ (x). for all y ∈ K, x ∈ ri(K), and v ∈ ∂f (v) K is convex, the set of points x ∈ ri(K) where ∇f (x) exists is dense in K and f is γ–tilted convex with respect to x 7→ {∇f (x)} : ri(K) ⇒ Rd , i.e., we have f (y) − f (x) ≥ ψγ (⟨∇f (x), y − x⟩), for all y ∈ K, x ∈ ri(K) where ∇f (x) exists. Here ∇f (x) is considered with respect to the affine hull of K. The proof is in Appendix C.4. For functions that are continuously differentiable on the relative interior, the equivalence of (i) and (v) gives the following test. Corollary 1. Let γ ∈ (0, 1] and let f : Rd → R be a proper, lsc, and radially continuous function with a convex effective domain K := dom(f ) such that f |ri(K) is continuously differentiable. Then f is γ–tilted convex if and only if f (y) − f (x) ≥ ψγ (⟨∇f (x), y − x⟩), for all x, y ∈ ri(K). Here ∇f (x) is considered with respect to the affine hull of K. Proof. Since K is non-empty and convex, ri(K) is non-empty and ri(K) = K (Theorems 6.2 and 6.3 in [Roc15]), so ri(K) is dense in K. By assumption, ∇f (x) exists at every x ∈ ri(K). Suppose that f is γ–tilted convex. By Theorem 8, (i) implies (v), so the inequality of the corollary holds for all x ∈ ri(K) and y ∈ K, and in particular for y ∈ ri(K). Conversely, suppose that the inequality holds for all x, y ∈ ri(K). Fix x ∈ ri(K) and y ∈ K, and for t ∈ (0, 1] put yt := (1 − t)y + tx. Then yt ∈ ri(K) by Theorem 6.1 in [Roc15], and yt − x = (1 − t)(y − x). The assumed inequality at the pair x, yt therefore reads
f (yt ) − f (x) ≥ ψγ (1 − t)⟨∇f (x), y − x⟩ , t ∈ (0, 1]. (11) As t → 0+ , the points yt on the segment [x, y] converge to y; since x, y ∈ K, radial continuity gives f (yt ) → f (y). The function ψγ is continuous by Lemma 1(1), so the right-hand side of (11) converges to ψγ (⟨∇f (x), y − x⟩). Passing to the limit in (11) gives f (y) − f (x) ≥ ψγ (⟨∇f (x), y − x⟩),

x ∈ ri(K), y ∈ K.

Together with the convexity of K and the density of ri(K) in K, this is condition (v) of Theorem 8, so f is γ–tilted convex. 20

5.1.4 Lines and limits A function is convex if and only if its restriction to every line is convex, and the same holds for γ–tilted convexity. Theorem 9. Let γ ∈ (0, 1] and let f : Rd → R be a proper lsc function. Then f is γ–tilted convex if and only if, for every line ℓ that intersects dom(f ), the restriction f |ℓ : ℓ → R is γ–tilted convex, where ℓ is regarded as a one-dimensional Euclidean space. The proof is in Appendix C.5. Together with Theorem 7, it gives radial continuity in Rd . Corollary 2. Let f : Rd → R be a proper lsc function that is γ–tilted convex for some γ ∈ (0, 1]. Then it is radially continuous. Proof. Fix x, y ∈ dom(f ). If x = y, the segment [x, y] is a single point and there is nothing to prove. Otherwise put d := (y − x)/∥y − x∥ and g(t) := f (x + td) for t ∈ R. The map t 7→ x + td is an isometry from R onto the line ℓ through x and y, so g is the restriction f |ℓ written in this coordinate. The function g is proper, since g(0) = f (x) ∈ R and g > −∞, and it is lsc as the composition of the lsc function f with a continuous map. By Theorem 9, g is γ–tilted convex, and hence, by Theorem 7, g is continuous on dom(g). The set dom(f ) is convex by Theorem 4, so [x, y] ⊆ dom(f ) and [0, ∥y − x∥] ⊆ dom(g). Hence t 7→ f (x + td) is continuous on [0, ∥y − x∥], i.e., f is continuous on [x, y]. The values of a proper lsc convex function on the relative interior of its domain determine its values on the closure. Theorem 10 extends this to γ–tilted convex functions, and its proof in Appendix C.6 uses continuity on lines for the finite boundary values and lower semicontinuity for the infinite ones. Theorem 10. Let γ ∈ (0, 1], and let f, g : Rd → R be proper, lsc, and γ–tilted convex. Set K := dom(f ). If f =g on ri(K), then on K.

f =g Moreover, we have

f = lsc f |ri(K) . If convex functions converge pointwise on a dense set, then they converge uniformly on compact subsets of the relative interior, and the limit is convex. Theorem 11, proved in Appendix C.7, gives the same for γ–tilted convex functions. Theorem 11. Let K ⊆ Rd be a non-empty convex set and γ ∈ (0, 1]. Let fn : Rd → R be proper, lsc, and γ–tilted convex functions that converge pointwise to finite values on a dense subset D ⊆ K. Then there is a continuous function f : ri(K) → R such that (i) fn → f uniformly on compact subsets of ri(K); (ii) the function lsc(f ) : Rd → R is γ–tilted convex. Moreover, if there is an lsc function g : K → R such that fn → g pointwise, then lsc(g) = lsc(f ) and so it is γ–tilted convex. 21

Applied to a constant sequence, Theorem 11 shows that restrictions to convex sets stay in the class. Corollary 3 (Stability under convex restrictions). Let γ ∈ (0, 1], let f : Rd → R be proper, lsc, and γ–tilted convex, and let C ⊆ Rd be a convex set such that C ∩ dom(f ) ̸= ∅. Then lsc(f |C ) : Rd → R is proper, lsc, and γ–tilted convex. In particular, if C is closed, then the function ( f (x), x ∈ C, fC (x) := +∞, x ∈ / C, is proper, lsc, and γ–tilted convex. Proof. The set K := C ∩ dom(f ) is non-empty, and it is convex because dom(f ) is convex by Theorem 4. The constant sequence fn := f converges pointwise on K to the finite-valued function g := f |K , which is lsc on K because for x ∈ K the lower semicontinuity of f gives lim inf y∈K,y→x f (y) ≥ lim inf y→x f (y) ≥ f (x). The final assertion of Theorem 11, applied with the dense set D := K, therefore shows that lsc(g) is γ–tilted convex. We next show that lsc(g) = lsc(f |C ). The effective domains of g and f |C are both K, so both envelopes equal +∞ outside K. For x ∈ K, the points y ∈ C \ K satisfy f (y) = +∞ and do not lower the lim inf, so lsc(f |C )(x) = lim inf f (y) = lim inf f (y) = lsc(g)(x). y∈C, y→x

y∈K, y→x

(12)

Hence lsc(f |C ) is γ–tilted convex, and it is lsc as an lsc envelope. It never takes the value −∞, because for x ∈ K, (12) and the lower semicontinuity of f give lsc(f |C )(x) ≥ lim inf y→x f (y) ≥ f (x) > −∞, while outside K it equals +∞. Since the envelope lies below f |C , which is finite on the non-empty set K, it is proper, and the first assertion follows. Let C be closed, so that fC is the proper extension of f |C . It is lsc at every x ∈ C, since fC ≥ f gives lim inf y→x fC (y) ≥ lim inf y→x f (y) ≥ f (x) = fC (x), and at every x ∈ / C, since Rd \ C is open and fC = +∞ on it. As noted after the definition of the lsc envelope, an lsc proper extension coincides with the lsc envelope, so fC = lsc(f |C ), and the second assertion follows from the first. 5.1.5 Closure under supremum A pointwise supremum of convex functions is convex, and the same holds in F γ . Theorem 12 (Closure under supremum). Let γ ∈ (0, 1], let {fi }i∈I ⊆ F γ be a non-empty family, and suppose that s := supi∈I fi is proper. Then s ∈ F γ . The proof in Appendix C.8 treats the maximum of two functions on each line with Lebourg’s mean value theorem, passes from finite maxima to countable suprema with Theorem 11, and replaces an arbitrary family by a countable subfamily with the same supremum. 5.2 Properties of γ–tilted monotone subdifferentials γ–tilted monotonicity of the subdifferential implies γ–tilted convexity, and γ–tilted convexity implies γ 2 –tilted monotonicity. The proof is in Appendix C.9.

22

Theorem 13. Assume f : Rd → R is a proper lsc function and F : Rd ⇒ Rd is a multi-valued operator where dom(F ) ⊆ dom(f ). Let γ ∈ (0, 1]. ˆ is γ–tilted monotone, then f is γ–tilted convex. (i) If ∂f (ii) If f is γ–tilted convex with respect to F , then F is γ 2 –tilted monotone. In particular, ∗ ∗ γmono (f ) ≤ γ ∗ (f ) ≤ (γmono (f ))1/2 . γ The class Fmono has characterizations analogous to Theorem 8, proved in Appendix C.10.

Theorem 14. Assume f : Rd → R is a proper lsc function with effective domain K := dom(f ) and γ ∈ (0, 1]. Then the following are equivalent. (i) ∂f is γ–tilted monotone. (ii) ∂f is γ–tilted monotone. ˆ is γ–tilted monotone. (iii) ∂f ˆ |ri(K) is γ–tilted (iv) K is convex, f is radially continuous, f |ri(K) is locally Lipschitz, and ∂f monotone. (v) K is convex, f is radially continuous, f |ri(K) is locally Lipschitz, and x 7→ {∇f (x)} : ri(K) ⇒ Rd is γ–tilted monotone, i.e., we have ⟨∇f (y), y − x⟩ ≥ ψγ (⟨∇f (x), y − x⟩), for all x, y ∈ ri(K) where ∇f (x) and ∇f (y) exist. Here ∇f is considered with respect to the affine hull of K. For functions that are continuously differentiable on the relative interior of their domain, the equivalence of (i) and (v) gives a first-order test. Corollary 4. Let γ ∈ (0, 1] and let f : Rd → R be a proper, lsc, and radially continuous function with a convex effective domain K := dom(f ) such that f |ri(K) is continuously differentiable. Then ∂f is γ–tilted monotone if and only if ⟨∇f (y), y − x⟩ ≥ ψγ (⟨∇f (x), y − x⟩), for all x, y ∈ ri(K). Here ∇f is considered with respect to the affine hull of K. Proof. By assumption, K is convex and f is radially continuous. The function f |ri(K) is locally Lipschitz, because every x ∈ ri(K) has a closed ball B of the affine hull of K centered at x with B ⊆ ri(K). The continuous map ∇f is bounded on the compact set B, say by L, and the mean value theorem on the segment between any two points of the convex set B gives |f (u)−f (w)| ≤ L∥u−w∥ for all u, w ∈ B. Finally, ∇f (x) exists at every x ∈ ri(K), so the inequality in condition (v) of Theorem 14 is required for all x, y ∈ ri(K) and coincides with the inequality of the corollary. Hence the inequality of the corollary is equivalent to condition (v), and by that theorem it is equivalent to condition (i), i.e., to γ–tilted monotonicity of ∂f . Remark 2. By Theorems 13, 8, and 14, conditions (i)–(v) of Theorem 1 remain equivalent when ˆ , which is also classical [RW98, Theorem 12.17]. ∂f is replaced by ∂f or ∂f 23

γ Membership in Fmono can also be checked on lines, where a restriction to a line is regarded as a function on a one-dimensional Euclidean space, extended by +∞ outside its effective domain. The proof in Appendix C.11, like that of Theorem 16 below, uses the approximate supports of γ Section 5.1.2, which members of Fmono have because they lie in F γ (Theorem 13).

Theorem 15 (Line restrictions of γ–tilted monotone subdifferentials). Let γ ∈ (0, 1] and let f : Rd → R be proper and lsc. Then ∂f is γ–tilted monotone if and only if, for every line ℓ meeting dom(f ), the operator ∂(f |ℓ ) is γ–tilted monotone on ℓ. Moreover, when these equivalent conditions hold, the same conclusion is true for the restriction of f to any nonempty closed interval contained in ℓ and meeting dom(f ). For closed convex sets, the line criterion also gives stability under restriction. Corollary 5 (Stability of γ–tilted monotone subdifferentials under closed convex restrictions). Let γ ∈ (0, 1], let f : Rd → R be proper and lsc with ∂f γ–tilted monotone, and let C ⊆ Rd be a closed convex set such that C ∩ dom(f ) ̸= ∅. Then the function fC that equals f on C and +∞ outside C is proper and lsc, and ∂fC is γ–tilted monotone. Proof. The function fC is finite on the non-empty set C ∩ dom(f ) and never equals −∞, so it is proper, and it is lsc by the argument in the last paragraph of the proof of Corollary 3. Let ℓ be a line meeting dom(fC ) = C ∩ dom(f ). Then ℓ ∩ C is a nonempty closed interval contained in ℓ and meeting dom(f ), and fC |ℓ is the restriction of f to this interval, so the final assertion of Theorem 15 shows that ∂(fC |ℓ ) is γ–tilted monotone. Since ℓ was arbitrary, Theorem 15, applied to fC , shows that ∂fC is γ–tilted monotone. γ has the same stability under limits as F γ , with proof in Appendix C.12. The class Fmono

Theorem 16 (Stability of γ–tilted monotone subdifferentials). Let K ⊆ Rd be a nonempty convex set and let γ ∈ (0, 1]. Let fn : Rd → R be proper lsc functions such that ∂fn is γ–tilted monotone for every n. Suppose that fn converges pointwise to finite values on a dense subset D ⊆ K. Then there is a continuous function f : ri(K) → R such that (i) fn → f uniformly on compact subsets of ri(K); (ii) if f¯ := lsc(f ), then ∂ f¯ is γ–tilted monotone. Moreover, if there is an lsc function g : K → R such that fn → g pointwise on K, then lsc(g) = f¯, and hence ∂(lsc(g)) is γ–tilted monotone. γ The class Fmono inherits the properties of Sections 5.1.1–5.1.4 through Theorem 13(i).

Corollary 6. Let γ ∈ (0, 1] and let f : Rd → R be proper and lsc. Suppose that ∂f is γ–tilted monotone, and set K := dom(f ). Then: (i) f is γ–tilted convex, K is convex, and f is quasiconvex; (ii) f |ri(K) is locally Lipschitz, and f is radially continuous; (iii) for every x ∈ K, the following are equivalent: x is a global minimizer of f , x is a local ˆ (x), and 0 ∈ ∂f (x); minimizer of f , 0 ∈ ∂f
(iv) f = lsc f |ri(K) . 24

ˆ (x) ⊆ Proof. Since f is lsc on Rd , it is locally lsc at every point of K, and Proposition 1(i) gives ∂f ∗ ˆ ∂f (x) for every x ∈ K. Inequality (6) for ∂f therefore holds in particular for all x ∈ ∂f (x) and ˆ (y), so ∂f ˆ is γ–tilted monotone. Theorem 13(i) now shows that f is γ–tilted convex, so y∗ ∈ ∂f Theorem 4 gives that K is convex and f is quasiconvex, which proves (i). Assertion (ii) follows from Theorem 6 and Corollary 2, assertion (iii) from Theorem 5, and assertion (iv) from Theorem 10. 5.3 Properties of γ–tilted secant convex functions Convexity can be stated through increasing secant slopes or through Jensen’s inequality, which bounds the value at a convex combination by the weighted arithmetic mean of the two values. Tilted secant convexity also has a Jensen form, in which the arithmetic mean is replaced by a mean that multiplies by γ the weight of the smaller value (Proposition 2). The right side of this form is nondecreasing in both values, which makes it the convenient form for suprema and lower semicontinuous envelopes (Theorem 17), and it extends to convex combinations of any finite number of points (Theorem 18). For γ ∈ (0, 1], λ ∈ [0, 1], and a, b ∈ R, define the mean   γλa + (1 − λ)b    γλ + (1 − λ) , a ≤ b, Mγ,λ (a, b) := (13)  λa + γ(1 − λ)b    λ + γ(1 − λ) , a > b. We extend this definition to γ = 0 by setting M0,λ (a, b) := max{a, b},

λ ∈ [0, 1].

(14)

Thus, M1,λ (a, b) = λa + (1 − λ)b, whereas M0,λ (a, b) = max{a, b}. For γ > 0, this mean is the upper expectile of a random variable taking a and b with probabilities λ and 1 − λ, at level 1/(1 + γ) [NP87, BKMRG14]. Equivalently, it uniquely minimizes     λ (a − t)2+ + γ(a − t)2− + (1 − λ) (b − t)2+ + γ(b − t)2− over t ∈ R, where u+ := max{u, 0} and u− := max{−u, 0}. Proposition 2 (Jensen form). Let γ ∈ [0, 1] and f : K → R ∪ {+∞}. Then f is γ–tilted secant convex if and only if
f (λy + (1 − λ)z) ≤ Mγ,λ f (y), f (z) for all y, z ∈ dom(f ) and λ ∈ [0, 1]. (15) Proof. We write a := f (y) and b := f (z) whenever y, z ∈ dom(f ), so that a, b ∈ R. For γ ∈ (0, 1], the denominators in (13) are positive, since γλ + (1 − λ) ≥ γλ + γ(1 − λ) = γ and likewise λ + γ(1 − λ) ≥ γ. So each branch of (13) is a weighted average of a and b with nonnegative weights that sum to one. Hence Mγ,λ (a, a) = a,

Mγ,1 (a, b) = a,

Mγ,0 (a, b) = b.

(16)

For λ = 1, the first branch is γa/γ = a and the second is a/1 = a; for λ = 0 the first branch is b/1 = b and the second is γb/γ = b. For γ = 0, (14) gives M0,λ (a, b) = max{a, b}, which equals a when a = b and is at least a and at least b. In all cases, Mγ,λ (a, a) = a, Mγ,1 (a, b) ≥ a, and Mγ,0 (a, b) ≥ b. 25

Let y, z ∈ dom(f ) and λ ∈ [0, 1]. If y = z, then λy + (1 − λ)z = y, and (15) reads a ≤ Mγ,λ (a, a) = a. If λ = 1, it reads a ≤ Mγ,1 (a, b), and if λ = 0, it reads b ≤ Mγ,0 (a, b). By the facts above, all three hold, and Definition 6 imposes nothing on these degenerate triples. So it suffices to fix y ̸= z in dom(f ) and λ ∈ (0, 1), put x := λy + (1 − λ)z, and show that h i x ∈ dom(f ) and (8) holds whenever ψγ (s− ) is defined ⇐⇒ f (x) ≤ Mγ,λ (a, b). (17) The right side Mγ,λ (a, b) of (17) is a real number, since a and b are real. If x ∈ / dom(f ), then f (x) = +∞, so the right side of (17) fails; the left side fails as well, so both sides of (17) are false and the equivalence holds. From now on let x ∈ dom(f ), and put m := f (x) ∈ R. It remains to show that (8), required whenever ψγ (s− ) is defined, is equivalent to m ≤ Mγ,λ (a, b). Since y − x = (1 − λ)(y − z) and x − z = λ(y − z), we have ∥y − x∥ = (1 − λ)∥y − z∥ and ∥x − z∥ = λ∥y − z∥. Put c := λ(1 − λ)∥y − z∥ > 0. Multiplying the definitions (7) of s+ and s− by c gives c s+ = λ(a − m), c s− = (1 − λ)(m − b). (18) In particular, s+ has the sign of a − m, and s− has the sign of m − b. Suppose first that γ > 0 and a > b, so that the second branch of (13) applies. Multiplying m ≤ Mγ,λ (a, b) by the positive denominator λ + γ(1 − λ) and rearranging, we get m ≤ Mγ,λ (a, b) ⇐⇒ λ(a − m) ≥ γ(1 − λ)(m − b) ⇐⇒ s+ ≥ γs− , where the last step uses (18) and divides by c > 0. If s− ≥ 0, then ψγ (s− ) = γs− , so s+ ≥ γs− is exactly (8). If s− < 0, then m < b < a, so s+ > 0 by (18). Then s+ > 0 > γs− , so m ≤ Mγ,λ (a, b) holds, and s+ > 0 > s− /γ = ψγ (s− ), so (8) holds. In both cases the two conditions agree. Suppose next that γ > 0 and a ≤ b, so that the first branch of (13) applies. Multiplying by the positive denominator γλ + (1 − λ) and rearranging, we get m ≤ Mγ,λ (a, b) ⇐⇒ γλ(a − m) ≥ (1 − λ)(m − b) ⇐⇒ γs+ ≥ s− ⇐⇒ s+ ≥ s− /γ. If s− ≤ 0, then ψγ (s− ) = s− /γ, so s+ ≥ s− /γ is exactly (8). If s− > 0, then m > b ≥ a, so s+ < 0 by (18). Then s+ < 0 < s− /γ, so m ≤ Mγ,λ (a, b) fails, and s+ < 0 < γs− = ψγ (s− ), so (8) fails. In both cases the two conditions agree. Finally, let γ = 0. Then M0,λ (a, b) = max{a, b}, and ψ0 (s− ) equals 0 if s− > 0, equals −∞ if s− < 0, and is undefined if s− = 0. We use the signs of s± from (18). If s− > 0, then m > b, so m ≤ max{a, b} holds exactly when m ≤ a, that is, when s+ ≥ 0 = ψ0 (s− ); the two conditions agree. If s− < 0, then m < b ≤ max{a, b}, and s+ ≥ −∞ = ψ0 (s− ); both conditions hold. If s− = 0, then m = b ≤ max{a, b}, and (8) is not required because ψ0 (0) is undefined; both conditions hold. These three cases cover all possibilities, which proves (17) and hence the proposition. The next theorem collects basic properties of γ–tilted secant convex functions, including closure under suprema in item (iii) and a reduction to lines in item (vi). Theorem 17. Let γ ∈ [0, 1], X ⊆ Rd be a convex set, and f : X → R∪{+∞} with effective domain K := dom(f ). Then the following are true. (i) If f is γ–tilted secant convex, then it is quasiconvex, and K is convex. (ii) If f is convex, then it is γ–tilted secant convex. (iii) The set of γ–tilted secant convex functions over X is closed under arbitrary non-empty suprema. 26

(iv) If f is γ–tilted secant convex and C ⊆ X is convex, the function f |C : C → R ∪ {+∞} is γ–tilted secant convex. (v) If γ ∈ (0, 1] and f is γ–tilted secant convex, then lsc(f ) : Rd → R ∪ {+∞} is γ–tilted secant convex and f = lsc(f ) on ri(K). (vi) f is γ–tilted secant convex if and only if, for every line ℓ that intersects X , the restriction f |ℓ : ℓ ∩ X → R ∪ {+∞} is γ–tilted secant convex. The proof is in Appendix C.13. A supremum of lsc functions is lsc, so by item (iii) a proper γ γ supremum of members of Fsec lies in Fsec . The binary mean in (13) extends to m values. For an integer m ≥ 2, let   m   X ∆m := λ = (λ1 , . . . , λm ) ∈ [0, 1]m λi = 1   i=1

denote the probability simplex. Definition 13 (Mean of m values; upper expectile). Fix m ≥ 2, λ ∈ ∆m , and γ ∈ (0, 1]. For a = (a1 , . . . , am ) ∈ Rm , define P P (m) / λi ai i∈S λi ai + γ P Pi∈S . (19) Mγ,λ (a) := max S⊆{1,...,m} i∈S λi + γ i∈S / λi For γ = 0, define (m)

M0,λ (a) := max ai .

(20)

1≤i≤m

For γ > 0, this is the expectile at level 1/(1 + γ) of the finite distribution assigning mass λi to ai [NP87, BKMRG14]. Equivalently, it is the unique minimizer over t ∈ R of m X

  λi (ai − t)2+ + γ(t − ai )2+ ,

r+ := max{r, 0};

i=1

see Lemma 11. Although (19) maximizes over subsets, the threshold characterization in that lemma computes the mean in O(m log m) arithmetic and comparison operations by sorting the values and scanning the resulting intervals. At γ = 0, (20) is a separate convention: it takes the maximum over all coordinates, whereas the positive-parameter limit ignores coordinates of weight zero. For m = 2 and λ = (λ, 1 − λ), this definition recovers the binary mean (by Lemma 11 for γ > 0, and by the definitions for γ = 0): (2)

γ ∈ [0, 1].

Mγ,(λ,1−λ) (a, b) = Mγ,λ (a, b), (m)

(m)

Moreover, M1,λ is the weighted arithmetic mean, and M0,λ is the maximum. 27

(21)

Theorem 18 (Characterizations of γ–tilted secant convexity). Let f : Rd → R be a proper lsc function with effective domain K := dom(f ), and let γ ∈ (0, 1]. Then the following are equivalent. (i) The function f is γ–tilted secant convex. (ii) For every y, z ∈ K and λ ∈ (0, 1), one has

λ ψ √γ f (y) − f (x) + (1 − λ) ψ √γ f (z) − f (x) ≥ 0,

x = λy + (1 − λ)z.

(iii) For every m ≥ 2, every y1 , . . . , ym ∈ K, and every λ ∈ ∆m , one has   m X
(m) f λi yi  ≤ Mγ,λ f (y1 ), . . . , f (ym ) .

(22)

(23)

i=1

(iv) For every m ≥ 2, every y1 , . . . , ym ∈ K, and every λ1 , . . . , λm > 0 where has m m X X
x= λi yi . λi ψ √γ f (yi ) − f (x) ≥ 0,

Pm

i=1 λi = 1, one

(24)

i=1

i=1

(v) K is convex and there is a multi-valued operator F : Rd ⇒ Rd such that ri(K) ⊆ dom(F ) and √ f is γ–tilted convex with respect to F . Furthermore, the equivalence of (i), (ii), (iii), and (iv) holds even if we do not assume f is lsc. The proof is in Appendix C.14. The next theorem, proved in Appendix C.15, places γ–tilted secant convexity between the other two notions: it implies γ–tilted convexity and follows from γ–tilted monotonicity of the subdifferential. Theorem 19. Let f : Rd → R be a proper lsc function and let γ ∈ (0, 1]. (i) If f is γ–tilted secant convex, then f is γ–tilted convex. (ii) If ∂f is γ–tilted monotone, then f is γ–tilted secant convex. In particular, ∗ ∗ γmono (f ) ≤ γsec (f ) ≤ γ ∗ (f ). γ The class Fsec inherits the properties of Sections 5.1.1–5.1.4 through Theorem 19(i).

Corollary 7. Let γ ∈ (0, 1] and let f : Rd → R be a proper lsc γ–tilted secant convex function with effective domain K := dom(f ). Then the following assertions hold: (i) f |ri(K) is locally Lipschitz, and f is radially continuous; (ii) for every x ∈ K, the following are equivalent: x is a global minimizer of f , x is a local ˆ (x), and 0 ∈ ∂f (x); minimizer of f , 0 ∈ ∂f
(iii) f = lsc f |ri(K) . Moreover, f |ri(K) is locally Lipschitz even if we do not assume that f is lsc.

28

Proof. By Theorem 19(i), f is γ–tilted convex. Assertion (i) follows from Theorem 6 and Corollary 2; assertion (ii) follows from Theorem 5; and assertion (iii) follows from Theorem 10. For the final claim, assume only that f is proper and γ–tilted secant convex, so that K is convex by Theorem 17(i) and ri(K) ̸= ∅. By Theorem 17(v), the envelope g := lsc(f ), a function from Rd to R ∪ {+∞}, is γ–tilted secant convex and g=f

on ri(K).

(25)

It is lsc, and it is proper because it never takes the value −∞ and is finite on ri(K) by (25). Its effective domain K′ := dom(g) satisfies K ⊆ K′ ⊆ K, because g ≤ f on K and g = +∞ outside K, so K′ = K. The set K′ is convex by Theorem 17(i) applied to g, and two convex sets with the same closure have the same relative interior (Corollary 6.3.1 in [Roc15]), so ri(K′ ) = ri(K). Assertion (i), applied to the proper lsc γ–tilted secant convex function g, shows that g|ri(K) is locally Lipschitz, and by (25) so is f |ri(K) . γ The class Fsec is stable under limits as well, with proof in Appendix C.16.

Theorem 20 (Stability of γ–tilted secant convexity under limits). Let K ⊆ Rd be a non-empty convex set and let γ ∈ (0, 1]. Let fn : Rd → R be proper γ–tilted secant convex functions that converge pointwise to finite values on a dense subset D ⊆ K. Then there is a continuous function f : ri(K) → R such that (i) fn → f uniformly on compact subsets of ri(K); (ii) f is γ–tilted secant convex. Moreover, if there is a function g : K → R such that fn → g pointwise on K, then g never takes the value −∞ and is γ–tilted secant convex. 5.4 Envelopes of tilted affine functions The tilted affine functions of Hγ play the role for γ 2 –tilted secant convexity that affine functions play for convexity. Theorem 21. Let f : Rd → R be a proper lsc function and let γ ∈ (0, 1]. Then f is Hγ –convex if and only if it is γ 2 –tilted secant convex. The proof is in Appendix C.17. Each member of Hγ is γ–tilted convex with respect to a constant operator, hence γ 2 –tilted secant convex by Theorem 18(v), and Theorem 17(iii) passes to suprema. Conversely, at each x in the relative interior of the domain, the operator of Theorem 18(v) gives a member of Hγ that minorizes f and equals f (x) at x, and f is determined by its values on the relative interior (Corollary 7(iii)). 5.5 Separations and negative results Three functions separate the classes from one another, and three properties of convex functions fail γ in them: closure under addition, classical second-order regularity, and, for Fmono , closure under maxima.

29

5.5.1 Separating the three classes √

γ γ , Fsec , and F γ . Example 1 attains For γ ∈ (0, 1), three functions separate the classes F γ , Fmono ∗ ∗ ∗ ∗ 1/2 ∗ < γ ∗ . Their proofs are γmono = γ , Example 2 attains γ = (γmono ) , and Example 3 gives γsec in Appendices C.18.1, C.18.2, and C.18.3.

Example 1. Fix γ ∈ (0, 1) and define f : R → R by    −1 ≤ x ≤ 0, x, f (x) := γx, 0 ≤ x ≤ 1,   +∞, x ∈ / [−1, 1]. Then f is proper and lsc, and ∗ ∗ γmono (f ) = γsec (f ) = γ ∗ (f ) = γ.

Therefore f∈ /F

√

γ

γ γ f ∈ Fmono ⊆ Fsec ⊆ Fγ.

,

: Example 2. Fix γ ∈ (0, 1) and define c := 1−γ γ . Define f R → R by f (x) :=

x 1 + cx

for x ∈ [0, 1] and f (x) = +∞ elsewhere. Then f is proper and lsc, its restriction to [0, 1] is continuously differentiable, and
1/2 ∗ ∗ γsec (f ) = γ ∗ (f ) = γmono (f ) = γ. Therefore

2

γ f∈ / Fmono ,

γ γ f ∈ Fsec ⊆ F γ ⊆ Fmono .

s Example 3. Fix γ ∈ (0, 1) and set s := γ and a := 2−s . Define the continuous function q : [0, 1] → (0, ∞) by   1 − 2(1 − s) x, 0 ≤ x ≤ s, s(2 − s) q(x) :=  a, s ≤ x ≤ 1,

and define f : R → R by

Z x f (x) :=

q(t) dt 0

for x ∈ [0, 1] and f (x) = +∞ elsewhere. Then f is proper and lsc, its restriction to [0, 1] is continuously differentiable, and  !2  γ 1−γ ∗ ∗  < γ. p γ ∗ (f ) = γ, γmono (f ) = , γsec (f ) = γ 1 − 2−γ 1 + (1 − γ)(2 − γ) In particular, γ f∈ / Fsec ,

2

γ f ∈ F γ ⊆ Fsec .

30

5.5.2 Closure under sums For every 0 < γ < 1, none of the three classes is closed under addition. Both results below use the concave quadratic hc (t) = −t2 − ct, whose parameters in the three classes are explicit. Example 4 (A concave quadratic). For c > 0, the function hc (t) = −t2 − ct on [0, 1] satisfies ∗ γ ∗ (hc ) = γsec (hc ) =

c , c+1

∗ γmono (hc ) =

c . c+2

Equivalently, for γ ∈ (0, 1), hc is γ–tilted convex, and also γ–tilted secant convex, if and only if c ≥ γ/(1 − γ), and its Clarke subdifferential is γ–tilted monotone if and only if c ≥ 2γ/(1 − γ). γ The proof is in Appendix C.19. The parameter of hc in Fmono is smaller than in the other two γ γ classes, so hc also separates Fmono from Fsec .

Example 5 (Non-closure under addition). For any γ ∈ (0, 1), there exist two functions f1 , f2 : γ γ [0, 1] → R that lie in Fmono , Fsec , and F γ , whose sum is not quasiconvex and therefore lies in none of the three classes for any positive parameter. γ Proof. Let c = 2γ/(1 − γ), and let f1 = hc and f2 (t) = hc (1 − t) on [0, 1], so that f1 ∈ Fmono by Example 4. Both defining formulas are polynomials on R, so their restrictions to [0, 1] have proper, lsc, and radially continuous extensions. For s, t ∈ (0, 1), Corollary 4 applied to f1 gives
h′c (t)(t − s) ≥ ψγ h′c (s)(t − s) . (26)

Let x, y ∈ (0, 1), and put s = 1 − x and t = 1 − y. Since f2′ (r) = −h′c (1 − r) and y − x = −(t − s), f2′ (y)(y − x) = h′c (t)(t − s), f2′ (x)(y − x) = h′c (s)(t − s).
γ . Theorem 19 then gives By (26), f2′ (y)(y − x) ≥ ψγ f2′ (x)(y − x) , and Corollary 4 gives f2 ∈ Fmono γ γ f1 , f2 ∈ Fsec ⊆ F . The linear terms −ct and −c(1 − t) add up to −c, so f1 (t) + f2 (t) = −t2 − (1 − t)2 − c

(t ∈ [0, 1]).

(27)

By (27), f1 + f2 equals −1 − c at t = 0 and at t = 1, and it equals − 12 − c > −1 − c at t = 12 . The sublevel set {t : f1 (t) + f2 (t) ≤ −1 − c} thus contains 0 and 1 but not their midpoint, so f1 + f2 is not quasiconvex. For η ∈ (0, 1], every member of F η is quasiconvex by Theorem 4, and η η Fmono ⊆ Fsec ⊆ F η by Theorem 19. Hence f1 + f2 lies in none of the three classes with parameter η. Every smooth function on the cube whose curvature is bounded below is a convex function plus n concave quadratics that lie in all three classes. In Theorem 22, 1 denotes the all-ones vector. Theorem 22 (Decomposition with controlled negative curvature). Fix 0 < γ < 1 and c = 2γ/(1 − γ). Suppose that f is twice continuously differentiable on a neighborhood of [0, 1]n and that ∇2 f (x) ⪰ −τ I on the cube for some τ ≥ 0. Then f =q+

n X i=1

pi ,

pi (x) = −

τ 2 xi + cxi , 2

τ τc ⊤ q(x) = f (x) + ∥x∥2 + 1 x, 2 2

(28)

γ γ where q is convex and every pi on [0, 1]n lies in Fmono , Fsec , and F γ . If ∥∇f ∥ ≤ L and |f | ≤ B on the cube, then √ ∥∇pi ∥ ≤ τ2 (c + 2), |pi | ≤ τ2 (c + 1), ∥∇q∥ ≤ L + τ2 (c + 2) n, |q| ≤ B + τ2n (c + 1).

31

The proof is in Appendix C.20. Given an evaluation oracle for f and a bound ∇2 f ⪰ −τ I, the theorem reduces minimizing f on the cube to minimizing a sum of n + 1 members of any of the three classes, whose terms are evaluated with one evaluation of f and O(n) arithmetic operations. Minimizing such sums is therefore as hard as minimizing a general smooth function on the cube with curvature bounded below. 5.5.3 Second-order conditions and their limitations For a C 2 function on an open convex set, the classical second-order characterization of convexity is f is convex

∇2 f (x) ⪰ 0

⇐⇒

for every x.

For a C 1 convex function on an open interval, the first derivative is monotone and therefore differentiable almost everywhere. More generally, Alexandrov’s theorem gives a second-order expansion almost everywhere for a finite convex function on an open convex set [RW98, Theorem 13.51]. An almost-everywhere condition on the classical Hessian alone does not characterize convexity: for example, −|x| has classical second derivative zero almost everywhere but is not convex. The next example shows that the one-dimensional C 1 assertion fails for every γ < 1, even in the smallest of the three classes; “second derivative” there means the classical derivative of the first derivative. Example 6 (Failure of classical second-order regularity). For every γ ∈ [0, 1) there is a proper lsc function fγ : R → R with effective domain [0, 1] such that γ γ fγ ∈ Fmono ∩ Fsec ∩ Fγ,

and fγ |(0,1) is continuously differentiable but its classical second derivative exists at no point of (0, 1). In fact, all three memberships hold with κγ = (1 + 3γ)/(3 + γ) > γ. Proof. Let w : [0, 1] → [−1, 1] be continuous and nowhere differentiable. Set δγ :=

1−γ , 2(1 + γ)

and define fγ (x) :=

hγ := 1 + δγ w,

Z x   h (t) dt, γ

x ∈ [0, 1],

0

x∈ / [0, 1].

 +∞,

Extend hγ continuously to R by keeping it constant on each of (−∞, 0] and [1, ∞). Its antiderivative is C 1 on R, and on [0, 1] its derivative satisfies 0 < m := 1 − δγ ≤ hγ ≤ M := 1 + δγ . Let κ = m/M . For x, y ∈ (0, 1) put d = y − x, u = fγ′ (x)d, and v = fγ′ (y)d. If d ≥ 0, then v ≥ md = κM d ≥ κu = ψκ (u). If d < 0, then u ≤ md < 0 and v ≥ M d = md/κ ≥ u/κ = ψκ (u). The proper extension of fγ is proper, lsc, and radially continuous, because fγ is continuous on [0, 1], m/M so Corollary 4 gives fγ ∈ Fmono . Since m 1 + 3γ = , M 3+γ

m 1 − γ2 −γ = > 0, M 3+γ

Theorem 19 at the positive parameter m/M , followed by nesting in the parameter, gives the asserted memberships, including those at γ = 0. The function is proper and lsc because it is continuous on its closed effective domain. On (0, 1) its first derivative is 1 + δγ w, which is nowhere differentiable because δγ > 0; this proves the final assertion. 32

Example 6 concerns classical second derivatives of C 1 functions. At γ = 0, second-order characterizations of quasiconvexity are classical [CM09]. γ 5.5.4 Maxima in Fmono γ γ For every 0 < γ < 1, the class Fmono is not closed under maxima, whereas F γ and Fsec are closed under proper suprema. γ Example 7 (A maximum outside Fmono ). Let γ ∈ (0, 1) and σ ∈ (γ, 1), put δ := σ − γ, and let K := [L, U ] with δ δ L := − , U := 2 + . 2(1 − σ) 2γ(1 − σ)

The functions  f1 (t) := min σt − δ, γσ(t − 2) + 2σ − δ ,

f2 (t) := min{t, γt}

γ , while h := max{f1 , f2 } on K satisfies on K belong to Fmono ∗ γmono (h) = γσ < γ,

∗ γsec (h) = γ ∗ (h) = γ.

γ In particular, h ∈ / Fmono .

Each of f1 and f2 is concave with one kink, at which its slope drops by the factor γ. Their maximum has slope 1 near the left endpoint and slope γσ near the right endpoint, and this drop by the factor γσ < γ violates (6). The proof is in Appendix C.21. γ Theorem 13 bounds how far the monotonicity parameter can drop under suprema in Fmono . γ ˆ ˆ Let {fi }i∈I ⊆ Fmono have a proper supremum s. Each ∂fi is γ–tilted monotone, since ∂fi ⊆ ∂fi by Proposition 1(i), so Theorem 13(i) gives fi ∈ F γ , Theorem 12 gives s ∈ F γ , and Theorem 13(ii) γ2 with F = ∂s gives s ∈ Fmono . The exponent 2 cannot be lowered: as σ ↓ γ, the maximum h in ∗ Example 7 has γmono (h) = γσ, which tends to γ 2 . 5.6 Relations between the classes Theorems 13 and 19 combine into one chain, in which the parameter drops by at most a square. Corollary 8. For γ ∈ (0, 1], F

√

γ

γ γ ⊆ Fmono ⊆ Fsec ⊆ Fγ.

Equivalently, if f : Rd → R is a proper lsc function, then ∗ ∗ (γ ∗ (f ))2 ≤ γmono (f ) ≤ γsec (f ) ≤ γ ∗ (f ). √

γ . Theorem 19(ii) gives Proof. If f ∈ F γ , then applying Theorem 13(ii) to F = ∂f gives f ∈ Fmono the next inclusion, and part (i) of that theorem gives the last one. Thus

F

√

γ

γ γ ⊆ Fmono ⊆ Fsec ⊆ Fγ,

γ ∈ (0, 1].

(29)

∗ ∗ (f ) ≤ γ ∗ (f ) For a fixed proper lsc function f , the same-parameter inclusions imply γmono (f ) ≤ γsec by taking suprema over admissible positive parameters. The first inclusion, applied with γ = θ2 , θ2 gives f ∈ Fmono whenever f ∈ F θ and θ > 0. Taking suprema and using continuity of the ∗ increasing map θ 7→ θ2 yields (γ ∗ (f ))2 ≤ γmono (f ). If an admissible positive-parameter set is empty, its parameter is zero by definition and the corresponding inequality is immediate.

33

γ By Theorem 21, the class Fsec in the chain is also the class of proper suprema of members of √ H γ. When γ = 1, all classes in the chain equal the class of proper lsc convex functions by Theorem 1. For γ ∈ (0, 1), Examples 1, 2, and 3 show that each inclusion in the chain is strict. Each of the three families is also strictly ordered in its parameter, and all of them lie strictly between the convex and the pseudoconvex functions. θ Theorem 23 (Strict hierarchy). Let 0 < γ ′ < γ < 1, and let C θ stand for Fmono for every θ, for θ θ 1 Fsec for every θ, or for F for every θ. Then C is the class of proper lsc convex functions, and [ [ ′ + C1 ⊊ Cγ ⊊ Cγ ⊊ Cθ = F θ ⊊ F 0 ⊊ F 0. (30) 0<θ≤1

0<θ≤1

1 1 , Proof. By Theorem 1, F 1 and Fmono consist of the proper lsc convex functions, and so does Fsec because 1–tilted secant convexity is convexity. Each of the three defining conditions with parameter ′ θ implies the same condition with every smaller parameter (Section 4.3), so C θ ⊆ C θ for θ′ < θ. √ Corollary 8 gives F θ ⊆ C θ ⊆ F θ for θ ∈ (0, 1], so the two unions in (30) agree, and (10) gives S θ 0+ ⊆ F 0 . θF ⊆F For θ ∈ (0, 1), let fθ be the function of Example 1 with parameter θ. Its three parameters equal ′ θ and are attained (Section 4.3), so fθ ∈ C θ and fθ ∈ / C η for η > θ. Hence fγ ∈ C γ \C 1 , fγ ′ ∈ C γ \C γ , ′ ′ and fγ ′ /2 ∈ C γ /2 \ C γ , which proves the first three strict inclusions. √ Let s(t) = t on [0, 1]. Its proper extension is proper and lsc, and s is continuous on [0, 1], hence √ radially continuous. For x√∈ (0, 1), s is continuously differentiable near x, so ∂s(x) = {1/(2 x)} by Proposition 1(iv). Since t is continuously differentiable near 1, Proposition 1(v) with K = [0, 1] gives ∂s(1) = 1/2 + N[0,1] (1) = [1/2, ∞). Every v ∈ ∂s(x) with x ∈ (0, 1] is therefore positive, so v(y − x) ≥ 0 forces y ≥ x and s(y) ≥ s(x), while s(y) ≥ 0 = s(0) for every y ∈ [0, 1]. This + is condition (i) of Theorem 3, so s ∈ F 0 . For θ ∈ (0, 1] and x ∈ (0, 1), inequality (5) with √ v = 1/(2 x) and y = 1 reads √ θ(1 − x) √ 1− x ≥ , (31) 2 x

which fails for small x, because its right side tends to +∞ as x ↓ 0. Hence s ∈ / F θ for every θ ∈ (0, 1]. Let g(t) = (t − 1/2)3 on [0, 1]. Its proper extension is proper and lsc, and g is continuous and nondecreasing, so it is radially continuous and quasiconvex, and Theorem 2 gives g ∈ F 0 . Proposition 1(iv) gives ∂g(1/2) = {0}, and with x = 1/2, v = 0, and y = 1/4 we have v(y − x) = 0 + but g(y) = −1/64 < 0 = g(x). So condition (i) of Theorem 3 fails, and g ∈ / F0 .

6. Examples Throughout this section, every feasible set is nonempty. We group the examples by the result that establishes their membership: Lemma 2 and Example 9 in Section 6.1, function values and envelopes in Section 6.2, and increasing transformations of convex costs in Section 6.3. Several examples of the last two subsections also have γ–tilted monotone generalized gradients and therefore lie in all three classes by Corollary 8, whereas membership through function values can hold at a larger parameter than monotonicity, as the function t/(1 + ct) of Example 2 shows. Table 2 lists the parameters of the examples by subsection, together with those of the separating functions of Section 5.5, and the proofs are in Appendix D. Algorithmic uses of these examples rely on the access model of Section 7.

34

6.1 Examples from monotone generalized gradients θ A function of one linear coordinate lies in Fmono when its slope changes by a bounded factor, with θ the ratio of the smallest to the largest slope. The proof is in Appendix D.1.

Lemma 2 (Slope ratios). Let K ⊆ Rd be nonempty, closed, and convex, let a ∈ Rd , and let J = {⟨a, x⟩ : x ∈ K}. Let H be continuously differentiable on an open interval containing J, and suppose that H ′ has constant sign on J with 0 < A ≤ |H ′ (s)| ≤ B for s ∈ J. Then f (x) = H(⟨a, x⟩) A/B on K belongs to Fmono . A concave polynomial in one coordinate with a large enough linear term is an example. Example 8. Let k ≥ 2 be an integer, a > 0, and 1 ≤ i ≤ n. The function f (x) = −xki − axi on ∗ [0, 1]n has exact monotonicity parameter γmono (f ) = a/(a + k). In particular, for γ ∈ (0, 1) and a ≥ kγ/(1 − γ), it lies in all three classes with parameter γ. Radial functions give nonconvex examples with a single minimizer surrounded by regions of negative curvature. Composing with an affine map covers functions of a residual. Example 9 (Radial functions). Let φ : [0, ∞) → R be continuously differentiable with φ′ (0) = 0, let R ∈ (0, ∞], suppose that φ′ (r) > 0 for every real r ∈ (0, R], and let δ = inf

n φ′ (s) φ′ (r)

o :0<r≤s≤R .

Let K ⊆ Rd be closed and convex, let A ∈ Rm×d and b ∈ Rm , and suppose that ∥Ax − b∥ ≤ R δ on K. If δ > 0, then f (x) = φ(∥Ax − b∥) on K belongs to Fmono . In particular, if R = ∞, then d φ(∥x∥) lies in all three classes on R with parameter δ. The number δ measures how far the radial slope can drop when moving outward, and it equals 1 when φ′ is nondecreasing, that is, when φ(∥x∥) is convex. Example 10 (A quadratic bowl with a radial oscillation). Let b, ω > 0, let κ = bω 2 , and let f (x) = 12 ∥x∥2 − b cos(ω∥x∥) on Rd . If κ ≤ 1, then f is convex. Let κ0 > 1 be the unique solution √ of 2π − arccos(−1/κ) = κ2 − 1. If 1 < κ < κ0 , then f is not convex, because along a line through γκ the origin f ′′ (t) = 1 + κ cos(ωt) is negative at t = π/ω, and f belongs to Fmono , and hence to all three classes, with √ 2π − sκ − κ2 − 1 √ γκ = , sκ = arccos(−1/κ). sκ + κ2 − 1 ∗ The parameter is exact: γmono (f ) = γκ . For κ = 2, sκ = 2π/3 and γ2 = (4π − √ monotonicity √ 1 2 3 3)/(2π + 3 3). This covers f (t) = 2 t − 2 cos t of Figure 1, for which d = 1, b = 2, and ω = 1.

6.2 Examples from function values and envelopes By Theorem 21, a proper lsc function is γ–tilted secant convex exactly when it is a pointwise supremum of members of H√γ , the tilted affine functions x 7→ c + ψ√γ (⟨x − w, z⟩). Membership in γ Fsec can therefore be shown either through an inequality on function values, as in the first examples below, or by writing the function as such a supremum. Ratios of a cost to a positive output, such as cost per unit of throughput, are classical nonconvex objectives [CC62, Din67, Sch76]. Their parameter is the ratio of the smallest to the largest value of the denominator. 35

Example 11 (Fractional objectives). Let K be compact and convex, and let W and Z be continuously differentiable on an open set containing K, with Z > 0 on K and γ = minK Z/ maxK Z. Suppose that W is convex and Z is affine, or that W ≥ 0 is convex and Z is concave. Then f = W/Z on K is γ–tilted secant convex, hence γ–tilted convex, with a γ 2 –tilted monotone Clarke subdifferential, and for all x, y ∈ K, f (y) − f (x) ≥

Z(x) ⟨∇f (x), y − x⟩, Z(y)

(32)

with equality when W and Z are both affine. ∗ = γ and The bound can be exact: t 7→ t/(1 + ct) on [0, 1] with c = (1 − γ)/γ has γ ∗ = γsec ∗ 2 γmono = γ , and here min[0,1] Z/ max[0,1] Z = γ for Z(t) = 1 + ct (Example 2 in Section 5.5.1).

Example 12 (Risk per unit of return and energy per bit). (a) Let Σ ≻ 0 be a covariance matrix, and let µi > 0 denote expected excess returns relative to a deterministic risk-free benchmark. On the probability simplex P ∆n = {x ∈ Rn+ : i xi = 1}, √ the inverse Sharpe ratio f (x) = x⊤ Σx/⟨µ, x⟩ is γ–tilted secant convex with γ = mini µi / maxi µi . (b) Let K ⊆ Rn+ be compact and convex, let c ≥ 0 and P0 > 0, and let R be concave and continuously differentiable on an open convex neighborhood of K, with R > 0 on K. The energy per bit f (x) = (⟨c, x⟩ + P0 )/R(x) is γ–tilted secant convex with γ = minK R/ maxK R [ZJ15]. Examples include sums of log(1 + SNRj (x)) with affine SNRj on a neighborhood where the logarithms are defined, provided the feasible set ensures a positive total rate. For sequences of either type, the hindsight objective is a sum of ratios and need not retain the structure of a single fractional objective. General sum-of-ratios optimization is NP-hard [SS03]. The remaining examples are suprema of members of Hγ , much as convex functions arise as suprema of affine functions. Example 13. Let ∅ ̸= {hi }i∈I ⊆ Hγ , let f = supi∈I hi , and suppose that f is finite at some point. Then f is Hγ –convex, hence γ 2 –tilted secant convex and γ 2 –tilted convex. Example 14 (Distorted convex functions and worst-margin costs). Let ϕ : Rd → R be convex. For each x0 and g0 ∈ ∂ϕ(x0 ), superadditivity and monotonicity of ψγ (Lemma 1) give ψγ (ϕ(x)) ≥ ψγ (ϕ(x0 )) + ψγ (⟨g0 , x − x0 ⟩) for all x, with equality at x = x0 . Hence ψγ ◦ ϕ is a supremum of members of Hγ and is γ 2 –tilted secant convex. Its Clarke subdifferential is also γ 2 –tilted monotone, so it lies in all three classes with parameter γ 2 . For a nonempty closed convex set C and r0 > 0, the choice ϕ = dist(·, C) − r0 penalizes distance at rate 1/γ within distance r0 of C and at rate γ beyond it. For labeled data (ai , bi ) with bi ∈ {−1, 1}, let ℓ(x) = maxi {−bi ⟨ai , x⟩}. This function is convex, and monotonicity of ψγ gives ψγ (ℓ(x)) = max ψγ (−bi ⟨ai , x⟩). i

Thus the displayed distortion has the class memberships proved above. For minimization, fix a bounded feasible set such as K = {x : ∥x∥ ≤ 1}, on which minimizing ℓ is equivalent to maximizing the minimum signed margin. The distortion has slope 1/γ at negative arguments and slope γ at positive arguments, and since it is strictly increasing, it preserves the set of minimizers of ℓ on every fixed K. Restricting to K keeps the three memberships: for Fsec by the definition, for F by Corollary 3, and for Fmono by Corollary 5. 36

Example 15 (Two-slope costs). Let 0 < b ≤ a and f (t) = at for t ≤ 0 and f (t) = bt for t ≥ 0. With p √ θ = b/a and ν = ab, f (t) = ψθ (νt), so f ∈ Hθ , and Example 14 with ϕ(t) = νt shows that ∂f ∗ ∗ (f ) = γ ∗ (f ) = b/a. is (b/a)–tilted monotone. On R, γ ∗ (f ) = b/a (Appendix D), so γmono (f ) = γsec Such a function models a marginal cost that drops from a to b beyond a threshold, as in a quantity discount. It is convex only when a = b, and because its derivative drops at 0, adding a quadratic does not make it convex either, in contrast with the additive relaxation of weak convexity [DD19]. 6.3 Increasing transformations of convex costs Robust estimation applies an increasing transformation to a convex cost, which keeps its minimizers and bounds the influence of large residuals. Example 16 bounds γ ∗ of such a composition through secants of the transformation relative to its derivative. Example 16 (Increasing transformations of convex costs). Let K be closed and convex, let g be convex and continuously differentiable on an open set containing K with g(K) ⊆ J, and let ϕ be continuously differentiable with ϕ′ > 0 on an open interval containing J. Suppose that 0 < a ≤ 1 ≤ b and, for all u, v ∈ J, ( a ϕ′ (u)(v − u), v ≥ u, ϕ(v) − ϕ(u) ≥ (33) b ϕ′ (u)(v − u), v < u. Then ϕ ◦ g is θ–tilted convex on K with θ = min{a, 1/b}. Examples 9 and 16 apply to two standard robust regression losses, which are radial functions of a residual. Example 17 (Cauchy and Geman–McClure losses). Let r(x) = ⟨a, x⟩ − β be an affine residual with |r| ≤ R on a closed convex set K, let σ > 0, and let q = R2 /σ 2 . √ θ with θ = 2 q/(1+q) (a) The Cauchy loss log(1+r(x)2 /σ 2 ) is convex if q ≤ 1 and belongs to Fmono if q ≥ 1. θ with (b) The Geman–McClure loss
r(x)2 /(σ 2 + r(x)2 ) is convex if q ≤ 1/3 and belongs to Fmono √ √ θ = 16 q/ 3 3 (1 + q)2 if q ≥ 1/3. It is also 1/(1 + q)–tilted convex (Example 16). For √ q ≥ 1/3, this second bound is at least θ exactly when q ≥ (101 + 16 37)/27.

As functions of the scalar residual r, the √ Cauchy and Geman–McClure profiles have negative second derivative for |r| > σ and |r| > σ/ 3, respectively. Their affine compositions therefore have negative curvature along any feasible line segment on which the residual varies and remains beyond the corresponding threshold. Both are standard bounded-influence losses in robust regression and vision [BR96, Bar19], and summing them over data points gives the robust M-estimation objective.

7. Polynomial-Time Offline Optimization An L–smooth function from any of the three classes can be minimized to additive accuracy ε with a number of gradient and projection calls polynomial in L, D, 1/γ, and 1/ε. We work with the largest class F γ , and Corollary 8 transfers the results to the two smaller classes. Smooth members are tilted convex in the sense of Martı́nez-Rubio [MR22], hence quasar-convex [MR26], and the bounds of this section are those of known first-order methods for quasar-convex functions. The access model is shared with the online problem in Sections 8 and 9.

37

Table 2: Parameters of the examples and of the separating functions of Section 5.5. Each entry is a lower bound on the parameter of its column, and the entries marked exact are equalities. By ∗ ∗ and γ ∗ , and a lower bound θ on γ ∗ Corollary 8, a lower bound θ on γmono is also a lower bound on γsec √ √ √ 2 gives the lower bound θ on the other two. Here θC = 2 q/(1 + q) and θGM = 16 q/(3 3(1 + q)2 ). Function Section 6.1 −xki − axi on [0, 1]n φ(∥Ax − b∥) 2 1 2 ∥x∥ − b cos(ω∥x∥), 1 < κ < κ0

∗ γmono

∗ γsec

γ∗

Reference

a/(a + k), exact δ γκ , exact

a/(a + k)

a/(a + k)

Ex. 8

δ γκ

δ γκ

Ex. 9 Ex. 10

γ2

γ

γ

Ex. 11

γ2

γ

γ

Ex. 12

Section 6.2 W/Z, convex over affine, γ = min Z/ max Z Inverse Sharpe ratio; energy per bit Supremum of members of Hγ ψγ ◦ ϕ, ϕ convex; worst-margin costs Two-slope cost, slopes a ≥ b

γ4 γ2

γ2 γ2

γ2 γ2

Ex. 13 Ex. 14

b/a, exact

b/a, exact

b/a, exact

Ex. 15

Section 6.3 ϕ ◦ g as in Ex. 16

θ2

θ2

Ex. 16

Cauchy loss, q ≥ 1 Geman–McClure loss, q ≥ 1/3

θC θGM

θC θGM

θ = min{a, 1/b} θC larger of θGM and 1/(1 + q)

γ, exact c/(c + 2), exact γ 2 , exact γ/(2 − γ), exact

γ, exact c/(c + 1), exact γ, exact γ(1−r2 ) < γ, exact, where r = √ 1−γ

γ, exact c/(c + 1), exact γ, exact γ, exact

Ex. 1 Ex. 4

γ, exact

γ, exact

Ex. 7

Separating functions, Section 5.5 x on [−1, 0], γx on [0, 1] −t2 − ct on [0, 1] t/(1 + ct) on [0, 1], c = (1 − γ)/γ Function of Example 3

1+

Maximum h of Example 7, σ ∈ (γ, 1)

γσ, exact

Ex. 17 Ex. 17

Ex. 2 Ex. 3

(1−γ)(2−γ)

Let K ⊆ Rd be a nonempty compact convex set of diameter at most a known bound D, with a known feasible rational starting point, and we assume efficient Euclidean projection onto K. A

38

value-and-support oracle returns, at a query point z ∈ K,
Qf (z) = f (z), qf (z) .

(34)

The vector qf (z) is a support vector : inequality (5) holds at x = z with v = qf (z) for every y ∈ K, as it does for every element of ∂f (z). We assume that f is G–Lipschitz on K and that qf (z) ≤ G at every feasible point. If f is differentiable on an open set containing K, then ∇f (z) is a Fréchet subgradient of the proper extension of f |K , hence a Clarke subgradient by Proposition 1(i), and we use qf = ∇f , so that (34) is a first-order oracle. We call f L–smooth on K when ∇f is L–Lipschitz on K. A linear minimization oracle returns, for a query vector c, a point of arg miny∈K ⟨c, y⟩; it is used only by the Frank–Wolfe method below. Query complexity counts oracle and projection calls. b with f (b Write f ⋆ = minK f ; an ε–minimizer is a feasible point x x) ≤ f ⋆ + ε. For smooth objectives, the results of Martı́nez-Rubio [MR26] apply directly. Corollary 9 (Query Complexity). Suppose that f is differentiable on an open set containing K, that ∇f is L–Lipschitz on K, and that f |K ∈ F γ . (i) Let x0 ∈ K, and let D0 ≤ D be the diameter
of {x ∈ K : f (x) ≤ f (x0 )}. The projected gradient iterates xt+1 = ΠK xt − L−1 ∇f (xt ) satisfy f (xt ) − f ⋆ ≤

20LD02 (t + 1)γ 2

(t ≥ 1),

(35)

so O(1 + LD2 /(γ 2 ε)) gradient evaluations and projections give an ε–minimizer. The step size does not depend on γ. (ii) With a linear minimization oracle, the Frank–Wolfe iterates of [MR26, Theorem 9] satisfy
⋆ 2 2 f (xt ) − f ≤ 6LD / (t + 1)γ for t ≥ 1. (iii) The accelerated method of [MR26, Theorem 5] finds an ε–minimizer with r  1 LD2  e O γ ε

(36)

e hides logarithmic factors. first-order queries, where O Proof. Since f is continuous and K is compact, f attains its minimum on K at some x⋆ . As noted above, ∇f (x) is a support vector at every x ∈ K, so f (y) − f (x) ≥ ψγ (⟨∇f (x), y − x⟩) for all x, y ∈ K. This is tilted convexity on K with γn = γp = γ [MR22, Eq. (3)], and tilted-convex functions are γ–quasar-convex in K with center x⋆ [MR26, Section 2]. The other hypotheses of Theorems 5, 8, and 9 of [MR26], namely L–smoothness on K and compactness and convexity of K, hold by assumption. Theorem 8 gives (35), Theorem 9 gives (ii), and Theorem 5 gives (iii). The right side of (35) is at most ε once t + 1 ≥ 20LD2 /(γ 2 ε), and each iteration in (i) uses one gradient evaluation and one projection.

8. γ-Approximate Online Optimization We give a√polynomial-time online algorithm for nonpositive γ–tilted convex losses whose γ–regret is O(DG T ); as in Section 7, results stated for F γ apply to the two smaller classes. The algorithm passes one vector per round to an online learner for linear losses, following the reduction of Pedramfar and Aggarwal [PA24] (Section 8.2). We prove that a suitable vector with coefficient γ exists for every continuous nonpositive loss in F γ at every point with a support vector (Theorem 24), compute it with polynomially many oracle calls (Theorem 25), and derive the regret bound (Corollary 10). 39

8.1 Online model and approximate regret A learner and an adversary interact for T rounds on K. In round t the learner chooses xt ∈ K from its past observations and incurs the loss ft (xt ); it then receives the oracle (34) for ft , which it may query at any points of K before choosing xt+1 . Our upper bounds hold for every realized loss sequence, including sequences that the adversary adapts to the learner’s past actions, whereas the lower bound of Section 9 uses independent losses drawn from a fixed distribution. The losses satisfy P −B ≤ ft ≤ 0 on K for some B > 0. Under this normalization, multiplying the benchmark minu t ft (u) ≤ 0 by a coefficient α < 1 raises it, so a smaller α gives a weaker benchmark. For α ∈ (0, 1], an interval I = [s, e] of rounds, and comparators us , . . . , ue ∈ K, let Regα (I; us:e ) =

e X   ft (xt ) − αft (ut ) .

(37)

t=s

The static α–regret on I is SRegα (I) =

X

ft (xt ) − α min u∈K

t∈I

X

ft (u),

(38)

t∈I

and we write SRegα (T ) for SRegα ([1, T ]), which is the α–regret Regα (T ) of Section 1. Since the lower bound of Section 9 concerns computation, the online results are stated in the bit model. Query points are rational, and values, support vectors, and feasible approximate projections are available to any requested accuracy in time polynomial in the input length and the requested number of bits. Numerical parameters are rational or are given by procedures that output certified rational intervals of width at most 2−b in time polynomial in the input length and b, and a nonrational γ also comes with a rational lower bound 0 < γlb ≤ γ of polynomial encoding length; we then call γ effectively specified. Running times are polynomial in the numerical quantities D, G, and 1/γ, not only in their encoding lengths. If D = 0 or G = 0, every feasible point has the same value; otherwise D and G are positive rational upper bounds. A coefficient α ∈ (0, 1] is achievable for a class of losses if an online algorithm, possibly randomized, that is given the horizon T and runs in time polynomial in N and T satisfies   (39) E SRegα (T ) ≤ P (N ) T 1−ρ for every horizon T and every sequence of losses from the class. Here P is a fixed polynomial and ρ > 0 is a fixed constant. The number N bounds the encoding length of K and of a single loss, the descriptions of the parameter-approximation procedures, the encoding lengths of all numerical bounds, and the numerical values B, D, G of the normalization and of Section 7. It does not depend on T or on the number of losses, and the oracle (34) answers to any requested accuracy, as in the bit model above. The online coefficient is sup({0} ∪ {α ∈ (0, 1] : α is achievable}). 8.2 Lower-linearizability and the reduction to linear losses Pedramfar and Aggarwal [PA24] reduce online optimization of nonlinear losses to online linear optimization whenever each loss admits a comparison vector that is valid against all comparators. In the minimization form, a class F of functions on K is lower-linearizable if there are maps g : F × K → Rd and χ : K → K and constants α ∈ (0, 1] and β > 0 such that f (χ(x)) − αf (y) ≤ β ⟨g(f, x), x − y⟩

40

for all f ∈ F and x, y ∈ K.

(40)

Algorithm 1 Reduction to online linear optimization (minimization form of OMBQ [PA24]) Require: Online algorithm A for linear losses on K; procedure G that maps a loss oracle and a point to a vector 1: for t = 1, 2, . . . , T do 2: Play the action xt proposed by A and incur ft (xt ). 3: Compute at = G(Qft , xt ). 4: Feed the linear loss z 7→ ⟨at , z⟩ to A. 5: end for The map χ transforms the point proposed by the linear learner into the point that is played. We use the special case χ = id and β = 1, and call F lower-linearizable with coefficient α if f (x) − αf (y) ≤ ⟨g(f, x), x − y⟩

for all f ∈ F and x, y ∈ K.

(41)

Convex functions satisfy (41) with α = 1 and g(f, x) ∈ ∂f (x). Let A be a deterministic online algorithm for linear losses z 7→ ⟨at , z⟩ on K with ∥at ∥ ≤ G. A pathwise regret bound for A is a function BA such that X ⟨at , xt − ut ⟩ ≤ BA (I; us:e ) (42) t∈I

for every realized sequence a1 , a2 , . . . , every interval I = [s, e], and every comparator path. The vector at may depend on the action xt that A plays in round t; Algorithm 1 implements this reduction. If at = g(ft , xt ) for a map g satisfying (41), then applying (41) with x = xt and y = ut and summing over t ∈ I gives Regα (I; us:e ) ≤ BA (I; us:e ). Every regret bound of A for linear losses thus becomes an α–regret bound for F with the same rate. For nonpositive losses in F γ we show that α = γ is possible with ∥g∥ ≤ G, and that an additive approximation of g can be computed from the oracle. 8.3 A comparison vector with coefficient γ In this subsection K is any convex set, f : K → R is continuous and nonpositive with f ∈ F γ , qf (x) is a support vector at the base point x, and r = −f ≥ 0. Lemma 3, which integrates (5) along a segment, is the source of the coefficient γ. Lemma 3 (Interpolation for nonnegative rewards). For y0 , y1 ∈ K and t ∈ [0, 1],

r (1 − t)y0 + ty1 ≥ γ (1 − t)r(y0 ) + t r(y1 ) .

(43)

More precisely, if A = r(y0 ) ≤ r(y1 ) = B0 , then
r (1 − t)y0 + ty1 ≥ B0 − (B0 − A)(1 − t)γ .

(44)

Proof. If y0 = y1 , inequality (43) reads r(y0 ) ≥ γr(y0 ), which holds because r ≥ 0 and γ ≤ 1, and (44) reads r(y0 ) ≥ r(y0 ). Assume y0 ̸= y1 . Let fe be the proper extension of f , let gb(t) = fe(y0 + t(y1 − y0 )) for t ∈ R, and let ϕ(t) = r((1 − t)y0 + ty1 ) for t ∈ [0, 1], so that gb = −ϕ on [0, 1]. By Theorem 9, the restriction of fe to the line through y0 and y1 is γ–tilted convex. Let c = ∥y1 − y0 ∥ > 0. We identify this line with R through the isometry τ 7→ y0 + τ (y1 − y0 )/c, so that k(τ ) = fe(y0 + τ (y1 − y0 )/c) is γ–tilted convex on R and gb(t) = k(ct). The map t 7→ ct is a 41

linear bijection of R, so the definitions of the subdifferentials give ∂b g (t) = c ∂k(ct). Let t, t′ lie in the domain of gb and let w ∈ ∂b g (t), so that w = cu with u ∈ ∂k(ct). Inequality (5) for k gives

gb(t′ ) − gb(t) = k(ct′ ) − k(ct) ≥ ψγ (ct′ − ct)u = ψγ (t′ − t)w . (45) By (45), gb is γ–tilted convex on R. Its domain {t : y0 + t(y1 − y0 ) ∈ K} is an interval containing [0, 1], because K is convex. By Theorem 6, gb is locally Lipschitz on the interior of its domain, which contains (0, 1). A locally Lipschitz function is Lipschitz on each compact subset, so ϕ = −b g is Lipschitz on every interval [t0 , t1 ] ⊂ (0, 1) and, by Rademacher’s theorem, differentiable at almost every t ∈ (0, 1). Since f is continuous on K, ϕ is continuous on [0, 1]. Let t ∈ (0, 1) be a point where ϕ is differentiable. Then gb = −ϕ on a neighborhood of t, so gb is differentiable at t with derivative −ϕ′ (t). Proposition 1(iii) gives −ϕ′ (t) ∈ ∂b g (t), and ′ ′ ′ Proposition 1(i) gives −ϕ (t) ∈ ∂b g (t). We apply (45) at t with t = 1 and w = −ϕ (t). Since gb(1) = −ϕ(1) = −B0 and gb(t) = −ϕ(t), it reads
ψγ −(1 − t)ϕ′ (t) ≤ ϕ(t) − B0 . (46) If ϕ(t) < B0 , the right side of (46) is negative. Since ψγ ≥ 0 on [0, ∞), this forces −(1 − t)ϕ′ (t) < 0. On negative arguments ψγ (z) = z/γ, so the left side of (46) equals −(1 − t)ϕ′ (t)/γ. Multiplying (46) by −γ/(1 − t) < 0 gives ϕ′ (t) ≥

γ (B0 − ϕ(t)) 1−t

whenever ϕ(t) < B0 .

(47)

Let w(t) = max{B0 − ϕ(t), 0}. It is continuous on [0, 1], Lipschitz on every [t0 , t1 ] ⊂ (0, 1), and w(0) = B0 − A. At every t ∈ (0, 1) where ϕ and w are both differentiable, w′ (t) ≤ −

γ w(t) . 1−t

(48)

Indeed, if w(t) > 0, then ϕ(t) < B0 and, by continuity, w = B0 − ϕ near t, so (47) gives w′ (t) = −ϕ′ (t) ≤ −γw(t)/(1 − t). If w(t) = 0, then t is an interior minimizer of the nonnegative function w, so w′ (t) = 0, which equals the right side of (48). Since ϕ and w are Lipschitz on every [t0 , t1 ] ⊂ (0, 1), both are differentiable almost everywhere, so (48) holds at almost every t ∈ (0, 1). The function u(t) = w(t)(1 − t)−γ is a product of two Lipschitz functions on every [t0 , t1 ] ⊂ (0, 1), hence absolutely continuous there. By the product rule and (48),  γ w(t)  u′ (t) = (1 − t)−γ w′ (t) + γ(1 − t)−γ−1 w(t) = (1 − t)−γ w′ (t) + ≤0 1−t

(49)

almost R t ′ everywhere. For 0 < t0 < t < 1, absolute continuity and (49) give u(t) − u(t0 ) = t0 u (σ) dσ ≤ 0, that is, w(t)(1 − t)−γ ≤ w(t0 )(1 − t0 )−γ . (50) The right side of (50) tends to w(0) = B0 − A as t0 ↓ 0, because w is continuous at 0. Hence w(t) ≤ (B0 − A)(1 − t)γ . Since w ≥ B0 − ϕ, we obtain ϕ(t) ≥ B0 − (B0 − A)(1 − t)γ for t ∈ (0, 1). At t = 0 and t = 1 the right side equals ϕ(0) = A and ϕ(1) = B0 , which proves (44). The function σ 7→ σ γ is concave on [0, ∞) because γ ∈ (0, 1], so it lies below its tangent 1 + γ(σ − 1) at σ = 1. With σ = 1 − t this gives (1 − t)γ ≤ 1 − γt. 42

(51)

Since B0 − A ≥ 0, substituting (51) into (44) gives

ϕ(t) ≥ B0 − (B0 − A)(1 − γt) = A + γt(B0 − A) = γ (1 − t)A + tB0 + (1 − γ)A ≥ γ (1 − t)A + tB0 , where the last step uses A ≥ 0 and γ ≤ 1. This is (43) when r(y0 ) ≤ r(y1 ). If r(y0 ) > r(y1 ), applying this case to the pair (y1 , y0 ) with parameter 1 − t gives the same point (1 − t)y0 + ty1 and the bound γ(t r(y1 ) + (1 − t)r(y0 )). For vectors p and p′ , let [p, p′ ] denote the closed segment between them. The comparison vector is a rescaling of the support vector qf (x) by a factor in [γ, 1]. Theorem 24 (Comparison vector with coefficient γ). Let f : K → R be continuous and nonpositive with f ∈ F γ , and let x ∈ K. The set  v ∈ [γqf (x), qf (x)] : f (x) − γf (y) ≤ ⟨v, x − y⟩ for every y ∈ K (52) is nonempty, convex, and compact. Let g(f, x) be its element of minimum norm. Then f (x) − γf (y) ≤ ⟨g(f, x), x − y⟩

(y ∈ K),

g(f, x) ≤ qf (x) .

(53)

In particular, the continuous nonpositive functions on K in F γ form a lower-linearizable class with coefficient γ, and g(f, x) ≤ G under the oracle bound of Section 7. Proof. Write q = qf (x), p = −q, and r = −f ≥ 0, and for y ∈ K let sy = ⟨p, y − x⟩ and cy = γr(y) − r(x). Since ⟨y − x, q⟩ = −sy , inequality (5) at x with the vector q reads f (y) − f (x) ≥ ψγ (−sy ), that is, r(y) − r(x) ≤ −ψγ (−sy ). If sy ≥ 0, then −sy ≤ 0 and −ψγ (−sy ) = sy /γ. If sy < 0, then −sy > 0 and −ψγ (−sy ) = γsy . Hence ( sy /γ, sy ≥ 0, r(y) − r(x) ≤ (54) γsy , sy < 0. For a real number t, the identities f = −r and ⟨q, x − y⟩ = sy give f (x) − γf (y) − ⟨t q, x − y⟩ = −r(x) + γr(y) − t sy = cy − t sy ,

(55)

so t q satisfies the constraint of (52) for y exactly when cy ≤ t sy . If q = 0, then sy = 0 for all y, (54) gives r(y) ≤ r(x), and cy ≤ γr(x)−r(x) = −(1−γ)r(x) ≤ 0. So (52) equals {0} = [γq, q], and all claims hold with g(f, x) = 0. Assume from now on that q ̸= 0, so that [γq, q] = {t q : t ∈ [γ, 1]}, and sort the comparators by the sign of sy , using (54) in each case. • If sy = 0, then r(y) ≤ r(x), so cy ≤ −(1 − γ)r(x) ≤ 0 = t sy , and the constraint holds for every t. • If sy > 0, dividing cy ≤ t sy by sy shows that the constraint reads t ≥ cy /sy . Since r(y) ≤ r(x) + sy /γ and r(x) ≥ 0, cy ≤ γr(x) + sy − r(x) = sy − (1 − γ)r(x) ≤ sy ,

so

cy ≤ 1. sy

(56)

• If sy < 0, dividing cy ≤ t sy by sy < 0 shows that the constraint reads t ≤ cy /sy . Since r(y) − r(x) ≤ γsy and r(y) ≥ 0,
cy ≥ γ. (57) cy = r(y) − r(x) − (1 − γ)r(y) ≤ γsy , so sy 43

To compare lower and upper bounds, let y+ , y− ∈ K with s+ = sy+ > 0 > s− = sy− , let λ = −s− /(s+ − s− ) ∈ (0, 1), and let z = λy+ + (1 − λ)y− ∈ K. Since y 7→ sy is affine and 1 − λ = s+ /(s+ − s− ), sz = λs+ + (1 − λ)s− =

−s− s+ + s+ s− = 0, s+ − s−

so r(z) ≤ r(x) by the first case. Lemma 3 with y0 = y− , y1 = y+ , and t = λ gives r(z) ≥ γ(λr(y+ ) + (1 − λ)r(y− )). Combining the two bounds on r(z) with the definition of cy ,
(58) λcy+ + (1 − λ)cy− = γ λr(y+ ) + (1 − λ)r(y− ) − r(x) ≤ r(z) − r(x) ≤ 0. Multiplying (58) by s+ − s− > 0 and using λ(s+ − s− ) = −s− and (1 − λ)(s+ − s− ) = s+ gives −s− cy+ + s+ cy− ≤ 0, that is, s+ cy− ≤ s− cy+ . Dividing by s+ s− < 0 reverses the inequality and gives cy cy+ ≤ −. (59) s+ s− So every lower bound is at most every upper bound. Let n cy o ∗ t = max γ, sup , (60) sy >0 sy with the supremum over the empty set equal to −∞. By (56), t∗ ∈ [γ, 1], and by (60), t∗ is at least every lower bound. By (57) and (59), both γ and every lower bound are at most every upper bound, so t∗ is at most every upper bound. Since the first case imposes no restriction, (55) shows that t∗ q lies in (52). The set (52) is the intersection of the compact segment [γq, q] with the closed half-spaces {v : cy ≤ ⟨v, x − y⟩}, which are all of Rd when y = x because cx ≤ 0. So it is convex and compact. The continuous function ∥·∥2 attains its minimum on it, and the minimizer is unique: if v1 ̸= v2 2 both attain the minimum m, then (v1 + v2 )/2 = 21 ∥v1 ∥2 + 12 ∥v2 ∥2 − 41 ∥v1 − v2 ∥2 < m. This defines g(f, x), and its membership in (52) is the first part of (53). Every t q with t ∈ [γ, 1] has norm t∥q∥ ≤ ∥q∥, which gives the second part. The final sentence of the theorem follows from (41) with α = γ and qf (x) ≤ G. The factor in [γ, 1] that defines g(f, x) depends on the values of f on all of K, so in general it cannot be read off from the oracle at x. Because one vector serves every comparator, it can be passed to the linear learner before the comparator is known. 8.4 Computing the comparison vector The set (52) is cut out by one constraint for each comparator y ∈ K, so we search the segment [γqf (x), qf (x)] by bisection on the scale factor and test each candidate by approximately maximizing its constraint violation over y. In this subsection K is compact with diameter at most D, and f : K → R is nonpositive and G–Lipschitz with f ∈ F γ , with the access of Section 7. For a base point x and a vector v, the violation of v is   Φx (v) = max f (x) − γf (y) − ⟨v, x − y⟩ . (61) y∈K

It is convex and D–Lipschitz in v, and a vector v ∈ [γqf (x), qf (x)] lies in (52) exactly when Φx (v) ≤ 0. The function maximized in (61) is not concave in y, so convex optimization does not evaluate Φx directly. Lemma 4 instead splits the range of f into levels and, on each level a, maximizes the linear function ⟨v, y − x⟩ over the sublevel set {y ∈ K : f (y) ≤ a}. 44

Algorithm 2 Computing a comparison vector with coefficient γ Require: Oracle Qf , point x ∈ K, parameter γ ∈ (0, 1], tolerance ε > 0, bounds D, G > 0 1: Query f (x) and qf (x). Return 0 if qf (x) = 0, and return qf (x) if γ = 1. 2: Set t = γ, t = 1, and h = ε/16. 3: while DG(t − t) > ε/4 do 4: Set t = (t + t)/2 and v = t qf (x). 5: Obtain yv and Vv from Lemma 4 with accuracy h. 6: if Vv ≤ ε/2 then 7: return v. 8: end if 9: if ⟨qf (x), x − yv ⟩ > 0 then 10: Set t = t. 11: else 12: Set t = t. 13: end if 14: end while t+t 15: return 2 qf (x). Lemma 4 (Evaluating the violation). Let x ∈ K, ∥v∥ ≤ G, and h > 0. A deterministic procedure returns a point yv ∈ K whose value Vv = f (x) − γf (yv ) − ⟨v, x − yv ⟩ satisfies Φx (v) − 4h ≤ Vv ≤ Φx (v),

(62)

 
O (1 + γDG/h) 1 + (DG/h)2

(63)

using value, support, and projection calls. In the proof, given in Appendix E.1, a projected method at level a moves against qf (zj ) when f (zj ) exceeds a + h/γ and along v otherwise. Inequality (5) shows that either move makes progress at least h toward a maximizer of ⟨v, y⟩ on {f ≤ a}, and the projected-subgradient potential [BTN01] bounds the number of steps. The levels are spaced h/γ apart because f (y) enters (61) with weight γ, and this spacing keeps inverse powers of γ out of (63). Algorithm 2 combines Lemma 4 with bisection on the scale factor. It maintains an interval of scale factors that contains a feasible one, which exists initially by Theorem 24. When a candidate t qf (x) has a large violation, the witness yv gives a constraint that every feasible scale factor satisfies and t violates, and the sign of ⟨qf (x), x − yv ⟩ determines on which side of t the feasible scale factors lie. Theorem 25 (Computing the comparison vector). Let x ∈ K and ε > 0, and let f : K → R be nonpositive and G–Lipschitz with f ∈ F γ . (i) Exact oracle. Algorithm 2 returns gb ∈ [γqf (x), qf (x)] with f (x) − γf (y) ≤ ⟨b g , x − y⟩ + ε

(y ∈ K),

(64)

using  
O (1 + γDG/ε) log(2 + DG/ε) 1 + (DG/ε)2 oracle calls. 45

(65)

(ii) Inexact oracle. Suppose that the oracle returns values fb(z) with |fb(z) − f (z)| ≤ ζ and vectors az with ∥az ∥ ≤ G and
f (y) − f (z) ≥ ψγ ⟨az , y − z⟩ − ξ (y, z ∈ K), (66) where ξ, ζ ≥ 0 are known. A variant of Algorithm 2 with the same number of calls returns gb ∈ [γax , ax ] with f (x) − γf (y) ≤ ⟨b g , x − y⟩ + ε + (1 + γ)ξ + (2 + 4γ)ζ

(y ∈ K).

(67)

(iii) Bit model. Suppose that B ≥ supK |f | is known. For rational γ, part (ii) has a rational implementation returning gb ∈ [γa, a], where a is rational, ∥a − ax ∥ ≤ γε/(16D), ∥a∥ ≤ G, and the output satisfies (67). For an effectively specified γ (Section 8.1), the implementation returns a rational vector satisfying (67) at the original γ, with ∥b g ∥ ≤ G. If B > 0, it uses a rational γ ′ ∈ [γ/2, γ] and guarantees gb ∈ [γ ′ a, a]; if B = 0, it may return zero. Running time is polynomial when 1/ε, D, G are polynomially bounded and the required numerical bounds and precisions have polynomial encoding length. The number of oracle calls has no inverse-power dependence on γ. For γ = 1, the support vector qf (x) itself satisfies (64) with ε = 0. The proof is in Appendix E.2. In part (ii), the defect ξ measures how far az is from satisfying the support inequality: if az lies within distance ν of a support vector, then (66) holds with ξ = Dν/γ because ψγ is (1/γ)–Lipschitz, and the rational implementation in part (iii) relies on this conversion. 8.5 Regret bounds We run Algorithm 1 with the procedure G of Theorem 25 at tolerance εt in round t. If the oracle for ft has support defect ξt and value error ζt as in Theorem 25(ii), let X errt = εt + (1 + γ)ξt + (2 + 4γ)ζt , Err(I) = errt . (68) t∈I

Errors from rational arithmetic are included in εt and in the bound BA of the linear learner. Corollary 10 (Static γ–regret). Let f1 , . . . , fT : K → R be nonpositive and G–Lipschitz with ft ∈ F γ . For every interval I = [s, e] and every comparator path, Algorithm 1 with the vectors of Theorem 25 satisfies Regγ (I; us:e ) ≤ BA (I; us:e ) + Err(I). (69) √ With projected online gradient descent with step size D/(G T ) as A, this gives SRegγ (T ) ≤ √ √ DG T + Err([1, T ]). If errt ≤ c/ t for every t, then every interval I = [s, e] satisfies p √
√ Err(I) ≤ 2c e − s − 1 ≤ 2c |I|, (70) √ and with an exact oracle the tolerances εt = c/ t satisfy this condition. Round t uses the number of oracle calls in (65) with ε = εt . With inverse-polynomial tolerances, the algorithm runs in polynomial time in the bit model. Inequality (69), proved in Appendix E.5, follows by summing (67) with x = xt and y = ut over t ∈ I and applying (42). For √ a known horizon and DG > 0, choose √ rational per-round comparison at most another DG T /2 to rational updates of the errors with errt ≤ DG/(2 T ) and charge √ linear learner, so that SRegγ (T ) ≤ 2DG T . The required precisions have polynomial encoding length under Section 8.1, so γ is achievable on F γ . If D = 0 or G = 0, every feasible point has the same loss and the approximate regret is nonpositive. 46

Remark 3 (Moving comparators and intervals). Since inequality (69) holds for every interval and every comparator path, every deterministic linear learner whose regret bound holds for all realized sequences of vectors gives the same bound on γ–regret, up to Err(I). Projected online graP dientpdescent with a
step size tuned to a known bound P on the path length t ∥ut+1 − ut ∥ gives O G T (D2 + DP ) against all such paths [Zin03]. The strongly adaptive learner of Daniely, Gonen,pand Shalev-Shwartz [DGSS15], played at the weighted average of its experts’ points, gives
O DG |I| log T on every interval I against a fixed comparator, because the loss of a linear function at the weighted average equals the weighted average of the losses.

9. Hardness Beyond γ: A Common Framework We give a class-independent criterion for hardness of approximate online optimization and verify it for the smallest of our three classes at a fixed parameter. By Corollary 8, γ γ Fmono ⊆ Fsec ⊆ Fγ.

(71)

For every fixed γ ∈ (0, 1), we prove that no coefficient α > γ is achievable on any of these classes unless NP = RP. The proof has two parts: a gap property that does not mention generalized convexity, and a scalar-profile criterion that places the same hard family in all three classes. Each loss in this family is smooth and can be minimized by one linear optimization. 9.1 The gap property The losses are continuous functions ft : K → [−1, 0] on a compact rational convex set K, and regret is the static α–regret (38). Computations use the bit model of Section 8.1, with rational actions of polynomial bit length, and P , ρ, and N are as in (39). Lipschitz continuity and smoothness of a loss refer to its restriction to K. Definition 14 (Online gap property). Let β ∈ [0, 1), and let L(K) be a class of continuous losses f : K → [−1, 0] for each compact rational convex set K. The class has the β–online gap property if, for every fixed α ∈ (β, 1], a deterministic polynomial-time reduction from SAT produces the following data: (i) a set K, a nonempty polynomial-size index set I, and explicit losses fi ∈ L(K) for i ∈ I; (ii) rational numbers Λ > 0 and 0 ≤ τ < α, and a score Q : K → [0, 1], such that 1 X Q(x) ; fi (x) = − |I| Λ

(72)

i∈I

(iii) on a satisfiable formula, a point x∗ ∈ K exists with Q(x∗ ) = 1; (iv) on an unsatisfiable formula, Q(x) ≤ τ for every rational x ∈ K. All descriptions have polynomial encoding length; the numerical quantities Λ and (α − τ )−1 are polynomially bounded in the SAT input size. The score is rational and exactly computable in polynomial time at rational points, so comparison with τ is exact. Loss evaluation and every stipulated feasibility or loss oracle are polynomial-time rational computations. Uniform indices from I can be sampled in expected polynomial time, with a polynomially bounded implementation obtained by capping rejection sampling.

47

Theorem 26 (Class-independent hardness criterion). Suppose L has the β–online gap property. Fix α ∈ (β, 1]. If a randomized polynomial-time online algorithm with rational actions satisfies either E[SRegα (T )] ≤ P (N )T 1−ρ , (73) which says that α is achievable, or P{SRegα (T ) ≤ P (N )T 1−ρ } ≥ 8/9,

(74)

for every horizon and every admissible loss sequence, then NP = RP. It suffices to assume the guarantee for independent uniform samples from each finite family produced by the reduction. In the proof (Appendix F.1), independent samples on a satisfiable instance satisfy  α − Q(xt )  E fJt (xt ) − αfJt (x∗ ) past and learner’s random tape = . Λ Summed over the rounds, this identity shows that a low-regret guarantee forces some iterate to have score above τ with inverse-polynomial probability after polynomially many rounds. An unsatisfiable instance never produces such an iterate, so testing the scores of the iterates gives a one-sided randomized algorithm for SAT. The argument uses no property of generalized convexity. 9.2 A scalar-profile criterion Theorem 27 reduces the gap property to a one-dimensional inequality and to membership of lifted losses in the class. For p, q ≥ 1, let   q p   X Y xu,a ≤ 1 . (75) Kp,q = xu ∈ Rq+ :   a=1

u=1

For ℓ ≥ 1, block indices uj ∈ [p], and subsets Rj ⊆ [q], define ℓ

1X X zu,R (x) = xuj ,a . ℓ

(76)

j=1 a∈Rj

Repeated indices are allowed, and 0 ≤ z ≤ 1 on Kp,q . Writing z(x) = ⟨a, x⟩, we have a ≥ 0 and ∥a∥22 ≤ q. Theorem 27 (Direct hard-profile criterion). Let β ∈ [0, 1). Suppose that for each rational θ ∈ (β, 1) there are a nondecreasing Lipschitz function Hθ : [0, 1] → [0, 1] and rational constants 0 ≤ aθ < θ and bθ > 0 such that Hθ (0) = 0,

Hθ (1) = 1,

Hθ (s) ≤ s,

Hθ (s) ≤ aθ s + bθ s2

(0 ≤ s ≤ 1).

(77)

Suppose also that for every p, q, ℓ there is a rational scale Sθ,p,q,ℓ ≥ 1, polynomially bounded in p + q + ℓ for fixed θ, such that all losses fu,R (x) = −

Hθ (zu,R (x)) Sθ,p,q,ℓ

(78)

belong to L(Kp,q ). The profiles, scales, losses, and stipulated oracles must have polynomial-time rational implementations, including exact profile evaluation on rational inputs. Then L has the β–online gap property. 48

The proof in Appendix F.2 reduces from Label Cover. An instance Γ of Label Cover consists of a bipartite graph with left vertices VL , right vertices VR , and edges E, finite label sets ΣL and ΣR , and a map πe : ΣL → ΣR for each edge e. A labeling assigns a label in ΣL to each left vertex and a label in ΣR to each right vertex, and it satisfies an edge e = (u, v) when πe sends the label of u to the label of v. The value val(Γ) is the largest fraction of edges that one labeling satisfies. For every fixed δ ∈ (0, 1), a polynomial-time reduction maps satisfiable SAT formulas to instances of value 1 and unsatisfiable ones to instances of value at most δ [MR10, Mos15]. P L The feasible set KΓ is a product of simplices, with one block {xu ∈ RΣ + : a xu,a ≤ 1} for each left vertex u, and a labeling of the left vertices corresponds to the one-hot point that puts weight one on the label of each u. For a right vertex v and a label b ∈ ΣR , the projected mass zv,b (x) averages, over the edges e = (u, v) at v, the weight that xu places on labels sent to b by πe , so that P z ≤ 1. With wv the fraction of edges incident to v, the score is b v,b QΓ (x) =

X

wv

X

v

Hθ (zv,b (x)).

b

It equals one at the one-hot point of a labeling that satisfies every edge, and rounding x deterministically to a labeling (Appendix F.2) gives QΓ (x) ≤ aθ + bθ val(Γ). On an unsatisfiable instance, the linear term aθ s of (77) still contributes up to aθ , since it does not depend on whether labels agree across edges; the quadratic term contributes only through agreement, and rounding bounds that contribution by bθ val(Γ). Choosing δ small enough that aθ + bθ δ < α establishes the gap for any α > θ, and θ can be taken arbitrarily close to β. The only class-specific step is membership of (78). 9.3 Profiles with a sharp monotonicity parameter The quadratic profile hθ (s) = θs+(1−θ)s2 satisfies −hθ = (1−θ)hc with c = θ/(1−θ). Multiplying a function by a positive constant does not change its parameters, so Example 4 gives ∗ γmono (−hθ + δ[0,1] ) =

c θ = . c+2 2−θ

(79)

γ Consequently, requiring this quadratic to belong to Fmono forces θ ≥ 2γ/(1 + γ), which is too large for a gap approaching γ. The score gap is controlled by the initial slope of the profile, whereas membership in the monotonicity class is controlled by the ratio of its smallest and largest slopes; for the quadratic these quantities differ by the factor 2 − θ. We choose a polynomial profile whose largest slope approaches one while its smallest-to-largest slope ratio remains fixed at κ, so its initial slope approaches κ and the score threshold can approach γ from above. When the extreme slopes occur at the endpoints, the slope-ratio criterion is exact: ∗ Lemma 5 (Derivative ratios determine γmono ). Let H be C 1 on a neighborhood of [0, 1], and ′ ′ suppose 0 < A ≤ H (s) ≤ B on [0, 1] with H (0) = A and H ′ (1) = B. Then Ψ = −H + δ[0,1] has ∗ γmono (Ψ) = A/B.

Within the normalized convex-profile construction considered here, the threshold cannot fall below γ. Let H be a convex profile that is C 1 near [0, 1], with H(0) = 0, H(1) = 1, and H ′ (0) > 0. ∗ Convexity makes H ′ nondecreasing, so Lemma 5 gives γmono (−H +δ[0,1] ) = H ′ (0)/H ′ (1). Convexity

49

γ also gives 1 = H(1) − H(0) ≤ H ′ (1) and H(s) ≥ H(0) + H ′ (0)s = H ′ (0)s. Membership in Fmono means H ′ (0) ≥ γH ′ (1) ≥ γ, and therefore

H(s) ≥ γs

(0 ≤ s ≤ 1). (80) P At a point P whose P blocks each carry total weight one, b zv,b (x) = 1 for every v, so (80) gives QΓ (x) ≥ γ v wv b zv,b (x) = γ, on satisfiable and unsatisfiable instances alike. The reduction can lower the unsatisfiable score to just above γ but no further, which matches the coefficient γ achieved by Corollary 10. For rational κ ∈ (0, 1) and a positive odd integer m, define   (m + 1)s − (1 − κ) 1 − (1 − s)m+1 Hκ,m (s) = , s ∈ [0, 1], (81) m+κ and set B=

m+1 , m+κ

A = κB,

C=

m(1 − κ)B . 2

(82)

Lemma 6 (A polynomial hard profile). The profile (81) is increasing and convex on [0, 1], satisfies H(0) = 0,

H(1) = 1,

0 ≤ H(s) ≤ s,

H(s) ≤ As + Cs2 ,

(83)

and has A ≤ H ′ (s) ≤ B and 0 ≤ H ′′ (s) ≤ 2C there. For Ψ = −H + δ[0,1] , ∗ γmono (Ψ) = κ.

(84)

Moreover, A ↓ κ as m → ∞ through odd integers. The polynomial H is convex on all of R. The proof is in Appendix F.3. To apply Theorem 27 with β = γ, fix a rational θ ∈ (γ, 1), choose rational κ ∈ (γ, θ), and choose an odd integer m>

κ(1 − θ) . θ−κ

(85)

Since A = κ(m + 1)/(m + κ), the inequality A < θ is equivalent to m(θ − κ) > κ(1 − θ), so (85) gives A < θ. Use Hθ = Hκ,m , aθ = A, bθ = C, and Sθ,p,q,ℓ = 1 + q(B + 2C).

(86)

The lifted losses are 1–Lipschitz and 1–smooth on Kp,q , and their proper extensions have κ–tilted γ monotone Clarke subdifferentials, including at the boundary, and therefore belong to Fmono . Theorem 28 (The common hardness threshold). Fix γ ∈ (0, 1) and α = γ + ε ≤ 1, where ε > 0 is fixed. Consider the classes γ γ Fmono , Fsec , Fγ. For any one of them, suppose a randomized polynomial-time algorithm satisfies either (73) or (74) for all sequences of nonpositive losses with Lipschitz constant at most one. Then NP = RP. The conclusion already holds on products of simplices for restrictions of rational, C ∞ , concave polynomials whose Hessians have rank at most one. On the feasible set, each loss is [−1, 0]-valued, 1–Lipschitz, 1–smooth, and coordinatewise nonincreasing. Its degree and number of relevant coordinates are bounded by constants depending only on (γ, α), and one linear optimization gives an exact minimizer. The losses may be independent uniform samples from a polynomial-size family. For F γ alone, the same hardness conclusion holds even when every loss is a single concave quadratic. 50

The proof is in Appendix F.4. The profile and its lifted losses verify the common criterion for γ Fmono ; inclusion (71) transfers it to the other two classes. The final assertion follows from the same criterion using hθ (s) = θs + (1 − θ)s2 . All parameters are fixed before the SAT reduction: the

polynomial degree and the Label Cover alphabet and degree bounds may depend on (γ, α), but not on the SAT input size. Corollary 11 (Exact coefficients and the endpoint classes). Under the oracle assumptions of Corollary 10, the online coefficient of each of the three classes is γ, unless NP = RP. For pseudoconvex and for quasiconvex nonpositive smooth losses, no positive coefficient is achievable unless NP = RP. For the first claim, Corollary 10 achieves the coefficient γ on F γ , and by (71) the same algorithm applies to the two subclasses, while Theorem 28 excludes every larger coefficient on each of the three classes. The products of simplices used by the reduction meet the domain assumptions of √ Section 8: projection decomposes over the blocks, and the diameter is at most pq (Appendix F.4). For the second claim, given any proposed positive coefficient on pseudoconvex or quasiconvex losses, choose 0 < γ < α; the constructed losses also belong to these endpoint classes, as verified directly in Appendix F.4. All lower bounds are for fixed (γ, α); no uniform polynomial dependence on 1/(α − γ) is asserted. At γ = 1 there is no coefficient in (γ, 1] to exclude.

10. Conclusion For every fixed, effectively specified 0 < γ < 1, the three formulations define distinct classes with the same optimal polynomial-time online coefficient: an online algorithm on the largest class achieves coefficient γ, and the hardness result on the smallest class rules out every larger coefficient unless NP = RP, already for smooth losses, whereas known methods for constrained quasarconvex functions minimize smooth members to any accuracy. The smooth support inequality is the equal-parameter tilted-convex condition, and its proper lower semicontinuous extension has convex effective domains, local regularity on relative interiors, and equivalent formulations through different generalized gradients. The secant class is exactly the class of proper H√γ –convex functions, and the examples show that the parameter losses in the inclusions are sharp and that none of the classes is closed under addition. The online algorithm uses polynomially many queries to each revealed loss, and whether coefficient γ can be achieved with one gradient query per round, or with only a value at the played point, remains open. Offline, it is open whether the accelerated rate of [MR26] is optimal on the three classes, since the lower bound for quasar-convex functions [HSS20] need not apply to them.

Use of AI Tools ChatGPT (OpenAI) assisted with checking the proofs, reviewing the manuscript, and proposing revisions. Claude Opus (Anthropic) assisted with exposition. The authors take full responsibility for the statements, proofs, references, and other content of the submitted paper.

51

Appendix A. Useful lemmas We use the following chain rule for subdifferentials; see, e.g., Theorem 10.6 in [RW98]. Theorem 29. Let X ⊆ Rm . Suppose g = f (F (x)) for a proper lsc function f : X → R and a continuously differentiable function F : Rn → Rm . Then, at any point x ∈ dom(g) = F −1 (dom(f )), we have ˆ ˆ (F (x)). ∂g(x) ⊇ ∇F (x)⊤ ∂f Moreover, if f is locally Lipschitz 7 at F (x), then ∂g(x) ⊆ ∇F (x)⊤ ∂f (F (x)). A common Lipschitz constant turns pointwise convergence on a dense set into uniform convergence on bounded sets; see, e.g., Proposition 9.59 in [RW98]. Theorem 30. Let X ⊆ Rm . Suppose that the functions fk : X → R, k ≥ 1, are Lipschitz continuous with the same constant L ≥ 0 and converge pointwise on a dense subset D ⊆ X . Then they converge uniformly on bounded subsets of X to a function f : X → R which is Lipschitz continuous on X with the same constant L. Moreover, for any point x ∈ int(X ) and v ∈ ∂f (x), there are sequences xk → x and vk ∈ ∂fk (xk ) such that vk → v. We also use Lebourg’s mean value theorem for Lipschitz functions; see, e.g., Theorem 2.3.7 in [Cla90]. Theorem 31 (Lebourg’s mean value theorem for Lipschitz functions). Let K ⊆ Rd be convex, let x, y ∈ K be distinct, and let f : K → R be Lipschitz on an open neighborhood of the line segment [x, y]. Then there is a point z ∈ (x, y) and v ∈ ∂f (z) = conv(∂f (z)) such that 8 f (y) − f (x) = ⟨v, y − x⟩. Lebourg’s theorem extends to the case where f is only continuous at the end-points. Corollary 12. Let K ⊆ Rd be convex, let x, y ∈ K be distinct, and let f : K → R be continuous on [x, y] and locally Lipschitz on an open set containing (x, y). Then there is a point z ∈ (x, y) and v ∈ ∂f (z) = conv(∂f (z)) such that f (y) − f (x) = ⟨v, y − x⟩. Proof. For λ ∈ [0, 1], let zλ := (1 − λ)x + λy. Define the function ϕ : [0, 1] → R by ϕ(λ) := f (zλ ) − (1 − λ)f (x) − λf (y). Then ϕ is continuous because f is continuous on [x, y], and ϕ(0) = ϕ(1) = 0. We claim that there are real numbers a, b with 0 < a < b < 1, ϕ(a) = ϕ(b). (87) If ϕ ≡ 0, any 0 < a < b < 1 satisfies (87). Otherwise, choose c ∈ (0, 1) with ϕ(c) ̸= 0. The value ϕ(c)/2 lies strictly between ϕ(0) = 0 and ϕ(c), and also strictly between ϕ(c) and ϕ(1) = 0. By The statement of Theorem 10.6 in [RW98] uses the assumption that the only vector v ∈ ∂ ∞ f (F (x)) with ∇F (x)⊤ v = 0 is v = 0. However, if f is locally Lipschitz at F (x), then ∂ ∞ f (F (x)) = {0} (e.g., see Theorem 9.13 in the same book) and this condition is trivially satisfied. 8 See Proposition 1 for the equality ∂f (z) = conv(∂f (z)). 7

52

the intermediate value theorem there are a ∈ (0, c) and b ∈ (c, 1) with ϕ(a) = ϕ(b) = ϕ(c)/2, which gives (87). Writing out ϕ(a) = ϕ(b) gives f (za ) − (1 − a)f (x) − af (y) = f (zb ) − (1 − b)f (x) − bf (y).
Rearranging, f (zb ) − f (za ) = (b − a) f (y) − f (x) , that is, f (y) − f (x) =

f (zb ) − f (za ) . b−a

(88)

Because [za , zb ] is compact and lies in (x, y), there is r > 0 such that its closed r–neighborhood Cr lies in the open set on which f is locally Lipschitz. The set Cr is compact and convex, so covering it by finitely many balls on which f is Lipschitz and splitting each segment in Cr at the boundaries of these balls shows that f is Lipschitz on Cr , and hence on the open r–neighborhood of [za , zb ]. Lebourg’s theorem (Theorem 31) applied to f on [za , zb ] therefore gives z ∈ (za , zb ) ⊆ (x, y) and v ∈ conv(∂f (z)) such that f (zb ) − f (za ) = ⟨v, zb − za ⟩. (89) By definition of zλ , we have zb − za = (b − a)(y − x). Substituting (89) into (88) gives   f (zb ) − f (za ) zb − za f (y) − f (x) = = v, = ⟨v, y − x⟩. b−a b−a For general lsc functions, an approximate mean-value theorem holds [Mor06]; we include a finite-dimensional proof. Theorem 32 (Approximate mean values for lsc functions). Let f : Rd → R be a proper lsc function. Let x ̸= y in X := dom(f ) and, for λ ∈ [0, 1], let zλ = λx + (1 − λ)y. Then there is λ ∈ [0, 1) with f ˆ (xk ) such that zλ ∈ dom(f ) and sequences xk − → zλ and vk ∈ ∂f 1 lim sup⟨vk , xk − x⟩ ≤ f (y) − f (x), 1 − λ k→∞ and lim sup⟨vk , y − x⟩ ≤ f (y) − f (x). k→∞ f ˆ (x′ ) such that Similarly, there is λ′ ∈ (0, 1] with zλ′ ∈ dom(f ) and sequences x′k − → zλ′ and vk′ ∈ ∂f k

f (y) − f (x) ≤

1 lim inf ⟨v′ , y − x′k ⟩, λ′ k→∞ k

and f (y) − f (x) ≤ lim inf ⟨vk′ , y − x⟩. k→∞

Proof. Put d := y − x, S := [x, y], and q := (f (y) − f (x))/∥d∥2 . Define φ(u) := f (u) − f (x) − q⟨d, u − x⟩, so that φ(x) = φ(y) = 0. The function φ is proper and lsc. It therefore has a finite minimum m ≤ 0 on the compact segment S. For ϵ > 0, write φϵ (u) := φ(u) − ϵ⟨d, u − x⟩. 53

For all sufficiently small ϵ, the minimizers of φϵ on S stay a fixed positive distance from x. Indeed, if m < 0, lower semicontinuity at x gives a neighborhood where φ > m/2, while minS φϵ ≤ m; the affine perturbation is uniformly smaller than −m/4 on S for small enough ϵ. If m = 0, the unique minimizer of φϵ on S is y, since φ ≥ 0 there and ⟨d, u − x⟩ ≤ ∥d∥2 , with equality only at y. Let C := {u : dist(u, S) ≤ 1}, a compact neighborhood of S. For fixed ϵ > 0 and each integer n ≥ 1, choose a minimizer uϵ,n over C of φϵ (u) + n dist(u, S)2 . These minimizers exist because the objective is lsc, bounded below on C, and finite at y. Comparison with a minimizer on S shows dist(uϵ,n , S) → 0 as n → ∞. Every cluster point minimizes φϵ on S: lower semicontinuity and the nonnegative penalty give this after passing to the limit in the same comparison. Choose ϵk ↓ 0. By the preceding two paragraphs, we can choose nk → ∞ so that uk := uϵk ,nk lies in the interior of C, dist(uk , S) → 0, and uk stays a fixed positive distance from x. Passing to a subsequence, assume uk → z ∈ S \ {x}. The minimizing inequality and minS φϵk ≤ m imply φ(uk ) ≤ m + ϵk ⟨d, uk − x⟩. f

Since φ(z) ≥ m and φ is lsc, it follows that φ(uk ) → φ(z) = m and hence uk − → z. Let pk := ΠS (uk ). For large k, pk ̸= x. The squared distance to a nonempty closed convex set is continuously differentiable, with gradient 2(u − ΠS (u)). The necessary condition for the local minimum at uk , followed by the smooth-sum rule in Proposition 1(v), gives ˆ rk := ϵk d − 2nk (uk − pk ) ∈ ∂φ(u k ). Projection onto the segment gives ⟨uk − pk , d⟩ ≥ 0,

⟨uk − pk , pk − x⟩ ≥ 0.

Both inner products vanish if pk lies in the open segment; if pk = y, they are nonnegative by the projection inequality. Consequently, ⟨rk , d⟩ ≤ ϵk ∥d∥2 ,

⟨rk , uk − x⟩ ≤ ϵk ⟨d, uk − x⟩.

ˆ (uk ) therefore gives Writing z = λx + (1 − λ)y with λ ∈ [0, 1) and setting vk := rk + qd ∈ ∂f lim sup⟨vk , d⟩ ≤ f (y) − f (x), k

lim sup⟨vk , uk − x⟩ ≤ (1 − λ)(f (y) − f (x)). k

Taking xk := uk proves the first pair of inequalities. Interchanging x and y proves the second pair. The lim sup and lim inf versions are equivalent: each follows from the other by swapping x and y.

Appendix B. Approximate γ–tilted convexity This appendix studies a weaker notion than γ–tilted convexity. Its results use only Section 3, and the proofs in Appendix C rely on them: those of Theorems 6, 7, 8, 9, 11, 15, 16, and 18, and through them the rest of Section 5. Section 5.1.2 summarizes the notion and its role in the main text. 54

Definition 15 (Approximate γ–tilted convexity). Let γ ∈ (0, 1], let f : Rd → R be a proper lsc function, and let F : Rd ⇒ Rd be a multivalued operator. We say that f is approximately γ–tilted convex with respect to F if, for every x ∈ dom(f ) ∩ dom(F ) and every ϵ > 0, there exists v ∈ F (x) such that f (y) − f (x) ≥ ψγ (⟨y − x, v⟩) − ϵ ∀y ∈ dom(f ). Here the operator F need not be a subdifferential, and the supporting vector may depend on ϵ; in particular, exact γ–tilted convexity with respect to F implies approximate γ–tilted convexity with respect to F . If the domain is convex and approximate supports exist on a dense subset of it, then the function is locally Lipschitz on the relative interior of its domain and continuous along every line meeting that relative interior. B.1 Local boundedness Approximately γ–tilted convex functions are bounded above on finitely generated subsets of their domain. Lemma 7. Let γ ∈ (0, 1], let f : Rd → R be proper and lsc, and set K := dom(f ). Assume that K is convex. Let F : Rd ⇒ Rd be such that K ∩ dom(F ) is dense in K, and suppose that f is approximately γ–tilted convex with respect to F in the sense of Definition 15. If y1 , . . . , ym ∈ K satisfy aff{y1 , . . . , ym } = aff(K), then f (x) ≤ max f (yi )

∀x ∈ conv{y1 , . . . , ym }.

1≤i≤m

Proof. Let C := conv{y1 , . . . , ym }, Each yi

M + := max f (yi ). 1≤i≤m

lies in K, so M + ∈ R. Convexity of K gives C ⊆ K. We first prove that

f (x) ≤ M + ∀x ∈ K ∩ dom(F ) ∩ ri(C). (90) P P + Fix such an x and write x = m i=1 λi yi , where λi ≥ 0 and i λi = 1. Suppose that f (x) > M . + Choose 0 < ϵ < f (x)−M . Definition 15, applied at x ∈ dom(f )∩dom(F ), gives a vector v ∈ F (x) with f (y)−f (x) ≥ ψγ (⟨y−x, v⟩)−ϵ for all y ∈ K. Taking y = yi and using f (yi ) ≤ M + < f (x)−ϵ gives ψγ (⟨yi − x, v⟩) ≤ f (yi ) − f (x) + ϵ < 0 (1 ≤ i ≤ m). (91) Applying the increasing inverse ψ γ to (91) and using ψ γ (t) = γt for t < 0, we obtain
⟨yi − x, v⟩ ≤ γ f (yi ) − f (x) + ϵ (1 ≤ i ≤ m). (92) P Multiply (92) by λ and sum over i. The left-hand sides add up to ⟨ i i λi yi − x, v⟩ = 0, and using P + i λi = 1 and f (yi ) ≤ M on the right gives   m X
0≤γ λi f (yi ) − f (x) + ϵ ≤ γ M + − f (x) + ϵ < 0, i=1

where the last inequality is the choice of ϵ. This contradiction proves (90).

55

Next we show that D := K ∩ dom(F ) ∩ ri(C) is dense in C. The set C is nonempty and convex, so ri(C) is dense in C (Theorem 6.3 in [Roc15]). It therefore suffices to show that every r ∈ ri(C) lies in the closure of D. By the definition of the relative interior, there is δ > 0 with B(r, δ) ∩ aff(C) ⊆ C. If w ∈ B(r, δ/2) ∩ aff(C), then B(w, δ/2) ∩ aff(C) ⊆ C, so w ∈ ri(C). By hypothesis, aff(C) = aff(K) ⊇ K, and hence B(r, δ/2) ∩ K ⊆ ri(C).

(93)

Let 0 < η < δ/2. Since r ∈ C ⊆ K and K ∩ dom(F ) is dense in K, the ball B(r, η) contains a point of K ∩ dom(F ). By (93), this point lies in D. As η was arbitrary, r lies in the closure of D. Finally, the sublevel set {y ∈ Rd : f (y) ≤ M + } is closed by lower semicontinuity. By (90) it contains D, so it contains the closure of D, which contains C. B.2 Local Lipschitz continuity The local Lipschitz constant depends only on a local bound for f and on the local geometry of its effective domain. Theorem 33 (Local Lipschitz regularity from approximate γ–tilted convexity). Let γ ∈ (0, 1], let f : Rd → R be proper and lsc, and set K := dom(f ). Assume that K is convex. Let F : Rd ⇒ Rd be such that K ∩ dom(F ) is dense in K, and suppose that f is approximately γ–tilted convex with respect to F in the sense of Definition 15. Then f is locally Lipschitz on ri(K). More precisely, put A := aff(K). If x0 ∈ ri(K) and R > 0 satisfy B(x0 , 3R) ∩ A ⊆ K, then n o M := sup |f (x)| : x ∈ B(x0 , 2R) ∩ A < ∞, and |f (x) − f (y)| ≤

2M ∥x − y∥ γ2R

∀x, y ∈ B(x0 , R) ∩ A.

(94)

∆ ∥x − y∥, γ2R

x, y ∈ B(x0 , R) ∩ A,

(95)

The translation-invariant form is |f (x) − f (y)| ≤ where ∆ :=

f−

sup B(x0 ,2R)∩A

inf

f.

B(x0 ,2R)∩A

Proof. If dim A = 0, then A = {x0 }, so M = |f (x0 )| < ∞ and (94) only concerns x = y = x0 . Suppose therefore that k := dim A ≥ 1. Let V := A − A, a linear subspace of dimension k, and fix an orthonormal basis e1 , . . . , ek of V . First we verify that f is bounded on B(x0 , 2R) ∩ A. Choose finitely many points y1 , . . . , ym ∈ B(x0 , 3R) ∩ A such that the polytope C := conv{y1 , . . . , ym } has affine hull A and contains B(x0 , 2R) ∩ A; we build them from cubes. For p ∈ A, consider the cube k n o X √ Qp := p + si ei : |si | ≤ R/ k (1 ≤ i ≤ k) ⊆ A. (96) i=1

56

√ P It is the convex hull of its 2k vertices p + (R/ k) i σi ei with σi ∈ {−1, 1}. By orthonormality, each vertex has distance exactly R from p. The √ vertices have affine hull A, because the difference of two vertices that differ only in σ is ±(2R/ k)ei . The cube Qp contains the set Op := {x ∈ i √ A : |⟨x − p, ei ⟩| < R/ k (1 ≤ i ≤ k)}, which is open relative to A and contains p. The sets Op with p ∈ B(x0 , 2R) ∩ A cover the compact set B(x0 , 2R) ∩ A, so finitely many of them, say for p1 , . . . , pN , already cover it. Let y1 , . . . , ym be the vertices of Qp1 , . . . , QpN . Every such vertex y of Qpj lies in A and satisfies ∥y − x0 ∥ ≤ ∥y − pj ∥ + ∥pj − x0 ∥ ≤ R + 2R = 3R. The hull C contains every Qpj , and hence B(x0 , 2R) ∩ A. Its affine hull lies in A and contains the affine hull A of the vertices of Qp1 . Since yi ∈ B(x0 , 3R) ∩ A ⊆ K and aff{y1 , . . . , ym } = A = aff(K), Lemma 7 gives f (x) ≤ max f (yi )

∀x ∈ C.

1≤i≤m

Thus f is bounded above on B(x0 , 2R) ∩ A. On the other hand, this set is compact and contained in K, and f is finite and lsc on K, so f attains a finite minimum on it and is bounded below there. Consequently, M < ∞. Next, fix x ∈ dom(F ) ∩ B(x0 , R) ∩ A, and let ϵ > 0. The point x lies in K. Choose v ∈ F (x) as in Definition 15, so that f (y) − f (x) ≥ ψγ (⟨y − x, v⟩) − ϵ

∀y ∈ K.

(97)

For y ∈ K ⊆ A we have y − x ∈ V . Hence replacing v by its orthogonal projection onto V does not change any inner product in (97), and we may assume v ∈ V . If v ̸= 0, then let w := x + R

v ∈ B(x0 , 2R) ∩ A; ∥v∥

indeed w ∈ A because v ∈ V , and ∥w − x0 ∥ ≤ ∥x − x0 ∥ + R < 2R. The point x also lies in B(x0 , 2R) ∩ A, so |f (w)|, |f (x)| ≤ M . Since ⟨w − x, v⟩ = R∥v∥ > 0 and ψγ (s) = γs for s > 0, applying (97) with y = w yields 2M ≥ f (w) − f (x) ≥ ψγ (R∥v∥) − ϵ = γR∥v∥ − ϵ. The same bound on ∥v∥ is automatic when v = 0, and therefore ∥v∥ ≤

2M + ϵ . γR

(98)

We also use the elementary estimate ψγ (s) ≥ −γ −1 |s|

∀s ∈ R,

(99)

which holds because ψγ (s) = γs ≥ 0 for s ≥ 0 and ψγ (s) = γ −1 s = −γ −1 |s| for s < 0. Let y ∈ B(x0 , R) ∩ A. Then y ∈ K. By (97), (99), the Cauchy–Schwarz inequality, and (98), 1 2M + ϵ f (y) − f (x) ≥ − ∥v∥ ∥y − x∥ − ϵ ≥ − 2 ∥y − x∥ − ϵ. γ γ R The right-hand side no longer depends on v. Letting ϵ ↓ 0, we conclude that 2M f (y) − f (x) ≥ − 2 ∥y − x∥ γ R 57

(100)

whenever x ∈ dom(F ) ∩ B(x0 , R) ∩ A and y ∈ B(x0 , R) ∩ A. Finally, take arbitrary x, y ∈ B(x0 , R) ∩ A. Because B(x0 , R) ∩ A ⊆ K ⊆ A, we have B(x0 , R) ∩ K = B(x0 , R) ∩ A. For each n, the open set B(x, 1/n) ∩ B(x0 , R) contains x ∈ K, so by density it contains a point xn of K ∩ dom(F ). Then xn ∈ dom(F ) ∩ B(x0 , R) ∩ A and xn → x. By lower semicontinuity, f (x) ≤ lim inf n f (xn ). Applying (100) to xn and y gives
2M 2M f (y)−f (x) ≥ f (y)−lim inf f (xn ) = lim sup f (y)−f (xn ) ≥ − 2 lim ∥y−xn ∥ = − 2 ∥y−x∥. n→∞ γ R n→∞ γ R n→∞ Interchanging x and y proves (94). For every x0 ∈ ri(K) there is ε > 0 with B(x0 , ε)∩A ⊆ K, and R := ε/4 satisfies B(x0 , 3R)∩A ⊆ K. So (94) holds on a neighborhood of x0 relative to A. Since x0 was arbitrary, f is locally Lipschitz on ri(K). Approximate tilted convexity is unchanged by adding a constant, so we may apply (94) to f − c, where c is the midpoint of the supremum and infimum of f on B(x0 , 2R) ∩ A. The corresponding absolute-value bound is ∆/2, which proves (95). B.3 Dense approximate γ–tilted convexity implies exact γ–tilted convexity over the relative interior Proposition 3. Let γ ∈ (0, 1], let f : Rd → R be proper and lsc, and set K := dom(f ). Assume that K is convex. Let F : Rd ⇒ Rd be such that K ∩ dom(F ) is dense in K, and suppose that f is approximately γ–tilted convex with respect to F in the sense of Definition 15. Then there is an operator Fe : Rd ⇒ Rd with dom(Fe) = ri(K) such that f is γ–tilted convex with respect to Fe. Proof. Put A := aff(K) and V := A − A. By Theorem 33, f is locally Lipschitz, and hence continuous, on ri(K). Fix z ∈ ri(K), and choose R > 0 such that B(z, 3R) ∩ A ⊆ K; this is possible by the definition of the relative interior. For each j, the open ball B(z, min{R, 1/j}) contains z ∈ K. Since K ∩ dom(F ) is dense in K, this ball contains a point zj ∈ K ∩ dom(F ). Then zj → z and zj ∈ B(z, R) ∩ A. Moreover B(zj , 2R) ∩ A ⊆ B(z, 3R) ∩ A ⊆ K, so zj ∈ ri(K). Choose a sequence ϵj ↓ 0. Definition 15, applied at zj with ϵj , gives vectors uj ∈ F (zj ) such that f (y) − f (zj ) ≥ ψγ (⟨y − zj , uj ⟩) − ϵj

∀y ∈ K.

Let vj be the orthogonal projection of uj onto V . Since y − zj ∈ V for every y ∈ K, we have ⟨y − zj , uj ⟩ = ⟨y − zj , vj ⟩, and therefore f (y) − f (zj ) ≥ ψγ (⟨y − zj , vj ⟩) − ϵj

∀y ∈ K.

By Theorem 33, applied with x0 := z, M := sup{|f (w)| : w ∈ B(z, 2R) ∩ A} < ∞. If vj ̸= 0, then wj := zj + R

vj ∈ B(z, 2R) ∩ A, ∥vj ∥

58

(101)

because vj ∈ V and ∥wj − z∥ ≤ ∥zj − z∥ + R < 2R. The point zj lies in the same set, so f (wj ) − f (zj ) ≤ 2M . Since ⟨wj − zj , vj ⟩ = R∥vj ∥ > 0 and ψγ (s) = γs for s > 0, inequality (101) with y = wj gives 2M ≥ f (wj ) − f (zj ) ≥ γR∥vj ∥ − ϵj . Hence ∥vj ∥ ≤ (2M + ϵj )/(γR) ≤ (2M + ϵ1 )/(γR). This bound is automatic when vj = 0, so (vj ) is bounded. After passing to a subsequence, vj → vz for some vz ∈ Rd . Fix y ∈ K and let j → ∞ in (101) along this subsequence. Since f is continuous on ri(K) and zj , z ∈ ri(K), we have f (zj ) → f (z). Also ⟨y − zj , vj ⟩ → ⟨y − z, vz ⟩, ψγ is continuous, and ϵj → 0. Therefore f (y) − f (z) ≥ ψγ (⟨y − z, vz ⟩) ∀y ∈ K. (102) Define Fe(z) := {vz } on ri(K) and Fe(z) := ∅ otherwise. Then dom(Fe) = ri(K). If x, y ∈ dom(f ) and v ∈ Fe(x), then x ∈ ri(K) and v = vx , so (102) with z = x is inequality (5). Thus f is exactly γ–tilted convex with respect to Fe. B.4 Continuity along lines Theorem 34. Let γ ∈ (0, 1], let f : Rd → R be proper and lsc, and set K := dom(f ). Assume that K is convex. Let F : Rd ⇒ Rd be such that K ∩ dom(F ) is dense in K, and suppose that f is approximately γ–tilted convex with respect to F in the sense of Definition 15. Then, for every line ℓ that intersects ri(K), the restriction f |ℓ∩K is continuous. Proof. By Theorem 33, f is locally Lipschitz, and hence continuous, on ri(K). By Proposition 3, there is an operator Fe : Rd ⇒ Rd with dom(Fe) = ri(K) such that f is γ–tilted convex with respect to Fe, that is, f (u) − f (w) ≥ ψγ (⟨u − w, v⟩)

∀w ∈ ri(K), v ∈ Fe(w), u ∈ K.

(103)

We also use the line segment principle (Theorem 6.1 in [Roc15]): w ∈ ri(K), u ∈ K

=⇒

(1 − t)u + tw ∈ ri(K)

∀t ∈ (0, 1].

(104)

Now fix a line ℓ meeting ri(K) at a point p. The set ℓ ∩ K is a convex subset of ℓ, that is, an interval. If it is a singleton, there is nothing to prove. Otherwise, let q ∈ ℓ ∩ K be a point that is not an endpoint of this interval. If q ̸= p, the interval contains a point r such that q lies strictly between p and r, so q = (1−t)r+tp for some t ∈ (0, 1). By (104), q ∈ ri(K). Thus all non-endpoint points of ℓ ∩ K belong to ri(K). Near such a point q, all points of ℓ ∩ K are non-endpoint points, so the restriction of f is continuous at q by continuity of f on ri(K). Now let x be an endpoint of this interval that belongs to K. If x ∈ ri(K), continuity at x is again known. Otherwise x ̸= p; put y := p ∈ ℓ ∩ ri(K) and zt := (1 − t)x + ty, g(t) := f (zt ). Since x is an endpoint, ℓ ∩ K ⊆ {zt : t ≥ 0}, so continuity of f |ℓ∩K at x means g(t) → g(0) as t ↓ 0. By (104), zt ∈ ri(K) for t ∈ (0, 1], so g is finite and continuous on (0, 1]. Lower semicontinuity of f at x gives, in the extended real sense, a := lim sup g(t) ≥ b := lim inf g(t) ≥ g(0). t↓0

t↓0

59

Here g(0) = f (x) ∈ R. We show that both inequalities are equalities. First suppose that a > b. Then b is finite. Choose a strictly decreasing sequence tn ↓ 0 in (0, 1) such that g(tn ) → b. We claim that, for every n, g(s) ≤ max{g(tn+1 ), g(tn )}

∀s ∈ [tn+1 , tn ].

(105)

This is immediate at the endpoints. Suppose that it fails at some s ∈ (tn+1 , tn ), and choose vzs ∈ Fe(zs ), which is possible because zs ∈ ri(K). By (103) with w = zs and u = zti , ψγ (⟨zti − zs , vzs ⟩) ≤ g(ti ) − g(s) < 0

(i ∈ {n, n + 1}).

(106)

Since ψγ (r) has the sign of r, both inner products in (106) are negative. Put λ := (s − tn+1 )/(tn − tn+1 ) ∈ (0, 1), so that s = (1 − λ)tn+1 + λtn . The map t 7→ zt is affine, so zs = (1 − λ)ztn+1 + λztn , and hence 0 = (1 − λ)(ztn+1 − zs ) + λ(ztn − zs ), vzs < 0, a contradiction. This proves (105). On the other hand, choose sn ↓ 0 in (0, t1 ] such that g(sn ) → a. The intervals [tk+1 , tk ], k ≥ 1, cover (0, t1 ] because tk ↓ 0. So for each n there is an index k(n) ≥ 1 such that sn ∈ [tk(n)+1 , tk(n) ]. Since tk(n)+1 ≤ sn → 0 and every tk is positive, k(n) → ∞. Equation (105), with n replaced by k(n), then yields g(sn ) ≤ max{g(tk(n)+1 ), g(tk(n) )} −→ b. Hence a = limn g(sn ) ≤ b, contradicting a > b, so a = b. Finally, suppose that a = b > g(0), so that g(t) → a ∈ (g(0), +∞] as t ↓ 0. Put ρ := 2 γ /(1 + γ 2 ) ∈ (0, 1), and choose t0 ∈ (0, 1) such that g(t) > g(0) for all t ∈ (0, t0 ]. Let tn := ρn t0 . Then tn ∈ (0, t0 ], so g(tn ) > g(0) ∀n ≥ 0. (107) The definition of ρ gives (1 + γ 2 )tn+1 = γ 2 tn , that is, tn+1 = γ(tn − tn+1 ). γ

(108)

For each n, choose vn+1 ∈ Fe(ztn+1 ) and put qn+1 := ⟨y−x, vn+1 ⟩. We have x−ztn+1 = −tn+1 (y−x) and ztn − ztn+1 = (tn − tn+1 )(y − x). So (103) with w = ztn+1 , and with u = x or u = ztn , gives g(0) − g(tn+1 ) ≥ ψγ (−tn+1 qn+1 ),

(109)

g(tn ) − g(tn+1 ) ≥ ψγ (tn − tn+1 )qn+1 .

(110)

By (107), the left-hand side of (109) is negative. Hence ψγ (−tn+1 qn+1 ) < 0, which implies qn+1 > 0. So the argument of ψγ is negative in (109) and positive in (110). Using ψγ (r) = r/γ for r < 0 and ψγ (r) = γr for r > 0, we get tn+1 qn+1 , γ g(tn ) − g(tn+1 ) ≥ γ(tn − tn+1 )qn+1 . g(0) − g(tn+1 ) ≥ −

When the last two inequalities are added, the terms with qn+1 cancel by (108), which yields g(0) + g(tn ) − 2g(tn+1 ) ≥ 0, that is, g(tn+1 ) − g(0) ≤

1 g(tn ) − g(0) . 2

60

(111)

All values here are finite because ztn ∈ K. Induction on n with (111) and (107) gives
0 < g(tn ) − g(0) ≤ 2−n g(t0 ) − g(0) ∀n ≥ 0. So g(tn ) → g(0), contrary to g(tn ) → a > g(0). Thus a = b = g(0), proving continuity at x. Applying the same argument to each endpoint of ℓ ∩ K completes the proof.

Appendix C. Proofs for Section 5 C.1 Proof of Theorem 4 Each point of the domain is approximated by points with controlled regular subgradients. Lemma 8. Let f : Rd → R be proper and lsc. For every z ∈ dom(f ), there are sequences f

zk − → z,

b (zk ), vk ∈ ∂f

⟨vk , zk − z⟩ −→ 0.

(112)

Proof. For each integer k ≥ 1, minimize f (w) + k∥w − z∥2 over B(z, 1), and let zk be a minimizer. A minimizer exists because the objective is lsc on a compact set and finite at z. The function f is bounded below on that set, so comparison with z gives zk → z. The same comparison gives f (zk ) + k∥zk − z∥2 ≤ f (z). Lower semicontinuity at z now implies f (zk ) → f (z) and k∥zk − z∥2 → 0. For all sufficiently large k, the point zk is in the interior of the ball and hence is an unconstrained local minimizer of the regularized objective. The Fréchet necessary condition and the smooth-sum rule give b (zk ). vk := −2k(zk − z) ∈ ∂f Discarding the finitely many initial indices, we obtain ⟨vk , zk − z⟩ = −2k∥zk − z∥2 → 0, which proves the claim. b ⊆ ∂f . Hence every γ–tilted convex function in the sense of Definition 3 By Proposition 1-(i), ∂f b . It therefore suffices to prove the following slightly stronger is γ–tilted convex with respect to ∂f version of Theorem 4. ˆ for some Theorem 35. Let f : Rd → R be proper, lsc, and γ–tilted convex with respect to ∂f γ ∈ (0, 1]. Then dom(f ) is convex and f is quasiconvex. Proof. Fix distinct x, y ∈ dom(f ), and put zt := tx + (1 − t)y and M := max{f (x), f (y)}. First we show that f (zt ) ≤ M whenever zt ∈ dom(f ). (113) f ˆ (wk ) with Suppose f (zt ) > M , which forces 0 < t < 1. Lemma 8 gives wk − → zt and vk ∈ ∂f ⟨vk , wk − zt ⟩ → 0. Apply the support inequality at wk with targets x and y, and then apply the increasing inverse ψ γ of ψγ . Taking the corresponding convex combination gives

⟨vk , zt − wk ⟩ ≤ tψ γ (f (x) − f (wk )) + (1 − t)ψ γ (f (y) − f (wk )). The left-hand side tends to zero, whereas the right-hand side tends to a strictly negative number, because both value differences at zt are negative. This proves (113). Set I := {t ∈ [0, 1] : zt ∈ dom(f )}. By (113), I = {t ∈ [0, 1] : f (zt ) ≤ M }, so lower semicontinuity makes I closed. It contains 0 and 1. If I ̸= [0, 1], choose a component (a, b) of its 61

complement. Then za , zb ∈ dom(f ), while (za , zb ) contains no point of dom(f ). Put d := zb −za ̸= 0 and ∆ := f (zb ) − f (za ). Choose A ∈ R so large that ψγ (A − ∆) > −∆. Define τ (w) :=

⟨w − za , d⟩ , ∥d∥2

(114)

q(w) := Aτ (w)(τ (w) − 1),

g := f + q.

The function g is proper and lsc, has the same domain as f , and satisfies g(za ) = f (za ) and g(zb ) = f (zb ). Apply Theorem 32 to g with endpoints za , zb . Its intermediate point belongs to g ˆ dom(g) ∩ (za , zb ], so it must be zb . Thus there are wk → − zb and uk ∈ ∂g(w k ) such that lim sup⟨uk , wk − za ⟩ ≤ ∆. k

ˆ (wk ). Since q is continuously differentiable, The smooth-sum rule gives vk := uk − ∇q(wk ) ∈ ∂f f

wk − → zb and ⟨∇q(wk ), wk − za ⟩ −→ ⟨∇q(zb ), d⟩ = A. Consequently, lim inf ⟨vk , za − wk ⟩ ≥ A − ∆. k

The support inequality for f at wk with target za now yields −∆ = lim(f (za ) − f (wk )) ≥ lim inf ψγ (⟨vk , za − wk ⟩) ≥ ψγ (A − ∆), k

k

contrary to (114). Therefore I = [0, 1]. Together with (113), this proves that the effective domain and every sublevel set are convex. The equal-endpoint case is immediate. C.2 Proof of Theorem 5 Proof. A global minimizer is a local minimizer, so (i) implies (ii). Since f is lsc, Proposition 1-(i) ˆ (x) ⊆ ∂f (x), so (iii) implies (iv). gives ∂f (ii) ⇒ (iii). If x is a local minimizer of the extended-real-valued function f , then there is δ > 0 such that f (y) ≥ f (x) whenever ∥y − x∥ < δ. Hence, for every y ∈ dom(f ) with 0 < ∥y − x∥ < δ, f (y) − f (x) − ⟨0, y − x⟩ f (y) − f (x) = ≥ 0. ∥y − x∥ ∥y − x∥ Consequently, lim inf y→x y∈dom(f ), y̸=x

f (y) − f (x) − ⟨0, y − x⟩ ≥ 0. ∥y − x∥

ˆ (x). By the definition of the Fréchet subdifferential, 0 ∈ ∂f (iv) ⇒ (i). Assume 0 ∈ ∂f (x). Then the γ–tilted convexity of f , applied at x with v = 0, yields f (y) − f (x) ≥ ψγ (⟨y − x, 0⟩) = ψγ (0) = 0

∀y ∈ dom(f ).

If y ∈ / dom(f ), then f (y) = +∞ ≥ f (x) because f is proper. Thus x is a global minimizer of f .

62

C.3 Proofs of Theorems 6 and 7 Both theorems are special cases of the results of Appendix B. Proof of Theorem 6. Let f be as in the theorem and take F := ∂f . The set K is convex by ˆ , and hence F , Theorem 4. By Proposition 1-(i), every point of K is a limit of points where ∂f is nonempty, so K ∩ dom(F ) is dense in K. Exact γ–tilted convexity with respect to F implies approximate γ–tilted convexity with respect to F , with the same v for every ϵ. Theorem 33 therefore shows that f is locally Lipschitz on ri(K). Proof of Theorem 7. Let f be as in the theorem. Its domain K is convex by Theorem 4. The γ–tilted convexity of f implies exact, and hence approximate, γ–tilted convexity with respect to b , whose domain is dense in K by Proposition 1-(i). For d = 1 the only line is ℓ = R, which meets ∂f ri(K) because K is nonempty and convex, and f |ℓ∩K = f |K . Theorem 34 therefore shows that f |K is continuous. C.4 Proof of Theorem 8: Alternative First-order Characterization Since f is lsc, Proposition 1-(i) gives b (x) ⊆ ∂f (x) ⊆ ∂f (x), ∂f

x ∈ K.

(115)

If f is γ–tilted convex with respect to an operator, then it is γ–tilted convex with respect to every operator whose values are subsets of the values of the first one. Hence (115) gives the implications (i) =⇒ (ii) =⇒ (iii). If (iii) holds, then Theorem 35 shows that K is convex, and the inequality in (iv) is the inequality of (iii) restricted to x ∈ ri(K). This proves (iii) =⇒ (iv). To see that (i) =⇒ (v), assume (i) and put A := aff(K). Theorem 4 shows that K is convex, and Theorem 6 shows that f |ri(K) is locally Lipschitz. By Rademacher’s theorem in the Euclidean space A, the points of ri(K) where the relative gradient ∇f exists have full measure in the relatively open set ri(K), so they are dense in ri(K). Since K is convex, ri(K) is dense in K (Theorem 6.3 in [Roc15]), and hence these points are dense in K. Let x ∈ ri(K) be such a point. Every y ∈ dom(f ) = K lies in A, so the differentiability of f |A at x gives lim

y→x y∈K, y̸=x

f (y) − f (x) − ⟨∇f (x), y − x⟩ = 0. ∥y − x∥

b and by (115), this yields By the definition of ∂f b (x) ⊆ ∂f (x). ∇f (x) ∈ ∂f

(116)

Condition (i) with v = ∇f (x) now gives the inequality in (v). It remains to prove (ii) =⇒ (i), (iv) =⇒ (ii), and (v) =⇒ (iv). C.4.1 Horizon subdifferentials and normal cones We use normal cones to control the horizon subgradients of γ–tilted convex functions. Definition 16. Let X ⊆ Rd and x ∈ X . A vector v ∈ Rd is regular normal to X , written v ∈ N̂X (x), if ⟨v, y − x⟩ lim sup ≤ 0. y→x ∥y − x∥ y∈X , y̸=x

63

Furthermore, a vector v ∈ Rd is normal to X , written v ∈ NX (x), if there are sequences xk ∈ X and vk ∈ Rd such that xk → x, vk → v, vk ∈ N̂X (xk ). For functions that are γ–tilted convex with respect to their regular subdifferentials, horizon subgradients are normal to the effective domain. b for some γ ∈ (0, 1], Lemma 9. Let f : Rd → R be proper, lsc, and γ–tilted convex with respect to ∂f and set K := dom(f ). Then, for any x ∈ K, we have ∂ ∞ f (x) ⊆ {v | ∀y ∈ K, ⟨v, y − x⟩ ≤ 0} = N̂K (x) = NK (x). In particular, if V = aff(K) − aff(K), then V ∩ ∂ ∞ f (x) ∩ −NK (x) = {0}. Proof. Theorem 35 shows that K is convex. Let v ∈ ∂ ∞ f (x). By the definition of the horizon subdifferential, there are sequences f

xk − → x,

λk > 0,

λk → 0,

b (xk ), vk ∈ ∂f

λk vk → v.

(117)

b gives Fix y ∈ K. Each xk lies in dom(f ) = K, so the γ–tilted convexity of f with respect to ∂f f (y) − f (xk ) ≥ ψγ (⟨vk , y − xk ⟩). By Lemma 1, ψ γ is increasing and ψ γ (ψγ (a)) = a. Applying ψ γ to both sides and multiplying by λk > 0 yields
⟨λk vk , y − xk ⟩ ≤ λk ψ γ f (y) − f (xk ) . (118) Since f (xk ) → f (x) and ψ γ is continuous, ψ γ (f (y)−f (xk )) converges to the real number ψ γ (f (y)− f (x)). As λk → 0, the right-hand side of (118) tends to zero, while by (117) the left-hand side tends to ⟨v, y − x⟩. Hence ⟨v, y − x⟩ ≤ 0 ∀y ∈ K. (119) For x′ ∈ K, put P (x′ ) := {v : ⟨v, y − x′ ⟩ ≤ 0 ∀y ∈ K}. By (119), ∂ ∞ f (x) ⊆ P (x). According to Theorem 6.9 in [RW98], we have bK (x) = NK (x). P (x) = N These equalities also follow directly from the definitions and the convexity of K. If v ∈ P (x′ ), then bK (x′ ) is nonpositive, so v ∈ N bK (x′ ). Conversely, let v ∈ N bK (x′ ) every quotient in the definition of N and y ∈ K \ {x′ }. By convexity, the points yt := x′ + t(y − x′ ) with t ∈ (0, 1] lie in K, they tend to x′ as t ↓ 0, and ⟨v, yt − x′ ⟩ ⟨v, y − x′ ⟩ = . ∥yt − x′ ∥ ∥y − x′ ∥ bK (x′ ) is at least this constant, so ⟨v, y − x′ ⟩ ≤ 0. Thus The limit superior in the definition of N ′ ′ ′ bK (x ) = P (x ) for every x ∈ K. Constant sequences show that N bK (x) ⊆ NK (x). Finally, let N bK (xk ) = P (xk ), we have v ∈ NK (x), with xk → x and vk → v as in the definition. Since vk ∈ N ⟨vk , y − xk ⟩ ≤ 0 for every y ∈ K, and letting k → ∞ gives v ∈ P (x). This completes the proof of the first claim. 64

For the second claim, let v ∈ V ∩ ∂ ∞ f (x) ∩ (−NK (x)). The first claim gives ⟨v, y − x⟩ ≤ 0 for all y ∈ K. Since −v ∈ NK (x) = P (x), we also have ⟨−v, y − x⟩ ≤ 0 for all y ∈ K. Hence ⟨v, y − x⟩ = 0

∀y ∈ K.

Because x ∈ K, the vectors y − x with y ∈ K span V . Thus v ∈ V ⊥ . Since also v ∈ V , we obtain ⟨v, v⟩ = 0, that is, v = 0. Conversely, 0 lies in V , in the cone NK (x), and in the closed cone ∂ ∞ f (x) (Theorem 8.6 in [RW98]), which proves the equality. C.4.2 Proof of (ii) =⇒ (i) b , so Lemma 9 applies to Proof. Assume (ii). By (115), f is also γ–tilted convex with respect to ∂f f . Fix x ∈ K. Let V ⊆ Rd be the set of vectors v such that f (y) − f (x) ≥ ψγ (⟨y − x, v⟩),

∀y ∈ K.

By Lemma 1, ψγ and ψ γ are increasing bijections that are inverse to each other. Hence ψγ (a) ≤ b holds if and only if a ≤ ψ γ (b), and n o V = v ∈ Rd | ψ γ (f (y) − f (x)) ≥ ⟨y − x, v⟩, ∀y ∈ K o \ n v ∈ Rd | ψ γ (f (y) − f (x)) ≥ ⟨y − x, v⟩ . =

(120)

y∈K

For each y ̸= x, the set on the right of (120) is a closed half-space, and for y = x it is Rd . So V is convex and closed. Condition (ii) states that ∂f (x) ⊆ V , so the convexity of V gives conv(∂f (x)) ⊆ V.

(121)

Now let v ∈ V and n ∈ N̂K (x). Lemma 9 gives ⟨y − x, n⟩ ≤ 0 for every y ∈ K, and therefore ⟨y − x, v + n⟩ ≤ ⟨y − x, v⟩ ≤ ψ γ (f (y) − f (x)),

∀y ∈ K.

Thus V + N̂K (x) ⊆ V . Together with (121), this gives conv(∂f (x)) + N̂K (x) ⊆ V . Since V is closed, we obtain conv(∂f (x)) + N̂K (x) ⊆ V. (122) Lemma 9 also gives ∂ ∞ f (x) ⊆ N̂K (x), and N̂K (x) is convex, being an intersection of closed halfspaces. Hence conv(∂ ∞ f (x)) ⊆ N̂K (x). By the definition of ∂f and by (122), ∂f (x) = conv(∂f (x)) + conv(∂ ∞ f (x)) ⊆ conv(∂f (x)) + N̂K (x) ⊆ V. Hence, by the definition of V , for all v ∈ ∂f (x) we have f (y) − f (x) ≥ ψγ (⟨y − x, v⟩), Since x ∈ K was arbitrary, f is γ–tilted convex.

65

∀y ∈ K.

C.4.3 Proof of (iv) =⇒ (ii) We first approximate regular subgradients at points of the domain by limiting subgradients at nearby points of the relative interior. ˆ (x), and y ∈ ri(K), Lemma 10. Under the assumptions of Theorem 8-(iv), for any x ∈ K, v ∈ ∂f f

there exist sequences zn ∈ ri(K) and vn ∈ ∂f (zn ) such that zn − → x and ⟨v, y − x⟩ ≤ lim inf ⟨vn , y − x⟩. n→∞

ˆ (x) ⊆ ∂f (x) by (115). Choosing zn = x and vn = v Proof. If x = y, then x ∈ ri(K), and v ∈ ∂f for all n ≥ 1 makes both sides of the required inequality zero. Assume x ̸= y, and put A := aff(K). For t ∈ (0, 1], define yt := x + t(y − x). Since x ∈ K and y ∈ ri(K), Theorem 6.1 in [Roc15] shows that (1 − s)x + sy ∈ ri(K),

s ∈ (0, 1].

(123)

ˆ (xi ) ̸= ∅. Since Let x′ ∈ ri(K). Proposition 1-(i) gives points xi → x′ in dom(f ) = K with ∂f ri(K) is open in A and K ⊆ A, these points lie in ri(K) for all large i. Therefore the set of points of ˆ is non-empty is dense in ri(K). Since K is convex, ri(K) is dense in K, so this set is ri(K) where ∂f b |ri(K) . Then K ∩ dom(F ) is dense in K, and condition (iv) states that f is dense in K. Put F := ∂f γ–tilted convex, hence approximately γ–tilted convex, with respect to F . Thus the hypotheses of Theorems 33 and 34 hold, and they give f is locally Lipschitz on ri(K), f |ℓ∩K is continuous for every line ℓ that meets ri(K).

(124)

The line through x and y meets ri(K) at y, so f is continuous on [x, y]. Fix t ∈ (0, 1]. By (123), the open segment (x, yt ) lies in ri(K), where f is locally Lipschitz, and f is continuous on [x, yt ] ⊆ [x, y]. We apply Corollary 12 to f |A in the Euclidean space A, at the points x and yt . It gives a point zt ∈ (x, yt ) ⊆ ri(K) and a vector vt in the convex hull of the limiting subdifferential of f |A at zt such that f (yt ) − f (x) = ⟨vt , yt − x⟩.

(125)

Since dom(f ) = K ⊆ A, the quotients defining the regular subdifferentials of f and of f |A at a point of K are the same. Hence every regular subgradient of f |A is a regular subgradient of f , and, passing to limits, every limiting subgradient of f |A is a limiting subgradient of f . In particular, vt ∈ conv(∂f (zt )). ˆ (x), we have Since v ∈ ∂f f (z) − f (x) − ⟨v, z − x⟩ ≥ 0. ∥z − x∥ z∈K, z̸=x lim inf z→x

(126)

The points yt lie in K \ {x} and tend to x as t ↓ 0, so the lower limit along yt is at least the left-hand side of (126). Using (125) and yt − x = t(y − x), we see that 0 ≤ lim inf t↓0

f (yt ) − f (x) − ⟨v, yt − x⟩ ⟨vt − v, yt − x⟩ ⟨vt − v, y − x⟩ = lim inf = lim inf . t↓0 t↓0 ∥yt − x∥ ∥yt − x∥ ∥y − x∥ 66

Multiplying by ∥y − x∥ > 0 and adding ⟨v, y − x⟩ gives ⟨v, y − x⟩ ≤ lim inf ⟨vt , y − x⟩.

(127)

t↓0

Pm Fix Pmt ∈ (0, 1] again. Since vt ∈ conv(∂f (zt )), write vt = i=1 θi vt,i , where vt,i ∈ ∂f (zt ), θi > 0, and i=1 θi = 1. Then m X ⟨vt , y − x⟩ = θi ⟨vt,i , y − x⟩. i=1

If every vt,i satisfied ⟨vt,i , y−x⟩ < ⟨vt , y−x⟩, then multiplying these inequalities by θi and summing would give ⟨vt , y − x⟩ < ⟨vt , y − x⟩. Hence we can choose one of the vt,i , denoted by vt ∈ ∂f (zt ), such that ⟨vt , y − x⟩ ≥ ⟨vt , y − x⟩. (128) Combining (127) and (128) gives ⟨v, y − x⟩ ≤ lim inf ⟨vt , y − x⟩ ≤ lim inf ⟨vt , y − x⟩. t↓0

t↓0

Choose any sequence tn ↓ 0 and set zn := ztn ∈ ri(K) and vn := vtn ∈ ∂f (zn ). The lower limit along the sequence tn is at least the lower limit as t ↓ 0, so the last display gives the required inequality for (vn ). Moreover, ∥zn − x∥ ≤ ∥ytn − x∥ = tn ∥y − x∥ → 0, and the continuity of f on f

[x, y] gives f (zn ) → f (x). Thus zn − → x. Proof of (iv) =⇒ (ii). Assume (iv). We show that f (y) − f (x) ≥ ψγ (⟨v, y − x⟩), for all x, y ∈ K and v ∈ ∂f (x). Put A := aff(K). The proof of Lemma 10 shows that condition (iv) implies (124). Case 1: x, y ∈ ri(K) and v ∈ ∂f (x). By definition of the limiting subdifferential, there exist xk → x,

ˆ (xk ), vk ∈ ∂f

f (xk ) → f (x),

vk → v.

Each xk lies in dom(f ) = K ⊆ A. Since x ∈ ri(K) and ri(K) is open in A, after discarding finitely ˆ (xk ), gives many terms we have xk ∈ ri(K). Condition (iv), applied at xk with vk ∈ ∂f f (y) − f (xk ) ≥ ψγ (⟨vk , y − xk ⟩). As k → ∞, we have f (xk ) → f (x) and ⟨vk , y − xk ⟩ → ⟨v, y − x⟩. Since ψγ is continuous, we obtain f (y) − f (x) ≥ ψγ (⟨v, y − x⟩). ˆ (x). Case 2: x ∈ K, y ∈ ri(K) and v ∈ ∂f f

Set d := y − x. By Lemma 10, there exist zn ∈ ri(K) and vn ∈ ∂f (zn ) such that zn − → x and ⟨v, d⟩ ≤ lim inf ⟨vn , d⟩. n→∞

(129)

Define yn := zn + d. Since zn ∈ A and d ∈ A − A, we have yn ∈ A, and yn → x + d = y. Since y ∈ ri(K) and ri(K) is open in A, after discarding finitely many terms we may assume that yn ∈ ri(K). Case 1, applied to zn , yn ∈ ri(K) and vn ∈ ∂f (zn ), together with yn − zn = d, gives f (yn ) − f (zn ) ≥ ψγ (⟨vn , d⟩). 67

(130)

f

By (124), f is continuous on ri(K), so f (yn ) → f (y). Also f (zn ) → f (x), because zn − → x. Hence the left-hand side of (130) converges to the real number f (y) − f (x). Let L := lim inf n→∞ ⟨vn , d⟩. By (129), L ≥ ⟨v, d⟩. If L = +∞, then ψγ (⟨vn , d⟩) → +∞, which contradicts (130). So L is finite. Choose a subsequence along which ⟨vn , d⟩ → L. Passing to the limit in (130) along it, and using that ψγ is continuous and increasing, we get f (y) − f (x) ≥ ψγ (L) ≥ ψγ (⟨v, d⟩) = ψγ (⟨v, y − x⟩). ˆ (x) whenever y ∈ ri(K). Thus the desired inequality holds for all v ∈ ∂f Case 3: x ∈ K, y ∈ ri(K) and v ∈ ∂f (x). By definition of the limiting subdifferential, there exist xk → x,

ˆ (xk ), vk ∈ ∂f

f (xk ) → f (x),

vk → v.

Each xk lies in dom(f ) = K. Applying Case 2 to each (xk , vk ) and the fixed y ∈ ri(K) yields f (y) − f (xk ) ≥ ψγ (⟨vk , y − xk ⟩). Letting k → ∞ and using continuity of ψγ , we obtain f (y) − f (x) ≥ ψγ (⟨v, y − x⟩). Case 4: x, y ∈ K and v ∈ ∂f (x). Fix y0 ∈ ri(K); this set is non-empty because K is non-empty and convex. If y = y0 , Case 3 applies. Otherwise, set yk := (1 − k −1 )y + k −1 y0 for k ≥ 1. By Theorem 6.1 in [Roc15], yk ∈ ri(K). All yk lie on the line through y and y0 , which meets ri(K) at y0 . By (124), f is continuous on the intersection of this line with K, so f (yk ) → f (y). Case 3, applied to x, yk , and v, gives f (yk ) − f (x) ≥ ψγ (⟨v, yk − x⟩). Letting k → ∞ and using continuity of ψγ gives f (y) − f (x) ≥ ψγ (⟨v, y − x⟩). Thus f is γ–tilted convex with respect to ∂f , proving (ii). C.4.4 Proof of (v) =⇒ (iv) Proof. Assume (v). Put A := aff(K) and W := A − A, and interpret gradients relative to A. Let G be the operator x 7→ {∇A f (x)}, defined at the points x ∈ ri(K) where ∇A f (x) exists. By condition (v), K is convex, K ∩dom(G) is dense in K, and f is γ–tilted convex, hence approximately γ–tilted convex, with respect to G. Theorem 33 therefore shows that f is locally Lipschitz on ri(K). b (x), and let vA be the orthogonal projection of v onto W . Fix x ∈ ri(K), y ∈ K, and v ∈ ∂f Every z ∈ dom(f ) = K satisfies z − x ∈ W , so ⟨v, z − x⟩ = ⟨vA , z − x⟩,

z ∈ K.

(131)

Hence the quotients in the definition of the regular subdifferential do not change when v is replaced by vA , and vA ∈ ∂bA f (x), where the subdifferential is taken relative to A. Let ∇A f (x) be the set of Proposition 1-(ii) for f |A in the Euclidean space A, and let w ∈ ∇A f (x). Then there is a sequence xk → x in A such that the relative gradients ∇A f (xk ) exist and converge to w. Since ri(K) is open in A, we have xk ∈ ri(K) for all large k. Condition (v) gives f (y) − f (xk ) ≥ ψγ (⟨∇A f (xk ), y − xk ⟩). 68

Since f is continuous at x and ψγ is continuous, letting k → ∞ gives f (y) − f (x) ≥ ψγ (⟨w, y − x⟩). Applying the increasing inverse ψ γ (Lemma 1) yields ψ γ (f (y) − f (x)) ≥ ⟨w, y − x⟩,

w ∈ ∇A f (x).

(132)

The set of w ∈ W satisfying (132) is a closed half-space of W or all of W , hence convex. It contains ∇A f (x), so it contains conv(∇A f (x)). Since f |A is locally Lipschitz at x, Proposition 1-(i) and (ii), applied in the Euclidean space A, give vA ∈ ∂bA f (x) ⊆ ∂ A f (x) = conv(∇A f (x)). Therefore (132) holds with w = vA . By (131) with z = y, we have ⟨vA , y − x⟩ = ⟨v, y − x⟩, and hence ψ γ (f (y) − f (x)) ≥ ⟨v, y − x⟩. Applying the increasing function ψγ to this inequality gives f (y) − f (x) ≥ ψγ (⟨v, y − x⟩). Together with the convexity of K, this proves (iv). C.5 Proof of Theorem 9 Proof. Forward direction. Suppose first that f is γ–tilted convex, and put K := dom(f ). By Theorem 4, the set K is convex. Fix a line ℓ meeting K, write ℓ = {b + td : t ∈ R},

∥d∥ = 1,

and set I := {t ∈ R : b + td ∈ K},

g(t) := f (b + td).

Since ∥d∥ = 1, the map t 7→ b + td is an isometry of R onto ℓ, and g is f |ℓ written in this coordinate; so it suffices to show that g is γ–tilted convex. The function g : R → R is proper and lsc, because it is the composition of f with a continuous affine map, and I = dom(g) is an interval, because it is the preimage of the convex set K under an affine map. If I = {t0 } is a singleton, the only instance of (5) for g compares t0 with itself, and both sides vanish; so g is γ–tilted convex. Suppose therefore that I is nondegenerate. Then ℓ contains two distinct points of K, so ℓ ⊆ aff(K). In what follows we identify aff(K) with a Euclidean space and take subdifferentials of f relative to this space. The representative of a relative subgradient in the direction space of aff(K) is also an ambient subgradient, because f = +∞ outside aff(K). Conversely, any normal component of an ambient subgradient is orthogonal to differences of points in K and hence does not affect the support inequalities below. Choose a ∈ ri(K). For ϵ ∈ (0, 1) define τϵ (t) := (1 − ϵ)(b + td) + ϵa,

gϵ (t) := f (τϵ (t))

(t ∈ I),

and extend gϵ by +∞ outside I. We use the standard relative-interior property of the convex set K: (1 − ϵ)z + ϵa ∈ ri(K) for all z ∈ K and ϵ ∈ (0, 1]. (133) For t ∈ I the point b + td is a limit of points of K, so it lies in K. Hence (133) gives τϵ (I) ⊆ ri(K). 69

Theorem 6 shows that f is locally Lipschitz, and hence continuous, on ri(K). Thus gϵ is continuous on the closed interval I, and its extension by +∞ is proper and lsc, with dom(gϵ ) = I. If t ∈ int(I), this extension agrees near t with the composition f ◦ τϵ , and f is locally Lipschitz at τϵ (t). The derivative of τϵ is (1 − ϵ)d. Hence the second part of Theorem 29 shows that every p ∈ ∂gϵ (t) has the form p = (1 − ϵ)⟨v, d⟩ for some v ∈ ∂f (τϵ (t)). (134) For every s ∈ I we have τϵ (s) − τϵ (t) = (1 − ϵ)(s − t)d,

(135)

and both τϵ (s) and τϵ (t) lie in ri(K) ⊆ dom(f ). The vector v in (134) belongs to ∂f (τϵ (t)) ⊆ ∂f (τϵ (t)) by Proposition 1-(i). Consequently, the γ–tilted convexity of f , followed by (135) and (134), gives   gϵ (s) − gϵ (t) ≥ ψγ v, τϵ (s) − τϵ (t) = ψγ ((s − t)p) (t ∈ int(I), p ∈ ∂gϵ (t), s ∈ I). (136) ˆ ϵ (t) ⊆ ∂gϵ (t) by Proposition 1The effective domain I of gϵ is convex, ri(dom(gϵ )) = int(I), and ∂g (i). Thus (136) is condition (iv) of Theorem 8 for gϵ , and gϵ is γ–tilted convex. Local uniform convergence. Fix t ∈ I and put x := b + td. If x = a, then τϵ (t) = a and gϵ (t) = g(t) for every ϵ. Otherwise, τϵ (t) = (1 − ϵ)x + ϵa lies on the segment [x, a] ⊆ K and tends to x as ˆ , ϵ ↓ 0. The line through x and a meets ri(K) at a, and f is γ–tilted convex with respect to ∂f whose domain is dense in K by Proposition 1-(i). So Theorem 34 shows that f is continuous on the intersection of this line with K. In both cases, gϵ (t) −→ g(t)

(ϵ ↓ 0, t ∈ I).

(137)

We next show directly that this convergence is locally uniform on int(I). Fix t0 ∈ int(I) and R > 0 such that [t0 − 3R, t0 + 3R] ⊆ I. Pointwise convergence (137) at t0 and t0 ± 2R gives U > 0 such that, for all sufficiently small ϵ, |gϵ (t0 )| ≤ U,

|gϵ (t0 − 2R)|, |gϵ (t0 + 2R)| ≤ U.

(138)

By Theorem 4, gϵ is quasiconvex. Every point of [t0 − 2R, t0 + 2R] is a convex combination of t0 − 2R and t0 + 2R, so (138) gives gϵ (t) ≤ max{gϵ (t0 − 2R), gϵ (t0 + 2R)} ≤ U

(t ∈ [t0 − 2R, t0 + 2R]).

(139)

Choose pϵ ∈ ∂gϵ (t0 ). Such a subgradient exists by Proposition 1-(ii), because gϵ is locally Lipschitz at t0 ∈ ri(dom(gϵ )) by Theorem 6. If pϵ ̸= 0, put wϵ := t0 + R

pϵ . |pϵ |

Then (wϵ − t0 )pϵ = R|pϵ | > 0, so ψγ ((wϵ − t0 )pϵ ) = γR|pϵ |. Since wϵ ∈ [t0 − R, t0 + R], (139) gives gϵ (wϵ ) ≤ U , and (138) gives gϵ (t0 ) ≥ −U . Combining these bounds with (136) at t = t0 and s = wϵ gives 2U ≥ gϵ (wϵ ) − gϵ (t0 ) ≥ γR|pϵ |, that is, |pϵ | ≤ 70

2U . γR

(140)

This bound is automatic if pϵ = 0. Next, the definition of ψγ gives ψγ (z) = z/γ for z ≤ 0 and ψγ (z) = γz ≥ 0 for z > 0, hence ψγ (z) ≥ −

|z| γ

(z ∈ R).

(141)

Let t ∈ [t0 − 2R, t0 + 2R], so that |t − t0 | ≤ 2R. By (136) with s = t, then (141), then (138) and (140), 1 1 2U 4U gϵ (t) ≥ gϵ (t0 ) − |pϵ | |t − t0 | ≥ −U − · · 2R = −U − 2 . γ γ γR γ Together with (139), and with M0 := U + 4U/γ 2 , this gives |gϵ (t)| ≤ M0

(t ∈ [t0 − 2R, t0 + 2R])

(142)

for all sufficiently small ϵ. Theorem 33 applies to gϵ at t0 with radius R, because the domain I ˆ ϵ , whose is convex and contains [t0 − 3R, t0 + 3R], and gϵ is γ–tilted convex with respect to ∂g domain is dense in I by Proposition 1-(i). By (142), the bound M in that theorem is at most M0 . Therefore, with L := 2M0 /(γ 2 R), |gϵ (t) − gϵ (t′ )| ≤ L|t − t′ |

(t, t′ ∈ X := (t0 − R, t0 + R))

(143)

for all sufficiently small ϵ. Let ϵk ↓ 0. By (143) and (137), the restrictions of gϵk to the bounded interval X are L–Lipschitz for large k and converge pointwise on X to g. Theorem 30 therefore shows that g is L–Lipschitz on X and sup |gϵk (t) − g(t)| −→ 0 (k → ∞). (144) t∈X

The whole family converges in the same way: if supX |gϵ − g| did not tend to 0 as ϵ ↓ 0, there would be η > 0 and ϵk ↓ 0 with supX |gϵk − g| ≥ η for all k, contradicting (144). Every compact subset of int(I) is covered by finitely many intervals X of this form. Hence gϵ → g locally uniformly on int(I). Passing to the limit. Fix t ∈ int(I), s ∈ I, and p ∈ ∂g(t). Take t0 := t in the previous paragraph, with the corresponding R, L, and X = (t − R, t + R), and fix a sequence ϵk ↓ 0. Since X is open, the subdifferentials of the restrictions to X agree with those of gϵk and g at points of X. The last part of Theorem 30, applied on X, gives tk → t,

pk ∈ ∂gϵk (tk ),

pk → p.

(145)

After discarding finitely many terms, tk ∈ X ⊆ int(I), so (136) gives gϵk (s) − gϵk (tk ) ≥ ψγ ((s − tk )pk ).

(146)

We let k → ∞ in (146). At the fixed point s we have gϵk (s) → g(s) by (137). Since g is L–Lipschitz on X, |gϵk (tk ) − g(t)| ≤ sup |gϵk − g| + L|tk − t|, X

and the right-hand side tends to 0 by (144) and (145). The right-hand side of (146) tends to ψγ ((s − t)p) by (145) and the continuity of ψγ (Lemma 1). We obtain g(s) − g(t) ≥ ψγ ((s − t)p). 71

(147)

The function g is proper and lsc, and its effective domain I is convex with ri(I) = int(I). By ˆ ˆ Proposition 1-(i), ∂g(t) ⊆ ∂g(t), so (147) holds for every p ∈ ∂g(t) with t ∈ ri(I) and every s ∈ I. This is condition (iv) of Theorem 8, which shows that g is γ–tilted convex. Reverse direction. Suppose every line restriction of f meeting its effective domain is γ–tilted convex. The effective domain of each such restriction is convex by Theorem 4. Thus, for any two points of dom(f ), the line through them contains the entire segment between them in its effective ˆ (x). If x = y, the domain; hence dom(f ) is convex. Fix x ∈ ri(dom(f )), y ∈ dom(f ), and v ∈ ∂f required inequality reads 0 ≥ ψγ (0) = 0. Otherwise, define d :=

y−x , ∥y − x∥

q(t) := f (x + td).

The function q is the restriction of f to the line through x and y in a unit-speed parametrization, so it is γ–tilted convex by assumption. The first part of Theorem 29, applied with F (t) := x + td and ∇F (0) = d, gives ˆ ⟨v, d⟩ ∈ ∂q(0). ˆ Since ∂q(0) ⊆ ∂q(0) by Proposition 1-(i), the γ–tilted convexity of q between the points 0 and ∥y − x∥ of its domain gives
f (y) − f (x) = q(∥y − x∥) − q(0) ≥ ψγ ∥y − x∥⟨v, d⟩ = ψγ (⟨v, y − x⟩). This is condition (iv) of Theorem 8, which proves that f is γ–tilted convex. C.6 Proof of Theorem 10 Proof. By Theorem 4, K is convex. It is nonempty because f is proper, so ri(K) ̸= ∅. Fix x ∈ K. If x ∈ ri(K), then f (x) = g(x) by assumption. Otherwise, choose y ∈ ri(K) and define zt := (1 − t)x + ty,

t ∈ (0, 1].

By the line segment principle (104), zt ∈ ri(K) for every t ∈ (0, 1], and hence f (zt ) = g(zt )

∀t ∈ (0, 1].

(148)

We claim that lim f (zt ) = f (x) t↓0

and

lim g(zt ) = g(x). t↓0

(149)

We give the argument for f ; the one for g is the same, since g is also proper, lsc, and γ–tilted convex, and g(y) = f (y) ∈ R. Let ℓ be the line through x and y; note that x ̸= y since x ∈ / ri(K). The line meets dom(f ) at y. By Theorem 9, the restriction f |ℓ , regarded as a function on a one-dimensional Euclidean space, is γ–tilted convex. It is also proper and lsc. Theorem 7 therefore shows that f |ℓ is continuous on its effective domain ℓ ∩ dom(f ). If x ∈ dom(f ), the points zt belong to ℓ ∩ dom(f ) and tend to x, so f (zt ) → f (x). If x ∈ / dom(f ), then f (x) = +∞, and lower semicontinuity of f at x gives lim inf t↓0 f (zt ) ≥ f (x) = +∞, so again f (zt ) → f (x). This proves (149). By (149) and (148), f (x) = lim f (zt ) = lim g(zt ) = g(x). t↓0

t↓0

Since x ∈ K was arbitrary, this proves f = g on K. For the final assertion, set
h := lsc f |ri(K) . 72

The domain of f |ri(K) is ri(K), and its closure is K (Theorem 6.3 in [Roc15]). By the definition of the lsc envelope, h(x) =

lim inf

f (w)

(x ∈ K),

h(x) = +∞

(x ∈ / K).

(150)

w∈ri(K), w→x

Let x ∈ K. The lower limit in (150) is taken over the subset ri(K), so it is at least lim inf w→x f (w), which is at least f (x) by lower semicontinuity of f . Hence f (x) ≤ h(x)

∀x ∈ K.

(151)

Now let x ∈ K, choose y ∈ ri(K), and set zt = (1 − t)x + ty. The points zt lie in ri(K) for t ∈ (0, 1] by (104), and zt → x as t ↓ 0. If x ∈ / ri(K), then (149) gives f (zt ) → f (x). If x ∈ ri(K), the same limit holds because f is continuous on ri(K) by Theorem 6. By (150), in both cases h(x) ≤ lim f (zt ) = f (x). t↓0

Together with (151), this gives h = f on K. If x ∈ K \ K, then f (x) = +∞, and (151) gives h(x) = +∞. Finally, both f and h equal +∞ outside K: the function f because these points lie outside K, and h by (150). Hence
f = lsc f |ri(K) . C.7 Proof of Theorem 11 Proof. Local uniform convergence. Put A := aff(K) and V := span(K − K), and identify A with a Euclidean space. For x0 ∈ A and r > 0 we write BV (x0 , r) := {x0 + w : w ∈ V, ∥w∥ < r} = B(x0 , r) ∩ A for the open ball of A, and BV (x0 , r) for its closure. Choose d0 ∈ D. Since fn (d0 ) converges to a finite value, d0 belongs to dom(fn ) for all sufficiently large n. Thus, after discarding finitely many terms, every restriction fn |A is proper and lsc. To see that it is γ–tilted convex, consider any line in A meeting dom(fn ). The forward implication of Theorem 9 shows that the restriction of fn to this line is γ–tilted convex. These are precisely the line restrictions of fn |A , so the reverse implication of the same theorem, applied in the Euclidean space A, shows that fn |A is γ–tilted convex. We continue to denote these restrictions by fn . Fix a compact set C ⊆ ri(K) and a point x0 ∈ C. Since ri(K) is open relative to A, there is R > 0 with BV (x0 , 4R) ⊆ ri(K). (152) We use the density of D to choose points y1 , . . . , ym ∈ D and a ∈ D ∩ BV (x0 , R/2) such that BV (x0 , 3R) ⊆ P := conv{y1 , . . . , ym }.

(153)

If V = {0}, then K = D = {x0 }, and we take m = 1 and y1 = a = x0 . Otherwise, the relative sphere S := {x0 + w : w ∈ V, ∥w∥ = 4R} is compact and lies in K by (152). Cover S by finitely many open balls of radius R/4 centered at points q1 , . . . , qm ∈ S, and choose yi ∈ D with ∥yi − qi ∥ < R/4. Choose a ∈ D ∩ BV (x0 , R/2), which is possible because x0 ∈ K. Let u ∈ V with ∥u∥ = 1. The point x0 + 4Ru ∈ S lies within R/4 of some qi , hence within R/2 of yi , and therefore max ⟨yi − x0 , u⟩ > 4R −

1≤i≤m

73

R > 3R. 2

(154)

If some z ∈ BV (x0 , 3R) were not in the compact convex set P ⊆ A, separation in A would give a unit vector u ∈ V with ⟨z − x0 , u⟩ > maxi ⟨yi − x0 , u⟩. By (154) this would force ∥z − x0 ∥ > 3R, which is impossible. This proves (153). The relatively open balls BV (x0 , R), x0 ∈ C, cover C, and a finite subcollection still covers C. It therefore suffices to work on one such ball B := BV (x0 , R),

K0 := BV (x0 , 2R).

Since fn converges to finite values at the selected points, after discarding finitely many terms there is M > 0 such that |fn (a)| ≤ M, |fn (yi )| ≤ M (1 ≤ i ≤ m). (155) For these indices, the points yi lie in dom(fn ), which is convex by Theorem 4, so P ⊆ dom(fn ). The same theorem shows that fn is quasiconvex. Applying quasiconvexity repeatedly to a convex combination of y1 , . . . , ym and using (155) gives fn (z) ≤ max fn (yi ) ≤ M 1≤i≤m

∀z ∈ P.

(156)

By (153), the set dom(fn ) ⊆ A contains the ball BV (x0 , 3R), which is open in A. Hence A is the affine hull of dom(fn ), and the point a lies in ri(dom(fn )). The function fn is locally Lipschitz at a by Theorem 6, so Proposition 1-(ii) provides pn ∈ ∂fn (a) ⊆ V . If pn ̸= 0, put wn := a + R

pn . ∥pn ∥

Then ∥wn −x0 ∥ ≤ ∥a−x0 ∥+R < 3R/2, so wn ∈ P by (153). Moreover, ⟨pn , wn −a⟩ = R∥pn ∥ > 0, so ψγ (⟨pn , wn − a⟩) = γR∥pn ∥. The support inequality at a, together with (156) at z = wn and (155), gives 2M ≥ fn (wn ) − fn (a) ≥ γR∥pn ∥, that is, ∥pn ∥ ≤

2M . γR

(157)

This bound is automatic when pn = 0. Let z ∈ K0 . Then z ∈ P by (153), and ∥z − a∥ ≤ ∥z − x0 ∥ + ∥x0 − a∥ ≤ 5R/2. The support inequality at a, followed by (141), the Cauchy–Schwarz inequality, (155), and (157), gives fn (z) ≥ fn (a) + ψγ (⟨pn , z − a⟩) ≥ fn (a) −

1 1 2M 5R 5M ∥pn ∥ ∥z − a∥ ≥ −M − · · = −M − 2 . γ γ γR 2 γ

Combining this with (156), and with M0 := M + 5M/γ 2 , we get sup |fn (z)| ≤ M0 .

(158)

z∈K0

Because BV (x0 , 3R) is contained in the effective domain of fn , and fn is γ–tilted convex with respect ˆ n , whose domain is dense in dom(fn ) by Proposition 1-(i), Theorem 33 applies with center to ∂f x0 and radius R. By (158), the bound M in that theorem is at most M0 , so it gives the common Lipschitz bound 2M0 |fn (x) − fn (y)| ≤ 2 ∥x − y∥ ∀x, y ∈ B. (159) γ R 74

The ball B is open in A and contained in K, and D is dense in K; hence D ∩ B is dense in B. Since fn converges on D ∩ B, Theorem 30 gives a function fB on B, Lipschitz with the constant in (159), such that fn → fB uniformly on B. (160) Every x ∈ ri(K) lies in such a ball (take C = {x}), and pointwise limits are unique. Hence f (x) := limn fn (x) defines a function f : ri(K) → R with f = fB on each ball B constructed above. In particular, f is Lipschitz on a neighborhood in A of each point, and so it is continuous. Applying (160) to the finite collection of balls covering C proves uniform convergence of fn to f on C. This proves (i). γ–tilted convexity of lsc(f ). Extend f by +∞ outside ri(K) and set f¯ := lsc(f ). At x ∈ ri(K), only points of ri(K) contribute finite values to lim inf y→x f (y), and f is continuous there; hence f¯ = f on ri(K). We first verify the support inequality on ri(K), taking all subdifferentials relative to A until stated otherwise. Fix x, y ∈ ri(K),

v ∈ ∂ f¯(x) = ∂f (x),

where the equality holds because f¯ = f on the set ri(K), which is open in A. Apply the construction of the first part with x0 := x. It gives a ball B = BV (x, R) on which, after discarding finitely many terms, the functions fn share the Lipschitz constant in (159) and converge uniformly to f by (160). By the last part of Theorem 30, applied on B, there exist xn → x,

vn ∈ ∂fn (xn ),

vn → v.

(161)

For large n we have xn ∈ B and y ∈ dom(fn ), because fn (y) → f (y) ∈ R. Moreover, vn ∈ ∂fn (xn ) by Proposition 1-(i). Since each fn is γ–tilted convex,
fn (y) − fn (xn ) ≥ ψγ ⟨vn , y − xn ⟩ . (162) We let n → ∞ in (162). First, fn (y) → f (y). Second, |fn (xn ) − f (x)| ≤ sup |fn − f | + |f (xn ) − f (x)| −→ 0 B

by (160), (161), and the continuity of f . Third, the right-hand side converges by (161) and the continuity of ψγ . This yields
f (y) − f (x) ≥ ψγ ⟨v, y − x⟩ (x, y ∈ ri(K), v ∈ ∂f (x)). (163) This inequality also shows that f¯ does not take the value −∞. Fix x ∈ ri(K) and v ∈ ∂f (x); such v exists by Proposition 1-(ii), since f is Lipschitz near x. Let z ∈ Rd . If z ∈ / K, some neighborhood ¯ ¯ of z misses ri(K), so f (z) = +∞. If z ∈ K, then f (z) is the lim inf of f (y) as y → z with y ∈ ri(K), and (163) together with the continuity of ψγ gives f¯(z) ≥ f (x) + ψγ (⟨v, z − x⟩) > −∞. Hence f¯ is proper, since it is finite on the nonempty set ri(K), and it is lsc by construction. Each fn is quasiconvex by Theorem 4. If x′ , y′ ∈ ri(K) and t ∈ [0, 1], then (1 − t)x′ + ty′ ∈ ri(K), and letting n → ∞ in fn ((1 − t)x′ + ty′ ) ≤ max{fn (x′ ), fn (y′ )} gives f ((1 − t)x′ + ty′ ) ≤ max{f (x′ ), f (y′ )}. 75

(164)

Thus f is quasiconvex on ri(K). Consequently, dom(f¯) is convex: if a, b ∈ dom(f¯), then, since f = +∞ outside ri(K), we can choose sequences ak , bk ∈ ri(K) such that ak → a,

f (ak ) → f¯(a),

bk → b,

f (bk ) → f¯(b).

Then, for t ∈ [0, 1], the points (1 − t)ak + tbk converge to (1 − t)a + tb. The lsc envelope is bounded by the lim inf along any convergent sequence, so (164) gives f¯((1 − t)a + tb) ≤ lim inf f ((1 − t)ak + tbk ) ≤ lim inf max{f (ak ), f (bk )} = max{f¯(a), f¯(b)} < +∞. k→∞

k→∞

Moreover, if y ∈ dom(f¯), choose yk ∈ ri(K) such that yk → y and f (yk ) → f¯(y). Applying (163) at yk , letting k → ∞, and using f¯(x) = f (x) gives
f¯(y) − f¯(x) ≥ ψγ ⟨v, y − x⟩ (165) for every x ∈ ri(K) and v ∈ ∂ f¯(x). We have ri(K) ⊆ dom(f¯) ⊆ K, and dom(f¯) is convex. A convex set lying between a convex set and its closure has the same relative interior, so ri(dom(f¯)) = ri(K). Therefore (165) also verifies the ambient support inequality in Theorem 8-(iv). To check this, let u ∈ ∂ˆf¯(x) be an ambient regular subgradient and y ∈ dom(f¯). Since dom(f¯) ⊆ A, we have y − x ∈ V , and hence ⟨u, y − x⟩ = ⟨PV u, y − x⟩. (166) The definition of ∂ˆf¯(x) involves only points of dom(f¯), so (166) shows that PV u is a relative regular subgradient, and it lies in ∂ f¯(x) by Proposition 1-(i). Then (165) with v = PV u and (166) give f¯(y) − f¯(x) ≥ ψγ (⟨u, y − x⟩). Theorem 8 shows that f¯ is γ–tilted convex. This proves (ii). The additional pointwise limit. Finally, consider the case where fn (x) → g(x) for all x ∈ K for some lsc function g. Both g and f are pointwise limits of fn on ri(K), so g = f on ri(K). Extend g by +∞ outside K and set ḡ := lsc(g). Because g is lsc on K, we have ḡ = g on K; in particular, ḡ = f = f¯ on ri(K). γ–tilted convexity of lsc(g). Fix x ∈ ri(K),

v ∈ ∂ḡ(x) = ∂f (x),

where the equality holds because ḡ = f on the set ri(K), which is open in A. The sequences xn and vn in (161) depend only on x, on v ∈ ∂f (x), and on the convergence fn → f near x, so they are available here. For any y ∈ K with g(y) < +∞, we have fn (y) → g(y) < +∞, so y ∈ dom(fn ) for all sufficiently large n. For these n, the γ–tilted convexity of fn gives (162). We let n → ∞ in (162), using pointwise convergence at y, the limit fn (xn ) → f (x) = g(x) obtained above, (161), and the continuity of ψγ . This yields
g(y) − g(x) ≥ ψγ ⟨v, y − x⟩ , (167) which in particular rules out g(y) = −∞, since the right-hand side is finite. Here g(x) = f (x) = ḡ(x). If y ∈ dom(ḡ), then, since g = +∞ outside dom(g), we can choose yk ∈ dom(g) such that yk → y and g(yk ) → ḡ(y). Applying (167) at yk and letting k → ∞, we obtain
ḡ(y) − ḡ(x) ≥ ψγ ⟨v, y − x⟩ . (168)

76

Next we verify the domain condition in Theorem 8. Let y′ , z′ ∈ dom(g) and t ∈ [0, 1], and put w := (1 − t)y′ + tz′ ∈ K. Letting n → ∞ in fn (w) ≤ max{fn (y′ ), fn (z′ )} gives g(w) ≤ max{g(y′ ), g(z′ )} < +∞. Thus g is quasiconvex and dom(g) is convex. The sequence argument used for f¯, with ri(K) replaced by dom(g), then shows that dom(ḡ) is convex. The function ḡ cannot take the value −∞: for fixed x and v as above, (167) bounds g(y) below by g(x) + ψγ (⟨v, y − x⟩), and this bound stays bounded below as y approaches any fixed point of K; outside K we have ḡ = +∞. Thus ḡ is proper and lsc. Moreover, ri(dom(ḡ)) = ri(K), because ri(K) ⊆ dom(ḡ) ⊆ K and dom(ḡ) is convex. As for f¯, projecting ambient regular subgradients onto V as in (166) shows that (168) verifies condition (iv) of Theorem 8. Hence ḡ is γ–tilted convex. Equality of the lsc envelopes. The functions f¯ and ḡ are therefore proper, lsc, γ–tilted convex, and agree on ri(K). Since ri(K) ⊆ dom(f¯) ⊆ K, we have ri(dom(f¯)) = ri(K),

dom(f¯) = K.

Theorem 10, applied to the pair f¯, ḡ, therefore gives ḡ = f¯ on K. Moreover, both f¯ = lsc(f ) and ḡ = lsc(g) equal +∞ outside K, because f and g equal +∞ outside K. Hence lsc(g) = ḡ = f¯ = lsc(f )

on Rd ,

as claimed. C.8 Proof of Theorem 12 Proof. Two functions on the real line. Let f, g : R → R belong to F γ and suppose that h := max{f, g} is proper. The epigraph of h is the intersection of the closed epigraphs of f and g, so h is lsc. Its domain I := dom(h) = dom(f ) ∩ dom(g) is non-empty, and it is an interval because dom(f ) and dom(g) are intervals by Theorem 4. If I = {x} is a single point, then the only pair in (5) is y = x, for which both sides vanish, so h ∈ F γ . Assume that I has non-empty interior. Then ri(I) = int(I) = int(dom(f )) ∩ int(dom(g)). By Theorem 6, f and g are locally Lipschitz on the interiors of their domains, so h is locally Lipschitz on int(I). A locally Lipschitz function on an open interval is differentiable almost everywhere, so the points of int(I) at which h′ exists are dense in int(I) and hence in I. Fix x ∈ int(I) at which h′ (x) exists, and fix y ∈ I. Choose τ > 0 with [x, x + τ ] ⊆ int(I) and put tn := τ /n. For each n, the value h(x + tn ) equals f (x + tn ) or g(x + tn ). After exchanging f and g and passing to a subsequence, we may assume that h(x + tn ) = f (x + tn )

for all n.

(169)

The functions f and h are continuous at x, so letting n → ∞ in (169) gives h(x) = f (x). Together with (169), this gives f (x + tn ) − f (x) h(x + tn ) − h(x) = −→ h′ (x) tn tn

(n → ∞).

(170)

The compact segment [x, x+τ ] lies in the open interval int(dom(f )), on which f is locally Lipschitz, so f is Lipschitz on an open neighborhood of [x, x+τ ]. Theorem 31 on [x, x+tn ] gives ξn ∈ (x, x+tn ) and wn ∈ ∂f (ξn ) with f (x + tn ) − f (x) = wn tn , (171) 77

where we use that the Clarke subdifferential at an interior point of the domain depends only on the values of f near that point. Since f is γ–tilted convex and ξn , y ∈ dom(f ),
f (y) − f (ξn ) ≥ ψγ wn (y − ξn ) . (172) By (170) and (171), wn → h′ (x). Moreover ξn → x, f (ξn ) → f (x) by continuity of f at x, and ψγ is continuous by Lemma 1. Letting n → ∞ in (172) and using h ≥ f and h(x) = f (x), we obtain
h(y) − h(x) ≥ f (y) − f (x) ≥ ψγ h′ (x)(y − x) . (173) The domain I is convex, the points of ri(I) at which h′ exists are dense in I, and (173) holds for each such point x and every y ∈ I. This is condition (v) of Theorem 8, so h ∈ F γ . Finitely many functions on Rd . Let f, g ∈ F γ be functions on Rd such that h := max{f, g} is proper. As before, h is lsc. Let ℓ be a line that meets dom(h) = dom(f ) ∩ dom(g). Then ℓ meets dom(f ) and dom(g), and the forward implication of Theorem 9 shows that f |ℓ and g|ℓ are γ–tilted convex. Both restrictions are proper and lsc, so they belong to F γ on the line ℓ. Their maximum h|ℓ is finite at a point of ℓ, so it is proper, and the previous paragraphs show that h|ℓ is γ–tilted convex. The reverse implication of Theorem 9 then gives h ∈ F γ . Let f1 , . . . , fm ∈ F γ have a proper maximum. For k ≤ m, the maximum max{f1 , . . . , fk } lies between f1 > −∞ and max{f1 , . . . , fm }, so it is also proper, and induction on k shows that max{f1 , . . . , fm } ∈ F γ .

(174)

Countable families. Let (fn )n≥1 ⊆ F γ have a proper supremum s. Set sn := max{f1 , . . . , fn }. Each sn is proper, since f1 ≤ sn ≤ s, so sn ∈ F γ by (174). Let K := dom(s), which is non-empty. To see that K is convex, let y, z ∈ K and w ∈ [y, z]. Each sn is quasiconvex by Theorem 4, so sn (w) ≤ max{sn (y), sn (z)} ≤ max{s(y), s(z)} < +∞,

(175)

and taking the supremum over n in (175) gives s(w) < +∞. For x ∈ K we have f1 (x) ≤ sn (x) ≤ s(x) < +∞, so the values sn (x) are finite and increase to s(x). The function g := s|K is lsc on K, because s is lsc as a supremum of lsc functions. Theorem 11, applied to the sequence (sn ) with the convex set K and the dense subset D := K, shows in its final assertion that lsc(g) is γ–tilted convex. The proper extension of g is s, which is lsc, so lsc(g) = s. Hence s ∈ F γ . Arbitrary families. Let {fi }i∈I ⊆ F γ have a proper supremum s. For each i the set Ui := Rd+1 \ epi(fi ) is open, because fi is lsc, and [ Rd+1 \ epi(s) = Ui , (176) i∈I

T because epi(s) = i∈I epi(fi ). Let B be the countable collection of open balls in Rd+1 with rational centers and rational radii. For each ball B ∈ B that lies in some Ui , choose one such index iB . Every point of an open set Ui lies in a ball of B contained in Ui , so the countably many sets UiB also cover the set in (176). This set is non-empty, since it contains (x, s(x) − 1) for any x ∈ dom(s), so at least one index iB is chosen. EnumerateTthe chosen indices as i1 , i2 , . . ., repeating one index if there are only finitely many. Then epi(s) = n epi(fin ) = epi(supn fin ), so s = supn fin , and the countable case gives s ∈ F γ .

78

C.9 Proof of Theorem 13 ˆ is γ–tilted monotone. Fix x, y ∈ dom(f ) and x∗ ∈ ∂f ˆ (x). If x = y, Proof. (i): Assume that ∂f both sides of (5) are 0, so suppose x ̸= y. By Theorem 32, there exist λ ∈ [0, 1) and sequences ˆ (xk ) such that xk → λx + (1 − λ)y and vk ∈ ∂f 1 f (y) − f (x) ≥ lim sup⟨vk , xk − x⟩. (177) 1 − λ k→∞ ˆ (x) and vk ∈ ∂f ˆ (xk ), inequality (6) for ∂f ˆ with (y, y∗ ) replaced by (xk , vk ) gives Since x∗ ∈ ∂f
⟨vk , xk − x⟩ ≥ ψγ ⟨x∗ , xk − x⟩ ∀k. (178) The limit of xk is λx + (1 − λ)y = x + (1 − λ)(y − x), so ⟨x∗ , xk − x⟩ → (1 − λ)⟨x∗ , y − x⟩. Since
ψγ is continuous (Lemma 1(1)), the right-hand side of (178) converges to ψγ (1 − λ)⟨x∗ , y − x⟩ . Combining (177) and (178), we obtain 1 f (y) − f (x) ≥ lim sup⟨vk , xk − x⟩ 1 − λ k→∞ 1 lim sup ψγ (⟨x∗ , xk − x⟩) ≥ 1 − λ k→∞ 1 = ψγ (⟨x∗ , (1 − λ)(y − x)⟩) 1−λ = ψγ (⟨x∗ , y − x⟩). The last equality holds because 1 − λ > 0 and ψγ is positively homogeneous (Lemma 1(1)). Since ˆ (x) were arbitrary, f is γ–tilted convex with respect to ∂f ˆ . Thus, by x, y ∈ dom(f ) and x∗ ∈ ∂f the implication (iii)⇒(i) of Theorem 8, it is γ–tilted convex. (ii): Assume that f is γ–tilted convex with respect to F . Fix x, y ∈ dom(F ), x∗ ∈ F (x), and y∗ ∈ F (y). Since dom(F ) ⊆ dom(f ), we may apply (5) at the pair (x, y) with v = x∗ and at the pair (y, x) with v = y∗ . This gives f (y) − f (x) ≥ ψγ (⟨x∗ , y − x⟩)

(179)

f (x) − f (y) ≥ ψγ (⟨y∗ , x − y⟩).

(180)

and Multiplying (180) by −1 and using Lemma 1(3) gives f (y) − f (x) ≤ −ψγ (−⟨y∗ , y − x⟩) = ψ γ (⟨y∗ , y − x⟩). Together with (179), this yields ψ γ (⟨y∗ , y − x⟩) ≥ ψγ (⟨x∗ , y − x⟩).

(181)

Applying the increasing function ψγ to both sides of (181), and using ψγ ◦ ψ γ = id (Lemma 1(6)) on the left and ψγ ◦ ψγ = ψγ 2 (Lemma 1(4)) on the right, gives ⟨y∗ , y − x⟩ = ψγ (ψ γ (⟨y∗ , y − x⟩)) ≥ ψγ (ψγ (⟨x∗ , y − x⟩)) = ψγ 2 (⟨x∗ , y − x⟩), which is (6) with parameter γ 2 . Hence F is γ 2 –tilted monotone. ∗ For the inequalities between the optimal parameters, let γ0 := γmono (f ). If γ0 > 0, then ∂f ˆ ˆ is also is γ0 –tilted monotone. By Proposition 1(i), ∂f (x) ⊆ ∂f (x) at every x ∈ dom(f ), so ∂f ∗ ∗ γ0 –tilted monotone. Part (i) then shows that f is γ0 –tilted convex, so γmono (f ) ≤ γ (f ). This inequality is trivial if γ0 = 0. Next let γ1 := γ ∗ (f ). If γ1 > 0, then part (ii) with F = ∂f , whose ∗ domain is contained in dom(f ), shows that ∂f is γ12 –tilted monotone. Hence γ ∗ (f )2 ≤ γmono (f ), ∗ ∗ 1/2 and this is again trivial if γ1 = 0. Taking square roots gives γ (f ) ≤ (γmono (f )) .

79

C.10 Proof of Theorem 14: Alternative First-order Characterization If an operator is γ–tilted monotone, then so is every operator whose values are subsets of its values. Since f is lsc, (115) holds, and it gives the implications (i) =⇒ (ii) =⇒ (iii). If (iii) holds, Theorem 13(i) shows that f is γ–tilted convex. Hence K is convex by Theorem 4, f |ri(K) is locally ˆ |ri(K) is Lipschitz by Theorem 6, and f is radially continuous by Corollary 2. The restriction ∂f ˆ is. Thus condition (iv) holds. For (iv) =⇒ (v), let x ∈ ri(K) γ–tilted monotone because ∂f be a point where the relative gradient ∇f (x) exists. The argument that gave (116) uses only ˆ (x). Hence the inequality in (v) is a special case of dom(f ) ⊆ aff(K), so it shows that ∇f (x) ∈ ∂f ˆ the γ–tilted monotonicity of ∂f |ri(K) in (iv), and the other requirements in (v) are part of (iv). It remains to prove (ii) =⇒ (i), (iii) =⇒ (ii), (iv) =⇒ (iii), and (v) =⇒ (iv). C.10.1 Proof of (ii) =⇒ (i) ˆ is γ–tilted monotone, so Theorem 13(i) Proof. Assume ∂f is γ–tilted monotone. By (115), ∂f ˆ , and shows that f is γ–tilted convex. By (115) again, f is γ–tilted convex with respect to ∂f ′ Theorem 4 shows that K is convex. Thus Lemma 9 applies to f and gives, for every x ∈ K, ∂ ∞ f (x′ ) ⊆ N̂K (x′ ) = {n : ⟨n, y − x′ ⟩ ≤ 0 ∀y ∈ K}.

(182)

The set N̂K (x′ ) is an intersection of closed half-spaces, hence convex, so conv(∂ ∞ f (x′ )) ⊆ N̂K (x′ ). By the definition of ∂f , this gives ∂f (x′ ) = conv(∂f (x′ )) + conv(∂ ∞ f (x′ )) ⊆ conv(∂f (x′ )) + N̂K (x′ ),

x′ ∈ K.

(183)

Fix x ∈ K and define n o V := v ∈ Rd | ⟨y − x, u⟩ ≥ ψγ (⟨y − x, v⟩), ∀y ∈ K, u ∈ ∂f (y) n o = v ∈ Rd | ψ γ (⟨y − x, u⟩) ≥ ⟨y − x, v⟩, ∀y ∈ K, u ∈ ∂f (y) o n \ = v ∈ Rd | ψ γ (⟨y − x, u⟩) ≥ ⟨y − x, v⟩ , y∈K,u∈∂f (y)

where the second equality follows from Lemma 1: ψγ and ψ γ are increasing and inverse to each other. Each set in the last intersection is a closed half-space or Rd , so V is closed and convex. The γ–tilted monotonicity of ∂f , applied to the pairs (x, v) with v ∈ ∂f (x) and (y, u), gives ∂f (x) ⊆ V , and hence conv(∂f (x)) ⊆ V . Now let v ∈ V and n ∈ N̂K (x). By (182), ⟨y − x, n⟩ ≤ 0 for every y ∈ K. Since ψγ is increasing, ψγ (⟨y − x, v + n⟩) ≤ ψγ (⟨y − x, v⟩) ≤ ⟨y − x, u⟩,

y ∈ K, u ∈ ∂f (y),

so v + n ∈ V . Consequently conv(∂f (x)) + N̂K (x) ⊆ V , and since V is closed, conv(∂f (x)) + N̂K (x) ⊆ V. By (183) with x′ = x, this gives ∂f (x) ⊆ V . Since x ∈ K was arbitrary, we have shown ⟨y − x, u⟩ ≥ ψγ (⟨y − x, v⟩),

∀x, y ∈ K, v ∈ ∂f (x), u ∈ ∂f (y).

80

(184)

Next, fix y ∈ K and define n o U := u ∈ Rd | ⟨y − x, u⟩ ≥ ψγ (⟨y − x, v⟩), ∀x ∈ K, v ∈ ∂f (x) n o \ = u ∈ Rd | ⟨y − x, u⟩ ≥ ψγ (⟨y − x, v⟩) . x∈K,v∈∂f (x)

Each set in the intersection is a closed half-space or Rd , so U is closed and convex. By (184), ∂f (y) ⊆ U , and hence conv(∂f (y)) ⊆ U . Now let u ∈ U and n ∈ N̂K (y). By (182) with x′ = y, we have ⟨x − y, n⟩ ≤ 0, that is, ⟨y − x, n⟩ ≥ 0, for every x ∈ K. Hence ⟨y − x, u + n⟩ ≥ ⟨y − x, u⟩ for every x ∈ K, and u + n ∈ U . Since U is closed, it follows that conv(∂f (y)) + N̂K (y) ⊆ U, and (183) with x′ = y gives ∂f (y) ⊆ U . Hence, for all y ∈ K and u ∈ ∂f (y), we have ⟨y − x, u⟩ ≥ ψγ (⟨y − x, v⟩),

∀x ∈ K, v ∈ ∂f (x).

This is the γ–tilted monotonicity of ∂f . C.10.2 Proof of (iii) =⇒ (ii) Proof. Assume (iii). Fix x, y ∈ K, v ∈ ∂f (x), and u ∈ ∂f (y). By the definition of the limiting subdifferential, there are sequences f

xk − → x,

ˆ (xk ), vk ∈ ∂f

f

vk → v,

yk − → y,

ˆ (yk ), uk ∈ ∂f

uk → u.

Applying (iii) to the pairs (xk , vk ) and (yk , uk ) gives ⟨uk , yk − xk ⟩ ≥ ψγ (⟨vk , yk − xk ⟩).

(185)

As k → ∞, we have uk → u, vk → v, and yk − xk → y − x. Since ψγ is continuous, passing to the limit in (185) gives ⟨u, y − x⟩ ≥ ψγ (⟨v, y − x⟩), which proves that ∂f is γ–tilted monotone. C.10.3 Proof of (iv) =⇒ (iii) Proof. Assume (iv). Put A := aff(K) and W := A − A, and work in the Euclidean space A. This does not change any of the inner products below, because all differences of points of K lie in W . We write f |ri(K) for the restriction of f to the relatively open set ri(K); its subdifferentials are taken in A and consist of vectors of W . Let x′ ∈ ri(K) and let a be a regular subgradient of f |ri(K) at x′ . Since ri(K) is open in A and K ⊆ A, all points of K close to x′ lie in ri(K). So the quotients in the definitions of the regular ˆ (x′ ). Thus the monotonicity subdifferentials of f |ri(K) and of f at x′ agree near x′ , and a ∈ ∂f assumed in (iv) also holds for the regular subdifferential of f |ri(K) . The proof of (iii) =⇒ (ii) above uses only the definitions of the regular and limiting subdifferentials and the continuity of ψγ . Applied to f |ri(K) in A, it shows that ⟨b, y′ − x′ ⟩ ≥ ψγ (⟨a, y′ − x′ ⟩) 81

(186)

for all x′ , y′ ∈ ri(K), a ∈ ∂(f |ri(K) )(x′ ), and b ∈ ∂(f |ri(K) )(y′ ). By condition (iv), f is locally Lipschitz on ri(K) and radially continuous. In particular, f is continuous on every line segment joining two points of K. ˆ (x), and u ∈ ∂f ˆ (y). If x = y, both sides of the required inequality are Fix x, y ∈ K, v ∈ ∂f zero. So assume x ̸= y and set d := y − x ̸= 0. Choose z ∈ ri(K) and, for ϵ ∈ (0, 1), define pϵ (t) := (1 − ϵ)(x + td) + ϵz,

t ∈ R.

For t ∈ [0, 1], the point x + td lies in K, so Theorem 6.1 in [Roc15] gives pϵ (t) ∈ ri(K). Thus the segment pϵ ([0, 1]) is contained in ri(K). The set {t ∈ R : pϵ (t) ∈ ri(K)} is open, because pϵ is continuous and ri(K) is open in A. It contains [0, 1], so its connected component containing [0, 1] is an open interval Jϵ ⊇ [0, 1] with pϵ (Jϵ ) ⊆ ri(K). Define hϵ (t) := f (pϵ (t)),

t ∈ Jϵ .

We now verify the assumptions of Theorem 29. The function f |ri(K) is proper and lsc on ri(K) (indeed continuous), while pϵ : R → A is affine, and hence continuously differentiable, with ∇pϵ (s)a = (1 − ϵ)ad

(a ∈ R).

On the open interval Jϵ , the function hϵ coincides with f |ri(K) ◦ pϵ , so the two functions have the same subdifferentials there. Moreover, the local Lipschitz property in condition (iv) shows that f |ri(K) is locally Lipschitz at pϵ (s) for every s ∈ Jϵ . Thus the limiting subdifferential part of Theorem 29 applies and gives the inclusion n o ∂hϵ (s) ⊆ ∇pϵ (s)⊤ ∂(f |ri(K) )(pϵ (s)) = ⟨a, (1 − ϵ)d⟩ a ∈ ∂(f |ri(K) )(pϵ (s)) , s ∈ Jϵ . (187) Also, since a composition of a locally Lipschitz function with an affine map is locally Lipschitz, hϵ is locally Lipschitz on Jϵ . We claim that ∂hϵ is γ–tilted monotone on Jϵ . Fix s, t ∈ Jϵ , a ∈ ∂hϵ (s), and b ∈ ∂hϵ (t). By (187), a = ⟨as , (1 − ϵ)d⟩, b = ⟨at , (1 − ϵ)d⟩ for suitable as ∈ ∂(f |ri(K) )(pϵ (s)) and at ∈ ∂(f |ri(K) )(pϵ (t)). Since pϵ (t) − pϵ (s) = (t − s)(1 − ϵ)d, the inequality (186) with x′ = pϵ (s) and y′ = pϵ (t) reads
⟨at , (t − s)(1 − ϵ)d⟩ ≥ ψγ ⟨as , (t − s)(1 − ϵ)d⟩ , that is, (t − s)b ≥ ψγ ((t − s)a).

(188)

This is precisely the required monotonicity inequality for ∂hϵ . Because hϵ is locally Lipschitz on Jϵ , Proposition 1-(ii) gives ∇hϵ ⊆ ∂hϵ ⊆ ∂hϵ = conv(∇hϵ ), and therefore ∂hϵ = conv(∂hϵ ) on Jϵ . Fix s, t ∈ Jϵ . For fixed b ∈ ∂hϵ (t), by Lemma 1 the set of a ∈ R satisfying (188) is {a : (t − s)a ≤ ψ γ ((t − s)b)}, a closed half-line or R. It is convex and contains ∂hϵ (s), so it contains ∂hϵ (s). Next, for fixed a ∈ ∂hϵ (s), the set of b ∈ R satisfying (188) is again a closed half-line or R. It contains ∂hϵ (t), so it contains ∂hϵ (t). Hence (188) holds for all a ∈ ∂hϵ (s) and b ∈ ∂hϵ (t), that is, ∂hϵ is also γ–tilted monotone on Jϵ . Fix 0 < r < s < 1. The function hϵ is continuous on [0, 1] and locally Lipschitz on Jϵ ⊇ [0, 1]. Applying Corollary 12 on [0, r] and [s, 1], respectively, gives points cϵ ∈ (0, r) and eϵ ∈ (s, 1) and numbers aϵ ∈ ∂hϵ (cϵ ) and bϵ ∈ ∂hϵ (eϵ ) such that aϵ =

hϵ (r) − hϵ (0) , r

bϵ = 82

hϵ (1) − hϵ (s) . 1−s

Since cϵ < r < s < eϵ , the monotonicity of ∂hϵ , applied to (cϵ , aϵ ) and (eϵ , bϵ ), and the positive homogeneity of ψγ give
(eϵ − cϵ )bϵ ≥ ψγ (eϵ − cϵ )aϵ = (eϵ − cϵ )ψγ (aϵ ). Dividing by eϵ − cϵ > 0 and substituting the values of aϵ and bϵ yields   hϵ (1) − hϵ (s) hϵ (r) − hϵ (0) . ≥ ψγ 1−s r

(189)

For every fixed t ∈ [0, 1], the point pϵ (t) lies on the segment joining x + td ∈ K to z ∈ K, and it tends to x + td as ϵ ↓ 0. Continuity of f on this segment gives hϵ (t) −→ f (x + td)

as ϵ ↓ 0.

Applying this with t ∈ {0, r, s, 1}, letting ϵ ↓ 0 in (189), and using the continuity of ψγ , we obtain   f (x + rd) − f (x) f (y) − f (x + sd) ≥ ψγ (0 < r < s < 1). (190) 1−s r Set a := ⟨v, d⟩,

b := ⟨u, d⟩.

ˆ (x), the definition of the regular subdifferential gives Since v ∈ ∂f f (w) − f (x) − ⟨v, w − x⟩ ≥ 0. ∥w − x∥ w∈K, w̸=x lim inf w→x

The points w = x + rd with r ∈ (0, 1] lie in K \ {x}. Evaluating this lower limit along them as r ↓ 0 yields f (x + rd) − f (x) − r⟨v, d⟩ 0 ≤ lim inf r↓0 r∥d∥   1 f (x + rd) − f (x) = lim inf −a . r↓0 ∥d∥ r Consequently, a ≤ lim inf r↓0

f (x + rd) − f (x) . r

ˆ (y) means that Similarly, u ∈ ∂f lim inf w→y w∈K, w̸=y

f (w) − f (y) − ⟨u, w − y⟩ ≥ 0. ∥w − y∥

This time, evaluating along w = x + sd = y − (1 − s)d as s ↑ 1 gives 0 ≤ lim inf s↑1

=

1 ∥d∥

f (x + sd) − f (y) + (1 − s)⟨u, d⟩ (1 − s)∥d∥ ! f (y) − f (x + sd) b − lim sup , 1−s s↑1 83

(191)

so b ≥ lim sup s↑1

f (y) − f (x + sd) . 1−s

(192)

Fix r ∈ (0, 1). By (190), every quotient in (192) with s ∈ (r, 1) is at least ψγ ((f (x + rd) − f (x))/r). Hence (192) gives   f (x + rd) − f (x) , r ∈ (0, 1). (193) b ≥ ψγ r Let L be the lower limit in (191), so that L ≥ a, and choose rn ↓ 0 such that (f (x+rn d)−f (x))/rn → L. If L = +∞, then the right-hand side of (193) with r = rn tends to +∞, which is impossible. So L is finite. Letting n → ∞ in (193) with r = rn , and using the continuity and monotonicity of ψγ , gives ⟨u, y − x⟩ = b ≥ ψγ (L) ≥ ψγ (a) = ψγ (⟨v, y − x⟩). ˆ is γ–tilted monotone. Thus ∂f C.10.4 Proof of (v) =⇒ (iv) Proof. Condition (v) includes all requirements of condition (iv) except the monotonicity assertion, so we prove only that assertion. Put A := aff(K) and W := A − A, and interpret gradients relative b (x), and u ∈ ∂f b (y). Let vA and uA be the orthogonal projections of to A. Fix x, y ∈ ri(K), v ∈ ∂f v and u, respectively, onto W . Every point of dom(f ) = K differs from x and from y by a vector of W . Hence the quotients in the definition of the regular subdifferential do not change when v and u are replaced by vA and uA , which gives vA ∈ ∂bA f (x),

uA ∈ ∂bA f (y),

where the subdifferentials are taken relative to A. For the same reason, since y − x ∈ W , ⟨v, y − x⟩ = ⟨vA , y − x⟩,

⟨u, y − x⟩ = ⟨uA , y − x⟩.

(194)

Let ∇A f denote the set of Proposition 1-(ii) for f |A in the Euclidean space A. Let w ∈ ∇A f (x) and z ∈ ∇A f (y). Then there are sequences xk → x and yk → y in A such that the relative gradients ∇A f (xk ) and ∇A f (yk ) exist and converge to w and z, respectively. Since ri(K) is open in A, the points xk and yk lie in ri(K) for all large k. Condition (v) gives
⟨∇A f (yk ), yk − xk ⟩ ≥ ψγ ⟨∇A f (xk ), yk − xk ⟩ . Letting k → ∞ and using the continuity of ψγ yields
⟨z, y − x⟩ ≥ ψγ ⟨w, y − x⟩ , and applying the increasing inverse ψ γ (Lemma 1) gives
ψ γ ⟨z, y − x⟩ ≥ ⟨w, y − x⟩,

w ∈ ∇A f (x), z ∈ ∇A f (y).

(195)

Since f |A is locally Lipschitz at x and at y, Proposition 1-(i) and (ii), applied in the Euclidean space A, give

vA ∈ ∂bA f (x) ⊆ conv ∇A f (x) , uA ∈ ∂bA f (y) ⊆ conv ∇A f (y) . (196)

84

Fix z ∈ ∇A f (y). The set of w ∈ W satisfying (195) is a closed half-space of W or all of W , hence convex. It contains ∇A f (x), so by (196) it contains vA . Applying the increasing function ψγ gives
⟨z, y − x⟩ ≥ ψγ ⟨vA , y − x⟩ , z ∈ ∇A f (y). The set of z ∈ W satisfying this inequality is again a closed half-space of W or all of W . It contains ∇A f (y), so by (196) it contains uA . Combining this with (194), we obtain

⟨u, y − x⟩ = ⟨uA , y − x⟩ ≥ ψγ ⟨vA , y − x⟩ = ψγ ⟨v, y − x⟩ . b |ri(K) is γ–tilted monotone, proving condition (iv). Thus ∂f C.11 Proof of Theorem 15 Proof. Forward direction. Assume that ∂f is γ–tilted monotone and put K := dom(f ). Since ˆ ⊆ ∂f by Proposition 1-(i), the operator ∂f ˆ is γ–tilted monotone, and Theorem 13-(i) shows ∂f that f is γ–tilted convex. Hence K is convex by Theorem 4. Fix a line ℓ meeting K, write ℓ = {b + td : t ∈ R},

∥d∥ = 1,

and set I := {t ∈ R : b + td ∈ K},

h(t) := f (b + td).

If I is a singleton, the only instance of (6) for ∂h compares a point with itself, and both sides vanish; so ∂h is γ–tilted monotone. Suppose, therefore, that I is nondegenerate. Then ℓ ⊆ aff(K). We work relative to aff(K); relative limiting subgradients embed as ambient limiting subgradients because f = +∞ outside this affine hull. Thus the relative limiting subdifferential of f is also γ–tilted monotone. Choose a ∈ ri(K) and a sequence ϵn ↓ 0. Define τn (t) := (1 − ϵn )(b + td) + ϵn a,

hn (t) := f (τn (t)).

For t ∈ I, the point τn (t) belongs to ri(K) by (133). Let s, t ∈ int(I). The function f is locally Lipschitz at τn (s) and τn (t) by Theorem 6, and the derivative of τn is (1 − ϵn )d. Hence the second part of Theorem 29 shows that every p ∈ ∂hn (s) and q ∈ ∂hn (t) have the form p = (1 − ϵn )⟨v, d⟩,

q = (1 − ϵn )⟨u, d⟩

for some v ∈ ∂f (τn (s)) and u ∈ ∂f (τn (t)), where the subdifferentials of f are relative to aff(K). Since τn (t) − τn (s) = (1 − ϵn )(t − s)d, the relative γ–tilted monotonicity of ∂f gives
(t − s)q = ⟨u, τn (t) − τn (s)⟩ ≥ ψγ ⟨v, τn (t) − τn (s)⟩ = ψγ ((t − s)p). (197) Thus ∂hn |int(I) is γ–tilted monotone. Each hn is a reparametrized line restriction of f , and it is γ–tilted convex. To see this, put cn := 1 − ϵn and h̃n (r) := f (τn (0) + rd), so that hn (t) = h̃n (cn t)

(t ∈ R).

(198)

The function h̃n is the restriction of f to the line τn (R) in a unit-speed parametrization, and this line meets K; so Theorem 9 shows that h̃n is γ–tilted convex. The first part of Theorem 29, applied 85

ˆ n (t) ⊆ cn ∂ˆh̃n (cn t). Hence, if t, t′ ∈ dom(hn ) and to the inverse relation h̃n (r) = hn (r/cn ), gives ∂h ˆ n (t), then p = cn w with w ∈ ∂ˆh̃n (cn t) ⊆ ∂ h̃n (cn t), and p ∈ ∂h hn (t′ ) − hn (t) = h̃n (cn t′ ) − h̃n (cn t) ≥ ψγ (cn (t′ − t)w) = ψγ ((t′ − t)p). By Theorem 8-(iii), hn is γ–tilted convex. Moreover, for t ∈ I the value hn (t) equals gϵn (t) in the proof of Theorem 9 (with the same b, d, a), and h(t) equals g(t) there. So (137), which was derived from Theorem 34, gives hn (t) −→ h(t) (t ∈ I). Applying Theorem 11 in one dimension, with the convex set I and the dense subset I, shows that hn → h locally uniformly on int(I) and that h is continuous there. This local uniform convergence gives a common bound for the hn on compact subintervals, and hence a common local Lipschitz constant. To see this, fix t0 ∈ int(I) and R > 0 with [t0 − 3R, t0 + 3R] ⊆ I ⊆ dom(hn ). Since [t0 − 2R, t0 + 2R] is a compact subset of int(I), after discarding finitely many terms, |hn (t)| ≤ M := 1 +

sup |t−t0 |≤2R

max

|t−t0 |≤2R

|h(t)|.

(199)

ˆ n , whose domain is dense in dom(hn ) by Proposition 1Each hn is γ–tilted convex with respect to ∂h (i). So Theorem 33 and (199) give |hn (t) − hn (t′ )| ≤

2M |t − t′ | γ2R

(t, t′ ∈ (t0 − R, t0 + R)).

(200)

Fix s, t ∈ int(I), p ∈ ∂h(s), and q ∈ ∂h(t). By (200) with t0 = s and with t0 = t, the functions hn are equi-Lipschitz on small open intervals around s and around t, and they converge pointwise there to h. Applying the last part of Theorem 30 on these intervals gives sn → s,

pn ∈ ∂hn (sn ),

pn → p,

tn → t,

qn ∈ ∂hn (tn ),

qn → q.

and After discarding finitely many terms, sn , tn ∈ int(I), and hence (197) gives (tn − sn )qn ≥ ψγ ((tn − sn )pn ). Passing to the limit, using the continuity of ψγ , shows that (t − s)q ≥ ψγ ((t − s)p).

(201)

Thus ∂h|int(I) is γ–tilted monotone. Finally, h is γ–tilted convex by Theorem 9, and we check condition (iv) of Theorem 14 for h. The domain I is convex and ri(dom(h)) = int(I). The restriction h|I is continuous by Theorem 7, so h is radially continuous. The function h is locally Lipschitz on int(I) by Theorem 6. Since ˆ ⊆ ∂h, (201) shows that ∂h| ˆ int(I) is γ–tilted monotone. Theorem 14 now shows that ∂h is ∂h γ–tilted monotone. Now let J ⊆ R be a nonempty closed interval such that J ∩ I ̸= ∅, and restrict h to J, with the value +∞ outside J. The function h|J is proper and lsc, because J is closed, and its domain J ∩ I is convex. If J ∩ I is a singleton, ∂(h|J ) is γ–tilted monotone, because the only instance of the inequality compares a point with itself. Otherwise, ri(J ∩ I) ⊆ ri(I), 86

ˆ J ) = ∂h ˆ on ri(J ∩ I), and (201) and h|J agrees locally with h at every point of ri(J ∩ I). Thus ∂(h| shows that the regular subdifferential of h|J is γ–tilted monotone on the relative interior of its ˆ J )(t), and s ∈ J ∩ I, we have p ∈ ∂h(t) ˆ domain. For t ∈ ri(J ∩ I), p ∈ ∂(h| ⊆ ∂h(t), so the γ–tilted convexity of h gives h(s) − h(t) ≥ ψγ ((s − t)p). So the support characterization in Theorem 8-(iv) shows that h|J is γ–tilted convex. Moreover, h|J is continuous on J ∩ I because h is continuous on I, and it is locally Lipschitz on ri(J ∩ I) because h is. Hence Theorem 14-(iv) shows that ∂(h|J ) is γ–tilted monotone. This also proves the additional assertion and completes the forward direction. Reverse direction. Assume that ∂(f |ℓ ) is γ–tilted monotone for every line ℓ meeting dom(f ). ˆ (x), and u ∈ ∂f ˆ (y). If x = y, both sides of the required inequality Fix x, y ∈ dom(f ), v ∈ ∂f vanish. Otherwise, put y−x d := , q(t) := f (x + td), ∥y − x∥ which is f |ℓ for the line ℓ through x and y, in a unit-speed parametrization. The first part of Theorem 29, applied with F (t) := x + td at t = 0 and at t = ∥y − x∥, gives ˆ ⟨v, d⟩ ∈ ∂q(0),

ˆ ⟨u, d⟩ ∈ ∂q(∥y − x∥).

These sets are contained in ∂q(0) and ∂q(∥y − x∥) by Proposition 1-(i). Since ∂q is γ–tilted monotone, applying (6) to the points 0 and ∥y − x∥ and multiplying out ∥y − x∥d = y − x gives
⟨u, y − x⟩ ≥ ψγ ⟨v, y − x⟩ . ˆ is γ–tilted monotone, and Theorem 14 shows that ∂f is γ–tilted monotone. Thus ∂f C.12 Proof of Theorem 16 ˆ n ⊆ ∂fn , the operator ∂f ˆ n is γ–tilted monotone, and Theorem 13-(i) shows that Proof. Since ∂f each fn is γ–tilted convex. Theorem 11 therefore gives a continuous function f satisfying (i), and it shows that f¯ := lsc(f ) is γ–tilted convex. For the monotonicity in (ii), put V := A − A,

A := aff(K),

and work in the Euclidean space A, with the balls BV of A as in the proof of Theorem 11. Choose d0 ∈ D. After discarding finitely many terms, d0 ∈ dom(fn ), so fn |A is proper and lsc. For every line ℓ ⊆ A meeting its effective domain, the forward direction of Theorem 15 shows that ∂(fn |ℓ ) is γ–tilted monotone. The reverse direction, now applied in the Euclidean space A, shows that the relative Clarke subdifferential ∂(fn |A ) is γ–tilted monotone. Hence, by Theorem 14, the relative limiting subdifferential ∂(fn |A ) is also γ–tilted monotone. By Theorem 13, these restrictions, which we continue to denote by fn , are also γ–tilted convex. On O := ri(K), the convergence fn → f is locally uniform and locally equi-Lipschitz. To see the latter, fix x0 ∈ O and choose R > 0 such that BV (x0 , 3R) ⊆ O. The function f is continuous on this compact ball, and fn → f uniformly on it. Hence, after discarding finitely many terms, |fn | ≤ M := 1 +

max BV (x0 ,3R)

87

|f |

on BV (x0 , 3R).

In particular, this ball is contained in dom(fn ), so aff(dom(fn )) = A, and M bounds |fn | on ˆ n , whose domain is dense in dom(fn ) BV (x0 , 2R). Since each fn is γ–tilted convex with respect to ∂f by Proposition 1-(i), Theorem 33 gives the common Lipschitz constant 2M/(γ 2 R) on BV (x0 , R). Fix x, y ∈ O, v ∈ ∂f (x), and u ∈ ∂f (y), where these subdifferentials are relative to A. Applying the last part of Theorem 30 on balls BV (x, R) and BV (y, R′ ) of the kind just constructed gives xn → x,

vn ∈ ∂fn (xn ),

vn → v,

yn → y,

un ∈ ∂fn (yn ),

un → u.

and After discarding finitely many terms, xn , yn ∈ O, so the relative γ–tilted monotonicity of ∂fn gives
⟨un , yn − xn ⟩ ≥ ψγ ⟨vn , yn − xn ⟩ . Passing to the limit and using the continuity of ψγ yields
⟨u, y − x⟩ ≥ ψγ ⟨v, y − x⟩ .

(202)

Thus the relative subdifferential ∂f |O is γ–tilted monotone. The functions f and f¯ agree on O. If x ∈ O and v ∈ ∂ˆf¯(x), then the orthogonal projection PV v belongs to the relative regular subdifferential of f at x, as in (166), and hence to the relative subdifferential ∂f (x) by Proposition 1-(i). Let x, y ∈ O, v ∈ ∂ˆf¯(x), and u ∈ ∂ˆf¯(y). Since y − x ∈ V , (202) applied to PV v and PV u gives ⟨u, y − x⟩ = ⟨PV u, y − x⟩ ≥ ψγ (⟨PV v, y − x⟩) = ψγ (⟨v, y − x⟩). So ∂ˆf¯|O is γ–tilted monotone. Moreover, ri(dom(f¯)) = O, because O ⊆ dom(f¯) ⊆ K and dom(f¯) is convex by Theorem 4. The γ–tilted convex function f¯ is radially continuous by Corollary 2 and locally Lipschitz on O by Theorem 6. Hence condition (iv) of Theorem 14 holds for f¯, and that theorem proves that ∂ f¯ is γ–tilted monotone. Finally, under the additional pointwise-convergence assumption, Theorem 11 gives lsc(g) = f¯. Hence ∂(lsc(g)) = ∂ f¯ is γ–tilted monotone. C.13 Proof of Theorem 17 Proof. Throughout, for y ̸= z in K and λ ∈ (0, 1) we put x := λy + (1 − λ)z and use the slopes s± of (7). Since ∥x − z∥ and ∥y − x∥ are positive, s− > 0 exactly when f (x) > f (z), and s+ ≥ 0 exactly when f (y) ≥ f (x). (i): Quasiconvexity and convexity of the effective domain. Suppose that f is γ–tilted secant convex. The definition requires x ∈ K for all y ̸= z in K and λ ∈ (0, 1), so K is convex. If f (x) > f (z), then s− > 0, so ψγ (s− ) ≥ 0 is defined, and (8) gives s+ ≥ 0, that is, f (y) ≥ f (x). Hence f (x) ≤ max{f (y), f (z)}. This inequality is immediate when y = z, when λ ∈ {0, 1}, and when y or z lies in X \ K, so f is quasiconvex. (ii): Convex functions. Let f be convex. Then K is convex, and the slopes of the secants of f increase, so s+ ≥ s− . By Lemma 1(2), s− ≥ ψγ (s− ) whenever ψγ (s− ) is defined. Hence (8) holds, and f is γ–tilted secant convex. 88

(iii): Suprema. Here and in (v) we use the Jensen form of Proposition 2, whose right side is nondecreasing in the values. For every γ ∈ [0, 1] and λ ∈ [0, 1], the map (a, b) 7→ Mγ,λ (a, b) is nondecreasing in each argument and satisfies Mγ,λ (a, b) ≤ max{a, b}.

(203)

For γ = 0, these facts follow immediately from (14). Now let γ ∈ (0, 1], and put D1 := γλ + (1 − λ) and D2 := λ + γ(1 − λ). Since γ ≤ 1, we have D1 ≥ γλ + γ(1 − λ) = γ and likewise D2 ≥ γ, so D1 ≥ γ > 0, By (13), with ω := γλ/D1 and ν := λ/D2 , ( ωa + (1 − ω)b, a ≤ b, Mγ,λ (a, b) = νa + (1 − ν)b, a > b,

D2 ≥ γ > 0.

1−ω =

1−λ , D1

(204)

1−ν =

γ(1 − λ) , D2

(205)

and ω, ν ∈ [0, 1]. Each case of (205) is a weighted arithmetic mean of a and b with nonnegative coefficients, so it is at most max{a, b}. Both cases give Mγ,λ (a, a) = a. For monotonicity in a, fix b and let a1 < a2 . If a1 and a2 fall in the same case of (205), then Mγ,λ (a1 , b) ≤ Mγ,λ (a2 , b) because ω, ν ≥ 0. If a1 ≤ b < a2 , then Mγ,λ (a1 , b) ≤ max{a1 , b} = b, while Mγ,λ (a2 , b) = νa2 + (1 − ν)b ≥ b. For monotonicity in b, fix a and let b1 < b2 . If b1 and b2 fall in the same case, the claim follows from 1−ω, 1−ν ≥ 0. If b1 < a ≤ b2 , then Mγ,λ (a, b1 ) ≤ max{a, b1 } = a, while Mγ,λ (a, b2 ) = ωa + (1 − ω)b2 ≥ a. Hence Mγ,λ is nondecreasing in each variable. Let {fi : X → R ∪ {+∞}}i∈I be a non-empty family of γ–tilted secant convex functions and g := supi∈I fi . Since I ̸= ∅ and each fi > −∞, g takes values in R ∪ {+∞}. Let y, z ∈ dom(g) and λ ∈ [0, 1]. Since −∞ < fi ≤ g, the values fi (y) and fi (z) are finite for every i ∈ I. The γ–tilted secant convexity of fi and the monotonicity in (203), applied with fi (y) ≤ g(y) and fi (z) ≤ g(z), give

fi (λy + (1 − λ)z) ≤ Mγ,λ fi (y), fi (z) ≤ Mγ,λ g(y), g(z) (i ∈ I). Taking the supremum over i proves that g is γ–tilted secant convex. (iv): Convex restrictions. Now let f be γ–tilted secant convex and let C ⊆ X be convex. If y, z ∈ dom(f |C ) = C ∩ K, then their convex combinations belong to C, and the defining condition for f is exactly the defining condition for f |C . Thus f |C is γ–tilted secant convex. (v): The lsc envelope. Assume that γ ∈ (0, 1] and that f is γ–tilted secant convex. By (i), K is convex. If K = ∅, then f ≡ +∞, lsc(f ) ≡ +∞, and both claims are immediate. So assume K ̸= ∅, and put A := aff(K) and k := dim A. We use two more properties of Mγ,λ . First, by (205), Mγ,λ (a, b) − a equals (1 − ω)(b − a) or (1 − ν)(b − a). By (205) and (204), 1 − ω ≤ (1 − λ)/γ and 1 − ν ≤ 1 − λ ≤ (1 − λ)/γ. Hence Mγ,λ (a, b) − a ≤

1−λ |b − a| γ

∀a, b ∈ R.

(206)

Second, for fixed λ, the map (a, b) 7→ Mγ,λ (a, b) is continuous, because both cases in (205) are continuous and they agree when a = b. We first show a local upper bound: for every w ∈ ri(K) there are r > 0 and C ∈ R such that B(w, r) ∩ A ⊆ K

f ≤ C on B(w, r) ∩ A.

and 89

(207)

If k = 0, then A = {w}, and r := 1 and C := f (w) work. Let k ≥ 1. Choose√δ > 0 with k B(w, δ) ∩ A P ⊆ K, choose an orthonormal basis e1 , . . . , ek of √ A − A, and put r := δ/(2 k). The 2 points w+r i σi ei with σi ∈ P {−1, 1} lie in A at distance r k = δ/2 from w, so they lie in K. Their convex hull is the cube {w + i si ei : |si | ≤ r}, which contains B(w, r) ∩ A. Quasiconvexity from (i) Pinduction on the number of points: if µm < 1, then Pmextends to finite convex combinations by := µ p = (1 − µ )q + µ p with q m m m j=1 j j j<m µj pj /(1 − µm ). Hence f ≤ C on the cube, where C is the largest value of f at its vertices. This proves (207), since B(w, r) ∩ A ⊆ B(w, δ) ∩ A ⊆ K. To show that f is lsc on ri(K), fix x ∈ ri(K), and take r and C from (207) with w = x. Let u ∈ B(x, r) ∩ A. The reflected point v := 2x − u lies in A and satisfies ∥v − x∥ = ∥u − x∥ < r, so f (u) ≤ C and f (v) ≤ C. Since x = 12 u + 21 v, the inequality (15) and monotonicity of Mγ,1/2 in its second argument give

γf (u) + C f (x) ≤ Mγ,1/2 f (u), f (v) ≤ Mγ,1/2 f (u), C = , γ+1 where the equality is the case a ≤ b of (13) with λ = 1/2. Solving for f (u) gives L ≤ f (u) ≤ C

∀u ∈ B(x, r) ∩ A,

L :=

(1 + γ)f (x) − C . γ

(208)

Let un ∈ K tend to p x. Since K ⊆ A, after discarding finitely many terms we have un ∈ := ∥un − x∥ + 1/n. For all sufficiently large n, τn ∈ (0, 1); moreover τn → 0 B(x, r) ∩ A. Put τn p and ∥un − x∥/τn ≤ ∥un − x∥ → 0. Define vn :=

1 − τn x − (1 − τn )un =x+ (x − un ). τn τn

Then vn ∈ A and ∥vn − x∥ ≤ ∥un − x∥/τn → 0. So for all sufficiently large n, vn ∈ B(x, r) ∩ A, and by (208) both f (un ) and f (vn ) lie in [L, C]. Since x = (1 − τn )un + τn vn , inequality (15) and (206) with λ = 1 − τn give
τn τn f (x) ≤ Mγ,1−τn f (un ), f (vn ) ≤ f (un ) + f (vn ) − f (un ) ≤ f (un ) + (C − L). γ γ Since τn → 0, taking the lower limit gives f (x) ≤ lim inf f (un ). n→∞

Now choose wn ∈ X with wn → x and f (wn ) → lsc(f )(x). Since lsc(f )(x) ≤ f (x) < +∞, eventually wn ∈ K, and the last display, applied to un := wn , gives f (x) ≤ lsc(f )(x). This proves f = lsc(f )

on ri(K).

(209)

Set f¯ := lsc(f ). We next show that f¯ never takes the value −∞. Outside K, the function f¯ equals +∞ by definition. Fix x ∈ ri(K) and y ∈ K. Choose δ > 0 with B(x, δ) ∩ A ⊆ K, and choose θ ∈ (0, 1) so small that θ∥x − y∥/(1 − θ) < δ/2. Then z :=

x − θy θ =x+ (x − y) ∈ A, 1−θ 1−θ

∥z − x∥ < δ/2,

so B(z, δ/2) ∩ A ⊆ B(x, δ) ∩ A ⊆ K and z ∈ ri(K). Take rz and C from (207) with w = z. Choose yn ∈ X with yn → y and f (yn ) → f¯(y). If f¯(y) = +∞, there is nothing to prove. Otherwise yn ∈ K for all large n. The points x − θyn zn := 1−θ 90

lie in A and tend to z, so for large n they lie in B(z, rz ) ∩ A and satisfy f (zn ) ≤ C. Since x = θyn + (1 − θ)zn , inequality (15) and monotonicity give

f (x) ≤ Mγ,θ f (yn ), f (zn ) ≤ Mγ,θ f (yn ), C .
If f (yn ) ≤ C, the case a ≤ b of (13) turns this into f (x) ≤ γθf (yn ) + (1 − θ)C /(γθ + 1 − θ). Solving for f (yn ) shows that, for all large n,   (γθ + 1 − θ)f (x) − (1 − θ)C . f (yn ) ≥ min C, γθ Letting n → ∞ gives f¯(y) > −∞. Thus f¯ : Rd → R ∪ {+∞}. For the γ–tilted secant convexity of f¯, fix y, z ∈ dom(f¯) and λ ∈ [0, 1]. By the definition of f¯ as a lower limit, there are sequences in X , which lie in K for large n because f¯(y), f¯(z) are finite, such that yn → y, f (yn ) → f¯(y), zn → z, f (zn ) → f¯(z). The points λyn + (1 − λ)zn lie in K and tend to λy + (1 − λ)z. The definition of f¯, inequality (15) for f , and the continuity of Mγ,λ give f¯(λy + (1 − λ)z) ≤ lim inf f (λyn + (1 − λ)zn ) n→∞

≤ lim Mγ,λ f (yn ), f (zn ) = Mγ,λ f¯(y), f¯(z) . n→∞

Thus f¯ is γ–tilted secant convex, and (209) proves the remaining claim about the relative interior. (vi): Restrictions to lines. Finally, suppose that f is γ–tilted secant convex and let ℓ be a line intersecting X . The set ℓ∩X is convex, so f |ℓ is γ–tilted secant convex by (iv). Conversely, suppose all the stated line restrictions are γ–tilted secant convex. Any two distinct points y, z ∈ K lie on a line ℓ, and the condition for f |ℓ at y and z is the condition for f at y and z, so f is γ–tilted secant convex. C.14 Proof of Theorem 18: Characterizations of γ–tilted secant convexity C.14.1 A Scalar Characterization of the Finite-dimensional Mean (m)

We collect the sublevel, balance, threshold, max, and variational descriptions of the mean Mγ,λ , together with its binary specialization and its limit as γ ↓ 0. Lemma 11 (Equivalent formulations of the finite-dimensional mean). Let m ≥ 2, λ ∈ ∆m , a ∈ Rm , and γ ∈ (0, 1]. Set (m) µ := Mγ,λ (a). For r ∈ R, write r+ := max{r, 0} and r− := max{−r, 0}. Then the following equivalent characterizations hold. (i) For every c ∈ R, c≤µ

⇐⇒

m X

λi ψ √γ (ai − c) ≥ 0.

i=1

(ii) The number µ is the unique t ∈ R satisfying m X

λi ψ √γ (ai − t) = 0.

i=1

91

(210)

(iii) The number µ is the unique t ∈ R such that, upon setting I− (t) := {i : ai < t}, one has

γ

t=

P γ

(iv) One has

I+ (t) := {i : ai ≥ t},

i∈I− (t) λi ai +

P

i∈I+ (t) λi ai

P

P

i∈I+ (t) λi

i∈I− (t) λi +

.

Pm λ i wi a i µ = max Pi=1 . m m w∈[γ,1] i=1 λi wi

(v) One has µ = argmin t∈R

m X

  λi (ai − t)2+ + γ(ai − t)2− ,

i=1

where the minimizer is unique. The sets in (iii) can be identified by sorting the coordinates of a and selecting the interval containing the corresponding value on the right-hand side. In the binary case, with m = 2 and λ = (λ, 1 − λ),   γλa + (1 − λ)b λa + γ(1 − λ)b Mγ,λ (a, b) = max , . γλ + (1 − λ) λ + γ(1 − λ) Finally, (m)

lim Mη,λ (a) = max{ai : λi > 0} ≤ max ai . i

η↓0

Thus this limit is maxi ai when every λi is positive; in particular, for m = 2 and λ ∈ (0, 1), it is max{a, b}. P m Proof. Since γ > 0 and m i=1 λi = 1, every w ∈ [γ, 1] satisfies m X

λ i wi ≥ γ

i=1

m X

λi = γ > 0.

(211)

i=1

For S ⊆ {1, . . . , m}, let RS denote the corresponding ratio in (19), and for c ∈ R put X X GS (c) := λi (ai − c) + γ λi (ai − c). i∈S

i∈S /

The denominator of RS is the sum in (211) with wi = 1 for i ∈ S and wi = γ otherwise, so it is positive. Multiplying the inequality c ≤ RS by this denominator shows that c ≤ RS if and only if GS (c) ≥ 0. Since µ = maxS RS , we get c≤µ

⇐⇒

max S⊆{1,...,m}

GS (c) ≥ 0.

(212)

The term of index i in GS (c) is λi (ai − c) if i ∈ S and γλi (ai − c) if i ∈ / S. Since λi ≥ 0, the choice : S = {i ai ≥ c} maximizes every term at once, and the maximal term is λi max{ai − c, γ(ai − c)}. By the definition of ψ √γ , ( γr, r < 0, √ √ max{r, γr} = γ ψ γ (r) = r ∈ R. (213) r, r ≥ 0, 92

Consequently, max S⊆{1,...,m}

m √ X GS (c) = γ λi ψ √γ (ai − c). i=1

√

Since γ > 0, substituting this into (212) proves (210). Let m X Q(t) := λi ψ √γ (ai − t). i=1

Each function t 7→ ψ √γ (ai − t) is continuous and strictly decreasing, with limits +∞ and −∞ as t tends to −∞ and +∞, respectively. The weights are nonnegative and at least one of them is positive. Hence Q is continuous and strictly decreasing, with the same limits, and it has a unique root t0 . By (210), Q(c) ≥ 0 ⇐⇒ c ≤ µ. Since Q(t0 ) = 0, this gives t0 ≤ µ. For every t > t0 we have Q(t) < 0, so this gives t > µ; hence µ ≤ t0 . Thus µ = t0 is the unique root of Q. This proves the equivalence of the sublevel characterization (i) and the balance characterization (ii). By (213), for every t ∈ R, X X √ γ Q(t) = γ λi (ai − t) + λi (ai − t). i∈I− (t)

i∈I+ (t)

Rearranging, Q(t) = 0 holds if and only if   X X X X t γ λi + λi  = γ λi ai + λi ai . i∈I− (t)

i∈I+ (t)

i∈I− (t)

i∈I+ (t)

The factor multiplying t is positive by (211). So Q(t) = 0 if and only if t satisfies the identity in (iii). Since µ is the unique root of Q, it is the unique t satisfying (iii). For the box-max representation, set S∗ := {i : ai ≥ µ}. The balance equation Q(µ) = 0, √ multiplied by γ and rewritten with (213), gives X

λi (ai − µ) + γ

i∈S∗

X

λi (ai − µ) = 0.

(214)

i∈S / ∗

P P := m For w ∈ [γ, 1]m , let Nw := m i=1 λi wi denote the numerator and denomii=1 λi wi ai and Dw P nator of the ratio in (iv), so that Nw − µDw = m λ i=1 i wi (ai − µ). Choose wi = 1 for i ∈ S∗ and wi = γ otherwise. Then (214) states that Nw − µDw = 0, so the corresponding ratio equals µ. Now let w ∈ [γ, 1]m be arbitrary. For i ∈ S∗ , we have ai − µ ≥ 0 and wi ≤ 1, so λi wi (ai − µ) ≤ λi (ai − µ). For i ∈ / S∗ , we have ai − µ < 0 and wi ≥ γ, so λi wi (ai − µ) ≤ γλi (ai − µ). Summing and using (214) gives X X Nw − µDw ≤ λi (ai − µ) + γ λi (ai − µ) = 0. i∈S∗

i∈S / ∗

Since Dw > 0 by (211), every ratio is at most µ. Thus its maximum is µ, proving (iv).

93

Denote by Φ(t) the objective in condition (v). For every a ∈ R, the functions t 7→ (a − t)2+ and t 7→ (a − t)2− are continuously differentiable, with derivatives −2(a − t)+ and 2(a − t)− . Hence Φ is continuously differentiable, and (213) gives   X X √ Φ′ (t) = −2  λi (ai − t) + γ λi (ai − t) = −2 γ Q(t). i:ai ≥t

i:ai <t

Since Q is strictly decreasing with unique root µ, we have Φ′ < 0 on (−∞, µ) and Φ′ > 0 on (µ, ∞). So Φ is strictly decreasing on (−∞, µ] and strictly increasing on [µ, ∞), and µ is its unique minimizer. This proves (v). For m = 2, write R1 :=

γλa + (1 − λ)b , D1

D1 := γλ + (1 − λ),

R2 :=

λa + γ(1 − λ)b , D2

D2 := λ + γ(1 − λ)

for the two ratios in the displayed single-max formula. Both denominators are positive by (211). Since the weights of a and b in each ratio sum to one, R1 = a + β1 (b − a),

β1 :=

1−λ , D1

R2 = a + β2 (b − a),

β2 :=

γ(1 − λ) . D2

A direct computation gives D2 − γD1 = λ(1 − γ 2 ), and therefore R1 − R2 = (β1 − β2 )(b − a) =

λ(1 − λ)(1 − γ 2 ) (1 − λ)(D2 − γD1 ) (b − a) = (b − a). D1 D2 D1 D2

The coefficient of b − a is nonnegative. Hence R1 ≥ R2 when a ≤ b, and R2 ≥ R1 when a > b. By (13), Mγ,λ (a, b) equals R1 when a ≤ b and R2 when a > b, so in both cases it equals max{R1 , R2 }. The subset-max (19) for m = 2 has four ratios: S = {1} gives R2 , S = {2} gives R1 , and S = ∅ and S = {1, 2} both give the arithmetic mean a + (1 − λ)(b − a). Since D1 ≤ 1 and D2 ≥ γ, we have β1 ≥ 1 − λ ≥ β2 , so the arithmetic mean lies between R1 and R2 . Hence (2) Mγ,(λ,1−λ) (a, b) = max{R1 , R2 } = Mγ,λ (a, b). : i ∈ P }. For For the limit as η ↓ 0, let P := {i : λi > 0}, which is non-empty, and a∗ := max{ai P S ⊆ {1, . . . , m} and η ∈ (0, 1], let RS (η) be the ratio in (19) with η in place of γ. If i∈S λi > 0, then, as η ↓ 0, P λi ai RS (η) −→ Pi∈S , i∈S λi P which is a weighted average of the P ai with i ∈ S ∩ P and hence at most a∗ . If i∈S λi = 0, then the factor η cancels and RS (η) = i∈S / λi ai , a weighted average of the ai with i ∈ P that does not depend on η and is at most a∗ . Thus every ratio has a limit bounded above by a∗ . Choose j ∈ P with aj = a∗ . Then R{j} (η) → aj = a∗ . The maximum of finitely many convergent functions (m)

converges to the maximum of their limits, so Mη,λ (a) → a∗ . Finally, a∗ ≤ maxi ai , with equality when P = {1, . . . , m}. C.14.2 Equivalence of (i) and (ii) Proof. Fix y, z ∈ K and λ ∈ (0, 1), and set x := λy + (1 − λ)z. If y = z, then x = y, and (22) holds with equality, so let y ̸= z. We first show that x ∈ K under either condition. Under condition (i), this is part of Definition 6. Under condition (ii), suppose that f (x) = +∞. Then 94

f (y)−f (x) = f (z)−f (x) = −∞, both transformed differences in (22) equal −∞, and the left-hand side of (22) is −∞. This contradicts the asserted inequality, so again x ∈ K. Therefore a := f (y), b := f (z), and m := f (x) are real, and the slopes (7) satisfy (18). Put √ ρ := γ > 0 and c := λ(1 − λ)∥y − z∥ > 0. By (18), λ(a − m) = cs+ and (1 − λ)(b − m) = −cs− . Positive homogeneity of ψ ρ and Lemma 1(3) give     λ ψ ρ (a − m) + (1 − λ) ψ ρ (b − m) = c ψ ρ (s+ ) + ψ ρ (−s− ) = c ψ ρ (s+ ) − ψρ (s− ) . So (22) holds if and only if ψ ρ (s+ ) ≥ ψρ (s− ). Since ψρ is increasing and inverts ψ ρ , this is equivalent to s+ ≥ ψρ (ψρ (s− )) = ψγ (s− ) by Lemma 1(1), (4), and (6), which is (8). C.14.3 Equivalence of (iii) and (iv) Proof. (iii). Let m ≥ 2, let y1 , . . . , ym ∈ K, and let λ1 , . . . , λm > 0 satisfy Pm Assume condition P m := λ = 1. Put x i i=1 λi yi . Every f (yi ) is real, and every ratio in (19) has a positive i=1 denominator, so the right-hand side of (23) is a real number. Hence f (x) < +∞, and properness implies that x ∈ K. Lemma 11–(i), applied with   m X ai = f (yi ), c=f λ i yi  , i=1

shows that (23) is equivalent to (24). Thus condition (iii) implies condition (iv). P Conversely, assume condition (iv), fix the data in condition (iii), and put x := m i=1 λi yi and P := {i : λi > 0}. A zero-weight index contributes nothing to the numerator and to the denominator of any ratio in (19). So each ratio for S ⊆ {1, . . . , m} equals the ratio for S ∩ P formed with the (m) indices in P only, and zero-weight coordinates do not affect Mγ,λ . If P = {j}, then λj = 1, the (m)

barycenter is yj , and every ratio in (19) equals f (yj ). So Mγ,λ (f (y1 ), . . . , f (ym )) = f (yj ) = f (x), and the desired inequality is an equality. If P has at least two elements, we apply condition (iv) to the points yi , i ∈ P , with the positive weights λi , i ∈ P , whose barycenter is x. It first implies that x ∈ K: otherwise f (x) = +∞, all the transformed differences would equal −∞, and (24) would fail. (|P |) Lemma 11–(i), applied to the indices in P with c = f (x), then gives f (x) ≤ Mγ,(λi )i∈P ((f (yi ))i∈P ). (m)

By the observation on zero-weight indices, the right-hand side equals Mγ,λ (f (y1 ), . . . , f (ym )), which is the desired inequality. C.14.4 Equivalence of (ii) and (iv) Proof. The implication (iv) =⇒ (ii) follows by taking m = 2. Conversely, assume condition (ii) holds, and recall that K = dom(f ). By the definition of ψ √γ , the function ψ √γ is strictly increasing, ψ √γ (0) = 0, and ψ √γ (u) has the same sign as u. Fix u, v ∈ K and t ∈ (0, 1), and set w := (1 − t)u + tv. Condition (ii), applied with y = u, z = v, and λ = 1 − t, entails that w ∈ K, since otherwise properness would give f (w) = +∞, so both transformed differences in (ii) would equal −∞, contradicting the asserted inequality. Since u, v, and t were arbitrary, K is convex. Condition (ii) also gives

(215) (1 − t)ψ √γ f (u) − f (w) + tψ √γ f (v) − f (w) ≥ 0. 95

Since the same inequality is automatic for t ∈ {0, 1}, (215) holds for every u, v ∈ K and every t ∈ [0, 1]. The inequality (215) has a strict consequence. Let c ∈ R, and suppose that f (u) ≤ c,

f (v) ≤ c,

at least one of these inequalities is strict, and 0 < t < 1. We claim that f ((1 − t)u + tv) < c.

(216)

Suppose to the contrary that f (w) ≥ c, where w = (1 − t)u + tv. Then f (u) − f (w) ≤ 0,

f (v) − f (w) ≤ 0.

At least one of these two inequalities is strict: if, for example, f (u) < c, then f (u) < c ≤ f (w) and hence f (u) − f (w) < 0; the argument is identical if f (v) < c. Since ψ √γ is strictly increasing and vanishes at zero, it follows that ψ √γ (f (u) − f (w)) ≤ 0,

ψ √γ (f (v) − f (w)) ≤ 0,

with at least one strict inequality. The weights 1 − t and t are both positive, so the left-hand side of (215) is strictly negative. This contradicts (215), and proves (216). This strict conclusion extends to finitely many points: if v1 , . . . , vN ∈ K and µr > 0,

N X

f (vr ) ≤ c

µr = 1,

(1 ≤ r ≤ N ),

r=1

f (vr0 ) < c then

 f

(217)

for at least one r0 ,

N X

 µr vr  < c.

(218)

r=1

To prove this implication, relabel the points so that r0 = 1, and define Mk :=

k X

k

µr ,

zk :=

1 X µ r vr Mk

(1 ≤ k ≤ N ).

r=1

r=1

Each zk belongs to K because it is a convex combination of points in the convex set K. Since M1 = µ1 , we have z1 = v1 , and hence f (z1 ) < c. Suppose inductively that f (zk ) < c for some k < N . The identities Mk+1 = Mk + µk+1 and zk+1 =

Mk µk+1 zk + vk+1 Mk+1 Mk+1

show that zk+1 is a convex combination of zk and vk+1 : both coefficients are strictly positive, and their sum is (Mk + µk+1 )/Mk+1 = 1. Since f (zk ) < c and f (v Pk+1 ) ≤ c, (216) implies that f (zk+1 ) < c. Induction gives f (zN ) < c. Finally, MN = 1, so zN = N r=1 µr vr . This proves (218).

96

For the finite-pointPinequality in condition (iv), let m ≥ 2, let y1 , . . . , ym ∈ K, and let λ1 , . . . , λm > 0 satisfy m i=1 λi = 1. Put x :=

m X

λ i yi ,

c := f (x).

i=1

The convexity of K proved above implies that x ∈ K, so c is finite. Suppose, toward a contradiction, that m X

λi ψ √γ (f (yi ) − c) < 0.

(219)

i=1

Define ai := ψ √γ (f (yi ) − c),

L := {i : ai < 0},

H := {j : aj ≥ 0},

and A :=

X

λi (−ai ),

B :=

i∈L

X

λj aj .

j∈H

Since the sets L and H partition {1, . . . , m}, the left-hand side of (219) equals X X λi ai + λj aj = −A + B. i∈L

j∈H

Consequently, (219) implies that −A + B < 0, or equivalently, A > B ≥ 0.

(220)

Here B ≥ 0 because aj ≥ 0 for every j ∈ H and all the weights are positive. The inequality A > B ≥ 0 gives A > 0; since the sum defining A would equal zero if L were empty, L is nonempty. If i ∈ L, then ai < 0. Since ψ √γ has the same sign as its argument, ai < 0 implies that f (yi )−c < 0, and hence f (yi ) < c. If H were empty, then every yi would satisfy f (yi ) < c. The positive weights λi sum to one and have barycenter x, so (218) would imply that f (x) < c, contradicting the definition c = f (x). Therefore H is nonempty. For i ∈ L and j ∈ H, define pij :=

λi λj aj , A

qij :=

λi (−ai )λj , A

zij :=

pij yi + qij yj . sij

and set

sij := pij + qij ,

These quantities are well defined: since A > 0, λi > 0, λj > 0, aj ≥ 0, and −ai > 0, we have pij ≥ 0 and qij > 0, and hence sij > 0. Thus pij ≥ 0, sij

qij > 0, sij

pij qij + = 1. sij sij

It follows from the convexity of K that zij ∈ K. Direct substitution also gives pij ai + qij aj =

λi λj aj ai + (−ai )aj = 0. A 97

(221)

We claim that f (zij ) ≤ c. Suppose instead that f (zij ) > c. For h ∈ {i, j}, this strict inequality implies that f (yh ) − f (zij ) < f (yh ) − c. Strict monotonicity of ψ √γ and the definition of ah therefore give ψ √γ (f (yi ) − f (zij )) < ai ,

ψ √γ (f (yj ) − f (zij )) < aj .

Multiplication of the first inequality by pij /sij ≥ 0 gives a weak inequality, which is strict when pij > 0. Multiplication of the second inequality by qij /sij > 0 gives a strict inequality. Adding the resulting inequalities and using (221) therefore yields qij √ pij ai + qij aj pij √ = 0. ψ γ (f (yi ) − f (zij )) + ψ γ (f (yj ) − f (zij )) < sij sij sij On the other hand,

pij qij yi + yj . sij sij

zij =

Applying (215) to the endpoints yi , yj with t = qij /sij shows that the left-hand side of the preceding strict inequality is nonnegative, a contradiction. Hence f (zij ) ≤ c,

i ∈ L, j ∈ H.

(222)

For i ∈ L, set   B ri := 1 − λi > 0. A The asserted inequality ri > 0 holds because λi > 0 and the strict inequality (220) implies 0 ≤ B/A < 1. For each i ∈ L, the definition of pij gives X

pij =

j∈H

λi X B λj aj = λi . A A j∈H

Similarly, for each j ∈ H, the definition of qij gives X

qij =

i∈L

λj X λi (−ai ) = λj . A i∈L

Using these two identities and sij = pij + qij , we obtain   X BX B X sij + ri = λi + λj + 1 − λi A A i∈L j∈H i∈L i∈L j∈H i∈L X X λi + λj =

XX

X

=

i∈L m X

j∈H

λi = 1.

i=1

98

(223)

Furthermore, sij zij = pij yi + qij yj . Hence, by the same two identities and the definition of ri , XX X XX X sij zij + ri yi = (pij yi + qij yj ) + ri yi i∈L j∈H

i∈L

i∈L j∈H

=

=

X B i∈L m X

A

i∈L

 λ i + r i yi +

X

λj yj

j∈H

λi yi = x.

(224)

i=1

By (223) and (224), x is a convex combination of the finitely many points zij , for i ∈ L and j ∈ H, and yi , for i ∈ L, with respective coefficients sij > 0 and ri > 0. By (222), every zij has value at most c. Every yi in this combination has i ∈ L and hence value strictly less than c; since L is nonempty, at least one point in the combination has value strictly less than c. Therefore the implication (217) ⇒ (218) gives f (x) < c, which again contradicts c = f (x). Hence (219) is impossible, and since the points and weights were arbitrary, for every finite convex combination, m m X X
√ λi ψ γ f (yi ) − f (x) ≥ 0, x= λ i yi . i=1

i=1

Thus condition (iv) holds, which together with the preceding two subsections proves the equivalence of conditions (i)–(iv); none of these arguments uses lower semicontinuity. C.14.5 (v) =⇒ (iv) Proof. Assume (v). Let y1 , . . . , ym ∈ K and λ1 , . . . , λm ∈ [0, 1] satisfy m X

λi = 1,

x :=

i=1

m X

λ i yi .

i=1

Since K is convex, x ∈ K, so f (x) is real. First suppose that x ∈ ri(K) and choose v ∈ F (x), which √ is possible because ri(K) ⊆ dom(F ). For every i, the γ–tilted convexity of f with respect to F gives f (yi ) − f (x) ≥ ψ√γ (⟨yi − x, v⟩). By Lemma 1, ψ √γ is increasing and inverse to ψ√γ , so
⟨yi − x, v⟩ ≤ ψ √γ f (yi ) − f (x) . P Multiplying by λi ≥ 0, summing over i, and using m i=1 λi (yi − x) = x − x = 0, we obtain *m + m X X
0= λi (yi − x), v ≤ λi ψ √γ f (yi ) − f (x) . i=1

(225)

i=1

To remove the assumption that x belongs to the relative interior, choose z ∈ ri(K) and, for t ∈ (0, 1], set m X yi,t := (1 − t)yi + tz, xt := λi yi,t = (1 − t)x + tz. i=1

99

By Theorem 6.1 in [Roc15], all these points belong to ri(K). So (225), applied to the points yi,t with barycenter xt , gives m X
(226) λi ψ √γ f (yi,t ) − f (xt ) ≥ 0. i=1

The hypotheses of Theorem 34 are satisfied by f : it is proper and lsc, its domain K is convex, √ √ exact γ–tilted convexity implies approximate γ–tilted convexity, and K ∩ dom(F ) is dense in K because it contains ri(K). For each i, the points yi,t lie on the line through yi and z, which meets ri(K) at z; the same holds for xt , x, and z. (If yi = z, then yi,t = yi for all t, and likewise for x.) Hence the theorem gives f (yi,t ) −→ f (yi ),

f (xt ) −→ f (x)

(t ↓ 0).

Letting t ↓ 0 in (226) and using the continuity of ψ √γ gives the finite-point inequality (24), and hence condition (iv). C.14.6 (iv) =⇒ (v) Proof. Assume condition (iv). Let y, z ∈ K and λ ∈ (0, 1), and set x := λy + (1 − λ)z. If x ∈ / K, then properness gives f (x) = +∞, so both transformed differences in condition (iv) with m = 2 equal −∞, contradicting its asserted inequality. So x ∈ K for every λ ∈ (0, 1), and for λ ∈ {0, 1} the point x is z or y. Hence K is convex. Set k := dim(K) and V := span(K − K), a k–dimensional subspace. For every x ∈ K, the affine hull of K is x + V . Define a multi-valued operator F : Rd ⇒ Rd as follows. For x ∈ / ri(K), set F (x) := ∅. For x ∈ ri(K), define the closed halfspaces in the k–dimensional space V by n
o Hy := v ∈ V : ⟨y − x, v⟩ ≤ ψ √γ f (y) − f (x) , y ∈ K, where Hx = V , and set \

F (x) :=

Hy .

y∈K

We show that this intersection is nonempty. If k = 0, then K = {x}, V = {0}, and F (x) = Hx = {0}. Assume k ≥ 1. Consider any finite family Hy1 , . . . , Hym . By Farkas’ lemma in the space V , its intersection is nonempty provided that θi ≥ 0,

m X

θi (yi − x) = 0

=⇒

i=1

m X

θi ψ √γ f (yi ) − f (x) ≥ 0.

(227)

i=1

P If A := i θi = 0, every coefficient vanishes and the implication is immediate. Otherwise discard P zero coefficients and put λi := θi /A on the remaining indices. Then x = i λi yi , and condition (iv) proves the required inequality after multiplication by A. If only one coefficient is positive, its point is x and the inequality follows from ψ √γ (0) = 0. Thus every finite subfamily has nonempty intersection. Choose an orthonormal basis e1 , . . . , ek of V . Since x ∈ ri(K), there is ε > 0 such that x ± εej ∈ K for every j. The finite intersection P :=

k \

Hx+εej ∩ Hx−εej

j=1

100

is nonempty by the preceding paragraph and is compact: these inequalities bound every coordinate ⟨v, ej ⟩ above and below by finite numbers. The closed subsets P ∩ Hy , y ∈ K, have the finiteintersection property, again by the preceding paragraph. Compactness of P therefore gives \ Hy ̸= ∅. y∈K

Hence ri(K) ⊆ dom(F ). Let x ∈ ri(K), v ∈ F (x), and y ∈ K. By the definition of Hy , ⟨y − x, v⟩ ≤ ψ √γ (f (y) − f (x)). Since ψ√γ is increasing and ψ√γ ◦ ψ √γ is the identity (Lemma 1), applying ψ√γ gives f (y) − f (x) ≥ ψ√γ (⟨y − x, v⟩). This holds for every x ∈ ri(K), v ∈ F (x), and y ∈ K, and F (x) = ∅ for all other x. Thus f is √ γ–tilted convex with respect to F , and condition (v) holds. C.15 Proof of Theorem 19 C.15.1 (i) Proof. Suppose that f is γ–tilted secant convex and put K := dom(f ). The set K is convex by b (x), and write d := y − x and ∆ := f (y) − f (x). The case Theorem 17(i). Fix x, y ∈ K and v ∈ ∂f d = 0 is immediate. For 0 < t < 1, put xt := x + td = ty + (1 − t)x. Definition 6, applied to the points y and x with weight t on y, gives xt ∈ K and  f (x ) − f (x)  f (y) − f (xt ) t ≥ ψγ . (1 − t)∥d∥ t∥d∥ Applying the increasing inverse ψ γ of ψγ (Lemma 1(1) and (6)) and multiplying by ∥d∥, which commutes with ψ γ by positive homogeneity, gives  f (y) − f (x )  f (xt ) − f (x) t ≤ ψγ . t 1−t

(228)

Regular subdifferentiability gives f (xt ) ≥ f (x)+t⟨v, d⟩+o(t) as t ↓ 0. Hence lim inf t↓0 f (xt ) ≥ f (x) and f (xt ) − f (x) ⟨v, d⟩ ≤ lim inf . t↓0 t Since ψ γ is continuous and increasing, the right side of (228) has limit superior at most ψ γ (∆) as t ↓ 0. Therefore ⟨v, d⟩ ≤ ψ γ (∆). Since ψγ is the increasing inverse of ψ γ , this is f (y) − f (x) ≥ ψγ (⟨v, y − x⟩). b . The equivalence of (iii) and (i) in Theorem 8 proves Thus f is γ–tilted convex with respect to ∂f the assertion. C.15.2 (ii) Proof. Suppose that ∂f is γ–tilted monotone. Put K := dom(f ); it is convex by Corollary 6. Fix distinct y0 , y1 ∈ K, put ℓ := ∥y1 − y0 ∥ and e := (y1 − y0 )/ℓ, and define ( f (y0 + te), 0 ≤ t ≤ ℓ, g(t) := +∞, otherwise. 101

Convexity of K gives dom(g) = [0, ℓ], and g is proper and lsc. The closed-interval assertion of Theorem 15, in the unit-speed coordinate t, shows that ∂g is γ–tilted monotone. Corollary 6(ii) gives continuity on [0, ℓ] and local Lipschitz continuity on (0, ℓ). Fix t ∈ (0, ℓ) and set A :=

g(t) − g(0) , t

B :=

g(ℓ) − g(t) . ℓ−t

Applying Corollary 12 to g on [0, t] and [t, ℓ] gives points u ∈ (0, t) and v ∈ (t, ℓ) with A ∈ ∂g(u) and B ∈ ∂g(v). Tilted monotonicity and positive homogeneity imply
(v − u)B ≥ ψγ (v − u)A = (v − u)ψγ (A), B ≥ ψγ (A). The point x := y0 + te = λy1 + (1 − λ)y0 , where λ := t/ℓ, lies in K, and A and B are the slopes s− and s+ of (7) for y = y1 and z = y0 . So B ≥ ψγ (A) is (8). Since y0 , y1 , and t were arbitrary, f is γ–tilted secant convex. C.16 Proof of Theorem 20 Proof. Set K0 := ri(K),

D0 := D ∩ K0 .

Then D0 is dense in K0 , because K0 is open relative to aff(K) and D ⊆ K ⊆ aff(K) is dense in K. Choose a point of D0 ; the values of fn there are finite for large n. So, after discarding finitely many terms, the restrictions hn := fn |K0 are proper. By Theorem 17(iv), each hn is γ–tilted secant convex. By Theorem 17(v), so is its lsc envelope h̄n := lsc(hn ) : Rd → R ∪ {+∞}, and hn = h̄n

on ri(dom(hn )).

(229)

Thus h̄n is lsc, never equal to −∞, and finite on the nonempty set ri(dom(hn )); hence it is proper. Theorem 19(i) shows that it is γ–tilted convex. For every compact set C ⊆ K0 , one can choose finitely many points d1 , . . . , dm ∈ D0 such that
C ⊆ ri conv{d1 , . . . , dm } ⊆ K0 , aff{d1 , . . . , dm } = aff(K). (230) To see this, let V := span(K − K) and use the balls BV of the proof of Theorem 11. For x0 ∈ C, choose R > 0 with BV (x0 , 5R) ⊆ K0 . The construction leading to (153) gives points y1 , . . . , yk ∈ D with ∥yi − x0 ∥ < 4R + R/4, hence yi ∈ D0 , whose convex hull contains BV (x0 , 3R). Finitely many balls BV (x0 , R) with x0 ∈ C cover C. Let d1 , . . . , dm be all the points yi obtained for these finitely many centers, and put Q := conv{d1 , . . . , dm }. Then Q ⊆ K0 because K0 is convex. The set Q contains a ball of aff(K), so aff(Q) = aff(K), and each covering ball is open in aff(K) and contained in Q, hence contained in ri(Q). This proves (230). Since fn (di ) converges to a finite value for each i, all these points belong to dom(hn ) for sufficiently large n. That domain is convex by Theorem 17(i), so Q ⊆ dom(hn ) ⊆ aff(K). As aff(Q) = aff(K), the two convex sets Q and dom(hn ) have the same affine hull, so ri(Q) ⊆ ri(dom(hn )). By (230), C ⊆ ri(dom(hn )), and (229) then gives fn = hn = h̄n 102

on C

(231)

for all sufficiently large n. Taking C to be a singleton in (231) shows that h̄n converges to finite values on the dense set D0 . Theorem 11, applied to h̄n with parameter γ and convex set K0 , therefore gives a continuous function f : ri(K0 ) = K0 → R such that h̄n → f uniformly on compact subsets of K0 . Applying (231) once more, now to an arbitrary compact C ⊆ K0 , proves that fn → f uniformly on C. This proves (i). Let y ̸= z in K0 and λ ∈ (0, 1), and put x := λy + (1 − λ)z ∈ K0 . By (i), fn converges to the finite values of f at x, y, and z, so the slopes (7) of fn converge to those of f . Since γ > 0, ψγ is continuous, and passing to the limit in (8) for fn gives (8) for f . Hence f is γ–tilted secant convex, proving (ii). Now suppose that fn → g pointwise on K. The function g cannot take the value −∞. To see this, observe that g = f on ri(K), since both are pointwise limits of fn there, and suppose that g(y) = −∞ for some y ∈ K. Choose x ∈ ri(K); then x ̸= y because g(x) = f (x) is finite. As θ ↓ 0, the point (x − θy)/(1 − θ) of aff(K) tends to x, so one may choose θ ∈ (0, 1) such that z :=

x − θy ∈ ri(K). 1−θ

Then x = θy + (1 − θ)z and y ̸= z. For large n, fn (y) is finite because fn (y) → −∞ and fn is proper, while fn (x) → f (x) and fn (z) → f (z). The slope s− of fn over [z, x] therefore converges to a finite limit, so ψγ (s− ) stays bounded, while the slope s+ = (fn (y) − fn (x))/∥y − x∥ tends to −∞. This contradicts (8) for fn . Finally, let y ̸= z in dom(g) and λ ∈ (0, 1), and put x := λy + (1 − λ)z ∈ K. Since g(y) and g(z) are finite, y, z ∈ dom(fn ) for large n, and Theorem 17(i) gives fn (x) ≤ max{fn (y), fn (z)}. Passing to the limit, g(x) ≤ max{g(y), g(z)} < +∞, and g(x) > −∞ by the previous paragraph, so x ∈ dom(g). The slopes of fn then converge to those of g, and the continuity of ψγ gives (8) for g. Hence g is γ–tilted secant convex. C.17 Proof of Theorem 21 Proof. First let h(y) = c + ψγ (⟨y − w, z⟩) be a member of Hγ . Superadditivity of ψγ gives, for every x, y ∈ Rd , h(y) − h(x) ≥ ψγ (⟨y − x, z⟩). Thus h is γ–tilted convex with respect to the constant operator F (x) = {z}, so Theorem 18(v), applied with parameter γ 2 , shows that h is γ 2 –tilted secant convex. By Theorem 17(iii), every proper nonempty pointwise supremum of members of Hγ is γ 2 –tilted secant convex, which proves the forward implication. Conversely, let f be proper, lsc, and γ 2 –tilted secant convex, and put K := dom(f ). Theorem 18 gives an operator F whose domain contains ri(K) and with respect to which f is γ–tilted convex. Define V := {h ∈ Hγ : h ≤ f on Rd }, q := sup h. h∈V

For each x ∈ ri(K) and z ∈ F (x), the function y 7→ f (x) + ψγ (⟨y − x, z⟩) belongs to V. Since K is nonempty and convex, its relative interior is nonempty. Hence V ̸= ∅, and q≤f

on Rd ,

q=f

on ri(K).

(232)

A fixed member of V bounds q below everywhere, and q is finite on ri(K), so q is proper; it is lsc as a supremum of continuous functions. By the forward implication, q is γ 2 –tilted secant convex. 103

We claim that K ⊆ dom(q) ⊆ K.

(233)

The first inclusion follows from q ≤ f and q > −∞. For the second, fix y ∈ / K. Strict separation gives an affine function a such that a ≤ 0 on K and a(y) > 0. Choose h0 = c + ψγ ◦ L ∈ V, with L affine, and for r > 0 set hr := c + ψγ ◦ (L + ra). Every such function belongs to Hγ : an affine function with nonzero linear part can be written as ⟨ · − w, z⟩, and a constant argument can be absorbed into the outer constant. On K, monotonicity of ψγ gives hr ≤ h0 ≤ f ; outside K the inequality hr ≤ f is automatic. Thus hr ∈ V. Because a(y) > 0 and γ > 0, hr (y) → +∞ as r → ∞, so q(y) = +∞, proving the second inclusion. Both K and dom(q) are convex, so the inclusions in (233) show that their closures, and hence their relative interiors, coincide. Corollary 7(iii), applied to f and q, gives

q = lsc q|ri(dom(q)) = lsc f |ri(K) = f. Thus f is Hγ –convex. C.18 Proofs of the separating examples C.18.1 Proof of Example 1 Proof. The function f is continuous on [−1, 1] and equals +∞ outside this closed interval, so it is proper and lsc. For −1 ≤ x < y ≤ 1 we write Ax,y :=

f (y) − f (x) y−x

for the secant slopes of f . The following three one-dimensional reductions are also used in the proofs of Example 2 and Lemma 12. Let g : R → R be proper and lsc with dom(g) an interval, let η ∈ (0, 1], and write Ax,y for the secant slopes of g. Fix a < b in dom(g), u ∈ ∂g(a), and w ∈ ∂g(b). By positive homogeneity and Lemma 1(3), ψη (−sr) = −s ψ η (r) for s > 0. For monotonicity, inequality (6) with (x, x∗ , y, y∗ ) = (a, u, b, w) reads (b − a)w ≥ ψη ((b − a)u) = (b − a)ψη (u), that is, w ≥ ψη (u). With (x, x∗ , y, y∗ ) = (b, w, a, u) it reads (a − b)u ≥ ψη ((a − b)w) = −(b − a)ψ η (w), that is, u ≤ ψ η (w). Applying the increasing function ψη and using Lemma 1(6) turns the second condition into w ≥ ψη (u) as well. For a = b both sides of (6) vanish. Hence ∂g is η–tilted monotone ⇐⇒ w ≥ ψη (u)

∀a < b in dom(g), u ∈ ∂g(a), w ∈ ∂g(b).

(234)

For η–tilted convexity, inequality (5) with (x, y, v) = (a, b, u) reads g(b) − g(a) ≥ (b − a)ψη (u). With (x, y, v) = (b, a, w) it reads g(a) − g(b) ≥ −(b − a)ψ η (w). Dividing both by b − a > 0, we obtain g is η–tilted convex ⇐⇒ ψη (u) ≤ Aa,b ≤ ψ η (w)

∀a < b in dom(g), u ∈ ∂g(a), w ∈ ∂g(b). (235) For η–tilted secant convexity, assume in addition that g is nondecreasing on dom(g), and let a < x < b in dom(g). With z = a and y = b in Definition 6, the slopes (7) are s− = Aa,x ≥ 0 and s+ = Ax,b , so (8) reads Ax,b ≥ ηAa,x . With z = b and y = a, they are s− = −Ax,b ≤ 0 and s+ = −Aa,x , so (8) reads −Aa,x ≥ −Ax,b /η, which is the same inequality. Since dom(g) is an interval, every point between two points of the domain lies in it. Therefore g is η–tilted secant convex ⇐⇒ Ax,b ≥ ηAa,x 104

∀a < x < b in dom(g).

(236)

We return to f . The Clarke subdifferential of f is    (−∞, 1], x = −1,      −1 < x < 0, {1}, ∂f (x) = [γ, 1], x = 0,    {γ}, 0 < x < 1,     [γ, +∞), x = 1.

(237)

On (−1, 0) and (0, 1) the function is differentiable on a neighborhood of each point, so Proposition 1(iv) applies. Near 0, f is Lipschitz with gradients 1 on the left and γ on the right, so Proposition 1(ii) gives ∂f (0) = conv{1, γ} = [γ, 1]. At x = ±1, write f = δ[−1,1] + f0 , where f0 (x) = x for x ≤ 0 and f0 (x) = γx for x ≥ 0. The function f0 is continuously differentiable near ±1, with f0′ (−1) = 1 and f0′ (1) = γ. By Proposition 1(v), ∂f (±1) = f0′ (±1) + N[−1,1] (±1), where the normal cone N[−1,1] (±1) is (−∞, 0] at −1 and [0, +∞) at 1. By (237), every point a < 1 satisfies ∂f (a) ⊆ (−∞, 1], and every point b > −1 satisfies ∂f (b) ⊆ [γ, +∞). So for a < b, u ∈ ∂f (a), and w ∈ ∂f (b), we have u ≤ 1,

w ≥ γ.

(238)

Since ψγ is increasing, (238) gives w ≥ γ = ψγ (1) ≥ ψγ (u), so ∂f is γ–tilted monotone by (234). For η > γ, take a = −1/2, u = 1, b = 1/2, and w = γ. Then ψη (u) = η > γ = w, so (234) fails and ∂f is not η–tilted monotone. Hence ∗ γmono (f ) = γ. Since f is piecewise linear with slopes 1 and γ, every secant slope Ax,y is a weighted average of these slopes, so γ ≤ Ax,y ≤ 1 ∀ − 1 ≤ x < y ≤ 1. (239) For a < b, u ∈ ∂f (a), and w ∈ ∂f (b), the bounds (238) and the monotonicity of ψγ and ψ γ give ψγ (u) ≤ ψγ (1) = γ,

ψ γ (w) ≥ ψ γ (γ) = 1.

Combined with (239), this gives ψγ (u) ≤ Aa,b ≤ ψ γ (w), so f is γ–tilted convex by (235). On the other hand, take a = 0, u = 1 ∈ ∂f (0), and b > 0. Then A0,b = γ, and (235) with parameter η would require γ ≥ ψη (1) = η. So f is not η–tilted convex for η > γ. Consequently, γ ∗ (f ) = γ. Finally, f is nondecreasing, so (236) applies. For −1 ≤ a < x < b ≤ 1, (239) gives Ax,b ≥ γ ≥ γAa,x , so f is γ–tilted secant convex. Choosing a < 0, x = 0, and b > 0 gives Aa,0 = 1 and A0,b = γ. Then (236) with parameter η would require γ ≥ η, so no larger parameter is possible. Therefore ∗ γsec (f ) = γ. √ √ γ ∗ / F γ . Since γmono (f ) = γ, we have f ∈ Fmono , and the Since γ ∗ (f ) = γ < γ, we have f ∈ remaining inclusions are parts (ii) and (i) of Theorem 19.

105

C.18.2 Proof of Example 2 Proof. Note that 1 + c = 1/γ. The formula x/(1 + cx) defines a continuously differentiable function on the open set (−1/c, +∞) ⊇ [0, 1], so f is proper and lsc. On [0, 1], the quotient rule gives f ′ (x) =

(1 + cx) − cx 1 = . 2 (1 + cx) (1 + cx)2

For 0 ≤ x < y ≤ 1, f (y) − f (x) =

y(1 + cx) − x(1 + cy) y−x = , (1 + cx)(1 + cy) (1 + cx)(1 + cy)

and hence Ax,y :=

f (y) − f (x) 1 = . y−x (1 + cx)(1 + cy)

(240)

By Proposition 1(v), the Clarke subdifferential of the proper extension is f ′ (x) plus the normal cone of [0, 1] at x:   (−∞, 1], x = 0,  ∂f (x) = {f ′ (x)}, (241) x ∈ (0, 1),   [γ 2 , +∞), x = 1. Here f ′ (0) = 1 and f ′ (1) = (1 + c)−2 = γ 2 . Since f ′ is positive on [0, 1], (241) shows that, for a < b in [0, 1], min ∂f (b) = f ′ (b) > 0. (242) max ∂f (a) = f ′ (a) > 0, To compute γ ∗ (f ), let η ∈ (0, 1] and 0 ≤ x < y ≤ 1. Since ψη and ψ η are increasing, (242) shows that the two inequalities in (235) hold for all u ∈ ∂f (x) and w ∈ ∂f (y) if and only if they hold for u = f ′ (x) and w = f ′ (y). Because f ′ (x), f ′ (y) > 0, these two inequalities are Ax,y ≥ ηf ′ (x) and Ax,y ≤ f ′ (y)/η. By (240), they are equivalent to η≤

Ax,y 1 + cx = f ′ (x) 1 + cy

and

η≤

f ′ (y) 1 + cx , = Ax,y 1 + cy

respectively. Consequently, f is η–tilted convex if and only if η≤

1 + cx 1 = = γ. 0≤x<y≤1 1 + cy 1+c inf

The infimum is attained at x = 0 and y = 1, because the ratio increases in x and decreases in y. Thus γ ∗ (f ) = γ. ∗ To compute γmono (f ), let x < y in [0, 1]. Since ψη is increasing, (242) shows that the condition in (234) holds for all u ∈ ∂f (x) and w ∈ ∂f (y) if and only if f ′ (y) ≥ ψη (f ′ (x)) = ηf ′ (x). This is equivalent to   f ′ (y) 1 + cx 2 η≤ ′ = . f (x) 1 + cy It follows that ∗ γmono (f ) =

 inf

0≤x<y≤1

1 + cx 1 + cy

106

2 =

1 = γ2. (1 + c)2

∗ (f ), we use (236), which applies because f is increasing on [0, 1]. For 0 ≤ a < x < b ≤ 1, For γsec (240) gives Ax,b (1 + ca)(1 + cx) 1 + ca = = . Aa,x (1 + cx)(1 + cb) 1 + cb

So the condition Ax,b ≥ ηAa,x in (236) is equivalent to η≤ Therefore ∗ γsec (f ) =

1 + ca . 1 + cb

1 + ca 1 = = γ. 0≤a<b≤1 1 + cb 1+c inf

∗ Combining the three computations proves the claimed equalities. Since γmono (f ) = γ 2 < γ = γ γ ∗ γsec (f ), we also obtain f ∈ Fsec and f ∈ / Fmono . The remaining inclusions are Theorem 19(i) and Theorem 13(ii) with F = ∂f .

C.18.3 Proof of Example 3 The following lemma is tailored to the two-piece, piecewise-linear derivative used in the next example. Lemma 12 (Two-piece affine integral test). Let I = [ℓ, r] with ℓ < r, fix m ∈ (ℓ, r), and let q : I → (0, ∞) be continuous and nonincreasing. Suppose that q is affine on each of [ℓ, m] and [m, r]. For c ∈ R, define f : R → R by  Z x  c + q(t) dt, x ∈ I, f (x) := ℓ  +∞, x∈ / I. For ℓ ≤ x < y ≤ r, write Ax,y :=

f (y) − f (x) . y−x

Then (

Ax,r q(y) inf γ ∗ (f ) = min , inf ℓ≤x<r q(x) ℓ<y≤r Aℓ,y ∗ γmono (f ) =

) ,

(243)

q(r) , q(ℓ)

∗ γsec (f ) = inf

(244) Ax,r

ℓ<x<r Aℓ,x

.

(245)

The secants in these formulas are explicit: q(ℓ) + q(x) , 2 q(x) + q(r) Ax,r = , 2 (m − x)(q(x) + q(m)) + (r − m)(q(m) + q(r)) Ax,r = , 2(r − x) (m − ℓ)(q(ℓ) + q(m)) + (y − m)(q(m) + q(y)) Aℓ,y = , 2(y − ℓ) Aℓ,x =

107

ℓ < x ≤ m, m ≤ x < r, ℓ ≤ x ≤ m, m ≤ y ≤ r.

Proof. Extend q to R by q(ℓ) on (−∞, ℓ) and by q(r) on (r, +∞). The extension is continuous, so Rx x 7→ c + ℓ q(t) dt is continuously differentiable on R and agrees with f on I. Hence f is proper and lsc. By Proposition 1(v), its Clarke subdifferential is q(x) plus the normal cone of I at x:    (−∞, q(ℓ)], x = ℓ, ∂f (x) = {q(x)}, ℓ < x < r,   [q(r), +∞), x = r. In particular, for x < y in I, max ∂f (x) = q(x) > 0,

min ∂f (y) = q(y) > 0.

(246)

The secant Ax,y is the average of q over [x, y]. Since q is nonincreasing, this average does not increase when the right end point moves to the right, and does not decrease when the left end point moves to the left. Therefore Ax,r ≤ Ax,y ≤ Aℓ,y

∀ ℓ ≤ x < y ≤ r.

(247)

Since q(r) ≤ q(t) ≤ q(ℓ) on I, all the ratios in (243)–(245) lie in [q(r)/q(ℓ), 1], so the infima are positive. Fix η ∈ (0, 1] and x < y in I. Since ψη and ψ η are increasing, (246) shows that the two inequalities in (235) hold for all subgradients at x and y if and only if they hold for u = q(x) and w = q(y). Since q is positive, these two inequalities are Ax,y ≥ ηq(x),

Ax,y ≤ η −1 q(y).

By (247), the first inequality holds for every y ∈ (x, r] if and only if it holds for y = r. Likewise, the second inequality holds for every x ∈ [ℓ, y) if and only if it holds for x = ℓ. Thus f is η–tilted convex if and only if η≤

Ax,r q(x)

∀x ∈ [ℓ, r)

and

η≤

q(y) Aℓ,y

∀y ∈ (ℓ, r].

The largest such η is the right-hand side of (243), which proves (243). For η–tilted monotonicity, (246) and the monotonicity of ψη show that the condition in (234) for x < y reduces to q(y) ≥ ψη (q(x)) = ηq(x). Since q is nonincreasing, q(y)/q(x) ≥ q(r)/q(ℓ) for all x < y, with equality at x = ℓ and y = r. So the minimum of q(y)/q(x) is q(r)/q(ℓ), which proves (244). Next, f is strictly increasing because q > 0, so by (236), f is η–tilted secant convex if and only if Ax,b ≥ ηAa,x ∀ ℓ ≤ a < x < b ≤ r. By (247), Ax,b ≥ Ax,r ,

Aa,x ≤ Aℓ,x .

Hence Ax,b /Aa,x ≥ Ax,r /Aℓ,x , with equality when a = ℓ and b = r. Taking the largest admissible η therefore proves (245).

108

Finally, on an interval where q is affine, its integral is the length of the interval times the average of its endpoint values, which gives the first two secant formulas directly. For the last two, we split the interval [x, r] or [ℓ, y] at m and apply the same rule on each piece. For example, for ℓ ≤ x ≤ m, Z r Z m (m − x)(q(x) + q(m)) (r − m)(q(m) + q(r)) q(t) dt = q(t) dt + f (r) − f (x) = + , 2 2 m x and dividing by r − x gives the third formula. The fourth formula follows in the same way. Proof of Example 3. Apply Lemma 12 with ℓ = 0, m = s, r = 1, and c = 0. The function q is continuous, positive, nonincreasing, and affine on [0, s] and on [s, 1]. Put k :=

2(1 − s) , s(2 − s)

so that q(x) = 1 − kx on [0, s]. Since 1 − a = 2(1−s) 2−s , we have k = (1 − a)/s, and q(0) = 1,

q(s) = 1 − ks = a = q(1),

2a 2s/(2 − s) = = s. 1+a 2/(2 − s)

(248)

By the first secant formula of Lemma 12 and (248), f (s) = s

1+a s = = a, 2 2−s

f (y) = a + a(y − s)

(s ≤ y ≤ 1),

f (1) = a(2 − s) = s. (249)

The monotonicity formula (244) gives ∗ γmono (f ) =

q(1) γ =a= . q(0) 2−γ

We next compute γ ∗ (f ) from (243), starting with its second infimum. If 0 < y ≤ s, then A0,y = (1 + q(y))/2. Since the map u 7→ 2u/(1 + u) is increasing on (0, ∞) and q(y) ≥ q(s) = a, (248) gives q(y) 2q(y) 2a = ≥ = s. A0,y 1 + q(y) 1+a If s ≤ y ≤ 1, then q(y) = a and A0,y = f (y)/y = a(1 + y − s)/y by (249). Hence q(y) y = ≥ s, A0,y 1+y−s where the inequality is equivalent to y(1 − s) ≥ s(1 − s), that is, to y ≥ s. Equality holds at y = s, so the second infimum in (243) equals s. For the first infimum, Ax,1 = a = q(x) and hence Ax,1 /q(x) = 1 when s ≤ x < 1. For 0 ≤ x ≤ s, we have f (x) = x − kx2 /2 and f (1) = s, so    1 x s − 1 + sk + kx − s 2 2 s − x + kx /2 − (s − skx)(1 − x) Ax,1 − sq(x) = = . 1−x 1−x Using sk = 1 − a, this becomes "  # x 1 Ax,1 − sq(x) = s − a + kx −s . 1−x 2 109

(250)

Here s − a = s(1−s) 2−s > 0. If s ≤ 1/2, both terms in the bracket of (250) are nonnegative and the first is positive. If s > 1/2, the bracket is decreasing in x, and its value at x = s is   1 s(1 − s) (1 − s)(1 − 2s) (1 − s)2 s − a + (1 − a) −s = + = > 0. 2 2−s 2−s 2−s In both cases the bracket is positive on [0, s], so (250) gives Ax,1 ≥ sq(x) for every x ∈ [0, s], with equality at x = 0. Together with the case s ≤ x < 1, this shows that the first infimum in (243) equals s, and formula (243) yields γ ∗ (f ) = s = γ. ∗ (f ) from (245). For s ≤ x < 1, (249) gives A It remains to compute γsec x,1 = a and A0,x = a(1 + x − s)/x. So the ratio in (245) is

Ax,1 x = ≥ s, A0,x 1+x−s as in the computation of the second infimum above. For 0 < x ≤ s, write x = sz with z ∈ (0, 1], and set 1−s h := , so that 1 − a = 2h, kx = (1 − a)z = 2hz. 2−s Then A0,x = (1 + q(x))/2 = 1 − kx/2 = 1 − hz. Also f (1) − f (x) = s − sz + kx2 /2 = s(1 − z + hz 2 ) and 1 − x = 1 − sz. Hence
Ax,1 1 − z + hz 2 = s 1 − h(1 − s) ϕ(z) , =s A0,x (1 − sz)(1 − hz)

ϕ(z) :=

z(1 − z) . (1 − sz)(1 − hz)

(251)

The second equality in (251) holds because (1 − sz)(1 − hz) − h(1 − s)z(1 − z) = 1 − (s + 2h − hs)z + hz 2 = 1 − z + hz 2 , where s + 2h − hs = s + h(2 − s) = s + (1 − s) = 1. To maximize ϕ on (0, 1), put P := (1 − s)(1 − h). Polynomial division and partial fractions give 1 (1 − sz)(1 − hz) 1 P = = −sh + + , ϕ(z) z(1 − z) z 1−z

z ∈ (0, 1).

By the Cauchy–Schwarz inequality,   √
2
1 P + z + (1 − z) ≥ 1 + P , z 1−z √ with equality if and only if (1 − z)/z = P , that is, at z∗ =

1 p ∈ (0, 1). 1 + (1 − s)(1 − h)

Hence the unique maximum of ϕ on (0, 1) is attained at z∗ , and max ϕ(z) =

0<z<1

1 1 √ = √ √
2 . 2 (1 + P ) − sh 1−s+ 1−h

110

(252)

√ √ √
2 √ The last equality uses (1 + P )2 − sh = 2 − s − h + 2 P = 1 − s + 1 − h . At z = 1 we have ϕ(1) = 0, so (252) is also the maximum over (0, 1]. Since 1 − h = (2 − s)−1 and h(1 − s) = (1 − s)2 /(2 − s), h(1 − s) max ϕ(z) = √ 0<z≤1

(1 − s)2 /(2 − s) 1 − s + (2 − s)−1/2

2 =

1−s 1+

!2

p (1 − s)(2 − s)

.

By (251), the infimum of the ratio over 0 < x ≤ s is attained at x = sz∗ and equals  !2  1−s  < s. p s 1 − 1 + (1 − s)(2 − s) The ratio is at least s on [s, 1), so this value is the infimum in (245), and  !2  1 − s ∗  < s. p γsec (f ) = s 1 − 1 + (1 − s)(2 − s) ∗ (f ) < γ. Since γ ∗ (f ) < Substituting s = γ proves the asserted formula and the strict inequality γsec sec γ ∗ γ γ, we have f ∈ / Fsec . Since γ (f ) = γ, we have f ∈ F , and Theorem 13(ii) with F = ∂f and γ2 γ2 Theorem 19(ii) with parameter γ 2 give F γ ⊆ Fmono ⊆ Fsec .

C.19 Proof of Example 4 Proof. Write h = hc . It is a polynomial, so it is continuously differentiable on R, and h′ (x) = −(2x + c) < 0 on [0, 1]. For x, y ∈ [0, 1], h(y) − h(x) = −(y − x)(x + y + c),

h′ (x)(y − x) = −(2x + c)(y − x).

(253)

Let γ ∈ (0, 1]. The proper extension e h is proper, lsc, and radially continuous, because h is continuous on the closed interval [0, 1]. By Corollary 1, h ∈ F γ if, for all x, y ∈ [0, 1],
h(y) − h(x) ≥ ψγ h′ (x)(y − x) . (254) Conversely, h′ (x) ∈ ∂ e h(x) by Proposition 1(v), so if e h is γ–tilted convex, then (5) with v = h′ (x) γ is (254). Thus h ∈ F exactly when (254) holds for all x, y ∈ [0, 1]. If y = x, both sides of (254) vanish. If y < x, then h′ (x)(y − x) = (2x + c)(x − y) > 0, so the right side of (254) is γ(2x + c)(x − y), and by (253) the left side is (x − y)(x + y + c). Dividing by (x − y)(2x + c) > 0 shows that (254) is equivalent to c+x+y γ≤ . (255) c + 2x The function x 7→ (c + x)/(c + 2x) has derivative −c/(c + 2x)2 < 0, so its minimum over [0, 1] is (c + 1)/(c + 2), attained at x = 1. Since y ≥ 0, this gives c+x+y c+x c+1 ≥ ≥ , c + 2x c + 2x c+2 with equality throughout at (x, y) = (1, 0).

111

(256)

If y > x, then h′ (x)(y −x) = −(2x+c)(y −x) < 0, so the right side of (254) is −(2x+c)(y −x)/γ, and by (253) the left side is −(y − x)(x + y + c). Dividing by −(y − x) < 0 reverses the inequality and gives x + y + c ≤ (2x + c)/γ, so (254) is equivalent to γ≤

c + 2x . c+x+y

(257)

The function x 7→ (c + 2x)/(c + x + 1) has derivative (c + 2)/(c + x + 1)2 > 0, so its minimum over [0, 1] is c/(c + 1), attained at x = 0. Since y ≤ 1, this gives c + 2x c c + 2x ≥ ≥ , c+x+y c+x+1 c+1

(258)

with equality throughout at (x, y) = (0, 1). Since (c + 1)2 − c(c + 2) = 1 > 0, we have c/(c + 1) < (c + 1)/(c + 2). If γ ≤ c/(c + 1), then (256) and (258) show that (255) and (257) hold for all pairs, so h ∈ F γ . If γ > c/(c + 1), then (257) fails at (x, y) = (0, 1), where its right side equals c/(c + 1), so h ∈ / F γ . Hence γ ∗ (h) = c/(c + 1). ∗ , let γ ∈ (0, 1], let 0 ≤ s < u ≤ 1, and let x ∈ (s, u). By (253), the slope of h over [s, x] is For γsec −(x + s + c), and its slope over [x, u] is −(u + x + c). In Definition 6 the point x lies between z and y, and the two orders of s and u give two conditions. If z = s and y = u, then s− = −(x+s+c) < 0 and s+ = −(u + x + c), and (8) reads −(u + x + c) ≥ −(x + s + c)/γ. If z = u and y = s, then s− = u + x + c > 0 and s+ = x + s + c, and (8) reads x + s + c ≥ γ(u + x + c). Both conditions are equivalent to x+s+c γ≤ . (259) u+x+c Since s, x ≥ 0 and u ≤ 1, (x + s + c)(c + 1) − c(u + x + c) = x + s + cs + c(1 − u) ≥ 0, so the right side of (259) is at least c/(c + 1). Hence (259) holds for all such s, u, x when γ ≤ c/(c + 1). At s = 0 and u = 1 the right side equals (x + c)/(x + c + 1), which tends to c/(c + 1) as x ↓ 0, so for γ > c/(c + 1), (259) fails for x close to 0. Therefore e h is γ–tilted secant convex exactly when ∗ (h) = c/(c + 1). γ ≤ c/(c + 1), and γsec For the monotonicity parameter, let κ = c/(c + 2), let x, y ∈ (0, 1), and put u = h′ (x)(y − x) and v = h′ (y)(y − x). Since h′ (t) = −(2t + c) is negative with c ≤ |h′ (t)| ≤ c + 2 on [0, 1], either u = v = 0 or u and v are nonzero with the same sign, and κ|u| ≤ κ(c + 2)|y − x| = c|y − x| ≤ |v| ≤ (c + 2)|y − x| =

c |u| |y − x| ≤ . κ κ

(260)

If u ≥ 0, then v ≥ 0, and (260) gives v ≥ κu = ψκ (u). If u < 0, then v < 0, and (260) gives κ v = −|v| ≥ −|u|/κ = u/κ = ψκ (u). Corollary 4 therefore gives e h ∈ Fmono . For η ≤ c/(c + 2), ′ e Lemma 1 gives ψη ≤ ψc/(c+2) , so ∂ h is also η–tilted monotone. Conversely, h (0) = −c ∈ ∂ e h(0) and ′ e h (1) = −(c + 2) ∈ ∂ h(1) by Proposition 1(v). The monotonicity inequality (6) with parameter η at x = 0 and y = 1 reads −(c + 2) ≥ ψη (−c) = −c/η, which is equivalent to η ≤ c/(c + 2). Hence ∂ e h ∗ is η–tilted monotone exactly when η ≤ c/(c + 2), and γmono (h) = c/(c + 2). Finally, for γ ∈ (0, 1), c γ ≥ γ ⇐⇒ c(1 − γ) ≥ γ ⇐⇒ c ≥ , c+1 1−γ

c 2γ ≥ γ ⇐⇒ c(1 − γ) ≥ 2γ ⇐⇒ c ≥ . c+2 1−γ

112

C.20 Proof of Theorem 22 Proof of Theorem 22. Let hc (σ) = −σ 2 − cσ on [0, 1], so that pi (x) = τ2 hc (xi ). The identity (28) holds because X τ X 2 τc X τX 2 q(x) + pi (x) = f (x) + xi + xi − (xi + cxi ) = f (x). 2 2 2 i

i

i

i

The Hessian of q is ∇2 f + τ I ⪰ 0 on [0, 1]n . For x, y in the cube, φ(t) = q(x + t(y − x)) has φ′′ (t) = (y − x)⊤ ∇2 q(x + t(y − x))(y − x) ≥ 0 on [0, 1], so φ is convex on [0, 1]. Hence q(x + t(y − x)) ≤ (1 − t)q(x) + tq(y) for t ∈ [0, 1], and q is convex on the cube. If τ = 0, every pi vanishes on the cube, and its proper extension is the indicator δ[0,1]n , which is convex, proper, and lsc. By Theorem 1, its Clarke subdifferential is 1–tilted monotone, and by (9) γ γ it is γ–tilted monotone, so pi ∈ Fmono . If τ > 0, then pi (x) = τ2 hc (xi ). Example 4 gives hc ∈ Fmono , and the proper extension of hc is proper, lsc, and radially continuous, so Corollary 4 gives
h′c (t)(t − s) ≥ ψγ h′c (s)(t − s) (s, t ∈ (0, 1)). (261) Let x, y ∈ (0, 1)n , and put s = xi and t = yi . Since ∇pi (z) = τ2 h′c (zi )ei , ⟨∇pi (y), y − x⟩ =

τ ′ h (t)(t − s), 2 c

⟨∇pi (x), y − x⟩ =

τ ′ h (s)(t − s). 2 c

Since
ψγ (λz) = λψγ (z) for λ > 0, multiplying (261) by τ /2 gives ⟨∇pi (y), y − x⟩ ≥ ψγ ⟨∇pi (x), y − x⟩ . The function pi is continuous on the cube, so its proper extension is proper, lsc, and radially γ γ continuous, and Corollary 4 gives pi ∈ Fmono . In both cases pi ∈ Fmono , and Theorem 19 gives γ ⊆ Fγ. pi ∈ Fsec On the cube, 0 ≤ xi ≤ 1, so ∇pi (x) = τ2 (2xi +c) ≤ τ2 (c+2) and |pi (x)| = τ2 (x2i +cxi ) ≤ τ2 (1+c). √ Differentiating (28) gives ∇q(x) = ∇f (x) + τ2 (2x + c1), and ∥2x + c1∥ ≤ 2∥x∥ + c∥1∥ ≤ (c + 2) n √ √ √ because ∥x∥ ≤ n and ∥1∥ = n. Hence ∇q(x) ≤ L + τ2 (c + 2) n. Finally, every term x2i + cxi τ P lies in [0, 1 + c], so |q(x)| ≤ |f (x)| + 2 i (x2i + cxi ) ≤ B + τ2n (c + 1). C.21 Proof of Example 7 Proof. Concave functions with one kink. Let a > 0, b := γa, t0 ∈ (L, U ), and c ∈ R, and let  φ(t) := min a(t − t0 ), b(t − t0 ) + c, so that φ has slope a on (−∞, t0 ] and slope b < a on [t0 , ∞). Let φ e be the proper extension of φ|K . Its Clarke subdifferential is    (−∞, a], t = L,      L < t < t0 , {a}, ∂ φ(t) e = [b, a], (262) t = t0 ,    {b}, t0 < t < U,     [b, ∞), t = U. At t ∈ (L, U ) \ {t0 }, φ e is affine near t, and Proposition 1(iv) applies. At t0 , φ e is Lipschitz near t0 , the limits of derivatives at nearby points are a and b, and Proposition 1(ii) gives conv{a, b} = [b, a]. Near L, φ e coincides with the proper extension to K of the affine function t 7→ a(t − t0 ) + c, and Proposition 1(v) with NK (L) = (−∞, 0] gives (−∞, a]. The case t = U is symmetric, with NK (U ) = [0, ∞). By (262), t∗ ≤ a if t < U,

t∗ ≥ b if t > L, 113

for t ∈ K, t∗ ∈ ∂ φ(t). e

(263)

Let x, y ∈ K, x∗ ∈ ∂ φ(x), e y ∗ ∈ ∂ φ(y), e and d := y − x. If d = 0, both sides of (6) vanish. If d > 0, then x < U and y > L, so (263) gives x∗ ≤ a and y ∗ ≥ b. Since ψγ is nondecreasing and ad > 0, y ∗ d ≥ bd = γad = ψγ (ad) ≥ ψγ (x∗ d). (264) If d < 0, then x > L and y < U , so x∗ ≥ b and y ∗ ≤ a. Since bd < 0, we have ψγ (bd) = bd/γ = ad, and y ∗ d ≥ ad = ψγ (bd) ≥ ψγ (x∗ d). (265) By (264) and (265), ∂ φ e is γ–tilted monotone. The function φ e is proper and lsc, since φ is continuous γ and K is closed, so φ|K ∈ Fmono . The function f2 is φ with a = 1, t0 = 0, and c = 0, and f1 is φ with a = σ, t0 = 2, and c = 2σ − δ, since both affine pieces of f1 equal 2σ − δ at t = 2. Both kinks lie in (L, U ) because γ L < 0 and U > 2, so f1 , f2 ∈ Fmono . The maximum. For t ∈ [L, 0] we have f2 (t) = t, f1 (t) = σt − δ, and |t| ≤ |L|, so δ f1 (t) − f2 (t) = (1 − σ)|t| − δ ≤ (1 − σ)|L| − δ = − < 0. 2

(266)

For t ∈ [2, U ] we have f2 (t) = γt, f1 (t) = γσ(t − 2) + 2σ − δ, and 2σ − 2γ − δ = δ, so f1 (t) − f2 (t) = δ − γ(1 − σ)(t − 2) ≥ δ − γ(1 − σ)(U − 2) =

δ > 0. 2

(267)

By (266), h(t) = t on [L, 0], and by (267), h(t) = γσ(t − 2) + 2σ − δ on [2, U ]. Let e h be the proper extension of h, and take x := L/2 ∈ (L, 0) and y := (2 + U )/2 ∈ (2, U ). The function e h is affine near x with slope 1 and affine near y with slope γσ, so Proposition 1(iv) gives 1 ∈ ∂ e h(x) and γσ ∈ ∂ e h(y). Suppose that ∂ e h is γ ′ –tilted monotone for some γ ′ ∈ [0, 1]. Since d := y − x > 0, ∗ ∗ inequality (6) with x = 1 and y = γσ reads γσ d ≥ ψγ ′ (d) = γ ′ d,

(268)

∗ (h) ≤ γσ < γ. so γ ′ ≤ γσ and γmono For the reverse bound, the middle branches cross at t = 1, because (σt − δ) − γt = δ(t − 1) on [0, 2]. Together with (266) and (267), this gives   t, L ≤ t ≤ 0,    γt, 0 ≤ t ≤ 1, h(t) =  σt − δ, 1 ≤ t ≤ 2,    γσ(t − 2) + 2σ − δ, 2 ≤ t ≤ U.

All interior slopes lie in [γσ, 1]; at a kink the Clarke subdifferential is the interval between the adjacent slopes, and at L, U it includes the respective normal ray. Consequently, for a < b in K, every u ∈ ∂ e h(a) and w ∈ ∂ e h(b) satisfy u ≤ 1,

w ≥ γσ,

w ≥ γσ = ψγσ (1) ≥ ψγσ (u).

γσ ∗ By (234), this proves h ∈ Fmono and hence γmono (h) = γσ. γ γ γ Since f1 , f2 ∈ Fmono , Theorem 19 gives f1 , f2 ∈ Fsec . The class Fsec is closed under proper γ suprema by Theorem 17(iii), and Theorem 19 again gives h ∈ Fsec ⊆ F γ . On the other hand, 1 ∈ ∂e h(0) and h(t) − h(0) = γt for 0 < t < 1. The support inequality at 0 with support vector 1 and parameter η > 0 would require γt ≥ ψη (t) = ηt, so γ ∗ (h) ≤ γ. Together with the inclusions ∗ (h) = γ ∗ (h) = γ. and the lower bounds established above, this gives γsec

114

Appendix D. Proofs for Section 6 D.1 Proof of Lemma 2 Proof of Lemma 2. Let κ = A/B ∈ (0, 1], and let I be the open interval on which H is continuously differentiable. The set U = {x ∈ Rd : ⟨a, x⟩ ∈ I} is open and contains K, and f (x) = H(⟨a, x⟩) is continuously differentiable on U with ∇f (x) = H ′ (⟨a, x⟩)a. Proposition 1(v) therefore gives ∂ fe(x) = ∇f (x) + NK (x) for x ∈ K, where fe is the extension of f by +∞. Fix x, y ∈ K, let d = y − x, and write arbitrary elements x∗ ∈ ∂ fe(x) and y∗ ∈ ∂ fe(y) as x∗ = ∇f (x) + nx and y∗ = ∇f (y) + ny with nx ∈ NK (x) and ny ∈ NK (y). The definition of the normal cone, applied at x with the point y and at y with the point x, gives ⟨nx , d⟩ ≤ 0,

⟨ny , d⟩ ≥ 0.

(269)

Let s = ⟨a, x⟩ and t = ⟨a, y⟩, so that s, t ∈ J and ⟨a, d⟩ = t − s. Let u = ⟨∇f (x), d⟩ = H ′ (s)(t − s),

v = ⟨∇f (y), d⟩ = H ′ (t)(t − s).

(270)

By (270), |u| = |H ′ (s)| |t − s| and |v| = |H ′ (t)| |t − s|. The slope bounds A ≤ |H ′ (s)|, |H ′ (t)| ≤ B and the identity A = κB then give A |u| |t − s| ≤ . (271) κ κ Since H ′ (s) and H ′ (t) are nonzero with the same sign, (270) shows that either u = v = 0 (when t = s) or u and v are nonzero with the same sign. Together with (271), this gives κ|u| ≤ κB|t − s| = A|t − s| ≤ |v| ≤ B|t − s| =

v ≥ ψκ (u).

(272)

If u ≥ 0, then v ≥ 0, and (271) gives v = |v| ≥ κ|u| = κu = ψκ (u). If u < 0, then v < 0, and (271) gives v = −|v| ≥ −|u|/κ = u/κ = ψκ (u). By (269), ⟨y∗ , d⟩ = v + ⟨ny , d⟩ ≥ v and ⟨x∗ , d⟩ = u + ⟨nx , d⟩ ≤ u. Since ψκ is nondecreasing (Lemma 1), these two bounds and (272) give
⟨y∗ , d⟩ ≥ v ≥ ψκ (u) ≥ ψκ ⟨x∗ , d⟩ . This is (6) with parameter κ, and dom(∂ fe) ⊆ K, so ∂ fe is κ–tilted monotone. The extension fe is κ proper, and it is lsc because f is continuous and K is closed, so fe ∈ Fmono . D.2 Proof of Example 8 Proof of Example 8. Apply Lemma 2 with K = [0, 1]n , the vector a = ei , and H(s) = −sk − as, which is continuously differentiable on R. Then J = {xi : x ∈ [0, 1]n } = [0, 1]. For s ∈ [0, 1] we have 0 ≤ ksk−1 ≤ k, so −(a + k) ≤ H ′ (s) = −(ksk−1 + a) ≤ −a < 0.

(273)

By (273), H ′ is negative on J with a ≤ |H ′ | ≤ a + k, and Lemma 2 with A = a and B = a + k a/(a+k) gives f ∈ Fmono . For γ ∈ (0, 1), a kγ ≥ γ ⇐⇒ a ≥ γ(a + k) ⇐⇒ a(1 − γ) ≥ kγ ⇐⇒ a ≥ . a+k 1−γ

(274)

To see that the monotonicity parameter is exact, take x = 0, y = ei , and the ordinary gradients, which belong to the constrained subdifferentials because the zero normal is admissible. For any admissible parameter η > 0, the monotonicity inequality gives −a/η ≤ −(a+k), hence η ≤ a/(a+k). Suppose that a ≥ kγ/(1−γ), so that γ ≤ a/(a+k) by (274). Lemma 1 gives ψγ (z) ≤ ψa/(a+k) (z) for every real z, so each instance of (6) with parameter a/(a + k) implies the same instance with γ parameter γ. Hence f ∈ Fmono , and Corollary 8 places f in the two other classes. 115

D.3 Proof of Example 9 and Examples 10 and 17 Proof of Example 9. Taking r = s in the definition of δ gives δ ≤ 1, so δ ∈ (0, 1] and ψδ is defined. Let g(X) = φ(∥X∥) on Rm . For X ̸= 0 the chain rule gives ∇g(X) = φ′ (∥X∥)X/∥X∥. At 0, the mean value theorem gives |g(X) − g(0)| = |φ(∥X∥) − φ(0)| ≤ ∥X∥ max[0,∥X∥] |φ′ |, and the maximum tends to |φ′ (0)| = 0 as X → 0, so g is differentiable at 0. Hence X ∇g(X) = φ′ (∥X∥) ∥X∥

(X ̸= 0),

∇g(0) = 0.

(275)

By (275), ∇g(X) = |φ′ (∥X∥)| → 0 as X → 0, so ∇g is continuous. Hence f (x) = g(Ax − b) is continuously differentiable on Rd with ∇f (x) = A⊤ ∇g(Ax − b), and Proposition 1(v) gives ∂ fe(x) = ∇f (x) + NK (x) for x ∈ K. As in the proof of Lemma 2, write x∗ = ∇f (x) + nx and y∗ = ∇f (y) + ny with normal vectors satisfying (269), where d = y − x. Then ⟨y∗ , d⟩ ≥ ⟨∇f (y), d⟩ and ⟨x∗ , d⟩ ≤ ⟨∇f (x), d⟩. Since ψδ is nondecreasing, it suffices to show
⟨∇f (y), y − x⟩ ≥ ψδ ⟨∇f (x), y − x⟩ (x, y ∈ K). (276) Let X = Ax − b, Y = Ay − b, p = ∥X∥, and q = ∥Y ∥, so that p, q ≤ R and A(y − x) = Y − X. Then ⟨∇f (x), y − x⟩ = ⟨∇g(X), Y − X⟩ and ⟨∇f (y), y − x⟩ = ⟨∇g(Y ), Y − X⟩. If p = 0, then by (275) the right side of (276) is ψδ (0) = 0 and the left side is φ′ (q)q ≥ 0. If q = 0, the left side is 0 and the right side is ψδ (−φ′ (p)p) ≤ 0. Otherwise p, q ∈ (0, R], and we let ξ = ⟨X/p, Y − X⟩ and ζ = ⟨Y /q, Y − X⟩. By (275), (276) reads
φ′ (q)ζ ≥ ψδ φ′ (p)ξ . (277) The Cauchy–Schwarz inequality gives ⟨X, Y ⟩ ≤ pq, so ξ = ⟨X, Y ⟩/p − p ≤ q − p and ζ = q − ⟨X, Y ⟩/q ≥ q − p, that is, ξ ≤ q − p ≤ ζ. (278) Since p, q ∈ (0, R], φ′ (p), φ′ (q) > 0. By (278), ξ ≤ ζ, so exactly one of the following three cases occurs. If ξ ≥ 0, then q ≥ p by (278), and the definition of δ with (r, s) = (p, q) gives φ′ (q) ≥ δφ′ (p). Since ζ ≥ ξ ≥ 0, φ′ (q)ζ ≥ φ′ (q)ξ ≥ δφ′ (p)ξ = ψδ (φ′ (p)ξ). If ξ < 0 ≤ ζ, the left side of (277) is nonnegative and the right side is negative. If ζ < 0, then q < p by (278), and the definition of δ with (r, s) = (q, p) gives φ′ (p) ≥ δφ′ (q), that is, φ′ (q) ≤ φ′ (p)/δ. Multiplying this by ζ < 0 and then using ζ ≥ ξ gives φ′ (q)ζ ≥

φ′ (p) φ′ (p) ζ≥ ξ = ψδ (φ′ (p)ξ), δ δ

where the last equality holds because ξ < 0. This proves (277), hence (276), so ∂ fe is δ–tilted monotone. The extension fe is proper, and it is lsc because f is continuous and K is closed, so δ fe ∈ Fmono . The last sentence of the example is the case A = I, b = 0, K = Rd , and R = ∞, together with Corollary 8. Proof of Example 10. Here f (x) = φ(∥x∥) with φ(r) = 12 r2 − b cos(ωr). With g(u) = u + κ sin u and κ = bω 2 , φ′ (r) = r + bω sin(ωr) =

g(ωr) , ω

φ′′ (r) = g ′ (ωr) = 1 + κ cos(ωr), 116

(279)

so φ′ (0) = 0. We use Example 9 with A = I, b = 0, K = Rd , and R = ∞. Once we show that g > 0 on (0, ∞), (279) gives φ′ > 0 on (0, ∞), and the substitution u = ωr, v = ωs in the definition of δ gives δ = inf{g(v)/g(u) : 0 < u ≤ v}. (280) If κ ≤ 1, then g ′ (u) = 1 + κ cos u ≥ 1 − κ ≥ 0, so g is nondecreasing with g(0) = 0. By (279), φ′ ≥ 0 and φ′′ ≥ 0 on [0, ∞), so φ is nondecreasing and convex there. For x, y ∈ Rd and λ ∈ [0, 1], the triangle inequality and these two properties give f (λx + (1 − λ)y) ≤ φ(λ∥x∥ + (1 − λ)∥y∥) ≤ λf (x) + (1 − λ)f (y), so f is convex. Let κ > 1, s = sκ = arccos(−1/κ), and s′ = 2π − s. Since −1/κ ∈ (−1, 0), we have s ∈ (π/2, π) and s′ ∈ (π, 3π/2). On [0, 2π], cos u > −1/κ exactly on [0, s) ∪ (s′ , 2π], so g ′ = 1 + κp cos u is positive ′ ′ 2 on [0, s), negative on (s, s ), √ and positive on ′(s , 2π]. Since s ∈ (0, π), sin s = 1 − 1/κ , and ′ 2 sin s = − sin s, so κ sin s = κ − 1 = −κ sin s and p p g(s) = s + κ2 − 1, g(s′ ) = 2π − s − κ2 − 1. (281) √ Let G(κ) = 2π − arccos(−1/κ) − κ√2 − 1 for κ ≥ 1, so that G(κ) = g(s′ ) by (281). For κ > 1, the derivative of arccos(−1/κ) is −1/(κ κ2 − 1), so √ κ2 − 1 1 κ 1 − κ2 ′ < 0. G (κ) = √ −√ = √ =− κ κ κ2 − 1 κ2 − 1 κ κ2 − 1 √ Also G is continuous on [1, ∞), G(1) = 2π − π = π, and G(κ) ≤ 2π − π/2 − κ2 − 1 → −∞ as κ → ∞. Hence G vanishes at exactly one point of (1, ∞), which is the number κ0 of Example 10, and g(s′ ) = G(κ) > 0 (1 < κ < κ0 ). (282) From now on let 1 < κ < κ0 . Let ak = s + 2πk and bk = s′ + 2πk for k ≥ 0, and b−1 = 0. Since g(u + 2π) = g(u) + 2π and g ′ has period 2π, the sign pattern of g ′ on [0, 2π] shows that, for every k ≥ 0, g is increasing on [bk−1 , ak ] and decreasing on [ak , bk ], with g(bk ) = g(s′ ) + 2πk.

g(ak ) = g(s) + 2πk,

(283)

Since g is decreasing on [s, s′ ], we have g(s′ ) < g(s). The function g is positive on (0, a0 ], because it increases there from g(0) = 0. On [ak , bk+1 ] it is at least min{g(bk ), g(bk+1 )} = g(bk ), which is positive by (283) and (282). Hence g > 0 on (0, ∞), and (280) holds. The monotonicity pattern also gives, for every j ≥ 0, min g = min{g(p), g(p′ )}

[p,p′ ]

for bj−1 ≤ p ≤ p′ ≤ bj ,

max g = g(aj ).

(284)

[bj−1 ,bj ]

Let 0 < u ≤ v, and let k ≥ 0 be such that u ∈ (bk−1 , bk ]. The second part of (284) with j = k gives g(u) ≤ g(ak ). Next, g(v) ≥ min{g(u), g(bk )}: if v ≤ bk , this is the first part of (284) on [u, bk ], and if v > bk , then v ∈ [bj , bj+1 ] for some j ≥ k, and the first part of (284) on [bj , bj+1 ] together with (283) gives g(v) ≥ min{g(bj ), g(bj+1 )} = g(bj ) ≥ g(bk ). Since 0 < g(u) ≤ g(ak ) and g(bk ) > 0, n g(b ) o n g(b ) o g(s′ ) + 2πk g(v) g(s′ ) k k ≥ min 1, ≥ min 1, = ≥ = γκ . g(u) g(u) g(ak ) g(s) + 2πk g(s) The equality uses (283) and g(s′ ) < g(s), and the last inequality holds because t 7→ (g(s′ ) + t)/(g(s) + t) has derivative (g(s) − g(s′ ))/(g(s) + t)2 > 0 for t ≥ 0. The identity g(s′ )/g(s) = γκ 117

is (281). At (u, v) = (s, s′ ) the ratio equals γκ , so (280) gives δ = γκ > 0. Example 9 then gives γκ f ∈ Fmono , and Corollary 8 places f in the two other classes. To prove exactness, fix a unit vector e, put r− = sκ /ω and r+ = (2π − sκ )/ω, and take x = r− e, y = r+ e. Their gradients project onto y − x as the positive numbers φ′ (r− )(r+ − r− ) and φ′ (r+ )(r+ − r− ). Monotonicity at parameter η therefore requires η φ′ (r− )(r+ − r− ) ≤ φ′ (r+ )(r+ − r− ), ∗ so η ≤ φ′ (r+ )/φ′ (r− ) = γκ . Together with membership at γκ , this proves γmono (f ) = γκ . 1 2 For nonconvexity, fix a unit vector e. Since cos is even, f (te) = 2 t − b cos(ωt) for t ∈ R, whose second √ derivative √ 1 + κ cos(ωt) equals 1 − √κ < 0 at t = π/ω. For κ = 2, s2 = arccos(−1/2) = 2π/3 and κ2 − 1 = 3. Then G(2) = 4π/3− 3 > 0, so 2 < κ0 because G is decreasing with G(κ0 ) = 0, and √ √ 4π/3 − 3 4π − 3 3 √ = √ . γ2 = 2π/3 + 3 2π + 3 3

Proof of Example 17. Both losses have the form φ(|r(x)|), which is the case m = 1, A = a⊤ , and b = β of Example 9, because ∥Ax − b∥ = |r(x)| ≤ R on K. In both cases φ is continuously differentiable on [0, ∞), φ′ (0) = 0, φ′ > 0 on (0, ∞), and for some σ0 > 0, φ′ is increasing on [0, σ0 ] and decreasing on [σ0 , ∞). For such φ we first show that δ=

φ′ (R) φ′ (σ0 )

if R ≥ σ0 .

(285)

Let 0 < r ≤ s ≤ R. If s ≤ σ0 , then φ′ (s) ≥ φ′ (r), so φ′ (s)/φ′ (r) ≥ 1 ≥ φ′ (R)/φ′ (σ0 ), where the last step holds because φ′ (σ0 ) is the maximum of φ′ . If s > σ0 , then s ∈ (σ0 , R], so φ′ (s) ≥ φ′ (R), and φ′ (r) ≤ φ′ (σ0 ). In both cases φ′ (s)/φ′ (r) ≥ φ′ (R)/φ′ (σ0 ), with equality at (r, s) = (σ0 , R), which proves (285). If instead R ≤ σ0 , then φ′ ≥ 0 is nondecreasing on [0, R], so φ is nondecreasing and convex on [0, R]. The function |r| is convex on K with values in [0, R], so for x, y ∈ K and λ ∈ [0, 1],

φ |r(λx + (1 − λ)y)| ≤ φ λ|r(x)| + (1 − λ)|r(y)| ≤ λφ(|r(x)|) + (1 − λ)φ(|r(y)|), and the loss is convex on K. For the Cauchy loss, φ(s) = log(1 + s2 /σ 2 ), and φ′ (s) =

2s , 2 σ + s2

φ′′ (s) =

2(σ 2 − s2 ) , (σ 2 + s2 )2

so the above holds with σ0 = σ and φ′ (σ) = 1/σ. The condition R ≥ σ is q ≥ 1, and then (285) gives √ 2 q φ′ (R) 2Rσ δ= ′ = 2 = , φ (σ) σ + R2 1+q √ where the last step divides numerator and denominator by σ 2 and uses R/σ = q. If q ≤ 1, then R ≤ σ = σ0 , and the loss is convex. For the Geman–McClure loss, φ(s) = s2 /(σ 2 + s2 ), and φ′ (s) =

2σ 2 s , (σ 2 + s2 )2

φ′′ (s) =

118

2σ 2 (σ 2 − 3s2 ) , (σ 2 + s2 )3

√ √ so the√above holds with√ σ0 = σ/ 3. Since σ 2 + σ02 √ = 4σ 2 /3, the maximum is φ′ (σ/ 3) = (2σ 3 / 3) · 9/(16σ 4 ) = 3 3/(8σ). The condition R ≥ σ/ 3 is q ≥ 1/3, and then (285) gives √ 16 q 8σ φ′ (R) 2σ 2 R √ = 2 √ √ · δ= = , (σ + R2 )2 3 3 φ′ (σ/ 3) 3 3 (1 + q)2 where the last step divides numerator and denominator by σ 4 . If q ≤ 1/3, then R ≤ σ0 , and the loss is convex. For the second claim in (b), take g = r2 , which is convex and continuously differentiable because r is affine, J = [0, R2 ], and ϕ(u) = u/(σ 2 + u) in Example 16. Then ϕ ◦ g is the Geman–McClure loss, and ϕ′ (u) = σ 2 /(σ 2 + u)2 > 0 on the open interval (−σ 2 , ∞) ⊇ J. For u ̸= v in J, let Λ(u, v) = (ϕ(v) − ϕ(u))/(ϕ′ (u)(v − u)). Dividing (33) by ϕ′ (u)(v − u), which is positive for v > u and negative for v < u, shows that (33) holds exactly when Λ(u, v) ≥ a for v > u and Λ(u, v) ≤ b for v < u (the case v = u is trivial). Here ϕ(v) − ϕ(u) =

σ 2 (v − u) (σ 2 + u)(σ 2 + v)

,

so

Λ(u, v) =

σ2 + u . σ2 + v

For v > u in J, Λ(u, v) ≥ σ 2 /(σ 2 +R2 ) = 1/(1+q), and for v < u in J, Λ(u, v) ≤ (σ 2 +R2 )/σ 2 = 1+q. Example 16 with a = 1/(1 + q) and b = 1 + q gives the parameter min{a, 1/b} = 1/(1 + q). For q ≥ 1/3, comparison of the two parameters gives √ √ √ 16 q 1 101 + 16 37 √ ≥ √ . ⇐⇒ 3 3(1 + q) ≥ 16 q ⇐⇒ q ≥ 1+q 27 3 3(1 + q)2 √ √ √ √ √ For the last equivalence, set t = q: the roots of 3 3t2 − 16t + 3 3 are (8 ± 37)/(3 3), and the √ smaller root is below 1/ 3. Finally, as functions of r, the second derivatives of the two losses are 2(σ 2 − r2 )/(σ 2√+ r2 )2 and 2σ 2 (σ 2 − 3r2 )/(σ 2 + r2 )3 , which are negative exactly when |r| > σ and when |r| > σ/ 3, respectively. D.4 Proofs of Example 11 and Example 12 Proof of Example 11. We first prove γ–tilted secant convexity. Since Z is continuous and positive on the compact set K, γ ∈ (0, 1], and for all y, z ∈ K, Z(z) minK Z ≥ = γ. Z(y) maxK Z

(286)

Let y ̸= z in K and λ ∈ (0, 1), and let x = λy + (1 − λ)z, which lies in the convex set K, wy = λZ(y) > 0, and wz = (1 − λ)Z(z) > 0. Convexity of W and W = f Z give W (x) ≤ λW (y) + (1 − λ)W (z) = wy f (y) + wz f (z). If Z is affine, then Z(x) = wy + wz , and dividing by Z(x) gives (287) below. If Z is concave and W ≥ 0, then Z(x) ≥ wy + wz > 0 and W (x) ≥ 0, so f (x) = W (x)/Z(x) ≤ W (x)/(wy + wz ). In both cases f (x) ≤

wy f (y) + wz f (z) . wy + w z

(287)

Put a = f (y), b = f (z), and m = f (x). Multiplying (287) by wy + wz gives λZ(y)(a − m) ≥ (1 − λ)Z(z)(m − b), and by (18) this is Z(y)s+ ≥ Z(z)s− . By (286), applied to both orders of y and z, the ratio r := Z(z)/Z(y) lies in [γ, 1/γ]. If s− ≥ 0, then s+ ≥ rs− ≥ γs− = ψγ (s− ). If s− < 0, then s+ ≥ rs− ≥ s− /γ = ψγ (s− ). Hence the extension of f by +∞ is γ–tilted secant 119

convex. It is proper and lsc, because f is continuous on the compact set K, so Theorem 19(i) γ2 makes it γ–tilted convex. The inclusion F γ ⊆ Fmono of Corollary 8 makes its Clarke subdifferential 2 γ –tilted monotone. For (32), differentiating W = f Z gives ∇f (x) = (∇W (x) − f (x)∇Z(x))/Z(x). For x, y ∈ K,
Z(y) f (y) − f (x) = W (y) − f (x)Z(y). (288) Convexity of W gives W (y) ≥ W (x) + ⟨∇W (x), y − x⟩.

(289)

If Z is affine, Z(y) = Z(x) + ⟨∇Z(x), y − x⟩. If Z is concave and W ≥ 0, then Z(y) ≤ Z(x) + ⟨∇Z(x), y − x⟩ and f (x) ≥ 0. In both cases
−f (x)Z(y) ≥ −f (x) Z(x) + ⟨∇Z(x), y − x⟩ . (290) Inserting (289) and (290) in (288) and using W (x) = f (x)Z(x) gives
Z(y) f (y) − f (x) ≥ ⟨∇W (x) − f (x)∇Z(x), y − x⟩ = Z(x)⟨∇f (x), y − x⟩, and dividing by Z(y) > 0 gives (32). If W and Z are affine, (289) and (290) are equalities, so (32) is an equality. Proof of the claims after Example 11 and of Example 12. For f (t) = t/(1 + ct) on [0, 1] with c = (1 − γ)/γ, take W (t) = t and Z(t) = 1 + ct. Both are affine, Z is positive on (−1/c, ∞) ⊇ [0, 1], and Z([0, 1]) = [1, 1 + c] = [1, 1/γ], so min[0,1] Z/ max[0,1] Z = γ and Example 11 gives f ∈ F γ . At x = 0 and y = 1, s = f ′ (0) = 1 > 0, and (32) holds with equality, so f (1) − f (0) = (Z(0)/Z(1))s = γs. For θ ∈ (γ, 1] this gives f (1) − f (0) < θs = ψθ (s). Since f ′ (0) ∈ ∂ fe(0) by Proposition 1(v), (5) ∗ (f ) = γ and γ ∗ 2 with parameter θ fails, so f ∈ / F θ and γ ∗ (f ) = γ. Example 2 gives γsec mono (f ) = γ . √ In Example 12(a), W (x) = x⊤ Σx = Σ1/2 x is convex as a norm of a linear map. It is continuously differentiable on the open set Rn \ {0}, because Σ1/2 x ̸= 0 for x ̸= 0, and this set contains P the probability simplex ∆n . The function Z(x) = ⟨µ, x⟩ is affine, and for x ∈ ∆n the value ⟨µ, x⟩ = i µi xi is a convex combination of the µi , so mini µi ≤ Z(x) ≤ maxi µi , with equality at vertices. Hence Z > 0 on ∆n , γ = mini µi / maxi µi , and the first case of Example 11 applies. In Example 12(b), W (x) = ⟨c, x⟩ + P0 is affine, hence convex, and it is positive on {x ≥ 0} because c ≥ 0 and P0 > 0. The denominator R is concave and continuously differentiable on an open convex neighborhood of K, and it is positive on K, so the second case of Example 11 applies. D.5 Proofs of Example 13 and Example 14 Proof of Example 13. Each hi is continuous, so f is lsc. It is proper, because f ≥ hi > −∞ everywhere and f is finite at some point. By definition f is Hγ –convex, Theorem 21 makes it γ 2 –tilted secant convex, and Theorem 19(i) makes it γ 2 –tilted convex. Proof of the monotonicity claim in Example 14. Let f = ψγ ◦ ϕ. Since ϕ : Rd → R is finite and convex, it is locally Lipschitz, and its Clarke subdifferential equals its convex subdifferential ∂ϕ. The function f is finite and continuous, hence proper and lsc. Since ∂ψγ (u) ⊆ [γ, 1/γ] for every u, the Clarke chain rule [Cla90, Theorem 2.3.9] gives ∂f (x) ⊆ co{sg : s ∈ [γ, 1/γ], g ∈ ∂ϕ(x)} = [γ, 1/γ] ∂ϕ(x). P P For the i λi si gi equals sg, where s = i λi si ∈ [γ, 1/γ] and P equality, a convex combination g = i (λi si /s)gi ∈ ∂ϕ(x) by convexity of the subdifferential. 120

Fix x, y, put d = y − x, and write arbitrary x∗ ∈ ∂f (x) and y∗ ∈ ∂f (y) as x∗ = sg and y∗ = th, with s, t ∈ [γ, 1/γ], g ∈ ∂ϕ(x), and h ∈ ∂ϕ(y). Monotonicity of ∂ϕ and t/s ∈ [γ 2 , 1/γ 2 ] give t ⟨y∗ , d⟩ = t⟨h, d⟩ ≥ t⟨g, d⟩ = ⟨x∗ , d⟩ s   1 2 ∗ ∗ ≥ min γ ⟨x , d⟩, 2 ⟨x , d⟩ = ψγ 2 (⟨x∗ , d⟩). γ Thus ∂f is γ 2 –tilted monotone. D.6 Proof of the exactness claim in Example 15 Proof of the exactness claim in Example 15. If a = b, then b/a = 1 is the largest possible parameter, so let 0 < b < a and θ ∈ (b/a, 1]. The function f is differentiable at x = 1 with f ′ (1) = b, so ∂f (1) = {b}. For y = −T with T > 0, f (y) − f (x) = −aT − b, and b(y − x) = −b(T + 1) < 0, so ψθ (b(y − x)) = −b(T + 1)/θ. The inequality (5) with parameter θ requires −aT − b ≥ −b(T + 1)/θ, that is, θ(aT + b) ≤ b(T + 1), equivalently T (θa − b) ≤ b(1 − θ). (291) Since θa − b > 0, (291) fails for T > b(1 − θ)/(θa − b), so f is not θ–tilted convex and γ ∗ (f ) ≤ b/a. ∗ ∗ ∗ (f ) ≤ γ ∗ (f ), so all three Example 15 shows γmono (f ) ≥ b/a, and Corollary 8 gives γmono (f ) ≤ γsec parameters equal b/a. D.7 Proof of Example 16 Proof of Example 16. Let U ⊇ K be an open set on which g is continuously differentiable, and I ⊇ J an open interval on which ϕ is continuously differentiable. Then ϕ ◦ g is continuously differentiable on the open set U ∩ g −1 (I) ⊇ K, with gradient ϕ′ (g(x))∇g(x). The parameter θ = min{a, 1/b} lies in (0, 1] and satisfies θ ≤ a and b ≤ 1/θ. Fix x, y ∈ K, and let u = g(x), v = g(y), and s = ⟨∇g(x), y − x⟩. Convexity of g gives v − u ≥ s.

(292)

The proper extension of (ϕ ◦ g)|K is proper, lsc, and radially continuous, because ϕ ◦ g is continuous on the closed set K. The vector y − x lies in the linear subspace parallel to the affine hull of K, so the gradient of ϕ ◦ g relative to that hull has the same inner product with y − x as ϕ′ (u)∇g(x). By Corollary 1, it therefore suffices to show ϕ(v) − ϕ(u) ≥ ψθ (ϕ′ (u)s). Recall ϕ′ (u) > 0. If s ≥ 0, then v − u ≥ s ≥ 0 by (292). The first line of (33), then (292) with aϕ′ (u) > 0, then θ ≤ a with ϕ′ (u)s ≥ 0, give ϕ(v) − ϕ(u) ≥ aϕ′ (u)(v − u) ≥ aϕ′ (u)s ≥ θϕ′ (u)s = ψθ (ϕ′ (u)s). If s < 0 and v ≥ u, then ϕ(v) − ϕ(u) ≥ 0 > ψθ (ϕ′ (u)s) because ϕ is increasing. If s < 0 and v < u, the second line of (33), then (292) with bϕ′ (u) > 0, then b ≤ 1/θ with ϕ′ (u)s < 0, give ϕ(v) − ϕ(u) ≥ bϕ′ (u)(v − u) ≥ bϕ′ (u)s ≥ ϕ′ (u)s/θ = ψθ (ϕ′ (u)s). Corollary 1 therefore gives (ϕ ◦ g)|K ∈ F θ .

121

Appendix E. Proofs for Section 8 Throughout this appendix, K is compact and convex with diameter at most D, a feasible point z1 ∈ K is known, f : K → R is nonpositive and G–Lipschitz with f ∈ F γ , and every vector qf (z) returned by the oracle is a support vector with qf (z) ≤ G. We use two facts about ψγ repeatedly: ψγ (σ) ≥ 0 for σ ≥ 0, and ψγ (σ) = σ/γ for σ < 0. E.1 Proof of Lemma 4 Fix x ∈ K, a vector v with ∥v∥ ≤ G, and h > 0. Nonpositivity of f is not used in this subsection. Define ℓ(y) = ⟨v, y − x⟩, W (y) = ℓ(y) − γf (y) (y ∈ K). Proof of Lemma 4. If D = 0, then K = {x} and yv = x attains Φx (v). If G = 0, then v = 0 and f is constant, so the maximand in (61) is the constant (1 − γ)f (x), and again yv = x attains Φx (v). In both cases one oracle call suffices. Assume D, G > 0. For y ∈ K we have −⟨v, x − y⟩ = ⟨v, y − x⟩ = ℓ(y), so f (x) − γf (y) − ⟨v, x − y⟩ = f (x) + W (y)

and

Φx (v) = f (x) + max W, K

(293)

where the maximum exists because W is continuous and K is compact. Since |f (y) − f (x)| ≤ G∥y − x∥ ≤ GD for y ∈ K, f (K) ⊆ [f (x) − DG, f (x) + DG]. (294) The procedure uses the grid A = {f (x) − DG + 2jDG/J : j = 0, . . . , J},

J = max{1, ⌈2γDG/h⌉}.

It contains both end points of the interval in (294). Since 2γDG/h > 0, we have J = ⌈2γDG/h⌉, so 2γDG/h ≤ J < 2γDG/h + 1. The first inequality shows that consecutive points are 2DG/J ≤ h/γ apart, and the second gives |A| √ = J + 1 ≤ 2 + 2γDG/h. For each level a ∈ A, let N = max{1, ⌈(2DG/h)2 ⌉} and η = D/(G N ), start from the known feasible point z1 , and for j = 1, . . . , N query the oracle at zj and set ( qf (zj ), f (zj ) > a + h/γ, bj = zj+1 = ΠK (zj − ηbj ). (295) −v, f (zj ) ≤ a + h/γ, We call zj recorded if the second case occurs, so every recorded point satisfies f (zj ) ≤ a + h/γ. If some point is recorded, let ya be a recorded point with the largest value of ℓ. After all levels are processed, the procedure returns, among the points ya , a point yv with the largest value of W , which is known from the stored oracle values. The analysis rests on the standard projected-subgradient potential. For u ∈ K, projection onto K does not increase distances and fixes u, so zj+1 − u

2

≤ zj − ηbj − u

2

= zj − u

2

− 2η⟨bj , zj − u⟩ + η 2 bj

2

.

Since qf (zj ) ≤ G and ∥v∥ ≤ G, we have bj ≤ G. Rearranging and dividing by 2η gives ⟨bj , zj − u⟩ ≤

zj − u

2

− zj+1 − u 2η

122

2

+

ηG2 . 2

(296)

Summing (296) over j = 1, . . . , N , the squared distances telescope, and ∥z1 − u∥2 ≤ D2 gives N X

⟨bj , zj − u⟩ ≤

j=1

D2 N ηG2 ∥z1 − u∥2 −∥zN +1 − u∥2 N ηG2 + ≤ + . 2η 2 2η 2

(297)

√ Dividing (297) by N and substituting η = D/(G N ) makes each of the two resulting terms equal √ √ to DG/(2 N ), and since N ≥ 2DG/h, this gives N

h 1 X D2 ηG2 DG ⟨bj , zj − u⟩ ≤ + =√ ≤ . N 2ηN 2 2 N

(298)

j=1

Fix a level a with Sa = {y ∈ K : f (y) ≤ a} ̸= ∅. The set Sa is a closed subset of the compact set K, hence compact, so ℓ attains its maximum pa on Sa at some u. We claim that some recorded point has ℓ ≥ pa − h, so that ya exists and f (ya ) ≤ a + h/γ,

ℓ(ya ) ≥ pa − h.

(299)

Suppose instead that every recorded point has ℓ < pa − h (this includes the case that no point is recorded). We show that ⟨bj , zj − u⟩ > h for every j; averaging over j then contradicts (298). If zj is not recorded, then f (zj ) > a + h/γ and f (u) ≤ a, so ∆ = f (zj ) − f (u) > (a + h/γ) − a = h/γ.

(300)

With σ = ⟨qf (zj ), zj − u⟩, inequality (5) at zj with comparator u gives ψγ (−σ) ≤ f (u) − f (zj ) = −∆ < 0. Since ψγ is negative only on negative arguments, −σ < 0, so ψγ (−σ) = −σ/γ. Then −σ/γ ≤ −∆ and (300) give ⟨bj , zj − u⟩ = σ ≥ γ∆ > h. If zj is recorded, then bj = −v and ⟨bj , zj − u⟩ = ⟨v, u − zj ⟩ = ℓ(u) − ℓ(zj ) = pa − ℓ(zj ) > h. This proves the claim. The recorded point with ℓ ≥ pa − h satisfies f ≤ a + h/γ, and ya has the largest ℓ among recorded points, which gives (299). Now let y∗ maximize W on K, and let a be the smallest point of A with a ≥ f (y∗ ). It exists because the largest point of A is f (x) + DG ≥ f (y∗ ) by (294). If a is the smallest point f (x) − DG of A, then a ≤ f (y∗ ) by (294). Otherwise the preceding point of A is below f (y∗ ) and at most h/γ below a. In both cases f (y∗ ) ≤ a ≤ f (y∗ ) + h/γ. (301) Since y∗ ∈ Sa , (299) applies, and pa ≥ ℓ(y∗ ). Combining (299) with (301) and pa ≥ ℓ(y∗ ) gives f (ya ) ≤ f (y∗ ) +

2h , γ

ℓ(ya ) ≥ ℓ(y∗ ) − h,

and therefore W (ya ) = ℓ(ya ) − γf (ya ) ≥ ℓ(y∗ ) − h − γf (y∗ ) − 2h = W (y∗ ) − 3h. The returned point satisfies W (yv ) ≥ W (ya ). By (293), Vv = f (x)+W (yv ) ≥ f (x)+W (y∗ )−3h = Φx (v) − 3h ≥ Φx (v) − 4h, and Vv ≤ Φx (v) because yv ∈ K. This proves (62). There are at most 2 + 2γDG/h levels, each with N ≤ 1 + (2DG/h)2 ≤ 2 + 4(DG/h)2 iterations, and each iteration uses one oracle call and one projection. Together with the query at x, this gives at most 1 + 2(2 + 2γDG/h)(2 + 4(DG/h)2 ) calls, which is (63).

123

E.2 Proof of Theorem 25(i) Parts (ii) and (iii) are proved in Appendices E.3 and E.4. In this subsection f is nonpositive and the oracle is exact. Proof of Theorem 25(i). Write q = qf (x). If q = 0, Algorithm 2 returns 0, and (5) gives f (y) − f (x) ≥ ψγ (0) = 0 for every y, and since f (y) ≤ 0 and γ ≤ 1,
f (x) − γf (y) = f (x) − f (y) + (1 − γ)f (y) ≤ 0 = ⟨0, x − y⟩, which is (64) with ε = 0. If γ = 1, the algorithm returns q, and since ψ1 (σ) = σ, (5) reads f (x) − f (y) ≤ ⟨q, x − y⟩, which is again (64) with ε = 0. Assume from now on that q ̸= 0 and γ < 1, so that every vector of [γq, q] is sq for some s ∈ [γ, 1]. Let T ∗ = {s ∈ [γ, 1] : Φx (sq) ≤ 0}. (302) The vector sq lies in (52) exactly when s ∈ T ∗ , so Theorem 24 gives T ∗ ̸= ∅. We also use that Φx is D–Lipschitz. For vectors v, v′ , the two maximands in (61) differ at each y by ⟨v′ − v, x − y⟩, whose absolute value is at most D v − v′ , and maxima of two functions that differ by at most a constant differ by at most that constant: |Φx (v) − Φx (v′ )| ≤ D v − v′ .

(303)

The algorithm maintains an interval [t, t] ⊆ [γ, 1] with [t, t] ∩ T ∗ ̸= ∅,

(304)

which holds initially with [t, t] = [γ, 1]. Consider a candidate v = t q with t = (t + t)/2. Since ∥v∥ ≤ ∥q∥ ≤ G, Lemma 4 applies with h = ε/16. If the candidate is accepted, that is, Vv ≤ ε/2, then (62) gives ε ε Φx (v) ≤ Vv + 4h ≤ + < ε, 2 4 which by (61) is (64) for gb = v, and v ∈ [γq, q]. If the candidate is rejected, let d = ⟨q, x − yv ⟩. Since ⟨v, x − yv ⟩ = t d, the witness satisfies f (x) − γf (yv ) − t d = Vv > ε/2,

(305)

while every s ∈ T ∗ satisfies, by the definition of Φx with y = yv , f (x) − γf (yv ) − s d ≤ Φx (sq) ≤ 0.

(306)

Subtracting (306) from (305) gives (s − t) d > ε/2 > 0

(s ∈ T ∗ ).

(307)

By (307), if d > 0, every s ∈ T ∗ satisfies s > t, so [t, t] ∩ T ∗ = [t, t] ∩ T ∗ , and setting t = t keeps (304). If d < 0, every s ∈ T ∗ satisfies s < t, and setting t = t keeps (304). The case d = 0 cannot occur, since then (s − t)d = 0 for the element s provided by (304). Each rejection halves t − t, so after j rejections t − t = (1 − γ)2−j ≤ 2−j .

124

(308)

A test is performed after j rejections only if DG(t−t) > ε/4, which by (308) requires DG 2−j > ε/4, that is, j < log2 (4DG/ε). So the loop performs at most max{0, ⌈log2 (4DG/ε)⌉} = O(log(2 + DG/ε)) tests. If it ends without acceptance, then DG(t − t) ≤ ε/4. Take s ∈ [t, t] ∩ T ∗ , which t−t t+t exists by (304), and gb = t+t g − sq∥ ≤ t−t 2 q. Then | 2 − s| ≤ 2 and ∥q∥ ≤ G, so ∥b 2 G. By (303) and Φx (sq) ≤ 0, t−t ε Φx (b g ) ≤ Φx (sq) + DG ≤ 0 + < ε, 2 8 which is (64); also gb ∈ [γq, q] because t+t 2 ∈ [γ, 1].
By Lemma 4 with h = ε/16, each test uses O (1 + γDG/ε)(1 + (DG/ε)2 ) calls. Multiplying by the number of tests and adding the query at x gives (65). The parameter γ ≤ 1 enters only through the factor 1 + γDG/ε, so no inverse power of γ appears. E.3 Proof of Theorem 25(ii): inexact oracle In this subsection f : K → R is nonpositive and G–Lipschitz with f ∈ F γ . The oracle returns fb(z) with |fb(z) − f (z)| ≤ ζ and az with ∥az ∥ ≤ G satisfying (66). The first lemma replaces Theorem 24 when the base vector has a support defect. Lemma 13 (Comparison from a defective base vector). Let f : K → R be continuous and nonpositive on a convex set K with f ∈ F γ , let x ∈ K, and suppose a vector q satisfies f (y) − f (x) ≥ ψγ (⟨q, y − x⟩) − ξ for all y ∈ K, with ξ ≥ 0. Then some v ∈ [γq, q] satisfies f (x) − γf (y) − ⟨v, x − y⟩ ≤ ξ

(y ∈ K).

Proof. We follow the proof of Theorem 24 with the defect carried along. Write r = −f ≥ 0, p = −q, sy = ⟨p, y − x⟩, cy = γr(y) − r(x), and c′y = cy − ξ. Since ⟨q, y − x⟩ = −sy , the hypothesis reads r(y) − r(x) ≤ −ψγ (−sy ) + ξ. Evaluating ψγ on the two signs as in the derivation of (54) gives ( sy /γ + ξ, sy ≥ 0, r(y) − r(x) ≤ (309) γsy + ξ, sy < 0. By (55), for a real t, f (x) − γf (y) − ⟨t q, x − y⟩ = cy − t sy . So t q satisfies the claim for y exactly when c′y ≤ t sy . If sy = 0, (309) gives r(y) ≤ r(x) + ξ, so cy ≤ γr(x) + γξ − r(x) = −(1 − γ)r(x) + γξ ≤ ξ,

(310)

that is, c′y ≤ 0, and the constraint holds for every t. In particular, if q = 0, then sy = 0 for every y and v = 0 proves the lemma, so assume q ̸= 0. If sy > 0, (309) gives cy ≤ γr(x) + sy + γξ − r(x) = sy − (1 − γ)r(x) + γξ ≤ sy + ξ,

(311)

so c′y ≤ sy , and the constraint t ≥ c′y /sy is a lower bound that is at most 1. If sy < 0, (309) and r(y) ≥ 0 give
cy = r(y) − r(x) − (1 − γ)r(y) ≤ γsy + ξ, (312) so c′y ≤ γsy , and dividing by sy < 0 shows that the constraint t ≤ c′y /sy is an upper bound that is at least γ.

125

Let y+ , y− ∈ K with s+ = sy+ > 0 > s− = sy− , let λ = −s− /(s+ − s− ), and let z = λy+ + (1 − λ)y− , so sz = 0 as in the proof of Theorem 24. Then r(z) ≤ r(x) + ξ by (309) with sz = 0, and r(z) ≥ γ(λr(y+ ) + (1 − λ)r(y− )) by Lemma 3. Hence
(313) λcy+ + (1 − λ)cy− = γ λr(y+ ) + (1 − λ)r(y− ) − r(x) ≤ r(z) − r(x) ≤ ξ, and subtracting ξ = λξ + (1 − λ)ξ gives λc′y+ + (1 − λ)c′y− ≤ 0. This is (58) with c′ in place of c, and the computation that leads from (58) to (59) gives c′y+ /s+ ≤ c′y− /s− . So every lower bound is at most every upper bound. As in (60), let t∗ = max{γ, supsy >0 c′y /sy }. By (311), t∗ ∈ [γ, 1]. By (312) and the comparison of bounds, t∗ is at most every upper bound, and by its definition it is at least every lower bound. So t∗ satisfies every constraint, and the vector v = t∗ q ∈ [γq, q] proves the lemma. Proof of Theorem 25(ii). Query fb(x) and q = ax , let h = ε/16, and let ℓ, W , and Φx be as in Appendix E.1. Three cases are handled directly. If γ = 1, return q: with ψ1 (σ) = σ, (66) at z = x gives f (x) − f (y) ≤ ⟨q, x − y⟩ + ξ. If q = 0, return 0: Lemma 13 gives f (x) − γf (y) ≤ ξ. If ζ > DG, return q: for y ∈ K, the Lipschitz bound and f (y) ≤ 0 give f (x) − γf (y) = (f (x) − f (y)) + (1 − γ)f (y) ≤ GD, and −⟨q, x − y⟩ ≤ ∥q∥∥x − y∥ ≤ GD, so f (x) − γf (y) − ⟨q, x − y⟩ ≤ 2DG < 2ζ ≤ (2 + 4γ)ζ. In each case (67) holds. From now on γ < 1, q ̸= 0, and ζ ≤ DG. Each candidate v = tq with t ∈ [γ, 1] is evaluated by the procedure of Appendix E.1 with two changes. The grid A covers [fb(x) − DG − ζ, fb(x) + DG + ζ] with gaps at most h/γ and at most 2 + 2γ(DG + ζ)/h ≤ 2 + 4γDG/h points; by (294) and |fb(x) − f (x)| ≤ ζ, this interval contains f (K). The update (295) is replaced by ( azj , fb(zj ) > a + ξ + ζ + h/γ, bj = (314) −v, fb(zj ) ≤ a + ξ + ζ + h/γ, with the same N and η. Points in the second case are recorded, and they satisfy f (zj ) ≤ fb(zj )+ζ ≤ a + ξ + 2ζ + h/γ. Let ya be a recorded point with the largest ℓ. Fix a level a with Sa ̸= ∅, and let u ∈ Sa maximize ℓ on Sa , with pa = ℓ(u). Inequality (298) uses only bj ≤ G, so it holds here. We claim that some recorded point has ℓ ≥ pa − h. Otherwise every j has ⟨bj , zj − u⟩ > h, contradicting (298). For a recorded zj this is the same computation as in Appendix E.1. For an unrecorded zj , the value error and f (u) ≤ a give f (zj ) ≥ fb(zj ) − ζ > a + ξ + h/γ ≥ f (u) + ξ + h/γ.

(315)

With σ = ⟨azj , zj −u⟩, (66) at zj with y = u and (315) give ψγ (−σ) ≤ f (u)−f (zj )+ξ < −h/γ < 0. Hence −σ < 0, so ψγ (−σ) = −σ/γ, and −σ/γ < −h/γ gives ⟨bj , zj − u⟩ = σ > h. The recorded point with ℓ ≥ pa − h satisfies the bound on f for recorded points, and ya has the largest ℓ among recorded points. Hence f (ya ) ≤ a + ξ + 2ζ + h/γ,

ℓ(ya ) ≥ pa − h.

(316)

Let y∗ maximize W , and let a be the smallest grid point with a ≥ f (y∗ ). The grid covers an interval containing f (K) with gaps at most h/γ, so the argument for (301) gives a ≤ f (y∗ ) + h/γ. Since y∗ ∈ Sa and pa ≥ ℓ(y∗ ), (316) gives f (ya ) ≤ f (y∗ ) + ξ + 2ζ + 2h/γ and ℓ(ya ) ≥ ℓ(y∗ ) − h. Therefore W (ya ) = ℓ(ya ) − γf (ya ) ≥ ℓ(y∗ ) − h − γf (y∗ ) − γξ − 2γζ − 2h, that is, W (ya ) ≥ W (y∗ ) − 3h − γξ − 2γζ. 126

(317)

For each produced point the report R(ya ) = fb(x) − γ fb(ya ) − ⟨v, x − ya ⟩ is computable. By (293), the true value is f (x) − γf (ya ) − ⟨v, x − ya ⟩ = f (x) + W (ya ), and the value errors give R(ya ) − f (x) − W (ya ) ≤ |fb(x) − f (x)| + γ|fb(ya ) − f (ya )| ≤ (1 + γ)ζ.

(318)

The procedure returns the produced point yv with the largest report and sets Vv = R(yv ). Using Vv ≥ R(ya ), (318), (317), and Φx (v) = f (x) + W (y∗ ) for the first inequality below, and (318) at yv for the second, Φx (v) − 3h − γξ − (1 + 3γ)ζ ≤ Vv ,

f (x) + W (yv ) ≥ Vv − (1 + γ)ζ.

(319)

The outer bisection uses Tξ∗ = {s ∈ [γ, 1] : Φx (sq) ≤ ξ}, which is nonempty by Lemma 13, and maintains [a, b] ∩ Tξ∗ ̸= ∅ from [a, b] = [γ, 1]. A candidate v = tq with t = (a + b)/2 is accepted if Vv ≤ ξ + (1 + γ)ζ + ε/2. Then the first inequality of (319) and 3h = 3ε/16 give Φx (v) ≤ ξ + (1 + γ)ζ +

11ε ε 3ε + + γξ + (1 + 3γ)ζ = + (1 + γ)ξ + (2 + 4γ)ζ, 2 16 16

which implies (67) by (61). If the candidate is rejected, then Vv > ξ + (1 + γ)ζ + ε/2. With d = ⟨q, x − yv ⟩, the identity f (x) + W (yv ) = f (x) − γf (yv ) − t d and the second inequality of (319) give f (x) − γf (yv ) − t d ≥ Vv − (1 + γ)ζ > ξ + ε/2, (320) while every s ∈ Tξ∗ satisfies f (x) − γf (yv ) − s d ≤ Φx (sq) ≤ ξ. Subtracting this from (320) gives (s − t) d > ε/2, which is (307) with Tξ∗ in place of T ∗ , and the bisection step is justified exactly as in Appendix E.2. If the loop ends with DG(b − a) ≤ ε/4, then for s ∈ [a, b] ∩ Tξ∗ , (303) gives Φx ( a+b 2 q) ≤ Φx (sq) + DG(b − a)/2 ≤ ξ + ε/8, which implies (67). Every output lies in [γq, q]. Each test uses at most 2 + 4γDG/h levels with N iterations each, and the count of tests in Appendix E.2 gives O(log(2 + DG/ε)) tests, which gives (65). E.4 Proof of Theorem 25(iii): bit model Proof of Theorem 25(iii). First suppose that γ, D, and G are positive rationals and that a bound |f | ≤ B on K is known. Let ξ0 , ζ0 ≥ 0 be the defects of the supplied oracle in the sense of Theorem 25(ii); for an exact oracle, ξ0 = ζ0 = 0. The rational implementation runs the algorithm of part (ii) with tolerance ε/2 and with the rounding described next. Values are rounded to rationals within ε/48, so the rounded values have value error at most ζ = ζ0 + ε/48.

(321)

To round vectors, let ν = γε/(16D). Approximate az by a rational a with ∥a − az ∥ ≤ ν/4, and compute a rational ra with ∥a∥ ≤ ra ≤ ∥a∥ + ν/2. If ra ≤ G, keep a. Otherwise replace a by a′ = (G/ra )a, which has a′ = G∥a∥ /ra ≤ G and  G ν ν ν 3ν a′ − a = ∥a∥ 1 − ≤ ra − G ≤ ∥a∥ + − G ≤ ∥az ∥ + + − G ≤ , ra 2 4 2 4 using ∥a∥ ≤ ra and ∥az ∥ ≤ G, and the triangle inequality gives a′ − az ≤ a′ − a +∥a − az ∥ ≤ 3ν/4 + ν/4 = ν. In both cases the stored vector, again called a, has ∥a∥ ≤ G and ∥a − az ∥ ≤ ν. For y ∈ K, |⟨a − az , y − z⟩| ≤ νD, and ψγ is (1/γ)–Lipschitz because it is piecewise linear with slopes γ and 1/γ. So ψγ (⟨az , y − z⟩) ≥ ψγ (⟨a, y − z⟩) − νD/γ, (322) 127

and (66) for az with defect ξ0 , combined with (322), yields (66) for a with defect ξ = ξ0 + Dν/γ = ξ0 + ε/16.

(323)

Projections are computed approximately: zj+1 is a rational point of K within distance κ of Pj = ΠK (zj − ηbj ). For u ∈ K, Pj − u ≤ D because both points lie in K, so zj+1 − u

2

≤ ( Pj − u + κ)2 ≤ Pj − u

2

+ 2Dκ + κ2 .

(324)

2

2

In the derivation of (296), the bound Pj − u ≤ zj − ηbj − u still holds, so (324) adds (2Dκ + κ2 )/(2η) to the right side of (296). Averaging over j as in (297) and (298) gives N

1 X D2 ηG2 2Dκ + κ2 ⟨bj , zj − u⟩ ≤ + + . N 2ηN 2 2η

(325)

j=1

√ √ 2 }⌉, a rational η with D/(2G N ) ≤ η ≤ D/(G N ), and κ ≤ Choose N = ⌈max{1, (16DG/h) √ min{D, ηh/(64D)}. Since N ≥ 16DG/h, the three terms on the right of (325) satisfy D2 h DG ≤√ ≤ , 2ηN 16 N

h ηG2 DG 2Dκ + κ2 3Dκ 3h ≤ √ ≤ , ≤ ≤ , (326) 2 32 2η 2η 128 2 N √ √ where the first bound uses η ≥ D/(2G N ), the second uses η ≤ D/(G N ), and the third uses κ2 ≤ Dκ and κ ≤ ηh/(64D). By (326), the right side of (325) is at most (8 + 4 + 3)h/128 = 15h/128 < h/2, so (298) holds with approximate projections and the argument for (316) applies unchanged. The number of iterations changes by an absolute factor and has no inverse power of γ. By (321) and (323), part (ii) applies with tolerance ε/2, and (67) gives the comparison error   ε ε ε  3ε + (1 + γ) ξ0 + + (2 + 4γ) ζ0 + ≤ + (1 + γ)ξ0 + (2 + 4γ)ζ0 , 2 16 48 4 using 1 + γ ≤ 2 and 2 + 4γ ≤ 6, which give (1 + γ)ε/16 + (2 + 4γ)ε/48 ≤ ε/8 + ε/8. The right side is at most ε + (1 + γ)ξ0 + (2 + 4γ)ζ0 , so (67) holds with the defects ξ0 , ζ0 of the supplied oracle. The output is a rational multiple t a of the stored base vector with t ∈ [γ, 1], so it lies in [γa, a] and has norm at most G. All grid points, thresholds, step sizes, and reports are obtained from the input by polynomially many arithmetic operations, every oracle input is a stored rational point, and the accuracies ε/48, ν, and κ have bit length polynomial in log(2 + 1/ε) + log(1/γ) + log(2 + D + D−1 + G + G−1 + B) + log d. With 1/ε, D, and G polynomially bounded, the number of calls (65) is polynomial, so the running time is polynomial. Finally, let γ be given by certified rational approximations. If the known bound is B = 0, then f ≡ 0 and 0 is exact. Otherwise choose a rational γ ′ ∈ [γ/2, γ] with γ − γ ′ ≤ ε/(4B). By Lemma 1(7) when γ ′ < γ, and trivially when γ ′ = γ, ψγ ′ (z) ≤ ψγ (z)

(z ∈ R).

(327)

′

By (327), inequality (5) for γ implies (5) for γ ′ , so f ∈ F γ , and (66) also holds with γ ′ . Running the rational construction at γ ′ with tolerance 3ε/4 in place of ε, the bound above gives comparison ′ ′ error at most 43 · 3ε 4 + (1 + γ )ξ0 + (2 + 4γ )ζ0 , so f (x) − γ ′ f (y) ≤ ⟨b g , x − y⟩ +

3ε + (1 + γ ′ )ξ0 + (2 + 4γ ′ )ζ0 4 128

(y ∈ K).

(328)

Since −f (y) ≤ B and γ − γ ′ ≤ ε/(4B), ε f (x) − γf (y) = f (x) − γ ′ f (y) + (γ − γ ′ )(−f (y)) ≤ f (x) − γ ′ f (y) + . 4 Adding this to (328) and using γ ′ ≤ γ gives (67) at the original γ. For B > 0 the segment guarantee is [γ ′ a, a], while the norm bound and comparison hold at the original parameter; the separate B = 0 branch returns zero and needs no segment assertion. E.5 Proof of Corollary 10 For an interval I = [s, e] and comparators us , . . . , ue , let PI (u) =

Pe

t=s+1 ∥ut − ut−1 ∥.

Lemma 14 (Projected online gradient descent). Let ∥at ∥ ≤ G and xt+1 = ΠK (xt −ηat ) with η > 0. For every interval I = [s, e] and us , . . . , ue ∈ K, X D2 + 2DPI (u) ηG2 |I| ⟨at , xt − ut ⟩ ≤ + . 2η 2

(329)

t∈I

Proof. The derivation of (296) gives, for each t, ⟨at , xt − ut ⟩ ≤

∥xt − ut ∥2 −∥xt+1 − ut ∥2 ηG2 + . 2η 2

(330)

Summing the numerators over t ∈ I and regrouping, e X

e X

∥xt − ut ∥2 −∥xt+1 − ut ∥2 = ∥xs − us ∥2 −∥xe+1 − ue ∥2 + ∥xt − ut ∥2 −∥xt − ut−1 ∥2 .

t=s

t=s+1

The first term is at most D2 and the second is nonpositive. For each t in the last sum, with m = xt − ut and m′ = xt − ut−1 , ∥m∥2 − m′

2

= (∥m∥ − m′ )(∥m∥ + m′ ) ≤ m − m′ · 2D = 2D∥ut − ut−1 ∥ .

So the numerators sum to at most D2 + 2DPI (u), and summing (330) over t ∈ I gives (329). Proof of Corollary 10. In round t, Theorem 25(ii) applied to ft at xt , with tolerance εt , defect ξt , and value error ζt , gives for every ut ∈ K ft (xt ) − γft (ut ) ≤ ⟨b gt , xt − ut ⟩ + errt ,

(331)

where an exact oracle is the case ξt = ζt = 0. The returned vector satisfies ∥b gt ∥ ≤ G by Theorem 25, including its irrational-parameter implementation, and Algorithm 1 feeds at = gbt to A, which played xt . Summing (331) over t ∈ I and applying (42) gives X Regγ (I; us:e ) ≤ ⟨b gt , xt − ut ⟩ + Err(I) ≤ BA (I; us:e ) + Err(I), t∈I

P which is (69). For static regret, the function t∈I ft is continuous on the compact set K and has a minimizer u, and X X SRegγ (I) = ft (xt ) − γ ft (u) = Regγ (I; u, . . . , u). (332) t∈I

t∈I

129

√ A fixed comparator has P[1,T ] (u) = 0, so Lemma 14 with I = [1, T ] and η = D/(G T ) gives √ √ √ D2 ηG2 T DG T DG T BA = (333) + = + = DG T . 2η 2 2 2 √ Then (332), (69), and (333) give SRegγ (T ) ≤ DG T + Err([1, T ]). √ √ √ √ √ √ P −1/2 . For t ≥ 1, If err√ then Err(I) ≤ c √et=s t√ t− t − 1 = 1/( t+ t − 1) ≥ 1/(2 t) t ≤ c/ t, √ because t − 1 ≤ t, so t−1/2 ≤ 2( t − t − 1), and summing over t = s, . . . , e telescopes: Err(I) ≤ c

e X

t−1/2 ≤ 2c

√

e−

√

s−1 .

t=s

p √ √ √ √ √ s + 1)2 = e + 2 (s − 1)(e − s + 1) ≥ e, we have e − √s − 1 ≤ e − s + 1 = Since ( s − 1 + e −p p |I|, so Err(I) ≤ 2c |I|, which is (70). With an exact oracle and εt = c/ t, errt = εt . Finally, round t calls the procedure of Theorem 25 once, with the number of calls in (65) for ε = εt , and performs one update of A. With inverse-polynomial tolerances, Theorem 25(iii) and the rational implementation of projected online gradient descent below give polynomial time per round. Rational implementation of the linear learners. For static regret, choose one rational step size 5D D √ ≤η≤ √ . G T 4G T √ Certified approximation of T supplies such a step √size with polynomial encoding length, and the exact-projection potential is then at most (9/8)DG T . If the update in round t is a feasible rational point within distance κt of the exact projection, the squared-distance estimate in Appendix E.4 adds at most 2Dκt + κ2t 3Dκt ≤ 2η 2η to the one-round bound whenever κt ≤ D. Taking κt ≤ min{D, ηG/(4T 2 )} makes the total addition on the implemented static linear regret√is at most √ every interval at most 3DG/(8T ). Thus√ (3/2)DG T , and comparison errors at most DG/(2 T ) per round give the stated 2DG T bound. The constant-step potential for moving comparators admits the same rational approximation and projection-error accounting. For a strongly adaptive learner, rational convex combinations and the linear experts’ errors can likewise be charged to its explicit linear-regret bound.

Appendix F. Proofs for Section 9 F.1 Proof of Theorem 26 Proof of Theorem 26. Fix α ∈ (β, 1] and apply Definition 14 to a SAT formula of size n. Write N for the resulting encoding length, which does not depend on the horizon, and ∆ = α − τ > 0. We may assume P (N ) ≥ 1. For a horizon T chosen below, draw indices J1 , . . . , JT independently and uniformly from I, independently of the learner’s random tape ω. In round t, obtain the learner’s rational action xt , then reveal fJt and all the prescribed feedback. Accept the SAT formula if Q(xt ) > τ for some t ≤ T . The score is computed exactly at rational points, so this test is exact. On an unsatisfiable formula, item (iv) of Definition 14 gives Q(xt ) ≤ τ for every t, so an unsatisfiable formula is never accepted. If the guarantee is stated for deterministic loss sequences, condition on the complete index sequence. The learner’s tape remains independent of it, so the assumed regret bound applies; 130

averaging over the index sequence gives the corresponding expected or high-probability bound for this simulation. If the guarantee is given directly for the independent-sampling model, this conditioning step is unnecessary. Suppose the formula is satisfiable, and fix x∗ with Q(x∗ ) = 1, which exists by item (iii) of Definition 14. Put Zt = fJt (xt ) − αfJt (x∗ ), At = σ(ω, J1 , . . . , Jt ). P P ∗ is fixed, so ∗ ) ≥ min The comparator x f (x J x∈K t t t fJt (x). Multiplying this by −α < 0 and P adding t fJt (xt ) gives, by (38), T X Zt ≤ SRegα (T ). (334) t=1

The action xt is a function of ω and of the feedback in rounds 1, . . . , t − 1, so it is At−1 -measurable, while Jt is uniform on I and independent of At−1 . Hence, applying (72) at xt and at x∗ , µt := E[Zt | At−1 ] =

1 X α X Q(xt ) αQ(x∗ ) α − Q(xt ) fi (xt ) − fi (x∗ ) = − + = . |I| |I| Λ Λ Λ i∈I

(335)

i∈I

Summing (335) over t gives T X t=1

T  X 1 αT − Q(xt ) . µt = Λ

(336)

t=1

Expected regret. By the tower property, EZt = Eµt for every t. Taking expectations in (334) and using (336) and (73), T

T

t=1

t=1

 X X 1 αT − E Q(xt ) = E Zt ≤ E[SRegα (T )] ≤ P (N )T 1−ρ . Λ Multiplying by Λ/T and rearranging yields   T X 1 E Q(xt ) ≥ α − ΛP (N )T −ρ . T

(337)

t=1

Choose a polynomially bounded integer horizon satisfying  T ≥

2ΛP (N ) ∆

1/ρ .

(338)

Such a horizon exists because N , Λ, and ∆−1 are polynomially bounded in n, and ρ is fixed. Any integer polynomial in n that exceeds the right-hand side can be used, so no irrational power has to be computed. By (338), ΛP (N )T −ρ ≤ ∆/2. Let X = maxt≤T Q(xt ) ∈ [0, 1]. Since X is at least the average score, (337) gives ∆ α+τ EX ≥ α − = . (339) 2 2 On the other hand, X ≤ τ on the event {X ≤ τ } and X ≤ 1 on its complement, so EX ≤ τ P(X ≤ τ ) + P(X > τ ) = τ + (1 − τ )P(X > τ ).

131

(340)

Since τ < α ≤ 1, we have 1 − τ > 0. Combining (339) with (340) and dividing by 1 − τ gives α−τ =: q0 . 2(1 − τ )

P(X > τ ) ≥

(341)

Since 1−τ ≤ 1, we have q0 ≥ ∆/2, so q0 is inverse-polynomial. Repeat the simulation independently k times with k ≥ (log 4)/q0 , and accept if any run accepts. All runs fail with probability at most (1 − q0 )k ≤ e−q0 k ≤ 1/4, so this procedure accepts a satisfiable formula with probability at least 3/4. Since 1/q0 ≤ 2/∆, an integer polynomial upper bound for k can again be used. The argument uses only Q ∈ [0, 1]; it does not need SRegα (T ) ≥ 0. High-probability regret. Let Dt = µt − Zt . Since Zt is At -measurable and µt = E[Zt | At−1 ], the sequence (Dt ) is a martingale difference sequence with respect to (At ). Each loss takes values in [−1, 0] and α ∈ (0, 1], so Zt ∈ [−1, α] ⊆ [−1, 1]; hence µt ∈ P [−1, 1] and |Dt | ≤ 2. The Azuma– 2 Hoeffding inequality with increments bounded by 2 gives P{ t Dt ≥ λ} ≤ exp(−λ /(8T )) for √ λ > 0. The choice λ = 2 2T log 9 has λ2 = 8T log 9, so   T  1 X p (342) P Dt ≥ 2 2T log 9 ≤ .  9  t=1

Let E1 be the event in (74), and let E2 be the complement of the event ByP the union P P in (342). bound, P(E1 ∩ E2 ) ≥ 1 − 1/9 − 1/9 = 7/9. On E1 ∩ E2 , the identity t µt = t Zt + t Dt and (334) give T T X X p µt ≤ SRegα (T ) + Dt < P (N )T 1−ρ + 2 2T log 9. t=1

t=1

Substituting (336), multiplying by Λ/T , and rearranging shows that on E1 ∩ E2 T

1X Q(xt ) ≥ α − ΛP (N )T −ρ − 2Λ T

r

t=1

2 log 9 . T

(343)

Choose an integer polynomial horizon satisfying  T ≥

4ΛP (N ) ∆

1/ρ

 T ≥ 2 log 9

,

8Λ ∆

2 .

(344)

p −ρ ≤ ∆/4, and the second gives The 2 log 9/T ≤ ∆/(8Λ), hence pfirst condition gives ΛP (N )T 2Λ 2 log 9/T ≤ ∆/4. By (343), the average score on E1 ∩ E2 is at least α − ∆/2 = (α + τ )/2 > τ . The largest score is at least the average, so some Q(xt ) exceeds τ and the rule accepts. One run therefore accepts with probability at least 7/9 > 3/4; put k = 1. Finite sampling and running time. Let M = |I|. If M = 1, the unique index is deterministic. Otherwise, let ℓ = ⌈log2 M ⌉, so that 2ℓ−1 < M ≤ 2ℓ . Sample uniformly from the integers in [0, 2ℓ − 1] using ℓ random bits, and retain the sample if it is below M . A retained sample is uniform on {0, . . . , M − 1}, and each attempt succeeds with probability M/2ℓ > 1/2. Cap the number of attempts for each of the at most kT draws at c = ⌈log2 (16kT )⌉, and reject the SAT formula if all c attempts for some draw fail. A single draw fails with probability at most 2−c ≤ 1/(16kT ), so by the union bound some draw fails with probability at most 1/16. Run on the same random bits, the capped sampler and uncapped independent sampling produce the same indices unless some draw fails, so the acceptance probabilities differ by at most 1/16. Thus the bounded-time algorithm 132

accepts a satisfiable formula with probability at least 3/4 − 1/16 > 1/2, and it never accepts an unsatisfiable formula because acceptance still requires an iterate with Q(xt ) > τ . The simulation, feedback, and exact score tests all take polynomial time. This proves SAT ∈ RP, hence NP ⊆ RP; the reverse inclusion always holds. F.2 Proof of Theorem 27 The proof reduces from Label Cover, using the gap theorem below and a rounding lemma for the projected masses. A Label Cover instance Γ = (VL , VR , E, ΣL , ΣR , (πe )e∈E ) has a bipartite graph with no isolated vertices and projection maps πe : ΣL → ΣR . A labeling assigns au ∈ ΣL to each u ∈ VL and bv ∈ ΣR to each v ∈ VR . An edge e = (u, v) is satisfied when πe (au ) = bv . The value val(Γ) is the largest fraction of satisfied edges, with parallel edges counted separately. Theorem 36 (Fixed-gap Label Cover [MR10, Mos15]). For every fixed rational δ ∈ (0, 1) there is a deterministic polynomial-time reduction from SAT to Label Cover that has value 1 on satisfiable formulas and value at most δ on unsatisfiable formulas. The alphabets and the two degrees of the biregular graph are bounded by constants depending only on δ. The statement follows by specializing the soundness parameter in Moshkovitz [Mos15, Theorem 2.1] to a fixed constant; it relies on a proved Label Cover theorem, not on the Projection Games Conjecture. Write kL = |ΣL |, kR = |ΣR |, and let    Y  X L : KΓ = xu ∈ RΣ x ≤ 1 . (345) u,a +   u∈VL

a∈ΣL

For a right vertex of degree ℓv , set wv = ℓv /|E| and zv,b (x) = For every feasible point, zv,b ≥ 0 and

1 ℓv

X

X

xu,a .

(346)

e=(u,v)∈E a:πe (a)=b

P

b zv,b ≤ 1; also

P

v wv = 1.

Lemma 15 (Pointwise rounding and the profile gap). Given a rational x ∈ KΓ , one can compute in polynomial time a labeling of value L(x) such that X L(x) ≥ wv max zv,b (x). (347) b

v

If H(0) = 0, H(1) = 1, 0 ≤ H(s) ≤ s, and H(s) ≤ as + bs2 with a ≥ 0 and b > 0, then X X QH wv H(zv,b (x)) ∈ [0, 1] Γ (x) := v

(348)

b

and QH Γ (x) ≤ a + bL(x) ≤ a + b val(Γ). If Γ is fully satisfiable, a one-hot point has QH Γ = 1.

133

(349)

Proof. We first verify the facts stated before the lemma. Each a ∈ ΣL has exactly one image πe (a), so summing (346) over b ∈ ΣR gives X

zv,b (x) =

b∈ΣR

1 ℓv

X

X

xu,a ≤ 1,

(350)

e=(u,v)∈E a∈ΣL

because each of the ℓv inner sums is at most one. Every coordinate isP nonnegative, so zP v,b ≥ 0, and therefore zv,b ∈ [0, 1]. Every edge has exactly one right endpoint, so v ℓv = |E| and v wv = 1. For each v, choose bv maximizing zv,b (x) over b. For u ∈ VL and a ∈ ΣL , let Nu (a) be the number of edges e = (u, v) at u with πe (a) = bv , and choose au maximizing Nu (a). Under the labeling (au )u , (bv )v , an edge e = (u, v) is satisfied exactly when πe (au ) = bv , so exactly Nu (au ) edges at u are satisfied and 1 X L(x) = Nu (au ). (351) |E| u∈VL P Since xu,a ≥ 0, a xu,a ≤ 1, and 0 ≤ Nu (a) ≤ Nu (au ), X X xu,a Nu (a) ≤ Nu (au ) xu,a ≤ Nu (au ). (352) a∈ΣL

a∈ΣL

Writing Nu (a) as a sum over the edges at u, exchanging the order of summation, and grouping the edges by their right endpoint gives, by (346), X X X X X xu,a Nu (a) = (353) xu,a = ℓv zv,bv (x). u∈VL a∈ΣL

v∈VR

e=(u,v)∈E a:πe (a)=bv

Summing (352) over u and using (351) and (353), we get X |E|L(x) ≥ ℓv zv,bv (x). v∈VR

Dividing by |E| and using wv = ℓv /|E| and zv,bv = maxb zv,b proves (347). All the choices use exact rational comparisons and edge counts, so the labeling is computed in polynomial time. Now let H satisfy the hypotheses P of the second part. Since zv,b (x) ∈ [0, 1], the bounds 0 ≤ H(s) ≤ s, together with (350) and v wv = 1, give X X X 0 ≤ QH (x) ≤ w z (x) ≤ wv = 1. v v,b Γ v

v

b

For each v, by (350), X

 X 2 zv,b ≤ max zv,b zv,b ≤ max zv,b . b

b

(354)

b

b

Applying H(s) ≤ as + bs2 to every term of (348) and then using (350) with a ≥ 0, (354) with b > 0, and finally (347), we get X X X X X X 2 QH wv zv,b + b wv zv,b ≤a wv + b wv max zv,b ≤ a + bL(x). Γ (x) ≤ a v

b

v

v

b

v

b

The value val(Γ) is the maximum over all labelings, so L(x) ≤ val(Γ). This proves (349). For a satisfying labeling (au )u , (bv )v , take xu,au = 1 and every other coordinate zero; this point lies in KΓ . Every edge e = (u, v) has πe (au ) = bv , so (346) gives P zv,bv = 1 and zv,b = 0 for b ̸= bv . The endpoint values H(1) = 1 and H(0) = 0 then give QH = v wv = 1. Γ 134

Proof of Theorem 27. Fix α ∈ (β, 1], and choose a rational θ ∈ (β, α). Take the profile Hθ from the hypothesis, and write a = aθ and b = bθ . Since a < θ and b > 0, every rational δ with 0 < δ < min{1, (θ − a)/b} satisfies τ := a + bδ < θ < α. (355) Fix such a δ and use Theorem 36 with it. Enumerate VL = [p] and ΣL = [q], so that KΓ = Kp,q , and let ℓ be the common right degree. For v ∈ VR with incident edges ej = (uj , v), j = 1, . . . , ℓ, and for b ∈ ΣR , put Rj = {a : πej (a) = b}. Then (346) reads ℓ

zv,b (x) =

1X X xuj ,a , ℓ j=1 a∈Rj

which is the local form (76). By the lifting hypothesis, all losses −Hθ (zv,b )/S belong to L(KΓ ), where S = Sθ,p,q,ℓ is common to the family. Index the family by I = E × ΣR and, for e = (u, v), define fe,b (x) = −

Hθ (zv,b (x)) . S

(356)

The same (v, b) loss thus occurs once for each of the ℓv edges at v. Grouping the edges by their right endpoint and using wv = ℓv /|E|, a uniform sample has mean X X QHθ (x) 1 X 1 fe,b (x) = − ℓv Hθ (zv,b (x)) = − Γ . |E|kR |E|kR S v SkR e,b

(357)

b

θ Take Λ = SkR and Q = QH Γ , so (357) is (72). The profile Hθ maps into [0, 1], so Hθ ≥ 0; by (77) it satisfies the hypotheses of the second part of Lemma 15 with a ≥ 0 and b > 0, and that lemma gives Q : KΓ → [0, 1]. On a satisfiable formula, val(Γ) = 1, and the lemma gives a point with Q = 1. On an unsatisfiable formula, val(Γ) ≤ δ, so (349) and (355) give Q(x) ≤ a + bδ = τ for every x ∈ KΓ . Also τ ≥ 0 is rational and τ < α. For fixed θ and δ, the alphabets and degrees are constants, so the graph, domain, loss family, and rational data have polynomial encoding length and S, hence Λ, is polynomially bounded. By (355), the gap α − τ is at least the fixed positive number α − θ. The exact score consists of polynomially many rational profile evaluations, and the other required computations are polynomial by hypothesis, which verifies every condition in Definition 14.

F.3 Proofs of Lemmas 5 and 6 Proof of Lemma 5. Put κ = A/B ∈ (0, 1]. The function −H is continuously differentiable near [0, 1], and its derivative −H ′ is negative with absolute value in [A, B] on [0, 1]. Lemma 2 with κ ∗ K = [0, 1] and a = 1 therefore gives Ψ ∈ Fmono , so γmono (Ψ) ≥ κ. By Proposition 1(v) and ′ ′ H (0) = A, H (1) = B,    (−∞, −A], s = 0, ′ ∂Ψ(s) = −H (s) + N[0,1] (s) = {−H ′ (s)}, 0 < s < 1,   [−B, +∞), s = 1. Conversely, suppose ∂Ψ is η–tilted monotone for some η ∈ (0, 1]. Take s = 0, t = 1, u = −A ∈ ∂Ψ(0), and v = −B ∈ ∂Ψ(1). Then (6) reads −B ≥ ψη (−A) = −A/η, hence η ≤ A/B = κ. Thus ∗ γmono (Ψ) ≤ κ, which proves exactness. 135

Proof of Lemma 6. Differentiate (81) and use (m + 1)/(m + κ) = B from (82) to obtain   H ′ (s) = B 1 − (1 − κ)(1 − s)m , ′′

m−1

H (s) = m(1 − κ)B(1 − s)

.

(358) (359)

For s ∈ [0, 1], we have 0 ≤ (1 − s)m ≤ 1, with value 1 at s = 0 and value 0 at s = 1. Hence (358) and A = κB give A ≤ H ′ (s) ≤ B

H ′ (0) = A,

(0 ≤ s ≤ 1),

H ′ (1) = B.

(360)

Similarly, 0 ≤ (1 − s)m−1 ≤ 1 on [0, 1], so (359) and C = m(1 − κ)B/2 give 0 ≤ H ′′ ≤ 2C there.
Substitution in (81) gives H(0) = 0 and H(1) = (m + 1) − (1 − κ) /(m + κ) = 1. By (360), H ′ ≥ A > 0, so H is increasing and H(s) ≥ H(0) = 0. Since H ′′ ≥ 0, H is convex on [0, 1], and
H(s) = H (1 − s) · 0 + s · 1 ≤ (1 − s)H(0) + sH(1) = s. Taylor’s formula with integral remainder at 0, together with H(0) = 0, H ′ (0) = A, and H ′′ ≤ 2C, gives Z Z s

H(s) = As +

s

(s − t)H ′′ (t) dt ≤ As + 2C

0

(s − t) dt = As + Cs2 .

0

This proves (83). The polynomial H is C 1 on R, and (360) holds with A > 0, so Lemma 5 gives ∗ γmono (Ψ) = A/B = κ, which is (84). Next, κ(m + 1) κ(1 − κ) =κ+ , m+κ m+κ and the last term is positive and decreases to 0 as m → ∞, so A ↓ κ. Finally, m − 1 is even, so (1 − s)m−1 ≥ 0 for every real s, and (359) gives H ′′ (s) ≥ 0 on R. Thus the ambient polynomial is convex. A=

F.4 Proofs of Theorem 28 and Corollary 11 Proof of Theorem 28. We first verify the profile criterion for the class of nonpositive, 1–Lipschitz γ losses whose proper extensions lie in Fmono . Given rational θ ∈ (γ, 1), choose rational κ ∈ (γ, θ) and a positive odd integer m satisfying (85). By (82), A = κ(m + 1)/(m + κ). Since m + κ > 0 and θ > κ, κ(1 − θ) . θ−κ Hence (85) gives A < θ. Write H = Hκ,m . By Lemma 6, H maps [0, 1] into [0, 1], is nondecreasing and B–Lipschitz there, and satisfies (77) with aθ = A and bθ = C. Also 0 ≤ A < θ, C > 0, and both are rational. A < θ ⇐⇒ κm + κ < θm + θκ ⇐⇒ κ(1 − θ) < m(θ − κ) ⇐⇒ m >

Geometry of the local map. Fix p, q, ℓ and a map z = ⟨a, ·⟩ of (76). The coordinate of a at (u, a) is the number of indices j with uj = u and a ∈ Rj , divided by ℓ, so it belongs to [0, 1]. Also ℓ

X i

1X ai = |Rj | ≤ q, ℓ

 X ∥a∥22 ≤ max ai ai ≤ q. i

j=1

(361)

i

For x ∈ Kp,q , all coordinates are nonnegative and every block sums to at most one, so ℓ

q

1 XX 0 ≤ z(x) ≤ xuj ,a ≤ 1. ℓ j=1 a=1

Set S = 1 + q(B + 2C) and let f = −H ◦ z/S on Kp,q . 136

(362)

Monotonicity. The finite-valued ambient formula is the polynomial −H(⟨a, x⟩)/S, and its gradient is −H ′ (z(x))a/S. Proposition 1(v) gives, for the proper extension fe = −H ◦ z/S + δKp,q , ∂ fe(x) = −

H ′ (z(x)) a + NKp,q (x). S

(363)

By Lemma 6, A ≤ H ′ ≤ B on [0, 1], and z(Kp,q ) ⊆ [0, 1] by (362). So the derivative of the polynomial −H/S is negative with absolute value in [A/S, B/S] on z(Kp,q ), and Lemma 2 gives A/B κ fe ∈ Fmono = Fmono . Since γ < κ, Lemma 1(7) shows that each instance of (6) with parameter γ κ implies the same instance with parameter γ, so fe ∈ Fmono . A lifted loss can have a parameter γ larger than κ, for example when z is constant; the reduction uses only membership in Fmono . Normalization and individual minimization. By (83), 0 ≤ H ≤ 1 on [0, 1], which together with (362) and S ≥ 1 gives f ∈ [−1, 0]. Lemma 6 gives H ′ ≤ B and H ′′ ≤ 2C on [0, 1]. Together √ with (361), ∥aa⊤ ∥op = ∥a∥22 , q ≤ q, and S ≥ max{Bq, 2Cq}, this gives, on the feasible set, √ B q H ′ (z(x)) ∥a∥2 ≤ ≤ 1, S S H ′′ (z(x)) ⊤ ∇2 f (x) = − aa , S

∥∇f (x)∥2 =

∥∇2 f (x)∥op ≤

2Cq ≤ 1. S

The feasible set is convex, so it contains the segment between any two of its points, and the mean value inequality along that segment turns these bounds into 1–Lipschitz continuity of both f and its gradient on the feasible set. The gradient −H ′ (z(x))a/S is coordinatewise nonpositive on the set, because H ′ ≥ A > 0 and a ≥ 0. For odd m, H ′′ ≥ 0 on all of R, so the ambient polynomial is concave, and its Hessian −H ′′ (z(x))aa⊤ /S has rank at most one everywhere. These Lipschitz and Hessian bounds are stated on the feasible set only. Finally, H ′ > 0 on [0, 1], so H is strictly increasing there, and by (362) any maximizer of z over Kp,q is an exact minimizer of f . The linear function z is a sum of one term ⟨au , xu ⟩ per block, and the blocks are constrained separately. Each block maximizes its term by putting mass one on a coordinate of largest positive coefficient, or by putting mass zero if all coefficients are zero. Rational implementations. For fixed θ, the rational constants κ, A, B, C and the degree m + 1 are fixed, and the coefficients of a have polynomial bit length. Evaluating the displayed polynomial in its factored form and evaluating its derivatives require a fixed number of rational arithmetic operations, with polynomial output bit length. The resulting gradient is a valid generalized gradient of the proper extension by (363), because zero belongs to every normal cone. The normalization S is rational and polynomially bounded. Linear optimization and Euclidean projection over the product of simplices are also exact polynomial-time rational operations. For projection onto a block, first clip negative coordinates to zero. If the clipped vector has sum at most one, it is the projection, since P it is the projection onto the nonnegative orthant and lies in the block. Otherwise the constraint a xu,a ≤ 1 is active at the projection, since an inactive constraint would make the projection equal to the clipped vector. In that case, sort the input coordinates and find the active set on which subtraction of a common threshold makes the sum one; the threshold is a rational average. Blockwise projection and the exact linear minimizers therefore have polynomial bit complexity. Every coordinate of a feasible point lies in [0, 1], so two feasible points differ by at most one in each of the pq coordinates, and √ the diameter is at most pq. All hypotheses of Theorem 27 are thus verified, so the class of nonpositive 1–Lipschitz losses in γ Fmono has the γ–online gap property, and Theorem 26 proves its hardness. The inclusions in (71) 137

prove the two other conclusions. For each fixed target α, choose one rational θ ∈ (γ, α) and use the corresponding fixed profile and fixed Label Cover soundness. The right degree and alphabet size are then constants. A loss −H(zv,b )/S involves the ℓ edges at v and at most kL labels per edge, so it depends on at most ℓkL coordinates, and its polynomial degree m + 1 is also constant. This establishes all the claimed restrictions on the common hard family. Preserving the single-quadratic result for F γ . For the last assertion, take the class of nonpositive, 1–Lipschitz losses whose proper extensions lie in F γ , and use quadratic profiles hθ (s) = θs + (1 − θ)s2 . The strict inequality aθ < θ in the profile criterion rules out aθ = θ, so at a criterion index θ ∈ (γ, 1) we choose any rational η ∈ (γ, θ) and use hη , with aθ = η and bθ = 1−η. Then hη (0) = 0, hη (1) = 1, and h′η (s) = η + 2(1 − η)s ∈ [η, 2 − η] (0 ≤ s ≤ 1), (364) so hη is increasing and 2–Lipschitz on [0, 1] with values in [0, 1]. Since (1 − η)s2 ≤ (1 − η)s on [0, 1], we get hη (s) ≤ s, and hη (s) ≤ ηs + (1 − η)s2 holds with equality. Also 0 ≤ aθ = η < θ and bθ = 1 − η > 0. Writing c = η/(1 − η), we have −hη (s) = (1 − η)(−s2 − cs), which is (1 − η) times the function hc of Example 4. That example gives γ ∗ (hc ) = c/(c + 1) = η. Multiplying a function by a positive constant multiplies its Clarke subdifferential by the same constant, and ψη is positively homogeneous, so γ ∗ (−hη ) = η. Since zero lies in every normal cone, −h′η (s) belongs to the Clarke subdifferential of the proper extension at s, and (5) gives, for s, t ∈ [0, 1],
−hη (t) + hη (s) ≥ ψη −h′η (s)(t − s) . (365) Let S = 2q, f = −hη ◦ z/S, and fe = f + δKp,q . As in (363), every v ∈ ∂ fe(x) has the form v = −h′η (z(x))a/S + nx with ⟨nx , y − x⟩ ≤ 0 for y ∈ Kp,q . Put s = z(x) and t = z(y), so t − s = ⟨a, y − x⟩. By (365), positive homogeneity of ψη , the sign of ⟨nx , y − x⟩ with monotonicity of ψη , and Lemma 1(7) with γ < η, ! h′η (s) −hη (t) + hη (s) f (y) − f (x) = ≥ ψη − ⟨a, y − x⟩ ≥ ψη (⟨y − x, v⟩) ≥ ψγ (⟨y − x, v⟩). S S Thus the lifted losses are η–tilted convex, hence γ–tilted convex. Since 0 ≤ hη ≤ 1 on [0, 1] and S ≥ 1, they have values in [−1, 0]. By (364), h′′η = 2(1 − η), and (361), √ 2 q ∥∇(−hη ◦ z/S)∥2 ≤ ≤ 1, 2q

∥∇2 (−hη ◦ z/S)∥op ≤

2(1 − η)q ≤ 1. 2q

They are coordinatewise nonincreasing on the domain, because h′η > 0 there and a ≥ 0, and they are concave quadratics with Hessian −2(1 − η)aa⊤ /S of rank at most one. Since hη is increasing on [0, 1], they are minimized by maximizing z, and all implementations are rational. Theorem 27 with β = γ and Theorem 26 then give the hardness conclusion for F γ with single concave quadratic losses. Finally, hθ is a polynomial, and h′θ (s) = θ + 2(1 − θ)s satisfies θ ≤ h′θ ≤ 2 − θ on [0, 1] with h′θ (0) = θ > 0 and h′θ (1) = 2 − θ. Lemma 5 with A = θ and B = 2 − θ therefore proves (79). Endpoint assertion in Corollary 11. For a proposed α > 0, choose 0 < γ < min{α, 1} and use Theorem 28. Every constructed loss is of the form f = −H ◦ z/S with S > 0 and H ′ > 0 on [0, 1], 138

which contains the range of z by (362); here H is Hκ,m or hη . Let x, y be feasible with f (y) < f (x). Then H(z(y)) > H(z(x)), and H is strictly increasing on [0, 1], so z(y) > z(x). Hence ⟨∇f (x), y − x⟩ = −

H ′ (z(x)) z(y) − z(x) < 0. S

Every element of the Clarke subdifferential of the proper extension fe at x is ∇f (x) + nx with ⟨nx , y − x⟩ ≤ 0, so its inner product with y − x is also negative. Thus, if ⟨y − x, v⟩ ≥ 0 for an element v ∈ ∂ fe(x), then f (y) ≥ f (x). This is pseudoconvexity on the feasible set. For c ∈ R, the sublevel set {x ∈ Kp,q : f (x) ≤ c} equals {x ∈ Kp,q : H(z(x)) ≥ −cS}. Since H is continuous and strictly increasing on [0, 1] with H(0) = 0 and H(1) = 1, this set is the whole feasible set if −cS ≤ 0, empty if −cS > 1, and {x ∈ Kp,q : z(x) ≥ H −1 (−cS)} otherwise. Each case is the intersection of the feasible set with a halfspace or is empty, so the losses are quasiconvex as well. This covers the quadratic specialization, which uses H = hη . An algorithm for either endpoint class would thus contradict the hardness theorem unless NP = RP.

139

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