Almost-Orthogonality in Lp Spaces: A Case Study with Grok Ziang Chen∗ 4 , Jaume de Dios Pont† 1 , Paata Ivanisvili† 2 , José Madrid† 3 , and Haozhu Wang∗ 4 1
arXiv:2605.05192v1 [math.CA] 6 May 2026
2
ETH Zürich, [email protected] University of California, Irvine, [email protected] 3 Virginia Tech, [email protected] 4 xAI, {zchen,hwang}@x.ai
†
Authors are listed alphabetically. Mathematical contributor, ∗ Engineering contributor. Abstract
Carbery proposed the following sharpened form of triangle inequality for many functions: for any p ≥ 2 and any finite sequence (fj )j ⊂ Lp we have !1/p′ X
fj
j
≤
p
sup j
X
c αjk
X j
k
∥fj ∥pp
1/p
,
q ∥fj fk ∥p/2 where c = 2, 1/p + 1/p′ = 1, and αjk = ∥fj ∥p ∥fk ∥p . In the first part of this paper we construct a counterexample showing that this inequality fails for every p > 2. We then prove that if an estimate of the above form holds, the exponent must satisfy c ≤ p′ . Finally, at the critical exponent c = p′ , we establish the inequality for all integer values p ≥ 2. In the second part of the paper we obtain a sharp three-function bound 3 X j=1
fj
p
≤
1 + 2Γc(p)
3 1/p′ X j=1
∥fj ∥pp
1/p
,
2 ln(2) where p ≥ 3, c(p) = (p−2) ln(3)+2 ln(2) and Γ = Γ(f1 , f2 , f3 ) ∈ [0, 1] quantifies the degree of orthogonality among f1 , f2 , f3 . The exponent c(p) is optimal, and improves upon the power 6 r(p) = 5p−4 obtained previously by Carlen, Frank, and Lieb. Some intermediate lemmas and inequalities appearing in this work were explored with the assistance of the large language model Grok.
1
Introduction and the main results
Let (X, µ) be a measure space. For p ≥ 1 let Lp (X, dµ) be the space of p-integrable functions on X. In his study of almost orthogonality in Schatten-von Neumann classes [1] Carbery was interested in the following question: Question 1. Under what conditions on the sequence of functions (fj )j ⊂ Lp do we have P p j fj ∈ L ? P P Notice that by the triangle inequality if j ∥fj ∥p < ∞ then necessarily j fj ∈ Lp without any further assumptions on the sequence (fj )j . On the other hand if fj ’s have disjoint P 1/p P p n support then ∥ nj=1 fj ∥p = for any n ≥ 1, and, therefore, the finiteness of j=1 ∥fj ∥p 1
P 1/p P P p p p . Since ∥f ∥ f ∈ L implies ∥f ∥ ≤ j ∥fj ∥p a natural baseline assumption j p j p j j j j P p is j ∥fj ∥p < ∞. In what follows let p ≥ 2. In [1] Carbery introduced nonnegative numbers s ∥fj fk ∥p/2 αjk := ∈ [0, 1], ∥fj ∥p ∥fk ∥p P
where we set αkj = 0 if ∥fj ∥p ∥fk ∥p = 0. The numbers αjk measure the “orthogonality” between functions fj and fk . Clearly if the fj , fk have disjoint support then one can take αjk = 0. In the L2 case the Question 1 has a simple answer in terms of αjk . Indeed, for any finite sequence (fj )j ⊂ L2 we have X X XZ X X X 2 2 2 fj ≤ |fj fk |dµ ≤ αjk ∥fj ∥2 ∥fk ∥2 ≤ ∥A∥ ∥fj ∥22 ≤ sup αjk ∥fj ∥22 , j
2
j,k
X
k
j
j,k
j
j
2 , and ∥A∥ denotes its spectral radius. One where A is the symmetric matrix with entries αjk 2 “naive” way of interpolation between L case and Cotlar–Stein lemma suggests that the following inequality may be true:
Question 2. Is it true that for p ≥ 2 the inequality !1/p′ X 1/p X X 2 fj ≤ sup αjk ∥fj ∥pp , p
j
j
(1)
j
k
where 1/p + 1/p′ = 1, holds for all finite sequences (fj )j ⊂ Lp ?
2 are replaced In [1] Carbery obtained a weaker form of (1) where p ≥ 2 is any integer and αjk by αjk . Since αjk ∈ [0, 1] therefore, having a larger power on αjk only makes the inequality stronger. However, the advantage of Carbery’s argument is that for even integer values of p the 2 . inequality (1), in fact, holds in Schatten-von Neumann Cp classes with αjk in place of αjk Recently [2] Carlen–Frank–Ivanisvili–Lieb verified the inequality (1) for two functions. In fact they proved a stronger estimate: for any f, g ∈ Lp one has 1/p′ 1/p ∥f + g∥p ≤ 1 + Γ2/p ∥f ∥pp + ∥g∥pp , (2) p 2∥f g∥
p/2
2/p
p/2 2 , therefore, (2) implies (1) for two functions. where Γp = Γp (f, g) = ∥f ∥p +∥g∥ ≤ α12 p . Clearly Γp p p Later the estimate (2) was further refined by Ivanisvili–Mooney in [4] by showing that one has the bound q q 1 ∥f + g∥p ≤ 1/p (1 + 1 + Γ2p )1/p + (1 + 1 − Γ2p )1/p (∥f ∥pp + ∥g∥pp )1/p , 2
and the right-hand side of this inequality is the best possible one can have in terms of ∥f ∥p , ∥g∥p , and ∥f g∥p/2 . In a subsequent work [3] Carlen–Frank–Lieb showed that for any n ≥ 2 and any p ≥ 2 we have n n 1/p′ X 1/p X fj ≤ 1 + (n − 1)Γp (f1 , . . . , fn )r ∥fj ∥pp (3) j=1
p
j=1
where Γp (f1 , . . . , fn ) =
p/2 n −1 P 1≤j<k≤n |fj fk | 2 p/2 Pn p −1 n j=1 ∥fj ∥p
2
2n is n-function extension of the Γp introduced in (2), and r = r(n, p) = 2n+(p−2)(2n−1) . Notice that Γp ∈ [0, 1]. The inequalities (3) and (1) do not imply each other. The goal of (3) was to find a possible n-function analog of (2) though we should point out that 2-function version of 4 (3) is actually weaker than (2) because r(2, p) = 4+3(p−2) < p2 . In [3] Carlen–Frank–Lieb asked about the optimal (largest) power r = r(n, p) one can have in (3). They provided an example showing that if (3) holds for some r > 0 then necessarily
r≤
ln(n − 1) − ln(n − 2)
:= β(n, p). ( p2 − 1) ln(n) + ln(n − 1) − p2 ln(n − 2)
The value β(2, p) should be understood in the “limiting” sense: limn→2 β(n, p) = p2 . We provide a counterexample showing that for each p > 2 the inequality (1) fails. In fact, c instead of α2 in the right hand side the example shows that if the inequality holds with αjk jk ′ of (1) for some c > 0 then necessarily c ≤ p . Our first theorem verifies the inequality at the critical power c = p′ for integers p ≥ 2. Theorem 1.1. For any integer p ≥ 2 the inequality !1/p′ X 1/p X X p′ fj ≤ sup αjk ∥fj ∥pp j
p
j
(4)
j
k
holds for all finite sequences (fj )j ⊂ Lp . In the second part of this paper we prove a more general theorem stating that Carlen–Frank– Lieb’s inequality (3) holds with power r = β(n, p) if and only if it holds for indicator functions fj = aj 1A for some common set A of finite positive measure, and all aj ≥ 0, j = 1, . . . , n. Theorem 1.2. Let p ≥ 2 and n > 2. The inequality (3) holds with r = β(n, p) if and only if n X j=1
aj
p ′
" ≤ 1 + (n − 1) n
holds for all a1 , . . . , an ≥ 0 satisfying
n−1 2
X
ai aj
# p/2 β(n,p)
(5)
1≤i<j≤n
p j=1 aj = 1
Pn
The inequality (5) numerically seems correct. The difficulty of verifying it rigorously lies in Pn the fact that the expression a j=1 j achieves several local maxima in the interior of the compact P p P set { aj = 1} ∩ { i<j ai aj = b}, see Lemma 2.8 in Section 2.3. Moreover, the numerics show that for different values p, n, and b, sometimes the global maximum is on the boundary and sometimes it is in the interior of that set. Remark 1.3. In Lemma 2.8 we prove that we can reduce inequality (5) further, to the case where aj ∈ {x, y, 0}, for some x, y ≥ 0. In the special case n = 3 there is a certain symmetry which helps to decrease the number of parameters and local maxima points involved in verification of (5). In particular, we manage to sharpen Carlen–Frank–Lieb’s inequality for three functions up to optimal power β(3, p). Theorem 1.4. For n = 3 and p ≥ 3 the inequality (3) holds with the optimal power r = β(3, p). Remark 1.5. There is nothing intrinsic about the assumption p ≥ 3. In principle, the arguments presented here could likely be extended to cover the range p ∈ [2, 3) as well. However, while the proof initially proceeds in essentially the same way for p ∈ [2, 3) as for p ≥ 3, the two cases begin to slightly diverge at a certain stage of the argument. In the interval p ∈ [2, 3) one must analyze a larger number of subcases, which would lengthen the already technical parts of the paper. For this reason, we restrict the presentation to the case p ≥ 3. 3
1.1
Structure of the paper and the use of the AI model Grok
This paper developed in part through interactions with the AI model Grok. In particular, an early turning point was Grok Heavy’s construction of a counterexample to Carbery’s question. The version originally produced by Grok was substantially longer than the one presented here and contained several numerical inaccuracies; see the accompanying chat transcript in Appendix C. Nevertheless, its underlying idea was correct. In the present paper, we give a more polished and streamlined version of this counterexample. We also emphasize that, prior to Grok’s construction, we were already aware that a counterexample should exist. However, the example we had previously found arose from a random brute-force search and did not provide conceptual insight, nor did it indicate the correct optimal exponent. By contrast, Grok’s construction revealed a clear structural pattern, which in turn led to the optimal power p′ . In Section 2.2, we prove the optimality of the exponent p′ . For this part, the main ingredient of the argument, Lemma 2.3, is essentially due to Grok; our contribution here consisted in reducing the proof of Theorem 1.1 to this lemma (refining Carbery’s ideas), and carefully checking the details and editing the presentation. The proofs of Theorems 1.2 and 1.4 in Section 2.3, as well as the subsequent arguments in Section 3, were developed through a genuine human–AI collaboration. In these sections, we prove the Carlen–Frank–Lieb conjectured inequality for three functions for all p ≥ 3, which appears to be the most technically difficult part of the paper. Grok 4.20 Heavy was used to generate a number of plots that were very helpful in guiding our intuition; many of these plots are included in the paper. Grok 4.20 was also used to verify numerous local inequalities arising in the argument; in total, we estimate that roughly thirty polynomial inequalities were checked in this way. To improve readability, we have moved these technical computations to Appendices A and B, and in Section 3 we state only the corresponding lemmas. This allows the reader to first see the overall logical structure of the argument without interruption, and then, if desired, verify each technical computation separately in the appendices. Without AI assistance, we likely would have chosen not to pursue such a lengthy and technical proof, since the computational verification alone would have required a substantial amount of time and effort. With Grok, however, this computational component became inexpensive and routine. At the same time, we wish to stress that the main conceptual idea of the proof, namely, the analysis based on counting roots of certain complicated functions, originated on the human side. The role of Grok was to help break this idea into a collection of simpler verifiable inequalities and to efficiently confirm those reductions.
2
Proofs
2.1
Counterexamples
Proposition 2.1. Let p ≥ 2. If the inequality !1/p′ X j
fj
p
≤
sup j
X
c αjk
X j
k
∥fj ∥pp
1/p
(6)
holds for some c > 0 and all finite sequences (fj )j ∈ Lp then c ≤ p′ . Proof. Consider a measure space (X, dµ) with n ≥ 3 pairwise disjoint “private” sets Pj each of measure 1, and one “common” set C of measure 1. Define functions fj (j = 1, . .P . , n) by fj = 1 p on Pj , fj = 1 on C, and fj = 0 elsewhere. Observe that ∥fj ∥p = 2. Let S := nj=1 fj . Then S = 1 on each Pj and S = n on C, so ∥S∥pp = n + np . 4
p/2
1 Next, notice that αjj = 1. For each j ̸= k, ∥fj fk ∥p/2 = 1, hence αjk = 21/p . Thus
sup j
n X
c αjk =1+
k=1
n−1 . 2c/p
Thus the inequality (6) takes the form p 1/p
(n + n )
′ n − 1 1/p ≤ 1 + c/p (2n)1/p . 2
p−1 1/(p−1) 1+np−1 ≤ This is the same as n+np ≤ 1 + n−1 2n which further can be rewritten as 2 2c/p
1 + n−1 . Thus we obtain 2c/p
2c/p ≤
n−1 1/(p−1)
1+np−1 2
.
(7)
−1
p Taking n → ∞ the inequality (7) yields 2c/p ≤ 21/(p−1) , i.e., c ≤ p−1 . This finishes the proof of Proposition 2.1.
Remark 2.2. The counterexample presented above is essentially the same as the one proposed to us by Grok, with two small modifications: we allow both n and p to be arbitrary parameters rather than fixed values, and we polish the presentation. Treating n as a free parameter and passing to the limit n → ∞ enables us to recover the optimal exponent p/(p − 1).
2.2
The proof of Theorem 1.1
Without loss of generality assume fj ≥ 0 for all j ≥ 1. We start with the following lemma. Lemma 2.3. Let (Ω, dx) be any measure space and let K(x, y) be a nonnegative symmetric integral kernel defined on Ω × Ω. Suppose κ is defined by Z p κ = sup K(x, y) p−1 dy. x
Ω
For any integer p > 1 we have Z
2
Y
K(xi , xj ) p−1
Ωp 1≤i<j≤p
p Y s=1
Fs (xs ) dx1 · · · dxp ≤ κ p−1
p Y s=1
∥Fs ∥p ,
for all Fs ∈ Lp , s = 1, . . . , p. Remark 2.4. The proof below was proposed by Grok; see Appendix C for a link to the conversation. After verifying its correctness, we present it essentially without modification. Proof. Since K is symmetric and nonnegative, we rewrite the product in the integrand as Y
2
K(xi , xj ) p−1 =
1≤i<j≤p
p Y Y k=1
K(xk , xj )
1
p−1
.
j̸=k
Thus, the left-hand side is Z
p h Y 1 i Y p−1 Fk (xk ) K(xk , xj ) dx1 · · · dxp .
Ωp k=1
j̸=k
5
Define functions hk : Ωp → R for k = 1, . . . , p by Y 1 p−1 . hk (x1 , . . . , xp ) = Fk (xk ) K(xk , xj ) j̸=k
Qp
Q The integrand is k=1 hk (x), so the integral is Ωp pk=1 hk (x) dx. ByP the generalized Hölder’s p p inequality on the space L (Ω ) (with each exponent equal to p, since pk=1 p1 = 1), R
p Y
Z
hk (x) dx ≤
Ωp k=1
p Y k=1
∥hk ∥Lp (Ωp ) .
Without loss of generality, compute ∥h1 ∥p (the others follow by symmetry, relabeling the Fk ). ∥h1 ∥pp =
Z Ωp
Z p Z p h Y 1 ip Y p p−1 p K(x1 , xj ) p−1 dxj dx1 , F1 (x1 ) · dx1 · · · dxp = F1 (x1 ) K(x1 , xj )
since the integrals separate. Each inner integral is p Z Y
j=2 Ω
Ω
j=2
K(x1 , xj )
p p−1
dxj =
Z
R
Ω K(x1 , y) p
p p−1
K(x1 , y) p−1 dy
dy ≤ κ, so
p−1
Ω
j=2
Thus,
Z
∥h1 ∥pp ≤ κp−1
Ω
≤ κp−1 .
F1 (x1 )p dx1 = κp−1 ∥F1 ∥pp ,
p−1
p−1
so ∥h1 ∥p ≤ κ p ∥F1 ∥p . Similarly for each k, ∥hk ∥p ≤ κ p ∥Fk ∥p . Therefore, p Y
Z
Ωp k=1
hk (x) dx ≤
p p Y Y p−1 κ p ∥Fk ∥p = κp−1 ∥Fk ∥p .
k=1
k=1
This completes the proof of Lemma 2.3. Now we are ready to prove Theorem 1.1. Remark 2.5. The proof below was obtained with the assistance of Grok (see Appendix C), after providing several specific directions to try. We present it here in a slightly polished form. Assume without loss of generality that the functions fj are nonnegative. Then, X j
fj
p p
=
Z X
fj (x)
p
j
dx =
p X Z Y j1 ,...,jp
fjs (x) dx.
s=1
For R Qpa fixed multi-index (j1 , . . . , jp ), set gs = fjs for s = 1, . . . , p. The integral becomes s=1 gs (x) dx. Rewrite this as Z Y (gi (x)gj (x))1/(p−1) dx, 1≤i<j≤p
since each gs appears in exactly p − 1 pairs, contributing an exponent of (p − 1) · (1/(p − 1)) = 1. Apply Hölder’s inequality with equal exponents: Z Y Y hij (x) dx ≤ ∥hij ∥(p) , 1≤i<j≤p
1≤i<j≤p
6
2
where
1 1/(p−1) . Then, 1≤i<j≤p (p) = 1. Set hij = (gi gj )
P
2
1/(p−1)
1/(p−1)
∥hij ∥(p) = ∥gi gj ∥ p /(p−1) = ∥gi gj ∥p/2 (2) 2 since
,
p 2 /(p − 1) = p/2. By the given condition,
∥gi gj ∥p/2 ≤ αj2i jj ∥gi ∥p ∥gj ∥p , so Y i<j
1/(p−1) ∥gi gj ∥p/2 ≤
p Y Y 2/(p−1) Y 2 1/(p−1) (αji jj ∥gi ∥p ∥gj ∥p ) = αji jj · ∥gs ∥p , i<j
s=1
i<j
where the last equality follows because each ∥gs ∥p appears in p − 1 terms, yielding exponent (p − 1) · (1/(p − 1)) = 1. Thus, Z Y p s=1
fjs (x) dx ≤
2/(p−1) αji jj
Y
p Y s=1
1≤i<j≤p
∥fjs ∥p .
Summing over all multi-indices gives X
fj
j
p p
≤
X j1 ,...,jp
2/(p−1) αji jj
Y
p Y s=1
1≤i<j≤p
∥fjs ∥p .
(8)
The right-hand side of (8) is the discrete analog of the left-hand side of Lemma 2.3 on the counting measure space over the indices j, with kernel K(m, n) = αmn and functions P p/(p−1) P p′ Fs (j) = ∥fj ∥p for each s = 1, . . . , p. Here, κ = supj k αjk = supj k αjk , and ∥Fs ∥p =
X j
∥fj ∥pp
1/p
By Lemma 2.3, X Y j1 ,...,jp
2/(p−1) α ji jj
,
s=1
p Y s=1
i<j
Thus, X
fj
j
p Y
so
p p
∥Fs ∥p =
∥fjs ∥p ≤ κp−1
≤ κp−1
X j
X
X j
j
∥fj ∥pp .
∥fj ∥pp .
∥fj ∥pp ,
and taking p-th roots yields X j
fj
p
≤ κ(p−1)/p
X j
∥fj ∥pp
1/p
=
sup j
X
′
p αjk
1/p′ X
k
This completes the proof of Theorem 1.1.
2.3
Proof of Theorems 1.2 and part of Theorem 1.4
We will need several technical lemmas. Lemma 2.6. For any p ≥ 2, and all n > 2 the map n−p/2 β(n,p) p−1 φn,p (t) = 1 + (n − 1) tn 2 h p/2 i is concave on 0, n1 n2 . 7
j
∥fj ∥pp
1/p
.
Remark 2.7. The proof of the lemma presented below is due to Grok. It is correct and we present it here in a slightly polished form. For the full chat link conversation see Appendix C. Proof. Let β := β(n, p). By rescaling it suffices to show that ψ(t) = (1 + (n − 1)tβ )p−1 is concave on [0, 1]. Set u = n − 1 > 1 and v = p − 1 ≥ 1. We have ψ(t) = (1 + utβ )v , so ψ ′ (t) = vuβtβ−1 (1 + utβ )v−1 ; ψ ′′ (t) = vuβ(β − 1)tβ−2 (1 + utβ )v−1 + v(v − 1)u2 β 2 t2(β−1) (1 + utβ )v−2 = uvβtβ−2 (1 + utβ )v−2 [(β − 1)(1 + utβ ) + u(v − 1)βtβ ] = uvβtβ−2 (1 + utβ )v−2 [β − 1 + utβ (vβ − 1)].
Thus the sign of ψ ′′ is determined by the sign of h(t) := β − 1 + utβ (vβ − 1). Next, we claim β≤
n . 1 + (n − 1)(p − 1)
(9)
n Assuming the claim, notice that 1+(n−1)(p−1) ≤ 1. Therefore h(0) = β − 1 ≤ 0. Since h is monotone it suffices to verify h(1) ≤ 0. We have h(1) = β − 1 + uvβ − u. Therefore h(1) ≤ 0 if u+1 and only if β ≤ 1+uv . This inequality coincides with (9).
. Substituting the expression for β yields To verify (9) we show that β1 ≥ 1+(n−1)(p−1) n p 2 ln
2 1 + n−2 + ln 1 ln 1 + n−2
n−1 n
≥
1 + (n − 1)(p − 1) . n
ln(1+ 2 ) Both sides are affine in p, and equal at p = 2. The coefficient of p on the left is 12 ln 1+ n−2 , and 1 ( n−2 ) on the right is n−1 n . 1 1+z Set z = n−2 > 0. The inequality on coefficients simplifies to ln(1+2z) ln(1+z) ≥ 2 1+2z . Define
ln(1+z) 1+z ln(1 + z). Then m(0) = 0 and m′ (z) = 2(1+2z) m(z) = ln(1 + 2z) − 2 1+2z 2 > 0 for z > 0, so m(z) > 0 for z > 0. Thus, the coefficient on the left exceeds that on the right for p > 2, implying the desired inequality for p ≥ 2.
Lemma 2.8. Let p > 2, and let n > 2 be an integer. The following are equivalent (i) The inequality n X j=1
xj
p
≤ φn,p
holds for all x1 , . . . , xn ≥ 0 satisfying
X
xj xk
p/2
(10)
1≤j<k≤n
p j xj = 1
P
(ii) The inequality (kx + (m − k)y)p ≤ φn,p (x2 k(k − 1)/2 + y 2 (m − k)(m − k − 1)/2 + k(m − k)xy)p/2 holds for all x, y ≥ 0, and all integers k = 0, . . . , m, and m = 1, . . . , n satisfying kxp + (m − k)y p = 1. Before we proceed with the proof of the lemma notice that the inequality (10) is the same as the one in Theorem 1.2.
8
Proof. The implication (i) ⇒ (ii) is trivial, take x1 = x2 = . . . = xk = x, xk+1 = . . . = xm = y, and the rest to be zero. So in what follows we focus on the implication (ii) ⇒ (i). Clearly φn,p is increasing. Therefore it suffices to show that T p ≤ φn,p (bp/2 ) where T = P sup j xj , wherePthe supremum is taken P over all nonnegative variables x1 , . . . , xn ≥ 0 satisfying p n the constraints j=1 xj = 1 and 1≤i<j≤n xi xj = b. Our goal is to understand on which variables x1 , . . . , xn the value T is achieved. We claim that this happens when (x1 , . . . , xn ) = (x, . . . , x, y, . . . , y , 0, . . . , 0) | {z } | {z } | {z } k
m−k
(11)
n−m
for some x, y > 0 and some integers k, m satisfying 0 ≤ k ≤ m ≤ n. We should point out that equality in (11) should be considered up to permutations of coordinates in the right-hand side of (11), i.e., the equality (11) simply says that k of the variables x1 , . . . , xn equal to x; m − k of them equal to y, and the rest is zero. First notice that b ∈ [0, n2 n−2/p ]. Indeed, the value b = 0 is achieved if and only if xk = 1 for some integer k, P and xj = 0 for all j ̸= k. On thePother hand by Lagrange multipliers the local maximum of i<j xi xj under the constraint j xpj = 1, xj > 0 for all j = 1, . . . , n is P achieved when j̸=k xj − λpxp−1 = 0 for all k = 1, . . . , n, and some λ > 0 P (notice that λ ≤ 0 k does not have a solution unless x1 = . . . = xn = 0 which would violate j xpj = 1). Thus P s := nj=1 xj = λpxp−1 + xk . Notice that the function h(t) = λptp−1 + t is strictly increasing k on [0, ∞) implying that λpxp−1 + xk = λpxp−1 + xj can happen for all pairs j, k if and only if j k Pn P p x1 = . . . = xn = x. In this case 1 = j=1 xj = nxp , and 1≤i<j≤n xi xj = n2 x2 = n2 n−2/p . The value n2 n−2/p is also a global maximum. Indeed, on the boundary where some of the xj ’s vanish we can argue in a similar way and conclude that the rest of the variables have to P be equal to each other. This leads to 1≤i<j≤n xi xj = k2 k −2/p for some integer k, 1 ≤ k ≤ n. On the other hand the map k 7→ k2 k −2/p is increasing on [1, n], hence its maximal value is n −2/p . 2 n Remark 2.9. An alternative way: By AM-GM inequality and Hölder inequality we have n n n − 1 X 2 n − 1 X p 2/p 1−2/p n −2/p b≤ xi ≤ xi n = n . 2 2 2 i=1
i=1
1 The equality happens if and only if x1 = x2 = · · · = xn = n1/p . Thus the implication (ii) ⇒ (i) is trivial if b = 0 or b = n2 n−2/p . In what follows assume P b ∈ (0, n2 n−2/p ). Let the structure of local maximum of j xj in the interior of the domain, i.e., x1 , . . . , xn > 0 and later we will explain what happens when part of the xj are equal to zero. Consider Lagrangian function X X X Ψ(x1 , . . . , xn ) = xj − λ xpj − 1 − µ xj xk − b . j
j
j<k
Partial derivatives of Ψ vanish if and only if 1 − λpxp−1 − µ j̸=k xj = P 0 for all k = 1, . . . , n. k Adding −µxk to both sides of the equation, and denoting as before s := nj=1 xj we obtain P
1 − µs = λpxp−1 − µxk k
for all
k = 1, . . . , n.
Let g(t) = λptp−1 − µt. If λ = 0 then the conclusion is trivial. If λ = ̸ 0 then since either This shows that at the interior local maximum, the variables x1 , . . . , xn take at most two values. If we are on the boundary, say part of the xj are zero, then we can argue similarly as before. So we conclude that at the global maximum variables x1 , . . . , xn take at most 3 values, these are 0, u, v for some u, v > 0. g ′′ > 0 or g ′′ < 0 the graph of g(t) intersects any horizontal line in at most two points.
9
Now we are ready to prove Theorem 1.2. Proof of Theorem 1.2. The implication (3) with r = β(n, p) implies (5) is trivial, just take fj = aj 1A , j = 1, . . . , n, where A is a common measurable set with finite positive measure. So in what follows we focus on the implication (5) implies (3). Recall that the inequality (10) is the same as (5). Pick nonnegative functions f1 , . . . , fn in Lp . Then the validity of (10) implies n X
gj
p
j=1
on a set X ′ = {
≤ φn,p
p j fj > 0}, where gk =
X
gi gj
p/2
(12)
1≤i<j≤n fk
P
P
measure
p 1/p j fj
, k = 1, . . . , n. Consider the probability
p j fj
P
p dµ j ∥fj ∥p
dν = P
p/2 on a set X ′ . Recall that the function φn,p is concave on [0, n1 n2 ], and the quantity P p/2 p/2 is at most n1 n2 (see the proof of Lemma 2.8). So, integrating the 1≤j<i≤n gi gj inequality (12) over the set X ′ with respect to the probability measure dν and applying Jensen’s inequality (here we invoke Lemma 2.6) we obtain Z X Z X p/2 p gj dν ≤ φn,p gi gj dν . (13) j
i<j
It is a straightforward calculation to verify that the inequality (13) coincides with (3) r = β(n, p). This finishes the proof of Theorem 1.2.
3
Completing the proof of Theorem 1.4
Notice an identity that is only valid for n = 3: (n − 1)
−p/2 !β(n,p) n n · = 1. 2 2
(14)
p−1 Therefore, in this special case we have φ3,p (t) = 1 + (2t)β(3,p) . Thus the inequality (10), after raising both sides to the power 1/(p − 1), can be rewritten as follows β(3,p) ′ (x + y + z)p ≤ 1 + 2(xy + xz + yz)p/2 . 2 ln(2) Lemma 3.1 (main inequality). Let p ≥ 3, and let c(p) = (p−2) ln(3)+2 ln(2) . The inequality
c(p) ′ (x + y + z)p ≤ 1 + 2(xy + xz + yz)p/2
(15)
holds for all x, y, z ≥ 0 satisfying xp + y p + z p = 1. In what follows by continuity we assume p > 3. Step 1. First we verify that the inequality holds on the boundary, i.e., when at least one of the variables is zero. Without loss of generality assume z = 0. In this case the inequality takes the form ′
(x + y)p ≤ 1 + (2xp/2 y p/2 )c(p) . 10
(16)
2 ln(2) 2 Notice that c(p) ≤ p2 . Indeed, to verify the inequality (p−2) ln(3)+2 ln(2) ≤ p divide both sides by 2 and multiply by the denominators, we arrive at p ln(2) ≤ (p − 2) ln(3) + 2 ln(2) which follows from (p − 2)(ln(3) − ln(2)) ≥ 0. Next we have 2xp/2 y p/2 ≤ xp + y p = 1, and hence the right hand side of (16) is at least 1 + (2xp/2 y p/2 )2/p = 1 + 22/p xy. On the other hand the stronger inequality ′
(x + y)p ≤ 1 + 22/p xy under the constraint xp + y p = 1 was proved recently in Theorem 1.1 (case p ≥ 2) in [2].
Step 2. By Lemma 2.8 applied to n = 3 and Step 1 it suffices to verify the inequality (15) when in the variables x, y and z are positive and two of them are equal to each other. Without loss of generality we can assume that z = y. So we want to show ′
(x + 2y)p ≤ 1 + 2c (2xy + y 2 )pc/2
(17)
under the constraint xp + 2y p = 1 where c = c(p). Set t := y/x. From xp + 2y p = 1 we obtain xp (1 + 2tp ) = 1 and hence x = (1 + 2tp )−1/p . Next, we have x + 2y = x(1 + 2t) = (1 + 2tp )−1/p (1 + 2t), therefore, the left hand side of (17) takes the form ′
′
(x + 2y)p = (1 + 2tp )−1/(p−1) (1 + 2t)p . Similarly we have 2xy + y 2 = x2 (2t + t2 ) = (1 + 2tp )−2/p (2t + t2 ). So the right hand side of (17) takes the form 1 + 2c (2xy + y 2 )pc/2 = 1 + 2c (1 + 2tp )−c (2t + t2 )pc/2 . Thus to prove (17) it suffices to show 2c
(2t + t2 )pc/2 (1 + 2t)p/(p−1) − +1≥0 (1 + 2tp )c (1 + 2tp )1/(p−1)
(18)
for all t ≥ 0. Next we make change of variables t = 1/s, s > 0. The expression in the left hand side of (18) rewrites as follows !c 1 2(2s + 1)p/2 (s + 2)p p−1 h(s) := − + 1. sp + 2 sp + 2
Plot of h(s) for p = 4, s ∈ [0, 8] 0.14 h(s)
h(s) y=0 Zeros of h(s)
0.12 0.1 8 · 10−2 6 · 10−2 4 · 10−2 2 · 10−2
s 1
2
3
4
11
5
6
7
8
We have " p −c h′ (s) = p s +2
0.25
2(2s + 1)p/2 sp + 2
!c
sp + sp−1 − 2 +2 2s + 1
(s + 2)p sp + 2
1
p−1
# sp−1 − 1 . (p − 1)(s + 2)
Plot of h′ (s) for p = 4, s ∈ [0, 8] h′ (s)
h′ (s) y=0 Zeros of h′ (s)
0.2 0.15 0.1 5 · 10−2
s 1
2
3
4
5
6
7
8
p(p−2)(3 ln 3−4 ln 2)c 9 ln 2
≥ 0, and
−5 · 10−2
Lemma 3.2. We have h(0) = h(1) = h′ (1) = 0, h′′ (1) = lims→∞ h(s) = 0. Proof. See Appendix A.
Lemma 3.3. For each p > 3 function h′ changes sign 3 times as follows: (+ − +−). More precisely from + to − at point q1 ∈ (0, 1); from − to + at point 1; and from + to − at point q2 > 1. Clearly Lemma 3.3 and Lemma 3.2 imply h ≥ 0 on [0, ∞) and hence the inequality (18). The proof of Lemma 3.3 will be divided into several parts.
It is enough to study the sign of the expression !c 1 sp + sp−1 − 2 (s + 2)p p−1 sp−1 − 1 2(2s + 1)p/2 + 2 . −c sp + 2 2s + 1 sp + 2 (p − 1)(s + 2) The fact that h′ (s) changes sign from − to + follows from h′′ (1) > 0. Multiply the expression (19) by the factor c (p − 1) 2s + 1 sp + 2 2 sp + sp−1 − 2 (2s + 1)p/2 12
(19)
(20)
Then the resulting expression takes the form −2
c−1
c− 1 1 pc p−1 (s + 2) p−1 · sp−1 − 1 · (2s + 1) 1− 2 · sp + 2 c · (p − 1) + sp + sp−1 − 2
(21)
Since the factor (20) has sign − on (0, 1), and sign + on (1, ∞) therefore the lemma is reduced to showing that the expression (21) changes sign two times, namely, from − to + at point q1 ∈ (0, 1), and from + to − at point q2 > 1. As s 7→ log(s) is increasing on (0, ∞) we can replace the expression (21) by c− 1 1 1− pc p−1 p p−1 p−1 2 (s + 2) · s − 1 · (2s + 1) · s +2 , ψ(s) := − log 2c−1 c · (p − 1) + log sp + sp−1 − 2 (22) and study its sign change.
Plot of ψ(s) for p = 4, s ∈ [0, 8] ψ(s)
ψ(s) y=0 Zeros of ψ(s)
0.1 5 · 10−2
s 1
2
3
4
5
6
7
8
−5 · 10−2 −0.1 −0.15 Lemma 3.4. We have ψ(0) = − log((p−1)c) < 0, ψ(1) = log −∞
3 (2p−1)c
> 0, and lims→∞ ψ(s) =
Proof. See Appendix A.
Thus to show that ψ changes sign from − to + at q1 ∈ (0, 1), and from + to − at point q2 > 1 it follows from Lemma 3.4 that one needs to show ψ ′ changes sign once from + to − on (0, ∞). A direct differentiation shows that p−1 1 (p − 1)sp−2 2 − pc p s psp−1 + (p − 1)sp−2 ψ ′ (s) = + p−1 + + cp − − p (p − 1)(s + 2) s −1 1 + 2s p−1 s +2 sp + sp−1 − 2 P (s) , =− p−1 (p − 1) (s + 2) (2s + 1) s − 1 sp + 2 sp + sp−1 − 2
13
where P (s) := (p − 1)(2 − cp) s3p + (−4cp2 + 4cp + 6p − 2) s3p−1 + (−5cp2 + 5cp + 5p + 1) s3p−2 − 2p(cp − c − 1) s3p−3 + p(c − 2)(p − 1) s2p+1 + (8cp2 − 8cp − 3p2 − 5) s2p
+ (17cp2 − 17cp + 3p2 − 15p − 8) s2p−1 + (10cp2 − 10cp + 2p2 − 14p + 4) s2p−2 − 4p(cp − c + p − 2) sp+1 + (−16cp2 + 16cp − 6p2 + 24p + 4) sp
+ (−16cp2 + 16cp + 6p2 + 12p − 8) sp−1 + 4(p − 1)2 sp−2 + 4p(cp − c − 2) s + (8cp2 − 8cp − 16p + 12).
Clearly sign(ψ ′ ) = −sign(P (s)). So the goal reduces to showing that P changes sign once from − to + on (0, ∞).
Plot of P (s) for p = 4, s ∈ [0, 1.65] P (s)
P (s) y=0 Zeros of P (s)
40
20 s 0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
−20 −40 Lemma 3.5. For p ≥ 3 we have P (0) < 0, P ′ (0) < 0, P (1) = P ′ (1) = 0, P ′′ (1) < 0, and lims→∞ P (s) = +∞. Proof. See Appendix A. Thus to obtain a sign change from − to + for P it suffices to show (thanks to the previous lemma) the following lemma Lemma 3.6. P ′′ (s) changes sign as (+ − +) on (0, ∞). The proof of Lemma 3.6 will be split into several parts. First notice that we have P ′′ (s) = sp−4 Q(s), where
14
Q(s) := 3p(p − 1)(3p − 1)(2 − cp) s2p+2 + (−4cp2 + 4cp + 6p − 2)(3p − 1)(3p − 2) s2p+1
+ (−5cp2 + 5cp + 5p + 1)(3p − 2)(3p − 3) s2p − 2p(3p − 3)(3p − 4)(cp − c − 1) s2p−1
+ p(c − 2)(p − 1)(2p + 1) 2p sp+3 + (8cp2 − 8cp − 3p2 − 5) 2p(2p − 1) sp+2
+ (17cp2 − 17cp + 3p2 − 15p − 8)(2p − 1)(2p − 2) sp+1 + (10cp2 − 10cp + 2p2 − 14p + 4)(2p − 2)(2p − 3) sp
− 4p(cp − c + p − 2)(p + 1)p s3 + (−16cp2 + 16cp − 6p2 + 24p + 4) p(p − 1) s2
+ (−16cp2 + 16cp + 6p2 + 12p − 8)(p − 1)(p − 2) s + 4(p − 1)2 (p − 2)(p − 3).
Plot of Q(s) for p = 4, s ∈ [0, 1.45] 1,000 Q(s)
Q(s) y=0 Zeros of Q(s)
800 600 400 200
s 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
−200 −400
Lemma 3.7. We have Q(0) > 0, and lims→∞ Q(s) = +∞. Proof. See Appendix A. Since sign(Q) = sign(P ′′ ), invoking Lemma 3.7 it suffices to show that Q changes sign as (+ − +) on (0, ∞). Notice that Q(1) = −9(p − 1)2 (p − 2) < 0, thus it suffices to show that Q′ has sign change (+ − +). Lemma 3.8. We have Q′ (0) > 0, Q′ (1) < 0, and Q′ (∞) > 0. Proof. See Appendix A. Thus to conclude that Q′ has sign change (+ − +) it suffices to show the following key lemma Lemma 3.9. If Q′′ (s) = 0 then Q′ (s) < 0. Clearly if the lemma is true then it implies that Q′ has (+ − +) sign change.
15
Plot of Q′ (s) and Q′′ (s) for p = 4, s ∈ [0, 1.3]
·104
Q′ (s) Q′′ (s) y=0 Zeros of Q′ (s) Zeros of Q′′ (s)
1
0.5
s 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
−0.5
The proof of Lemma 3.9 will be split into several technical lemmas. Lemma 3.10. For any p > 3 we have 5 4 < c(p) < . 4p − 3 3p − 2 1 In particular, we have p−1 < c(p) < 2/p.
Proof. See Appendix A. Next let g(s) := sQ′′ (s), and set H(s) := sQ′′ (s) − (p + 1)Q′ (s). Remark 3.11. Notice that the statement “if Q′′ (s) = 0 then Q′ (s) < 0” is equivalent to the statement “if g(s) = 0 for some s ̸= 0 then H(s) > 0 (or equivalently Q′ (s) < 0)”. Our first goal will be the following. Lemma 3.12. For p > 3 function g has at most 3 positive real zeros (counted with multiplicities). Proof. See Appendix A. Our second goal will be to show that g has exactly 3 simple positive zeros (here simple means that multiplicity is one) with sign pattern (− + −+). Indeed, this follows from the previous lemma and the following technical observation. Before we state that observation we denote sA =
p−1 , p+1
r=
p p+1
so that
0 < sA < r < 1.
Lemma 3.13 (g changes sign 3 times). We have (i) The inequality g(s) < 0
holds in a neighborhood of zero; 16
g(1) < 0;
lim g(s) = +∞.
s→∞
(ii) Moreover, we have g(sA ) > 0;
g(r) < 0.
Proof. See Appendix A for the proof of (i) and Appendix B for the proof of (ii). Clearly these two lemmas combined with the previous lemmas show that g changes sign as (− + −+). Next we need to learn how to control the monotonicity of H. Let W = sH ′ . Observe that zeros of W and g interlace in the following sense: between any two positive consecutive zeros of g = sQ′′ (s) there is at least one zero of W . This follows from the observation that d W = s − (p + 1) g(s). ds Therefore, if we apply the Rolle’s theorem to s−(p+1) g(s) we will obtain the desired claim.
Since g has exactly 3 positive zeros, it follows that W has at least two different positive zeros. Lemma 3.14. W has exactly two positive simple zeros 0 < u1 < u2 . Moreover, W > 0 on (0, u1 ); W < 0 on (u1 , u2 ); and W > 0 on (u2 , ∞). Proof. See Appendix A. Lemma 3.15. We have W (r) < 0. Proof. See Appendix B. One more technical lemma Lemma 3.16. We have Q′ (sA ) < 0, and H(r) > 0. Proof. See Appendix B. Finally we are ready to show the complete proof of Lemma 3.9. Here is its reformulation: Lemma 3.17. For every p > 3 if g(s) = 0 (for some s > 0) then H(s) > 0 (equivalently f := Q′ (s) < 0). Proof. Fix p > 3. By previous lemmas g has exactly three simple positive zeros, denote them 0 < s1 < s2 < s3 . Moreover, since g changes sign at each si , the sign pattern is g < 0 on (0, s1 ),
g > 0 on (s1 , s2 ),
g < 0 on (s2 , s3 ),
g > 0 on (s3 , ∞).
(23)
Since g(s) = sf ′ (s) and s > 0, this is also the sign pattern of f ′ (s). Step 1: Recall that W has exactly two positive simple zeros 0 < u1 < u2 . We know that 0 < u1 < r < u2 .
(24)
and W > 0 on (0, u1 ),
W < 0 on (u1 , u2 ), 17
W > 0 on (u2 , ∞).
(25)
Step 2: interlacing s1 < u1 < s2 < u2 < s3 . This we know because of the Rolle’s theorem we mentioned before (between any two positive consecutive zeros of g there is at least one zero of W ). Thus we obtain s1 < u1 < s2 < u2 < s3 . (26) Step 3: monotonicity of H from W = sH ′ . Since W (s) = sH ′ (s) and s > 0, sign(H ′ ) = sign(W ). Thus, (25) implies H is strictly increasing on (0, u1 ),
H is strictly decreasing on (u1 , u2 ),
H is strictly increasing on (u2 , ∞). (This is the key reason for introducing W , it will help us to show H(s2 ) > 0). Step 4: prove H(si ) > 0 for i = 1, 2, 3. (i) The middle root s2 . By (26) and (24) we have u1 < s2 < r < u2 (to get this order we are using that both W (r) and g(r) are negative). Since H is decreasing on (u1 , u2 ), we obtain H(s2 ) > H(r) > 0. (ii) The first root s1 . From (23), f ′ (s) < 0 on (0, s1 ) and f ′ (s) > 0 on (s1 , s2 ), so s1 is a strict local minimum of f , and f is strictly increasing on (s1 , s2 ). Also recall that sA < r, g(sA ) > 0, and g(r) < 0. Therefore sA ∈ (s1 , s2 ). Hence f (s1 ) < f (sA ) = Q′ (sA ) < 0.
(iii) The third root s3 . Again, from (23), f ′ (s) < 0 on (s2 , s3 ) and f ′ (s) > 0 on (s3 , ∞), so s3 is a strict local minimum of f , and f is strictly decreasing on (s2 , s3 ). Since g(r) < 0 and r < 1, while g(1) < 0 and g(s) → +∞ as s → ∞, the third zero satisfies s3 > 1 (indeed g is negative at 1 but eventually positive). Thus, 1 ∈ (s2 , s3 ) and monotonicity on (s2 , s3 ) yields f (s3 ) < f (1) = Q′ (1) < 0.
This finishes the proof of the lemma.
Graph of f → (s)/5000 (blue) and W (s)/20000 (red) for p = 4 8
f → (s)/5000 W (s)/20000
W
6 f→ > 0 4
f→ < 0 f→ > 0
2 sA s1
u1
r s2
u2 f→ < 0
→2
18
1
s s3
So to summarize, we needed the following list of technical inequalities whose proofs are given in appendices A and B: Q′ (1) < 0; Q′ (sA ) < 0; g(sA ) > 0; g(r) < 0; g(1) < 0; W (r) < 0; H(r) = rQ′′ (r) − (p + 1)Q′ (r) > 0; sign pattern of coefficients for g and sign pattern for W .
A
An application of Descartes’ generalized rule of signs
Proof of Lemma 3.2. We first show h(0) = 0. Substituting s = 0 into the definition of h(s) gives !c 1 2(2 · 0 + 1)p/2 (0 + 2)p p−1 − + 1 = 1c − 2 + 1 = 0. 0+2 0+2 Next, h(1) = 0. Substituting s = 1 yields !c 1 c 2 · 3p/2 3p p−1 + 1 = 2 · 3(p−2)/2 − 3 + 1. − 3 3 By the choice of c = c(p), c · ln 2 · 3(p−2)/2 = ln 2,
c so 2 · 3(p−2)/2 = 2. Therefore h(1) = 2 − 3 + 1 = 0. For h′ (1) = 0, note that the prefactor p/(sp + 2) in the given expression for h′ (s) evaluates to p/3 > 0 at s = 1. The bracket, however, consists of two summands: the first contains the factor sp + sp−1 − 2, which vanishes at s = 1; the second contains the factor sp−1 − 1, which also vanishes at s = 1. Hence the bracket is zero at s = 1, and h′ (1) = 0. To obtain h′′ (1), write p h′ (s) = p M (s), s +2 where M (s) denotes the bracket in the given formula for h′ (s). Since M (1) = 0, p h′′ (1) = M ′ (1). 3 Denote F (s) =
2(2s + 1)p/2 sp + 2
Then M (s) = −c F (s)
!c
,
G(s) =
(s + 2)p sp + 2
1
p−1
.
sp + sp−1 − 2 sp−1 − 1 + 2 G(s) . 2s + 1 (p − 1)(s + 2)
We have F (1) = 2 and G(1) = 3. Moreover, both fractions vanish at s = 1. Therefore, when differentiating, the product rule applied to each term yields (at s = 1) d d sp + sp−1 − 2 sp−1 − 1 ′ +6· . M (1) = −c · 2 · ds 2s + 1 ds (p − 1)(s + 2) s=1 s=1 For a quotient num/den with num(1) = 0, the derivative at s = 1 simplifies to num′ (1)/den(1). For the first fraction: num′ (s) = psp−1 + (p − 1)sp−2 , so num′ (1) = 2p − 1 and den(1) = 3. Thus its derivative at 1 is (2p − 1)/3. For the second fraction: num′ (s) = (p − 1)sp−2 , so num′ (1) = p − 1 and den(1) = 3(p − 1). Thus its derivative at 1 is 1/3. Hence 2p − 1 1 2c(2p − 1) M ′ (1) = −2c · +6· =2− . 3 3 3 It follows that p 2c(2p − 1) p(p − 2)(3 ln 3 − 4 ln 2)c ′′ h (1) = 2− = . 3 3 9 ln 2 19
Finally, for the limit as s → ∞: 2(2s + 1)p/2 ∼ 21+p/2 s−p/2 → 0, sp + 2 so the first summand raised to the power c tends to 0. For the second summand, (s + 2)p → 1, sp + 2 hence its (1/(p − 1))-th power tends to 1. Therefore h(s) → 0 − 1 + 1 = 0 as s → ∞. Proof of Lemma 3.10. Observe that since 26 < 34 , then 8 ln 3 − 8 ln 2 − 4 ln 2 < 3[4 ln 3 − 2 ln 2 4 5 8 6 ln 2] < [4 ln 3 − 6 ln 2]p. Thus c(p) = 2 ln 2+(p−2) ln 3 < 3p−2 . Similarly, since 3 < 2 , then 5 6 ln 2 + 10 ln 2 − 10 ln 3 < 3[8 ln 2 − 5 ln 3] < [8 ln 2 − 5 ln 3]p, or equivalently 4p−3 < c(p). Proof of Lemma 3.4. We start observing that ψ(0) = − log(2c−1 c(p − 1)) + log(2c−1 ) = − log((p − 1)c) < 0
(27)
1 by Lemma 3.10. By L’Hopital’s rule and (14) we also have since c > p−1 pc
(p − 1)3c+1− 2 ψ(1) = − log(2c−1 c(p − 1)) + log 2p − 1 " !# 3(1−p/2)c 3 2 = log (2p − 1)c 2c 3 = log > 0, (2p − 1)c
4 3 where the last inequality follows from Lemma 3.10, since 2p−1 > 3p−2 > c. Finally, for the limit as s → ∞, we observe c− 1 1 pc p−1 pc p 1 (s + 2) p−1 · sp−1 − 1 · (2s + 1) 1− 2 · sp + 2 ∼ log s p−1 +p−1+1− 2 +pc− p−1 −p log p p−1 s +s −2 pc − 1 log s. ∼ 2
Then lims→+∞ ψ(s) = −∞, since c < p2 by Lemma 3.10. Proof of Lemma 3.5. By Lemma 3.10 c<
2 4p − 3 16p − 12 < = 2 , p 2p(p − 1) 8p − 8p
2 then P (0) = 8cp2 − 8cp − 16p + 12 < 0. Moreover, P ′ (0) = 4p(cp − c − 2) < 0 since c < p2 < p−1 . ′ ′′ 2 Direct evaluation also shows that P (1) = P (1) = 0, and P (1) = −9(p − 2)(p − 1) < 0. Finally, since c < p2 , then the leading coefficient of P is positive, thus lims→∞ P (s) = +∞.
Proof of Lemma 3.7. Observe that Q(0) = 4(p − 1)2 (p − 2)(p − 3) > 0 since p > 3. Moreover, since c < p2 , the leading coefficient 3p(p − 1)(2 − cp) is positive, thus lims→∞ Q(s) = +∞.
20
Proof of Lemma 3.8. We define F (x) := (ln 3)(3x3 − 16x + 8) + (ln 4)(−5x2 + 14x − 4),
for x ∈ (3, ∞). We observe that
F ′′ (x) = 18x ln 3 − 10 ln 4 > 54 ln 3 − 10 ln 4 > 0,
for all x ∈ (3, ∞). Then F ′ is increasing, moreover
F ′ (x) = (ln 3)(9x2 − 16) + (ln 4)(−10x + 14).
Then, F ′ (x) ≥ 65 ln 3 − 16 ln 4 > 0 for all x ∈ (3, ∞). Thus, F is increasing, in particular Therefore Q′ (0) > 0, since c=
F (x) ≥ 41 ln 3 − 7 ln 4 > 0.
ln 4 3p2 + 6p − 4 6p2 + 12p − 8 < = . (p − 2) ln 3 + ln 4 8p(p − 1) 16p(p − 1)
Moreover, by Lemma 3.10 we have c < p2 , hence 3p(p − 1)(3p − 1)(2 − cp) > 0. Therefore the leading coefficient of Q′ (s) is positive, so lim Q′ (s) = +∞.
s→∞
In particular, Q′ (∞) > 0. Finally, direct evaluation shows that Q′ (1) = −18 (p − 1)2 (2p − 1)(2cp2 − cp − 2p − 2) < 0, 2(p+1) 5 since by Lemma 3.10 we have c > 4p−3 > p(2p−1) , where in the last inequality we used that 2 2p − 7p + 6 = (2p − 3)(p − 2) > 0 for all p ∈ (3, ∞).
Proof of Lemma 3.12. Fix p > 3 and write c = c(p). Recall (from the bounds on c(p) proved above) that 4 1 2 5 <c< and in particular <c< . (28) 4p − 3 3p − 2 p−1 p Starting from the explicit formula for Q written above, a direct differentiation gives that g(s) = sQ′′ (s) is a pseudopolynomial (generalized polynomial) with the following 10 monomials: g(s) = A1 s + A2 s2 + Ap−1 sp−1 + Ap sp + Ap+1 sp+1 + Ap+2 sp+2 + A2p−2 s2p−2 + A2p−1 s2p−1 + A2p s2p + A2p+1 s2p+1 , where the coefficients are (all prefactors below are strictly positive for p > 3) A1 = 2p(p − 1) −16c p(p − 1) − 6p2 + 24p + 4 , A2 = −24 p2 (p + 1) c(p − 1) + p − 2 , Ap−1 = p(p − 1)(2p − 2)(2p − 3) 10c p(p − 1) + 2p2 − 14p + 4 , Ap = p(p + 1)(2p − 1)(2p − 2) 17c p(p − 1) + 3p2 − 15p − 8 , Ap+1 = 2p(2p − 1)(p + 1)(p + 2) 8c p(p − 1) − 3p2 − 5 , Ap+2 = 2p2 (p − 1)(p + 2)(p + 3)(2p + 1) (c − 2), A2p−2 = −2p(2p − 1)(2p − 2)(3p − 3)(3p − 4) c(p − 1) − 1 , A2p−1 = 2p(2p − 1)(3p − 2)(3p − 3) −5c p(p − 1) + 5p + 1 , A2p = 2p(2p + 1)(3p − 1)(3p − 2) −4c p(p − 1) + 6p − 2 , A2p+1 = 3(2p + 2)(2p + 1)p(p − 1)(3p − 1) (2 − cp). 21
(29)
We now determine the signs of these coefficients using (28): 1 A) The two lowest-degree terms. Since c > p−1 , we have −16c p(p − 1) < −16p, hence
−16c p(p − 1) − 6p2 + 24p + 4 ≤ −16p − 6p2 + 24p + 4 = −2(3p2 − 4p − 2) < 0, so A1 < 0. Also c(p − 1) + p − 2 > 0, so A2 < 0.
1 , we get B) The sp−1 and sp coefficients. Using c > p−1
10c p(p − 1) + 2p2 − 14p + 4 > 10p + 2p2 − 14p + 4 = 2(p2 − 2p + 2) > 0, hence Ap−1 > 0. Similarly, 17c p(p − 1) + 3p2 − 15p − 8 > 17p + 3p2 − 15p − 8 = 3p2 + 2p − 8 > 0, so Ap > 0. C) The sp+1 and sp+2 coefficients. From c < p2 we have 8c p(p − 1) < 16(p − 1), and for p > 3, 16(p − 1) < 3p2 + 5
=⇒
8c p(p − 1) − 3p2 − 5 < 0,
so Ap+1 < 0. Also c < p2 < 1 implies c − 2 < 0, hence Ap+2 < 0.
1 D) The s2p−2 and s2p−1 coefficients. Since c > p−1 , we have c(p − 1) − 1 > 0, thus A2p−2 < 0. 5 For A2p−1 we use the stronger lower bound c > 4p−3 :
5c p(p − 1) >
25p(p − 1) ≥ 5p + 1 4p − 3
(p ≥ 3),
so −5c p(p − 1) + 5p + 1 < 0 and therefore A2p−1 < 0.
4 E) The top two coefficients. Using the upper bound c < 3p−2 we obtain
4c p(p − 1) <
16p(p − 1) < 6p − 2, 3p − 2
hence −4c p(p − 1) + 6p − 2 > 0 and A2p > 0. Finally c < p2 gives cp < 2, so 2 − cp > 0 and thus A2p+1 > 0. Collecting the signs in the increasing-exponent order 1, 2, p − 1, p, p + 1, p + 2, 2p − 2, 2p − 1, 2p, 2p + 1, we get the sign pattern sign
(A1 , A2 , Ap−1 , Ap , Ap+1 , Ap+2 , A2p−2 , A2p−1 , A2p , A2p+1 ) = (−, −, +, +, −, −, −, −, +, +), which has exactly 3 sign changes. (When 3 < p < 4, the order of the two exponents p + 2 and 2p − 2 swaps, but both coefficients are negative, so the number of sign changes is unchanged.) By the generalized Descartes rule of signs for pseudopolynomials associated with a Chebyshev system of monomials {sα } on (0, ∞) the number of positive zeros of g counted with multiplicity is bounded by the number of sign changes of its coefficient sequence. Therefore g has at most 3 positive zeros (with multiplicity), as claimed.
22
Proof of Lemma 3.14. We show that W has at most 2 positive roots counted with multiplicities. Recall g(s) = sQ′′ (s) and H(s) = sQ′′ (s) − (p + 1)Q′ (s), and define W = sH ′ . Since g = sQ′′ , we can write d W (s) = sH ′ (s) = s − (p + 1) g(s). ds Using the expansion (29) from the previous lemma, we obtain X W (s) = Aα (α − (p + 1)) sα . α∈{1,2,p−1,p,p+1,p+2,2p−2,2p−1,2p,2p+1}
The term with α = p+1 vanishes identically, so W is again a pseudopolynomial with 9 monomials and exponents 1, 2, p − 1, p, p + 2, 2p − 2, 2p − 1, 2p, 2p + 1
(ordered increasingly).
Since α−(p+1) < 0 for α ∈ {1, 2, p−1, p} and α−(p+1) > 0 for α ∈ {p+2, 2p−2, 2p−1, 2p, 2p+1}, the coefficient signs of W follow from the signs of Aα proved in Lemma 3.12: (+, +, −, −, −, −, −, +, +), which has exactly 2 sign changes. By the generalized Descartes rule of signs for pseudopolynomials it follows that W has at most two positive zeros counted with multiplicity. Finally, observe that the leading coefficient of W (s) is pA2p+1 > 0 thus lims→∞ W (s) = +∞, and for small values of s (in a neighborhood of 0) we have W (s) ∼ −pA1 > 0. Then, by Lemma 3.15 and the intermediate value theorem, W has a zero u1 ∈ (0, r) and another zero u2 ∈ (r, +∞). Moreover, W (s) > 0 for all s ∈ (0, u1 ), W (s) < 0 for all s ∈ (u1 , u2 ), and W (s) > 0 for all s ∈ (u2 , +∞). Proof of Lemma 3.13 (i). The leading coefficient of g is 3p(p−1)(3p−1)(2−cp)(2p+2)(2p+1) > 0 (since c < p2 by Lemma 3.10), then lims→∞ g(s) = +∞. Moreover, g(1) = Q′′ (1) = −6p(p − 1)2 8c(2p − 1)(3p2 − 2p + 1) − (54p2 − p + 2) . Observe that ln 4 (48p3 − 56p2 + 32p − 8) − (54p2 − p + 2) (ln 3)(p − 2) + ln 4
= (p − 2)[(96 ln 2 − 54 ln 3)p2 + (ln 3 − 28 ln 2)p + (10 ln 2 − 2 ln 3)]
=: (p − 2)G(p) > 0
for all p > 3, since G′ (p) = 2(96 ln 2 − 54 ln 3)p + (ln 3 − 28 ln 2) ≥ 6(96 ln 2 − 54 ln 3) + (ln 3 − 28 ln 2) = 548 ln 2 − 323 ln 3 > 0 and G(3) = 5(158 ln 2 − 97 ln 3) > 0. Then c=
ln 4 54p2 − p + 2 > , (ln 3)(p − 2) + ln 4 48p3 − 56p2 + 32p − 8
for all p > 3, thus g(1) < 0.
B
Around Sturm’s algorithm
Here we discuss the proofs of the most technical ingredients of our argument: the proof of Lemma 3.13 (ii), Lemma 3.16 and Lemma 3.15.
23
Lemma B.1 (part of Lemma 3.16 Q′ (sA ) < 0). Let p > 3, c = c(p) = sA =
p−1 . Then p+1
2 ln 2 , and (p − 2) ln 3 + 2 ln 2
Q′ (sA ) < 0.
Proof. Throughout the proof we write x := sAp =
p−1 p+1
p ∈ (0, 1).
Recall that c(p) < c̄ :=
4 3p − 2
and
1 17 <x< 8 125
(p > 3).
Step 1: Q′ (sA ) is strictly increasing in the parameter c. For this step we temporarily regard c as a free parameter (so Q depends on c). A direct differentiation of the explicit formula for Q(s) and substitution s = sA gives an affine dependence on c; more precisely, ∂ ′ 2p up (x), Q (sA ) = − ∂c (p + 1)2
up (x) := Au (p) x2 + Bu (p) x + Cu (p),
(30)
where Au (p) := 108p6 + 114p5 − 154p4 − 186p3 + 14p2 + 48p + 8,
Bu (p) := −72p6 − 36p5 + 213p4 + 56p3 − 150p2 − 28p + 17, Cu (p) := 30p5 − 60p4 − 16p3 + 76p2 − 14p − 16.
∂ We claim that up (x) < 0 for the x in (32). This will imply ∂c Q′ (sA ) > 0, i.e. Q′ (sA ) is strictly increasing in c. 17 (i) up is strictly decreasing on [0, 125 ]. Indeed, u′p (x) = 2Au (p)x + Bu (p) is increasing in x 17 , (the polynomial q 7→ Au (3 + q) has positive coefficients for q > 0), hence for 0 ≤ x ≤ 125 17 ′ ′ up (x) ≤ up . 125
A direct computation yields 125 u′p
17 125
= 34Au (p) + 125Bu (p) =: D(p),
where D(p) = −5328p6 − 624p5 + 21389p4 + 676p3 − 18274p2 − 1868p + 2397. Let q := p − 3 ≥ 0. Expanding D(p) at p = 3 gives D(p) = −(5328q 6 + 96528q 5 + 707251q 4 + 2675936q 3 + 5499184q 2 + 5804192q + 2452656) < 0. 17 17 Hence u′p 125 < 0, so u′p (x) < 0 for all x ∈ [0, 125 ], and up is strictly decreasing there. (ii) up ( 81 ) < 0. Compute 1 64 up = Au (p) + 8Bu (p) + 64Cu (p) =: H(p), 8
where H(p) = −468p6 + 1746p5 − 2290p4 − 762p3 + 3678p2 − 1072p − 880. 24
With q = p − 3 ≥ 0 one finds H(p) = −(468q 6 + 6678q 5 + 39280q 4 + 123822q 3 + 224040q 2 + 222112q + 93952) < 0, hence up ( 81 ) < 0. 17 Since up is decreasing on [0, 125 ] and x > 81 by (32), we get 1 < 0. up (x) ≤ up 8
Therefore, by (30), ∂ ′ Q (sA ) > 0 ∂c so c 7→ Q′ (sA ) is strictly increasing.
(p > 3),
4 Step 2: estimate Q′ (sA ) at the upper bound c̄ = 3p−2 . A direct simplification of Q′ (sA ) at c = c̄ yields 2 Q′ (sA ) c=c̄ = − gp (x), (31) (p − 1)(p + 1)2 (3p − 2)
where
gp (x) := A(p) x2 + B(p) x + C(p) with A(p) := 27p8 + 15p7 − 301p6 − 110p5 + 465p4 + 127p3 − 171p2 − 8p + 4,
B(p) := −126p8 + 117p7 + 675p6 − 924p5 − 220p4 + 617p3 − 121p2 − 34p + 16, C(p) := 27p8 − 78p7 + 30p6 + 68p5 − 37p4 + 10p3 − 36p2 + 16.
Since (p − 1)(p + 1)2 (3p − 2) > 0 for p > 3, it suffices to show gp (x) > 0.
17 (i) gp is strictly decreasing on [0, 125 ]. Indeed, gp′ (x) = 2A(p)x + B(p). Next we notice that A(p) > 0 for p > 3 because the polynomial q 7→ A(q + 3) has nonnegative coefficients. Therefore, we can use the estimate 17 34 ′ ′ gp (x) ≤ gp = A(p) + B(p) < 4A(p) + B(p), 125 125 34 since 125 < 4. Thus it is enough to show 4A(p) + B(p) < 0, i.e. −B(p) − 4A(p) > 0. A direct computation gives
−B(p) − 4A(p) = 18p8 − 177p7 + 529p6 + 1364p5 − 1640p4 − 1125p3 + 805p2 + 66p − 32. Let q = p − 3 ≥ 0 and expand at p = 3: −B(p)−4A(p) = 18q 8 +255q 7 +1348q 6 +4649q 5 +25030q 4 +130764q 3 +377320q 2 +532800q+292288 > 0. 17 Hence gp′ (x) < 0 on [0, 125 ], and gp is strictly decreasing there. 17 (ii) gp ( 125 ) > 0. Compute 17 1 N (p) gp = 289A(p) + 2125B(p) + 15625C(p) = , 125 15625 15625
where N (p) =161928p8 − 965790p7 + 1816136p6 − 932790p5 − 911240p4 + 1504078p3 − 869044p2 − 74562p + 285156.
25
Let q = p − 3 ≥ 0. Expanding N (p) at p = 3 yields N (p) =161928q 8 + 2920482q 7 + 22340402q 6 + 94058484q 5 + 235735480q 4 + 352833112q 3 + 295060604q 2 + 111719616q + 6561792 > 0. 17 ) > 0. Therefore gp ( 125 17 17 Finally, since x < 125 by (32) and gp is decreasing on [0, 125 ], we have 17 > 0. gp (x) > gp 125
Plugging this into (31) gives Q′ (sA ) c=c̄ < 0. Step 3: conclude for c = c(p). By Step 1, c 7→ Q′ (sA ) is strictly increasing, and by Step 1, c(p) < c̄. Hence Q′ (sA ) c=c(p) < Q′ (sA ) c=c̄ < 0. This proves the lemma. Lemma B.2 (part of Lemma 3.13 (ii): g(sA ) > 0). For all p > 3 we have g(sA ) > 0. Proof. First we show g(sA ) > 0 i.e., Q′′ (sA ) > 0. Set p−1 p p ρ := sA = ∈ (0, 1). p+1 Step 1: Bounds on ρ. Define X(p) :=
p−1 p+1
p
for p > 1. Then
d p−1 2p p+1 1 1 log X(p) = log + 2 = − log + + . dp p+1 p −1 p−1 p−1 p+1 Since t 7→ 1t is strictly convex on (0, ∞), the trapezoid rule gives Z p+1
dt 2 1 1 1 1 < + = + . 2 p−1 p+1 p−1 p+1 p−1 t d But the left-hand side equals log p+1 p−1 , hence dp log X(p) > 0. Thus X(p) is strictly increasing on (1, ∞). In particular, for p > 3, 3 1 2 x = X(p) > X(3) = = . 4 8 Also lim X(p) = e−2 (standard limit), hence x < X(p) < e−2 for all finite p. Moreover p→∞ 17 −2 e < 125 . Therefore we have the uniform bounds
1 17 1 <x< < 8 125 7
(p > 3).
(32)
Step 2: An explicit formula for Q′′ (sA ). A direct differentiation of the explicit formula for Q(s) (given in the text) and substitution s = sA yields the following representation: Q′′ (sA ) =
i 2p h 2 (c a (p) + b (p))ρ + (c a (p) + b (p))ρ + (c a (p) + b (p)) , 2 2 1 1 0 0 p2 − 1 26
(33)
where a2 (p) := −(p − 1)(p + 1)2 216p4 − 60p3 − 173p2 + 23p + 24 , a1 (p) := p(p − 1)2 72p4 + 180p3 + 75p2 − 106p − 69 , a0 (p) := −28p(p − 1)3 (p + 1),
b2 (p) := 270p6 + 393p5 − 168p4 − 398p3 − 50p2 + 77p + 20, b1 (p) := −72p6 − 72p5 + 98p4 + 80p3 − 4p2 − 24p − 6, b0 (p) := −18p5 + 66p4 − 26p3 − 70p2 + 44p + 4.
Since p22p−1 > 0 for p > 3, it suffices to prove that the bracket in (33) is positive. Step 3: The bracket is increasing in c for ρ ∈ [1/8, 1/7]. Let F (ρ) := a2 (p)ρ2 + a1 (p)ρ + a0 (p). Then the bracket in (33) equals c F (ρ) + G(ρ), where G(ρ) := b2 ρ2 + b1 ρ + b0 . Because a2 (p) < 0 for p > 3, F is concave in ρ. Compute 1 a + 8a + 64a (p − 1) 2 1 0 F = = 360p6 + 492p5 − 2555p4 + 727p3 + 2191p2 − 1311p − 24 , 8 64 64 and 1 a + 7a + 49a 2(p − 1) 2 1 0 F = = 144p6 + 192p5 − 1015p4 + 244p3 + 867p2 − 480p − 12 . 7 49 49 Writing p = 3 + t with t > 0, 360p6 + 492p5 − 2555p4 + 727p3 + 2191p2 − 1311p − 24 = 360t6 + 6972t5 + 53425t4 + 208747t3 + 441004t2 + 479664t + 210432 > 0,
144p6 + 192p5 − 1015p4 + 244p3 + 867p2 − 480p − 12 = 144t6 + 2784t5 + 21305t4 + 83104t3 + 175053t2 + 189402t + 82356 > 0.
Hence F (1/8) > 0 and F (1/7) > 0 for all p > 3. By concavity, F (ρ) > 0 for every ρ ∈ [1/8, 1/7], and therefore the bracket in (33) is strictly increasing as a function of c on that interval. Step 4: A convenient lower bound for c(p). Set c0 :=
5 . 4p − 3
We claim that c(p) > c0 for all p > 3. Indeed, 2 ln 2 5 > ⇐⇒ (8 ln 2 − 5 ln 3)(p − 2) > 0, (p − 2) ln 3 + 2 ln 2 4p − 3 and 8 ln 2 − 5 ln 3 = ln(256/243) > 0, so the claim holds for p > 2, in particular for p > 3. Since the bracket in (33) is increasing in c (Step 3), it is enough to show that it is positive at c = c0 . Step 5: Positivity at c = c0 on ρ ∈ [1/8, 1/7]. Define E0 (ρ) := (c0 a2 + b2 )ρ2 + (c0 a1 + b1 )ρ + (c0 a0 + b0 ). Then E0 is a quadratic polynomial in ρ. Moreover, (p − 1) 72p6 + 180p5 − 277p4 − 631p3 + 203p2 + 279p − 18 c0 a1 + b1 = >0 4p − 3 27
(p > 3),
since with p = 3 + t, 72p6 + 180p5 − 277p4 − 631p3 + 203p2 + 279p − 18
= 72t6 + 1476t5 + 12143t4 + 51125t3 + 115646t2 + 132420t + 59400 > 0.
If c0 a2 + b2 ≥ 0, then for ρ ≥ 0 we have E0′ (ρ) = 2(c0 a2 + b2 )ρ + (c0 a1 + b1 ) ≥ c0 a1 + b1 > 0, so E0 is increasing on [0, ∞) and hence on [1/8, 1/7]. If c0 a2 + b2 ≤ 0, then E0 is concave and its minimum on [1/8, 1/7] is attained at an endpoint. Thus in all cases h1 1i 1 1 E0 (ρ) ≥ min E0 for every ρ ∈ , . , E0 8 7 8 7 A direct simplification gives 1 N8 (p) E0 = , N8 (p) = 288p7 −1881p6 +4065p5 −1899p4 −3331p3 +3490p2 −378p−282, 8 32(4p − 3) and 1 N7 (p) = , E0 7 49(4p − 3)
N7 (p) = 504p7 −2790p6 +5917p5 −3114p4 −4466p3 +5314p2 −819p−402.
Writing p = 3 + t with t > 0, N8 (3 + t) = 288t7 + 4167t6 + 24639t5 + 77301t4 + 140471t3 + 152764t2 + 99024t + 32640 > 0, N7 (3 + t) = 504t7 + 7794t6 + 50953t5 + 185271t4 + 412936t3 + 576616t2 + 474648t + 178320 > 0. Hence E0 (1/8) > 0 and E0 (1/7) > 0 for all p > 3, so E0 (ρ) > 0 for all ρ ∈ [1/8, 1/7].
Step 6: Conclusion. For p > 3 we have ρ ∈ (1/8, 1/7) (Step 1), c(p) > c0 (Step 4), and the bracket in (33) is increasing in c (Step 3). Therefore, (ca2 + b2 )ρ2 + (ca1 + b1 )ρ + (ca0 + b0 ) > (c0 a2 + b2 )ρ2 + (c0 a1 + b1 )ρ + (c0 a0 + b0 ) = E0 (ρ) > 0. Using (33) and p22p−1 > 0, we conclude Q′′ (sA ) > 0. Lemma B.3 (the remaining part of Lemma 3.13 (ii): g(r) < 0). Let p > 3, c = c(p) = Then
2 ln 2 , (p − 2) ln 3 + 2 ln 2
r=
p . p+1
Q′′ (r) < 0. Proof. Set
p p ∈ (0, 1). p+1 A direct differentiation of the explicit formula for Q(s) (given in the manuscript) shows that, after evaluating at s = r and grouping the terms with powers r2p and rp , one can write x := rp =
Q′′ (r) = U (p, c) x2 + V (p, c) x + W (p, c),
(34)
where the coefficients are (affine in c) 96 48 U (p, c) = 540p5 − 6p4 − 852p3 + 200p2 + 456p − 194 − + 2 (35) p p 132 48 + 2 , + c −432p6 + 408p5 + 694p4 − 898p3 − 178p2 + 622p − 132 − p p 2 V (p, c) = c(72p8 + 36p7 − 147p6 − 11p5 + 114p4 − 57p3 − 37p2 + 30p) (36) p(p + 1) + (−90p7 − 99p6 + 82p5 + 24p4 − 66p3 + 35p2 + 30p − 12) , W (p, c) = 4p −9p3 + 27p2 − 10p − 2 + c(−14p3 + 22p2 − 8p) . (37) 28
In particular, for fixed p and x, the right-hand side of (34) is affine in c. Step 1: bounds for x = rp . Define ϕ(p) = p ln(1 + 1/p), so that x(p) = e−ϕ(p) . Then ϕ′ (p) = ln(1 + 1/p) −
1 >0 p+1
u (using ln(1 + u) > 1+u for u > 0), hence ϕ is increasing and x(p) is decreasing. Therefore for p ≥ 3, 3 3 27 x(p) ≤ x(3) = (38) = . 4 64 Also, since ln(1 + u) < u for u > 0, we have p ln(1 + 1/p) < 1, hence (1 + 1/p)p < e and therefore
1 1 > > 9/25. (1 + 1/p)p e
(39)
27 9 < x(p) ≤ . 25 64
(40)
x(p) = Thus, for all p > 3,
Next recall that
5 2 ≤ c(p) ≤ 4p − 3 p
(p > 3).
(41)
Step 2: reduce to two endpoint values of c. Fix p > 3 and x = x(p). Because Q′′ (r) is affine in c by (34), and because c(p) ∈ [c1 , c2 ] where c1 :=
5 , 4p − 3
c2 :=
2 , p
it suffices to show Q′′ (r) c=c1 < 0
and
Q′′ (r) c=c2 < 0.
We prove these two inequalities separately. Step 3: endpoint c = c1 = 5/(4p − 3). Substitute c = c1 into (34) and write Q′′ (r) c=c1 = q1 (x) := U1 (p)x2 + V1 (p)x + W1 (p), where the coefficients simplify to 2 198p7 + 40p6 − 567p5 + 167p4 + 483p3 − 231p2 − 90p + 48 U1 (p) = , p2 (4p − 3) 2 54p7 − 110p6 − 205p5 + 234p4 + 53p3 − 170p2 + 12p + 36 V1 (p) = , p(4p2 + p − 3) 4p −36p4 + 65p3 − 11p2 − 18p + 6 W1 (p) = . 4p − 3 We claim U1 (p) > 0 and V1 (p) > 0 for all p ≥ 3. Indeed, setting p = 3 + y with y ≥ 0, 198p7 + 40p6 − 567p5 + 167p4 + 483p3 − 231p2 − 90p + 48
= 198y 7 + 4198y 6 + 37575y 5 + 184172y 4 + 534387y 3 + 919038y 2 + 868680y + 348672 > 0,
54p7 − 110p6 − 205p5 + 234p4 + 53p3 − 170p2 + 12p + 36
= 54y 7 + 1024y 6 + 8021y 5 + 33339y 4 + 78101y 3 + 99505y 2 + 57852y + 7020 > 0.
Since the denominators are also positive for p > 3, this proves U1 (p) > 0 and V1 (p) > 0. Therefore q1′ (x) = 2U1 (p)x + V1 (p) > 0 (x > 0), 29
so q1 is strictly increasing in x. Using (38) we get ′′
Q (r) c=c1 = q1 (x(p)) ≤ q1
27 64
.
A direct computation yields q1
27 64
=
N1 (p) , 2 2048 p (4p2 + p − 3)
where N1 (p) = − 57258p8 + 220990p7 − 296055p6 − 124816p5 + 467130p4 − 60900p3 − 213273p2 + 31590p + 34992.
The denominator is positive for p > 3, so it suffices to prove N1 (p) < 0 for p > 3. Differentiate: N1′ (p) = −2 G(p), where G(p) = 229032p7 − 773465p6 + 888165p5 + 312040p4 − 934260p3 + 91350p2 + 213273p − 15795. Again writing p = 3 + y (y ≥ 0) gives G(3 + y) =229032y 7 + 4036207y 6 + 30252843y 5 + 125651980y 4 + 314379690y 3 + 477328041y 2 + 409981824y + 154357488 > 0. Hence G(p) > 0 for p ≥ 3, so N1′ (p) < 0 for p ≥ 3, i.e. N1 is strictly decreasing on [3, ∞). Finally, N1 (3) = −104115456 < 0, so N1 (p) < 0 for all p > 3. Therefore q1 (27/64) < 0, hence Q′′ (r) c=c1 < 0. Step 4: endpoint c = c2 = 2/p. Substitute c = c2 into (34) and write Q′′ (r) c=c2 = q2 (x) := U2 (p)x2 + V2 (p)x + W2 (p). The leading coefficient simplifies and factors as U2 (p) = −
2(p + 1) A2 (p), p3
A2 (p) := 162p7 −567p6 +299p5 +499p4 −549p3 +24p2 +156p−48.
With p = 3 + y (y ≥ 0) we have A2 (3 + y) = 162y 7 + 2835y 6 + 20711y 5 + 81529y 4 + 185439y 3 + 240540y 2 + 160464y + 39840 > 0, so A2 (p) > 0 for p ≥ 3, hence U2 (p) < 0 for p > 3. Thus q2 is concave in x and q2′ (x) = 2U2 (p)x + V2 (p) is strictly decreasing in x. We claim q2′ (9/25) < 0 for p > 3. Indeed one computes q2′
9 25
=−
2 25 p3 (p + 1)
30
R2 (p),
where R2 (p) = 1566p9 − 3699p8 − 6814p7 + 9490p6 + 9414p5 − 8375p4 − 5110p3 + 3984p2 + 1080p − 864.
With p = 3 + y (y ≥ 0), R2 (3 + y) = 1566y 9 + 38583y 8 + 411794y 7 + 2485936y 6 + 9282096y 5 + 21949213y 4 + 32108144y 3 + 26596356y 2 + 9612792y + 79920 > 0, so R2 (p) > 0 for p ≥ 3 and hence q2′ (9/25) < 0 for p > 3. Since q2′ is decreasing in x, it follows that 9 9 q2′ (x) ≤ q2′ <0 for all x ≥ , 25 25 so q2 is strictly decreasing on [9/25, ∞). Thus Q′′ (r) c=c2 = q2 (x(p)) ≤ q2 Finally, one computes q2
9 25
=−
2 625 p3 (p + 1)
9 25
.
P2 (p),
where P2 (p) = 972p9 − 2358p8 + 5687p7 + 1230p6 + 4138p5 − 6300p4 − 18045p3 + 12528p2 + 4860p − 3888.
With p = 3 + y (y ≥ 0), P2 (3 + y) = 972y 9 + 23886y 8 + 264023y 7 + 1730937y 6 + 7456057y 5 + 21986871y 4 + 44499348y 3 + 59538402y 2 + 47542464y + 17126640 > 0, so P2 (p) > 0 for p ≥ 3 and therefore q2 (9/25) < 0 for p > 3. Hence Q′′ (r) c=c2 < 0. Step 5: conclude for c = c(p). By (41) we have c(p) ∈ [c1 , c2 ]. Since Q′′ (r) is affine in c by (34), and since it is negative at both endpoints c = c1 and c = c2 by Steps 3–4, it follows that Q′′ (r) c=c(p) < 0. This completes the proof. Lemma B.4 (the remainin part of Lemma 3.16: H(r) > 0). Let p > 3, set r=
p , p+1
c = c(p) =
2 ln 2 . (p − 2) ln 3 + 2 ln 2
Let Q be the pseudo–polynomial defined above (so that P ′′ (s) = sp−4 Q(s)). Then r Q′′ (r) − (p + 1) Q′ (r) > 0.
31
Proof. Define For a monomial asα one has
H(s) := sQ′′ (s) − (p + 1)Q′ (s). d2 (asα ) = aα(α − 1)sα−2 , ds2
d (asα ) = aαsα−1 , ds hence
d2 d (asα ) − (p + 1) (asα ) = aα(α − p − 2)sα−1 . 2 ds ds p Applying this term-by-term to the explicit formula for Q and then substituting r = p+1 , we obtain the exact identity s
H(r) =
2(p − 1) S(p, c, u), (p + 1)3
u := rp−1 =
p p−1 ∈ (0, 1), p+1
(42)
where S is a quadratic polynomial in u: S(p, c, u) = A(p, c) + B(p, c) u + C(p, c) u2 , with A(p, c) = 30cp7 + 30cp6 − 54cp5 − 70cp4 + 8cp3 + 40cp2 + 16cp + 9p7 − 24p6 − 66p5 + 57p3 + 16p2 − 16p − 8,
B(p, c) = − 72cp7 − 64cp6 + 157cp5 + 148cp4 − 80cp3 − 77cp2
− 18p7 + 45p6 + 156p5 + 8p4 − 148p3 − 51p2 + 16p,
C(p, c) = − 108cp8 + 138cp7 + 217cp6 − 307cp5 − 154cp4 + 199cp3 + 39cp2 − 36cp + 135p7 − 33p6 − 294p5 + 32p4 + 207p3 + 13p2 − 36p.
> 0 for p > 3, it suffices to prove Since 2(p−1) (p+1)3 S(p, c(p), u(p)) > 0.
(43)
3 Step 1: rational bounds for c(p). Write a := log2 3 = ln ln 2 . Since
35 = 243 < 256 = 28 we have
Using c(p) =
312 = 531441 > 524288 = 219 ,
and
19 8 <a< . 12 5 2 and that c is decreasing in a, it follows that 2 + (p − 2)a c− :=
5 24 < c(p) < c+ := . 4p − 3 19p − 14
(44)
Step 2: an explicit upper bound for u(p). Let n := p − 1 > 2 and x := 1/p ≥ 0. Consider n(n − 1) 2 f (x) := (1 + x)n − 1 + nx + x . 2 Then f (0) = f ′ (0) = f ′′ (0) = 0, and for x ≥ 0, f ′′′ (x) = n(n − 1)(n − 2)(1 + x)n−3 ≥ 0. 32
Hence f ′′ is increasing with f ′′ (0) = 0, so f ′′ ≥ 0; thus f ′ is increasing with f ′ (0) = 0, so f ′ ≥ 0; thus f is increasing with f (0) = 0, so f ≥ 0. Therefore, for x ≥ 0, (1 + x)n ≥ 1 + nx +
n(n − 1) 2 x . 2
With x = 1/p and n = p − 1 this gives
1+
1 p−1 p − 1 (p − 1)(p − 2) 5p2 − 5p + 2 ≥1+ = . + p p 2p2 2p2
Taking reciprocals yields the bound u=
p p−1 1 2p2 = ≤ U (p) := . p+1 (1 + 1/p)p−1 5p2 − 5p + 2
(45)
Step 3: S is decreasing in u on [0, U (p)] for c = c(p). Since S(p, c, u) = A + Bu + Cu2 is quadratic in u, ∂S (p, c, u) = B(p, c) + 2C(p, c) u ∂u is linear in u. Hence it suffices to show ∂S (p, c(p), 0) < 0 ∂u
∂S (p, c(p), U (p)) < 0. ∂u
and
(46)
(i) The endpoint u = 0. We have ∂S ∂u (p, c, 0) = B(p, c). The coefficient of c in B equals −72p7 − 64p6 + 157p5 + 148p4 − 80p3 − 77p2 = −p2 72p5 + 64p4 − 157p3 − 148p2 + 80p + 77 . For p = 3 + x with x ≥ 0, 72p5 +64p4 −157p3 −148p2 +80p+77 = 72x5 +1144x4 +7091x3 +21335x2 +31025x+17426 > 0, so B(p, c) is strictly decreasing in c for all p ≥ 3. Using c(p) ≥ c− from (44) gives B(p, c(p)) ≤ 5 B(p, c− ). A direct substitution c = c− = 4p−3 yields B(p, c− ) = −
p NB (p), 4p − 3
where NB (p) := 72p7 + 126p6 − 169p5 − 349p4 − 124p3 + 160p2 + 168p + 48. For p = 3 + x with x ≥ 0, NB (3 + x) = 72x7 + 1638x6 + 15707x5 + 82166x4 + 252638x3 + 455074x2 + 442767x + 178626 > 0, hence B(p, c− ) < 0, and therefore B(p, c(p)) < 0. (ii) The endpoint u = U (p). Set D(p, c) := B(p, c) + 2C(p, c) U (p) = A simplification gives D(p, c) = − with
∂S (p, c, U (p)). ∂u
p M (p) c + K(p) , 5p2 − 5p + 2
M (p) := 432p9 − 192p8 − 908p7 + 267p6 + 789p5 + 30p4 − 467p3 − 81p2 + 154p. 33
For p = 3 + x with x ≥ 0, M (3 + x) =432x8 + 10176x7 + 103924x6 + 600819x5 + 2150214x4 3+x + 4877544x3 + 6849487x2 + 5445906x + 1877824 > 0, so M (p) > 0 for p ≥ 3, and thus D(p, c) is strictly decreasing in c. Using again c(p) ≥ c− gives 5 yields D(p, c(p)) ≤ D(p, c− ). Substituting c = c− = 4p−3 D(p, c− ) = −
p ND (p), (4p − 3)(5p2 − 5p + 2)
where ND (p) := 360p9 − 342p8 − 1363p7 + 1452p6 + 939p5 − 982p4 − 256p3 + 8p2 + 96p + 96. For p = 3 + x with x ≥ 0, ND (3 + x) =360x9 + 9378x8 + 107069x7 + 703125x6 + 2926524x5 + 8004428x4 + 14383469x3 + 16369613x2 + 10703106x + 3061824 > 0, hence D(p, c− ) < 0, and therefore D(p, c(p)) < 0. Combining (i) and (ii) proves (46). Since ∂S/∂u is linear in u, it follows that ∂S/∂u < 0 for all u ∈ [0, U (p)], so S(p, c(p), u) is strictly decreasing on [0, U (p)]. Using u(p) ≤ U (p) from (45) we obtain S(p, c(p), u(p)) ≥ S(p, c(p), U (p)). (47) Step 4: replace c(p) by the upper bound c+ . Since S is affine in c, it suffices to compute its c–coefficient at u = U (p). A direct simplification gives ∂S 2p(p − 1) p, c, U (p) = − N2 (p), ∂c (5p2 − 5p + 2)2 where N2 (p) := 216p10 − 75p9 − 174p8 + 229p7 − 140p6 − 107p5 + 179p4 + 54p3 − 72p2 − 48p + 32. For p = 3 + x with x ≥ 0, N2 (3 + x) =216x10 + 6405x9 + 85281x8 + 671593x7 + 3464881x6 + 12239092x5 + 29980589x4 + 50292735x3 + 55296009x2 + 35983743x + 10524704 > 0, hence ∂S ∂c (p, c, U (p)) < 0 for all p > 3. Therefore S(p, c, U (p)) is strictly decreasing in c. Using c(p) ≤ c+ from (44) yields S(p, c(p), U (p)) ≥ S(p, c+ , U (p)),
c+ =
24 . 19p − 14
(48)
Step 5: positivity at (c+ , U (p)). Substituting c = c+ and u = U (p) into S = A + Bu + Cu2 simplifies to (p − 2)(p − 1) S(p, c+ , U (p)) = R(p), (19p − 14)(5p2 − 5p + 2)2
where
R(p) := 747p10 −2469p9 +102p8 +4954p7 −1385p6 −2997p5 +2124p4 +380p3 −1096p2 +128p+224. 34
For p = 3 + x with x ≥ 0, R(3 + x) =747x10 + 19941x9 + 235974x8 + 1627726x7 + 7235131x6 + 21607281x5 + 43792452x4 + 59291840x3 + 51146048x2 + 25304972x + 5451008 > 0, hence R(p) > 0 for all p ≥ 3, and therefore S(p, c+ , U (p)) > 0 for all p > 3. Finally, combining (47) and (48) gives S(p, c(p), u(p)) ≥ S(p, c(p), U (p)) ≥ S(p, c+ , U (p)) > 0, which proves (43). Returning to (42) yields H(r) > 0, i.e. rQ′′ (r) − (p + 1)Q′ (r) > 0. Lemma B.5 (the proof of Lemma 3.15: W (r) < 0). For all p > 3 we have W (r) < 0. Proof. Recall H(s) = sQ′′ (s) − (p + 1)Q′ (s) and W (s) = sH ′ (s). Differentiating gives H ′ (s) = sQ′′′ (s) − pQ′′ (s)
=⇒
W (s) = s2 Q′′′ (s) − psQ′′ (s).
Since each coefficient of Q(s) is affine linear in c (hence in particular linear in c after differentiation), it follows that for fixed p the quantity W (r) is an affine linear function of c. Define the two rational endpoints cL :=
5 , 4p − 3
cU :=
4 . 3p − 2
By Lemma 3.10 we have c(p) ∈ (cL , cU ). Since c 7→ W (r) is affine linear, it is enough to prove W (r) c=c < 0
and
L
Set x := rp =
p p+1
W (r) c=c < 0. U
(∗)
p . By (38) and (39), we have x∈
1 27 1 27 , ⊂ , . e 64 3 64
A direct (symbolic) computation from the explicit formula for Q yields the following closed forms (here p > 3 and x = rp ): W (r) c=c = L
2(p − 1) PL (p, x), p(p + 1)2 (4p − 3)
W (r) c=c = − U
2(p − 1) PU (p, x), p(p + 1)2 (3p − 2)
(49) (50)
where the (quadratic-in-x) polynomials are PL (p, x) := 414p8 + 100p7 − 1493p6 − 385p5 + 1685p4 + 201p3 − 882p2 − 48p + 144 x2
+ p −72p8 − 126p7 + 101p6 + 367p5 + 150p4 − 160p3 − 136p2 + 72p + 72 x + p4 72p4 − 34p3 − 118p2 + 12 , (51) PU (p, x) := 54p9 − 411p8 − 260p7 + 1422p6 + 492p5 − 1633p4 − 250p3 + 862p2 + 60p − 144 x2 + p 54p8 + 117p7 − 82p6 − 344p5 − 100p4 + 177p3 + 74p2 − 80p − 48 x + p4 −54p4 + 14p3 + 96p2 + 20p − 8 . (52) 35
Since the prefactor in (49) is positive and the prefactor in (50) is negative, to prove (∗) it suffices to show 1 27 . (∗∗) , PL (p, x) < 0 and PU (p, x) > 0 for all p > 3, x ∈ 3 64 Step 1: monotonicity in x. Because PL and PU are quadratic in x, their x-derivatives are affine in x. Define the six one-variable polynomials (in p): 1 27 1 , L1 (p) := 3∂x PL p, , L2 (p) := 32∂x PL p, , L0 (p) := 9PL p, 3 3 64 1 1 27 U0 (p) := 9PU p, , U1 (p) := 3∂x PU p, , U2 (p) := 32∂x PU p, . 3 3 64 Expanding gives L0 (p) = −216p9 + 684p8 + 97p7 − 1454p6 + 65p5 + 1313p4 − 207p3 − 666p2 + 168p + 144,
L1 (p) = −216p9 + 450p8 + 503p7 − 1885p6 − 320p5 + 2890p4 − 6p3 − 1548p2 + 120p + 288,
L2 (p) = − 2304p9 + 7146p8 + 5932p7 − 28567p6 − 5595p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888,
and U0 (p) = 216p9 − 546p8 − 380p7 + 1254p6 + 372p5 − 1174p4 − 28p3 + 622p2 − 84p − 144,
U1 (p) = 270p9 − 471p8 − 766p7 + 1812p6 + 684p5 − 2735p4 − 278p3 + 1484p2 − 24p − 288,
U2 (p) =3186p9 − 7353p8 − 9644p7 + 27386p6 + 10084p5 − 38427p4 − 4382p3 + 20714p2 + 84p − 3888.
Step 2: a Sturm-sign check for the one-variable polynomials. A Sturm chain computation (equivalently, root-counting via Sturm’s theorem) shows: L0 (p) < 0, L1 (p) < 0, L2 (p) < 0
and
U0 (p) > 0, U1 (p) > 0, U2 (p) > 0
for all p > 3. (†)
We also present an alternative way to verify these inequalities: Analysis of L0 (p): If p ≥ 4 then L0 (p) ≤ −864p8 + 684p8 + 97p7 − 1454p6 + 65p5 + 1313p4 − 207p3 − 666p2 + 168p + 144
= [−180p + 97]p7 + [(−5816 + 65)p + 1313]p4 + [−207p2 + 168]p + [−666p2 + 144] < 0,
and, if p ∈ (3, 4), then L0 (p) ≤ −648p8 + 684p8 + 97p7 − 1454p6 + 65p5 + 1313p4 − 207p3 − 666p2 + 168p + 144 ≤ [36(16) + 97(4) − 1454]p6 + 65p5 + 1313p4 − 207p3 − 666p2 + 168p + 144 = −490p6 + 65p5 + 1313p4 − 207p3 − 666p2 + 168p + 144 ≤ −468p6 + 1313p4 − 207p3 − 666p2 + 168p + 144
≤ [−468(9) + 1313]p4 + [−207p2 + 168]p + [−666p2 + 144] < 0.
Analysis of L1 (p): For all p > 3 we have L1 (p) ≤ −216p9 + 450p8 + 503p7 − 1885p6 − 320p5 + 2890p4 − 6p3 − 1548p2 + 120p + 288
≤ −648p8 + 450p8 + 503p7 − 1885p6 − 320p5 + 2890p4 − 6p3 − 1548p2 + 120p + 288 ≤ [−198(3) + 503]p7 − 1885p6 − 320p5 + 2890p4 − 6p3 − 1548p2 + 120p + 288
≤ [−198(3) + 503]p7 + [−1885(9) + 2890]p4 − 320p5 − 6p3 + [(−1548(3) + 120)p + 288] < 0 36
Analysis of L2 (p): If p ≥ 4 then L2 (p) ≤[−9216 + 7146]p8 + 5932p7 − 28567p6 − 5595p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888,
≤[−2070 + 5932]p7 + [−28567(16) − 5595(4) + 40375 + 1075/3]p4 + [(−21510(3) + 1008)p + 3888] < 0,
and, if p ∈ (3, 4), then L2 (p) ≤[−2304p2 + 7146p + 5932]p7 − 28567p6 − 5595p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888
≤[−2304(3)2 + 7146(3) + 5932]p7 − 28567p6 − 5595p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888
≤[6634p2 − 28567p − 5595]p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888
≤[6634(16) − 28567(4) − 5595]p5 + 40375p4 + 1075p3 − 21510p2 + 1008p + 3888 ≤[−13719(3) + 40375]p4 + 1075p3 − 21510p2 + 1008p + 3888 ≤[−782(3) + 1075]p3 + [(−21510(3) + 1008)p + 3888] < 0
since −2304p2 + 7146p + 5932 is decreasing in (3, 4) and 6634p2 − 28567p − 5595 is increasing in (3, 4). Analysis of U0 (p). If p ≥ 4, then U0 (p) ≥(864 − 546)p8 − 380p7 + 1254p6 + 372p5 − 1174p4 − 28p3 + 622p2 − 84p − 144
≥[318(4) − 380]p7 + [1254(16) − 1174]p4 + [372(16) − 28]p3 + [(622(4) − 84)(4) − 144] > 0
and, if p ∈ (3, 4), then U0 (p) ≥(648 − 546)p8 − 380p7 + 1254p6 + 372p5 − 1174p4 − 28p3 + 622p2 − 84p − 144 ≥[306 − 380]p7 + +1254p6 + 372p5 − 1174p4 − 28p3 + 622p2 − 84p − 144 ≥[(−74)4 + 1254]p6 + 372p5 − 1174p4 − 28p3 + 622p2 − 84p − 144
≥[958(9) − 1174]p4 + [372(9) − 28]p3 + [(622(3) − 84)(3) − 144] > 0
Analysis of U1 (p). For all p > 3 we have U1 (p) ≥[810 − 471]p8 − 766p7 + 1812p6 + 684p5 − 2735p4 − 278p3 + 1484p2 − 24p − 288
≥[339(3) − 766]p7 + [1812(9) − 2735]p4 + [684(9) − 278]p3 + [(1484(3) − 24)3 − 288] > 0.
Analysis of U2 (p). If p ≥ 4, then U2 (p) ≥[12744 − 7353]p8 − 9644p7 + 27386p6 + 10084p5 − 38427p4 − 4382p3 + 20714p2 + 84p − 3888
≥[5391(4) − 9644]p7 + [27386(16) − 38427]p4 + [10084(16) − 4382]p3 + [20714(16) + 84(4) − 3888] > 0
and, if p ∈ (3, 4), then U2 (p) ≥[9558 − 7353]p8 − 9644p7 + 27386p6 + 10084p5 − 38427p4 − 4382p3 + 20714p2 + 84p − 3888
≥[2205(3) − 9644]p7 + 27386p6 + 10084p5 − 38427p4 − 4382p3 + 20714p2 + 84p − 3888
≥[−3029(4) + 27386]p6 + 10084p5 − 38427p4 − 4382p3 + 20714p2 + 84p − 3888 ≥[15270(9) − 38427]p4 + [10084(9) − 4382]p3 + [20714(9) + 84(3) − 3888] > 0. 37
Step 3: conclude (49) and (50). Since ∂x PL (p, x) is affine in x, the inequalities L1 (p) < 0 and L2 (p) < 0 imply ∂x PL (p, x) < 0 for all x ∈ [ 13 , 27 64 ]. Thus PL (p, x) is decreasing in x on this interval, hence 1 1 = L0 (p) < 0. PL (p, x) ≤ PL p, 3 9 This proves PL (p, x) < 0. Similarly, U1 (p) > 0 and U2 (p) > 0 imply ∂x PU (p, x) > 0 on [ 13 , 27 64 ], so PU (p, x) is increasing in x and therefore 1 1 = U0 (p) > 0. PU (p, x) ≥ PU p, 3 9 This proves PU (p, x) > 0. Hence (∗∗) holds, and then (49), (50) give W (r) c=c < 0 and L W (r) c=c < 0. U
Finally, since c(p) ∈ (cL , cU ) and c 7→ W (r) is affine linear, we conclude W (r) c=c(p) < 0, i.e. W (r) < 0 for all p > 3.
C
Links to chat conversations with Grok
In this section we provide links to chat conversations with Grok. • Counterexample. https://grok.com/share/c2hhcmQtNA_fcd3923c-9c6f-4d33-ba77-f196890932b9 • Proof of Lemma 2.3. https://grok.com/share/c2hhcmQtNA_885159eb-bd8b-4fa6-b602-c673afb8825c?ri d=04645a03-2250-44b6-9903-b419077f817f • Proof of Lemma 2.6. https://grok.com/share/c2hhcmQtNA_17466e98-8494-438c-a0cf-43ded161b713 • Proof of Theorem 1.1 given some hints. https://grok.com/c/191f11d1-0596-43d7-8703-aec8c95b7408?rid=885aaf78-4e6 9-4fb5-8b5e-c43d19ee6775 • Plots of h(s) and h′ (s) for p = 4. https://grok.com/share/c2hhcmQtNA_5bd64c5c-b016-4a93-9543-dd2f107a73b8
Acknowledgments P.I. acknowledges partial support from the NSF CAREER grant DMS-2152401, a Simons Fellowship, and a Humboldt Research Fellowship for Experienced Researchers. P.I. also thanks the University of Würzburg and Université de Toulouse for their warm hospitality. J.M. was partially supported by the AMS Stefan Bergman Fellowship and the Simons Foundation Grant #453576. J.M. is thankful to Felipe Goncalves for interesting discussion.
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References [1] Anthony Carbery. Almost-orthogonality in the Schatten–von Neumann classes. J. Operator Theory, 62(1):151–158, 2009. [2] Eric A. Carlen, Rupert L. Frank, Paata Ivanisvili, and Elliott H. Lieb. Inequalities for Lp -norms that sharpen the triangle inequality and complement Hanner’s inequality. The Journal of Geometric Analysis, 31:4051–4073, 2021. [3] Eric A. Carlen, Rupert L. Frank, and Elliott H. Lieb. Inequalities that sharpen the triangle inequality for sums of N functions in Lp . Arkiv för Matematik, 58(1):57–69, 2020. [4] Paata Ivanisvili and Connor Mooney. Sharpening the triangle inequality: envelopes between L2 and Lp spaces. Analysis & PDE, 13(5):1591–1603, 2020.
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