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A direct proof of the functional Santalo inequality

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Submitted on 6 Jan 2017

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A direct proof of the functional Santalo inequality

Joseph Lehec

To cite this version:

Joseph Lehec. A direct proof of the functional Santalo inequality. Comptes Rendus. Mathématique, Académie des sciences (Paris), 2009, 347, pp.55 - 58. �10.1016/j.crma.2008.11.015�. �hal-00365764�

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A direct proof of the functional Santal´ o inequality

Joseph Lehec June 2008

Abstract

We give a simple proof of a functional version of the Blaschke- Santal´o inequality due to Artstein, Klartag and Milman. The proof is by induction on the dimension and does not use the Blaschke-Santal´o inequality.

Published in C. R. Acad. Sci. Paris, S´er. I 347 (2009) 55–58.

1 Introduction

For x, y ∈ Rn, we denote their inner product by hx, yi and the Euclidean norm of x by |x|. If A is a subset of Rn, we let A = {x ∈ Rn| ∀y ∈ A,hx, yi ≤ 1} be its polar body. The Blaschke-Santal´o inequality states that any convex bodyK inRn with center of mass at 0 satisfies

voln(K) voln(K)≤voln(D) voln(D) =vn2, (1) where volnstands for the volume,Dfor the Euclidean ball andvnfor its vol- ume. Letg be a non-negative Borel function on Rn satisfying 0<R

g <∞ and R

|x|g(x)dx <∞, then bar(g) = R

g−1 R

g(x)x dx

denotes its center of mass (or barycenter). The center of mass (or centroid) of a measurable subset of Rn is by definition the barycenter of its indicator function.

Let us state a functional form of (1) due to Artstein, Klartag and Mil- man [1]. Iff is a non-negative Borel function on Rn, the polar function of f is the log-concave function defined by

f(x) = inf

y∈Rn

e−hx,yif(y)−1

LAMA (UMR CNRS 8050) Universit´e Paris-Est.

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Theorem 1 (Artstein, Klartag, Milman). If f is a non-negative integrable function on Rn such thatf has its barycenter at 0, then

Z

Rn

f(x)dx Z

Rn

f(y)dy≤ Z

Rn

e12|x|2dx2

= (2π)n.

In the special case where the functionf is even, this result follows from an earlier inequality of Keith Ball [2]; and in [4], Fradelizi and Meyer prove something more general (see also [5]). In the present note we prove the following:

Theorem 2. Let f and g be non-negative Borel functions on Rn satisfying the duality relation

∀x, y∈Rn, f(x)g(y)≤e−hx,yi. (2) If f (org) has its barycenter at0 then

Z

Rn

f(x)dx Z

Rn

g(y)dy≤(2π)n. (3)

This is slightly stronger than Theorem 1 in which the function that has its barycenter at 0 should be log-concave. The point of this note is not really this improvement, but rather to present a simple proof of Theorem 1. The- orem 2 yields an improved Blaschke-Santal´o inequality, obtained by Lutwak in [6], with a completely different approach.

Corollary 3. Let S be a star-shaped (with respect to 0) body inRn having its centroid at 0. Then

voln(S) voln(S)≤vn2. (4) Proof. Let NS(x) = inf{r > 0|x ∈ rS} be the gauge of S and φS = exp −12NS2

. IntegratingφS and the indicator function ofS on level sets of NS, it is easy to see thatR

RnφS=cnvoln(S) for some constantcndepending only on the dimension. ReplacingS by the Euclidean ball in this equality yieldscn= (2π)n/2vn−1. Therefore it is enough to prove that

Z φS

Z

φS ≤(2π)n. (5)

Similarly, it is easy to see that bar(φS) =c0nbar(S) = 0. Besides, we have hx, yi ≤ NS(x)NS(y) ≤ 12NS(x)2+ 12NS(y)2, for all x, y ∈ Rn. Thus φS and φS satisfy (2), then by Theorem 2 we get (5).

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2 Main results

Theorem 4.Letf be a non-negative Borel function onRnhaving a barycen- ter. LetH be an affine hyperplane splitting Rn into two half-spacesH+ and H. Define λ∈ [0,1] by λR

Rnf = R

H+f. Then there exists z ∈ Rn such that for every non-negative Borel functiong

∀x, y∈Rn, f(z+x)g(y)≤e−hx,yi

⇒ Z

Rn

f Z

Rn

g≤ 1

4λ(1−λ)(2π)n. (6) In particular, in every medianH (λ= 12) there is a pointz such that for all g

∀x, y∈Rn, f(z+x)g(y)≤e−hx,yi

⇒ Z

Rn

f Z

Rn

g≤(2π)n. (7) A similar result concerning convex bodies (instead of functions) was obtained by Meyer and Pajor in [7].

Let us derive Theorem 2 from the latter. Letf, gsatisfy (2). Assume for example that bar(g) = 0, then 0 cannot be separated from the support of g by a hyperplane, so there existsx1, . . . , xn+1∈Rn such that 0 belongs to the interior of conv{x1. . . xn+1} andg(xi)>0 for i= 1. . . n+ 1. Then (2) implies thatf(x)≤Ce−kxk, for some C >0, where kxk= max hx, xii |i≤ n+ 1

. Assume also that R

f > 0, then f has a barycenter. Apply the

“λ= 1/2” part of Theorem 4 tof. There exists z∈Rnsuch that (7) holds.

On the other hand, by (2)

f(z+x)g(y)ehy,zi≤e−hz+x,yiehy,zi= e−hx,yi for allx, y∈Rn. Therefore

Z

Rn

f(x)dx Z

Rn

g(y)ehy,zidy≤(2π)n. (8) Integrating with respect to g(y)dy the inequality 1≤ehy,zi− hy, zi we get

Z

Rn

g(y)dy≤ Z

Rn

g(y)ehy,zidy− Z

Rn

hy, zig(y)dy.

Since bar(g) = 0, the latter integral is 0 and together with (8) we obtain (3).

Observe also that this proof shows that Theorem 4 in dimension n implies Theorem 2 in dimensionn.

In order to prove Theorem 4, we need the following logarithmic form of the Pr´ekopa-Leindler inequality. For details on Pr´ekopa-Leindler, we refer to [3].

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Lemma 5. Letφ1, φ2 be non-negative Borel functions onR+. Ifφ1(s)φ2(t)≤ e−st for everys, t in R+, then

Z

R+

φ1(s)ds Z

R+

φ2(t)dt≤ π

2. (9)

Proof. Let f(s) = φ1(es)es, g(t) = φ2(et)et and h(r) = exp(−e2r/2)er. For all s, t ∈ R we have p

f(s)g(t) ≤ h(t+s2 ), hence by Pr´ekopa-Leindler R

RfR

Rg≤ R

Rh2

. By change of variable, this is the same asR

R+φ1

R

R+φ2 ≤ R

R+e−u2/2du2

which is the result.

3 Proof of Theorem 4

Clearly we can assume thatR

f = 1. Letµ be the measure with densityf. In the sequel we letfz(x) =f(z+x) for allx, z.

We prove the theorem by induction on the dimension. Let f be a non- negative Borel function on the line, let r ∈ R and λ = µ [r,∞)

∈ [0,1].

Letg satisfyf(r+s)g(t)≤e−st, for all s, t. Apply Lemma 5 twice: first to φ1(s) =f(r+s) andφ2(t) =g(t) then toφ1(s) =f(r−s) andφ2(t) =g(−t).

Then Z

R+

fr Z

R+

g≤ π

2 and

Z

R

fr Z

R

g≤ π 2. ThereforeR

R+g≤ π and R

Rg≤ 2(1−λ)π , which yields the result in dimen- sion 1.

Assume the theorem to be true in dimensionn−1. Let H be an affine hy- perplane splitting Rn into two half-spacesH+ and H and let λ=µ(H+).

Provided that λ 6= 0,1 we can define b+ and b to be the barycenters of µ|H+ and µ|H, respectively. Since µ(H) = 0, the point b+ belongs to the interior ofH+, and similarly forb. Hence the line passing throughb+ and b intersects H at one point, which we call z. Let us prove that z satisfies (6), for all g. Clearly, replacing f by fz and H by H−z, we can assume thatz= 0. Let gsatisfy

∀x, y∈Rn, f(x)g(y)≤e−hx,yi. (10) Lete1, . . . , enbe an orthonormal basis ofRnsuch thatH =en andhb+, eni>

0. Letv=b+/hb+, eni and A be the linear operator onRn that mapsen to v andei to itself fori= 1. . . n−1 and letB = (A−1)t. Define

F+:y ∈H 7→

Z

R+

f(y+sv)ds and G+ :y0 ∈H7→

Z

R+

g(By0+ten)dt.

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By Fubini, and sinceAhas determinant 1,R

HF+=R

H+f◦A=µ(H+) =λ.

Also, lettingP be the projection with rangeH and kernelRv, we have bar(F+) = 1

λ Z

H+

P(Ax)f(Ax)dx= 1 λPZ

H+

xf(x)dx

=P(b+), and this is 0 by definition ofP. Since hAx, Bx0i=hx, x0i for allx, x0 ∈Rn, we havehy+sv, By0+teni =hy, y0i+stfor all s, t∈R and y, y0 ∈H. So (10) implies

f(y+sv)g(By0+ten)≤e−st−hy,y0i.

Applying Lemma 5 to φ1(s) =f(y+sv) and φ2(t) = g(By0 +ten) we get F+(y)G+(y0)≤ π2e−hy,y0i for everyy, y0 ∈H. Recall that bar(F+) = 0, then by the induction assumption (which implies Theorem 2 in dimensionn−1)

Z

H

F+

Z

H

G+≤ π

2(2π)n−1. (11)

henceR

H+g(Bx)dx≤ 1 (2π)n. In the same wayR

Hg(Bx)dx≤ 4(1−λ)1 (2π)n, adding these two inequalities, we obtain

Z

Rn

g(Bx)dx≤ 1

4λ(1−λ)(2π)n which is the result sinceB has determinant 1.

References

[1] S. Artstein-Avidan, B. Klartag, and V. Milman, The Santal´o point of a function, and a functional form of Santal´o inequality, Mathematika 51 (2005) 33–48.

[2] K. Ball,Isometric problems in `p and sections of convex sets, doctoral thesis, University of Cambridge, 1986.

[3] K. Ball, An elementary introduction to modern convex geometry, in Flavors of geometry, edited by S. Levy, Cambridge University Press, 1997.

[4] M. Fradelizi and M. Meyer,Some functional forms of Blaschke-Santal´o inequality, Math. Z. 256 (2007) (2) 379–395.

[5] J. Lehec, Partitions and functional Santal´o inequalities, Arch. Math.

92 (2009) (1) 89–94.

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[6] E. Lutwak, Extended affine surface area, Adv. Math. 85 (1991) (1) 39–68.

[7] M. Meyer and A. Pajor, On the Blaschke Santal´o inequality, Arch.

Math. (Basel) 55(1990) 82–93.

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