HAL Id: hal-00365764
https://hal.archives-ouvertes.fr/hal-00365764
Submitted on 6 Jan 2017
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires
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�
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.
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
e−12|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).
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].
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≤ 2λπ and R
R−g≤ 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 =e⊥n 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.
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≤ 4λ1 (2π)n. In the same wayR
H−g(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.
[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.