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

voln(K) voln(K^{◦})≤voln(D) voln(D^{◦}) =v_{n}^{2}, (1)
where vol_{n}stands for the volume,Dfor the Euclidean ball andv_{n}for its vol-
ume. Letg be a non-negative Borel function on R^{n} 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 R^{n} 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 R^{n}, the polar function of
f is the log-concave function defined by

f^{◦}(x) = inf

y∈R^{n}

e^{−hx,yi}f(y)^{−1}

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

Theorem 1 (Artstein, Klartag, Milman). If f is a non-negative integrable
function on R^{n} such thatf^{◦} has its barycenter at 0, then

Z

R^{n}

f(x)dx Z

R^{n}

f^{◦}(y)dy≤
Z

R^{n}

e^{−}^{1}^{2}^{|x|}^{2}dx2

= (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 R^{n} satisfying
the duality relation

∀x, y∈R^{n}, f(x)g(y)≤e^{−hx,yi}. (2)
If f (org) has its barycenter at0 then

Z

R^{n}

f(x)dx Z

R^{n}

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 inR^{n} having
its centroid at 0. Then

voln(S) voln(S^{◦})≤v_{n}^{2}. (4)
Proof. Let NS(x) = inf{r > 0|x ∈ rS} be the gauge of S and φS =
exp −^{1}_{2}N_{S}^{2}

. Integratingφ_{S} and the indicator function ofS on level sets of
N_{S}, it is easy to see thatR

R^{n}φ_{S}=c_{n}vol_{n}(S) for some constantc_{n}depending
only on the dimension. ReplacingS by the Euclidean ball in this equality
yieldscn= (2π)^{n/2}v_{n}^{−1}. Therefore it is enough to prove that

Z
φ_{S}

Z

φ_{S}^{◦} ≤(2π)^{n}. (5)

Similarly, it is easy to see that bar(φS) =c^{0}_{n}bar(S) = 0. Besides, we have
hx, yi ≤ N_{S}(x)N_{S}^{◦}(y) ≤ ^{1}_{2}N_{S}(x)^{2}+ ^{1}_{2}N_{S}^{◦}(y)^{2}, for all x, y ∈ R^{n}. Thus φ_{S}
and φ_{S}^{◦} satisfy (2), then by Theorem 2 we get (5).

### 2 Main results

Theorem 4.Letf be a non-negative Borel function onR^{n}having a barycen-
ter. LetH be an affine hyperplane splitting R^{n} into two half-spacesH+ and
H−. Define λ∈ [0,1] by λR

R^{n}f = R

H+f. Then there exists z ∈ R^{n} such
that for every non-negative Borel functiong

∀x, y∈R^{n}, f(z+x)g(y)≤e^{−hx,yi}

⇒ Z

R^{n}

f Z

R^{n}

g≤ 1

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

∀x, y∈R^{n}, f(z+x)g(y)≤e^{−hx,yi}

⇒ Z

R^{n}

f Z

R^{n}

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 existsx_{1}, . . . , x_{n+1}∈R^{n} such that 0 belongs to
the interior of conv{x_{1}. . . x_{n+1}} andg(x_{i})>0 for i= 1. . . n+ 1. Then (2)
implies thatf(x)≤Ce^{−kxk}, for some C >0, where kxk= max hx, x_{i}i |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∈R^{n}such that (7) holds.

On the other hand, by (2)

f(z+x)g(y)e^{hy,zi}≤e^{−hz+x,yi}e^{hy,zi}= e^{−hx,yi}
for allx, y∈R^{n}. Therefore

Z

R^{n}

f(x)dx Z

R^{n}

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

Z

R^{n}

g(y)dy≤ Z

R^{n}

g(y)e^{hy,zi}dy−
Z

R^{n}

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(e^{s})e^{s}, g(t) = φ2(e^{t})e^{t} and h(r) = exp(−e^{2r}/2)e^{r}.
For all s, t ∈ R we have p

f(s)g(t) ≤ h(^{t+s}_{2} ), 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^{−u}^{2}^{/2}du2

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
φ_{1}(s) =f(r+s) andφ_{2}(t) =g(t) then toφ_{1}(s) =f(r−s) andφ_{2}(t) =g(−t).

Then Z

R+

f_{r}
Z

R+

g≤ π

2 and

Z

R^{−}

f_{r}
Z

R^{−}

g≤ π 2. ThereforeR

R+g≤ _{2λ}^{π} 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 R^{n} 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 f_{z} and H by H−z, we can assume
thatz= 0. Let gsatisfy

∀x, y∈R^{n}, f(x)g(y)≤e^{−hx,yi}. (10)
Lete_{1}, . . . , e_{n}be an orthonormal basis ofR^{n}such thatH =e^{⊥}_{n} andhb_{+}, e_{n}i>

0. Letv=b+/hb_{+}, eni and A be the linear operator onR^{n} 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+ :y^{0} ∈H7→

Z

R^{+}

g(By^{0}+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, Bx^{0}i=hx, x^{0}i for allx, x^{0} ∈R^{n},
we havehy+sv, By^{0}+te_{n}i =hy, y^{0}i+stfor all s, t∈R and y, y^{0} ∈H. So
(10) implies

f(y+sv)g(By^{0}+ten)≤e^{−st−hy,y}^{0}^{i}.

Applying Lemma 5 to φ1(s) =f(y+sv) and φ2(t) = g(By^{0} +ten) we get
F_{+}(y)G_{+}(y^{0})≤ ^{π}_{2}e^{−hy,y}^{0}^{i} for everyy, y^{0} ∈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≤ _{4λ}^{1} (2π)^{n}. In the same wayR

H−g(Bx)dx≤ _{4(1−λ)}^{1} (2π)^{n},
adding these two inequalities, we obtain

Z

R^{n}

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.