aa r X i v : . [ m a t h . F A ] N ov 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 dueto Artstein, Klartag and Milman. The proof is by induction on the dimension and doesnot use the Blaschke-Santal´o inequality.Published in C. R. Acad. Sci. Paris, S´er. I 347 (2009) 55–58.
For x, y ∈ R n , we denote their inner product by h x, y i and the Euclidean norm of x by | x | .If A is a subset of R n , we let A ◦ = { x ∈ R n | ∀ y ∈ A, h x, y i ≤ } be its polar body. TheBlaschke-Santal´o inequality states that any convex body K in R n with center of mass at 0satisfies vol n ( K ) vol n ( K ◦ ) ≤ vol n ( D ) vol n ( D ◦ ) = v n , (1)where vol n stands for the volume, D for the Euclidean ball and v n for its volume. Let g be a non-negative Borel function on R n satisfying 0 < R g < ∞ and R | x | g ( x ) dx < ∞ , thenbar( g ) = (cid:0)R g (cid:1) − (cid:0)R g ( x ) x dx (cid:1) 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 indicatorfunction.Let us state a functional form of (1) due to Artstein, Klartag and Milman [1]. If f is anon-negative Borel function on R n , the polar function of f is the log-concave function definedby f ◦ ( x ) = inf y ∈ R n (cid:0) e −h x,y i f ( y ) − (cid:1) Theorem 1 (Artstein, Klartag, Milman) . If f is a non-negative integrable function on R n such that f ◦ has its barycenter at , then Z R n f ( x ) dx Z R n f ◦ ( y ) dy ≤ (cid:0)Z R n e − | x | dx (cid:1) = (2 π ) n . In the special case where the function f is even, this result follows from an earlier inequalityof Keith Ball [2]; and in [4], Fradelizi and Meyer prove something more general (see also [5]).In the present note we prove the following: ∗ LAMA (UMR CNRS 8050) Universit´e Paris-Est. heorem 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 −h x,y i . (2) If f (or g ) has its barycenter at then Z R n f ( x ) dx Z R n g ( y ) dy ≤ (2 π ) n . (3)This is slightly stronger than Theorem 1 in which the function that has its barycenter at0 should be log-concave. The point of this note is not really this improvement, but ratherto present a simple proof of Theorem 1. Theorem 2 yields an improved Blaschke-Santal´oinequality, obtained by Lutwak in [6], with a completely different approach. Corollary 3.
Let S be a star-shaped (with respect to ) body in R n having its centroid at .Then vol n ( S ) vol n ( S ◦ ) ≤ v n . (4) Proof.
Let N S ( x ) = inf { r > | x ∈ rS } be the gauge of S and φ S = exp (cid:0) − N S (cid:1) . Integrating φ S and the indicator function of S on level sets of N S , it is easy to see that R R n φ S = c n vol n ( S )for some constant c n depending only on the dimension. Replacing S by the Euclidean ball inthis equality yields c n = (2 π ) n/ v − n . Therefore it is enough to prove that Z φ S Z φ S ◦ ≤ (2 π ) n . (5)Similarly, it is easy to see that bar( φ S ) = c ′ n bar( S ) = 0. Besides, we have h x, y i ≤ N S ( x ) N S ◦ ( y ) ≤ N S ( x ) + N S ◦ ( y ) , for all x, y ∈ R n . Thus φ S and φ S ◦ satisfy (2), then byTheorem 2 we get (5). Theorem 4.
Let f be a non-negative Borel function on R n having a barycenter. Let H be an affine hyperplane splitting R n into two half-spaces H + and H − . Define λ ∈ [0 , by λ R R n f = R H + f . Then there exists z ∈ R n such that for every non-negative Borel function g (cid:0) ∀ x, y ∈ R n , f ( z + x ) g ( y ) ≤ e −h x,y i (cid:1) ⇒ Z R n f Z R n g ≤ λ (1 − λ ) (2 π ) n . (6) In particular, in every median H ( λ = ) there is a point z such that for all g (cid:0) ∀ x, y ∈ R n , f ( z + x ) g ( y ) ≤ e −h x,y i (cid:1) ⇒ Z R n f Z R n g ≤ (2 π ) n . (7)A similar result concerning convex bodies (instead of functions) was obtained by Meyerand Pajor in [7].Let us derive Theorem 2 from the latter. Let f, g satisfy (2). Assume for example thatbar( g ) = 0, then 0 cannot be separated from the support of g by a hyperplane, so there exists x , . . . , x n +1 ∈ R n such that 0 belongs to the interior of conv { x . . . x n +1 } and g ( x i ) > i = 1 . . . n + 1. Then (2) implies that f ( x ) ≤ C e −k x k , for some C >
0, where k x k =2ax (cid:0) h x, x i i | i ≤ n + 1 (cid:1) . Assume also that R f >
0, then f has a barycenter. Apply the“ λ = 1 /
2” part of Theorem 4 to f . There exists z ∈ R n such that (7) holds. On the otherhand, by (2) f ( z + x ) g ( y )e h y,z i ≤ e −h z + x,y i e h y,z i = e −h x,y i for all x, y ∈ R n . Therefore Z R n f ( x ) dx Z R n g ( y )e h y,z i dy ≤ (2 π ) n . (8)Integrating with respect to g ( y ) dy the inequality 1 ≤ e h y,z i − h y, z i we get Z R n g ( y ) dy ≤ Z R n g ( y )e h y,z i dy − Z R n h y, z i g ( y ) dy. Since bar( g ) = 0, the latter integral is 0 and together with (8) we obtain (3). Observe alsothat this proof shows that Theorem 4 in dimension n implies Theorem 2 in dimension n .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.
Let φ , φ be non-negative Borel functions on R + . If φ ( s ) φ ( t ) ≤ e − st for every s, t in R + , then Z R + φ ( s ) ds Z R + φ ( t ) dt ≤ π . (9) Proof.
Let f ( s ) = φ (e s )e s , g ( t ) = φ (e t )e t and h ( r ) = exp( − e r / r . For all s, t ∈ R we have p f ( s ) g ( t ) ≤ h ( t + s ), hence by Pr´ekopa-Leindler R R f R R g ≤ (cid:0)R R h (cid:1) . By change of variable,this is the same as R R + φ R R + φ ≤ (cid:0)R R + e − u / du (cid:1) which is the result. Clearly we can assume that R f = 1. Let µ be the measure with density f . In the sequel welet f z ( x ) = f ( z + x ) for all x, z .We prove the theorem by induction on the dimension. Let f be a non-negative Borelfunction on the line, let r ∈ R and λ = µ (cid:0) [ r, ∞ ) (cid:1) ∈ [0 , g satisfy f ( r + s ) g ( t ) ≤ e − st ,for all s, t . Apply Lemma 5 twice: first to φ ( s ) = f ( r + s ) and φ ( t ) = g ( t ) then to φ ( s ) = f ( r − s ) and φ ( t ) = g ( − t ). Then Z R + f r Z R + g ≤ π Z R − f r Z R − g ≤ π . Therefore R R + g ≤ π λ and R R − g ≤ π − λ ) , which yields the result in dimension 1.Assume the theorem to be true in dimension n −
1. Let H be an affine hyperplane splitting R n into two half-spaces H + and H − and let λ = µ ( H + ). Provided that λ = 0 , 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 of H + , and similarly for b − . Hence the line passing through b + 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 that z = 0. Let g satisfy ∀ x, y ∈ R n , f ( x ) g ( y ) ≤ e −h x,y i . (10)3et e , . . . , e n be an orthonormal basis of R n such that H = e ⊥ n and h b + , e n i >
0. Let v = b + / h b + , e n i and A be the linear operator on R n that maps e n to v and e i to itself for i = 1 . . . n − B = ( A − ) t . Define F + : y ∈ H Z R + f ( y + sv ) ds and G + : y ′ ∈ H Z R + g ( By ′ + te n ) dt. By Fubini, and since A has determinant 1, R H F + = R H + f ◦ A = µ ( H + ) = λ . Also, letting P be the projection with range H and kernel R v , we havebar( F + ) = 1 λ Z H + P ( Ax ) f ( Ax ) dx = 1 λ P (cid:16)Z H + xf ( x ) dx (cid:17) = P ( b + ) , and this is 0 by definition of P . Since h Ax, Bx ′ i = h x, x ′ i for all x, x ′ ∈ R n , we have h y + sv, By ′ + te n i = h y, y ′ i + st for all s, t ∈ R and y, y ′ ∈ H . So (10) implies f ( y + sv ) g ( By ′ + te n ) ≤ e − st −h y,y ′ i . Applying Lemma 5 to φ ( s ) = f ( y + sv ) and φ ( t ) = g ( By ′ + te n ) we get F + ( y ) G + ( y ′ ) ≤ π e −h y,y ′ i for every y, y ′ ∈ H . Recall that bar( F + ) = 0, then by the induction assumption(which implies Theorem 2 in dimension n − Z H F + Z H G + ≤ π π ) n − . (11)hence R H + g ( Bx ) dx ≤ λ (2 π ) n . In the same way R H − g ( Bx ) dx ≤ − λ ) (2 π ) n , adding thesetwo inequalities, we obtain Z R n g ( Bx ) dx ≤ λ (1 − λ ) (2 π ) n which is the result since B has determinant 1. References [1] S. Artstein-Avidan, B. Klartag, and V. Milman,
The Santal´o point of a function, and afunctional form of Santal´o inequality , Mathematika (2005) 33–48.[2] K. Ball, Isometric problems in ℓ p and sections of convex sets , doctoral thesis, Universityof 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. (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. (1991) (1) 39–68.[7] M. Meyer and A. Pajor, On the Blaschke Santal´o inequality , Arch. Math. (Basel)55