aa r X i v : . [ m a t h . F A ] N ov The symmetric property ( τ ) for the Gaussian measure Joseph Lehec ∗ December 2007
Abstract
We give a proof, based on the Poincar´e inequality, of the symmetric property ( τ ) forthe Gaussian measure. If f : R d → R is continuous, bounded from below and even, wedefine Hf ( x ) = inf y f ( x + y ) + | y | and we have Z e − f dγ d Z e Hf dγ d ≤ . This property is equivalent to a certain functional form of the Blaschke-Santal´o inequality,as explained in a paper by Artstein, Klartag and Milman.Published in Ann. Fac. Sci. Toulouse Sr. 6, 17 (2) (2008) 357–370.
The Blaschke-Santal´o inequality states that if K is a symmetric convex body of R d then | K | d | K ◦ | d ≤ | D | d | D ◦ | d = v d , (1)where |·| d stands for the volume, K ◦ is the polar body of K , D the Euclidean ball and v d its volume. It was first proved by Blaschke for dimension 2 and 3 and Santal´o [15] extendedthe result to any dimension. K. Ball in [2] was the first to prove a functional version of thisinequality. Since then, several authors found either improvements or other versions of Ball’sinequality, for example Lutwak and Zhang [10], Artstein, Klartag and Milman [1], or Fradeliziand Meyer [7].Ball’s inequality implies in particular the following: if F is a non negative even measurablefunction on R d then Z F ( x ) dx Z F ◦ ( x ) dx ≤ (2 π ) d , (2)where F ◦ is the polar function of F : F ◦ ( x ) = inf y ∈ R d e − x · y F ( y ) . Let K be a symmetric convex body and N K the associated norm. Defining F K = e − N K ,it is easy to see that ( F K ) ◦ = F K ◦ . Besides, a standard computation shows that v d R F K = ∗ LAMA (UMR CNRS 8050) Universit´e Paris-Est. π ) d/ | K | d and similarly for K ◦ . Therefore, applying (2) to F K , we get (1). So (2) is afunctional form of Santal´o’s inequality.In [1], Artstein, Klartag and Milman extend this inequality to the non even setting: theyprove that (2) holds as soon as the barycenter of F ◦ is at 0 (which is true when F is even).At the end of their paper is an interesting remark on the property ( τ ). This property wasintroduced by Talagrand and named after him by Maurey in [11]. Here is the definition: let µ be a probability measure on R d and w a weight (which is a non negative function equal to0 at 0). We say that ( µ, w ) satisfies property ( τ ) if for any non negative continuous function f , the following inequality holds: Z e − f dµ Z e f ⊡ w dµ ≤ , (3)where f ⊡ w is the infimum convolution of f and w : f ⊡ w ( x ) = inf y ∈ R d (cid:0) f ( x + y ) + w ( y ) (cid:1) . Clearly, we may replace “ f ≥
0” by “ f bounded from below”. Of course f ⊡ w ( x ) ≤ f ( x )(take y = 0). In order to find f ⊡ w ( x ), one is allowed to move from x to some x + y suchthat f ( x + y ) is smaller than f ( x ), but the displacement costs w ( y ).Let γ d be the standard Gaussian measure on R d . For x ∈ R d , we let | x | be its Euclideannorm and we define the quadratic cost c : x
7→ | x | . Maurey remarked that ( γ d , c ) satisfiesproperty ( τ ). It follows from the famous Pr´ekopa-Leindler inequality: let u , v and w bemeasurable functions satisfying w ( x + y ) ≤ ( u ( x ) + v ( y )) for all x and y in R d , then (cid:16)Z e − u dx (cid:17) (cid:16)Z e − v dx (cid:17) ≤ Z e − w dx. (4)This is a reverse form of H¨older’s inequality, we refer to [3] for a proof and selected applications.Applying this inequality to u = f + c , v = − ( f ⊡ c ) + c and w = c yields property ( τ ) forthe Gaussian measure.It is pointed out in [1] that there is a strong connection between property ( τ ) and the functionalversion of Santal´o’s inequality. Indeed, applying (2) to F = e − f − c we obtain easily Z e − f dγ d Z e f ⊡ c dγ d ≤ . Hence, if we restrict to even functions we have property ( τ ) for ( γ d , c ) (we gain a factor 2 inthe cost). This is what we call the symmetric property ( τ ), and our purpose is to prove itdirectly. We will show that it is related to the eigenvalues of the Laplacian in Gauss space.This will provide a new proof of the functional Santal´o inequality, and this proof avoids usingthe usual Santal´o inequality (for convex sets), which was not the case of Ball’s proof.From now on we denote by Hf the function f ⊡ c : Hf ( x ) = inf y ∈ R d (cid:0) f ( x + y ) + | y | (cid:1) . We restate the theorem we want to prove:
Theorem 1.
Let f be an even, bounded from below and continous function on R d . Then Z e − f dγ d Z e Hf dγ d ≤ . A “small ǫ ” inequality Definition 2.
Let
C > ǫ ∈ (0 , F ( C, ǫ ) be the class of functions on R d satisfyingthe following properties:(i) f is lipschitz with constant Cǫ (ii) for any x and h in R d : f ( x + h ) + f ( x − h ) − f ( x ) ≤ Cǫ | h | . This class F ( C, ǫ ) is stable under various operations. For example it is clearly a convex set,and, if f is in F ( C, ǫ ) and u ∈ R d then x f ( x + u ) and x f ( − x + u ) are also in F ( C, ǫ ).Besides, if ( f i ) i ∈ I is a family of functions in F ( C, ǫ ) and if for some x , inf i ∈ I f i ( x ) > −∞ ,then f = inf i f i is still in F ( C, ǫ ). Indeed, it is then clear that f is nowhere equal to −∞ andthat it is Cǫ -lipschitz. Let x, h ∈ R d and let i satisfy f i ( x ) ≤ f ( x ) + η ; we have f ( x + h ) + f ( x − h ) − f ( x ) ≤ f i ( x + h ) + f i ( x − h ) − f i ( x ) + 2 η. Applying property (ii) for f i and letting η go to 0, we get the result.Let X be a standard Gaussian vector on R d , the expectation of a random variable will bedenoted by E . Lemma 3.
Let
C > , for every ǫ ∈ (0 , , for every even function f ∈ F ( C, ǫ ) we have E e − f ( X ) E e Hf ( X ) ≤ Kǫ , with a constant K depending on C solely . The proof of this lemma is based on a Taylor expansion and on a symmetric Poincar´einequality.
Lemma 4.
Let f be a smooth function in L ( R d , γ d ) . Assume that f is orthogonal to constantsand to linear functions: E f ( X ) = 0 and E Xf ( X ) = 0 . (5) Then E f ( X ) ≤ E |∇ f ( X ) | . (6)It is a well-known result, we recall its proof for completeness, the reader can also have alook at [6] where a more general statement is proved. Proof.
We consider the sequence ( H α ) α ∈ N d of the Hermite polynomials on R d . It is an or-thonormal basis of L ( γ d ) in which the (unbounded) operator L : f ∆ f − x · ∇ f is diagonal: for all α we have − LH α = | α | H α , where | α | = P α i . We refer to [13, chapter 2]for details. Moreover, we have the following integration by parts formula: − E Lf ( X ) g ( X ) = E ∇ f ( X ) · ∇ g ( X ) . f = P f α H α . We have − Lf = P | α | f α H α and (5) implies that f α = 0 when | α | = 0 or 1. Therefore E |∇ f ( X ) | = − E Lf ( X ) f ( X ) = X | α |≥ | α | f α ≥ E f ( X ) , which concludes the proof. Proof of Lemma 3.
In what follows, K , K , . . . are constants that depend only on C . Let ǫ ∈ (0 ,
1] and f ∈ F ( C, ǫ ), even. Replace f by its convolution by some even regular Diracapproximation ( ρ n ) (Gaussian for example). Clearly ρ n ∗ f is even and belongs to F ( C, ǫ ).On the other hand, since f is lipschitz, ρ n ∗ f → f uniformly, which implies H ( ρ n ∗ f ) → Hf .Moreover, being lipschitz, f is bounded by a multiple of 1 + | x | , and the same holds for Hf . Then e − ρ n ∗ f and e H ( ρ n ∗ f ) are both dominated by some function x → A e B | x | . By thedominated convergence theorem, we obtain E e − ( ρ n ∗ f )( X ) E e H ( ρ n ∗ f )( X ) → E e − f ( X ) E e Hf ( X ) , when n goes to infinity. Hence it is enough to prove the inequality when f is C . Moreover,the inequality does not change if we add a constant to f . Hence we can also assume that E f ( X ) = 0. When f is C , property (ii) implies that Hess f ( x ) ≤ Cǫ Id for any x . Hence byTaylor’s formula(ii’) for any x and h , f ( x + h ) ≤ f ( x ) + ∇ f ( x ) · h + C ǫ | h | .First we estimate E e − f ( X ) . We write E e − f ( X ) = ∞ X k =0 ( − k k ! E f ( X ) k . To estimate the moments of f ( X ), we use the concentration property of the Gaussian measure:if φ is a real valued, mean 0 and 1-lipschitz function on R d , then φ ( X ) is closed to 0 withhigh probability. More precisely, for any t >
0, we haveProb (cid:8) | φ ( X ) | ≥ t (cid:9) ≤ − t / . We refer to [9, theorem 2.5] for a proof of this statement. This yields that there exists auniversal constant M such that for any p ≥ (cid:16) E | φ ( X ) | p (cid:17) /p ≤ M √ p. Since f is Cǫ -lipschitz, fCǫ is 1-lipschitz, and it has mean 0 so the preceding inequality applies.Thus, for k ≥
3, we can bound | ( − k E f ( X ) k | from above by ( M C √ k ) k ǫ . We obtain E e − f ( X ) ≤ E f ( X ) + K ǫ . (7)In the same way, we have E e f ( X ) ≤ E f ( X ) + K ǫ and E (cid:12)(cid:12) e f ( X ) − (cid:12)(cid:12) ≤ K ǫ . (8)4e now deal with Hf . Trying y = −∇ f ( x ) in the definition of Hf , we get Hf ( x ) ≤ f ( x − ∇ f ( x )) + |∇ f ( x ) | . Applying (ii’) to x and h = −∇ f ( x ), we obtain Hf ( x ) ≤ f ( x ) − |∇ f ( x ) | + |∇ f ( x ) | + C ǫ |∇ f ( x ) | . Remark that since f is Cǫ -lipschitz, we have |∇ f ( x ) | ≤ Cǫ for any x . We obtain Hf ( x ) ≤ f ( x ) − |∇ f ( x ) | + C ǫ . Taking the exponential, and using |∇ f | ≤ Cǫ again, we gete Hf ( x ) ≤ e f ( x ) (cid:0) − |∇ f ( x ) | + K ǫ (cid:1)(cid:0) K ǫ (cid:1) ≤ e f ( x ) − |∇ f ( x ) | + (1 − e f ( x ) ) |∇ f ( x ) | + K e f ( x ) ǫ . We take the expectation and we use (8), we obtain E e Hf ( X ) ≤ E f ( X ) − E |∇ f ( X ) | + K ǫ . (9)Multiplying (7) and (9) we get E e − f ( X ) E e Hf ( X ) ≤ E f ( X ) − E |∇ f ( X ) | + Kǫ . (10)Now we use Lemma 4: since f has mean 0 and is even, it satisfies (5), hence E f ( X ) ≤ E |∇ f ( X ) | , which concludes the proof. Remark.
Without the assumption “ f even” the Poincar´e inequality would only say E f ( X ) ≤ E |∇ f ( X ) | , and we would be able to prove E e − f ( X ) E e ( f ⊡ c X ) ≤ Kǫ . In order to prove the symmetric property ( τ ), we need to tensorize Lemma 3. This requiresvarious technical tools, most of which are well known to specialists. We begin with a verysimple property of the class F ( C, ǫ ). Lemma 5.
Let E and E be Euclidean spaces, and µ a probability measure on E . Let f : E × E → R be such that for any x ∈ E the function y f ( x, y ) is in F ( C, ǫ ) . Wedefine φ by e − φ ( y ) = Z E e − f ( x,y ) dµ ( x ) . Then, unless φ = −∞ , φ belongs to F ( C, ǫ ) .Proof. Property (i) is simple. We check (ii): multiplying by − the inequality f ( x, y + h ) + f ( x, y − h ) ≤ f ( x, y ) + Cǫ | h | , taking the exponential and integrating with respect to µ , weobtain Z E e − f ( x,y ) dµ ( x )e − Cǫ | h | ≤ Z E e − f ( x,y + h ) e − f ( x,y − h ) dµ ( x ) . Then we use Cauchy-Schwarz to bound the right hand side, we take the log and we end upwith the desired inequality for φ . 5 Symmetrisation
An important tool in geometry is Steiner’s symmetrisation, see [4] for definition, propertiesand applications. In particular, it is the main idea behind Meyer and Pajor’s proof of theBlashke-Santal´o inequality. It is thus natural, and as far as we know this idea dates back to[1] to introduce functional analogues of this symmetrisation.
Definition 6.
Let E be a Euclidean space and f : E → R be continuous. We define Sf : x inf u ∈ E (cid:16) (cid:0) f ( u + x ) + f ( u − x ) (cid:1) + | u | (cid:17) . The function Sf is even.Let E , E be Euclidean spaces and g : E × E → R . We define S g and S g to be thesymmetrisations of g with respect to the first and the second variable, respectively: S g : x, y inf u ∈ E (cid:16) (cid:0) f ( u + x, y ) + f ( u − x, y ) (cid:1) + | u | (cid:17) , and similarly for S . Lemma 7.
Let γ be the normal distribution on E . For any continuous f on E , we have Z E e − f dγ ≤ Z E e − Sf dγ. (11) Let γ , γ be the normal distributions on E and E , respectively. Any continuous g on E × E satisfies Z E × E e Hg dγ ⊗ dγ ≤ Z E × E e HS g dγ ⊗ dγ . (12) Proof.
Set ˜ f : x → f ( − x ). By definition of Sf , we have for every z and u ∈ R d Sf ( z ) + | z | ≤ f ( u + z ) + f ( − z + u ) + | u | + | z | = (cid:0) f ( u + z ) + | u + z | ) + (cid:0) ˜ f ( z − u ) + | z − u | (cid:1) . Let x, y ∈ R d , we apply this inequality to z = x + y and u = x − y . We obtain Sf ( x + y (cid:12)(cid:12) x + y (cid:12)(cid:12) ≤ (cid:0) f ( x ) + | x | ) + (cid:0) ˜ f ( y ) + | y | (cid:1) . Applying the Pr´ekopa-Leindler inequality we get (11).Inequality (12) is the functional version of the following: if K is a centrally symmetric convexbody and K u its Steiner symmetral with respect to some direction u then | ( K u ) ◦ | ≥ | K ◦ | . Thisis the key argument in Meyer and Pajor’s proof of the symmetric Santal´o inequality, see [12].Actually this idea dates back to Saint-Raymond [14] although he considered a symmetrisationof his own instead of Steiner’s. Let us prove (12). It follows from − S ( − Hg ) ≤ HS g. (13)Indeed, combining it with (11) we get for every y ∈ E Z e Hg ( x,y ) dγ ( x ) = Z e − ( − Hg )( x,y ) dγ ( x ) ≤ Z e − S ( − Hg ( x,y )) dγ ( x ) ≤ Z e HS g ( x,y ) dγ ( x ) , HS g ( x, y ) = inf u,v (cid:8) S g ( x + u, y + v ) + | u | + | v | (cid:9) = inf u,v,w (cid:8) g ( x + u, w + y + v ) + g ( x + u, w − y − v )+ | u | + | v | + | w | (cid:9) . Since g is even, the latest is the same asinf u,v,w (cid:8) g ( x + u, w + y + v ) + g ( − x − u, − w + y + v ) + | u | + | v | + | w | (cid:9) . Whereas − S ( − Hg )( x, y ) = − inf t (cid:8) ( − Hg )( t + x, y ) + ( − Hg )( t − x, y ) + | t | (cid:9) = sup t (cid:8) Hg ( t + x, y ) + Hg ( t − x, y ) − | t | (cid:9) . Hence we have to prove that for any t, u, v, wHg ( t + x, y ) + Hg ( t − x, y ) ≤ g ( x + u, w + y + v ) + g ( − x − u, − w + y + v )+ | u | + | v | + | w | + | t | , which is clear: use twice Hg ( a ) − g ( b ) ≤ | b − a | . Definition 8.
Let n ∈ N ∗ , we define the class F n ( C, ǫ ) by induction on n : let f be a functionon ( R d ) n , we say that f belongs to F n ( C, ǫ ) if- for every y ∈ R d , the function x ∈ ( R d ) n − f ( x, y ) is in F n − ( C, ǫ )- for every x ∈ ( R d ) n − , the function y ∈ R d f ( x, y ) is in F ( C, ǫ ).In other words F n ( C, ǫ ) is the class of functions on R d × · · · × R d that belong to F ( C, ǫ )with respect to each coordinate separately. The crucial point is that the class F n ( C, ǫ ) isstable under symmetrisation.
Lemma 9.
Let f belong to F n ( C, ǫ ) , set E = ( R d ) n − and E = R d . Then S f also belongsto F n ( C, ǫ ) .Proof. It follows from the stability properties of the class F ( C, ǫ ) that we mentionned earlier:for every x ∈ E and u ∈ E , the function y (cid:0) f ( x, u + y ) + f ( x, u − y ) (cid:1) + | u | is in F ( C, ǫ ) so the same is true for the infimum over u , provided that this infimum is not −∞ , which is clear, since (cid:0) f ( x, u + y ) + f ( x, u − y ) (cid:1) is lipschitz in u . Similarly, for any y, u ∈ E , the function x (cid:0) f ( x, u + y ) + f ( x, u − y ) (cid:1) + | u | is in F n − ( C, ǫ ) and the same is true for the infimum over u .7he next lemma is a tensorisation of Lemma 3, with the same constant K . Lemma 10.
Let X , . . . , X n be independent copies of X and let f be even and in F n ( C, ǫ ) .Then E e − f ( X ,...,X n ) E e Hf ( X ,...,X n ) ≤ (1 + Kǫ ) n . Proof.
It is done by induction. When n = 1, it is Lemma 3. Let n ≥
2, we assume thatthe resluts holds for n −
1. Let f : ( R d ) n → R be even and in F n ( C, ǫ ). Set E = ( R d ) n − , E = R d and ˜ X = ( X , . . . , X n − ), it is a normal vector on E . We have to prove E e − f ( ˜ X,X n ) E e Hf ( ˜ X,X n ) ≤ (1 + Kǫ ) n . Let g = S f . We use Lemma 7, for every x ∈ E we have: E e − f ( x,X n ) ≤ E e − S f ( x,X n ) , whichimplies that E e − f ( ˜ X,X n ) ≤ E e − S f ( ˜ X,X n ) . On the other hand, the second part of Lemma 7 gives E e Hf ( ˜ X,X n ) ≤ E e HS f ( ˜ X,X n ) . Hence it is enough to prove that E e − g ( ˜ X,X n ) E e Hg ( ˜ X,X n ) ≤ (1 + Kǫ ) n . (14)By Lemma 9, we have g ∈ F n ( C, ǫ ). By definition of S , for every x ∈ E , the function y g ( x, y ) is even. But as g is globally even we have also g ( − x, y ) = g ( x, − y ) = g ( x, y ).Hence, for any y ∈ E , the function g y : x → g ( x, y ) is even and in F n − ( C, ǫ ). By theinduction assumption E e − ( g y )( ˜ X ) E e H ( g y )( ˜ X ) ≤ (1 + Kǫ ) n − . (15)The following computation can be found in [11], we recall it for the sake of completeness. Wedefine the operator H by H g ( x, y ) = H ( g y )( x ) = inf u ∈ E (cid:0) g ( x + u, y ) + | u | (cid:1) . We define a new function φ by e − φ ( y ) = E e − g ( ˜ X,y ) . For any x ∈ E , the function y → g ( x, y ) is even and belongs to F ( C, ǫ ). By Lemma 5, thesame is true for φ . Hence we can apply Lemma 3 to get E e − φ ( X n ) E e Hφ ( X n ) ≤ Kǫ . (16)On the other hand, inequality (15) implies that for any y and u in R d we have E e H g ( ˜ X,y + u )+ | u | ≤ e φ ( y + u )+ | u | (1 + Kǫ ) n − . (17)Let ( x, y ) ∈ E × E . For any u ∈ R d we have H g ( x, y + u ) + | u | ≥ Hg ( x, y ). Hence,taking the infimum over u in (17), we get E e Hg ( ˜ X,y ) ≤ e Hφ ( y ) (cid:0) Kǫ (cid:1) n − , (18)which of course implies that E e Hg ( ˜ X,X n ) ≤ E e Hφ ( X n ) (cid:0) Kǫ (cid:1) n − . (19)Combining (19) with (16) yields (14) and the proof is complete.8 Proof of Theorem 1
Let n ∈ N , and f be an even function from R d to R belonging to F ( C, g : ( x , . . . , x n ) ∈ ( R d ) n f (cid:0) √ n ( x + · · · + x n ) (cid:1) . It is clear that g is even and belongs to F n ( C, √ n ). Applying Lemma 10, we obtain E e − g ( X ,...,X n ) E e Hg ( X ,...,X n ) ≤ (1 + K n / ) n ≤ K ′ √ n . (20)The convexity of the square of the norm implies that Hg ( x , . . . , x n ) = inf u ,...,u n f (cid:0) √ n P ( x i + u i ) (cid:1) + P | u i | ≥ inf u ,...,u n f (cid:0) √ n P x i + √ n P u i (cid:1) + (cid:12)(cid:12) √ n P u i (cid:12)(cid:12) , = Hf (cid:0) √ n P x i (cid:1) . We combine this inequality with (20), since the random vector √ n P X i has the same law as X , we obtain E e − f ( X ) E e Hf ( X ) ≤ K ′ √ n . Now we let n → ∞ and we get the result for f . Hence we have the inequality for any even f in F ( C, C , so we can take C as big as we want. In particular, weget the result for any even and C function with compact support. And a density argumentshows that the inequality is valid for any function that is even, continuous and bounded frombelow. Remarks
It is also possible to derive the usual property ( τ ) for the Gaussian measure from the Poincar´einequality. The function f is not assumed to be even anymore, we use the non-symmetricversion of Lemma 3 and then perform the tensorisation (just as in Section 4 but without thesymmetrisation). Of course, this is a bit more complicated than the usual proof (using thePr´ekopa-Leindler inequality) but we find it interesting to see that property ( τ ) has to do withthe first eigenvalue of the Laplacian whereas its symmetric version has to do with the secondeigenvalue.Unfortunately we are not able to go farther: what happens if f is orthogonal to constants,linear functionals and polynomials of degree 2? Actually even in the case of a function thatis orthogonal to constants and linear functionals our method does not give the result: we hadto suppose that our function was even, which is stronger.Lastly, following Klartag [8], we explain how to extend the symmetric property ( τ ) to measuresthat are even and log-concave with respect to the Gaussian. It is well known that property ( τ )is stable under 1-Lipschitz image. Clearly, the symmetric property ( τ ) will have the followingstable under pushforward by a 1-Lipschitz odd map. For example, it was proved by Caffarelli(see [5]) that if a probability measure µ is log-concave with respect to the Gaussian measure –meaning that dµ = e − V dγ for some V convex – then the Brenier map transporting the Gaus-sian measure to µ is 1-Lipschitz. Now suppose additionally that µ is even, then the Brenier9ap is automatically odd so µ is the pushfoward of the Gaussian measure by a 1-Lipschitzodd map. Therefore, probability measures that are even and log-concave with respect to theGaussian measure, in particular restrictions of the Gaussian measure to symmetric convexbodies, satisfy the symmetric property ( τ ). References [1] S. Artstein, B. Klartag, and V. Milman,
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