Blaschke-Santalo inequality for many functions and geodesic barycenters of measures
aa r X i v : . [ m a t h . F A ] O c t Blaschke–Santal´o inequality for many functionsand geodesic barycenters of measures ∗ Alexander V. Kolesnikov † and Elisabeth M. Werner ‡ Abstract
Motivated by the geodesic barycenter problem from optimal transporta-tion theory, we prove a natural generalization of the Blaschke–Santal´o inequal-ity for many sets and many functions. We derive from it an entropy boundfor the total Kantorovich cost appearing in the barycenter problem.
The Blaschke–Santal´o inequality, see [3, 30], states that every 0-symmetric convexbody K in R n satisfies vol n ( K )vol n ( K ◦ ) ≤ (vol n ( B n )) , where K ◦ = { y ∈ R n : h x, y i ≤ ∀ x ∈ K } is the polar body of K , B n = { x ∈ R n : | x | ≤ } is the Euclidean unit ball and | · | denotes the Euclidean norm on R n . The left-hand side of this inequality is called the Mahler volume. The sharplower bound for the Mahler volume is still open in dimensions 4 and higher. Thefamous Mahler conjecture suggests that this functional is minimized by the couple( B n , B n ∞ ). Partial results can be found in e.g., [18, 22, 27, 28].Here we ask: What is a natural generalization of the bounds for the Mahlervolume for multiple sets? While this is not obvious from the geometric viewpoint,we suggest in this paper a reasonable extension, which is naturally related to afunctional counterpart of the Blaschke–Santal´o inequality.The functional Blaschke–Santal´o inequality was discovered by K. Ball [5] andlater extended and generalized in [2], [15], [25]. In its simplest form it states thatfor every two measurable even functions V, W on R n we have that Z e − V ( x ) dx Z e − W ( y ) dy ≤ (2 π ) n , ∗ Keywords: Blaschke–Santal´o inequality, optimal transport, multimarginal Monge–Kantorovitch problem, geodesic barycenters of measures, Prekopa-Leindler inequality. 2020 Math-ematics Subject Classification: 52A20, 52A40, 60B. † Supported by RFBR project 20-01-00432; Section 7 results have been obtained with support ofRSF grant No 19-71-30020. The article was prepared within the framework of the HSE UniversityBasic Research Program ‡ Partially supported by NSF grant DMS-1811146 and by a Simons Fellowship V ( x ) + W ( y ) ≥ h x, y i . Equality is attained if and only if V ( x ) = | x | + c, W ( y ) = | y | − c , where c is a constant. An interesting link to optimaltransportation theory was discovered recently by M. Fathi [12]. He showed that forprobability measures µ = f · γ, ν = g · γ , where γ is the standard Gaussian measure,such that R xf dγ = 0, the following inequality holds,12 W ( µ, ν ) ≤ Ent γ ( µ ) + Ent γ ( ν ) (1.1)and that this inequality is equivalent to the functional Blaschke–Santal´o inequality.Here, W ( µ, ν ) is the L Kantorovitch distance (see Section 7 for the definition) andEnt γ ( µ ) = Z f log f dγ is the relative entropy with respect to Gaussian measure. Inequality (1.1) is a re-markable strengthening of the Talagrand transportation inequality and the startingpoint of our paper. We refer to, e.g., [4] for Talagrand’s inequality and it’s funda-mental importance in probability theory. In this context, please also note a veryrecent result of N. Gozlan about a transportational approach to the lower bound forthe Blaschke–Santal´o functional [21].We would like to point out an important connection of the Blaschke–Santal´oinequality to the K¨ahler–Einstein equation. Fathi’s result establishes, in particular,that the functional f → W ( µ, ν ) − Ent γ ( µ ) is bounded from below. The min-imum of this functional solves the so-called K¨ahler–Einstein equation. This wasestablished by F. Santambrogio [29]. The form of the functional presented herewas considered in [24]. The well-posedness of the K¨ahler–Einstein equation wasproved by D. Cordero-Erausquin and B. Klartag [9]. Generalization to the sphereand relations to the logarithmic Minkowski problem were established in [23]. Otherrelated transportation inequalities can be found in [14]. We do not pursue herethis viewpoint further, but will discuss in a subsequent paper the relation of theBlaschke–Santal´o inequality to the K¨ahler–Einstein equation.To analyze the case of k functions with k > c ( x , · · · , x n ) = X i,j =1 ,i
Theorem 3.1
Let f i : R n → R + , ≤ i ≤ k , be unconditional functions satisfying k Y i =1 f i ( x i ) ≤ ρ (cid:0) k X i,j =1 i Theorem 6.1 Let λ > and assume that the symmetric non-negative functions f i satisfy k Y i =1 f i ( x i ) ≤ e − λ | P ki =1 x i | . Then k Y i =1 Z f i ( x i ) dx i ≤ (cid:16)Z e − λ k | x i | dx i (cid:17) k . In Section 7 we study applications of our result to transportation inequalities forthe barycenter problem. Recall that for a family of probability measures µ , · · · , µ k ,its barycenter µ (with coefficients 1 /k ) is the minimum point of the functional F ( ν ) = 12 k k X i =1 W ( µ i , ν ) . We obtain the following bound which generalizes a result of Fathi [12] and, in par-ticular, a classical estimate of Talagrand. Theorem 7.2 Assume that µ i = ρ i · γ , where γ is the standard Gaussian measureand the ρ i are even. Then the barycenter µ of µ , · · · , µ k satisfies F ( µ ) ≤ k k X i =1 Z ρ i log ρ i dγ. f, in addition, the ρ i are unconditional, then F ( µ ) ≤ k − k k X i =1 Z ρ i log ρ i dγ. In this section we recall basic facts on duality relations for the transportation costappearing in the theory of barycenters of measures. The following result is knownin the theory of optimal transportation, see [1], Proposition 3.8. We give a proof forthe reader’s convenience. We will need the definition of the Legendre conjugate V ∗ ,which for a function V on R n is defined as V ∗ ( y ) = sup x ∈ R n h x, y i − V ( x ) . Lemma 2.1. Let V i , ≤ i ≤ k, be a family of convex Borel functions on R n . Thenthe following conditions are equivalent.1. For all x i ∈ R n , P ki =1 V i ( x i ) ≥ P ki,j =1 ,i Let V i , ≤ i ≤ k, be a family of Borel functions on R n . Thenthe functional S ( V , · · · , V k ) = k Y i =1 Z e − V i ( x i ) dx i is bounded on the set L n,k = n ( V , · · · , V k ) : V i is even ∀ i ∈ { , · · · , k } , k X i =1 V i ( x i ) ≥ k X i,j =1 ,i Let us fix arbitrary ( V , · · · , V k ) ∈ L n,k and estimate S ( V , · · · , V k ). First wenote that the functions V i can be assumed to be convex. Indeed, if V is not convex,replace it by the following convex function˜ V ( x ) = sup x i ,i =1 (cid:16) k X i,j =1 ,i 1, and V k by ˜ V k ( x k ) = V k ( x k )+ V (0)+ · · · + V k − (0) and this replacementdoes not influence the value of the integral functional. One has ˜ V i (0) = 0 , ≤ i ≤ k − 1. Next we note that k − X i =1 ˜ V i ( x i ) ≥ k − X i = j h x i , x j i + * k − X i =1 x i , x k + − ˜ V k ( x k ) , for all x k , is equivalent to k − X i =1 ˜ V i ( x i ) ≥ k − X i = j h x i , x j i + ( ˜ V k ) ∗ (cid:16) k − X i =1 x i (cid:17) , which in turn is equivalent to k − X i =1 | x i | V i ( x i ) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k − X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + ( ˜ V k ) ∗ (cid:16) k − X i =1 x i (cid:17) . We now define a function F by the following relation | t | F ( t ) = inf t = P k − i =1 x i k − X i =1 | x i | V i ( x i ) . Clearly ( ˜ V k ) ∗ ≤ F , hence ˜ V k ≥ F ∗ . Thus S ( ˜ V , · · · , ˜ V k ) ≤ S ( ˜ V , · · · , F ∗ ). Moreover,it follows immediately from the definition of F and the above inequalities that( ˜ V , · · · , F ∗ ) ∈ L n,k . Since ˜ V i ≥ V i (0) = 0, we immediately get F (0) = 0.Hence, F ∗ (0) = 0. Thus the tuple ( ˜ V , · · · , F ∗ ) satisfies ˜ V (0) = · · · = F ∗ (0) = 0and gives a larger value to S .Finally, we show that S is bounded. We observe that for every j = mV m ( x m ) ≥ sup x i ,i = m (cid:16) k X i,j =1 ,i Assumption (L) seems to be a natural generalization of the conditionthat two convex functions are related by the Legendre transform. Proposition 2.4. There exist functions V , V , V such that the triple ( V , V , V ) satisfies (L) with λ = 1 and S ( V , V .V ) = 0 .Proof. The desired functions are V ( x ) = ( x = 0+ ∞ else V ( x ) = | x | , V ( x ) = | x | . The reader can easily check the claim. In this section we verify our conjecture (inequality part) for the unconditional func-tions. A function f : R n → R is called unconditional, if f ( ε x , · · · , ε n x n ) = f ( x , x , . . . , x n ) , for every ( ε , · · · , ε n ) ∈ {− , } n and every ( x , · · · , x n ) ∈ R n . We show that theconjecture is true in the unconditional case.8 heorem 3.1. Let f i : R n → R + , ≤ i ≤ k , be measurable unconditional functionssatisfying k Y i =1 f i ( x i ) ≤ ρ k X i,j =1 ,i Clearly, for unconditional functions it is sufficient to check that k Y i =1 Z R n + f i ( x i ) dx i ≤ "Z R n + ρ k (cid:18) k ( k − | u | (cid:19) du k , provided that on R n + , k Y i =1 f i ( x i ) ≤ ρ k X i,j =1 ,i Theorem 4.1 (Pr´ekopa–Leindler) . Let f i , ≤ i ≤ k , and h be nonnegative in-tegrable real functions on R n such that for all x i , for all λ i , ≤ λ ≤ k , with P ki =1 λ = 1 , h k X i =1 λ i x i ! ≥ k Y i =1 f λ i i ( x i ) . Then k Y i =1 (cid:16)Z R n f i dx i (cid:17) λ i ≤ Z R n h dx. (4.1)10 quality holds in the Prekopa–Leindler inequality if and only if there exist vectors y , · · · , y k such that for all xf ( x − y ) R R n f dx = f ( x − y ) R R n f dx = · · · f k ( x − y k ) R R n f k dx = e − ψ ( x ) , (4.2) where ψ is a convex function such that R R ⋉ e − ψ ( x ) dx = 1 and h ( x ) = sup x = P ki =1 λ i x i k Y i =1 f λ i i ( x i ) = k Y i =1 (cid:16)Z R n f i dx i (cid:17) λ i e − ψ (cid:0) x + P ki =1 λ i y i (cid:1) . Proof. It is clear that equality holds in inequality (4.1), if the functions satisfy thecondition (4.2).The proof of the inequality is well known and can be found in e.g., [20, 30]. We givea proof of the inequality by induction on the number of functions. This allows toestablish the equality characterizations, as for two functions, those were establishedby Dubuc [11]. We havesup x = P ki =1 λ i x i k Y i =1 f λ i i ( x i ) = sup λ x +(1 − λ ) y f λ ( x ) g − λ ( y ) , where g ( y ) = sup y = − λ P ki =2 λ i x i k Y i =2 f λi − λ i ( x i ) . Applying the Pr´ekopa–Leindler inequality for two functions gives Z sup x = P ki =1 λ i x i k Y i =1 f λ i i ( x i ) ≥ (cid:18)Z f dx (cid:19) λ (cid:18)Z gdy (cid:19) − λ . Applying the induction step, one gets Z gdy ≥ k Y i =2 (cid:18)Z f i ( x i ) dx i (cid:19) λi − λ . This completes the proof of the inequality. The equality characterization followsfrom the equality characterization for two functions. Theorem 4.2. Let f i : R n → R + , ≤ i ≤ k , be measurable unconditional functionssatisfying k Y i =1 f i ( x i ) ≤ ρ k X i,j =1 ,i Proposition 5.2. Let K i , ≤ i ≤ k , be unconditional convex bodies in R n withradial functions r i = r K i . Assume that for all x i = (( x i ) , · · · , ( x i ) n ) ∈ R n , k Y i =1 r i ( x i ) ≤ (cid:16)P nj =1 (cid:0) | ( x ) j | k · · · | ( x k ) j | k (cid:1) (cid:17) k . (5.2) Then k Y i =1 vol n ( K i ) ≤ (vol n ( B n )) k . Equality holds if and only if K i is a Euclidean ball for all ≤ i ≤ k .Proof. Let m ∈ R , 1 ≤ m < n and put x i = e t i . Set w = k P ki =1 t i . Then k Y i =1 r mi ( e t i ) {| e ti |≤ } e P i,j ( t i ) j ≤ {| e w |≤ } e P i,j ( t i ) j (cid:16)P nj =1 e w j (cid:17) km = {| e w |≤ } e k P nj =1 ( w ) j (cid:16)P nj =1 e w j (cid:17) km . (5.3)We now apply again the change of variables x i = e t i , 1 ≤ i ≤ k , the Pr´ekopa–Leindler inequality and (5.3), k Y i =1 Z B n ∩ R n + r mi dx i ! k = k Y i =1 Z R n r mi ( e t i ) {| e ti |≤ } e P j ( t i ) j dt i ! k ≤ Z R n sup w = k P ki =1 t i " k Y i =1 r mk i ( e t i ) {| e ti |≤ } e k P i,j ( t i ) j dw ≤ Z R n {| e w |≤ } e P nj =1 ( w ) j (cid:16)P nj =1 e w j (cid:17) m dw = Z B n ∩ R n + dx | x | m . Hence by symmetry k Y i =1 Z B n r mi dx i ! k ≤ Z B n dx | x | m . Next we observe that every radial function r i satisfies r i ( x i ) = r i (cid:18) x i | x i | (cid:19) | x i | . For every 1 ≤ m < n , m ∈ R , we introduce the finite probability measure dµ m = Bn ( u ) R Bn du | u | m du | u | m . The inequality above can then be rewritten as follows, k Y i =1 Z B n r mi (cid:18) x i | x i | (cid:19) dµ m ≤ . µ m is rotational invariant, the above inequality can be rewritten as k Y i =1 Z S n − r mi ( θ ) dσ ( θ ) ≤ σ ( S n − ) k , (5.4)where σ is the ( n − m → n and applying the Fatou’s Lemma one gets that (5.4) holds for m = n . On the otherhand, for m = n one has for all i Z S n − r ni ( θ ) dσ ( θ ) = σ ( S n − ) vol n ( K i )vol n ( B n ) . (5.5)From this we derive the desired estimate.Now we address the equality characterizations. If equality holds in the inequality,then equality holds everywhere and we get with (5.4) for m = 1, k Y i =1 Z B n ∩ R n r i dx i = k Y i =1 Z S n − r i ( θ ) dσ ( θ ) = σ ( S n − ) k . (5.6)Equality holds in the Pr´ekopa–Leindler inequality. Thus there exist a convex func-tion ψ and y , · · · , y k such that for all xr ( x − y ) B n ( x − y ) R R n ∩ B n r dx = r ( x − y ) B n ( x − y ) R R n ∩ B n r dx = · · · r k ( x − y k ) B n ( x − y k ) R R n ∩ B n r k dx = e − ψ ( x ) . (5.7)As K i is in particular 0-symmetric, we have that r i ( x − y i ) = k x − y i k Ki . We put R i = R B n ∩ R n r i dx i and thus we get for all i, jR i k x − y i k K i B n ( x − y j ) = R j k x − y j k K j B n ( x − y i ) . We let x = y i in this inequality and get for all j = i ,0 = k y i − y j k K j or y = y = · · · = y k = y . With z = x − y , we get by (5.7) for all z ∈ B n , r ( z ) R = r ( z ) R = · · · r k ( z ) R k = e − ψ ( z + y ) . (5.8)With (5.6) k Y i =1 r k i ( z ) = k Y i =1 R i ! k e − k P ki =1 ψ ( z + y ) = σ ( S n − ) e − ψ ( z + y ) . Now we use (5.2) and get that Q ki =1 r k i ( z ) = | z | . Hence we have for all i ,1 σ ( S n − ) | z | = e − ψ ( z + y ) = r i ( z ) R i , which means that K i = R i σ ( S n − ) B n for all i .16 The symmetric case In the symmetric case, we prove the following theorem. Here we apply a symmetriza-tion technique, which is the standard tool for results of this kind, see, for instance,[3], [26]). Theorem 6.1. Let λ ∈ R + and assume that the symmetric non-negative functions f i satisfy the inequality k Y i =1 f i ( x i ) ≤ e − λ (cid:12)(cid:12) P ki =1 x i (cid:12)(cid:12) . (6.1) Then k Y i =1 Z f i ( x i ) dx i ≤ (cid:16)Z e − λ k | x | dx (cid:17) k . Remark 6.2. This statement is weaker than our conjectured inequality (3.1). In-deed, since (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ kk − k X i,j =1 ,i Lemma 6.3. Let t = ( t , · · · , t n ) ∈ R n and ˜ F ( t , t , · · · , t n ) = 12 (cid:16) F ( t , t , · · · , t n ) + F ( − t , t , · · · , t n ) (cid:17) be the linear symmetrization of a function F with respect to t . Then Z e − ( ˜ F ) ∗ dx ≥ Z e − F ∗ dx. Proof. Let us show that( ˜ F ) ∗ (cid:16) t + s (cid:17) ≤ F ∗ ( t , t , · · · , t n ) + 12 F ∗ ( − s , s , · · · , s n ) . It is suffucient to show that (cid:10) t + s , u (cid:11) − (cid:16) F ( u , u , · · · , u n ) + F ( − u , u , · · · , u n ) (cid:17) ≤ F ∗ ( t , t , · · · , t n ) + 12 F ∗ ( − s , s , · · · , s n ) . F ( u , u , · · · , u n ) + F ∗ ( t , t , · · · , t n ) ≥ h t, u i ,F ( − u , u , · · · , u n ) + F ∗ ( − s , s , · · · , s n ) ≥ h s, u i . Hence by the Pr´ekopa–Leindler inequality Z e − ( ˜ F ) ∗ dt ≥ (cid:18)Z e − F ∗ ( t ,t , ··· ,t n ) dt (cid:19) (cid:18)Z e − F ∗ ( − s ,s , ··· ,s n ) ds (cid:19) = Z e − F ∗ dt. Proof. [ Theorem 6.1 ] We define functions V i by f i ( x i ) = e − V i ( x i ) − λ | x i | . Note that the V i are even functions. The f i satisfy assumption (6.1), which isequivalent to the V i satisfying k X i =1 V i ( x i ) ≥ λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − | x i | = λ k X i,j =1 ,i 18o prove (6.3) let us consider symmetrizations˜ F i ( t ) = 12 (cid:0) F i ( t , · · · , t n ) + F i ( − t , · · · , t n ) (cid:1) with respect to the first coordinate t . Clearly, the symmetrized functions ˜ F i dosatisfy (6.2) as well. By Lemma 6.3, Z e − ˜ F ∗ i dx i ≥ Z e − F ∗ i dx i = Z e − λ | xi | − V i ( x i ) dx i . Since the functions ˜ F i satisfy (6.2), the functionsˆ V i = ˜ F ∗ i − λ | x i | satisfy the main assumption. In addition, they are even, symmetric with respect tothe first coordinate and satisfy the desired inequality (6.3). The proof is complete. Recall that for a given family of probability measures µ , · · · , µ k its barycenter µ (with coefficients 1 /k ) is the minimum point of the functional F ( ν ) = 12 k k X i =1 W ( µ i , ν ) . Here, W ( ν , ν ) = inf (cid:26)Z | x − y | dπ : π ∈ P ( R n × R n ) , π ( · , R n ) = ν , π ( R n , · ) = ν (cid:27) is the L Kantorovitch distance of probability measures ν , ν .Recall that γ = π ) n e − | x | dx is the standard Gaussian measure.We apply the following result of Agueh–Carlier [1], Proposition 4.2. Theorem 7.1. [1] k X i =1 W ( µ i , µ ) = Z k X i =1 | x i − T ( x ) | dπ, where T ( x ) = k ( x + · · · + x k ) and π is the solution to the multimarginal Monge–Kantorovich maximization problem with marginals µ i and cost function P ki,j =1 ,i Assume that for ≤ i ≤ k , µ i = ρ i · γ and the ρ i are unconditional.Then F ( µ ) ≤ k − k k X i =1 Z ρ i log ρ i dγ = k − k k X i =1 Ent γ ( µ i ) . (7.1) Proof. By the Kantorovich duality (see e.g., [31]), F ( µ ) = k − k Z k X i =1 f i ( x i ) dπ = k − k Z k X i =1 f i ( x i ) dµ i , for some f i satisfying k X i =1 f i ( x i ) ≤ k − k X i,j =1 ,i Fathi has shown that in the class of symmetric functions inequality (7.2) is equivalentto a Blaschke–Santal´o inequality involving two exponential functions. We follow hisapproach in [12] to show that the inequality of Theorem 7.2 is also equivalent to afunctional Blaschke–Santal´o for multiple exponential functions.Indeed, letting ρ ( t ) = e − tk − in Theorem 3.1, we get the following multifunctionalBlaschke–Santal´o inequality:Let f i : R n → R + , 1 ≤ i ≤ k , be measurable unconditional functions such that k X i =1 f i ( x i ) ≤ − k − k X i,j =1 ,i Inequality (7.1) is equivalent to the functional Blaschke–Santal´oinequality (7.4).Proof. One implication is just Theorem 7.2.For the other implication, we first rewrite inequality (7.1). Thus, let µ be thebarycenter of the µ i = ρ i · γ with coefficients k and unconditional ρ i . We recall thatfor a probability measure ν Ent γ ( ν ) = Ent dx ( ν ) + n π ) + 12 Z | x | dν and use this and the definition of the Kantorovitch distance to get that (7.1) isequivalent to − k inf π k X i,j =1 ,i 1) log(2 π ) n (7.5)Let now the f i be unconditional and such that they satisfy (7.3). We apply (7.5) to µ i = ρ i γ = e fi R e fi . We also use that for a probability measure ν Ent dx ( ν ) = sup f Z f dν − log Z e f dx (7.6)21nd get − k inf π k X i,j =1 ,i 1) log(2 π ) n . (7.7)By the Kantorovitch duality, the left hand side of this inequality equals − k inf π k X i,j =1 ,i Theorem 7.5. Assume that µ i = ρ i · γ , where γ is the standard Gaussian measureand that the ρ i are symmetric. Then F ( µ ) ≤ k k X i =1 Z ρ i log ρ i dγ. Proof. 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Kolesnikov Faculty of MathematicsNational Research Institute Higher School of EconomicsMoscow, Russian Federation [email protected] Elisabeth Werner Department of Mathematics Universit´e de Lille 1Case Western Reserve University UFR de Math´ematiqueCleveland, Ohio 44106, U. S. A. 59655 Villeneuve d’Ascq, France [email protected]@case.edu