Concentration inequalities and geometry of convex bodies

Transcription

1 Concentration inequalities and geometry of convex bodies Olivier Guédon 1, Piotr Nayar, Tomasz Tkocz 3 March 5, 14 Abstract Our goal is to write an extended version of the notes of a course given by Olivier Guédon at the Polish Academy of Sciences from April 11-15, 11. The course is devoted to the study of concentration inequalities in the geometry of convex bodies, going from the proof of Dvoretzky s theorem due to Milman [75] until the presentation of a theorem due to Paouris [78] telling that most of the mass of an isotropic convex body is contained in a multiple of the Euclidean ball of radius the square root of the ambient dimension. The purpose is to cover most of the mathematical stuff needed to understand the proofs of these results. On the way, we meet different topics of functional analysis, convex geometry and probability in Banach spaces. We start with harmonic analysis, the Brascamp-Lieb inequalities and its geometric consequences. We go through some functional inequalities like the functional Prékopa-Leindler inequality and the well-known Brunn-Minkowski inequality. Other type of functional inequalities have nice geometric consequences, like the Busemann Theorem, and we will present some of them. We continue with the Gaussian concentration inequalities and the classical proof of Dvoretzky s theorem. The study of the reverse Hölder inequalities also called reverse Lyapunov s inequalities) is very developed in the context of log-concave or γ-concave functions. Finally, we present a complete proof of the result of Paouris [78]. We will need most of the tools introduced during the previous lectures. The Dvoretzky theorem, the notion of Z p bodies and the reverse Hölder inequalities are the fundamentals of this proof. There are classical books or surveys about these subjects and we refer to [8, 9, 13, 48, 49, 55, 3, 8, 7] for further readings. The notes are accessible to people with classical knowledge about integration, functional and/or harmonic analysis and probability. 1 Université Paris Est, Laboratoire d Analyse et de Mathématiques Appliquées UMR 85), UPEM, F-77454, Marne-la-Vallée, France, olivier.guedon@u-pem.fr University of Warsaw, Institute of Mathematics, Ul. Banacha, -97 Warsaw, Poland, nayar@mimuw.edu.pl 3 University of Warwick, Mathematics Institute, Coventry CV4 7AL, United Kingdom, t.tkocz@warwick.ac.uk 1

2 Contents Contents 1 Introduction 3 Brascamp-Lieb inequalities in a geometric context 7.1 Motivation and formulation of the inequality The proof Consequences of the Brascamp-Lieb inequality Notes and comments Borell and Prékopa-Leindler type inequalities, the notion of Ball s bodies Brunn-Minkowski inequality Functional version of the Brunn-Minkowski inequality Functional version of the Blaschke-Santaló inequality Borell and Ball functional inequalities Consequences in convex geometry Notes and comments Concentration of measure. Dvoretzky s Theorem Isoperimetric problem Concentration inequalities Dvoretzky s Theorem Comparison of moments of a norm of a Gaussian vector Notes and comments Reverse Hölder inequalities and volumes of sections of convex bodies Berwald s inequality and its extensions Some concentration inequalities Kahane Khinchine type inequalities Notes and comments Concentration of mass of a log-concave measure The result The Z p -bodies associated with a measure The final step Notes, comments and further readings

3 1 Introduction In harmonic analysis, Young s inequalities tell that for a locally compact group G equipped with its Haar measure, if 1/p + 1/q = 1 + 1/s then f L p G), g L q G), f g s f p g q. The constant 1 is optimal for compact groups such that constant functions belong to each L p G). However, it is not optimal for example in the real case. During the seventies, Beckner [14] and Brascamp-Lieb [6] proved that the extremal functions in Young s inequality are among Gaussian densities. We discuss the geometric version of these inequalities introduced by Ball [7]. The problem of computing the value of the integrals for the maximizers disappears when we write these inequalities in a geometric context. The proof can be done via the transport argument that we will present. The geometric applications of this result are that the cube, among symmetric convex bodies, has several extremal properties. Indeed, Ball [7] proved a reverse isoperimetric inequality, namely that for every centrally symmetric convex body K in R n, there exists a linear transformation K of K such that Vol K = Vol B n and Vol K Vol B n. Moreover, in the case of random Gaussian averages, Schechtman and Schmuckenschläger [86] proved that for every centrally symmetric convex body K in R n which is in the so-called John position, E g 1,..., g n ) K E g 1,..., g n ) where g 1,..., g n are independent Gaussian standard random variables. Another powerful inequality in convex geometry is the Prékopa-Leindler inequality [8]. This is a functional version of the Brunn-Minkowski inequality which tells that for any non-empty compact sets A, B R n vola + B) 1/n vola) 1/n + volb) 1/n. We prove the Prékopa-Leindler inequality and we discuss a modified version of this inequality introduced by Ball [6], see also [4]. Ball [6] used it to create a bridge between probability and convex geometry, namely that one can associate a convex body with any log-concave measure. 1.1 Theorem. Suppose f : R n R + L 1 R n ) is an even log-concave function and p > 1. Then 1/p+1 x = r frx)dr) p, x, x = defines a norm on R n. 3

4 The result is seen as a generalisation of Busemann theorem [9]. Some properties of these bodies will be studied in Section 6. Dvoretzky s Theorem tells that l is finitely representable in any infinite dimensional Banach space. Its quantified version due to Milman [8] is one of the fundamental result of the local theory of Banach spaces. 1. Theorem. Let K be a symmetric convex body such that K B n. Define M K) = h K θ)dσθ). S n 1 Then for all ε > there exists a vector subspace E of dimension such that k = k K) = cnm K)) ε / log1/ε) 1 ε)m K)P E B n P E K 1 + ε)m K)P E B n. Instead of using the concentration of measure on the unit Euclidean sphere, this can be proved via the use of Gaussian operators. We will present some classical concentration inequalities of a norm of a Gaussian vector following the ideas of Maurey and Pisier [8]. The argument of the proof of Dvoretzky s theorem is now standard and is done in three steps: a concentration inequality for an individual vector of the unit sphere, a net argument and discretisation of the sphere, a union bound and optimisation of the parameters. The subject of the inverse Hölder inequalities is very wide. In the context of logconcave or s-concave measures, major tools were developed by Borell [1, ]. In particular, he proved that for every log-concave function f : [, ) R +, the function p 1 Γp + 1) t p ft)dt is log-concave on 1, + ). For p 1, the Z p -body associated with a log-concave density f is defined by it support function 1/p h Zpf)θ) = x, θ +fx)dx) p, where x, θ + is the positive part of x, θ. We present some basic properties of these bodies. It will be of particular interest to understand the behaviour of the bodies Z p π E f)) where π E f) is the marginal density of f on a k-dimensional subspace E. The inverse Hölder inequalities give some information and we will try to explain how it reflects geometric properties of the density f. The goal of the last Section is to present a probabilistic version of Paouris theorem [78] that appeared in []. 4

5 1.3 Theorem. There exists a constant C such that for any random vector X distributed according to a log-concave probability measure on R n, we have E X p ) 1/p C E X + σ p X)) for all p 1, where σ p X) = sup θ S n 1 E X, θ p + is the weak p-th moment associated with X. Moreover, if X is such that for any θ S n 1, E X, θ = 1, then for any t 1, where c is a universal constant. P X c t n) exp t n), Most of the tools presented in the first lectures are needed to make this proof : Dvoretzky s theorem, Z p -bodies, the inverse Hölder inequalities. The sketch of the proof is the following. Let G N, Id) be a standard Gaussian random vector in R n. Observe that for any random vector X distributed with a log-concave density f, E X p ) 1/p = γ + p ) 1 E X E G X, G p +) 1/p = γ + p ) 1 E G h Zpf)G) p) 1/p, where for a standard Gaussian random variable g N, 1), γ + p = Eg p +) 1/p. By a Gaussian concentration inequality, we see that for any 1 p ck Z p f)), EhZpf)G) p) 1/p EhZpf)G) = M Z p f))e G, where k Z p f)) is the Dvoretzky dimension of the convex Z p f). Looking at the conclusion of Dvoretzky s theorem, we also observe that M Z p f)) is the 1 -th power of k the volume of most of the k-dimensional projection of Z p f) where k k Z p f)). It remains to study the volume of these projections. For any k dimensional subspace E, let π E f denote the marginal of the density f on E, that is x E, π E fx) = fx + y)dy. E By the Prékopa-Leindler inequality, π E f is still log-concave on E. We can prove that for any p 1 and any k-dimensional subspace E P E Z p f)) = Z p π E f) = Z p K k+p π E f)), where K k+p π E f) is the convex body whose norm is x Kk+p π E f) = k + p) 5 ) 1 t k+p 1 k+p π E ftx)dt.

6 In a log-concave setting, we will see that for p k, Z p K k+p π E f)) is approximately K k+p π E f) so that the 1-th power of the volume of P k EZ p f)) is approximately the 1-th k power of the volume of K k+p π E f). The reverse Hölder inequalities will give several properties that will lead to the conclusion. Besides the standard notation, we adopt throughout the notes the common convention that universal constants sometimes change from line to line. Acknowledgements. We are grateful to the Institute of Mathematics Polish Academy of Sciences IMPAN) for its hospitality when the lectures were given and to Rafa l Lata la for his editorial work. We would like also to thank an anonymous referee who read carefully the preliminary version of these notes and proposed several improvements of the presentation. 6

7 Brascamp-Lieb inequalities in a geometric context.1 Motivation and formulation of the inequality Let G be a locally compact group with Haar measure µ. Let p, q, s 1 be such that 1/p + 1/q = 1 + 1/s, let f L p G, µ) and g L q G, µ). Then we have the following Young s inequality f g s f p g q,.1) where f g)x) = G fxy 1 )gy) dµy). The constant 1 in.1) is optimal when constant functions belong to L p G), p 1, but it is not optimal when G = R and µ is the Lebesgue measure. In the seventies, Beckner and independently Brascamp and Lieb proved that in R the equality case is achieved for sequences of functions f n and g n with Gaussian densities, i.e. functions of the form h a x) = a/πe ax. Note that if 1/r + 1/s = 1 then f g s = sup h L rr) h r 1 fx y)gy)hx) dy dx R R and we have f L p R), g L q R), h L r R) with = =. Let r p q r s v 1 = 1, 1), v =, 1) and v 3 = 1, ). Then fx y)gy)hx) dy dx = f X, v 1 )g X, v )h X, v 3 ) dx. R R R This is a type of expression studied by Brascamp and Lieb. Namely, they prove.1 Theorem. Let n, m 1 and let p 1,..., p m > be such that m i=1 v 1,..., v m R n and f 1,..., f m : R R + then R m m i=1 f i v i, x ) dx m i=1 f i pi 1 p i = n. If is maximized when f 1,..., f m are Gaussian densities. However, the supremum may not be attained in the sense that one has to consider Gaussian densities f a with a. In this context, it remains to compute the constants for the extremal Gaussian densities which is not so easy. In a geometric setting we have a version of the Brascamp-Lieb inequality due to Ball [7]. 7

8 . Theorem. Let n, m 1 and let u 1,..., u m S n 1, c 1,..., c m > be such that Id = m c ju j u j. If f 1,..., f m : R R + are integrable functions then m m ) cj f j x, u j )) c j dx f j..) R n R.3 Remark. The condition means that and is equivalent to Id = m c j u j u j.3) x R n, x = x R n, x = m c j x, u j u j m c j x, u j. We can easily construct examples of vectors satisfying condition.3). Let H be an n- dimensional subspace of R m. Let e 1,..., e m be the standard orthonormal basis in R m and let P : R m H be the orthogonal projection onto H. Clearly, Id R m = m e j e j and x = m x, e j e j, hence P x = m x, e j P e j. If x H then P x = x and x, e j = P x, e j = x, P e j, therefore x = m x, P e j P e j. Thus Id H R n = m c ju j u j, where c j = P e j and u j = P e j / P e j..4 Remark. Let f j t) = e αt for 1 j m. If.3) is satisfied then ) m m f j x, u j )) c j = exp αc j x, u j = exp α x ). Thus, R n m f j x, u j )) c j dx = exp α x ) dx = R n = m cj exp αt ) dt) = R exp αt ) dt R m ) n ) cj f j, R since we have n = trid) = m c j tru j u j ) = m c j u j = m c j. Therefore we have equality in.) when f j s are identical Gaussian densities. 8

9 . The proof We start the proof of Theorem. with a simple lemma..5 Lemma. Suppose u 1,..., u m S n 1 and c 1,..., c m are positive numbers. Assume that Id = m c ju j u j. Then 1) If x = m c jθ j u j for some numbers θ 1,..., θ m, then x m c jθ j. ) For all T LR n ) we have det T m a generalisation of Hadamard s inequality). T u j c j 3) For all α 1,..., α m > we have m ) det c j α j u j u j Moreover, if α 1 =... = α m, then equality holds. m α c j j. Proof. 1) Using the Cauchy-Schwarz inequality we obtain m m x = x, x = c j θ j u j, x = c j θ j u j, x m c j θj ) 1 m c j u j, x ) 1 m = c j θj ) 1 x. ) We can assume that T is symmetric and positive definite. Indeed, since T T is symmetric, for any T GL n R) we have the decomposition T T = U DU, where U is orthogonal and D is diagonal. Let S = U D 1 U. Clearly, S = T T and S is symmetric and positive definite. Suppose we can show ) for S. Then we have and det S = det D = det T T = det T T u j = T u j, T u j = u j, T T u j = u j, S u j = Suj, Su j = Su j. Thus ) is also true for T. 9

10 Assume that T is symmetric and positive definite. Then there exist λ 1,..., λ n > and an orthonormal basis v 1,..., v n of R n such that T = n λ i v i v i. i=1 Clearly, T u j = n i=1 λ i u j, v i v i and therefore T u j = n λ i u j, v i. i=1 Since u j = 1, we have n i=1 u j, v i = 1. Let λ i = a i and p i = u j, v i. Then n i=1 p i = 1 and therefore by the AM GM inequality, we get n a i p i i=1 n i=1 a p i i, which means that We obtain m T u j c j as m c j u j, v i = v i = 1. T u j n i=1 n i=1 λ u j,v i i. m λ c j u j,v i i = n λ i = det T, 3) We prove that for all symmetric positive definite matrices we have det S) 1/n = min T : det T =1 i=1 tr T ST )..4) n If λ 1,..., λ n are the eigenvalues of the symmetric and positive definite matrix T ST then tr T ST ) = 1 n n ) 1/n λ i λ i = dett ST )) 1/n = det S) 1/n. n n i=1 i=1 To find the equality case in.4) take the orthogonal matrix U such that S = U DU, where D is diagonal. Let T = D det S) 1/n ) 1 U = D1 U. 1

11 Clearly, det T = 1. We also have tr T ST ) n = tr D 1USU D 1 ) n = tr D 1D) n = det S) 1/n. Let S = m c jα j u j u j. Following our last observation we can find a matrix T with det T = 1 such that det S) 1/n = tr T ST ). Note that n T u j u j )T = T u j u jt = T u j T u j ) = T u j ) T u j ). Therefore m )) 1/n det c j α j u j u j = 1 m ) n tr c j α j T u j T u j = 1 n m c j α j T u j m αj T u j ) c j n m α c j n j. The second inequality follows from point ) of our lemma. Besides the lemma, we need the notion of mass transportation. Let us now briefly introduce it..6 Definition. Let µ be a finite Borel measure on R d and let T : R d R d be measurable. The pushforward of µ by T is a measure T µ on R d defined by T µ A) = µt 1 A)), A BR d ). If ν = T µ then we say that T transports µ onto ν. Note that if ν = T µ then for all bounded Borel functions h : R d R we have hy) dνy) = ht x)) dµx). R d R d If µ and ν are absolutely continuous with respect to the Lebesgue measure, i.e. dµx) = fx)dx and dνy) = gy)dy then hy)gy) dy = ht x))fx) dx. R d R d Assuming T is C 1 on R d, we obtain by changing the variable in the first integral hy)gy) dy = ht x))gt x)) det dt x) dx, R d R d 11

12 where dt is the differential of T. Therefore µ almost everywhere we have gt x)) det dt x) = fx). This is the so called transport equation or a Monge-Ampère equation). Assume that µ and ν are probabilistic measures absolutely continuous with respect to the Lebesgue measure on R, say measures with densities f, g. There exists a map T : R R which is non-decreasing and which transports µ onto ν. Indeed, define T by If then x ft) dt = Rx) = x T x) T x) = R 1 x gt) dt gu) du. ) ft) dt. The simplest case is when f and g are continuous and strictly positive. Then T is of class C 1 and T x)gt x)) = fx), x R. In higher dimensions for T we can set the so called Knöthe map [61] or Brenier map [8]. For instance, the Brenier map is of the form T = φ, where φ is a convex function. Proof of Theorem.. We have Id = m c ju j u j and u j = 1. We would like to prove m m ) cj f j x, u j, )) c j dx f j. R n By homogeneity we can assume that f j = 1. Moreover, let us suppose that each f j is continuous and strictly positive. Let gs) = e πs. Then g = 1. Let T j : R R be the map which transports f j x)dx onto gs)ds, i.e. t f j s) ds = Tj t) gs) ds. We have the transport equation f j t) = T jt)gt j t)). Hence, using 3) of Lemma.5 we obtain m m f j x, u j )) c j dx = T j x, u j ) ) m c j gt j x, u j ))) c j dx R n R n 1

13 m ) det c j T j x, u j ) u j u j exp π R n ) m c j T j x, u j )) dx. Note that T j > since f and g are strictly positive and continuous. Let y = m c j T j x, u j ) u j. Note that and therefore By 1) of Lemma.5 we have y x i = D y x) = m c j T j x, u j ) u j, e i u j m c j T j x, u j ) u j u j. m c j T j x, u j )) y, thus, changing variables we arrive at m f j x, u j, )) c j dx exp ) π y dy = 1. R n R n For general integrable functions f j : R R +, let ε > and define f ε) j is a centered Gaussian variable of variance ε. The new function f ε) j positive so the inequality holds true for the functions f ε) 1,..., f ε) classical Fatou lemma gives the inequality for f 1,..., f m )..3 Consequences of the Brascamp-Lieb inequality Let us state the reverse isoperimetric inequality. = f j g ε where g ε is C 1 and strictly m ). Letting ε, the.7 Theorem. Let K be a symmetric convex body in R n. Then there exists an affine transformation K of K such that K = B n, and K B n.5) or equivalently K K n 1 n Bn B n n 1 n = n..6) 13

14 Before we give a proof of Theorem.7 we introduce the notion of the volume ratio..8 Definition. Let K R n be a convex body. The volume ratio of K is defined as { K ) 1/n vrk) = inf, E K is an ellipsoid}. E The ellipsoid of maximal volume contained in K is called the John ellipsoid. If the John ellipsoid of K is equal to B n then we say that K is in the John position. We have the following two theorems..9 Theorem. For every symmetric convex body K R n we have vrk) vrb n ) =..7) B n 1/n ).1 Theorem. If B n K is the ellipsoid of maximal volume contained in a symmetric convex body K R n then there exist c 1,..., c m > and contact points u 1,..., u m R n such that u j = u j K = u j K = 1 for 1 j m and Id R n = m c j u j u j..8) Here we do not give a proof of Theorem.1. Originally, John [58] proved it with a simple extension of the Karush, Kuhn and Tucker theorem in optimisation to a compact set of constraints instead of finite number of constraints). We refer to [5] for a modern presentation, very close to the original approach of John. We only show how John s theorem implies Theorem.9. Proof of Theorem.9. The quantity vrk) is invariant under invertible linear transformations.we let as an exercise to check that the ellipsoid of maximal volume contained in K is unique. Therefore we may assume that the John ellipsoid of K is B n. Using Theorem.1 we find numbers c 1,..., c m > and unit vectors u 1,..., u m R n on the boundary of K such that m Id R n = c j u j u j. Since u j B n K and K is symmetric we get K K := {x R n, x, u j 1, for all 1 j m}. Let f j t) = 1 [ 1,1] t) for 1 j m. Note that f j = f c j j, 1 j m. From Theorem. we have K K = m R n f c j j x, u j ) dx 14 m f j ) cj = m c j = n = B n.

15 Clearly, this also yields that B n is the John ellipsoid for the cube B n. Therefore vrb n ) = B n ) 1/n. We finish our considerations on the reverse isoperimetric problem showing that Theorem.9 implies Theorem.7. Proof of Theorem.7. Let K be the linear image of K such that B n ellipsoid of K. By Theorem.9 we have K n. Hence, K is the John K + εb n K ε = n K n 1 1 = n K n K n K = lim inf ε + This finishes the proof as the ratio K K n 1 n lim inf ε + n K n 1 n. is affine invariant. K + ε K K ε We state yet another application of the Brascamp-Lieb inequality..11 Theorem. If K is a symmetric convex body in the John position then E G K E G, where G is the standard Gaussian vector in R n, i.e. the vector g 1,..., g n ) where g i ) i n are independent standard Gaussian random variables. Proof. As in the proof of Theorem.7 we consider numbers c 1,..., c m > and vectors u 1,..., u m satisfying the assertion of the Theorem.1. Note that Clearly, Moreover, K K = {x R n, x, u j 1 1 j m}. G K G K = max 1 j m G, u j. E G K = + ) P max G, u j t j We have G = max 1 j m G, e j so that + ) E G = P max G, e j t dt = j + dt. 1 P g t) n ) dt, where g is the standard Gaussian random variable. To get the conclusion, it suffices to prove ) P max G, u j t P g t)) n. j 15

16 Take h j s) = 1 [ t,t] s) e s / π, f j s) = 1 [ t,t] s). Since Theorem. implies that ) P max G, u j t = j = = x = m c j x, u j, 1 / π) n/ e x dx m f c j 1 j x, u j ) exp x, u ) j cj dx π) n/ m h j x, u j ) c j dx R n 1 {maxj x,u j ) t} R n R n m = P g t)) n, ) cj t h j = where we have used the fact that m c j = n..4 Notes and comments t ) n 1 e u / du π This section is devoted to the study of the Brascamp-Lieb inequalities [6] in a convex geometric setting. As we emphasized, this approach is due to Ball [7] where he proved Theorem.9 and Theorem.7. We refer to [9] for a large survey on this subject. The proof using mass transportation approach is taken from [1]. It is important to notice a significant development of this study, the reverse Brascamp-Lieb inequality due to Barthe [11]. Theorem.11 is due to Schechtman and Schmuckenschläger [86] and has a very nice application in the study of Dvoretzky s theorem, because it gives a Euclidean structure associated with a convex body where the minimum among convex bodies K of MK) = S n 1 x dσ n x) is known and attained for the cube. We refer to Section 4 to learn about it. A non-symmetric version of these results is known, see [7, 88, 1]. 16

17 3 Borell and Prékopa-Leindler type inequalities, the notion of Ball s bodies 3.1 Brunn-Minkowski inequality Brunn discovered the following important theorem about sections of a convex body. 3.1 Theorem. Let n and let K be a convex body in R n. Take θ S n 1 and define H r = {x R n, x, θ = r} = rθ + θ. Then the function is concave on its support. r volh r K)) 1 n 1 Minkowski restated this result providing a powerful tool. 3. Theorem. If A and B are non-empty compact sets then for all λ [, 1] we have vol 1 λ)a + λb) 1/n 1 λ)vol A) 1/n + λvol B) 1/n. 3.1) Note that if either A = or B =, this inequality does not hold in general since 1 λ)a + λb =. We can use homogeneity of volume to rewrite Brunn-Minkowski inequality in the form vol A + B) 1/n vol A) 1/n + vol B) 1/n. 3.) At this stage, there is always a discussion between people who prefer to state the Brunn- Minkowski inequality for Borel sets but it remains to prove that if A and B are Borel sets then A + B is a measurable set) and people who prefer to work with approximation and say that for any measurable set C, vol C is the supremum of the volume of the compact sets contained in C. We choose the second way in this presentation. The proof of the theorem of Brunn follows easily. For any t R, define A t = {x θ, x + tθ K}. Observe that when s = 1 λ)r + λt, only the inclusion A s λa t + 1 λ)a r is important. And inequality 3.1) applied in θ which is of dimension n 1 leads to the conclusion. We can also deduce from inequality 3.) the isoperimetric inequality. 3.3 Theorem. Among sets with prescribed volume, the Euclidean balls are the one with minimum surface area. 17

18 Proof. By a compact approximation of C, we can assume that C is compact and vol C = vol B n. We have volc + εb n ) volc) vol C = lim inf. ε + ε By the Brunn-Minkowski inequality 3.1), we get volc + εb n ) 1/n vol C) 1/n + εvol B n ) 1/n, hence so vol C) lim inf ε + volc + εb n ) 1 + ε) n vol C, 1 + ε) n 1) volc) ε = n volc) = n volb n ) = vol B n ). There is an a priori weaker statement of the Brunn-Minkowski inequality. Applying the AM GM inequality to the right hand side of 3.1) we get 1 λ)a + λb A 1 λ B λ, λ [, 1]. 3.3) Note that this inequality is valid for any compact sets A and B the assumption that A and B are non-empty is no longer needed). We can see that there is no appearance of dimension in this expression. The strong version of the Brunn-Minkowski inequality 3.1) tells us that the Lebesgue measure is a 1 -concave measure. The weaker statement 3.3) justifies that it is a logconcave n measure. 3.4 Definition. A measure µ on R n is log-concave if for all compact sets A and B we have µ1 λ)a + λb) µa) 1 λ µb) λ, λ [, 1]. 3.5 Definition. The function f : R n R is log-concave if for all x, y R n we have f1 λ)x + λy) fx) 1 λ fy) λ, λ [, 1]. Note that these definitions are dimension free. The weak form of the inequality 3.3) for the Lebesgue measure is in fact equivalent to the strong inequality 3.1). It is a consequence of the homogeneity of the Lebesgue measure. Indeed, if λvol B) 1/n µ = 1 λ)vol A) 1/n + λvol B) 1/n 18

19 then ) 1 λ)a + λb vol 1 λ)vol A) 1/n + λvol B) 1/n = vol vol ) A 1 µ) vol A) + µ B 1/n vol B) 1/n ) 1 µ ) µ A B = 1. vol A) 1/n vol B) 1/n 3. Functional version of the Brunn-Minkowski inequality If we take f = 1 A, g = 1 B and m = 1 1 λ)a+λb then 3.3) says that and obviously m, f, g satisfies m ) 1 λ f ) λ g m1 λ)x + λy) fx) 1 λ gy) λ. We will prove the following functional version of the Brunn-Minkowski inequality called the Prékopa-Leindler inequality. This will conclude the proof of inequality 3.1) and of Theorem Theorem. Let f, g, m be nonnegative measurable functions on R n and let λ [, 1]. If for all x, y R n we have then m1 λ)x + λy) fx) 1 λ gy) λ, m R n R n f ) 1 λ R n g We start with proving inequalities 3.1) and 3.3) in dimension Lemma. Let A, B be non-empty compact sets in R. Then 1 λ)a + λb 1 λ)a + λb, λ [, 1]. Moreover, for any compact sets A, B in R, 1 λ)a + λb A 1 λ B λ, λ [, 1]. ) λ. 3.4) Proof. Observe that the operations A A + v 1, B B + v where v 1, v R do not change the volumes of A, B and 1 λ)a + λb adding a number to one of the sets only 19

20 shifts all of this sets). Therefore we can assume that sup A = inf B =. But then, since A and B, we have 1 λ)a + λb 1 λ)a λb). But 1 λ)a and λb) are disjoint, up to the one point. Therefore 1 λ)a + λb 1 λ)a + λb, hence we have proved 3.1) in dimension 1. The log-concavity of the Lebesgue measure on R follows from the AM GM inequality. Proof of Theorem 3.6. Step 1. Let us now justify the Prékopa-Leindler inequality in dimension 1. We can assume, considering f1 f M and g1 g M instead of f and g, that f, g are bounded. Note also that this inequality possesses some homogeneity. Indeed, if we multiply f, g, m by numbers c f, c g, c m satisfying c m = c 1 λ f c λ g, then the hypothesis and the assertion do not change. Therefore, taking c f = f 1, c g = g 1 and c m = f 1 λ) g λ we can assume since we are in the situation when f and g are bounded) that f = g = 1. But then R m = R R f = g = {m s} ds, {f r} dr, {g r} dr. Note also that if x {f r} and y {g r} then by the assumption of the theorem we have 1 λ)x + λy {m r}. Hence, 1 λ){f r} + λ{g r} {m r}. Moreover, the sets {f r} and {g r} are non-empty for r [, 1). This is very important since we want to use the 1-dimensional Brunn-Minkowski inequality proved in Lemma 3.7! For any non empty compact subsets A {f r} and B {g r}, we have by Lemma 3.7, {m r} 1 λ) A + λ B. Since Lebesgue measure is inner regular, we get that {m r} 1 λ) {f r} + λ {g r}.

21 Therefore, we have + 1 m = {m r} dr {m r} dr 1 λ){f r} + λ{g r} dr λ) {f r} dr + λ {g r} dr = 1 λ) f + λ g ) 1 λ f g) λ. 1 Observe that we have actually proved a stronger inequality m 1 λ) f + λ g, but under the assumption f = g = 1, without which the inequality does not hold as it lacks homogeneity, in contrast to 3.6). Step the inductive step). Suppose our inequality is true in dimension n 1. We will prove it in dimension n. Suppose we have numbers y, y 1, y R satisfying y = 1 λ)y 1 + λy. Define m y, f y1, g y : R n 1 R + by m y x) = my, x), f y1 x) = fy 1, x), g y x) = y, x), where x R n 1. Note that since y = 1 λ)y 1 + λy we have m y 1 λ)x 1 + λx ) = m1 λ)y 1 + λy, 1 λ)x 1 + λx ) fy 1, x 1 ) 1 λ gy, x ) λ = f y1 x 1 ) 1 λ g y x ) λ, hence m y, f y1 and g y satisfy the assumption of the n 1)-dimensional Prékopa-Leindler inequality. Therefore we have m y R n 1 R n 1 f y1 ) 1 λ R n 1 g y ) λ. Define new functions M, F, G : R R + My ) = m y, R n 1 F y 1 ) = f y1, R n 1 Gy ) = g y. R n 1 We have seen the above inequality) that when y = 1 λ)y 1 + λy then there holds M1 λ)y 1 + λy ) F y 1 ) 1 λ Gy ) λ. 1

22 Hence, by 1-dimensional Prékopa-Leindler inequality proved in Step 1, we get ) 1 λ λ M F G). R R R But so we conclude that R M = m, R n R m R n F = f, R n R n f ) 1 λ R R n g G = g, R n ) λ. The next theorem will be useful in the sequel to prove the functional version of the so-called Blaschke-Santaló inequality, Theorem Theorem. Suppose f, g, m : [, ) [, ) are measurable and suppose there exists λ [, 1] such that mt) fr) 1 λ gs) λ, whenever t = r 1 λ s λ Then m ) 1 λ f g) λ. 3.5) Proof. This inequality has a lot of homogeneity. Again, if we multiply f, g, m by numbers c f, c g, c m satisfying c m = c 1 λ f c λ g, then the hypothesis and the assertion do not change. Moreover, we can rescale arguments of f, g, m by d f, d g, d m in such a way that d m = d 1 λ f d λ g. We can assume, by taking f1 f M 1 [ M,M], g1 g M 1 [ M,M] that f and g are bounded and have compact support. Moreover, by scaling we can assume that Let sup rfr) = sup rgr) = ) Mx) = e x me x ), F x) = e x fe x ), Gx) = e x ge x ). Clearly, changing variables we have + mt) dt = + Mω) dω, + ft) dt = + F ω) dω,

23 By 3.6), we get + gt) dt = + Gω) dω. + F ω) dω = 1 {F r} dr By the hypothesis of f, g and m we have and + Gω) dω = 1 {G r} dr. M1 λ)u + λv) = me u ) 1 λ e v ) λ )e u ) 1 λ e v ) λ fe u )) 1 λ ge v )) λ e u ) 1 λ e v ) λ = F u)) 1 λ Gv)) λ. 3.7) Hence, for any r [, 1), if x {F r} and y {G r}, then 1 λ)x+λy {M r}. The sets {F r} and {G r} are not empty therefore by Lemma 3.7 which is the 1- dimensional Brunn-Minkowski inequality), for any non empty compact sets A {F r} and B {G r}, {M r} 1 λ) A + λ B. Since Lebesgue measure is inner regular, we conclude that {M r} 1 λ) {F r} + λ {G r} and + Mω) dω 1 + {M r} dr 1 λ) + F ω) dω + λ ) 1 λ + λ F ω) dω Gω) dω). + Gω) dω Note that after establishing 3.7) we could have directly used the 1-dimensional Prékopa-Leindler inequality, Theorem 3.6. But we can also recover Theorem 3.6. Indeed, let Mt) = {m t}, F r) = {f r}, Gs) = {g s}. We have to prove that + Mt) dt + ) 1 λ + λ F r) dr Gs) ds). Note that if t = r 1 λ s λ, then from the hypothesis of Theorem 3.6 we have From Lemma 3.7, we get {m t} 1 λ){f r} + λ{g s}. Mt) F r) 1 λ Gs) λ even if the sets are empty, because we just use the log-concavity of the Lebesgue measure on R). We conclude by using Theorem

24 3.3 Functional version of the Blaschke-Santaló inequality We only recall the Blaschke-Santaló inequality without the proof). symmetric case. We discuss the 3.9 Definition. Let C be a compact and symmetric set in R n. We define the polar body C by C = {y R n, x C x, y 1}. 3.1 Theorem. Let C be a compact and symmetric set in R n. Then C C B n. B-S) Nevertheless, we will prove its functional version using B-S) inequality Theorem. Suppose f, g : R n [, ) and Ω : [, ) [, ) are integrable and f, g are even. Suppose that Ωt) fx)gy) whenever x, y t. Then R n Ω x ) dx = n B n + t n 1 Ωt) dt ) 1/ f g) 1/. 3.8) 3.1 Remark. We can recover the classical version of the B-S) inequality from the functional one. Take f = 1 C, g = 1 C and Ω = 1 [,1]. If x C and y C then x, y 1. Hence, if t > 1 then Ωt) = fx)gy) =. If t 1 then obviously 1 = Ωt) fx)gy). By Theorem 3.11 we get B-S). Proof of Theorem The first equality is just an integration in polar coordinates. It is enough to prove the statement for the function t sup{ fx)gy) : x, y t }, so that we can assume Ω non-increasing. For r, s, t R + we take φr) = {f r}, ψs) = {g s}, mt) = B n {Ω t} n. We claim that m rs) φr)ψs). Thanks to this we can apply Theorem 3.8 with λ = 1/ and obtain ) 1/ 1/ m φ ψ). Thus, the proof of 3.8) will be finished since + mt) dt = B n = B n + + {Ω t} n dt = B n + {Ω t} + nu n 1 1 [, {Ω t} ] u) dt du = B n nu n 1 du dt + nu n 1 Ωu) du. 4

25 Now we prove our claim. Let C = {f r} and C = {y, x C x, y 1}. Let α = sup{ x, y, fx) r, gy) s}. Using the definition of C, C and α we get {y, gy) s} α C. By the assumption on Ω, f, g we obtain Ωu) rs for u < α, hence {Ω rs} α. Therefore, m rs) α n B n. By the B-S) inequality we have C C B n. Thus, B n {f r} C {f r} {g s} α n, so φr)ψs) = {f r} {g s} B n α n m rs). 3.4 Borell and Ball functional inequalities The following is another type of functional inequality, in the spirit of Theorem 3.6. We will see in the next section its role in convex geometry Theorem. Suppose f, g, m :, ) [, ) are measurable and such that { mt) sup fr) s r 1 r+s gs) r+s, r + 1 s = } t for all t >. Then ) 1 mt)t p 1 p dt for every p >. ) 1 ft)t p 1 p ) 1 dt + gt)t p 1 p dt 3.9) Proof. Considering min{f i, M}1 M for f 1 = f, f = g, f 3 = m we can assume that f, g, m are bounded, compactly supported in, ) and not a.e. We do not have good homogeneity. Let θ > be such that Let Define A = sup r p+1 fr) = θ p+1 sup r p+1 gr). 3.1) ) 1 ft)t p 1 p ) 1 dt, B = gt)t p 1 p ) 1 dt, C = mt)t p 1 p dt. F u) = f 1 u ) ) p+1 1 1, Gu) = g u θu 5 ) ) p+1 1, u

26 ) p θ Mu) = m Hence, changing the variables we have + F u) du = A p, + Gu) du = θb) p, 1 u ) ) p+1 1. u + We want to prove that C 1 A + 1 B. Note that by virtue of equality 3.1) we have sup G = sup F. We claim that ) p θ Mu) du = C p. Mw) sup{f u) u u+θv Gv) θv u+θv, u + θv = w}, w, ). If u + θv = w, then setting r = 1/u, s = 1/θv), t = 1/w we have 1/r + 1/s = /t and hence F u) u u+θv Gv) θv u+θv 1 s r + s = θv u θv = u u + θv, ) s r = fr) r+s gs) r+s r s r p+1 r+s θs) r+s. We obtain Thus, r s r s r+s s r+s r + s r + r rs θs = 1 + θ) r + s r + s = 1 + θ 1 w. F u) u u+θv Gv) θv u+θv Summarizing, we have sup F = sup G and ) p+1 ) ) p θ 1 1 m = Mw). w w 1 {F ξ} + θ {G ξ} {M ξ}. Therefore, Lemma 3.7 which is nothing else but Brunn-Minkowski inequality in dimension 1) yields that M = 1 sup F {M ξ} dξ 1 F + θ G. sup F {F ξ} dξ + θ sup G {G ξ} dξ 6

27 In terms of A, B, C we have 1 + θ ) p+1 C p 1 Ap + θ θb)p, hence Define φ : [, ) [, ) by C p p Ap + θ p+1 B p 1 + θ) p+1. φθ) = Ap + θ p+1 B p 1 + θ) p+1. Since inf R+ φθ) = φa/b) calculate the derivative to see that φ is unimodal and φ A/B) = ), we get ) C p p Ap + A p+1 /B ) 1 + A p+1 = p A p p AB ) B 1 + A p =. A + B B Now the proof is complete. 3.5 Consequences in convex geometry Having the Prékopa-Leindler inequality at hand we can constitute a handful of basic properties of log-concave measures. We begin with a simple observation that a measure with a log-concave density is log-concave Proposition. If h : R n R + is log-concave and h L 1 loc then µa) = h defines a log-concave measure on R n. Proof. For compact sets A, B take mz) = 1 λa+1 λ)b z)hz), fx) = 1 A x)hx), gy) = 1 B y)hy). Then from log-concavity of h and by the definition of the Minkowski sum we have mλx + 1 λ)y) fx) λ gy) 1 λ. Therefore, by the Prékopa-Leindler inequality, i.e. Theorem 3.6, we have m f) λ g) 1 λ, which is exactly the desired inequality µλa + 1 λ)b) µa) λ µb) 1 λ. A 7

28 For example, the standard Gaussian measure and the standard symmetric exponential distribution on R n are log-concave measures. Another key example is the following. Let µ be the uniform measure on a convex body K R n, that is for any measurable set A R n, K A µa) =. K Since the function x 1 K x) is log-concave, µ is log-concave. It follows also from the weak form of the Brunn-Minkowski inequality, see Lemma 3.7. Now we show that marginal distributions of a log-concave density are again logconcave Theorem. If h : R n+p R + is a log-concave integrable function R n R p ) x, y) hx, y), then the function R n x hx, y) dy R p is log-concave on R n. Proof. We want to prove that for x, x 1 R n, y R p and λ [, 1] we have Let ) λ ) 1 λ h1 λ)x + λx 1, y) dy hx, y) dy hx 1, y) dy. R p R p R p my) = h1 λ)x + λx 1, y), fy) = hx, y), gy) = hx 1, y). Then log-concavity of h yields m1 λ)y + λy 1 ) = h1 λ)x, y ) + λx 1, y 1 )) hx, y ) 1 λ hx 1, y 1 ) λ = fy ) 1 λ gy 1 ) λ. Therefore, by the Prékopa-Leindler inequality, i.e. Theorem 3.6, we get ) 1 λ ) λ my) dy hx, y ) dy hx 1, y 1 ) dy 1. R p R p R p A simple consequence is that the class of log-concave distributions is also closed with respect to convolving Proposition. Let f, g : R n R + be log-concave. Then the convolution f g : x R n fx y)gy)dx is also log-concave. 8

29 Proof. Apply Theorem 3.15 to hx, y) = fx y)gy). Due to the Brunn-Minkowski inequality, the function of the measures of sections of a convex body is not completely arbitrary Theorem. Let K be a convex body in R n and let E be k-dimensional subspace of R n. Let F = E. Then the function f : F R + fy) = vol k y + E) K) is 1/k-concave on its support P F K), namely when fx)fy) >. f 1/k λx + 1 λ)y) λf 1/k x) + 1 λ)f 1/k y), Proof. As in the proof of Theorem 3.1, we deduce from convexity of K and Brunn- Minkowski inequality in R k, i.e. Theorem 3., f 1/k λx + 1 λ)y) vol 1/k k λk x + E)) + 1 λ)k y + E))) λ vol 1/k k K x + E)) + 1 λ) vol 1/k k K y + E)) = λf 1/k x) + 1 λ)f 1/k y) Remark. If K is symmetric with respect to then f is even and therefore f is maximal at. Moreover, it is known from a result of Fradelizi [4] that if K has center of mass at the origin then max fy) e k f). y 3.19 Remark. If K, L are convex bodies in R n, then the function fy) = voly + L) K) is 1 -concave on its support, that is K L. Moreover, Fradelizi [4] proved also that if n K L has barycentre at the origin, then Proof. It is enough to check that max voly + L) K) e n voll K) 3.11) y λx + 1 λ)y + L) K λx + L) K + 1 λ)y + L) K and then the same argument as in Theorem 3.17 finishes the proof. Suppose we have a point λa + 1 λ)b, where a x + L) K and b y + L) K. Then a, b K 9

30 and a = x + a, b = y + b, where a, b L. Therefore, from convexity of K we have λa + 1 λ)b K. Moreover, λa + 1 λ)b = λx + 1 λ)y + λa + 1 λ)b λx + 1 λ)y + L from convexity of L. Our next observation concerns the measure of both sections and projections of convex bodies. 3. Proposition. Let C be a convex body in R n with non-empty interior. Let E be k-dimensional subspace of R n and let F = E. Then 1 P F C) max C y + E) C y F n F C) max C y + E). 3.1) k) P Before giving a proof, we show the following corollary about two bodies, known as Rogers-Shephard inequalities. 3.1 Corollary. Let A, B be two convex bodies in R n. Then n A B max A x) B y) A B x,y Rn n n ) A B max A x) B y). x,y Rn n In particular, if A B has barycentre at the origin then up to a universal constant ) A B ) 1/n A B 1/n A B. Moreover, if A, B are symmetric, then n A + B n A B A B n ) A + B A B. n and Proof. Take C = A B R n and y F ) A B ) 1/n A + B 1/n A B. E = {x, y) R n, x = y}. Then F = E = {x, y) R n, x + y = }. 3

31 Note that x, y) = x + y 1, 1) + x y 1, 1). Therefore, P F x, y) = x y 1, 1), hence { } x y P F C) = 1, 1) R n x A, y B. Consider the linear function L : R n R n, Lx) = x, x). Clearly, L A B)/) = P F C). Therefore, P F C) = A B ) n. Moreover, [ ] A B) x, y) + E) = A x) B y)) E + x, y). If we consider Rx) = x, x), R : R n R n then RA x) B y)) = A x) B y)) E. Thus, C x, y) + E) = ) n A x) B y), and the conclusion follows from Proposition 3.. To prove the second inequality it suffices to observe that if A B has barycentre at the origin, we get from inequality 3.11) that A B max x,y R n A x) B y) en A B. Moreover, if A and B are symmetric, then A = A, B = B and A x) B y) is maximal when x = y =. Proof of Proposition 3.. Consider the function f : F R + given by fy) = y +E) C. Obviously, C = fy) dy P F C) max fy). y F P F C) The second estimate is more delicate. By translation we can assume that max y F fy) = f). Let PF C) be the gauge induced by P F C) on F. If y P F C) then y PF C) 1. Note that y = 1 y PF C) ) + y y P F C). y PF C) 31

32 Since, by Theorem 3.17, f is 1/k-concave on its support P F C), P F C), and y/ y PF C) P F C), we have f 1/k y) f 1/k )1 y PF C) ) + f 1/k y/ y PF C) ) y P F C) f 1/k )1 y PF C) ). Hence, C = P F C) k fy) dy f) 1 y PF C)) dy. P F C) It is clear that for a convex body K in R m, by integrating with respect to the cone measure, we have since K g y K ) dy = y K t 1 y K =t 1 dy = t m K, gt) dy dt = K y K =t 1 gt)mt m 1 dt, 1 dy = mt m 1 K. Applying this to the convex body P F C) which lives in dimension n k, we get C f) P F C) 1 which was our goal since f) = max y F fy). n k)1 t) k t n k 1 dt = f) P F C) n k), Just to illustrate the usefulness of the functional inequalities from the previous section, we show a one dimensional result which does not seem to be obvious at first glance. 3. Proposition. For A, B, ) we set { HA, B) = Then we have 1/a + 1/b, a A, b B HA, B) }. A B A + B. 3.13) Proof. Set f = 1 A, g = 1 B and m = 1 HA,B) and use Theorem 3.13 with p = 1. At the end of this section we consider how to construct a convex body out of a log-concave function. It is a crucial observation following from Theorem Let us emphasize its importance in the sequel Section 6) where we establish basic properties of the so-called Z p -bodies. 3

33 3.3 Theorem. Suppose that a function f : R n [, ) is log-concave, integrable and not a.e.. Then for p > + 1/p x = frx)r dr) p 1, x, x = is a gauge on R n. Proof. Obviously, λx = λ x if λ > and x = if and only if x =. Therefore, the main difficulty is to prove that x + y x + y. Fix x, y R n. Let us take gr) = frx), hs) = fsy) and mt) = f 1 tx + y)) for r, s, t. Suppose 1/r + 1/s = /t. Let λ = r/r + s) so that t/ = λs = 1 λ)r. By log-concavity of f ) 1 mt) = f tx + y) frx) 1 λ fsy) λ = gr) s r r+s hs) r+s. Now it suffices to use Theorem 3.13 for m, g and h. The previous theorem can be seen as a generalisation of a theorem due to Busemann from [9]. Choosing f and p suitably we obtain the following result. 3.4 Theorem. Let K be a symmetric convex body with in its interior. Then x = x x K is a norm on R n. 3.6 Notes and comments Most of the material of this section is taken from the PhD Thesis of Keith Ball [5]. Historically, the names of Prékopa and Leindler stay attached to Theorem 3.6. Indeed, Prékopa [81, 8, 83] studied a lot the notion of log-concave functions. Theorem 3.6 is the culmination in this theory, and yet it is a simple statement. Prékopa s proof uses an argument of transport of mass which can be traced back to Knöthe [61]. On the other hand, Borell [4] submitted his paper only six months after the paper of Prékopa and he presented a more general version of the inequality. But it seems that the general version of a Theorem of Borell [4] has been forgotten. This is why we would like to restate it here. 33

34 3.5 Theorem. Let Ω 1,..., Ω N be open subsets of R n and let φ : Ω 1 Ω N R n be a C 1 function such that φ = φ 1,..., φ n ) and φ j x k i > for all j {1,..., n} and all i {1,..., N}, k {1,..., n}. Define Ω = φω 1,..., Ω N ), and dµ i = f i x)dx where f i L loc 1 Ω i ), i =, 1,..., n. Suppose Φ : [, + ) N [, + ) is a continuous function homogeneous of degree one and increasing in each variable separately. Then the inequality µ φa 1,..., A N )) Φµ 1 A 1 ),..., µ N A N )), holds for all nonempty sets A 1 Ω 1,..., A N Ω N if and only if for almost all x 1,..., x N, for all i = 1,..., N, k = 1,... n and ρ k i >, we have ) n N φ j n n f φx 1,..., x N ) Φ f 1 x 1 ) ρ k 1,..., f N x N ) k=1 i=1 ρ k i x k i Of course, the sets are not necessarily measurable. This is why the measures have to be understood as inner measures. By the inner measure associated with µ we mean µ A) = sup{µk), K A, K compact} defined for any set A. Borell s proof followed the argument of Hadwiger and Ohman [57] and Dinghas [34]. The papers of Das Gupta [3] and of Prékopa [83] illuminate very much the situation. It is now well understood that we can prove the Prékopa-Leindler inequality Theorem 3.6) using a parametrisation argument like we have used in the proof of Theorem.. We refer to [13] for an exhaustive presentation. Fradelizi see [44]) kindly indicated to us that this argument can also be followed for proving Theorem 3.5. Theorem 3.5 is extremely important, not only in the log-concave case but also in the s-concave setting, s R. The case s < is also known in the literature as the case of convex measures or unimodal functions. The geometric consequences of these functional inequalities are now classical. Theorem 3.15 is due to Prékopa [8]. Proposition 3.16 appeared first in [33]. Proposition 3. and Corollary 3.1 are due to Rogers and Shephard [84] and Theorem 3.3 is due to Ball [6]. There was a big amount of work to develop the functional forms of some classical convex geometric inequalities and we refer the interested reader to [4, 45, 46, 66, 67, 47]. k=1 k=1 ρ k N. 34

35 4 Concentration of measure. Dvoretzky s Theorem. 4.1 Isoperimetric problem The Brunn-Minkowski inequality yields the isoperimetric inequality for the Lebesgue measure on R n. Indeed, suppose we have a compact set A R n and let B be a Euclidean ball of the radius r A such that B = A. Then from the Brunn-Minkowski inequality we have A ε 1/n = A + εb n 1/n A 1/n + εb n 1/n = B n 1/n r A + B n 1/n ε = B + εb n 1/n = B ε 1/n. In general, an isoperimetric problem reads as follows. Isoperimetric problem. Let Ω, d) be a metric space and let µ be a Borel measure on Ω. Let α > and ε >. We set A ε = {x Ω, dx, A) ε}. What are the sets A Ω of the measure α such that µa ε ) = inf µb ε). µb)=α This problem is very difficult in general. It has been solved in a few cases. For example, as we have seen, the case of R n equipped with the Lebesgue measure and the Euclidean distance follows from the Brunn-Minkowski inequality. For the spherical and the Gaussian settings the isoperimetry is also known. These two examples will lead us to the notion of the concentration of measure. We start with the spherical case S n 1, d, σ n ) where d is the geodesic metric and σ n is the Haar measure on S n Theorem. For all < α < 1 and all ε >, min{σ n A ε ), σ n A) = α} is attained for a spherical cap C = {x S n 1, dx, x ) r} with x S n 1, r >, such that σc) = α. A crucial consequence of Theorem 4.1 is the concentration of measure phenomenon on S n 1. Indeed, if α = 1 then the spherical cap of measure 1/ is a half sphere. A simple exercise consists in showing that π σ n Cx, r)) c ε) 8 exp n )ε /). It is now easy to deduce the following Corollary. 35

36 4. Corollary. If A is a Borel set on S n 1 such that σ n A) 1/ then π σ n A ε ) 1 8 exp n )ε / ). We can therefore deduce the concentration of Lipschitz functions on the Euclidean sphere. The statement of this result may be considered as the starting point of the concentration of measure phenomenon. It tells that any 1-Lipschitz function on the sphere of high dimension may be viewed as constant when looking at its behaviour on sets of overwhelming measure. Of course the statement is interesting in large dimension. 4.3 Corollary. Let f : S n 1 R be 1-Lipschitz with respect to the geodesic distance. If M is a median of f, namely σ n {f M}) 1 and σ n{f M}) 1, then for ε > π π σ n {f M + ε}) 8 exp nε /4), and σ n {f M ε}) 8 exp nε /4), Moreover, σ n { f M ε}) π exp nε /4). We also know the solution of the isoperimetric problem in the Gaussian setting. Let R n be equipped with the Euclidean distance and γ n be the standard Gaussian distribution dγ n x) = e x / dx π). n/ Let Φ be the distribution function of γ 1, i.e., we define for any u R Φu) = 1 π u e t / dt. 4.4 Theorem. Let a R and let A be a Borel set in R n such that γ n A) = Φa), then γ n A ε ) Φa + ε). The theorem tells that half spaces are solutions of the isoperimetric problem that is γ n A ε ) γ n H ε ), whenever γ n H) = γ n A) = Φa), and for some θ S n 1, H = {x R n, x, θ a} is a half space. As before, having this isoperimetric result at hand, we deduce results concerning the concentration of measure phenomenon in the Gaussian setting. Since for any r > we have 1 Φr) 1 e r / it is easy to deduce the following corollary. 36

37 4.5 Corollary. If A R n satisfies γ n A) 1/ then γ n A r ) 1 1 e r /. Moreover, if F : R n R is a 1-Lipschitz function with respect to and M is a median of F then γ n {F M + r}) 1 e r /, γ n {F M r}) 1 e r / and γ n { F M r}) e r /. 4. Concentration inequalities In many applications we just want concentration inequalities and we do not care much about the constants. This is why we are interested in presenting simpler proofs of these concentration inequalities, which may lead to more general results. We start off by proving the following simple and deep inequality. 4.6 Theorem. Let A R n and let γ n be the Gaussian measure. Then ) dx, A) exp dγ n x) 1 4 γ n A). 4.1) Moreover, if γ n A) 1/ then γ n A ε ) 1 exp ε /4). 4.) Proof. Let and We show that fx) = 1 π) n/ expdx, A) /4) exp x /), gy) = hz) = 1 π) n/ 1 Ay) exp y /) 1 π) n/ exp z /). ) x + y h fx) gy). 37