On the infimum convolution inequality

On the infim um con v olution inequal it y ∗ † ‡ R. Lata la and J. O. W o jtaszczyk Abstract In the pap er we study the infimum c onvolution inequalites. Su c h an inequality w as first introduced b y B. Maurey to gi ve the optimal concen- tration of measure b ehaviour for the p ro du ct exp onential measure. W e sho w ho w IC inequalities are tied to concentration and study th e optimal cost functions for an arbitrary probabilit y measure µ . In particular, we sho w the o p timal IC inequality for produ ct log–conca ve measures and for uniform measures on the ℓ n p balls. Such an optimal inequalit y implies, f or a given measure, in particular the Cen tral Limit Theorem o f Kla rtag and the tai l estimates of P aouris. 1 In tro d u ction and N otation In the semina l pap er [16], B. Maur ey introduced the so called prope rty ( τ ) for a probability measure µ with a co st function ϕ (see Definition 2.1 b elow) and established a v er y elegan t and simple pro of of T alagra nd’s tw o level concen tra- tion for the pro duct ex po nent ia l distribution ν n using ( τ ) for this distribution and an appropriate cost fu nctio n w . It is na tural to ask what other pairs ( µ, ϕ ) hav e pro pe rty ( τ )? As any µ satisfies ( τ ) with ϕ ≡ 0, o ne will ra ther ask how big a cost function can one take. In this pap er w e study the probability measur es µ that have prop er t y ( τ ) with resp ect to the lar gest (up to a m ultiplicative factor) po ssible conv ex co st function Λ ⋆ µ . This b ound comes from checking proper t y ( τ ) for linear funct io ns. W e say a mea sure satisfies the infimum c onvolution ine quality ( I C for shor t) if the pair ( µ, Λ ⋆ µ ) satisfies τ . It turns out that s uch an optimal infimum c onv olution inequality ha s v er y strong consequences. It gives the best po ssible concentration behaviour, gov- erned by the so–c alled L p -centroid b o dies (Corollary 3.10). This, in turn, implies in particular a w eak –strong momen t compar ison (Prop osition 3.12), the Central Limit Theo rem of Klartag [1 2] and the tail estimates e stimates of Paouris [18] (Prop ositio n 3.15). W e be lieve that I C holds for any log–c oncav e probability measure, whic h is the main motiv ation for this paper. ∗ Pa r tially supp orted b y t he F oundation for Polish Science and MEiN Grant 1 PO3A 012 29 † Keyw ords: infimum con vo l ution, c oncetration, log–co ncav e measure, isop erimetry , ℓ n p ball ‡ 2000 Mathematica l Sub ject Classification: 52A20 (52A40, 60E15) 1 Maurey’s ineq uality for t he exp onential measure is of this optimal t y pe. W e transp ort this to any log–co ncav e meas ure o n the r eal line, a nd as the inequal- it y tensoriz es, any pro duct lo g–concave mea sure s atisfies I C (Corolla ry 2.19). How ever, the main challenge is to provide non–pro duct examples o f measures satisfying I C . W e show how such an optimal result ca n be obtained from con- centration inequalites, and follow on to pr ov e I C for the uniform measur e on any ℓ n p ball for p ≥ 1 (Theorem 5.30). With the tech niq ues develope d w e also pr ov e a few other results. W e give a pro of of the log– Sob olev inequality fo r ℓ n p balls, wher e p ≥ 2 (Theor em 5.31) and provide a new co ncentration inequality f o r the exp onential measure fo r sets lying far awa y from the origin (Theo rem 4.6). Organization of the pap er. This se ction, apa rt from the ab ov e intro- duction, defines the notation used througho ut the pap er. The second s ection is devoted to studying the general prop erties of the ine quality I C . In subsection 2.1 we r ecall the definition o f prop erty ( τ ) and its ties to conce n tr ation fro m [16]. In subsec tion 2.2 we study the o ppo site implication — what additio nal as - sumptions o ne needs to infer ( τ ) from co ncentration inequa lities. In subs ection 2.3 we show that Λ ⋆ µ is indeed the larg est p oss ible cost function and define the inequality I C . In subsection 2 .4 we show that pro duct log– concav e measures satisfy I C . In the third section we give mor e attention to the conce n tr ation inequa lities tied to I C . In subsection 3.1 we sho w the connection to Z p bo dies. In subsection 3.2 we contin ue in this vein with the a dditional ass umption our measure is α – regular . In subsection 3.3 we show how I C implies a co mparison of weak and strong moments and the results of [12] and [18]. In the fourth section we give a mo difica tion of the t wo–level concentration for the exp onential measure, in whic h for sets lying far awa y fro m the origin only an enlargement by tB n 1 is used. This will b e used in the fifth s ection, which fo cuses o n the uniform mea sure o n the B n p ball. In s ubsection 5.1 we define a nd study tw o r ather standa rd tra nsp orts of mea sure use d further on. In subsec tion 5.2 we us e thes e tr ansp orts along with the concentration from section 4 and a Cheeger inequality from [19] to give a pro of of I C for p ≤ 2. In section 5.3 we show a pro of of I C for p ≥ 2 and a proof of the log–Sob olev inequalit y for p ≥ 2. W e co nclude with a few p oss ible extensions o f the r esults of the pap er in the sixth sectio n. Notation. By h· , ·i we denote the standard scala r pr o duct on R n . F o r x ∈ R n we put k x k p = ( P n i =1 | x i | p ) 1 /p for 1 ≤ p < ∞ and k x k ∞ = max i | x i | , we also use | x | for k x k 2 . W e s et B n p for a unit ba ll in l n p , i.e .. B n p = { x ∈ R n : k x k p ≤ 1 } . By ν we denote the symmetric exp onential distribution o n R , i.e . the pr ob- ability meas ure with the density 1 2 exp( −| x | ). F or p ≥ 1, ν p is the proba bilit y distribution on R with the density (2 γ p ) − 1 exp( −| x | p ), wher e γ p = Γ(1 + 1 /p ), in pa rticular ν 1 = ν . F or a pr obability measure µ we write µ n for a pro duct measure µ ⊗ n , th us ν n p has the density (2 γ p ) − n exp( −k x k p p ). F or a Bo rel set A in R n by | A | o r λ n ( A ) we mean the Lebes gue measure of A . W e choose num b er s r p,n in s uch a way that | r p,n B n p | = 1 and by µ p,n denote 2 the unifor m distribution o n B n p . The letters c, C denote absolute n umerical constants, which ma y change fro m line to line. By c ( p ) , C ( p ) we mean constants dependent on p (or , forma lly , a family of absolute cons tant s indexed by p ), these also may change fr om line to line. Other letter s, in particular gr eek letters, deno te co nstants fixed for a given pr o of or section. F or any s ets of p ositive real num b ers a i and b i , i ∈ I , by a i ∼ b i we mean there exist absolute numerical constants c, C > 0 such that ca i < b i < C a i for any i ∈ I . Similarly for co llections of sets A i and B i by A i ∼ B i we mean cA i ⊂ B i ⊂ C A i for any i ∈ I , where again c, C > 0 ar e absolute numerical constants. By ∼ p we mean the co nstants ab ov e c an dep end on p . 2 Infim um con v olution inequalit y 2.1 Prop erty ( τ ) Definition 2.1 . L et µ b e a pr ob ability me asur e on R n and ϕ : R n → [0 , ∞ ] b e a me asur able function. We say that the p air ( µ, ϕ ) has prop erty ( τ ) if for any b ounde d me asur able function f : R n → R , Z R n e f ✷ ϕ dµ Z R n e − f dµ ≤ 1 , (1) wher e for two fu n ctions f and g on R n , f ✷ g ( x ) := inf { f ( x − y ) + g ( y ) : y ∈ R n } denotes the infimu m co nv olution of f and g . The following tw o easy obser v ations are a lmost immediate (c.f. [16]): Prop ositi o n 2.2 (T enso rization) . If p airs ( µ i , ϕ i ) , i = 1 , . . . , k have pr op erty ( τ ) and ϕ ( x 1 , . . . , x k ) = ϕ 1 ( x 1 ) + . . . + ϕ k ( x k ) , then the c ouple ( ⊗ k i =1 µ i , ϕ ) also has pr op erty ( τ ) . Prop ositi o n 2.3 (T ransp or t of measure) . Supp ose t hat ( µ, ϕ ) has pr op erty ( τ ) and T : R n → R m is su ch that ψ ( T x − T y ) ≤ ϕ ( x − y ) for al l x, y ∈ R n . Then the p air ( µ ◦ T − 1 , ψ ) has pr op erty ( τ ) . Maurey noticed that prop er t y ( τ ) implies µ ( A + B ϕ ( t )) ≥ 1 − µ ( A ) − 1 e − t , where B ϕ ( t ) := { x ∈ R n : ϕ ( x ) ≤ t } . W e will need a slight mo difica tion of this estimate. 3 Prop ositi o n 2.4. Pr op erty ( τ ) for ( ϕ, µ ) implies for any Bor el set A and t ≥ 0 , µ ( A + B ϕ ( t )) ≥ e t µ ( A ) ( e t − 1) µ ( A ) + 1 . (2) In p articular for al l t > 0 , µ ( A ) > 0 ⇒ µ ( A + B ϕ ( t )) > min { e t/ 2 µ ( A ) , 1 / 2 } , (3) µ ( A ) ≥ 1 / 2 ⇒ 1 − µ ( A + B ϕ ( t )) < e − t/ 2 (1 − µ ( A ) ) (4) and µ ( A ) = ν ( −∞ , x ] ⇒ µ ( A + B ϕ ( t )) ≥ ν ( − ∞ , x + t/ 2] . (5) Pr o of. T ake f ( x ) = t 1 R n \ A . Then f ( x ) is no n–negative on R n , so f ✷ ϕ is non– negative (recall that by definition we consider o nly nonneg ative cost functions). F or x 6∈ A + B ϕ ( t ) we hav e f ✷ ϕ ( x ) = inf y ( f ( y ) + ϕ ( x − y )) ≥ t , for either y 6∈ A , and then f ( y ) = t , or y ∈ A , and then ϕ ( x − y ) ≥ t as x 6∈ A + B ϕ ( t ). Thu s from prop er t y ( τ ) for f w e hav e 1 ≥ Z e f ✷ ϕ ( x ) dµ ( x ) Z e − f ( x ) dµ ( x ) ≥ h µ  A + B ϕ ( t )  + e t  1 − µ ( A + B ϕ ( t ))  i  µ ( A ) + e − t (1 − µ ( A ))  , from which, extracting the co ndition up on µ ( A + B ϕ ( t )) b y direct calculatio n, we ge t (2). Let f t ( p ) := e t p/ (( e t − 1) p + 1), notice that f t is increasing in p and for p ≤ e − t/ 2 / 2, ( e t − 1) p + 1 ≤ e t/ 2 + 1 − 1 2 ( e t/ 2 + e − t/ 2 ) < e t/ 2 , hence f t ( p ) > min( e t/ 2 p, 1 / 2 ) and (3) follows. Moreover for p ≥ 1 / 2 1 − f t ( p ) = 1 − p ( e t − 1) p + 1 ≤ 1 − p ( e t + 1) / 2 < e − t/ 2 (1 − p ) and we get (4). Let F ( x ) = ν ( −∞ , x ] and g t ( p ) = F ( F − 1 ( p ) + t ). Previous calculations show that for t, p > 0, f t ( p ) ≥ g t/ 2 ( p ) if F − 1 ( p ) + t / 2 ≤ 0 or F − 1 ( p ) ≥ 0. Since g t + s = g t ◦ g s and f t + s = f t ◦ f s , we get that f t ( p ) ≥ g t/ 2 ( p ) for all t, p > 0, hence (2) implies (5). The main theorem of [16] states that ν satisfies ( τ ) with a sufficiently chosen cost function. Theorem 2.5 . L et w ( x ) = 1 36 x 2 for | x | ≤ 4 and w ( x ) = 2 9 ( | x | − 2) otherwise. Then the p air ( ν n , P n i =1 w ( x i )) has pr op erty ( τ ) . 4 Theorem 2 .5 together with Pr op osition 2.4 immediately gives the following t wo-level co ncentration: ν n ( A ) = ν ( −∞ , x ] ⇒ ∀ t ≥ 0 ν n ( A + 6 √ 2 tB n 2 + 18 tB n 1 ) ≥ ν ( −∞ , x + t ] , (6) that was firs t esta blished (with different universal, rather large constants) by T alagr and [2 1]. 2.2 F rom concen tration to prop ert y ( τ ) Prop ositio n 2.4 shows that prop erty ( τ ) implies c oncentration, the next result presents the fir st approach to the co nv erse implication. Corollary 2.6. Supp ose that the c ost function ϕ is r adius-wise nonde cr e asing, µ is a Bor el pr ob ability me asure on R n and β > 0 is such that for any t > 0 and A ∈ B ( R n ) , µ ( A ) = ν ( −∞ , x ] ⇒ µ ( A + β B ϕ ( t )) ≥ ν ( −∞ , x + max { t, √ t } ] . (7) Then the p air ( µ, 1 36 ϕ ( · β )) has pr op erty ( τ ) . In p articular if ϕ is c onvex, sym- metric and ϕ (0) = 0 t hen (7) implies pr op erty ( τ ) for ( µ, ϕ ( · 36 β )) . Pr o of. Let us fix f : R n → R . F o r any measurable function h on R k and t ∈ R we put A ( h, t ) := { x ∈ R k : h ( x ) < t } . Let g be a nondecr easing rig ht-con tinuous function on R such that µ ( A ( f , t )) = ν ( A ( g , t )). Then the distr ibution of g with r esp ect to ν is the same as the distribution o f f with r esp ect to µ and thus Z R n e − f ( x ) dµ ( x ) = Z R e − g ( x ) dν ( x ) . T o finish the pro of of the first assertion, by Theo rem 2.5 it is enough to show that Z R n e f ✷ 1 36 ϕ ( · β ) dµ ≤ Z R e g ✷ w dν. W e will establish stronger prop erty: ∀ u µ  A  f ✷ 1 36 ϕ  · β  , u   ≥ ν ( A ( g ✷ w, u )) . Since the set A ( g ✷ w , u ) is a halfline, it is enough to prove that g ( x 1 ) + w ( x 2 ) < u ⇒ µ  A  f ✷ 1 36 ϕ  · β  , u   ≥ ν ( −∞ , x 1 + x 2 ] . (8) Let us fix x 1 and x 2 with g ( x 1 ) + w ( x 2 ) < u and take s 1 > g ( x 1 ) s 2 = w ( x 2 ) with s 1 + s 2 < u . Put A := A ( f , s 1 ), then µ ( A ) = ν ( A ( g , s 1 )) ≥ ν ( −∞ , x 1 ]. 5 By the definition of w it easily follows that x 2 ≤ max { 6 √ s 2 , 9 s 2 } , hence by (7), µ ( A + β B ϕ (36 s 2 )) ≥ ν ( −∞ , x 1 + x 2 ] . Since A + β B ϕ (36 s 2 ) = A ( f , s 1 ) + B ϕ ( · β ) / 36 ( s 2 ) ⊂ A  f ✷ 1 36 ϕ  · β  , s 1 + s 2  , we obta in the prop er t y (8). The last par t of the statement immediately follows since any symmetric conv ex function ϕ is radius- wise nondecreasing and if additionally ϕ (0) = 0, then ϕ ( x/ 36 ) ≤ ϕ ( x ) / 36 for any x . The next prop ositio n shows that inequalities (3) and (4) are stro ngly rela ted. Prop ositi o n 2.7. The fol lowing two c onditions ar e e quivalent for any Bor el set K and γ > 1 , ∀ A ∈B ( R n ) µ ( A ) > 0 ⇒ µ ( A + K ) > min n γ µ ( A ) , 1 2 o , (9) ∀ ˜ A ∈B ( R n ) µ ( ˜ A ) ≥ 1 2 ⇒ 1 − µ ( ˜ A − K ) < 1 γ (1 − µ ( ˜ A )) . (10) Pr o of. (9) ⇒ (10). Suppo se that µ ( ˜ A ) ≥ 1 / 2 and 1 − µ ( ˜ A − K ) ≥ γ − 1 (1 − µ ( ˜ A )). Let A := R n \ ( ˜ A − K ), then ( A + K ) ∩ ˜ A = ∅ , so µ ( A + K ) ≤ 1 / 2 and µ ( A + K ) ≤ 1 − µ ( ˜ A ) ≤ γ (1 − µ ( ˜ A − K )) = γ µ ( A ) and this contradicts (9). (10) ⇒ (9). Let us take A ∈ R n with µ ( A ) > 0 such that µ ( A + K ) ≤ min { γ µ ( A ) , 1 / 2 } . Let ˜ A := R n \ ( A + K ), then µ ( ˜ A ) ≥ 1 / 2 . Moreov er ( ˜ A − K ) ∩ A = ∅ , thus 1 − µ ( ˜ A − K ) ≥ µ ( A ) ≥ 1 γ µ ( A + K ) = 1 γ (1 − µ ( ˜ A )) and we get the contradiction with (10). Corollary 2.8. Supp ose t hat t > 0 and K is a symmetric c onvex set in R n such t hat ∀ A ∈B ( R n ) µ ( A ) > 0 ⇒ µ ( A + K ) > min { e t µ ( A ) , 1 / 2 } . Then for any Bor el set A , µ ( A ) = ν ( −∞ , x ] ⇒ µ ( A + 2 K ) > ν ( −∞ , x + t ] . Pr o of. Let us fix the set A with µ ( A ) = ν ( −∞ , x ]. Notice that A + 2 K = A + K + K ⊃ A + K . If x + t ≤ 0, then µ ( A + K ) > e t µ ( A ) = ν ( −∞ , x + t ]. If x ≥ 0, Prop o sition 2.7 gives µ ( A + K ) > 1 − e − t (1 − µ ( A ) ) = ν ( −∞ , x + t ] . 6 Finally , if x ≤ 0 ≤ x + t , we get µ ( A + K ) ≥ 1 / 2 = ν ( −∞ , 0 ], hence by the previous ca se, µ ( A + 2 K ) = µ (( A + K ) + K ) > ν ( −∞ , t ] ≥ ν ( −∞ , x + t ] . Corollar y 2.8 shows that if the cost function ϕ is symmetr ic and conv ex , condition (7 ) (with 2 β instead of β ) for t ≥ 1 is implied by the following: ∀ A ∈B ( R n ) µ ( A ) > 0 ⇒ µ ( A + β B ϕ ( t )) > min { e t µ ( A ) , 1 / 2 } . (11) T o treat the ca se t ≤ 1 we will need Cheeger’s version of the Poincar´ e inequality . W e say that a probability measure µ o n R n satisfies Che e ger’s ine quality with co nstant κ if for any Borel set A µ + ( A ) := lim inf t → 0+ µ ( A + tB n 2 ) − µ ( A ) t ≥ κ min { µ ( A ) , 1 − µ ( A ) } . (12) It is not har d to chec k tha t Chee ger’s inequa lit y (cf. [6, Theor em 2.1]) implies µ ( A ) = ν ( −∞ , x ] ⇒ µ ( A + tB n 2 ) ≥ ν ( −∞ , x + κt ] . Finally , we may summarize this section with the following sta temen t. Prop ositi o n 2.9. Supp ose that the c ost function ϕ is c onvex, symmetric with ϕ (0) = 0 and 1 ∧ ϕ ( x ) ≤ ( α | x | ) 2 for al l x . If the me asur e µ satisfies Che e ger’s ine quality with the c onst ant β = 1 /δ and the c ondition (11) is satisfie d for al l t ≥ 1 and C = γ then ( µ, ϕ ( · /C )) has pr op erty ( τ ) with t he c onstant C = 36 min { 2 γ , αδ } . Pr o of. Notice that αB ϕ ( t ) ⊃ √ tB n 2 for all t < 1, hence Cheeg er’s inequality implies that condition (7) holds for t < 1 with C = αδ . Ther efore (7) holds for all t ≥ 0 with C = min { 2 γ , αδ } and the as sertion follows by Coro llary 2.6. 2.3 Optimal cost functions A natural ques tion arises: what other pairs ( µ, ϕ ) hav e prop erty ( τ )? First we hav e to choo se the rig h t cost function. T o do this let us r ecall the following definitions. Definition 2.10. L et f : R n → ( −∞ , ∞ ] . The Legendre trans form of f , denote d L f is define d by L f ( x ) := sup y ∈ R n {h x, y i − f ( y ) } . The Lege ndre tr ansform of any function is a convex function. If f is conv ex and low er semi-co n tinuous, then LL f = f , and other wise LL f ≤ f . In genera l, if f ≥ g , then L f ≤ L g . The L egendre transform sa tisfies L ( C f )( x ) = C L f ( x/ C ) and if g ( x ) = f ( x/C ), then L g ( x ) = L f ( C x ). F or this a nd o ther pr op erties of L , cf. [15]. The Legendre transfor m has b een previously us ed in the context of convex ge ometry , see fo r insta nce [2] a nd [13]. 7 Definition 2.1 1. L et µ b e a pr ob ability me asur e on R n . We define M µ ( v ) := Z R n e h v,x i dµ ( x ) , Λ µ ( v ) := lo g M µ ( v ) and Λ ⋆ µ ( v ) := L Λ µ ( v ) = s up u ∈ R n n h v , u i − ln Z R n e h u,x i dµ ( x ) o . The function Λ ⋆ µ plays a crucial role in the theor y o f la rge deviations cf. [9]. Remark 2 .12. L et µ b e a symmetric pr ob ability me asur e on R n and let ϕ b e a c onvex c ost function su ch that ( µ, ϕ ) has pr op erty ( τ ). Then ϕ ( v ) ≤ 2Λ ⋆ µ ( v / 2) ≤ Λ ⋆ µ ( v ) . Pr o of. T ake f ( x ) = h x, v i . Then f ✷ ϕ ( x ) = inf y ( f ( x − y ) + ϕ ( y )) = inf y ( h x − y , v i + ϕ ( y )) = h x, v i − L ϕ ( v ) . Prop erty ( τ ) yields 1 ≥ Z e f ✷ ϕ dµ Z e − f dµ = e −L ϕ ( v ) Z e h x,v i dµ Z e −h x,v i dµ = e −L ϕ ( v ) M 2 µ ( v ) , where the last equality uses the fact that µ is symmetric. Thus by ta king the logarithm we g et L ϕ ( v ) ≥ 2Λ µ ( v ), and by applying the Legendre transfor m we obtain ϕ ( v ) = LL ϕ ( v ) ≤ 2 Λ ⋆ µ ( v / 2) . The inequality 2Λ ⋆ µ ( v / 2) ≤ Λ ⋆ µ ( v ) follows by the convexit y of Λ ⋆ µ . The ab ov e rema rk motiv ates the following definition. Definition 2.13. We say that a symmetric pr ob ability me asur e µ satisfies t he infim um conv olution inequa lity with constant β ( IC( β ) in short), if the p air ( µ, Λ ∗ µ ( · β )) has pr op erty ( τ ) . Prop ositi o n 2.14. If µ i ar e symmetric pr ob ability me asur es on R n i , 1 ≤ i ≤ k satisfying IC( β i ) , then µ = ⊗ k i =1 µ i satisfies IC ( β ) with β = max i β i . Pr o of. By indep endence, Λ µ ( x 1 , . . . , x k ) = P k i =1 Λ µ i ( x i ) a nd Λ ∗ µ ( x 1 , . . . , x k ) = P k i =1 Λ ∗ µ i ( x i ). Since IC( β ) implies IC( β ′ ) with any β ′ ≥ β , the result immedi- ately follows by Pr op osition 2.2. Prop ositi o n 2.15. F or v = ( v 0 , v 1 , . . . , v n ) in R n +1 let ˜ v denote the ve ctor ( v 1 , v 2 , . . . , v n ) ∈ R n . A pr ob ability me asure µ on R n satisfies IC( β ) if and only if for any nonempty V ⊂ R n +1 and a b ounde d me asur able function f on R n , Z R n e f ✷ ψ V dµ Z R n e − f dµ ≤ s up v ∈ V  e v 0 Z R n e β h x, ˜ v i dµ ( x )  , (13) wher e ψ V ( x ) := sup v ∈ V { v 0 + h x, ˜ v i} . 8 Pr o of. If we put V = { ( v 0 , ˜ v ) : v 0 = − Λ µ ( β ˜ v ) } , then the r ight-hand side is equal to 1 and ψ V ( x ) = Λ ⋆ µ ( x/β ), so if µ satisfies (13) for this V , it s atisfies IC ( β ). On the other hand, suppo se µ satisfies IC( β ). T ake a n arbitra ry nonempty set V . If the right-hand side supre m um is infinite, the ineq uality is o bvious, so we may assume it is equal to some s < ∞ . This mea ns that for any ( v 0 , ˜ v ) ∈ V we have v 0 + Λ µ ( β ˜ v ) ≤ log s , that is v 0 ≤ log s − Λ µ ( β ˜ v ). Th us ψ V ( x ) = s up v ∈ V { v 0 + h x, ˜ v i} ≤ lo g s + sup v ∈ V {h x, ˜ v i − Λ µ ( β ˜ v ) } ≤ log s + sup ˜ v ∈ R n {h x, ˜ v i − Λ µ ( β ˜ v ) } = lo g s + Λ ⋆ µ ( x/β ) , which in turn means from IC( β ) that the left ha nd side is no larger than s . Prop ositi o n 2. 16. L et L : R n → R k b e a line ar map and supp ose that a pr ob- ability me asur e µ on R n satisfies IC( β ) . Then the pr ob abili t y me asur e µ ◦ L − 1 satisfies IC ( β ) . Pr o of. F or a n y set V ⊂ R × R k and any function f : R k → R put ¯ f ( x ) := f ( L ( x )) and ¯ V := { ( v 0 , L ⋆ ( ˜ v )) : ( v 0 , ˜ v ) ∈ V } , where L ⋆ is the Hermitian co njugate o f L . Then direct calculation sho ws ψ V ( L ( x )) = ψ ¯ V ( x ) and f ✷ ψ V ( L ( x )) ≤ ¯ f ✷ ψ ¯ V ( x ), th us Z R k e f ✷ ψ V d ( µ ◦ L − 1 ) ≤ Z R n e ¯ f ✷ ψ ¯ V dµ and Z R k e − f d ( µ ◦ L − 1 ) = Z R n e − ¯ f dµ and finally sup v ∈ V n e v 0 Z R k e β h x, ˜ v i d ( µ ◦ L − 1 ) o = sup v ∈ ¯ V n e v 0 Z R n e β h x, ˜ v i dµ o , which s ubstituted into (1 3) gives the thesis. Prop ositi o n 2 .17. F or any x ∈ R , 1 5 min( x 2 , | x | ) ≤ Λ ∗ ν ( x ) ≤ min( x 2 , | x | ) , in p articular the me asur e ν satisfies IC(9 ) . Pr o of. Direct calculation s hows that Λ ν ( x ) = − ln(1 − x 2 ) fo r | x | < 1 a nd Λ ⋆ ν ( x ) = p 1 + x 2 − 1 − ln  √ 1 + x 2 + 1 2  . Since a/ 2 ≤ a − ln(1 + a/ 2) ≤ a for a ≥ 0, we get 1 2 ( √ 1 + x 2 − 1 ) ≤ Λ ⋆ ν ( x ) ≤ √ 1 + x 2 − 1. Finally min( x, | x | 2 ) ≥ p 1 + x 2 − 1 = x 2 √ 1 + x 2 + 1 ≥ 1 √ 2 + 1 min( | x | , x 2 ) . The last statemen t follows b y Theorem 2.5, since min(( x/ 9 ) 2 , | x | / 9) ≤ w ( x ). 9 2.4 Logaritmically conca v e pro duct measures A mea sure µ on R n is lo gari t hmic al ly c onc ave (log–co ncav e for short) if for all nonempty co mpact sets A, B and t ∈ [0 , 1], µ ( tA + (1 − t ) B ) ≥ µ ( A ) t µ ( B ) 1 − t . By Borell’s theorem [7] a meas ure µ on R n with a full– dimensional supp ort is logarithmica lly concav e if and only if it is absolutely co nt inuous with res pe ct to the Lebes gue measure and has a lo garithmically concav e densit y , i.e. dµ ( x ) = e h ( x ) dx for some concave function h : R n → [ −∞ , ∞ ). Note that if µ is a pr obabilistic, symmetric a nd log –concave measure o n R n , then b o th Λ µ and Λ ⋆ µ are conv ex and symmetric, and Λ µ (0) = Λ ⋆ µ (0) = 0. Recall a lso tha t a pro bability mea sure µ on R n is calle d isotr opic if Z h u, x i dµ ( x ) = 0 and Z h u, x i 2 dµ ( x ) = | u | 2 for a ll u ∈ R n . It is ea sy to check that for a ny measure µ with a full–dimensional supp or t there exists a linear map L such that µ ◦ L − 1 is iso tropic. The next theorem (with a diff er ent universal, but rather large consta n t) may be deduced fr om the r esults of Go zlan [10]. W e give the following, relatively short pro of fo r the sa ke of completenes s. Theorem 2.18. Any symmetric lo g-c onc ave me asur e on R satisfies IC(4 8) . Pr o of. Let µ be a symmetr ic lo g–concave probability measure on R , we may assume µ is iso tropic by P rop osition 2.16. Denote the dens it y of µ by g ( x ) and let the ta il function b e µ [ x, ∞ ) = e − h ( x ) . Let a := inf { x > 0 : g ( x ) ≤ e − 1 g (0) } , then g ( x ) ≤ e − x/a g (0) for x > a and g ( x ) ≥ e − x/a g (0) for x ∈ [0 , a ). Therefor e Z ∞ 0 g ( x ) dx ≤ ag (0) + Z ∞ a g (0) e − x/a dx, that is 1 / 2 ≤ ag (0)(1 + e − 1 ) . (14) W e also have Z a 0 x 2 g (0) e − x/a dx ≤ Z ∞ 0 x 2 g ( x ) dx, so a 3 g (0)(2 − 5 e − 1 ) ≤ 1 / 2 . (15) F rom (14) a nd (15) we get in particular 1 8 ≤ s e 2 (2 e − 5 ) 4( e + 1 ) 3 ≤ g (0) . 10 Let T : R → R b e a function such tha t ν ( −∞ , x ) = µ ( − ∞ , T x ). Then µ = ν ◦ T − 1 , T is o dd and c oncav e o n [0 , ∞ ). In particula r, | T x − T y | ≤ 2 | T ( x − y ) | for a ll x, y ∈ R . Notice that T ′ (0) = 1 / (2 g (0)) ≤ 4 , thus by concavit y of T , T x ≤ 4 x for x ≥ 0. Mo reov er for x ≥ 0, h ( T x ) = x + ln 2. Define ˜ h ( x ) :=  x 2 for | x | ≤ 2 / 3 max { 4 / 9 , h ( x ) } for | x | > 2 / 3 . W e claim tha t ( µ, ˜ h ( · 48 )) has pr op erty ( τ ). Notice that ˜ h (( T x − T y ) / 48 ) ≤ ˜ h ( T ( | x − y | ) / 24) so by P rop osition 2 .3 it is enough to chec k that ˜ h  T x 24  ≤ w ( x ) for x ≥ 0 , (16) where w ( x ) is as in Theorem 2.5. W e have tw o cases. i) T x ≤ 16, then ˜ h  T x 24  =  T x 24  2 ≤ min n 4 9 ,  x 6  2 o ≤ w ( x ) . ii) T x ≥ 16, then x ≥ 4 and ˜ h  T x 24  = max n 4 9 , h  T x 24 o ≤ max n 4 9 , h ( T x ) 24 o = max n 4 9 , x + ln 2 24 o ≤ x 9 ≤ w ( x ) . So (16) holds in b oth cases. T o c onclude we need to show that Λ ∗ µ ( x ) ≤ ˜ h ( x ). F or | x | ≤ 2 / 3 it follows from the mo re g eneral P rop osition 3.3 below. Notice that for a ny t, x ≥ 0, Λ µ ( t ) ≥ tx + ln µ [ x, ∞ ) = tx − h ( x ), hence Λ ∗ µ ( x ) = Λ ∗ µ ( | x | ) = sup t ≥ 0  t | x | − Λ µ ( t )  ≤ h ( | x | ) ≤ ˜ h ( x ) for | x | > 2 / 3 . Using Corolla ry 2.14 we get Corollary 2 . 19. Any symmetric, lo g–c onc ave pr o duct pr ob ability m e asu r e on R n satisfies IC(48) . W e exp ect that in fact a mor e ge neral fact holds. Conjecture 1. Any symmetric lo g–c onc ave pr ob ability me asur e satisfies IC( C ) with a uniform c onstant C . 11 3 Concen tration inequalities. 3.1 L p -cen troid b o dies and related set s Definition 3.1 . Le t µ b e a pr ob ability me asur e on R n , for p ≥ 1 we define t he fol lowing sets M p ( µ ) := n v ∈ R n : Z | h v , x i | p dµ ( x ) ≤ 1 o , Z p ( µ ) := ( M p ( µ )) ◦ = n x ∈ R n :   h v , x i   p ≤ Z   h v , y i   p dµ ( y ) for al l v ∈ R n o and for p > 0 we put B p ( µ ) := { v ∈ R n : Λ ∗ µ ( v ) ≤ p } . Sets Z p ( µ K ) for p ≥ 1 , when µ K is the unifor m disribution on the conv ex bo dy K ar e called L p -c entr oid b o dies of K , their pro pe rties were inv estigated in [1 8]. Prop ositi o n 3 .2. F or any symmet ric pr ob ability me asur e µ on R n and p ≥ 1 , Z p ( µ ) ⊂ 2 1 /p eB p ( µ ) . Pr o of. Let us take v ∈ Z p ( µ ), we ne ed to show that Λ ⋆ µ ( v / (2 1 /p e )) ≤ p , that is h u, v i 2 1 /p e − Λ µ ( u ) ≤ p for all u ∈ R n . Let us fix u ∈ R n with R | h u, x i | p dµ ( x ) = β p , then u/β ∈ M p ( µ ). W e will consider tw o cases . i) β ≤ 2 1 /p ep . Then, sinc e Λ µ ( u ) ≥ R h u, x i dµ ( x ) = 0 , h u, v i 2 1 /p e − Λ µ ( u ) ≤ β 2 1 /p e  u β , v  ≤ p · 1 . ii) β > 2 1 /p ep . W e ha ve Z e h u,x i dµ ( x ) ≥ Z   e h u,x i /p   p I {h u,x i≥ 0 } dµ ( x ) ≥ Z    h u, x i p    p I {h u,x i≥ 0 } dµ ( x ) ≥ 1 2 Z    h u, x i p    p dµ ( x ) , th us Z e 2 1 /p ep h u,x i /β dµ ( x ) ≥ 1 2 Z    2 1 /p e h u, x i β    p dµ ( x ) = e p . Hence Λ µ (2 1 /p epu/β ) ≥ p a nd Λ µ ( u ) ≥ β 2 1 /p ep Λ µ (2 1 /p epu/β ) ≥ β 2 1 /p e . Therefor e h u, v i 2 1 /p e − Λ µ ( u ) ≤ β 2 1 /p e  u β , v  − β 2 1 /p e ≤ 0 . 12 Prop ositi o n 3. 3. If µ is a symmetric, isotr opic pr ob abili ty me asur e on R n , then min { 1 , Λ ∗ µ ( u ) } ≤ | u | 2 for al l u , in p articular √ pB n 2 ⊂ B p ( µ ) for p ∈ (0 , 1) . Pr o of. Using the sy mmetry and iso tropicity of µ , we get Z e h u,x i dµ ( x ) = 1 + ∞ X k =1 1 (2 k )! Z h u, x i 2 k dµ ( x ) ≥ 1 + ∞ X k =1 | u | 2 k (2 k )! = cosh( | u | ) . Hence for | u | < 1, Λ ∗ µ ( u ) ≤ L (ln cosh)( | u | ) = 1 2 h (1 + | u | ) ln(1 + | u | ) + (1 − | u | ) ln(1 − | u | ) i ≤ | u | 2 , where to get the last inequality we us ed ln(1 + x ) ≤ x for x > − 1. 3.2 α -regular measures. T o establish inlusions opp osite to thos e in the previous subsection, we in tro duce the fo llowing pr op erty: Definition 3.4. We say t hat a me asur e µ on R n is α -r e gu lar if for any p ≥ q ≥ 2 and v ∈ R n ,  Z | h v , x i | p dµ ( x )  1 /p ≤ α p q  Z | h v , x i | q dµ ( x )  1 /q . Prop ositi o n 3 .5. If µ is α -r e gular for some α ≥ 1 , then for any p ≥ 2 , B p ( µ ) ⊂ 4 eα Z p ( µ ) . Pr o of. First we will show that u ∈ M p ( µ ) ⇒ Λ µ  pu 2 eα  ≤ p. (17) Indeed if we fix u ∈ M p ( µ ) a nd put ˜ u := pu 2 eα , then  Z | h ˜ u, x i | k dµ ( x )  1 /k = p 2 eα  Z | h u, x i | k dµ ( x )  1 /k ≤  p 2 eα k ≤ p k 2 e k > p. Hence Z e h ˜ u,x i dµ ( x ) ≤ Z e |h ˜ u,x i| dµ ( x ) = ∞ X k =0 1 k ! Z | h ˜ u, x i | k dµ ( x ) ≤ X k ≤ p 1 k !    p 2 eα    k + X k>p 1 k !    k 2 e    k ≤ e p 2 eα + 1 ≤ e p 13 and (17) follows. T ake an y v / ∈ 4 eα Z p ( µ ), then we ma y find u ∈ M p ( µ ) such that h v , u i > 4 eα and obta in Λ ∗ µ ( v ) ≥ D v , pu 2 eα E − Λ µ  pu 2 eα  > p 2 eα 4 eα − p = p. Prop ositi o n 3 .6. If µ is symmetric, isotr opic α -r e gular for some α ≥ 1 , then Λ ∗ µ ( u ) ≥ min n | u | 2 αe , | u | 2 2 α 2 e 2 o , in p articular B p ( µ ) ⊂ max { 2 αep, αe p 2 p } B n 2 for al l p > 0 . Pr o of. W e hav e by the sy mmetry , isotropicity a nd regularity o f µ , Z e h u,x i dµ ( x ) = ∞ X k =0 1 (2 k )! Z h u, x i 2 k dµ ( x ) ≤ 1 + | u | 2 2 + ∞ X k =2 ( αk | u | ) 2 k (2 k )! ≤ 1 + | u | 2 2 + ∞ X k =2  αe | u | 2  2 k . Hence if αe | u | ≤ 1, Z e h u,x i dµ ( x ) ≤ 1 + | u | 2 2 + 4 3  αe | u | 2  4 ≤ 1 + α 2 e 2 | u | 2 + ( αe | u | ) 4 2 ≤ e α 2 e 2 | u | 2 / 2 so Λ µ ( u ) ≤ α 2 e 2 | u | 2 / 2 for αe | u | ≤ 1. Th us Λ ∗ µ ( u ) ≥ min { | u | 2 αe , | u | 2 2 α 2 e 2 } for all u . Remark 3.7. We always have for p ≥ q , M p ( µ ) ⊂ M q ( µ ) and Z q ( µ ) ⊂ Z p ( µ ) . If the me asur e µ is α -r e gular, then M q ( µ ) ⊂ αp q M p ( µ ) and Z p ( µ ) ⊂ αp q Z q ( µ ) for p ≥ q ≥ 2 . Mor e over for any symmetric me asur e µ , Λ ∗ µ (0) = 0 , henc e by the c onvexity of Λ ∗ µ , B q ( µ ) ⊂ B p ( µ ) ⊂ p q B q ( µ ) for al l p ≥ q > 0 . Prop ositi o n 3 .8. Symmetric lo g–c onc ave me asur es ar e 1-r e gular. Pr o of. If X is distr ibuted a ccording to a symmetric, log– concav e measur e µ and u ∈ R n , then the ra ndom v ar iable S = h u, X i has a log –concav e symmetric distribution on the rea l line. W e need to show that ( E | S | p ) 1 /p ≤ p q ( E | S | q ) 1 /q for p ≥ q ≥ 2. The pro o f of Remark 5 in [14 ] shows that ( E | S | p ) 1 /p ≤ (Γ( p + 1)) 1 /p (Γ( q + 1)) 1 /q ( E | S | q ) 1 /q , 14 so it is enough to show that the function f ( x ) := 1 x (Γ( x + 1)) 1 /x is nonincreasing on [2 , ∞ ). Binet’s form of the Stirling formula (cf. [1, Theore m 1.6.3 ]) gives Γ( x + 1 ) = x Γ( x ) = √ 2 π x x +1 / 2 e − x + µ ( x ) , where µ ( x ) = R ∞ 0 arctg( t/x )( e 2 π t − 1) − 1 dt is decrea sing function. Thus ln f ( x ) = µ ( x ) x + ln(2 π x ) 2 x − 1 is indeed nonincr easing on [2 , ∞ ). Let us int r o duce the following notion: Definition 3.9. We say that a me asur e µ satifies the c oncentration ineq uality with co nstant β ( CI( β ) in short) if ∀ p ≥ 2 ∀ A ∈B ( R n ) µ ( A ) ≥ 1 2 ⇒ 1 − µ ( A + β Z p ( µ )) ≤ e − p (1 − µ ( A )) . (18) This definition is motiv ated by the following Coro llary: Corollary 3. 10. L et µ b e an α –r e gular symmetric and isotr opic pr ob ability me asur e with α ≥ 1 . Then i) If µ s atisfies IC( β ) , then µ satisfies CI(8 eαβ ) , ii) If µ satisfies CI( β ) and additional ly satisfies Che e ger’s ine quality (12) with c onstant 1 /γ , t hen µ satisfies IC(36 min { 6 eβ , γ } ) . Pr o of. By Rema rk 3 .7, P rop osition 2 .4 and the definition of B p ( µ ) we hav e µ ( A + 2 β B p ( µ )) ≥ µ ( A + β B 2 p ( µ )) ≥ 1 − e − p (1 − µ ( A )) , so the fir st part of the statement immediately follows by P rop osition 3 .5. On the other ha nd, if µ satisfies CI( β ), then by Remar k 3.7 a nd P rop osition 3.2 we have for µ ( A ) ≥ 1 / 2 and p ≥ 1 e − p (1 − µ ( A ) ) > e − 2 p (1 − µ ( A ) ) ≥ 1 − µ ( A + β Z 2 p ( µ )) ≥ 1 − µ ( A + e 2 1 / 2 p β B 2 p ( µ )) ≥ 1 − µ ( A + 3 eβ B p ( µ )) . By P rop osition 2 .7 this implies pr op erty (11 ). Additionally Λ ⋆ µ is convex, s ym- metric and Λ ⋆ µ (0) = 0. Finally , from P rop ositio n 3 .3 we hav e min { 1 , Λ ⋆ µ ( u ) } ≤ | u | 2 . Thus, from Prop ositio n 2.9 we ge t the sec ond part of the statement. By Propo sition 2.7 in t he definition 3.9 w e could use the equiv alent condition µ ( A + β Z p ( µ )) ≥ min { e p µ ( A ) , 1 / 2 } . The next prop osition shows that for log- concav e measure s these conditions ar e satisfied for la rge p a nd for small sets. 15 Prop ositi o n 3.11. L et µ b e a symmetric lo g-c onc ave pr ob ability me asur e on R n and c ∈ (0 , 1] . Then µ  A + 40 c Z p ( µ )  ≥ 1 2 min { e p µ ( A ) , 1 } for p ≥ cn or µ ( A ) ≤ e − cn . Pr o of. Using a sta ndard volumetric es timate for any r > 0 we may choo se S ⊂ M r ( µ ) with # S ≤ 5 n such that M r ( µ ) ⊂ S u ∈ S ( u + 1 2 M r ( µ )). Then for t > 0, x / ∈ t Z r ( µ ) ⇒ max u ∈ S h u, x i ≥ t/ 2 and by the Chebyshev inequality , µ  R n \ t Z r ( µ )  ≤ X u ∈ S µ n x : h u, x i ≥ t 2 o ≤ X u ∈ S  2 t  r Z h u, x i r + dµ ≤ 1 2 5 n  2 t  r . Let µ ( A ) = e − q , we will cons ider tw o cases. i) p ≥ max { q , cn } . Then by Remark 3.7, µ (30 c − 1 Z p ( µ )) > µ (30 Z max { p,n } ) ≥ 1 − 1 2 e − max { p,n } ≥ 1 − µ ( A ) , so A ∩ 30 c − 1 Z p ( µ ) 6 = ∅ , hence 0 ∈ A + 30 c − 1 Z p ( µ ) and µ ( A + 4 0 c − 1 Z p ( µ )) ≥ µ (10 c − 1 Z p ( µ )) ≥ 1 / 2 . ii) q ≥ ma x { p, cn } . Le t ˜ q := max { q , n } ˜ A := A ∩ 3 0 c − 1 Z q ( µ ) , we have as in i), µ (30 c − 1 Z q ( µ )) > 1 − e − ˜ q / 2, th us µ ( ˜ A ) ≥ µ ( A ) / 2. Moreover,  1 − p q  ˜ A ⊂ A − p q 30 c − 1 Z q ( µ ) ⊂ A + 30 c − 1 Z p ( µ ) and µ  A + 4 0 c − 1 Z p ( µ )  ≥ µ  1 − p q  ˜ A + p q 10 c − 1 Z q ( µ )  ≥ µ  1 − p q  ˜ A + p q 10 Z ˜ q ( µ )  ≥ µ ( ˜ A ) 1 − p q µ (10 Z ˜ q ) p q ≥  1 2 µ ( A )  1 − p q  1 2  p q ≥ 1 2 µ ( A ) µ ( A ) − p q = 1 2 e − p µ ( A ) . The previous facts mo tiv ate the following. Conjecture 2. Any symmetric lo g–c onc ave pr ob ability me asur e satisfies CI( C ) for some universal c onstant C . 16 Prop ositio n 3 .10 shows that Conjecture 1 implies Conjecture 2. Bo th hy- po theses would be equiv alent pro vided that the following conjecture of Ka nnan, Lov´ asz and Simonovits holds. Conjecture 3 (K annan–Lov´ a sz–Simonovits [1 1]) . Ther e exists an absolute c on- stant C su ch that any symmet ric isotr opic lo g–c onc ave pr ob ability me asur e sat- isfies Che e ger’s ine quality with c onstant 1 /C . 3.3 Comparison o f wea k and strong momen ts Prop ositi o n 3.12. Su pp ose that a pr ob ability me asur e µ on R n is α - r e gular and satisfies CI( β ) . Then for any norm k · k on R n and p ≥ 2 ,  Z   k x k − Med µ ( k x k )   p dµ  1 /p ≤ 2 αβ sup k u k ∗ ≤ 1  Z | h u, x i | p dµ  1 /p , wher e k · k ∗ denotes the norm dual to k · k . Pr o of. F or p ≥ 2 we define m p := sup k u k ∗ ≤ 1  Z | h u, x i | p dµ  1 /p . Let M := Med µ ( k x k ), A := { x : k x k ≤ M } and ˜ A := { x : k x k ≥ M } . The n µ ( A ) , µ ( ˜ A ) ≥ 1 / 2 s o by CI( β ) and Rema rk 3 .7, ∀ t ≥ p 1 − µ  A + β αt p Z p ( µ )  ≤ 1 2 e − t , 1 − µ  ˜ A + β αt p Z p ( µ )  ≤ 1 2 e − t . Let y ∈ Z p , then ther e exists u ∈ R n with k u k ∗ ≤ 1 such that k y k = h u, y i ≤  Z | h u, x i | p dµ ( x )  1 /p ≤ m p , hence k x k ≤ M + tm p for x ∈ A + t Z p ( µ ). Thus for t ≥ p , µ n x : k x k ≥ M + αβ t p m p o ≤ 1 − µ  A + β αt p Z p ( µ )  ≤ 1 2 e − t . In a similar wa y we show k x k ≥ M − tm p for x ∈ ˜ A + t Z p ( µ ) and µ { x : k x k ≤ M − αβ tm p /p } ≤ e − t / 2, therefore µ n x :   k x k − M   ≥ αβ t p m p o ≤ e − t for t ≥ p. 17 Now integrating by parts,  Z |k x k − M | p dµ  1 /p ≤ αβ m p p h p +  p Z ∞ p t p − 1 µ n x :   k x k − M   ≥ αβ t p m p o dt  1 /p i ≤ αβ m p p h p +  p Z ∞ p t p − 1 e − t dt  1 /p i ≤ αβ m p  1 + Γ( p + 1) 1 /p p  ≤ 2 αβ m p . Remark 3 . 13. Under the assu mptions of Pr op osition 3.12 by the triangle in- e quality we get for γ = 4 αβ , ∀ p ≥ q ≥ 2  Z    k x k −  Z k x k q dµ  1 /q    p dµ  1 /p ≤ γ sup k u k ∗ ≤ 1  Z | h u, x i | p dµ  1 /p . (19) This motiv ates the following definition. Definition 3.1 4 . We say t hat a pr ob ability me asure µ on R n has compar able weak and strong momen ts with the c onstant γ ( CWSM( γ ) in short) if (19) holds for any norm k · k on R n . Conjecture 4. Every symmetric lo g–c onc ave pr ob ability on R n me asur e satis- fies CWSM( C ) . Prop ositi o n 3.15 . L et µ b e an isotr opic, pr ob ability me asur e on R n satisfying CWSM( γ ) . Then i) R |k x k 2 − √ n | 2 dµ ( x ) ≤ γ 2 , ii) if µ is also α –r e gular, t hen for al l p > 2 ,  Z k x k p 2 dµ  1 /p ≤ √ n + γ α 2 p. Pr o of. Notice that R k x k 2 2 dµ = n and k u k ∗ 2 = k u k 2 . Hence i) fo llows directly from (19) with p = q = 2. Moreov er (19) with q = 2 implies  Z k x k p 2 dµ  1 /p ≤ √ n + sup k u k 2 ≤ 1  Z | h u, x i | p dµ  1 /p ≤ √ n + γ α 2 p by the α -r egularity a nd isotr opicity of µ . Remark 3.16. Pr op erty i) plays the crucial r ole in the Klartag pr o of of the c entr al limit the or em for c onvex b o dies [12]. Paouris [18 ] showe d that moments of the Euclid e an norm for symmetric isotr opic lo g-c onc ave me asur es ar e b oun de d by C ( p + √ n ) . Thus Conje ctur e 4 would imply b oth Klartag CL T (with t he optimal sp e e d of c onver genc e) and Paouris c onc entr ation. 18 W e conclude this section with the es timate that shows compar ison of weak and str ong mo men ts for any probability mea sure and p > n/C . Prop ositi o n 3 .17. F or any p > 0 we have  Z   k x k − Med µ ( k x k )   p dµ  1 /p ≤  Z k x k p dµ  1 /p ≤ 2 · 5 n/p sup k u k ∗ ≤ 1  Z | h u, x i | p dµ  1 /p . Pr o of. As in the proo f of Prop osition 3.11 w e can find u 1 , . . . , u N with k u i k ∗ ≤ 1, N ≤ 5 n such that k x k ≤ 2 max i ≤ N h u i , x i for all x . Then Z k x k p dµ ≤ 2 p Z X i ≤ N | h u i , x i | p dµ ≤ 2 p 5 n sup k u k ∗ ≤ 1 Z | h u i , x i | p dµ. Moreov e r Z {k x k≥ M } ( k x k− M ) p dµ ( x ) ≤ Z {k x k≥ M } ( k x k p − M p ) dµ ( x ) ≤ Z k x k p dµ ( x ) − 1 2 M p and Z {k x k 0 then for any i ∈ { 1 , . . . , n } we have    A + tB n 1  ∩ nB n 1 ∩  x : | x i | ≥ u − t    ≥ e t/ 2   A ∩ nB n 1 ∩  x : | x i | ≥ u    . Pr o of. Obviously we may a ssume that i = 1 a nd u ≤ n . Let A 1 := A ∩ nB n 1 ∩ { x : x 1 ≥ u } and B := { x ∈ B n 1 : x 1 ≥ P i ≥ 2 | x i |} . F ro m the definition of B and A 1 we hav e A 1 − tB ⊂ nB n 1 . O n the other hand B = { x : | x 1 − 1 / 2 | + P i ≥ 2 | x i | ≤ 1 / 2 } , so | B | = 2 − n | B n 1 | = (2 r 1 ,n ) − n . Thus | ( A 1 + tB n 1 ) ∩ nB n 1 | ≥ | ( A 1 − tB ) ∩ n B n 1 | = | A 1 − tB | . 19 Now let us take s := 2 | A 1 | 1 /n r 1 ,n t + 2 | A 1 | 1 /n r 1 ,n . Then we easily c heck that | tB / (1 − s ) | = | A 1 /s | . Since A 1 ⊂ { x ∈ nB n 1 : x 1 ≥ t } we get | A 1 | 1 /n ≤ ( n − t ) /r 1 ,n and s ≤ 2 ( n − t ) / (2 n − t ). Now we can use the Brunn-Minko wsk i ineq uality to g et | A 1 − tB | =    s A 1 s + (1 − s ) − t 1 − s B    ≥    A 1 s    s    − t 1 − s B    1 − s =    A 1 s    = s − n | A 1 | ≥  2 n − t 2 n − 2 t  n | A 1 | =  1 1 − t 2 n − t  n | A 1 | ≥ e tn 2 n − t | A 1 | ≥ e t/ 2 | A 1 | . Notice that A 1 + tB n 1 ⊂ { x : x 1 ≥ u − t } , so we obtain    A + tB n 1 ) ∩ nB n 1 ∩  x : x 1 ≥ u − t    ≥ e t/ 2   A ∩ nB n 1 ∩  x : x 1 ≥ u    , in the s ame way we show    A + tB n 1  ∩ nB n 1 ∩  x : x 1 ≤ − u + t    ≥ e t/ 2   A ∩ nB n 1 ∩  x : x 1 ≤ − u    . Remark 4 .2. A similar r esult (although with a c onstant multiplic ative factor) c an b e obtaine d using the same te chnique and mor e c alculations for n 1 /p B n p inste ad of nB n 1 for p ∈ [1 , 2] . Lemma 4.3. If u ≥ t > 0 then for any i ∈ { 1 , . . . , n } we have ν n  A + tB n 1  ∩  x : | x i | ≥ u − t  ≥ e t/ 2 ν n  A ∩  x : | x i | ≥ u  . Pr o of. T ake an ar bitrary k ∈ N . Let P : R n + k → R n be the pro jection o n to first n co ordina tes. Let ρ k be the uniform pro bability measure on ( n + k ) B n + k 1 , and ˜ ν k the measure defined by ˜ ν k ( A ) = ρ k ( P − 1 ( A )). T ake an arbitrar y set A ⊂ R n . Notice that for any set C ⊂ R n we have C ∩ { x : | x i | ≥ s } = P  P − 1 ( C ) ∩ { x : | x i | ≥ s }  and als o P − 1 ( A ) + B n + k 1 ⊂ P − 1 ( A + B n 1 ). F ro m Lemma 4.1 we hav e ρ k  P − 1 ( A ) + tB n + k 1  ∩  x : | x i | ≥ u − t  ≥ e t/ 2 ρ k  P − 1 ( A ) ∩  x : | x i | ≥ u  , and thus ˜ ν k  A + tB n 1  ∩  x : | x i | ≥ u − t  ≥ e t/ 2 ˜ ν k  A ∩  x : | x i | ≥ u  . When k →∞ , we have ˜ ν k ( C ) → ν n ( C ) for a ny set C ∈ B ( R n ). Thus by g oing to the limit we get the assertio n. 20 Prop ositi o n 4 .4. F or any t > 0 and any n ∈ N we have Z A + tB n 1 | x | 2 dν n ( x ) ≥ e t/ 2 Z A ( | x | − t √ n ) 2 + dν n ( x ) . Pr o of. Let A t = A + tB n 1 . By Lemma 4.3 we get for any s ≥ 0 a nd any i : Z A t I {| x i |≥ s } dν n ( x ) ≥ e t/ 2 Z A I {| x i |≥ s + t } dν n ( x ) . Thu s Z A t x 2 i dν n ( x ) = Z A t Z ∞ 0 2 sI {| x i |≥ s } ds dν n ( x ) = Z ∞ 0 2 s Z A t I {| x i |≥ s } dν n ( x ) ds ≥ e t/ 2 Z ∞ 0 2 s Z A I {| x i |≥ s + t } dν n ( x ) ds = e t/ 2 Z A Z ∞ 0 2 sI {| x i |≥ s + t } ds dν n ( x ) = e t/ 2 Z A  | x i | − t  2 + dν n ( x ) . T o get the asse rtion it is e nough to ta ke the sum over a ll i and no tice that the function f ( y ) := ( √ y − t ) 2 + is conv ex o n [0 , ∞ ), hence n X i =1 ( | x i | − t ) 2 + = n X i =1 f ( x 2 i ) ≥ nf  1 n n X i =1 x 2 i  = ( | x | − t √ n ) 2 + Lemma 4.5. Supp ose that A ⊂ { x ∈ R n : | x | ≥ 5 t √ n } . Then ν n ( A + tB n 1 ) ≥ 1 8 e t/ 2 ν n ( A ) . Pr o of. Let A k := A ∩ { x : 5 t √ n + 2 t ( k − 1 ) ≤ | x | < 5 t √ n + 2 tk } , k = 1 , 2 , . . . . Then A k + tB n 1 ⊂ { x : 5 t √ n + t (2 k − 3 ) ≤ | x | < 5 t √ n + t (2 k + 1 ) } , hence ν n ( A + tB n 1 ) ≥ 1 2 X k ≥ 1 ν ( A k + tB n 1 ) . F rom Prop os ition 4.4 applied for A k we have  5 t √ n + t (2 k + 1 )  2 ν n ( A k + tB n 1 ) ≥ Z A k + tB n 1 | x | 2 dν n ( x ) ≥ e t/ 2 Z A k  | x | − t √ n ) 2 + dν n ( x ) ≥ e t/ 2  4 t √ n + 2 t ( k − 1 )  2 ν n ( A k ) . 21 Thu s ν n ( A k + tB n 1 ) ≥  4 t √ n + 2 t ( k − 1) 5 t √ n + t (2 k + 1)  2 e t/ 2 ν n ( A k ) ≥ 1 4 e t/ 2 ν n ( A k ) . and ν n ( A + tB n 1 ) ≥ 1 2 X k ≥ 1 1 4 e t/ 2 ν n ( A k ) = 1 8 e t/ 2 ν n ( A ) . Theorem 4.6. F or any A ∈ B ( R n ) and any t ≥ 10 , either ν n  ( A + tB n 1 ) ∩ 50 √ nB n 2  ≥ 1 2 ν n ( A ) or ν n ( A + tB n 1 ) ≥ e t/ 10 ν n ( A ) . (20) In p articular ( 20 ) holds if A ∩ (50 √ nB n 2 + tB n 1 ) = ∅ . Pr o of. Let A k denote A + 10 k B n 1 for k = 0 , 1 , . . . . If for any 0 ≤ k ≤ t/ 10 we hav e ν n ( A k ∩ 5 0 √ nB n 2 ) ≥ ν n ( A ) / 2, the thesis is proved. Thus w e assume otherwise. Let A ′ k := A k \ 50 √ nB n 2 . F r om Lemma 4 .5 we hav e ν n ( A k +1 ) ≥ ν n ( A ′ k + 10 B n 1 ) ≥ 1 8 e 5 ν n ( A ′ k ) ≥ 1 16 e 5 ν n ( A k ) ≥ e 2 ν n ( A k ) . By a simple induction we get ν n ( A k ) ≥ e 2 k ν n ( A ) for any k ≤ t/ 1 0. Thus w e get ν n ( A + tB n 1 ) ≥ ν n  A ⌊ t/ 10 ⌋  ≥ e 2 ⌊ t/ 10 ⌋ ν n ( A ) ≥ e t/ 10 ν ( A ) . 5 Uniform measure on B n p In this section we will prov e the infim um conv olution prop er t y IC( C ) for B n p balls. Recall that ν n p is a pro duct measure, while µ p,n denotes the uniform measure on r p,n B n p . W e hav e r − n p,n = | B n p | = 2 n Γ(1 + 1 /p ) n Γ(1 + n/ p ) ∼ (2Γ(1 + 1 /p )) n ( ep ) n/p n n/p ( p n/p + 1) , where the la st par t follows from Stirling’s formula. Thus r p,n ∼ n 1 /p . F or ν n p we hav e IC(48) b y Corollary 2.19. Let us first try to understand what sort of conce n tr ation this implies, that is, how do es the function Λ ⋆ behave for ν n p . 22 Prop ositi o n 5 .1. F or any p ≥ 1 and t ∈ R we have B t ( ν p ) ∼ { x : f p ( | x | ) ≤ t } , and Λ ⋆ ν p ( t/C ) ≤ f p ( | t | ) ≤ Λ ⋆ ν p ( C t ) , wher e f p ( t ) = t 2 for t < 1 and f p ( t ) = t p for t ≥ 1 . Pr o of. W e shall use the facts proved in Sectio n 3 to approximate B t ( ν p ). Note that ν p is log- concav e (as its density is log-concave) and symmetric. It is 1– regular from P rop osition 3 .8. Also σ 2 p := Z R x 2 dν p ( x ) = 1 2 γ p Z R x 2 e −| x | p dx = Γ(1 + 3 p ) 3Γ(1 + 1 p ) ∼ 1 for p ∈ [1 , ∞ ). The measur e ˜ ν p with the densit y σ p dν p ( σ p x ) is iso tropic, hence Prop ositio ns 3.3 and 3.6 yield B t ( ˜ ν p ) ∼ √ tB 1 2 = [ − √ t, √ t ] fo r t ≤ 1. Th us, as B t ( ν p ) = σ p B t ( ˜ ν p ), we get B t ( ν p ) ∼ [ − √ t, √ t ] for t ≤ 1. F or t ≥ 1 we hav e M t ( ν p ) = n u ∈ R : 1 2 γ p Z R | u | t | x | t e −| x | p dx ≤ 1 o = ( u ∈ R : | u | ≤ t v u u t ( t + 1 )Γ(1 + 1 p ) Γ(1 + t +1 p ) ) ∼ { u ∈ R : | u | ≤ t − 1 /p } . Thu s Z t ( ν p ) ∼ [ − t 1 /p , t 1 /p ] for | t | ≥ 1, so by Prop os itions 3 .2 a nd 3.5, B t ( ν p ) ∼ [ − t 1 /p , t 1 /p ]. Hence, for all t ≥ 0 w e hav e { x : f p ( | x | ) ≤ t } ∼ { x : Λ ⋆ ν p ( x ) ≤ t } , so Λ ⋆ ν p ( t/C ) ≤ f p ( t ) ≤ Λ ⋆ ν p ( C t ). As Λ ⋆ ν p is symmetr ic, the pr o of is finished. Corollary 5.2. F or any t > 0 and n ∈ N we have B t ( ν n p ) ∼  √ tB n 2 + t 1 /p B n p for p ∈ [1 , 2] √ tB n 2 ∩ t 1 /p B n p for p ≥ 2 . Pr o of. By P rop osition 5 .1, B t ( ν n p ) = { x ∈ R n : X Λ ⋆ ν p ( x i ) ≤ t } ∼ { x ∈ R n : X f p ( | x i | ) ≤ t } . Simple c alculations show that { x ∈ R n : P f p ( | x i | ) ≤ t } ∼ t 1 / 2 B n 2 + t 1 /p B n p for p ∈ [1 , 2] and { x ∈ R n : P f p ( | x i | ) ≤ t } ∼ t 1 / 2 B n 2 ∩ t 1 /p B n p for p ≥ 2. Prop ositi o n 5.3. F or any t ∈ [0 , n ] , p ≥ 1 and n ∈ N we have B t ( µ p,n ) ∼ B t ( ν n p ) . Pr o of. F or t < 1 we use Pr op ositions 3.3 and 3.6. Both µ p,n and ν n p are sym- metric, log –concav e measures, and b oth can b e resca led as in Prop o sition 5.1 to b e isotr opic, thus B t ( µ p,n ) ∼ √ tB n 2 ∼ B t ( ν n p ). Lemma 6 fr om [3] gives (after rescaling by r p,n ),  Z | h a, x i | t dµ p,n ( x )  1 /t ∼ r p,n (max { n, t } ) 1 /p  Z | h a, x i | t dν n p ( x )  1 /t (21) 23 for any p, t ≥ 1 and a ∈ R n . No te that as r p,n ∼ n 1 /p , this s imply means the equiv alence of t -th moments of µ p,n and ν p,n for t ∈ [0 , n ]. Thus M t ( µ p,n ) ∼ M t ( ν p,n ) fo r t ≤ n and therefore B t ( µ p,n ) ∼ B t ( ν p,n ). Remark 5.4 . It is not har d to verify t hat B t ( µ p,n ) ∼ r p,n B n p for t ≥ n . 5.1 T r ansp ort s of measure W e ar e now go ing to investigate tw o transp o rts of mea sure. They will co mbin e to tr ansp ort a measur e with k nown concentration pro p erties ( ν n or ν n 2 , that is the ex po nential or Gaus sian measure) to the uniform mea sure µ p,n . W e will inv estigate the con tr active prop erties of these trans po rts with respect to v arious norms. Our motiv ation is the following: Remark 5.5 . L et U : R n → R n b e a map such that k U ( x ) − U ( y ) k p p ≥ δ k x − y k q q for al l x ∈ R n , y ∈ A. Then U  A + t 1 /q B n q  ⊃ U  R n  ∩  U ( A ) + δ 1 /p t 1 /p B n p  . Analo gously if k U ( x ) − U ( y ) k p p ≤ δ k x − y k q q for al l x ∈ R n , y ∈ A then U  A + t 1 /q B n q  ⊂ U ( A ) + δ 1 /p t 1 /p B n p . Pr o of. Let us prove the first s tatement , the seco nd pr o of is almost identical. Suppo se U ( x ) ∈ U ( A ) + δ 1 /p t 1 /p B n p . Then there ex ists y ∈ A such that k U ( x ) − U ( y ) k p p ≤ δ t. F ro m the assumption we hav e t ≥ k x − y k q q , whic h means x ∈ A + t 1 /q B n q , and U ( x ) ∈ U ( A + t 1 /q B n q ). The first transp ort we intro duce is the radial transp ort T p,n which tra nsforms the pro duct measur e ν n p onto µ p,n – the uniform meas ure on r p,n B n p . W e will show this tr ansp ort is Lips chit z with resp ect to the ℓ p norm and Lipschitz on a large s et with r esp ect to the ℓ 2 norm fo r p ≤ 2. Definition 5.6 . F or p ∈ [1 , ∞ ) and n ∈ N let f p,n : [0 , ∞ ) → [0 , ∞ ) b e given by the e quation Z s 0 e − r p r n − 1 dr = (2 γ p ) n Z f p,n ( s ) 0 r n − 1 dr (22) and T p,n ( x ) := xf p,n ( k x k p ) / k x k p for x ∈ R n . Let us first show the following simple estimate. Lemma 5.7. F or any q > 0 and 0 ≤ u ≤ q/ 2 , q Z u 0 e − t t q − 1 dt ≤ e − u u q  1 + 2 u q  . 24 Pr o of. Let f ( u ) := e − u u q  1 + 2 u q  − q Z u 0 e − t t q − 1 dt. Then f (0) = 0 and f ′ ( u ) = e − u u q (1 − 2 u/q + 2 /q ) ≥ 0 for 0 ≤ u ≤ q/ 2. Now we a re rea dy to state the basic prop erties o f T p,n . Prop ositi o n 5.8. i) The map T p,n tr ansp orts the pr ob ability me asur e ν n p onto the me asure µ p,n . ii) F or al l t > 0 we have e − t p /n t ≤ 2 γ p f p,n ( t ) ≤ t and f ′ p,n ( t ) ≤ (2 γ p ) − 1 ≤ 1 . iii) F or any t > 0 , 0 ≤ f p,n ( t ) /t − f ′ p,n ( t ) ≤ min { 1 , 2 pt p /n } . iv) The function t 7→ f p,n ( t ) /t is de cr e asing on (0 , ∞ ) and for any s, t > 0 , | t − 1 f p,n ( t ) − s − 1 f p,n ( s ) | ≤ ( st ) − 1 | s − t | f p,n ( s ∧ t ) ≤ | s − t | max { s, t } . Pr o of. The definition of T p,n directly implies i). Differentiation of (22) gives e − s p s n − 1 = (2 γ p ) n f n − 1 p,n ( s ) f ′ p,n ( s ) . (23) By (22), e − t p t n ≤ n Z t 0 e − r p r n − 1 dr = (2 γ p ) n f n p,n ( t ) ≤ n Z t 0 r n − 1 dr = t n , which, w hen the n - th r o ot is taken, give the first part o f ii). F or the second part o f ii) we use (23) and the es timate ab ov e to get f ′ p,n ( s ) = e − s p (2 γ p ) − n  s f p,n ( s )  n − 1 ≤ e − s p (2 γ p ) − n  e s p /n 2 γ p  n − 1 = e − s p /n (2 γ p ) − 1 ≤ (2 γ p ) − 1 ≤ 1 . T o show iii) fir st notice that by (23) and ii), tf ′ p,n ( t ) f p,n ( t ) =  t f p,n ( t )  n e − t p (2 γ p ) − n ≤  e t p /n 2 γ p  n e − t p (2 γ p ) − n = 1 , th us f p,n ( t ) /t − f ′ p,n ( t ) ≥ 0. Mor eov er by ii), f p,n ( t ) /t − f ′ p,n ( t ) ≤ f p,n ( t ) /t ≤ 1, so we may ass ume that 2 pt p /n ≤ 1. By (22) a nd Le mma 5.7 we obtain (2 γ p ) n f n p,n ( t ) = n p Z t p 0 e − u u n/p − 1 du ≤ e − t p t n  1 + 2 pt p n  . Thu s using aga in (23) and pa rt ii) we ge t f p,n ( t ) t − f ′ p,n ( t ) = f p,n ( t ) t  1 − e − t p t n (2 γ p ) n f n p,n ( t )  ≤ 1 −  1 + 2 pt p n  − 1 ≤ 2 pt p n . 25 By iii) we get ( f p,n ( t ) /t ) ′ ≤ 0, which proves the first part of iv). F o r the second pa rt supp os e that s > t > 0, then 0 ≤ f p,n ( t ) t − f p,n ( s ) s ≤ f p,n ( t ) t − f p,n ( t ) s = s − t st f p,n ( t ) ≤ s − t s . The nex t Prop o sition may b e also deduced (with different constant) from the mo re g eneral fact prov ed in [1 7]. Prop ositi o n 5 .9. F or any x, y ∈ R n we have k T p,n x − T p,n y k p ≤ 2 k x − y k p . Pr o of. Assume s := k x k p ≥ t := k y k p , we apply Pro p osition 5.8 and get k T p,n x − T p,n y k p =  X i   ( T p,n x ) i − ( T p,n y ) i   p  1 /p =  X i    f p,n ( t ) t ( x i − y i ) +  f p,n ( s ) s − f p,n ( t ) t  x i    p  1 /p ≤  X i  | x i − y i | + | s − t | s | x i |  p  1 /p ≤  X i | x i − y i | p  1 /p + | s − t | s  X i | x i | p  1 /p = k x − y k p +   k x k p − k y k p   k x k p k x k p ≤ 2 k x − y k p . Prop ositi o n 5 .10. L et u ≥ 0 , p ∈ [1 , 2] and x ∈ R n b e su ch that k x k 2 n − 1 / 2 ≤ u k x k p n − 1 /p , then k T p,n x − T p,n y k p ≤ (1 + u ) k x − y k p for al l y ∈ R n . Pr o of. Let s = k x k p and t = k y k p , we use Pr op osition 5.8 as in the pro o f of Prop ositio n 5.9, and the H¨ older inequality , k T p,n x − T p,n y k 2 ≤  X i    | x i − y i | + | x i | | s − t | s    2  1 / 2 ≤ k x − y k 2 + | s − t | s k x k 2 ≤ k x − y k 2 + k x − y k p k x k p k x k 2 ≤ k x − y k 2 + k x k 2 k x k p n 1 p − 1 2 k x − y k 2 ≤ (1 + u ) k x − y k 2 . 26 The seco nd tra nsp ort we will use is a s imple pr o duct tra nsp ort which tr ans- po rts the measure ν n p onto ν n q . W e shall be particula rly interested in the cases p = 1 and p = 2, but most of the results can b e sta ted in the more general setting. Definition 5.1 1. F or 1 ≤ p, q < ∞ we define the map w p,q : R → R by 1 γ p Z ∞ x e − t p dt = 1 γ q Z ∞ w p,q ( x ) e − t q dt. (24) By v p we denote w p, 1 . We also define W n p,q : R n → R n by W n p,q ( x 1 , x 2 , . . . , x n ) = ( w p,q ( x 1 ) , w p,q ( x 2 ) , . . . , w p,q ( x n )) . Note tha t w − 1 p,q = w q,p and ( W n p,q ) − 1 = W n q,p . Differentiating equality (24) we ge t w ′ p,q ( x ) = γ q γ p e − x p + w q p,q ( x ) . (25) W e will pr ov e that w p,q behaves very m uch like x p/q for la rge x , and is mo re or less linea r for small x . W e b eg in with the b ound for q = 1. Lemma 5.12. F or p ≥ 1 we have i) v p ( x ) ≥ x p + ln( pγ p x p − 1 ) and v ′ p ( x ) ≥ px p − 1 for x ≥ 0 , ii) v p ( x ) ≤ e + x p + ln( pγ p x p − 1 ) and v ′ p ( x ) ≤ e e px p − 1 for x ≥ 1 , iii) | v p ( x ) − v p ( y ) | ≥ 2 1 − p | x − y | p . Pr o of. Note that γ 1 = 1. W e have for x ≥ 0, e − v p ( x ) = 1 γ p Z ∞ x e − t p dt ≤ 1 pγ p x p − 1 Z ∞ x pt p − 1 e − t p dt = e − x p pγ p x p − 1 (26) and for x ≥ 1, since (1 + r/p ) p ≤ e r ≤ 1 + er for r ∈ [0 , 1 ], we get e − v p ( x ) ≥ 1 γ p Z x + x 1 − p /p x e − t p ≥ 1 pγ p x p − 1 e − ( x + x 1 − p /p ) p ≥ e − e e − x p pγ p x p − 1 . Notice that by (25), v ′ p ( x ) = e − x p + v p ( x ) /γ p , hence we may estimate v ′ p using the just der ived b ounds on v p . The low er bo und on v ′ p yields | v p ( x ) − v p ( y ) | ≥ | x − y | p for x, y ≥ 0. The same estimate holds for x, y ≤ 0, since v p is o dd. Finally for x ≥ 0 ≥ y we hav e | v p ( x ) − v p ( y ) | = | v p ( x ) | + | v p ( y ) | ≥ | x | p + | y | p ≥ 2 1 − p | x − y | p . Lemma 5.13. i) F or p ≥ q ≥ 1 , | w p,q ( x ) | ≥ | x | p/q and w ′ p,q ( x ) ≥ γ q γ p ≥ 1 2 . ii) F or p ≥ 2 , w ′ p, 2 ( x ) ≥ 1 8 √ p | x | p/ 2 − 1 . 27 Pr o of. Since the function w p,q is o dd, we may and will a ssume that x ≥ 0. i) W e hav e by the monotonicity of u p/q − 1 on [0 , ∞ ), 1 γ p Z ∞ x e − t p dt = 1 γ q Z ∞ w p,q ( x ) e − t q dt = R ∞ w p,q ( x ) e − t q dt R ∞ 0 e − t q dt = R ∞ w p,q ( x ) q/p u p/q − 1 e − u p du R ∞ 0 u p/q − 1 e − u p du ≥ R ∞ w p,q ( x ) q/p e − u p du R ∞ 0 e − u p du = 1 γ p Z ∞ w p,q ( x ) q/p e − u p du, th us w p,q ( x ) q/p ≥ x and w p,q ( x ) ≥ x p/q . F ormula (25) gives w ′ p,q ( x ) ≥ γ q /γ p ≥ 1 / 2. ii) W e b egin by the following Gaussian tail estimate for z > 0: Z ∞ z e − t 2 dt ≥ 1 2 √ z 2 + 1 e − z 2 . (27) W e hav e equa lit y when z →∞ , and direc t calculatio n sho ws the der iv ative of the left–hand–side is no la rger than the deriv ative of the rig ht–hand–side. Let κ := 4 √ π , we will now show that for all x > 0 a nd p ≥ 2, w p, 2 ( x ) ≥ u p ( x ) := max n √ π 2 x, q  x p + ln( √ px p/ 2 − 1 /κ )  + o . (28) Suppo se on the contrary that w p, 2 ( x ) < u p ( x ) for some p ≥ 2 and x > 0. Note that by i) we hav e w ′ p, 2 ≥ γ 2 /γ p ≥ γ 2 = √ π / 2. Thus u p ( x ) is equal to the second part o f the ma ximum. This in pa rticular implies tha t x ≥ 2 / 3, since for x < 2 / 3 we have x p + ln( √ px p/ 2 − 1 /κ ) ≤ 4 9 +  p 2 − 1  ln 2 3 + √ p κ − 1 ≤ 0 . Therefore u p ( x ) ≥ √ π x/ 2 ≥ 1 / √ 3. Now by (26 ), (24) and (27), √ π 1 px p − 1 e − x p ≥ γ 2 γ p 1 px p − 1 e − x p ≥ γ 2 γ p Z ∞ x e − t p dt = Z ∞ w p, 2 ( x ) e − t 2 dt > Z ∞ u p ( x ) e − t 2 dt ≥ 1 2 q u 2 p ( x ) + 1 e − u 2 p ( x ) ≥ 1 4 u p ( x ) e − u 2 p ( x ) = 1 4 u p ( x ) e − ( x p +ln( √ px p/ 2 − 1 /κ )) = √ π √ pu p ( x ) x 1 − p/ 2 e − x p . After s implifying this gives u p ( x ) > √ px p/ 2 . Hence px p < u 2 p ( x ) = x p + 1 2 ln( px p ) + ln 1 κx ≤ p 2 x p + 1 2 px p = px p , which is impossible . This condratiction shows that (28) holds . 28 Thu s w e hav e w p, 2 ( x ) ≥ u p ( x ) and by (2 5) we obtain w ′ p, 2 ( x ) ≥ γ 2 γ p e − x p + u 2 p ( x ) ≥ √ π 2 1 κ √ px p/ 2 − 1 = 1 8 √ px p/ 2 − 1 . Remark 5. 14. By taking u p ( x ) = max { √ π x/ 2 , p ( x p + ln( px p/ 2 − 1 / ( κ ln p ))) + } for sufficiently lar ge κ and estimating c ar eful ly one may arrive at the b ound w ′ p, 2 ( x ) ≥ C − 1 px p/ 2 − 1 / ln p . One c annot, however, re c eive a b ound of t he or der of px p/ 2 − 1 . Prop ositi o n 5 .15. F or p ≥ q ≥ 1 we have i) ν n p ( W n q,p ( A )) = ν n q ( A ) for A ∈ B ( R n ) , ii) | w q,p ( x ) − w q,p ( y ) | ≤ 2 | x − y | for x ∈ R , iii) for x, y ∈ R n and r ≥ 1 , k W n q,p ( x ) − W n q,p ( y ) k r ≤ 2 k x − y k r , iv) for x, y ∈ R ,   w 1 ,p ( x ) − w 1 ,p ( y )   ≤ 2 min( | x − y | , | x − y | 1 /p ) ≤ 2 | x − y | 1 /q , v) k W n 1 ,p ( x ) − W n 1 ,p ( y ) k q q ≤ 2 q k x − y k 1 for x, y ∈ R n . Pr o of. Prop er t y i) follows from the definition of w q,p and W n q,p . Since w q,p = w − 1 p,q we ge t ii) by Lemma 5.1 3 i). P rop erty iii) is a direct consequence of ii). By Lemma 5 .12 iii), | w 1 ,p ( x ) − w 1 ,p ( y ) | = | v − 1 p ( x ) − v − 1 p ( y ) | ≤ 2 1 − 1 /p | x − y | 1 /p . The ab ove inequa lity tog ether with ii) gives iv) and iv) yie lds v). Now w e d e fine a transp o rt from the exp onential measure ν n to µ p,n for p ≥ 2: Definition 5.16. F or n ∈ N and 2 ≤ p < ∞ we define the map S p,n : R n → R n by S p,n ( x ) := T p,n ( W n 1 ,p ( x )) . This transp ort s atisfies the following b o und: Prop ositi o n 5.17. We have k S p,n ( x ) − S p,n y k 2 ≤ 4 k x − y k 2 for al l x, y ∈ R n and p ≥ 2 . Pr o of. It is enough to show that k D S p,n ( x ) k ≤ 4, wher e D S p,n is the deriv ative matrix, a nd the norm is the op erato r norm from ℓ n 2 int o ℓ n 2 . Let s = k W n 1 ,p ( x ) k p . By direct calculation we g et ( ∂ S p,n ) j ∂ x i ( x ) = δ ij f p,n ( s ) w ′ 1 ,p ( x i ) s + α ( s ) w 1 ,p ( x j ) β ( x i ) (29) 29 where α ( s ) := s − p − 1  sf ′ p,n ( s ) − f p,n ( s )  and β ( t ) := | w 1 ,p ( t ) | p − 1 sgn( w 1 ,p ( t )) w ′ 1 ,p ( t ) . Thu s we can bo und k D S p,n ( x ) k ≤ f p,n ( s ) s max i | w ′ 1 ,p ( x i ) | + | α ( s ) |   W n 1 ,p ( x )   2  n X i =1 β 2 ( x i )  1 / 2 . Since w 1 ,p = w − 1 p, 1 , Prop ositio n 5.13 i) implies | w ′ 1 ,p ( x j ) | ≤ 2, while by Prop o - sition 5.8 we hav e f p,n ( s ) /s ≤ 1 . Thus the fir st summand can b e b ounded by 2. F or the second summand note that by P rop osition 5 .8 iii), | α ( s ) | = s − p    f ′ p,n ( s ) − f p,n ( s ) s    ≤ s − p min n 1 , 2 ps p n o . (30) Moreov e r, k W n 1 ,p ( x ) k 2 ≤ n 1 / 2 − 1 /p s by the H¨ older inequality and | β ( t ) | = | w 1 ,p ( t ) | p − 1 | w ′ 1 ,p ( t ) | = | w 1 ,p ( t ) | p − 1 v ′ p ( w 1 ,p ( t )) ≤ 1 p . by Lemma 5.12. Th us k D S p,n ( x ) k ≤ 2 + s − p min n 1 , 2 ps p n o n 1 / 2 − 1 /p s n 1 / 2 p ≤ 2 + 2 sn − 1 /p min { ns − p , 1 } ≤ 4 . Prop ositi o n 5 .18. F or any y , z ∈ R n and p ≥ 2 we have k S p,n ( y ) − S p,n ( z ) k 2 ≤ k W n 1 ,p ( y ) − W n 1 ,p ( z ) k 2 + 2 n − 1 / 2 k y − z k 1 . Pr o of. Let u i ( t ) = ( y 1 , y 2 , . . . , y i − 1 , t, z i +1 , z i +2 , . . . , z n ) for i = 1 , . . . , n . Note that u i ( y i ) = u i +1 ( z i +1 ), u 1 ( z 1 ) = z and u n ( y n ) = y , hence S p,n ( z ) − S p,n ( y ) = n X i =1  S p,n ( u i ( z i )) − S p,n ( u i ( y i ))  . Let s i ( t ) := k w 1 ,p ( u i ( t )) k p . By vector–v alued integration and (29) we get S p,n  u i ( z i )  − S p,n  u i ( y i )  = Z z i y i ∂ S p,n ∂ x i ( u i ( t )) dt = a i + b i , where a i := Z z i y i f p,n ( s i ( t )) s i ( t ) w ′ 1 ,p ( t ) e i dt 30 and b i := Z z i y i α ( s i ( t )) β ( t ) W n 1 ,p ( u i ( t )) dt. As in the pro of of Pro po sition 5.17 we show that   α ( s i ( t )) β ( t ) W n 1 ,p ( u i ( t ))   2 ≤ 2 n − 1 / 2 s i ( t ) n − 1 /p min { ns i ( t ) − p , 1 } ≤ 2 n − 1 / 2 , th us    n X i =1 b i    2 ≤ n X i =1 k b i k 2 ≤ 2 n − 1 / 2 n X i =1 | y i − z i | = 2 n − 1 / 2 k y − z k 1 . T o deal wit h the sum of a i ’s we notice tha t, since f p,n ( s ) /s ≤ 1 and w ′ 1 ,p ( x ) ≥ 0,    D X j a j , e i E    = | h a i , e i i | =    Z z i y i f p,n ( s i ( t )) s i ( t ) w ′ 1 ,p ( t ) dt    ≤    Z z i y i w ′ 1 ,p ( t ) dt    =   w 1 ,p ( z i ) − w 1 ,p ( y i )   . Thu s k X i a i k 2 ≤ k X i  w 1 ,p ( z i ) − w 1 ,p ( y i )  e i k 2 = k W n 1 ,p ( z ) − W n 1 ,p ( y ) k 2 . Corollary 5.19. If x − y ∈ tB n 1 + t 1 / 2 B n 2 for some t > 0 , then for al l p ≥ 2 , S p,n x − S p,n y ∈ 1 0( t 1 / 2 B n 2 ∩ t 1 /p B n p ) . Pr o of. Let us fix x, y with x − y ∈ tB n 1 + t 1 / 2 B n 2 . By Pro po sition 5.15 iv), k W n 1 ,p ( x ) − W n 1 ,p ( y ) k p p = X i | w 1 ,p ( x i ) − w 1 ,p ( y i ) | p ≤ 2 p X i min( | x i − y i | p , | x i − y i | ) ≤ 2 p X i min( | x i − y i | 2 , | x i − y i | ) ≤ 2 p +2 t. Thu s by Pro po sition 5.9, k S p,n x − S p,n y k p ≤ 2 k W n 1 ,p x − W n 1 ,p y k p ≤ 8 t 1 /p . By H¨ older’s inequality k S p,n x − S p,n y k 2 ≤ n 1 / 2 − 1 /p k S p,n x − S p,n y k p ≤ 8 t 1 / 2 for t ≥ n . Assume now that t ≤ n . Let z be such that x − z ∈ t 1 / 2 B n 2 and z − y ∈ tB n 1 . Then S p,n x − S p,n z ∈ 4 t 1 / 2 B n 2 by Propo sition 5.17 and k W n 1 ,p z − W n 1 ,p y k 2 ≤ 2 √ t by Pr op osition 5 .15 v). Th us by Pr op osition 5.1 8, k S p,n x − S p,n z k 2 ≤ 4 t 1 / 2 + 2 n − 1 / 2 t ≤ 6 t 1 / 2 . Hence S p,n x − S p,n y ∈ 1 0 t 1 / 2 B n 2 . 31 The last function we define transp or ts the Gaussia n measur e ν n 2 to µ p,n for p ≥ 2. Definition 5.20. F or n ∈ N and 2 ≤ p < ∞ we define the map ˜ S p,n : R n → R n by ˜ S p,n ( x ) := T p,n ( W n 2 ,p ( x )) . Prop ositi o n 5.2 1. We have k ˜ S p,n ( x ) − ˜ S p,n y k 2 ≤ 18 k x − y k 2 for al l x, y ∈ R n and p ≥ 2 . Pr o of. W e ar gue in a similar way a s in the pro of of P rop osition 5.17. W e need to show that k D ˜ S p,n ( x ) k ≤ 18. Direct calculation gives ( ∂ ˜ S p,n ) j ∂ x i ( x ) = δ ij f p,n ( ˜ s ) w ′ 2 ,p ( x i ) ˜ s + α (˜ s ) w 2 ,p ( x j ) ˜ β ( x i ) (31) where ˜ s = k W n 2 ,p ( x ) k p , α ( s ) := s − p − 1  sf ′ p,n ( s ) − f p,n ( s )  and ˜ β ( t ) := | w 2 ,p ( t ) | p − 1 sgn( w 2 ,p ( t )) w ′ 2 ,p ( t ) . Thu s we can bo und k D ˜ S p,n ( x ) k ≤ f p,n ( ˜ s ) ˜ s max i | w ′ 2 ,p ( x i ) | + | α ( ˜ s ) |   W n 2 ,p ( x )   2  n X i =1 ˜ β 2 ( x i )  1 / 2 . (32 ) The fir st summand is b ounded b y 2 as in the pro of o f P rop osition 5.17. Since w 2 ,p = w − 1 p, 2 we ge t by Lemma 5.13 ii) | ˜ β ( x ) | = | w 2 ,p ( x ) | p − 1 | w ′ 2 ,p ( x ) | = | w 2 ,p ( x ) | p − 1 w ′ p, 2 ( w 2 ,p ( x )) ≤ 8 √ p | w 2 ,p ( x ) | p/ 2 , hence  n X i =1 ˜ β 2 ( x i )  1 / 2 ≤ 8 √ p ˜ s p/ 2 . Using (30) and k W n 2 ,p ( x ) k 2 ≤ n 1 / 2 − 1 /p ˜ s we b ound the second s ummand in (32) by ˜ s − p min n 1 , 2 p ˜ s p n o n 1 / 2 − 1 /p ˜ s 8 √ p ˜ s p/ 2 = 8 p − 1 /p min { u − 1 / 2 , 2 u 1 / 2 } u 1 /p ≤ 16 , where u := p ˜ s p /n . 5.2 Applying ν 1 results – p ≤ 2 W e start with the version of Theorem 4.6 for ν p . Lemma 5 . 22. F or any A ∈ B ( R n ) , p ∈ [1 , 2] and t ≥ 1 at le ast one of t he fol lowing holds: 32 • ν n p ( A + 2 0 t 1 /p B n p ) ≥ e t ν n p ( A ) or • ν n p  ( A + 2 0 t 1 /p B n p ) ∩ 10 0 √ nB n 2  ≥ 1 2 ν n p ( A ) . Pr o of. W e will use the transp o rt W n 1 ,p from ν n to ν n p . Prop ositio n 5.15 v) gives k W n 1 ,p ( x ) − W n 1 ,p ( y ) k p p ≤ 2 p k x − y k 1 . By Remar k 5.5 this means tha t A + 2 (10 t ) 1 /p B n p ⊃ W n 1 ,p ( W n p, 1 ( A ) + 10 tB n 1 ). Let us fix t ≥ 1 and apply T heorem 4.6 to W n p, 1 ( A ) and 1 0 t . If the seco nd case o ccurs , we hav e ν n p ( A + 2 0 t 1 /p B n p ) ≥ ν n p  W n 1 ,p ( W n p, 1 ( A ) + 10 tB n 1 )  = ν n ( W n p, 1 ( A ) + 10 tB n 1 ) ≥ e t ν n ( W n p, 1 ( A )) = e t ν n p ( A ) . If the firs t case of Theo rem 4.6 o ccur s, then due to P rop osition 5.15 iii) we hav e k W n 1 ,p ( x ) k 2 ≤ 2 k x k 2 , so 2 αB n 2 ⊃ W n 1 ,p ( αB n 2 ) fo r any α > 0. Th us ν n p  ( A + 2 0 t 1 /p B n p ) ∩ 100 √ nB n 2  ≥ ν n p  W n 1 ,p ( W n p, 1 ( A ) + 10 tB n 1 ) ∩ 100 √ nB n 2  = ν n p  W n 1 ,p  ( W n p, 1 ( A ) + 10 tB n 1 ) ∩ W n p, 1 (100 √ nB n 2 )  ≥ ν n p  W n 1 ,p  ( W n p, 1 ( A ) + 10 tB n 1 ) ∩ 50 √ nB n 2  = ν n  W n p, 1 ( A ) + 10 tB n 1  ∩ 50 √ nB n 2  ≥ 1 2 ν n ( W n p, 1 ( A )) = 1 2 ν n p ( A ) . Lemma 5.23. Ther e ex ist s a c onst ant C such that for any p ∈ [1 , 2 ] , t > 0 and n ∈ N we have ν n p  A + C ( t 1 /p B n p + t 1 / 2 B n 2 )  ≥ min n 1 2 , e t ν n p ( A ) o . Pr o of. Corolla ry 5 .2 gives B s ( ν n p ) ⊂ C ( s 1 /p B n p + s 1 / 2 B n 2 ) for s > 0. By Co rol- lary 2.1 9, ν n p satisfies I C (48), which, due to Prop ositio n 2.4 implies ν n p ( A + 48 B 2 t ( ν n p )) ≥ min { 1 / 2 , e t ν n p ( A ) } fo r any Bo rel set A . Thus we have ν n p  A + 9 6 C ( t 1 /p B n p + t 1 / 2 B n 2 )  ≥ min { 1 / 2 , e t ν n p ( A ) } . Prop ositi o n 5.24. F or any α > 1 t her e exists a c onstant c ( α ) such t hat for any n ∈ N and p ≥ 1 we have ν n p ( { x : k x k p < c ( α ) n 1 /p } ) < α − n . Pr o of. W e hav e ν n p ( { x : k x k p < c ( α ) n 1 /p } ) = n Γ(1 + n p ) Z c ( α ) n 1 /p 0 e − r p r n − 1 dr ≤ n Γ(1 + n p ) Z c ( α ) n 1 /p 0 r n − 1 = c ( α ) n n n/p Γ(1 + n p ) ≤ ( C c ( α )) n , 33 where in t he last step w e use the Stirling appr oximation and C a s alwa y s denotes a universal cons tant. Thus it is enough to ta ke c ( α ) < ( C α ) − 1 . Theorem 5.25. Ther e exist s a universal c onstant C such that µ p,n satisfies CI( C ) for any p ∈ [1 , 2] and n ∈ N . Pr o of. By P rop ositions 2.7, 3.11, 3.5 and 5.3 it is eno ugh to show µ p,n  A + C  t 1 /p B n p + t 1 / 2 B n 2  ≥ min { 1 / 2 , e t µ p,n ( A ) } . (33) for 1 ≤ t ≤ n a nd µ p,n ( A ) ≥ e − n . Recall that T p,n denotes the ma p transp or ting ν n p to µ p,n . Apply Lemma 5.22 to T − 1 p,n ( A ) and t . If the fir st cas e o ccur s, we hav e ν n p  T − 1 p,n ( A ) + 20 t 1 /p B n p  ≥ e t ν n p  T − 1 p,n ( A )  = e t µ p,n ( A ) . Prop ositio n 5.9 gives k T p,n x − T p,n y k p ≤ 2 k x − y k p , th us by Remark 5.5, µ p,n  A + 40 t 1 /p B n p  = ν n p  T − 1 p,n  A + 40 t 1 /p B n p  ≥ ν n p  T − 1 p,n ( A ) + 20 t 1 /p B n p  ≥ e t µ p,n ( A ) and we obtain (33) in this case. Hence we may ass ume that the second case of Lemma 5.2 2 holds , that is ν n p ( A ′ ) ≥ 1 2 ν n p ( T − 1 p,n ( A )) = 1 2 µ p,n ( A ) , where A ′ :=  T − 1 p,n ( A ) + 20 t 1 /p B n p  ∩ 100 √ nB n 2 . In particular ν n p ( A ′ ) ≥ e − n / 2. Let A ′′ := A ′ ∩ { x : k x k p ≥ ˜ cn 1 /p } , where ˜ c = c (4 e ) is a co nstant g iven by Pr op osition 5.2 4 for α = 4 e . Then ν n p ( A ′′ ) ≥ ν n p ( A ′ ) − (4 e ) − n ≥ 1 2 ν n p ( A ′ ) ≥ 1 4 µ p,n ( A ) . W e apply L emma 5.2 3 for A ′′ and 4 t to get µ p,n  T p,n  A ′′ + 4 C  t 1 /p B n p + t 1 / 2 B n 2  ≥ ν n p  A ′′ + C  (4 t ) 1 /p B n p + (4 t ) 1 / 2 B n 2  ≥ min n 1 2 , e 4 t ν n p ( A ′′ ) o ≥ min n 1 2 , e 4 t µ p,n ( A ) 4 o ≥ min n 1 2 , e t µ p,n ( A ) o . Prop ositio n 5.9 and Remar k 5.5 imply T p,n  A ′′ + 4 C t 1 / 2 B n 2 + 4 C t 1 /p B n p  ⊂ T p,n  A ′′ + 4 C t 1 / 2 B n 2  + 8 C t 1 /p B n p . 34 Moreov e r, for x ∈ A ′′ we hav e k x k 2 ≤ 1 00 √ n and k x k p ≥ ˜ cn 1 /p . Thu s n − 1 / 2 k x k 2 ≤ 100 ˜ c − 1 n − 1 /p k x k p , s o we can use Pr op osition 5.10 along with Re- mark 5.5 to get T p,n ( A ′′ + 4 C t 1 / 2 B n 2 ) ⊂ T p,n ( A ′′ ) + ˜ C t 1 / 2 B n 2 . Prop ositio n 5.9, Remark 5.5 and the definitions of A ′ and A ′′ yield T p,n ( A ′′ ) ⊂ T p,n ( A ′ ) ⊂ T p,n  T − 1 p,n ( A ) + 20 t 1 /p B n p  ⊂ A + 40 t 1 /p B n p . Putting the four estimates tog ether, we can write µ p,n  A +(40 + 8 C ) t 1 /p B n p + ˜ C t 1 / 2 B n 2  ≥ µ p,n  T p,n ( A ′′ ) + ˜ C t 1 / 2 B n 2 + 8 C t 1 /p B n p  ≥ µ p,n  T p,n  A ′′ + 4 C t 1 / 2 B n 2  + 8 C t 1 /p B n p  ≥ µ p,n  T p,n  A ′′ + 4 C  t 1 /p B n p + t 1 / 2 B n 2  ≥ min n 1 2 , e t µ p,n ( A ) o , which g ives (33) in the seco nd cas e and ends the pro of. Corollary 5. 2 6. Ther e exists an absolute c onstant C such that the me asur e µ p,n satisfies IC( C ) for any p ∈ [1 , 2] and n ∈ N . Pr o of. Let ˜ µ p,n ( A ) := µ p,n ( σ p,n A ) , wher e σ 2 p,n := Z x 2 1 dµ p,n , then ˜ µ p,n is isotropic . Both prop erties IC and CI are affine inv a riant, so b y Theorem 5.25, ˜ µ p,n has pr op erty CI( C ) and w e are to show that it satisfies IC( C ). By Corolla ry 3.10 we only need to show Cheeg er’s ineq uality for ˜ µ p,n with unifor m consta nt . A recent result of S. So din ([19, Theorem 1]) states (after r escaling fro m B n p to r p,n B n p ) tha t µ + p,n ( A ) ≥ c min { µ p,n ( A ) , 1 − µ p,n ( A ) } log 1 − 1 /p 1 min { µ p,n ( A ) , 1 − µ p,n ( A ) } (34) for some universal consta nt c . Thus ˜ µ p,n satisfies Cheeger ’s inequality with constant cσ n,p and it is enoug h to notice that by (21), σ p,n ∼ r p,n n 1 /p  Z x 2 1 dν p ( x )  1 / 2 ∼ 1 . 35 5.3 The easy case – p ≥ 2 This cas e will follow easily from the exp onential case a nd the facts fro m sub- section 5 .1. Theorem 5.27. Ther e ex ists a un iversal c onstant C such that for any A ⊂ R n , any t ≥ 1 , n ≥ 1 and p ≥ 2 we have µ p,n  A + C  t 1 /p B n p ∩ t 1 / 2 B n 2   ≥ min { 1 / 2 , e t µ p ( A ) } . Pr o of. In this case we will again use the transp ort S p,n . Assume A ⊂ r p,n B n p , let ˜ A := S − 1 p,n ( A ). B y T alagra nd’s inequality (6) we h ave ν n ( ˜ A + C tB n 1 + √ C tB n 2 ) ≥ min { e t ν n ( ˜ A ) , 1 / 2 } . How ever by Co rollar y 5.1 9 we hav e S p,n  ˜ A + C tB n 1 + √ C tB n 2 )  ⊂ S p,n ( ˜ A ) + 10 C  √ tB n 2 + t 1 /p B n p  . Thu s , as S p,n ( ˜ A ) = A a nd S p,n transp orts the measure ν n to µ p,n , we get the thesis. By Prop o sitions 2 .7, 3 .11, 3.5 and 5.3, Theorem 5.27 y ield the following. Corollary 5.28 . Ther e exists an absolute c onst ant C such t hat µ p,n satisfies CI( C ) for p ≥ 2 . Theorem 5.29. F or any p ≥ 2 and n ≥ 1 the m e asu r e µ p,n satisfies Che e ger’s ine quality (12) with t he c onstant 1 / 20 . Pr o of. Again we shall tra nsp ort this result from the exp onential mea sure. By [5] Cheeger’s ineq uality holds for ν n with the constant κ = 1 / (2 √ 6), thus by Prop ositio n 5.17 µ p,n satisfies (12) with the consta nt κ/ 4 ≥ 1 / 20. As in the pro o f o f Cor ollary 5 .26 we s how that Theorem 5 .29 and Cor ollary 5.28 imply infimum conv o lution inequality for µ p,n , p ≥ 2 . Adding the tw o results together we g et Theorem 5 .30. Ther e exists a universal c onstant C su ch t hat for any p ∈ [1 , ∞ ] and any n ∈ N the me asur e µ p,n satisfies IC ( C ) . W e conclude this section with the proof of logaritmic Sobo lev–type inequality for µ p,n . Theorem 5.31. L et Φ( x ) = (2 π ) − 1 / 2 R ∞ x exp( − y 2 / 2) b e a Gaussian distribu- tion fun ction, A ∈ B ( R n ) and p ≥ 2 , then µ p,n ( A ) = Φ( x ) ⇒ µ p,n ( A + 1 8 √ 2 tB n 2 ) ≥ Φ( x + t ) for al l t > 0 . In p articular ther e ex ists a universal c onstant C such t hat µ p,n + ( A ) ≥ 1 C min  µ p,n ( A ) s ln 1 µ p,n ( A ) , (1 − µ p,n ( A )) s ln 1 1 − µ p,n ( A )  . 36 Pr o of. By Prop osition 5.2 1, ˜ S p,n ( √ 2 · ) is 18 √ 2–Lipschitz and tra nsp orts the canonical Gaussian measure o n R n onto µ p,n . Hence the firs t part of theo- rem follows by the Gaussia n iso per imetric inequality o f Borell [8] and Sudako v, Tsirel’son [2 0]. The la st estimate immediately follows by a standar d estimate of the Gaussia n isop er imetric function. 6 Concluding Remarks 1. Wit h the notion of the IC pro p erty one may a sso ciate IC– domination of symmetric pr obability mea sures µ, ˜ µ on R n : we s ay that µ is IC –dominate d by ˜ µ with a c onstant β if ( µ, Λ ∗ ˜ µ ( · β )) has pro per ty τ . IC–domination has the tensorization pr op erty: if µ i are IC( β )–dominated by ˜ µ i , 1 ≤ i ≤ n , then ⊗ µ i is IC( β )–do minated b y ⊗ ˜ µ i . An ea sy mo dification of the pro of of Co rollary 3.1 0 gives that if µ is IC( β )–do minated by a n α – regular measure ˜ µ , then ∀ p ≥ 2 ∀ A ∈B ( R n ) µ ( A ) ≥ 1 2 ⇒ 1 − µ ( A + c ( α ) β Z p ( ˜ µ )) ≤ e − p (1 − µ ( A ) ) . F ollowing the pr o of o f P rop osition 3 .12 we also get for a ll p ≥ 2,  Z   k x k − Med µ ( k x k )   p dµ  1 /p ≤ ˜ c ( α ) β sup k u k ∗ ≤ 1  Z | h u, x i | p d ˜ µ  1 /p . 2. O ne may co nsider conv ex versions of prop erties CI and IC. W e say that a symmetric proba bilit y meas ure µ satisfies the c onvex infimum c onvolution ine quality with a c onstant β if the pair ( µ, Λ ∗ µ ( · β )) has conv ex prop erty ( τ ), i.e. the inequa lit y (1) ho lds for all co nv ex function and f with ϕ ( x ) = Λ ∗ µ ( x/β ). Analogously µ sa tisfies c onvex c onc entr ation ine quality with a c onstant β , if (18) holds for all co n vex Borel s ets A . W e do no t k now if convex IC implies conv ex CI, howev er for α –reg ular mea sures it implies a weaker version of co n vex CI, namely µ ( A ) ≥ 1 / 2 ⇒ µ ( A + c 1 ( α ) β Z p ( ˜ µ )) ≥ 1 − 2 e − p and this pro pe rty yie lds CWSM( c 2 ( α ) β ). F rom the r esults of [16] o ne may easily deduce that the uniform distr ibution on {− 1 , 1 } n satisfies conv ex IC( C ) with a universal constant C . 3. Pro p erty IC may be also inv estigated for nonsymmetric measur es. 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