Best Constants in Kahane-Khintchine Inequalities in Orlicz Spaces

J. Multivariate Aal. Vol. 45, No., 993, (83-6) Prerit Ser. No. 0, 99, Math. Ist. Aarhus Best ostats i Kahae-Khitchie Iequalities i Orlicz Saces GORAN PESKIR Several iequalities of Kahae-Khitchie s tye i certai Orlicz saces are roved. For this the classical symmetrizatio techique is used ad four basically differet methods have bee reseted. The first two are based o the well-kow estimates for subormal radom variables, see [9], the third oe is a cosequece of a certai Gaussia-Jese s majorizatio techique, see [6], ad the fourth oe is obtaied by Haageru-Youg-Stechki s best ossible costats i the classical Khitchie iequalities, see [4]. Moreover, by usig the cetral limit theorem it is show that this fourth aroach gives the best ossible umerical costat i the iequality uder cosideratio: If f " i j i g is a Beroulli sequece, ad k k deotes the Orlicz orm iduced by the fuctio (x) e x 0 for x R ; the the best ossible umerical costat satisfyig the followig iequality: X a i " i X j a i j for all a;... ; a R ad all, is equal to 83. Similarly, the best ossible estimates of that tye are also deduced for some other iequalities i Orlicz saces, discovered i this aer.. Itroductio Let f " i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k deote the gauge orm o (; F ; P ), that is: k X k if f a > 0 j E[ (Xa) ] g for all real valued radom variables X o (; F ; P ), where (x) e x 0 for x R, ad with if ;. The the followig iequality is satisfied: () a i " i j a i j for all a;... ; a R ad all, where is a umerical costat, see for istace [9], [3], [4], [9], [], [30]. I this aer we shall show that the best ossible umerical costat AMS 980 subject classificatios. Primary 4A44, 4A50, 46E30, 60E5, 60G50. Secodary 44A0. Key words ad hrases: Orlicz orm, Frechet orm, Khitchie iequality, Haageru-Youg-Stechki s costats, subormal, symmetrizatio. gora@imf.au.dk

i iequality () is equal to 83, ad the reset work is devoted to the study of various ways for rovig (), as well as to the study of the aalogous questio for some other Orlicz orms, see sectio below. Moreover, by usig the classical symmetrizatio techique, iequality (), as well as some other iequalities of that tye which will be deduced later, will be exteded i a aroriate way to more geeral cases. Let us say that the iequality give i () has a umber of alicatios. I articular, usig that result together with certai modulus of cotiuity results of Presto s tye, see [5], [0] ad [30], oe ca obtai a coectio betwee the cetral limit theorem i a Baach sace ad the uiform law of large umbers o the uit ball of its dual sace, see [7] ad [8]. For alicatios to the law of iterated logarithm i a Baach sace, see [30].. Prelimiary facts Orlicz fuctioals, orms ad saces. Let (X; A; ) be a measure sace, ad let (B; k k) be a Baach sace. Let ' be a icreasig left cotiuous fuctio from [0; [ ito [0; [ such that '(0) lim t#0 '(t) 0 : The the Orlicz fuctioals ' ; T ' ad 7 ' geerated by ' are defied as follows: ' (R f ) if f a > 0 j T ' (R f ) if f a > 0 j 7 ' (R f ) 0 0 0 R f (t) '(dt) R f (at) '(dt) g R f (at) '(dt) a g where R f (t) 3 f k f k > t g for f B X. Recall that R 0 f (t) '(dt) deotes the itegral of a fuctio f with resect to the Lebesgue-Stieltjes measure ' defied by ' ([0; x[) '(x) for all x 0, that 3 deotes the outer -measure, ad that B X deotes the set of all fuctios from X ito B. Also the followig formula is valid, see [5]: 3 ' d 0 3 f > t g '(dt) for every fuctio from [0; ] ito [0; ], where we ut '() lim t! '(t). The the fuctio orm ad the fuctio sace, see [5], iduced by fuctioals ', T ' ad 7 ' are give resectively by: k f k ' if f a > 0 j 3 '( a k f k ) d L ' ( 3 ; B) f f B X j lim "#0 k "f k ' 0 g k f k T' if f a > 0 j 3 '( g a k f k ) d a g L T' ( 3 ; B) f f B X j lim "#0 k "f k T' 0 g

k f k 7' 3 '( k f k ) d L 7' ( 3 ; B) f f B X j lim k "f k 7' 0 g "#0 R 3 f d for all f B X. Recall that deotes the uer -itegral of a fuctio f. It is show i [5] that the followig statemets are satisfied: () ( B X ; k k T' ) is a seudo-f 3 -sace, ad cosequetly ( L T' ( 3 ; B) ; k k T' ) is a seudo-frechet sace. () If ' is covex, the ( B X ; k k' ) is a seudo-b3 -sace, ad cosequetly ( L ' ( 3 ; B) ; k k ' ) is a seudo-baach sace. (3) If ' is subadditive ( for istace, cocave ), the ( B X ; k k 7' ) is a seudo-f 3 -sace, ad thus ( L 7' ( 3 ; B) ; k k 7' ) is a seudo-frechet sace. Moreover, the we have: r for all f B X. q k f k 7' ^ k f k 7' k f k T' k f k 7' _ k f k 7' (4) If ' is covex at 0, i.e. '(t) '(t) for all [0; ] ad all t 0, the ( B X ; q q k k' ) for all f B X. is a seudo-f 3 -sace, ad we have: k f k ' ^ k f k ' k f k T' q k f k ' _ k f k ' (5) If ' is cocave at 0, i.e. '(t) '(t) for all [0; ] ad all t 0, the ( B X ; q q k k 7' ) for all f B X. is a seudo-f 3 -sace, ad we have: k f k 7' ^ k f k 7' k f k T' q k f k 7' _ k f k 7' I the ext cosideratios we shall maily work with the fuctio defied by (x) e x 0 for x R. The Baach sace B will be equal to R, ad the measure will be a robability measure, that is (X). We shall write L () ; L T () ad L 7 () to deote the saces of all -measurable fuctios i L ( 3 ;R) ; L T ( 3 ;R) ad L 7 ( 3 ;R), resectively. Oe ca verify that the give saces are closed ad the atural ijectios are cotiuous. The fuctio orm k k will be shortly deoted by k k. The Orlicz sace (L (); k k ) will be called the gauge sace, ad the Orlicz orm k k will be called the gauge orm. For more iformatios i this directio we shall refer the reader to []. Subormal radom variables. Let X be a real valued radom variable defied o a robability sace (; F ; P ). The the Lalace trasform of X is give by: 3

L X (z) E(e zx ) for all z D(L X ), where D(L X ) f z j E je zx j < g is the comlex domai of L X. The real domai of L X is give by R(L X ) D(L X ) \ R. Let us recall that a ormal distributed radom variable N N (; ) with mea ad variace 0 has the Lalace trasform give by: L N (z) e z + z for all z D(L N ). Ad a real radom variable X is called subormal, if its Lalace trasform is domiated o the real lie by the Lalace trasform of some ormally distributed radom variable. I other words, X is subormal, if there exists R ad 0 such that: (6) L X (t) e t + t for all t R. It is well-kow that for a subormal radom variable X we have: (7) EX ad VarX. Ad if (6) is satisfied with 0 ad, the X is said to be a stadard subormal radom variable. The by (7) we have EX 0 ad VarX. A sequece of ( stadard ) subormal radom variables will be called a ( stadard ) subormal sequece. For more iformatios i this directio we shall refer the reader to [9] (.6). Gaussia-Jese s majorizatio techique. Recall that a fiite or ifiite sequece of ideedet ad idetically distributed radom variables " ; " ;... takig values 6 with the same robability is called a Beroulli sequece. ertai iequalities ivolvig Beroulli sequeces ca be easily obtaied by usig a techique which will be called the Gaussia-Jese s majorizatio techique ad which may be described as follows, see [6] (.-): Suose that we have a covex ad sig-symmetric fuctio g : R! ] 0 ; ], i.e. g is covex ad g (6x ;... ; 6x ) g (x ;... ; x ) for all choices of sigs 6 ad all (x ;... ; x ) R ; ad let X ;... ; X L (P ) be real valued radom variables. The we evidetly have: g (x ;... ; x ) g (jx j ;... ; jx j) for all (x ;... ; x ) R, ad thus by Jese s iequality we get: (8) g (x ;... ; x ) Eg (X ;... ; X ) where EjX i j jx i j, for i ;... ;. Moreover, suose that we have: (9) g (x ;... ; x ) E h x i " i is a Beroulli sequece ad h : R! R is a give fuctio such that the right where ";... ; " side i (9) is well-defied ad belogs to ] 0 ; ] for all (x;... ; x ) R. The takig X i;xi N (0; jx i j ) to be mutually ideedet for i ;... ;, as well as ideedet of 4

";... ; ", by (8), Fubii s theorem ad the fact that E j X i;xi j j x i j for i ;... ;, we get: (0) g (x;... ; x ) E h X i;xi " i for (x;... ; x ) R. A usefuless of above iequality lies i the fact that we obviously have P X i;x i " i N (0 ; P jx ij ), so it might be a "good" chace to comute exactly the right side i (0), or at least to make a "good" estimate for it, ad i this way to obtai a estimate for the left side i (0), that is for the fuctio g defied by (9). For a alicatio of this techique see roof 3 of theorem 3. below. lassical Khitchie iequalities. Let f " i j i g be a Beroulli sequece, ad let 0 < ; q < ad. The the Khitchie costat K (; q) is defied to be the smallest umber satisfyig the followig iequality: () E j " i x i j K (; q) for all x ;... ; x () K(; q) su jx i j qq R. Let us ut (; q) ( 0 q )+, ad defie: K (; q) (;q) for all 0 < ; q <. The we have, see [6] (.0): (3) K (; q) (0 q )+, for all ; q > 0 ad all (4) K (; q) ad [K (; q)] are icreasig i (; ; q) (5) K (; q) ad log K (; q) are covex i, for all q > 0 ad all (6) K (; q) ( r 0 q )+ K (; r), for all ; q; r > 0 ad all (7) K(; q) 8 >< >: (8) K(; q) 0( + ) if ; q < if 0 < or 0 < q 0( + ), for all < q < < Note that K(; q) EjNj for ; q, where N N (0; ) is a stadard ormal radom variable. U. Haageru i [4], R.M.G. Youg i [3] ad S.B. Stechki i [4] have show that the costats K(; q) i (7) are ideed the best ossible i () for ; q. Let us i additio remid that Stirlig s formula states: (9)! e 0 e r, + < r < for. Recall that a cocave fuctio ' : R +! R + is subadditive, i.e. we have '(x + y) '(x) + '(y) for all x; y 0. Usig this fact ad Jese s iequality we may easily 5

obtai the followig two iequalities: (0) () jx i j jx i j ( 0 ), jx i j for all 0 < beig valid for all x;... ; x R ad all., jx i j for all < < 3. Basic results o the gauge sace L (P ) This sectio is devoted to the study of a iequality of Kahae-Khitchie s tye i the gauge sace (L (P ) ; k k ), where we recall that (x) e x 0 for x R. The first ste i that directio is doe i the ext theorem, see iequality () i theorem below. The result is well-kow ad our attetio is maily directed to its roof i order to fid the best ossible umerical costat uder cosideratio, as well as to obtai aroriate tools for closely related questios o some other Orlicz saces i the ext two sectios. Usig these results we reach this first ambitio i corollary 4 below. The, usig the classical symmetrizatio techique, we exted the give results to more geeral cases, see remark 7 ad corollary 9 below. Theorem. ( A Kahae-Khitchie iequality i the gauge sace ). Let f " i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k deote the gauge orm o (; F ; P ). The there exists a umerical costat > 0 such that the followig iequality is satisfied: () a i " i X ja i j for all a ;... ; a R ad all. Proof. Give a ;... ; a R for some, we deote S P a i" i ; T (S ) P ad A ja ij. The by the defiitio of the gauge orm k k, it is eough to show the followig iequality: () ex T A dp I order to illustrate various ways for rovig (), as well as to make ossible their comarisos, we shall reset four basically differet roofs of (): Proof. The first roof is based o a classical Kahae-Khitchie iequality for subormal sequeces of radom variables ad Stirlig s formula, ad we shall show that (), ad thus () also, holds with + e 56 :0.... For this ut 0 + e 56, ad let us cosider the Lalace trasform L S of S. Sice " ;... ; " are ideedet ad cosh (x) e x for x R, the we have: 6

(3) L S (t) E [ ex (t S ) ] Y cosh (ta i ) Y E [ ex (ta i " i ) ] Y t ja i j t A ex ex for all t R. Thus if we ut D S A, the for its Lalace trasform L D we get: L D (t) e t, for all t R, or i other words: (4) 8 S A j 9 is a stadard subormal sequece Now we aly a idea reseted i Kahae s book [9], see. 43. Sice S is symmetric, the so is (S ) k+ ad thus E (S ) k+ 0 for all k 0; ;..., ad we have: for all L S (t) t R. Hece by (3) we have: X k0 t k (k)! E(S ) k E(S ) k (k)! ex t k t A for all k ; ;..., ad all t > 0. I that way we may obtai the followig estimate for the Lalace trasform L T of T : L T (u) E [ ex (u T ) ] E [ ex (u (S ) ) ] X k0 u k k! E (S ) k + X k u k k! (k)! (t k ) (tk ex ) A k where t k > 0 are arbitrary umbers for k ; ;.... I order to fid a suitable choice for t k s let us ut (t k ) A k for k ; ;.... The we get: L T (u) + + X X k k u k k! (k)! Usig Stirlig s formula (.9), oe ca easily obtai: (k)! k k! (k) k ex (t k ) k ex ( k) (k)! k k! ( k ) (ua ) k ex ( k k ) k 0 k + 0 k for all k ; ;.... Thus uttig k k for k ; ;..., we get: 7

L T (u) + + + X k (ua ) k ex k 0 k + e 56 e 56 X k (ua ) k 0 ua 0 rovided that j ua j <. Puttig u 0 A we may coclude: 0 ex T dp L T 0 A + Thus () is satisfied ad roof is comlete. e 56 0 A 0 0 0 Proof. The secod roof is based o a classical Kahae-Khitchie iequality for tail robabilities of symmetric subormal radom variables ad o the real reresetatio of the P - itegral, ad we shall show that (), ad thus () also, holds with 6 :44.... Put for this 0 6 ad E T A ; ad let us cosider the Lalace trasform L E of E. By the real reresetatio of the P -itegral we have: (5) L E (u) E [ ex (u E ) ] ex (u E ) dp 0 + P f ex (u E ) > t g dt + 0 e t P 8 T A > t u 9 dt + 0 P 8 ex u T A e t P 8 jd j > for all u > 0, where D is give as above by D S A. By (4) we have: q t u > t 9 dt 9 dt E [ ex (v D ) ] e v for all v R, ad sice S is symmetric, hece by Markov s iequality we get: P f jd j t g P f D t g P f ex (v D ) ex (vt) g v ex (0vt) E [ ex (v D ) ] ex 0 vt for all t; v 0. Now it is easy to check that the fuctio S(v) v 0 vt attais its miimal value o R + at the oit v mi t, ad S(v mi ) 0t. I this way we may coclude: P f jd j t g e 0t 8

for all t > 0. Isertig this last iequality ito (5) we fid that: L E (u) + + 0 e (0 u ) t dt u 0 + 4u 0 u for all 0 < u <. Sice 0 < <, thus we may coclude: 0 0 ex dp L E 0 0 A T + 4 0 0 0 Thus () is satisfied ad roof is comlete. Proof 3. The third roof is based o a alicatio of the Gaussia-Jese s majorizatio techique which is described i sectio, ad we show that (), ad thus () also, holds with 43 :04.... Put 0 43 ad defie: g(x ;... ; x ) E h 0 x for (x ;... ; x ) R, where h(x) ex 0 A for x R. The evidetly g is a covex fuctio from R ito R +, ad oe may easily observe that g is sig-symmetric. Thus by (.0) we get: ] g(a ;... ; a ) E h E [ ex 0 A T E [ ex x i " i 0 A X i " i X i " i ] where X i N (0; ja i j ) for i ;... ; are ideedet, as well as ideedet of " ;... ; " : Sice " i X i X i for i ;... ;, the by ideedece of X ;... ; X ; " ;... ; " we have P X i" i N (0; A ) ad thus: 0 A X i " i where Y N (0; ). Hece we get: 0 E [ ex T 0 A A X i " i ] E [ ex (Y ) ] 9 0 ex Y x 0 x 0 ex h 0 0 dx x i dx

where 0. Sice R 0 e 0ax dx a ex 0 0 rovided that roof 3 is comlete. 0 A T dp for a > 0, the we may coclude: s 0 q 0 0 < 0, or i other words that 0 >. Thus () is satisfied ad Proof 4. The idea of the fourth roof is very simle. Namely, we shall exad the itegrad i () ito Taylor s series ad the we shall aly the classical Khitchie iequalities (.) with (.7). Thus there is a good reaso to believe that the give result will bevery dee, see [4]. Ideed, we shall see i corollary 4 below that the soo give costat 83 :63... is really the best ossible. Let us say that durig author s comutatios o the subject, this fact was cojectured by J. Hoffma-Jørgese. So, we shall show that (), ad thus () also, holds with 83. Let us cosider the left side i (). The we have: ex T dp E [ ex A X k0 A (S ) ] k! ( A E (S ) k ) k By the classical Kahae-Khitchie iequalities (.) with (.7) we fid: E j where K(k; ) k 0(k + ) (6) ex A T dp a i " i j k K(k; ) X ja i j k for k ; ;.... Sice 0( ), the we have: X k0 k0 k! ( A k 0(k + ) (A ) k ) k X k0 k k! 0(k + ) Now oe ca easily check that 0(k + ) (k 0 )!! for k, where (k 0 )!! k (k 0 ) (k 0 3)... 3. Sice evidetly j j <, thus we may coclude: X (k 0 )!! k (7) ex T dp A k k! 0 0 0

Thus () is satisfied ad roof 4 is comlete. I order to rove that 83 is the best ossible costat i iequality () i theorem, we shall first tur out the ext two auxiliary results which are also of iterest i themselves. Lemma. Let f X i j i g be a sequece of ideedet idetically distributed radom X i ad j g is a symmetric stadard subormal variables such that E (X) <, ad let (S 0 ES ), where S P VarX, for. Suose that f sequece, that is, is symmetric ad we have L (t) e t for all t R ad all. The for every >, the sequece 8 ex [ ] j 9 is uiformly itegrable. Proof. Let > be give, the it is eough to show that for some > we have: su E [ ex ] < Thus let us deote I() su 0 E [ ex ]. The by the real reresetatio of the P -itegral we have: E f ex [ ] g I() su su 0 + su P f ex [ P f j j > ] > t g dt log t g dt Sice by our assumtio f j g is a symmetric stadard subormal sequece, the by the estimate established i roof of theorem we have: I() + ex 0 log t dt + t Thus I() <, if >. Sice by our assumtio >, the we see that there exists ] ; [ for which I() <, ad this fact comletes the roof. dt Proositio 3. Let f X i j i g be a sequece of ideedet idetically distributed

radom variables defied o a robability sace ( ; F; P ) such that E (X) <, let (S 0 ES ) ; where S P X i ad VarX for, let N N (0; ) be a stadard ormal radom variable, ad let k k deote the gauge orm o (; F; P ). If f j g is a symmetric stadard subormal sequece, the () k k 0! k N k as!, where k N k 83. Proof. Let us ut k k for, ad k N k. The oe ca easily comute that 83. I order to rove () we shall first rove: () lim su! Ideed suose < lim su!. Thus + " < k for some 0 < <... ad some " > 0. Sice >, the by lemma the sequece f ex [ +" ] j g is uiformly itegrable, ad therefore by the cetral limit theorem ad the defiitio of the gauge orm k k we get: lim! lim k! N N ex [ ] dp > ex [ ] dp + " ex [ ] dp + " k ex [ ] dp + " Thus < lim su! leads to a cotradictio, ad hece () is roved. Secod we show: (3) lim if! Agai suose > lim if!. Thus 0 " > k for some < <... ad some " > 0 with 0 " >. Hece by lemma, the cetral limit theorem, ad the defiitio of the gauge orm k k we get: k lim if ex [ ] dp k! k k lim if ex [ ] dp k! 0 " lim ex [ ] dp! 0 " N ex [ ] dp > 0 "

Thus > lim if! leads to a cotradictio, ad cosequetly (3) is roved. But the () ad (3) evidetly comlete the roof. orollary 4. The best ossible umerical costat to 83. i iequality () i theorem is equal Proof. We have show i roof 4 of theorem that iequality () i theorem is satisfied with 83. Thus the best ossible costat i iequality () i theorem, say D, is less tha 83. To rove D 83, let us take a... a for. The () i theorem gets the followig form: () " i D beig valid for all. Ad i order to aly roositio 3 we should kow that f P ( ) " i j g is a symmetric stadard subormal sequece. But let us ote that this fact is established i roof of theorem, see its relatio (4). Thus lettig! i (), by roositio 3 we get 83 D, ad hece we may coclude that the best ossible costat D is ideed equal to 83. This fact comletes the roof. Problem 5. I order to set u a atural questio related to the result i roositio 3, let us recall some basic facts from [5] which are i the backgroud of our sectio o Orlicz fuctioals, orms ad saces. Let L deote the set of all icreasig left cotiuous fuctios ' from [0; [ ito [0; [ with '(0) lim t!0+ '(t) 0 ; ad let R deote the set of all decreasig right cotiuous fuctios R from [0; [ ito [0; ]. Let Q deote the set of all fuctios q from R ito [0; ] such that q(0) 0 ad q(r) su q(r ) wheever R lim if! R for R; R ; R... R. Recall that q Q is said to be: (i) subadditive, if q(r) q(r ) + q(r ) wheever R; R ; R R ad R R 8 R, i.e. 8'; ; L with ' 8, which meas '(x + y) (x) + (y) 8x; y 0 ; we have: 0 R(x) '(dx) 0 R (x) (dx) + 0 R (x) (dx) (ii) strogly subadditive, if q(r) q(r ) + q(r ) wheever R; R ; R R ad R R 8 R, i.e. R(x + y) R (x) + R (y) 8x; y 0 (iii) weakly subadditive, if q(r) q(r ) + q(r ) wheever R; R ; R R ad R R + R, i.e. R(x) R (x) + R (x) 8x 0 (iv) homogeeous, if q(r () ) q(r) for all > 0, wheever R R (v) o-degeerate, if q(r) 0 imlies R 0 (vi) moderated, if q(r () ) q(r) for all 0 <, wheever R R Recall that R () (t) R(t) for t 0, wheever R R. Ad a atural questio related to 3

the result i roositio 3, see also roositio 4.3 ad () i theorem 5. below, may be stated as follows: What are ( ecessary ad ) sufficiet coditios for R (t)! R(t) (8t S) to imly q(r )! q(r), where R; R; R... R ; q Q ad S [0; [ is a give subset? Sice coditio (i) lays a imortat role i cases whe fuctio orms ad fuctio saces are iduced by a measure sace, see [5] (.8), from that oit of view, it is ot a big restrictio to assume that q i our questio satisfies that coditio also. Let us say that the aswer to that geerally stated roblem will have iterestig cosequeces related to Orlicz saces, as well as Lauret saces, ad for more iformatios i that directio we refer the reader to [5]. Usig the classical symmetrizatio techique we shall ow exted the result from theorem i a aroriate way to a more geeral case. First we shall look at the sig-symmetric case i the ext theorem, ad the we shall ass to the geeral case i theorem 8 below. Theorem 6. Let f X i j i g be a sequece of ideedet a:s: bouded symmetric real valued radom variables defied o a robability sace (; F ; P ), let k k deote the gauge orm, ad let k k deote the usual su-orm o (; F ; P ). The for every the followig iequality is satisfied: () X X i X 83 k X i k Moreover, the umerical costat 83 is the best ossible i (). Proof. Give, we deote X (X;... ; X ), ad let ";... ; " be a Beroulli sequece such that " (";... ; " ) is ideedet of X (X;... ; X ). It is o restrictio to assume that X ad " are defied o the same robability sace (; F ; P ). Put 0 83 ad let us defie a fuctio f from R R ito R as follows: " # f (x;... ; x ; ;... ; ) ex x i i 0 jx i j for (x;... ; x ; ;... ; ) R R. Sice X (X;... ; X ) ad (";... ; " ) are ideedet, the by Fubii s theorem we may coclude: where E ex " 0 jx i j g(x) E f (x; ") E ex # o X i " i " 0 jx i j E g(x) # o x i " i for x (x;... ; x) R. By roof 4 of theorem oe directly fids that g(x) for all 4

x R, ad thus we may obtai Eg(X). Sice by our assumtios X (X;... ; X ) is sig-symmetric, the by the iequality just established we may coclude: () E But the we have: E ex " E ex Thus we may coclude: " E 0 ex " jx i j 0 X # o X i 0 k X i k ex " 0 jx i j X # o X i " i X # o X i jx i j X X i X 0 X # o X i k X i k ad the roof of () is comlete. The last statemet follows directly by corollary 4. These facts comlete the roof. Remark 7. Note that () i the roof of theorem 6 states: If X;... ; X are ideedet symmetric real valued radom variables, the the followig iequality is satisfied: X X i 83 jx i j Moreover, the umerical costat 83 is the best ossible ( the smallest ) with that roerty. Theorem 8. Let f X i j i g be a sequece of ideedet a:s: bouded real valued radom variables defied o a robability sace (; F ; P ), let k k deote the gauge orm, ad let k k deote the usual su-orm o (; F ; P ). The for every the followig iequality is satisfied: () X (X i 0 EX i ) X 33 k X i 0 EX i k 5

Proof. Give, we deote X (X;... ; X ), ad let Y (Y;... ; Y ) be a radom vector such that X ad Y are ideedet ad idetically distributed. It is o restrictio to assume that X ad Y are defied o the same robability sace (; F ; P ), as well as that EX i 0 for i ;... ;. Put S P X i ad T P Y i, the we have: () k S k k S 0 T k For this it is eough to show that E ex h S () i o where () k S 0 T k. Thus defie: f (s; t) ex h s 0 t () i for s; t R. The evidetly t 7! f (s; t) is a covex fuctio o R, for all s R, ad moreover by our assumtios we have T L (P ) with ET 0. Therefore by Fubii s theorem ad Jese s iequality we may easily obtai: Ef (S ; ET ) Ef (S ; T ) Sice ET 0, the by the defiitio of the Orlicz orm k k we get: E 8 ex h S () i 9 Ef (S ; ET ) Ef (S ; T ) h S 0 T E 8 ex () i 9 Thus () is roved. Now sice X (X;... ; X ) ad Y (Y;... ; Y ) are ideedet ad idetically distributed, ad X;... ; X are ideedet, the obviously X 0 Y (X 0 Y;... ; X 0 Y ) is sig-symmetric, ad thus by () ad theorem 6 we may coclude: k S k k S 0 T k 83 83 33 X h X Thus () is showed ad the roof is comlete. X (X i 0 Y i ) k X i 0 Y i k 0 k X i k + k Y i k i k X i k 6

orollary 9. Let f X i j i g be a sequece of ideedet a:s: bouded real valued radom variables defied o a robability sace (; F ; P ), let k k deote the gauge orm, ad let k k deote the usual su-orm o (; F ; P ) : The for every > 0 ad all we have: () X where () is give by: () (X i 0 EX i ) 8 < : () X 33 ; if 0 < k X i 0 EX i k 33 0 ; if < < : Moreover, if X; X;... are symmetric, the for every > 0 ad all we have: () X X i X D () where D () is give by: D () 8 < : k X i k 83 ; if 0 < 83 0 ; if < < : Proof. Iequality () follows by theorem 8, (.0) ad (.), ad iequality () follows by theorem 6, (.0) ad (.). 4. Basic results o the Orlicz sace L T (P ) Recall that the gauge orm k k from the revious sectio is the orm k k iduced by the Orlicz fuctioal, as defied i sectio, where (x) e x 0 for x R. osequetly, oe ca be iterested to fid out iequalities ivolvig the orms k k T ad k k7 which corresod to the iequalities reseted i corollary 3.9 ad remark 3.7. Note that the iequality i (.4) ca be used for this urose as well as the iequalities i (.3) ad (.5) for related oes, but we shall try to rove it directly usig the facts obtaied i revious aroaches. The startig oit should obviously be the iequality reseted i theorem 3. for which we have tured out four basically differet roofs, or i other words four differet techiques. However, ote that the Orlicz orm k k T is ot ecessarily homogeeous, but oe ca easily check that we have: () k c X k T j c j k X k T, for all j c j () k c X k T j c j k X k T, for all j c j where X is a give radom variable, see also (.). Ad accordig to the result i corollary 3.4 we may coclude that the fourth aroach i the roof of theorem gives the best estimate for the left side uder cosideratio, so it is reasoable to aly that techique i order to get as otimal 7

result as ossible. Ad this is really doe i the roof of the ext theorem. Theorem. Let f "i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio. The the followig iequality is satisfied: () jaij ai"i T for all a;... a R ad all, where the umerical costat s () 0 + 7 + 0 7 3 0 3 s 9 54 0 r 3 08 Moreover, that umerical costat is the best ossible i (). + 3 s 9 54 +, with is give by: Proof. Give a;... P ; a R for some, we deote as before P S a i"i, T (S) ad A ja ij. The by the defiitio of the Orlicz orm k k T, it is eough to show the followig iequality: (3) ex T dp + A For this ote that by (7) i roof 4 of theorem 3. we have: (4) ex T dp 0 0 x x A for all x >. Put (x) ( 0 x ) 0 ad (x) + x for x >. The oe ca easily check that > o ] ; [, < o ] ; [ ad () (). Moreover, give satisfies the followig equatio: 4 + 3 0 0 4 0 0. Thus by usig Ferrari s formulas, see [6], it is a matter of routie to check that is give by () above. Hece (3) follows by (4) ad the roof of () is comlete. To rove that the give umerical costat is the best ossible i (), we shall follow the idea reseted i the roof of corollary 3.4. So, let D deote the best ossible umerical costat i (). If we take a... a for ; the () gets the followig form: (5) " i T D Now oe eeds a result similar to that reseted i roositio 3.3 but with k k T istead of k k. Ad this fact will be established i roositio 3 below, so lettig! i (5), by r 3 08 8

roositio 3 below we may coclude that D. Sice the iequality D follows by (), this fact comletes the roof. Remark. We have see i the roof of theorem that the umerical costat give by () i theorem is a uique solutio of the equatio: x 4 + x 3 0 x 0 4x 0 0 for x >. By the well-kow criterio for ratioal solutios for algebraic equatios with ratioal coefficiets, see [6], each ratioal solutio of the above equatio belogs to the set f 6; 6 g. Ad oe ca easily check that 6, as well as 6, does ot satisfy the above equatio, ad thus we may coclude that the above equatio has o ratioal solutios at all. Therefore the umerical costat give by () i theorem is ot a ratioal umber. But oe ca easily check that we have: :53865763... ; as well as that the followig ratioal aroximatios are valid: () 60 39 < < 77 50 ad q 7 30 < < q 9 8 where 7750 0 0:003..., ad 98 0 0:004.... Thus iequality () i theorem is satisfied with 7750, as well as with 98. Note that 83, see corollary 3.4. 98 < Proositio 3. Let f X i j i g be a sequece of ideedet idetically distributed radom variables defied o a robability sace ( ; F ; P ) such that E (X ) <, let (S 0 ES ) ; where S P X i ad VarX for, let N N (0; ) be a stadard ormal radom variable, ad let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio. If f j g is a symmetric stadard subormal sequece, the () k k T 0! k N k T as!, where k N k T is equal to the umerical costat give by () i theorem. Proof. Let us ut k k T for, ad k N k T. The by the defiitio of the Orlicz orm k k T oe ca easily check that > ad 4 + 3 0 0 4 0 0 : Thus by the roof of theorem we see that is give by () i theorem. Ad i order to rove that! as!, oe ca reeat the roof of () ad (3) reseted i the roof of roositio 3.3, where the umerical costat should be relaced by the umerical costat + o the right laces. These facts easily comlete the roof. Theorem 4. Let f X i j i g be a sequece of ideedet symmetric real valued radom variables defied o a robability sace (; F ; P ), ad let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio. The for every the followig iequality is satisfied: () jx i j X i T 9

where is the umerical costat give by () i theorem. Moreover, that umerical costat is the best ossible i (). Proof. The roof of () is comletely the same as the roof of () i theorem 3.6, where the umerical costat should be relaced by the umerical costat + o the right laces, with 0, ad oe should use theorem istead of roof 4 of theorem 3. to obtai the desired iequality. Also ote that the last statemet follows directly by the last statemet i theorem. These facts easily comlete the roof. We shall cotiue our cosideratios by searchig for a aalogous iequality to that reseted i theorem 3. where the gauge orm k k should be relaced by the Orlicz orm k k T as defied i sectio. For this, let us cosider a Beroulli sequece f "i j i g which is defied o a robability sace (; F ; P ). Desite the fact that the Orlicz orm k k T o (; F ; P ) is ot homogeeous, accordig to (4.) ad iequality () i theorem, we may coclude that the followig iequality is satisfied: (3) X ai"i for all a;... ; a R ad all T jaij for which P ja ij, where is the umerical costat give by () i theorem. Ad oe ca ask is this iequality true i geeral, or i other words, does (3) hold for all a;... ; a R ad all? We shall ow show that the aswer to this questio is egative. For this, suose that (3) is satisfied for all a;... ; a R ad all. The takig a... a for, we get: "i T for all. Hece by the defiitio of the Orlicz orm k k T ex " i dp + we may coclude: for all. Lettig!, by lemma 3. ad the cetral limit theorem we easily obtai: ex! N dp lim ex " i dp where N N (0; ) is a stadard ormal radom variable. Sice this iequality obviously does ot hold for ay real umber, thus (3) does ot hold i that case. By the way, let us ote that usig the fact established i lemma 3. together with the classical Hartma-Witer law of iterated logarithm oe ca easily coclude: lim! ex " i dp 0

for ay > ad ay > 0. However, ote that by the geeral iequality give i (.4) ad theorem 3. we get: X h a i " T i 83 ja i j i h _ 83 ja i j i for all a;... ; a R ad all. Hece we ca deduce the followig iequality with a uique costat: (4) X a i " T h i 83 ja i j _ for all a;... ; a R ad all. I articular, we have: (5) X a i " T i 83 ja i j 4 ja i j 4 i for all a;... ; a R ad P for which ja ij. Now oe ca ask, first of all, is the umerical costat 83 the best ossible i (4)? I other words we may ask, is it the P best ossible i (5), sice by (3) we kow that this is ot true for ja ij? I order to get some relimiary iformatios i this directio, let D deote the best ossible umerical costat i (5), that is, let (5) be satisfied for all a;... ; a R ad all for which P ja ij ; if we relace 83 by D, ad let D be the smallest umber with that roerty. Puttig a... a for all, by roositio 3 ad (5) we may easily coclude that D ; where deotes the umerical costat give by () i theorem. Thus D 83 ; ad i the ext lemma we show that the first iequality is actually equality. These facts aswer our first questio. Lemma 5. The best ossible umerical costat i iequality (4) above is equal to by () i theorem. give Proof. Give a;... ; a R for some, we deote S P P a i" i, T (S ) ad A ja ij. Accordig to (3) ad coclusios which recede lemma 5, the oly fact which remais to rove is iequality (5) with the umerical costat give by () i theorem, istead of 83. For this we follow roof 4 of theorem 3., ad usig the same argumets as for relatios (6) ad (7) there, we may obtai: () ex T dp A X X k0 k0 X k0 k! A k 0(k + ) (k 0 )!! k k! k! ( k 0(k + ) (A ) k A ) k A k 0 A 0

sice evidetly by our assumtios j A j <, where is the umerical costat give by () i theorem. Now oe ca easily check that: x 3 + 4 x + 0 x 0 0 for all () 0 x, ad thus the followig iequality is satisfied: 0 x 0 + x for all 0 x. Ad by () ad () we get: ex A T dp + 4 A Hece by the defiitio of the Orlicz orm k k T we may coclude that k S k T 4 A, ad this fact comletes the roof. Note that a mai reaso for exoet 4 aears i iequalities (4) ad (5) above is comig from the geeral iequality give i (.4). Ad oe ca ask is this exoet ideed the best ossible i that case? I order to aswer this questio, let f "i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio. Give a ;... ; a R for some, we deote S P a i" i, T (S ) P ad A ja ij. Suose that A, let > be a give umber, ad let q be the cojugate exoet of, that is: + q. The the same argumet as i () i the roof of lemma 5, yields the followig estimate: (6) ex (A ) X k0 T dp k! k 0(k + ) (A ( ) k (A ) ) k X q k! (A ) k 0(k + ) k0 X (k 0 )!! (A ) q k k k! 0 (A ) q 0 k0 sice evidetly j (A ) q j <, where is the umerical costat give by () i theorem. Let us ow defie: q 3 su 8 q j 0 xq 0 + x 0q ; 8x [0; ] 9 Recall that by the defiitio of, see the roof of theorem, we have:

0 0 + Thus the iequality i defiitio of q 3 is always satisfied for x, as well as for x 0. Let 3 be the cojugate exoet of q 3, that is: 3 + q 3. The we evidetly have: for all 0 xq3 0 + x 0q3 + x 3 0 x, ad thus by (6) we may obtai: ex (A) 3 T dp + (A) 3 By the defiitio of the Orlicz orm k k T hece we may coclude: (7) k S kt (A ) 3 Note that by lemma 5 we have 3, or i other words it is o restrictio to assume that q 3. Moreover, it is easy to check that the iequality i defiitio of q 3, for q 4, is equivalet to the followig iequality: x 6 0 x 4 + 4 x3 0 x + 0 for all 0 x, which is obviously ot satisfied, sice the left side takes the value > 0 at x 0. Thus q 3 < 4, or i other words 3 > 43. osequetly, we may coclude: (8) 8 3 < 3 8 Ad the aim of the ext lemma is to establish that q 3 3, or i other words that 3 3. Lemma 6. The largest umber q satisfyig the followig iequality: () 0 xq 0 + x 0q for all 0 x, where is the umerical costat give by () i theorem, is equal to 3. Proof. By (8) above we kow that the largest umber q satisfyig (), say q 3, is strictly less tha 4. Furthermore, oe ca easily verify that for q 3, iequality () is equivalet to the followig easy to check iequality: x + 4 0 x 0 + 0 for all 0 x. Thus we may deduce that 3 q 3 < 4. Let us therefore take 0 < " <, 3

ad let us cosider iequality () for q 3 + ". The oe ca easily verify that iequality () is equivalet to the followig iequality: x +" 0 x +" + 4 x+" 0 x " + 0 for all 0 < x. But ote that for every " > 0, the left side above takes the value > 0 at x 0, ad therefore we may coclude that () is ot satisfied for ay q 3 + " with some " > 0. This fact shows that the largest q satisfyig iequality () is equal exactly to 3, ad the roof is comlete. Theorem 7. Let f "i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio. The the followig iequality is satisfied: () X ai"i h 3 i T jaij _ jaij for all a ;... ; a R ad all, where is the umerical costat give by () i theorem. Moreover, that umerical costat is the best ossible i (). Proof. Iequality () follows directly by (3), (7) ad lemma 6, ad the last statemet follows straight forward by lemma 5. Problem 8. Note that the exoet 3 i iequality () i theorem 7 is otimal i the framework of our best estimate established i roof 4 of theorem 3. which i tur lies o the best ossible costats i classical Khitchie iequalities, see (6) ad (7) i roof 4 of theorem 3.. However, ote that we did ot rove that it is ideed the best ossible, maily because of the fact that somethig "wrog" may hae for "small" A s for which our basic iequality emloyed above, see (6), ossibly does ot work i the best way. Hece, we ca set u a atural questio: What is the best ossible exoet which ca take the lace of 3 i iequality () i theorem 7? Note that by results established above this umber is greater or equal to 3, ad strictly less tha. As usual, by usig the classical symmetrizatio techique, we shall exted the result of theorem 7 to more geeral cases i the ext two theorems. Theorem 9. Let f X i j i g be a sequece of ideedet a:s: bouded symmetric real valued radom variables defied o a robability sace (; F ; P ), let k k T deote the Orlicz orm as defied i sectio, ad let k k deote the usual su-orm o (; F ; P ). The for every the followig iequality is satisfied: () where X X T i h k X i k _ k X i k 3 i is the umerical costat give by () i theorem. Moreover, that umerical costat 4

is the best ossible i (). Proof. The roof of () is comletely the same as the 0P roof of () i theorem 3.6, where the umerical costat should be relaced by + jx ij 0P _ jx ij 3 3 0P ; ad the by + k X i k 0P _ k X i k 3 3 P o the right laces, as well as exressios jx ij P ad jx ij 0P by jx ij 0P _ jx ij 3 ad 0P jx ij 0P _ jx ij 3, ad where theorem 7 should be used istead of theorem 3. o the right laces. Also ote that the last statemet follows directly by the last statemet give i theorem 7. These facts easily comlete the roof. Theorem 0. Let f X i j i g be a sequece of ideedet a:s: bouded real valued radom variables defied o a robability sace (; F ; P ), let k k T deote the Orlicz orm as defied i sectio, ad let k k deote the usual su-orm o (; F ; P ). The for every the followig iequality is satisfied: () X h (X i 0 EX i ) T k X i 0 EX i k k X i 0 EX i k where is the umerical costat give by () i theorem. 3 i Proof. The roof of () is comletely the same to the roof of () i theorem 3.8, where the umerical costat should be relaced by the umerical costat + () + k S 0T k T o the right laces. I this way we may coclude: k S k T k S 0 T k T for all, ad i the rest oe should use theorem 9 istead of theorem 3.6 to deduce the fial coclusio. These facts easily comlete the roof. orollary. Let f X i j i g be a sequece of ideedet a:s: bouded real valued radom variables defied o a robability sace (; F ; P ), let k k T deote the Orlicz orm o (; F ; P ) as defied i sectio, let k k deote the usual su-orm o (; F ; P ), ad let be the umerical costat give by () i theorem. The for every > 0 ad all we have: () X (X i 0 EX i ) where () is give by: T () _ h X k X i 0 EX i k X k X i 0 EX i k 3 i _ 5

() 8 < : ; if 0 < 0 ; if < < : Moreover, if X; X;... are symmetric, the for every > 0 ad all we have: () X X T i D () where D () is give by: D () 8 < : h k X i k _ ; if 0 < 0 ; if < < : k X i k. 3 i Proof. Iequality () follows by theorem 0, (.0) ad (.), ad iequality () follows by theorem 9, (.0) ad (.). 5. Basic results o the Orlicz sace L 7 (P ) The cosideratios i this sectio are devoted to the study of the questios reseted i the last two sectios, but ow for the Orlicz orm k k7 as defied i sectio, where (x) ex 0 for x R. Similarly to the revious aroach we shall essetially use the estimate established i roof 4 of theorem 3.. The first result i this directio may be stated as follows: Theorem. Let f "i j i g be a Beroulli sequece defied o a robability sace (; F ; P ), ad let k k 7 deote the Orlicz orm o (; F ; P ) as defied i sectio. The for every > the followig iequality is satisfied: () jaij ai"i 7 0 0 for all a ;... ; a R ad all. Moreover, the estimate give by () is the best ossible i the sese described i the roof below. (S) Proof. Give a ;... ; a R for some, we deote as before S P a i"i ; T P ad A ja ij. Accordig to (7) i roof 4 of theorem 3. we may coclude: ex jaij T A ai"i dp T7 0 0 0 0 6

0 0 for all ad all >. Thus () is satisfied ad the first art of the roof is comlete. Ad for the last statemet, let us take a... a for, the () gets the followig form: " i 7 for all ad all > () as! P, for all > S " i, we have: (3) " i 7 ex 0 0. Ad we shall ow show that: 0! 0 0. For this, it is eough to show that for give > h S i dp 0! 0, with as!. I order to rove (3), by the cetral limit theorem it is eough to verify that the sequece f ex ( ) ] j g is uiformly itegrable, where S for. Ad by lemma 3. we may otice that this fact is satisfied, if f j g is a symmetric stadard subormal sequece. But this last fact is established i roof of theorem 3., see its relatio (4). Thus (3) follows, ad therefore () is satisfied also. Hece we may coclude that the estimate give by () is ideed the best ossible, i the sese that give > ad " > 0 with 0 " > ; we ca fid " ad a ;... ; a " R such that the followig two iequalities are satisfied: ( 0 ") " X " X ja i j ja i j X " X " a i " i 7 a i " i 7 > > 0 + " ( + ") 0 Note that the fuctio 7! 0 is decreasig o ] ; [. These facts comlete the roof. 0 0 Theorem. Let f X i j i g be a sequece of ideedet symmetric real valued radom variables defied o a robability sace (; F ; P ), ad let k k 7 deote the Orlicz orm o (; F ; P ) as defied i sectio. The for every > ad all the followig iequality is satisfied: () jx i j X i 7 0 0 7

Moreover, the estimate give by () is the best ossible i the sese described i the roof of theorem. Proof. Give, we deote X (X ;... ; X ), ad let " ;... ; " be a Beroulli sequece such that " (" ;... ; " ) is ideedet of X (X ;... ; X ). It is o restrictio to assume that X ad " are defied o the same robability sace (; F ; P ). Exactly as i the roof of theorem 3.6 we may coclude: for all > 8 h E ex jx i j X i 9 X i " i, where the fuctio g is give by: 8 h g(x; ) E ex for x (x ;... ; x ) R ad > g(x; ) jx i j Eg(X; ) X i 9 x i " i. By () i theorem we have: 0 for all x R ad all >. Sice by our assumtios X (X ;... ; X ) is sigsymmetric, the by the iequality above we may coclude: 8 h E ex E 8 ex h jx i j jx i j X i 9 X i X i 9 X i " i 0 for all >. Hece () follows directly, ad the first art of the roof is comlete. The last statemet follows straight forward by the last statemet i theorem. These facts comlete the roof. Remark 3. By (4.6) we may easily deduce the followig "dual" estimate which exteds the result of theorem : () a X i " i 0 ja i j 7 ja i j q 0 0 for all a;... a R, all, ad all > 0 for which 0 P ja ij < 0 8 0 ;

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