A characterization of the Radon-Nikodym property

Pub . Mat . UAB Vol . 29 Ns 1 Abril 1985 ACHARACTERIZATION OF THE RADON-NIKODYM PROPERTY OscarBlasco de la Cruz §l . INTRODUCTION . The aim of this paper is to give a new characterization of the Radon-Nikodym property in terms of martingales in X-valued Orliczspaces . Let X be a Banachspace andput Z : , the Lebesgue measurable sets in 00,11 . It is well known (see [l]) : (1 .1) . X has the Radon-Nikodym propertywithrespect to [0,1] if and only if every boundeduniformly integrable martingale in LX[0,1] , (f n ,B n ) where o(UB n ) _  , is convergent in  LX[0, 1] . We are interested in a generalizationof this fact . In this paper we shall prove the following Theorem (1 .2) . Let ~ be a Young function with the ¿S 2 -condition .  X has the Radon-Nikodym property if and only if every bounded martingale in  LX  ,  (fn ,B n )  where  o(U Bn )  _  ,  is  convergent in LX . The definitions and the main results relating to Xvalued martingales and Orliczspaces may be foi :nd in [1] and [2] respectively . We are going to denote by 0 a Young function, a .nd LX = {f : [0,11  -,>-X~ strongly rúcasiurable withrespect Lebesgue measure s .t . p(f,O) = f  0(11f(x)l1)dx < -} . 0 Let ~ be the complementary Youngfunction of ~ . We 1 ~~f(x)~~ ~g(x)1dx with {f :[0,11 -~ X strongly shallwrite  1 Ifl l o =  sup { g e Lyá , p(g,~) 1 1} and L o= measurablewith  U filo < W} It is well known that Lo is a vector space and llfli~ is a normon it . Besides, LX = LX if and only if 0 verifies the lá 2-  _ condition .  It is easy to prove that the convergente and the boundedness in L~ and L. are equivalentusing the X X following fact : (1 .3)  Suppose  ~  verifies 0 2 -condition, i .e . there exists  K > 0  and  T 'k 0  such  that  QS(2t)  4 K« t)  for all t ~! T then  f  Tm+2  See (2] 7 page 158 .,  for a proof . 2 §2 . PREVIOUS LEMMAS . Lemnia 1 . If (f n , n e N) is a bounded sequence in LX , thcss (f n , n e N) is a boundeduniformly integrablesequence in L 1 X If thereexists m, helongs to (N . witi  p(f,0) L' 1/K m Proof . For a Young function we have (2 .1) ~tt - :> °°  as t +  , and by (2 .1) we obtain lif n 11 1 ¢ p(f n ,~) + A , where . A is a eonstant . LX We have only  to show that  j E  Ilf n(x) Ildx  0  as m(E) -> 0 . Given E > 0 , by (2 .1) , thereexists T > 0 such that (2 .2)  ltt  > sc  for  t >T  where  sup  p(f n m  4 c . n Let d = E/2T . If m(E) <ó and denoting An ={x : j I f n (x) 11  :1 T} (1 E  and Bn = {x : jif n (x)11 > T} (1 E we obtain Lemma 2 . Proof : E 11f n (x)11 = 11f n (x)~~dx + A n + J lif (x)jIdx  < B  n  (2 .2) TM(E) + E 2c 0 ~(11f n (x)11)dx<E If 0 verifies the ~ 2 -condition then the simplefunc= : tions are dense in LX . Given t e LX, sünce f is strOngly measurable,there exists a sequence  (f n  , n e N)  of co untably valued functions such that (2 .3)  ~fn(x)-f(x)  <  ñ  for almost al l  x e  [0,1] andfor all n e N . Suppose fn = - 57 - x n m XE where m=o n,m x n m e X and  XE  are the characteristic functions of '  n,m disjoint measurables sets . Since 2JIf n (x)jj < 211f(x)II, ñ a .e . and 0 is a convexfunction .2 we have 2f n e LX . Therefore,there is a number pn e N such that Lemma 3 (2 .4)  Jl, .f  E  _  0(211f n (x) 11)dx < m=pn n,m We consider the simple function By (2 .3) and (2 .4) 1 n pn gn =  x  . n,m X m=o E n,m J o ~1 If(x) - gn (x)j - j)dx ¿ 2  Jo 0(21 j f(x) - f n (x) 1 jdx 1 21 (2~~f n (x) - g n (x)dx  2 0(ñ)  + ñ 0 Since «t) -> 0 as t 0+ the proof is finished . Let (BT , T e 1) be a family of sub-a-fields of  . Suppose ~ .with á2 -condition . If fn convergs to f in LX then E(f n /B T ) convergs to E(f/B T ) uniformly in BT , where E( ./B Z ) denotes the conditional expectation relative to Bz . Proof It may be proved , with-a slight, modification in the argument in [lj,pa :g° 122 that if B is a sub-a-field of then p(E(G/B)M  p(g,0) for all g e Lo . Now, given e > 0 , let m o be a number such that max  (~ m +2  ,  m )  < c  where  K,T  are the  constants  in 2 o  Ko the 0 2 -condition . Sinc .e  1 1 fn - f 1 1 ~ (n -) then p(f n - f,O) -> 0 (n -> -) so there is a number n o such that if n > n o we have p(fn-f,O) < m . This implies,by Ko (1 .3) and the first result in the proof,that 1 JE(f n - f/B T )110 < e  for  n ~ n o and it is true for all Te I . 93 . PROOF OF THE THEOREM (1 .2) . Suppose X has the Radon-Nikodym property and let (f n ,B n ) be a bounded martingale in LX with a((JB n ) _ By lemma 1 and (1 .1), there is a function f in LX such that f n -~ f in LX and f n = E(f/B n ) as it may be seen in [1] . Since the convergenteof martingales in LX implies the convergentealmost everywhere, we obtain,using the continuity of 0 that « llf n (x)il) -~ « lif(x)l1) a .e . and by Fatou's Lemma . 1« 11f(x)11)dx  l~  lim  inf  j ¢(11f n (x)11 dx  ~=  M Therefore f belongs to LX J which coincides with LX . We shallprove that f n --~ f in LX . From Lemma 2, we see that given e > 0 , there exísts a number m o and a sequence of simple functions such that (3 .1)  lis m - fil a <  e/2  for  m  >  mo and usingLemma 3 with Bn , there is m 1 in !N such that (3 .2)  11E(f-sm/BnM O .  <e/2  for  m ~ m l and for all n Since  o(U Bn )  = T-  , we can take the functions  s m on measurable sets from lJ B n . If s m = Em i G Bn for i = 1, . . .,p and in this case E(s m /B n ) = s m for n = n o . Therefore if n !> n o , by (3 .1) and (3 .2) To prove the converse we are going to use the characterization of the Radon-Nikodym property in termsof operators : For every T :L 1 [0,1]  ~p X there exists a function f in Bn the o-fixed generated by the dyadicintervalsof length 2 n  ,  i .e .  Bn =  o(l n  i  '  i  =  0, . . .,2n-1)  where Let m be a fixed number such that m > max (mo ,m l ) . let  n o be 'a number such that jif - f n110 `  lIf - s m 11 0  +  lis m-fnil~  = = jjf-s m 11 0 + 11E(sm -f/B n )jj < e/2 + c/2 = e 1 such that T(q) _  9P(x)f(x)dx for ~o e L*1 [0, 1]  (see [j, page (i3) . Let  T :L~ [0,1] -,oX be a boundedo , perator . We consider ii±l In,i = `2n . 2n) 2 n _1 Let  fn =  r  2n T (XI  _)XI  .  It is easy to prove  that i=o  n,i n,i E(f n+1 /B n )  = f n and obviously  o(U Bn ) = E-  . Since n 2ni 1If n 11  =  Z1  2  1 T(XI  x,  ,  it is clear that i=o  n,i n,i 11f n (x)jj 11 JITII for all x e [0,1] . Then p(f n ,¢) 11 0(11TI1)  for all n, and we can find a function f in LX such that f n -~ fin LX . This is equivalent to « lif n ID) --  « jifil)  in  L 1 and therefore there is a subsequence  ~ (l i f n (x) 1 J)  ->  (  1 f (x) 11)  ái :e .  Hencé . k f(x)~~)  ¡TI 1)  a .e .  and f belongs to To conclude the proof, we must only provethat . 1 (3 .3)  T (s)  _~  s(x)f(x)dx  for all simple function on 0 U B n -measurable sets .  -First, .we,Shall .prove that (3 .4)  f n =  E(f/B n ) . If E is a B n -measurable set, 1 f n (x)dx =í E f n+k (x)dx for k~,l lE and then it is sufficent to prove that f (x)dx -~~ f(x)dx as n-*oo .It is clear from the Holder S E n  '  E inequality ir 11 fn  - (x)  -  f(x) 11  dx~  11f n -  fl1XEU 'From  (3 .4) ~ J I  f(x)dx =~I  f n (x)dx = T(XIn,i) n,i  n,i and by linearity we obtain (3 .3) and finish the proof . SPAIN REFERENCES [1] J . DIESTEL and J .J . UHL (1977) . Vector Measures . Amer . Math . Soc . Mathematica l Surveys, 15 . [2]  A .  KUFNER et al .  (1977) .  Function spaces .  Noo rdhoff International Publishing . Rebut ee . 9 de 'novembne dei 1984 Departamento de Teoría de Funciones Facultad de Ciencias Zaragoza