Weak Convergence in Uniformly Convex Spaces,

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1 Weak Convergence in Uniformly Convex Spaces, Shiztlo by KAKUTANI, Osaka. S. B a n a e h and S. Saks proved(1) that the space Lp(p > 1) has the following property : (P) every weakly convergent sequence of points contains a subsequence whose arithmetic mean converges strongly to the same limit. As is shown by J. Schreier(3), this theorem does not hold in the space C. What then is the reason why the same theorem holds in L"(p>l), while it fails in C? As an answer to this question, we shall prove in the present paper the Theorem. If a Banach space is uniformly convex, then it has the property (P). We must first explain the meaning of uniform convexity. This notion was introduced by J. A. Clarkson(3) in his paper on the theory of integration of a function whose range lies in a Banaeh space. Definition. A B anach space is called to be uniformly convex, if it has the following property : (1) S. Banach and S. Saks : Sur la convergence forte dans lee champs Lip, Studla Math., 2 (1930). ($) J. Schreier : EinGegenbeispiel zur Theorie der schwachen Konvergenz, Studia Math., 2 (1930). It is to be mentioned that the following theorem holds true : Theorem (Z. Zalcwasser). If {fi(x)} is a sequence of continuous functions which converges weakly to a continuous function f(x) (that is, if { fn(x)} is a sequence of uniformly bounded continuous functions which converges pointwise. to a continuous function f(x) in asv b), then there is a schema of constants At(i=I,2,...n; n=1,2,...) such that and for which converges strongly (uniformly) to ft.r). This is also a consequence of a general theorem of S. Mazur. Z. Zalcwasser : Sur une proprietd du champ des fonetions continues, Studia Math., 2 (1930). S. Mazur : 'fiber konvexe Mengen in linearen normierten Riiumen, Studia Math., 4 (1933). (3) J A. C l a rks o n : Uniformly convex spaces, Trans. Amer. Math. Soc., 40 (1936).

2 WEAK CONVERGENCE IN UNIFORMLY CONVEX SPACES. 189 To each a>0 there corresponds a S(c)>0 such that the conditions (U) imply Since, as is shown by J. A. Clarkson(3), the space Lp(p>1) is uniformly convex, the result of S. B an ac h and S. Sak s is a direct consequence of our theorem; and the counter-example of J. S c h- reier(2) has its reason in the fact that the space C is not uniformly convex. That the converse of the theorem is not true (that is, that there is a Banach space which has the property (P) without being uniformly convex), may be easily seen from the fact that the same Banach space may become uniformly convex or not depending on its norm (-wihout changing the topology 1)(4). Before entering into the proof of the theorem, let us replace the condition (U) by the equivalont one : To each a>0 there corresponds a 81(e)>O such that the condition (U') implies Putting x = y =1 and V(s)=8(s) in (U1) we have (U), so, that (U') implies (U). To prove the converse we proceed as follows: (0 Since x and y are symmetric in (U'), we may assume (ii) Since II ax I I=Ia I x l I for any real number a and since the relations in (U') are homogeneous in x and y, we may assume without the loss of generality that y x I I =1(5). By (i) and (ii) we have only to prove that (U) implies (U11): To each r > 0 there corresponds a 8"(e) > 0 such that the conditions imply (U") (4) Indeed, the (x, y)-plane is not uniformly convex if we define I I (x, y) I I =max (I x 1, I y 1), although it is uniformly convex in the Euclidean norm : I I (x, y) I I (6) If x >0, replace x and y, by x/iixii and y x respectively; If x =0 then 11y11=0 and this case is trivial.

3 190 SHIZUO KAKUTANI: Proof of (U)(U") : Let (<e) be a sufficiently small positive number, which we shall determine later, and let x and y be such that IIxII =1, 1 y 1and IIx-- y l I e. Putting z y/ x the conditions imply (by (U)) and consequently Since the right hand side is <1 for a sufficiently small q>o, say =,q0 (e being fixed 1), we have for any x and y with x =, 1 y -:1-io and IIx-yII?e. Since in case I I y I I s 1-71o the inequality is evident, we have for any x and y with IIxII =1, I i y I I s_ 1 and IIx-y e the relation : S"(E) being a positive number, independent of x and y. Thus the proof that (U) implies (U:') (and also (U')) is completed. Now we shall proceed to the Proof of the theorem : Let { xyj }(n = 1, 2,...) be a sequence of points of a uniformly convex space, which converges weakly to x. We shall prove that there is a subsequence { x.j (k =1, 2,... ) of {xn} such that Without the loss of generality we may assume that x=0. Since the sequence 1x,, is weakly convergent, there is a constant M such that

4 WEAK CONVERGENCE IN UNIFORMLY CONVEX SPACES. 191 We shall show first that there is a subse- 1<m1<m2<... <m2n-1<m2n<...) of IIxnII s M for n=1,,2, quence kxmj (n=1, 2,...; { x,j such that for n =1, 2,..., where is a constant indepent of the given sequence (that is, depending only on the property of the space in question). Put m,=2 and consider the following two cases : (1) If I X2 I s 2, put m2= 3 ; then we have (ii) If x211> 2, take as ms the least integer n(>2) such that I x2--xn I > 2. Thexistence of such an n is a consequence of the fact, that if x2-xn 2 for all n>2, then IIx2II=IIx:--01 Jim I x:--xq Il s_ 2, which is contradictory to our assumption that I X21 I > 5-. For these m, and m: we have and therefore, by the property (U'), After ms is determined, put m3=m2+ 1 and determine m4 precisely in the same manner as m2 is determined from mi; and then put m, = mu + 1, and so on. It will be almost obvious, that, proceeding in this way, we shall have the desired subsequence {xmn}(n =1, 2,... ). For the sake of simplicity we put (i)1 Clearly.the sequence {xn} (n=1, 2,....) has the following properties:

5 192 SIIIZU0 KAKUTANI (ii)1 (iii)1 In general, after the sequence { x(p) }, (n = 1, 2,...) is already defined so as to satisfy the conditions : (i) (ii) (iii) we define a sequence {x(,p+1)}, (n =1, 2,...) so as to satisfy the conditions : (i)n+1 (ii)p+1 This may be done precisely in the same manner as {x(n)} is defined from Ix.}. Clearly i s satisfied. the condition : (iii) { 0'+1)} (= ) converges weakly to 0. Repeating the same method, we shall have a sequence of sequences f {.,v(p) }, (n = 1, 2,... )11 (p =1, 2,... ), which satisfies the conditions (i),,, (ii), and (iii),, for p=1, 2,... Now consider the sequence { x(jp) }, (p =1, 2,...). As is easily seen from the construction of {xnp)}, each xjp) is of the form : with 1<l;l)<l21)<l12)<... <l2)<h3)<... <ip-1)gl1(p)<... <l2p)<.... moreover, for any q< p and is an element of {x;;)} and hence of norm not greater than &.11r. If we plat nl-==i and n2"+:_1 =:l1(p) (j=1, 2,.,, 2P; p= 1, 2,... ) (that is, n1=.1, 'A2=1(1'), n3=l('), nl ='h2),..., n7== l ), ns=li3),..., n p_, l p~l', n,v=11''),..., nz~+l_1=12v,.. ), then nk is, uniquely

6 WEAK CONVERGENCE IN UNIFORMLY CONVEX SPACES. 193 determined for k=1. 2,..., and the subsequence { xnk } (k= ) of (x,,} (%=1p2,...) thus obtained is the required one. In order to prove that let us consider for r-2's k <(r+1) 21 the inequality : Since is an element of j x(101 U=2,3,...,r; for some p and 2 l lljq =nl.2q-1 1), we have and therefore Since q is arbitrary and since 9<1 is fixed, the proof is hereby completed. Mathematical Institute, Osaka Imperial University, Osaka, Japan. (Received July 20, 1933.)