
#1
Nov1305, 08:41 AM

P: 307

hello folks,
I've got a question about weak convergence. I'm sure I'm missing something but can't see what it is (<standard "I'm dumb apology") The problem concerns little l 1 and little l infinity (which is dual to little l 1) To make notation easier I'm going to denote these spaces by L1 and Linf. If a sequence in L1 weakly converges then it strongly converges. So we take a sequence in L1 {x_n} where each x_n is a inftuple (a_1,...) of real numbers such that Suma_k is bounded. Take an element of Linf, call it g. We given that limg(x_n)=g(x) for an x in L1. Now we need to show that limx_n=x. So my dilemma is that I essentially keep proving that strong convergence implies weak convergence. I keep trying to work expressions like g(x_m)g(x_n)<e and g(x)g(x_n)<e into the analogs for the x_n (using of course the appropriate norm). As per usual i can't get my inequalities going in the right direction. I need only a tiny push, I'm sure. So advise sparingly. As always, your help is much appreciated, Kevin 



#2
Nov2509, 02:38 AM

P: 13

By Schur's lemma every weakly Cauchy sequence converges. So your answer lies in the proof of Schur's lemma.



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