Weakly Cauchy

**rishirich** · January 18th 2011, 07:17 AM

Let X be a reflexive normed linear space.Then show that X is weakly complete..i'e;Every Cauchy sequence is weakly convergent..

**Jose27** · January 18th 2011, 08:30 AM

Pick a weak Cauchy sequence $(x_n)$ and call $(\hat{x}_n)$ its corresponding sequence in the double dual then for any $l\in X^*$ we have, since Cauchy sequences are bounded, $|\hat{x}_n(l)|=|l(x_n)|\leq C_l$ and by the uniform boundednes principle we get that $(x_n)$ is a bounded sequence, by reflexivity this gives a weakly convergent subsequence, call it $(x_k)$ and a limit, call it $x$ . Now just use the estimate $|l(x-x_n)|-|l(x_k-x_n)|\leq |l(x-x_k)|<\varepsilon$ and if $k,n$ are large enough then the result follows.

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