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