# weakly convergent subsequence implies strongly convergent

Statement: Let $X$ be a Banach space If $x_n \rightarrow x$ weakly and every subsequence of $\{x_n\}$ has a strongly convergent subsequence, then $x_n\rightarrow x$ strongly in $X$

Attempt: ?

• Hint: If you can show that weak limit is unique, then the statement become: Every subsequence of $x_n$ has a subsequence converges strongly to $x$. – user99914 Apr 6 '14 at 4:55

Suppose $\lim_{n \to \infty}x_n\neq x$. Then there is an $\epsilon \gt0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_k}-x\|\ge\epsilon$ for all $k\in\mathbb{N}$. Use this to derive a contradiction.