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: ?
[Math] weakly convergent subsequence implies strongly convergent
analysisconvergence-divergencefunctional-analysisweak-convergence
Related Solutions
No. For example, in $\ell_1$ ($=\ell_1(\Bbb N)$) every weakly convergent sequence is norm convergent. Thus the identity operator, $I$, on $\ell_1$ maps weakly convergent sequences to norm convergent sequences. But $I$ is not a compact operator.
Incidentally, operators that map weakly convergent sequences to norm convergent sequences are called completely continuous. If $X$ is a reflexive space, then every completely continuous operator on $X$ to a Banach space $Y$ is compact (as a result that the closed unit ball of $X$ is weakly compact for reflexive $X$).
I think that the argument in the linked post should filter through fine (in fact, it seems constructed for general reflexive spaces).
A reflexive Banach space is separable if and only if its dual is. (The standard result is that $X^*$ separable implies $X$ separable.) Fix a countable dense subsequence $\{\phi_1,\phi_2,\ldots\}\subseteq X^*$, and let $\{a_1,a_2,\ldots\}\subseteq A$ be any sequence. Let them all be bounded by $M$ in norm, since $A$ was bounded. Consider $\{\phi_1(a_1),\phi_2(a_2),\ldots\}\subseteq \mathbb{K}$. This is a bounded sequence in a locally compact space (bounded because all of them have norm less than or equal to $\|\phi_1\|\cdot M$), so it has a convergent subsequence $\phi_1(a_{11}),\phi_2(a_{12}),\ldots$. Now for each $k+1$, similarly extract a convergent subsequence from $\phi_{k+1}(a_{1k}),\phi_{k_1}(a_{2k}),\ldots$. Do a standard diagonal argument and $a_{11},a_{22},\ldots$ is a sequence for which each functional $\phi_i$ converges. Moreover, because the $\phi_i$ were dense, for each $\phi\in X^*$, $\phi(a_{11}),\phi(a_{22}),\ldots$ is convergent. Define the functional $a$ on $X^*$ by $a(\phi)=\lim\limits_{n\to\infty} \phi(a_{nn})$. This is continuous (needs checking) and therefore by reflexivity is represented by some element $x\in X$. Then the sequence $\{a_{nn}\}\xrightarrow{w} x$, and because $A$ was $w$-closed, $x$ is in $A$.
For general spaces, fix your sequence $\{a_1,\ldots\}$ and let $Y$ be its closed linear span. Then $Y$ is separable, and it is a closed subspace of a reflexive space, so itself reflexive (A closed subspace of a reflexive Banach space is reflexive) and the dual is separable. But now we're in the old case, so we extract $x$ with $y^*(a_n)\rightarrow y^*(x)$ for each $y^*\in Y^*$. Now this convergence holds for any linear functional in $X^*$, because the restriction to $Y$ is a linear functional on $Y$.
Best Answer
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.