TITLE: Bounded sequence in a Hilbert such that all subsequences that weakly converge do so to the same limit
QUESTION [5 upvotes]: Let $H$ be a Hilbert space, $a \in H$ and $(x_n)_n$ a bounded sequence in $H$ such that every subsequence of $H$ that converges weakly converges to $a$. How do you prove that $(x_n)_n$ converges weakly to $a$? The only thing I know is that there is indeed at least one subsequence of $(x_n)_n$ that converges weakly.

REPLY [4 votes]: This works in general in reflexive Banach spaces. Let $y^*\in X^*$ be a bounded functional. We have that  $y^{*}(x_n)$ is a bounded sequence, so if it is not convergent  it must have at least two distinct accumulation points. Say $(y_n)$ and $(z_n)$ are subsequences of $(x_n)$ such that $y^{*}(y_n)\to \alpha$ $y^{*}(z_n)\to \beta$. Since $(y_n)$ and $(z_n)$ are bounded and $X$ is reflexive (so bounded sets are relatively weakly compact), each must contain a $w$-convergent subsequence. However any $w$-convergent subsequence converges to $a$, hence $\alpha=\beta=y^*(a)$. This shows that $y^{*}(x_n)$ is convergent, and it must converge to $y^*(a)$. Since $y^*$ was arbitrary, this shows $(x_n)$ is $w$-convergent.