Bounded sequence in a Hilbert such that all subsequences that weakly converge do so to the same limit

Question

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.

Answer 1

If $(x_n)_n$ does not converge weakly to $a$, then there is a subsequence of $(x_n)_n$ with the property that for some $v \in H$, there is some $\varepsilon > 0$ such that $|\langle x_n ,v \rangle - \langle a , v \rangle| \geq \varepsilon$ for every $n$ indexing the subsequence. But that subsequence is bounded and therefore admits a weakly convergent subsubsequence that weakly converges to $a$, which is a contradiction to the fact that the subsequence stayed away from $a$.

Answer 2

Note that every subsequence of $(x_n)_{n\in \mathbb{N}}$ is also a bounded sequence and therefore has a weakly convergent subsequence which converges to $a$ by hypothesis. As every subsequence of $(x_n)_{n\in \mathbb{N}}$ has a further subsequence that converges to $a$, $x_n\rightharpoonup a$.

Answer 3

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.