Let E be a Banach space, $K \subset E$ a compact subset in the strong topology and $(x_n)_{n \geq 1} \subset K$, $x_n \rightharpoonup x$ weakly in $\sigma (E, E^*)$.
If there is a subseqence $(x_{n_k})_k$ of $(x_n)_n$ such that $$x_{n_k} \to x \;\; \text{strongly and} \;\; x \in K,$$ can we conclude that $$x_n \to x?$$
Thank you!
On
In your situation, you have $x_n \to x$. This can be proved by contradiction. Suppose $(x_n)$ does not converge to $x$. Then, there is $\varepsilon > 0$ and a subsequence $(x_{n_k})$ with $\|x_{n_k} - x\| \ge \varepsilon$ for all $k$. But this subsequence converges weakly to $x$ and, hence, is bounded. By compacity of $K$, a subsequence converges strongly, which is a contradiction.
In fact, we can directly conclude that $x_n \to x$ in the norm topology.
Compact metric spaces are sequentially compact, so every sequence in $K$ has a convergent (in the norm topology) subsequence. Thus every subsequence $(x_{n_k})$ of $(x_n)$ has a subsequence $\bigl(x_{n_{k_m}}\bigr)$ that converges to some $y \in K$. But that subsequence also converges weakly to $x$, since $x_n \rightharpoonup x$. Hence $y = x$.
Thus every subsequence of $(x_n)$ has a further subsequence converging to $x$, and that means the full sequence $(x_n)$ converges to $x$. [If it didn't, there'd be an $\varepsilon > 0$ such that there are infinitely many $n$ with $\lVert x_n -x \rVert \geqslant \varepsilon$. That gives you a subsequence $(x_{n_k})$ with $\lVert x_{n_k} - x\rVert \geqslant \varepsilon$, and hence no subsequence of that can converge to $x$.]