$ \lim\limits_{n \to \infty} \| x_{n}\| = \| x \|$ if $\left(x_{n}\right)$ weakly convergent

Question

I want to show that for a banach space $X, U \subset X$ non-empty and compact and $\left(x_{n}\right) \subset U$ a weakly convergent sequence with limit $x$, it converges in norm to $\mathrm{x}$

We know that $\quad f\left(x_{n}\right) \rightarrow f(x) \: \: \forall f \in X’$.
So

$$ \| x \| = sup_{\| f \| = 1} \| f(x) \| \\ = sup_{\| f \| = 1} \lim\limits_{n \to \infty} \| f(x_{n}) \| \\ = \lim\limits_{n \to \infty}sup_{\| f \| = 1} \| f(x_{n}) \| \\ = \lim\limits_{n \to \infty} \| x_{n}\|$$

However, is it valid that I can interchange the limit and supremum?

Answer 1

Indeed it is true. However, I would argue differently. Pick some subsequence $(x_{n_k})_{k}$. By compactness it admits a convergent subsequence. However, as the strong and the weak limit coincide, it comverges strongly to $x$. This implies that every subsequence of $(x_n)_n$ admits a subsequence converging strongly to $x$ and thus the whole sequence converges strongly to $x$.

Answer 2

Considering the two topological spaces $(U,\|{\cdot}\|)$ and $(U,w)$, we have that the former is compact and the latter is Hausdorff. Since we know that the identity map $$ (U,\|{\cdot}\|) \to (U,w) $$ is continuous, it must therefore be a homeomorphism by a well know Theorem of General Topology.

Thus the weak- and norm-convergent sequences are the same.