# A weakly convergent sequence in a compact set, is strongly convegnet

Let $$E$$ be a Banach space, and $$K \subset E$$, compact set for the strong topology.

And let $$(x_n)_n$$ converges for the weak topology $$\sigma(E,E^*)$$ to $$x$$.

Why $$(x_n)_n$$ converges for the strong topology ?

My idea :

Since $$K$$ is a compact set for the norm topology then $$(x_n)_n$$ has a convergent subsequence $$(x_{n_k})_k$$ for the norm topology to $$x$$ (Since $$(x_{n_k})_k$$ converges weakly to x).

How to prove that the sequence $$(x_n)_n$$ converges strongly to $$x$$ ?

I'm stuck in going from Since $$(x_{n_k})_k$$ converges weakly to x. then $$(x_{n_k})_k$$ to Since $$(x_{n_k})_k$$ converges weakly to x. then $$(x_{n})_n$$.

Suppose that $$(x_n)$$ doesn't converge in norm to $$x$$. Then there exists some $$\varepsilon>0$$ and a subsequence $$(x_{n_k})$$ such that $$\|x_{n_k}-x\|\geq\varepsilon$$ for all $$k$$. Since $$(x_n)$$ is contained in a compact set, we must have some sub-subsequence of $$(y_\ell)$$ of $$(x_{n_k})$$ which converges in norm. But this norm limit must be $$x$$, which contradicts $$\|y_\ell-x\|\geq\varepsilon$$ for all $$\ell$$.
• Or, do rephrase half of your argument: A sequence $(x_n)$ converges to $x$ if and only if every subsequence of $(x_n)$ has a subsequence that converges to $x$. Sounds a little complicated, but it comes in very handy in these kinds of situations. – MaoWao Jan 22 at 23:51
Here is another approach. Let $$X$$ denote $$K$$ equipped with the strong topology, and $$Y$$ denote $$K$$ equipped with the weak topology. Since the strong topology is finer than the weak topology, the identity map $$X\to Y$$ is continuous. But $$X$$ is compact and $$Y$$ is Hausdorff, so any continuous bijection $$X\to Y$$ is a homeomorphism. So the identity is a homeomorphism; that is, the weak topology and strong topology on $$K$$ are the same.