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 $(x_{n})_n$.
Best Answer
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.