2
$\begingroup$

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 $(x_{n})_n$.

$\endgroup$
4
$\begingroup$

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 $\ell$.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – MaoWao Jan 22 at 23:51
  • $\begingroup$ @MaoWao Yes, that's a great way to put it. I always forget about that characterization of convergence. $\endgroup$ – Aweygan Jan 23 at 3:01
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That's a great approach @Eric ! $\endgroup$ – Anas BOUALII Jan 24 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.