Strong convergence from weak convergence

I am trying to show that a sequence $(x_n)_n \subseteq \mathcal{H}$ converges strongly to $x$ if it converges weakly to $x \in \mathcal{H}$ and $\|x_n\| \to \|x\|$ as $n \to \infty$

$\mathcal{H}$ is an infinite dimensional hilbert space

Now my proof is clearly bogus because I do not even use the second condition to complete the proof. Can someone point out where the problem is :

Proof Since $x_n \to_{weakly} x$ and since $\mathcal{H}$ is a hilbert space we have that by the Riesz representation theorem that $\forall Y \in \mathcal{H}$

$\langle y,x_n \rangle \to \langle y,x \rangle \iff \langle y,x_n-x \rangle\to 0 \forall y\in \mathcal{H}$. In particular if we choose $y=x_n-x$ we get $\|x_n-x\|^2 \to 0$

This can't be right since otherwise there would be no difference between weak convergence and strong convergence on Hilbert space What is the mistake? Thanks

• @copper.hat I fixed it – user3503589 Jun 15 '15 at 22:35

The weak convergence tells you that, for every $y\in H$, $\langle x_n-x,y \rangle \rightarrow 0$. But, you can't chose $y=x_n-x$, because it depends on $n$.

To prove the strong convergence with $\|x_n\| \rightarrow \|x\|$, simply see that : $\|x_n-x\|^2 = \langle x_n-x,x_n-x \rangle = \|x_n\|^2+\|x\|^2 - 2\langle x_n,x \rangle$

$\|x_n\|^2 \rightarrow \|x\|^2$ by hypothesis and $\langle x_n,x \rangle \rightarrow \|x\|^2$ by weak convergence, so $\|x_n-x\|^2$ tends to $0$.

• Nice answer, just one minor thing, your code will look a lot nicer if you use \langle and \rangle for the angle brackets. – Ruvi Lecamwasam Jun 15 '15 at 23:38
• You're totally right, I'll edit my answer ;) – Sylvain L. Jun 16 '15 at 8:02

Here is a proof for a real Hilbert space.

You have $\|x-x_n\|^2 = \|x\|^2+\|x_n\|^2 - 2 \langle x_n, x \rangle$.

You are given that $\langle x_n, x \rangle \to \langle x, x \rangle = \|x\|^2$, from which the result follows.