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
Best Answer
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 $0$.