Weak convergence and convergence of norms imply strong convergence in Hilbert space

Let $$(x_n)$$ be a sequence in a Hilbert space $$H$$ which weakly converges to $$x$$. If $$\|x_n\| \rightarrow \|x\|$$ also, show that $$x_n$$ converges strongly to $$x$$.

So, this statement seems to be true. I was wondering how to show it.

I tried with that:

Since $$x_n \overset{w}{\rightarrow} x$$ weakly, that means that $$|\langle x_n - x, y\rangle| < \epsilon$$ for every $$y$$. Since its true for every $$y$$, I can pick $$y = x_n - x$$ , which proves the strong convergence is implied by only the weak. Why is that wrong? And how to fix it?

• You cannot pick $y=x_n-x$ as $y$ may not depend on $n$. Instead, expand $\|x_n-x\|^2$ and take the limit. – A.Γ. Feb 6 '18 at 7:30
• Why y shall not depend on n? – chuckyy Feb 6 '18 at 7:38
• By definition of weak convergence: for every $y$, but every partricular $y$, it cannot be a sequence of different $y$'s. Otherwise you get strong convergence. For example, $x_n\to 0$ weakly, then $\langle x_n,y\rangle\to 0$. Take $y=x_n$, and you get strong convergence. Nonsense. – A.Γ. Feb 6 '18 at 7:40