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?