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?
Best Answer
\begin{align*} \|x_{n}-x\|^{2}&=\left<x_{n}-x,x_{n}-x\right>\\ &=\left<x_{n},x_{n}\right>-\left<x_{n},x\right>-\left<x,x_{n}\right>+\left<x,x\right>\\ &=\|x_{n}\|^{2}+\|x\|^{2}-\left<x_{n},x\right>-\left<x,x_{n}\right>\\ &\rightarrow\|x\|^{2}+\|x\|^{2}-\left<x,x\right>-\left<x,x\right>\\ &=0. \end{align*}