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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?

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    $\begingroup$ You cannot pick $y=x_n-x$ as $y$ may not depend on $n$. Instead, expand $\|x_n-x\|^2$ and take the limit. $\endgroup$ – A.Γ. Feb 6 '18 at 7:30
  • $\begingroup$ Why y shall not depend on n? $\endgroup$ – chuckyy Feb 6 '18 at 7:38
  • $\begingroup$ 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. $\endgroup$ – A.Γ. Feb 6 '18 at 7:40
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\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*}

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