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?

  • 1
    $\begingroup$ 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

\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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.