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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, 2018 at 7:30
  • $\begingroup$ Why y shall not depend on n? $\endgroup$
    – chuckyy
    Feb 6, 2018 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, 2018 at 7:40
  • $\begingroup$ As no one has mentioned this, this is known as the Kadec-Klee property of norms. There is also a stronger property property of norms, known as the Kadec property, where the norm and weak topologies agree on the sphere (i.e. the two subspace topologies on the unit sphere, derived from the norm and weak topologies, are the same). $\endgroup$
    – Theo Bendit
    Nov 10, 2021 at 1:50

1 Answer 1

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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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