In some notes I found the following claim.Can someone please help me in proving this?

Let $H$ be a separable Hilbert space and $(e_n)$ be ONB for $H$. Define $T_n \in B(H)$ as $T_n(x)= \langle x,e_1\rangle e_n$ Show that $T_n$ converges in weak operator topology but not strongly.

Please help!

  • $\begingroup$ Do you know the definitions of the weak and strong operator topologies? $\endgroup$ Nov 27, 2015 at 17:56

1 Answer 1


These hints can help you, provided that you know the definitions:

  • First try to find what the limit $T$ is in the weak topology. By the Riesz representation theorem, any linear functional in $H$ is written as $\ell(x) = \langle x, \xi\rangle$ for some vector $\xi$. What happens with $\ell(T_n(x))$ when $n\rightarrow \infty$?
  • Now consider $x = e_1$. What happens with $T_n(x)$ as $n\to\infty$?
  • 1
    $\begingroup$ Since $ e_n \to 0$ weakly therefore $T=0$,correct? $\endgroup$ Nov 28, 2015 at 3:28
  • $\begingroup$ @Dontknowanything You have to analyse what happens to $\ell(T_n(x)) = \langle \langle x, e_1\rangle e_n, \xi\rangle$, not $e_n$, to find the weak limit. $\endgroup$ Nov 28, 2015 at 13:30
  • $\begingroup$ Sorry,I'm confused here.But again since $e_n \to 0$ weakly therefore $l(T_n(x)) \to 0 $ in Norm.Can you please add some details? $\endgroup$ Nov 29, 2015 at 1:57
  • $\begingroup$ @Dontknowanything You are right, just be careful with the wording. The sequence $\ell(T_n(x))$ is a sequence of scalars, not a vectors, so it is misleading to say it converges "in norm" (instead of assuming the usual topology coming from the absolute value). You can use the fact that $e_n$ converges weakly to 0, but be sure to have a clear understanding of why that implies that the former sequence converges to 0 too. $\endgroup$ Nov 29, 2015 at 22:55

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