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I am stuck on a question of convergence, and I am not sure it is true. Suppose $(x_n)$ is a sequence of vectors in a separable Hilbert space and $T_n$ is a sequence of bounded operators such that $||T_n||<K$ for all $n$.

If $x_n\to x\neq 0$, $T_n x_n\to y\neq 0$ (both convergence in norm), and $T_n\stackrel{WOT}{\to} T\neq 0$ (convergence in weak operator topology), does it follow that $Tx=y$? Does it make a difference if we require $T_n$ to converge in operator norm or SOT?

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Actually boundeness of $\|T_n\|$ need not be assumed. It is a consequence of convergence in WOT. For each $x$ we have $T_nx \to Tx$ weakly. This implies that $\sup_n \|T_nx\|<\infty$ for each $x$. By Uniform Boundedness Principle this implies that $\sup_n \|T_n\|<\infty$. Since $\|T_nx_n-T_nx\| \leq \|T_n\|\|x_n-x\|$ we see that $T_nx_n-T_nx \to 0$ in the norm,hance also weakly. But $T_nx_n \to y$ (in the norm, hence weakly) and $T_nx \to Tx$ weakly. Hence $y=Tx$.

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  • $\begingroup$ @reuns Please elaborate. I don't know what you are asking. $\endgroup$ Jul 4, 2020 at 5:15
  • $\begingroup$ The OP is introducing useless and unused vocabulary, define it in two words. $\endgroup$
    – reuns
    Jul 4, 2020 at 5:21
  • $\begingroup$ We say $T_n \to T$ in WOT if $T_nx \to Tx$ weakly for every $x$. $\endgroup$ Jul 4, 2020 at 5:29
  • $\begingroup$ @reuns I am not sure what useless and unused vocabulary are you referring. WOT and SOT topologies are standard in operator theory. $\endgroup$
    – Markus
    Jul 4, 2020 at 16:45
  • $\begingroup$ @KaviRamaMurthy It looks that it is sufficient $T_n x_n\to y$ weakly, not necessarily in norm. Can we get the same conclusion if we also require $x_n\to x$ weakly, and not in norm? $\endgroup$
    – Markus
    Jul 4, 2020 at 18:30

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