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Let $\{T_n\}$ be a sequence of bounded operators on a Hilbert space. Assume that $T_n\rightarrow T$ in weak operator topology. Is this sequence necessarily uniformly bounded? It seems that if this sequence converges in ultra-weak topology, then it is necessarily uniformly bounded but not for weak operator topology.

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2 Answers 2

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One can apply Banach-Steinhaus in this situation. Fix $x\in H$. Consider the linear maps $R_n:H\to\mathbb C$ given by $R_n(y)=\langle T_nx,y\rangle$. Each $R_n$ is bounded, and for each $y\in X$ the (numeric) sequence $\{|R_n(y)|\}_n$ is bounded, since it converges. Then Banach-Steinhaus applies, telling us that $$\tag{$*$} \sup\{|\langle T_nx,y\rangle|:\ n\in\mathbb N, \ \|y\|=1\}<\infty. $$ Now, given $n\in\mathbb N$ and $y$ with $\|y\|=1$, consider the linear maps $S_{n,y}:H\to\mathbb C$ given by $S_{n,y}(x)=\langle T_nx,y\rangle$. By $(*)$ we may apply Banach-Steinhaus to the family $\{S_{n,y}\}_{n,y}$. Thus $$\tag{$**$} \sup\{|\langle T_nx,y\rangle|:\ n\in\mathbb N, \ \|x\|=1, \|y\|=1\}=L<\infty. $$ Then $$ \|T_n\|=\sup\{|\langle T_nx,y\rangle:\ \|x\|=\|y\|=1\}\leq L $$ and the sequence $\{T_n\}$ is uniformly bounded.

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  • $\begingroup$ a question: since $<T_nx,y>\to <Tx,y>$, it follows that $<T_nx, y>$ is uniformly bounded because it is a sequence. However, how do you know $\sup _{n,y}|S_{n,y}(x)|<\infty$ for every fixed $x$? $\endgroup$
    – user92646
    Commented Oct 14, 2018 at 0:19
  • $\begingroup$ Because of Banach-Steinhaus. $\endgroup$ Commented Oct 14, 2018 at 0:37
  • $\begingroup$ Ah, it is just your eq(*). THX a lot. $\endgroup$
    – user92646
    Commented Oct 14, 2018 at 0:41
  • $\begingroup$ Glad I could help! :) $\endgroup$ Commented Oct 14, 2018 at 2:05
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$\langle T_nx,y\rangle \to \langle T_nx,y\rangle$ for all $x,y$ This implies that $\{Tx_n\}$ is weakly convergent. Weakly convergent sequence are norm bounded. Hence $\{T_nx\}$ is norm bounded for each $x$. Banach - Steinhaus Theorem implies that $sup_n\|T_n\| <\infty$.

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    $\begingroup$ but wot convergence is not weakly convergence. $\endgroup$
    – user92646
    Commented Oct 13, 2018 at 12:38

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