1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Oct 14 '18 at 0:19
  • $\begingroup$ Because of Banach-Steinhaus. $\endgroup$ – Martin Argerami Oct 14 '18 at 0:37
  • $\begingroup$ Ah, it is just your eq(*). THX a lot. $\endgroup$ – user92646 Oct 14 '18 at 0:41
  • $\begingroup$ Glad I could help! :) $\endgroup$ – Martin Argerami Oct 14 '18 at 2:05
1
$\begingroup$

$\langle T_nx,y\rangle \to \langle T_nx,y\rangle$ 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 $sup_n\|T_n\| <\infty$.

$\endgroup$
  • 1
    $\begingroup$ but wot convergence is not weakly convergence. $\endgroup$ – user92646 Oct 13 '18 at 12:38

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.