Can a wo-converging sequence of bounded operators on a Hilbert space be NOT uniformaly bounded？

Question

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.

Answer 1

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.

Answer 2

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