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

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.

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.
• 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 $x$? – user92646 Oct 14 '18 at 0:19
$$\langle T_nx,y\rangle \to \langle T_nx,y\rangle$$ 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 $$sup_n\|T_n\| <\infty$$.