On convergence of WOT

Question

Let $H$ Hilbert space. $\{A_n\}_{n\ge0}$ a sequence in $B(H)$ such that the sequence $\{\langle A_nx,y\rangle\}_{n\ge0}$ converges for all $x,y\in H$.

Can we show that there exist $A\in B(H)$ such that $$\lim_{n \to \infty}\langle A_nx,y\rangle=\langle Ax,y\rangle$$ for all $x,y\in H$.

I guessed one can use Uniform Bounded Principle but couldn't figure out a concrete argument.

This question is motivated from the fact (can be proved using Uniform Boundedness Principle)- If $\{A_n\}_{n\ge0}$ a sequence in $B(H)$ such that the sequence $\{ A_nx\}_{n\ge0}$ converges for all $x\in H$ then there exist $A\in B(H)$ such that $$\lim_{n \to \infty}A_nx= Ax$$

Thanks in advance.

Answer 1

Define $\sigma:H\times H\to\mathbb{C}$ by setting $$\sigma(x,y):=\lim_{n\to\infty}\langle A_nx,y\rangle $$

This is a well defined map and one can easily see that this is linear in the first argument and conjugate linear in the second argument. Moreover, note that $\{\|A_n\|\}_{n=1}^\infty\subset[0,\infty)$ is a bounded sequence: You can find a proof of this result (using the principle of uniform boundedness) here.

So now there exists $M>0$ such that $|\sigma(x,y)|\leq M\cdot\|x\|\cdot\|y\|$ for all $x,y\in H$. This proves that there exists a bounded operator $A\in B(H)$ so that $\sigma(x,y)=\langle Ax,y\rangle$, as you can deduce from this theorem.