I am stuck on a question of convergence, and I am not sure it is true. Suppose $(x_n)$ is a sequence of vectors in a separable Hilbert space and $T_n$ is a sequence of bounded operators such that $||T_n||<K$ for all $n$.
If $x_n\to x\neq 0$, $T_n x_n\to y\neq 0$ (both convergence in norm), and $T_n\stackrel{WOT}{\to} T\neq 0$ (convergence in weak operator topology), does it follow that $Tx=y$? Does it make a difference if we require $T_n$ to converge in operator norm or SOT?
Actually boundeness of $\|T_n\|$ need not be assumed. It is a consequence of convergence in WOT. For each $x$ we have $T_nx \to Tx$ weakly. This implies that $\sup_n \|T_nx\|<\infty$ for each $x$. By Uniform Boundedness Principle this implies that $\sup_n \|T_n\|<\infty$. Since $\|T_nx_n-T_nx\| \leq \|T_n\|\|x_n-x\|$ we see that $T_nx_n-T_nx \to 0$ in the norm,hance also weakly. But $T_nx_n \to y$ (in the norm, hence weakly) and $T_nx \to Tx$ weakly. Hence $y=Tx$.