# A question of convergence in Hilbert spaces

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 $$||T_n|| for all $$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 $$y=Tx$$.
• We say $T_n \to T$ in WOT if $T_nx \to Tx$ weakly for every $x$. Jul 4, 2020 at 5:29
• @KaviRamaMurthy It looks that it is sufficient $T_n x_n\to y$ weakly, not necessarily in norm. Can we get the same conclusion if we also require $x_n\to x$ weakly, and not in norm? Jul 4, 2020 at 18:30