Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x \in X$ and $y \in Y$, $(\langle x, T_ny \rangle )_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{C}$. Prove that exist a linear bounded operator $T \in \mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x \in X$ and $y \in Y$ $\langle x, T_ny \rangle \xrightarrow[]{n \to \infty } \langle x, Ty \rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=\sup_{||x||=||y||=1}|\langle x, T_ny\rangle |$) and $(\langle x, T_ny \rangle )_{n \in \mathbb{N}} \xrightarrow[]{n \to \infty} L_{x,y}$ for some $L_{x,y}\in \mathbb{C}$. But I don't know how construct a linear operator $T \in \mathcal{L}(X,Y) $ such that $L_{x,y}=\langle x, Ty \rangle$. How can I construct that if is possible?
Best Answer
Riesz representation can do it. Note that $$ x\mapsto L_{x,y} $$ defines a bounded linear functional from $X$ into $\mathbb{C}$ for each $y\in Y$ since it holds that $$ |L_{x,y}|\leq \left(\sup_n \lVert T_n\rVert\right)\lVert x\rVert \lVert y\rVert\quad\cdots(*). $$ By Riesz representation theorem, this implies that for each $y\in Y$, there exists unique $\xi_y\in X$ such that $$ L_{x,y} = \langle x,\xi_y \rangle, $$ and it holds that $$ |\xi_y| = \lVert L_{\cdot,y} \rVert\leq \sup_n \lVert T_n\rVert \cdot \lVert y\rVert, $$ by $(*)$. Finally, note that $y\mapsto \xi_y$ is linear in $y\in Y$. Therefore, by defining a bounded linear operator $T$ as $$ T : y\in Y\mapsto \xi_y \in X, $$ we have $L_{x,y} = \lim_{n\to\infty} \langle x,T_ny\rangle = \langle x,Ty\rangle$ and $T$ is a weak limit of $\{T_n\}$.