$\lim_{n \to \infty} \langle\bar A_nx,y\rangle = \langle Ax,y\rangle$

Question

Let $A$ be an operator on a separable Hilbert Space $H$. So we can consider $A=(a_{ij})_{1\leq i,j\leq \infty}$.

Let $\tilde A_n = $$ \begin{pmatrix} A_n & 0 \\ 0 & -I \end{pmatrix} $ where $\tilde A_n $ is an infinite matrix and $A_n=(a_{ij})_{1\leq i,j\leq n}$, the upper corner $n \times n$ submatrix of $A$ and $I$ is the infinite order identity matrix.

To prove $\displaystyle\lim_{n \to \infty} \langle\tilde A_nx,y\rangle = \langle Ax,y\rangle$. i.e. $\tilde A_n\to A$, in Weak Operator Topology.

Facing difficulty to prove the fact.

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Answer 1

Here is a solution assuming A is a contraction. Let $A=\begin{pmatrix} A_n & X_n\\ Y_n & Z_n\\ \end{pmatrix}$ and $x=(x_i)_{1\leq i\leq \infty},y=(y_i)_{1\leq i\leq\infty}\in \mathcal{H}.$ Then $\vert\langle (A-\tilde{A_n})x,y\rangle\vert\\ =\vert\langle X_n(x_i)_{n\leq i \leq \infty}, (y_i)_{1\leq i \leq n}\rangle+\langle Y_n(x_i)_{1\leq i \leq n}, (y_i)_{n\leq i \leq \infty}\rangle+\langle (Z_n+I)(x_i)_{n\leq i \leq \infty},(y_i)_{n\leq i \leq \infty}\rangle\vert\\\leq\vert\langle X_n(x_i)_{n\leq i \leq \infty}, (y_i)_{1\leq i \leq n}\rangle\vert+\vert\langle Y_n(x_i)_{1\leq i \leq n}, (y_i)_{n\leq i \leq \infty}\rangle\vert+\vert\langle (Z_n+I)(x_i)_{n\leq i \leq \infty},(y_i)_{n\leq i \leq \infty}\rangle\vert\\ \leq\Vert X_n\Vert\Vert (x_i)_{n\leq i \leq \infty}\Vert\Vert(y_i)_{1\leq i \leq n}\Vert+\Vert Y_n\Vert\Vert(x_i)_{1\leq i \leq n}\Vert\Vert (y_i)_{n\leq i \leq \infty}\Vert\Vert+\Vert(Z_n+I)\Vert(x_i)_{n\leq i \leq \infty}\Vert\Vert (y_i)_{n\leq i \leq \infty}\Vert\\ \leq(\sum\limits_{i=n}^\infty \vert x_i\vert^2)^{1/2}(\sum\limits_{i=1}^n \vert y_i\vert^2)^{1/2}+(\sum\limits_{i=1}^n \vert x_i\vert^2)^{1/2}(\sum\limits_{i=n}^\infty \vert y_i\vert^2)^{1/2}+2(\sum\limits_{i=n}^\infty \vert x_i\vert^2)^{1/2}(\sum\limits_{i=n}^\infty \vert y_i\vert^2)^{1/2}\\ \longrightarrow 0$ whenever $n\rightarrow \infty $ as $x,y\in \mathcal{H}$ So $\tilde{A_n}\xrightarrow{\text{WOT}} A$

Answer 2

Assuming that $A$ is a bounded operator I've proved the strong convergence. For the unbounded case the same trick fails, a similar trick seems to work for the weak topology and symmetric operators though (assuming $y\in D(A)$).

$\textbf{(Bounded case)}$ Assume $||A||<\infty$ and define the operators \begin{align*} P_n:H&\rightarrow H\\ (x_1,x_2,\dots)&\mapsto (x_1,\dots,x_n,0,0\dots), \end{align*} \begin{align*} P_n^{\perp}:&H\rightarrow H\\ (x_1,x_2,\dots)&\mapsto(0,\dots,0,x_{n+1},x_{n+2},\dots), \end{align*} so $Id-P_n=P_n^{\perp}$.

Now note that $\tilde{A}_n=P_nAP_n-P_n^{\perp}$ and $A=P_nA+P_n^{\perp}A$, so for every $x\in H$ \begin{align*} ||(A-\tilde{A}_n)x||&= ||(P_nA+P_n^{\perp}-P_nAP_n+P_n^{\perp})x||\\ &\leq ||P_nA(Id-P_n)x||+||P_n^{\perp}(A+Id)x||\\ &\leq ||P_n||\cdot||A||\cdot||P_n^{\perp}x||+||P_n^{\perp}(A+Id)x||, \end{align*} and we can conclude proving that $||P_n||\leq 1$ (which is actually easy!) and for every $y\in H$ $P_n^{\perp}y\rightarrow 0$, so taking $y=x$ and $y=(A+Id)x$ would do the job.

$\textbf{(Unbounded Symmetric case)}$ Let $x,y\in D(A)$ and note that

\begin{align*} |\langle (A-\tilde{A}_n) x,y\rangle|&=|\langle (P_nA+P_n^{\perp}-P_nAP_n+P_n^{\perp})x,y\rangle|\\ &\leq |\langle P_nAP_n^{\perp}x,y\rangle|+|\langle P_n^{\perp}(A+Id)x,y\rangle|\\ &= |\langle P_n^{\perp}x,AP_ny\rangle|+|\langle P_n^{\perp}(A+Id)x,y\rangle|\\ &\leq ||P_n^{\perp}x||\cdot||AP_ny||+||P_n^{\perp}(A+Id)x||\cdot||y||, \end{align*} where we have used that $P_n$ is also symmetric and $P_n^{\perp}\rightarrow 0$ strongly as I've mentioned before.

$\textbf{Remark:}$ Rigorously and In order to make the last inequality to have sense, we need that $y\in D(P_{n_k})$ for a sequence $\{n_k\}\in\mathbb{N}$ such that $n_k\rightarrow\infty$, but I'm convinced that if $y\in D(A)$ there is no problem with defining $AP_ny$.