Vtyjkyuk

Let $A$ be an operator on a separable Hilbert Space $H$. So we can consider $A=(a_ij)_1leq i,jleq infty$.

Let $tilde A_n = $$
beginpmatrix
A_n & 0 \
0 & -I
endpmatrix
$ where $tilde A_n $ is an infinite matrix and $A_n=(a_ij)_1leq i,jleq n$, the upper corner $n times n$ submatrix of $A$ and $I$ is the infinite order identity matrix.

To prove $displaystylelim_n to infty langletilde A_nx,yrangle = langle Ax,yrangle$. i.e. $tilde A_nto A$, in Weak Operator Topology.

Facing difficulty to prove the fact.

Here is a solution assuming A is a contraction.
Let $A=beginpmatrix
A_n & X_n\
Y_n & Z_n\
endpmatrix$ and $x=(x_i)_1leq ileq infty,y=(y_i)_1leq ileqinftyin mathcalH.$ Then
$vertlangle (A-tildeA_n)x,yranglevert\
=vertlangle X_n(x_i)_nleq i leq infty, (y_i)_1leq i leq nrangle+langle Y_n(x_i)_1leq i leq n, (y_i)_nleq i leq inftyrangle+langle (Z_n+I)(x_i)_nleq i leq infty,(y_i)_nleq i leq inftyranglevert\leqvertlangle X_n(x_i)_nleq i leq infty, (y_i)_1leq i leq nranglevert+vertlangle Y_n(x_i)_1leq i leq n, (y_i)_nleq i leq inftyranglevert+vertlangle (Z_n+I)(x_i)_nleq i leq infty,(y_i)_nleq i leq inftyranglevert\
leqVert X_nVertVert (x_i)_nleq i leq inftyVertVert(y_i)_1leq i leq nVert+Vert Y_nVertVert(x_i)_1leq i leq nVertVert (y_i)_nleq i leq inftyVertVert+Vert(Z_n+I)Vert(x_i)_nleq i leq inftyVertVert (y_i)_nleq i leq inftyVert\
leq(sumlimits_i=n^infty vert x_ivert^2)^1/2(sumlimits_i=1^n vert y_ivert^2)^1/2+(sumlimits_i=1^n vert x_ivert^2)^1/2(sumlimits_i=n^infty vert y_ivert^2)^1/2+2(sumlimits_i=n^infty vert x_ivert^2)^1/2(sumlimits_i=n^infty vert y_ivert^2)^1/2\
longrightarrow 0$ whenever $nrightarrow infty $ as $x,yin mathcalH$
So $tildeA_nxrightarrowtextWOT A$

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 $yin D(A)$).

$textbf(Bounded case)$ Assume $||A||<infty$ and define the operators
beginalign*
P_n:H&rightarrow H\
(x_1,x_2,dots)&mapsto (x_1,dots,x_n,0,0dots),
endalign*
beginalign*
P_n^perp:&Hrightarrow H\
(x_1,x_2,dots)&mapsto(0,dots,0,x_n+1,x_n+2,dots),
endalign*
so $Id-P_n=P_n^perp$.

Now note that $tildeA_n=P_nAP_n-P_n^perp$ and $A=P_nA+P_n^perpA$, so for every $xin H$
beginalign*
||(A-tildeA_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^perpx||+||P_n^perp(A+Id)x||,
endalign*
and we can conclude proving that $||P_n||leq 1$ (which is actually easy!) and for every $yin H$ $P_n^perpyrightarrow 0$, so taking $y=x$ and $y=(A+Id)x$ would do the job.

$textbf(Unbounded Symmetric case)$ Let $x,yin D(A)$ and note that

beginalign*
|langle (A-tildeA_n) x,yrangle|&=|langle (P_nA+P_n^perp-P_nAP_n+P_n^perp)x,yrangle|\
&leq |langle P_nAP_n^perpx,yrangle|+|langle P_n^perp(A+Id)x,yrangle|\
&= |langle P_n^perpx,AP_nyrangle|+|langle P_n^perp(A+Id)x,yrangle|\
&leq ||P_n^perpx||cdot||AP_ny||+||P_n^perp(A+Id)x||cdot||y||,
endalign*
where we have used that $P_n$ is also symmetric and $P_n^perprightarrow 0$ strongly as I've mentioned before.

$textbfRemark:$ Rigorously and In order to make the last inequality to have sense, we need that $yin D(P_n_k)$ for a sequence $n_kinmathbbN$ such that $n_krightarrowinfty$, but I'm convinced that if $yin D(A)$ there is no problem with defining $AP_ny$.