Vtyjkyuk

Let $A $ be an operator acting on an infinite dimensional complex Hibert space $ H $. Let $W(A)$ be
the set of all $langle Ax,x rangle $ with $xin H,~Vert x Vert=1$ and
$ w(A)=displaystyle sup_zin W(A)vert zvert. $
Let $(x_n)$ be a sequence of unit vectors in $H$ such that $$ displaystyle lim_nvertlangle Ax_n,x_n ranglevert=w(A).$$ Does exist a sequence $(y_n)$ of unit vectors in $H$ such that
$ langle x_n , y_n rangle=0 $ for all $n$ and
$$ displaystyle lim_nvert langle Ay_n,y_n ranglevert= w(A)?$$

Our approach is based on the following. By the Berberian-Orland trick,
there exist a complex Hilbert space $K supset H $ and an operator $A^o$ acting on $K$ with $A^o_vertH=A$ and $W(A^o)=adh(W(A))$, so we have $w(A)=w(A^o)$.
Put $$ displaystyle lim_nlangle Ax_n,x_n rangle=langle A^ox,x rangle_K quad and quad displaystyle lim_nlangle Ay_n,y_n rangle=langle A^oy,y rangle_K, $$ with $xin K$, $yin K$ and $Vert x Vert_K=Vert yVert_K=1$. Therefore, it suffices to show that $langle x,y rangle_K=0$.
If it is the case, since $langle A^ox,x rangle in W(A^o)cap partial W(A^o) $, then we get $vertlangle A^ox,x rangle_Kvert=vertlangle A^oy,y rangle_Kvert$. That is $ displaystyle lim_nvertlangle Ay_n,y_n ranglevert=w(A)$.

This is not possible. Take $H=l^2$, define
$$
Ax = x_1e_1,
$$
then $langle Ax,xrangle = |x_1|^2$. Let $(x_n)$ be a sequence of unit vectors such that $langle Ax_n,x_nrangle to 1$, which implies $|x_n,1|to 1$.
Since $|x_n|=1$, it follows $|x_n -x_n,1e_1|to0$. Let $(y_n)$ be another sequence with these properties. Then clearly $|langle x_n,y_nrangle|to1$:
$$
langle x_n,y_nrangle = x_n,1y_n,1 +langle x_n -x_n,1e_1,x_n -x_n,1e_1rangle,
$$
which converges to a limit with absolute value $1$, as the second addend is the inner product of two sequences converging to zero.