Semi-norm of bounded linear operators on Hilbert spaces I see the following paragraph in a paper


I think there is a wrong in the paper because, I think we have only
  $$\|T\|_A=\sup_{\substack{\xi,\eta\in \overline{R(A)},\\\|\xi\|_A \leq 1, \|\eta\|_A \leq  1}}|\langle T\xi\;|\;\eta\rangle _A| \,.$$

Thank you for your help.
 A: As you say, you need to restrict the $x$ and $y$ to the range of $A$. 
Take $H=\mathbb C^2$, with $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. Take $T=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.  Then for $\omega=(\omega_1,\omega_2)$,
$$
\|\omega\|_A=|\omega_1|
$$
and $$R(A)=\{(t,0):\ t\in\mathbb C\}.$$ Thus $T\omega=0$ for all $\omega\in R(A)$ and 
\begin{align}
\|T\|_A&=0.
\end{align}
On the other hand, since $AT=T$,
\begin{align}
\sup\{|\langle Tx,y\rangle_A|:\ \|x\|_A=\|y\|_A=1\}
&=\sup\{|\langle Tx,y\rangle|:\ \|x\|_A=\|y\|_A=1\}\\ \ \\
&=\sup\{|x_2\overline{y_1}|:\ |x_1|=|y_1|=1\}\\ \ \\
&=\infty
\end{align}
(on the other hand, if we force $x,y\in R(A)$, then $x_2=0$ and the two expressions agree). 
A: Because $|\langle Tx,y\rangle| \le \|Tx\|\|y\| \le \|T\|\|x\|\|y\|$, then
$$
               \sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \le \|T\|.
$$
Without losss of generality, assume $\|T\|\ne 0$. Let $0 < \epsilon < \|T\|$ be given. There exists $x'$ such that $\|x'\|=1$ and $\|Tx'\| \ge \|T\|-\epsilon$. Therefore,
$$
    \sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \ge |\langle Tx',\frac{1}{\|Tx'\|}Tx'\rangle| = \|Tx'\| \ge \|T\|-\epsilon.
$$
Because this is true of all $\epsilon > 0$, then $\sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \ge \|T\|$.
