Explain a proof from a paper: why the inequality $b<\|A\|$ is strict? For $A\in\mathcal{L}(E)$, we consider
$$K=\{x\in E:\;\|x\|=1,\;\Re e\langle Ax,x\rangle\leq\frac{a}{2}\;\},$$
where $a$ is such that $\Re e( W_{0}(A))\geq a>0$, with
\begin{eqnarray*}
W_{0}(A)
&=&\{\alpha\in \mathbb{C}:\;\exists\,(z_n)\subset E\;\;\hbox{such that}\;\|z_n\|=1,\displaystyle\lim_{n\rightarrow+\infty}\langle A z_n,z_n\rangle=\alpha,\\
&&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}\|Az_n\|= \|A\| \}.
\end{eqnarray*}
Set 
$$b:=\sup_{x\in K}\|Ax\|.$$

Why
  $$b<\|A\|?$$

This above strict inequality figures in the proof of the following theorem


 A: We have that,
$$\eta  = \sup\{ \|Tx\| : x \in \mathfrak{S}\}$$
Suppose for contradiction that $\eta = \| T \|$. 
It follows that all of the sequences $z_n \in \mathfrak{S}$ for which $\|Tz_n\| \to \eta$, also have the property that $\mathcal{Re}(Tz_n, z_n) < \tau / 2$. Let $\xi_n =(Tz_n, z_n) $ be such a sequence.
The complex part of $\xi_{n}$ is also bounded, since $\|z_n\| = 1$, and $T$ is a linear operator. By the Bolzano-Weierstrass theorem, there exists a convergent subsequence $\xi_{n_k}$ with a limit which we will call $\alpha$. We also have that $ \|Tz_{n_k}\| \to \eta $ as $k \to \infty$.
This is a contradiction since it would mean that $\mathcal{Re}(\alpha) \in \mathcal{Re}W_0(T)$, by the definition of $W_0(T)$; but $\mathcal{Re}(\alpha) \leq \frac{\tau}{2} < \tau $, and $ \tau $ is less than or equal to every element of $\mathcal{Re}W_0(T)$. It follows that $\eta \neq \|T\|$.
Since, 
$$\| T \| = \sup \{ \|Tx \| : \|x\| = 1 \}$$
and,
$$ \{ \|Tx\| : x \in \mathfrak{S}\} \subset \{ \|Tx \| : \|x\| = 1 \} $$
it follows that $\eta < \|T\|$.
