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
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 $\eta < \|T\|$.