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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

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  • $\begingroup$ What is $a$? If $a/2>\left\|A\right\|$ then $K$ can be all of the unit sphere and then $b=\left\|A\right\|$. $\endgroup$
    – user525761
    Commented Feb 3, 2018 at 11:30
  • $\begingroup$ @Student It seems $K$ is a subset of the unit sphere of $E$, since for $x \in K$ we are insisting $||x|| = 1$. We know that $||A||$ is the maximum of $|Ax|$ over the entire unit sphere, so its maximum over a subset of the unit sphere, namely $K$, has to be less than or equal to $||A||$. $\endgroup$ Commented Feb 3, 2018 at 12:28
  • $\begingroup$ Thank you but my problem is why the inequality is strict? $\endgroup$
    – Student
    Commented Feb 3, 2018 at 12:31
  • $\begingroup$ @астонвіллаолофмэллбэрг But in the proof the inequality is strict. $\endgroup$
    – Student
    Commented Feb 3, 2018 at 13:08
  • $\begingroup$ @Student Use the other fact, namely that $\langle Tx,x \rangle$ is less than or equal to $\frac \tau 2$, and the fact that the maximum of this over the unit sphere is also $||A||$. Now look at relations between $\tau$ and $||A||$ as given in the definition of $W_0$. $\endgroup$ Commented Feb 3, 2018 at 14:36

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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 $\eta < \|T\|$.

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  • $\begingroup$ Thank you for your answer but $ \mathcal{Re}(Tz_n, z_n) \to \alpha $ doesn't means $(Tz_n, z_n) \to \alpha $ $\endgroup$
    – Student
    Commented Feb 4, 2018 at 8:51
  • $\begingroup$ No that is true. It is a mistake, but $(Tz_n, z_n)$ does converge. If $\| Tz_n \|$ converges to a finite limit, then so does $z_n$ by the linearity of $T$ and continuity of $ \|\cdot \| $. We can then say that $(Tz_n, z_n)$ converges to a number $\alpha$ since $(\cdot, \cdot) $ is continuous. $\endgroup$
    – Damien
    Commented Feb 4, 2018 at 9:02
  • $\begingroup$ We have only $(Tz_n,z_n)$ is bounded and so has a convergent subsequence $\endgroup$
    – Student
    Commented Feb 4, 2018 at 9:07
  • $\begingroup$ I don't understand why $\alpha \in W_0(T)$? $\endgroup$
    – Student
    Commented Feb 4, 2018 at 9:08
  • $\begingroup$ $\alpha$ cannot be in $W_0(T)$ since it is strictly less that $\tau$. It is possible to prove that there exists a $z_n$ which converges. You are right that $(Tz_n,z_n)$ is bounded and therefore has a convergent subsequence. In fact, $z_n$ is also bounded and contains a convergent subsequence, $z_{n_k}$. The limit of $\| Tz_{n_k} \|$ is $\eta$, and say the limit of $ (Tz_{n_k}, z_{n_k}) $ is $ \alpha $, then $\alpha \in W_0(T)$ by the definition of $W_0$. $\endgroup$
    – Damien
    Commented Feb 4, 2018 at 9:28

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