$\omega(A)=\|A\|\Longleftrightarrow C_A\cap \overline{W(A)}\neq \varnothing $?

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

Let $A\in\mathcal{B}(F)$ and consider $$C_A:=\{z\in \mathbb{C}:\;|z|=\|A\| \}.$$

I want to prove that $$\omega(A)=\|A\|\Longleftrightarrow C_A\cap \overline{W(A)}\neq \varnothing .$$

Note that $$W(A)=\{\langle Ax,x\rangle:\;x \in F,\;\;\|x\|=1\}\quad\text{and}\quad \omega(A)=\sup_{\lambda\in W(A)}|\lambda|.$$

I inspired this question from this proof:

Answer 1

Note that you always have $\omega(A)\leq\|A\|$.

If $\omega(A)=\|A\|$, then there exists $\{z_j\}\subset W(A)$ with $|z_j|\to \|A\|$. Since $\overline{W(A)}$ is compact, there is a cluster point $z$ of $\{z_j\}$. Then $|z|=\lim|z_j|=\|A\|$, so $z\in C_A\cap \overline{W(A)}$.

Conversely, if $z\in C_A\cap \overline{W(A)}$, then $$\|A\|\geq\omega(A)\geq|z|=\|A\|.$$