Let $F$ be a complex Hilbert space. We recall that an operator $S\in\mathcal{B}(F)$ is said to be hyponormal if $S^\*S\geq SS^\*$ (i.e. $\langle (S^\*S-SS^\*)z,z \rangle\geq 0$ for all $z\in F$).



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> Assume that $S$ is hyponormal operator. Is $\omega(S)=\|S\|?$, with $\omega(S)$ denotes the numerical radius of $S$ and defined as:
>  $$\omega(S)=\displaystyle\sup\_{\|x\|=1}|\langle Sx, x\rangle |.$$
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Thank you.



numerical radius of hyponormal operator
Yes, the proof can be found in Stampfli, Joseph G., *Hyponormal operators*, Pacific J. Math. (12), no. 4 (1962), 1453--1458.


The proof is actually very simple. First of all, hyponormality may be stated as $\|Sx\|\geqslant \|S^{\ast}x\|$ for any $x\in F$. It follows that $\|Sx\|^2 = \langle Sx, Sx \rangle = \langle x, S^{\ast}Sx\rangle \leqslant \|x\|\cdot \|S^{\ast}Sx\| \leqslant \|x\|\cdot \|S^2x\|$, where we applied the definition to the vector $y:=Sx$. It follows that $\|S\|^2 \leqslant \|S^2\|$, so $\|S^2\|=\|S\|^2$. Then you proceed by induction to prove that $\|S^n\|=\|S\|^n$ for any $n$.


