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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 $z\in F$).

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

Thank you.

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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 $n$.

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  • $\begingroup$ Thank you for you answer but I don't mean the spaectral radius $\endgroup$
    – Schüler
    Commented Jan 16, 2018 at 14:46
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    $\begingroup$ (Added for completeness) So the spectral radius $\rho(S)=\lim ||S^n||^{1/n}$ satisfies $\rho(S)=||S||$. But [1] since $\rho(S)\leq\omega(S)\leq ||S||$ the same is true for the numerical radius. [1] Moshe Goldberg, Eitan Tadmor, On the numerical radius and its applications, Linear Algebra and its Applications, Volume 42, 1982, Pages 263-284 $\endgroup$
    – Luca Ghidelli
    Commented Jan 16, 2018 at 14:50
  • $\begingroup$ I am sorry; I actually misread the question but, fortunately, it was easy to complete the answer. $\endgroup$
    – Mateusz Wasilewski
    Commented Jan 16, 2018 at 17:42

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