Let $E$ be a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. Let $T\in \mathcal{L}(E)$. The numerical range of $T$ is defined as follow $$W({ T})=\{\langle Tx\;|\;x\rangle:\;x \in E,\;\;\|x\|=1\}.$$ The numerical radius $\omega(T)$ of $T$ is defined by \begin{eqnarray*} \omega(T) &=&\sup\left\{|\lambda|,\;\lambda\in W(T)\right\}\\ &=&\sup\{|\langle Tx\;|\;x\rangle|,\;x\in E,\;\|x\|=1\;\}. \end{eqnarray*}

Moreover, the spectral radius of $T$ is given by $$r(T)=\sup\{|\lambda|,\;\;{\lambda \in \sigma(T)}\},$$ where $\sigma(T)$ is the spectrum of $T$.

Why we have $$r(T)\leq \omega(T)??$$

Thank you, everyone !!!


It isn't a one-line proof. See Theorem 1.1 in Moshe Goldberg, Eitan Tadmor: On the numerical radius and its applications, doi: 10.1016/0024-3795(82)90155-0.

  • $\begingroup$ I fixed it. Try again please. $\endgroup$ – uniquesolution Aug 11 '17 at 8:21
  • $\begingroup$ Ok, good luck ! $\endgroup$ – uniquesolution Aug 11 '17 at 8:29
  • $\begingroup$ Yes, maybe a bounty will buy someone's time to sort out your reading problems. $\endgroup$ – uniquesolution Aug 13 '17 at 14:39

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