Showing that $r(T)\leq \omega(T)$?

Question

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 !!!

Theorem 1.1 in this reference could be helpful but the definition of the joint numerical radius is not the one given in my definition. https://arxiv.org/pdf/math/0410492.pdf — Student, Aug 16 '17 at 13:08
Now reposted to MathOverflow: Another concept to show that $r(T)\leq \omega(T)$. This answer gives, in my opinion, very reasonable advice about cross-posting. Of course, other discussions about [meta-tag:cross-posting] might be of interest, too. — Martin Sleziak, Aug 21 '17 at 10:14

score 2 · Accepted Answer · edited Sep 04 '17 at 11:49

2

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.

edited Sep 04 '17 at 11:49

Martin Sleziak

53,687

answered Aug 11 '17 at 08:18

uniquesolution

18,541

I fixed it. Try again please. – uniquesolution Aug 11 '17 at 08:21
Ok, good luck ! – uniquesolution Aug 11 '17 at 08:29
Yes, maybe a bounty will buy someone's time to sort out your reading problems. – uniquesolution Aug 13 '17 at 14:39

Showing that $r(T)\leq \omega(T)$?

1 Answers1