1
$\begingroup$

If $(X, \langle \cdot, \cdot \rangle)$ is a complex Hilbert-space and $T : X \rightarrow X$ a normal operator, i.e. an operator such that $T T^\ast = T^* T$ then I'd like to show that:

$$\sup_{\Vert x \Vert = 1} \mathrm{Re} \langle x, Tx \rangle = \sup_{\lambda \in \sigma(T)} \mathrm{Re} \lambda$$

where $\sigma (T)$ denotes the spectrum of $T$.

I do not have any idea on how to solve such kinds of problems. Can someone give me a hint on how to tackle this?

Thanks!

$\endgroup$
2
$\begingroup$

Let $\rho = \sup_{\|x\|=1}\Re (Tx,x)$. Suppose $\rho <\Re\lambda$. Then $\Re(Tx,x)\le \rho(x,x)$ for all $x$, and \begin{align} (\Re\lambda-\rho)\|x\|^2 &\le \Re ((\lambda I-T)x,x) \\ (\Re\lambda-\rho)\|x\|^2 &\le \|(\lambda I-T)x\|\|x\| \\ (\Re\lambda-\rho)\|x\| &\le \|(\lambda I-T)x\|. \end{align} Therefore $T-\lambda I$ is injective for $\Re\lambda > \rho$, and the above also implies that the range of $\lambda I -T$ is closed for such $\lambda$. Because $T$ is normal, $\|(T-\lambda I)x\|=\|(T-\lambda I)^{\star}x\|$ for all $x\in X$, which also gives $$ \mathcal{R}(\lambda I-T)=\mathcal{N}((\lambda I-T)^{\star})^{\perp}= \mathcal{N}(\lambda I-T)^{\perp}=X. $$ So $\Re\lambda > \rho$ implies $\lambda\in\rho(T)$, or $$ \sigma(T) \subseteq \{ \lambda \in\mathbb{C} : \Re\lambda \le \sup_{\|x\|=1}(Tx,x)\}. $$ For the opposite inclusion, the spectrum of $\sigma(T)$ is closed and bounded, which gives the existence of $\lambda_0\in\sigma(T)$ for which $\Re\lambda \le \Re\lambda_0$ for all $\lambda\in\sigma(T)$. Any point $\lambda_0\in\sigma(T)$ is either in the point spectrum or the approximate point spectrum of $T$ because $T$ is normal, which gives the existence of a sequence of unit vectors $\{ x_n\}$ such that $\|Tx_n-\lambda_0x_n\|\rightarrow 0$; hence $$ (Tx_n,x_n)=(Tx_n-\lambda_0x_n,x_n)-\lambda_0\rightarrow\lambda_0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.