# For a normal operator $T$ we have $\sup_{\Vert x \Vert = 1} \mathrm{Re} \langle x, Tx \rangle = \sup_{\lambda \in \sigma(T)} \mathrm{Re} \lambda$

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!

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 $$(Tx_n,x_n)=(Tx_n-\lambda_0x_n,x_n)-\lambda_0\rightarrow\lambda_0.$$