# Normal Operators: Numerical Range

Disclaimer: As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.!

Prove that for normal operators the spectrum is contained in the closure of the numerical range: $$\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$

If the distance from $\lambda$ to $\mathcal{W}(A)$ is $d > 0$, then $$|((A-\lambda I)\phi,\phi)| \ge d\|\phi\|^{2},\;\;\; \phi \in \mathcal{D}(A)\\ \implies d\|\phi\| \le \|(A-\lambda I)\phi\|,\;\;\; \phi \in\mathcal{D}(A).$$ If $(A\phi,\phi)=(\phi,A^{\star}\phi)$ for $\phi$ in a core domain of $A^{\star}$, then you also get $$d\|\phi\| \le \|(A^{\star}-\overline{\lambda}I)\phi\|,\;\;\;\phi \in \mathcal{D}(A^{\star}).$$ So, in this case $\mathcal{N}(A^{\star}-\overline{\lambda}I)=\{0\}$. Using the adjoint relation, $\mathcal{R}(A-\lambda I)^{\perp}=\mathcal{N}(A^{\star}-\overline{\lambda}I)$, it follows that $(A-\lambda I)$ has dense range and bounded inverse. So, under these assumptions, $\lambda \in \rho(A)$.

• What if $\mathcal{D}(A)\cap\mathcal{D}(A^*)=\{0\}$? – C-Star-W-Star Jun 7 '14 at 10:28
• I cannot think of an argument that will work in that case. There's not much to work with. Assuming that some subspace of $\mathcal{D}(A)$ is a core for $\mathcal{D}(A^{\star})$ is a stretch, too. I'd say that the argument almost requires a bounded $A$. – DisintegratingByParts Jun 7 '14 at 10:43
• Hmm I was hoping to expand the proof from Werner's book to closed ones with dense range but I thought so too that this probably won't work - to bad... – C-Star-W-Star Jun 7 '14 at 11:10
• Btw your argument fails for nonreal lambda since the numerical range is not closed under conjugation in general. You will note this if you go through the proof for the self adjoint case where (there you'll have to consider two separate cases for lambda). – C-Star-W-Star Jun 7 '14 at 18:19
• Assume $\|Ax\| \ge d\|x\|$ for some $d > 0$ and for all $x$ in a core $C$ for $A$. To say that $C$ is a core for $A$ means that $\{ (x,Ax)\in X\times X : x \in C\}$ is dense in the graph of $A$. So, if $x \in \mathcal{D}(A)$, there exists $\{ c_{n} \}\subset C$ such that $c_{n}\rightarrow x$ and $Ac_{n}\rightarrow Ax$. So $\|Ax\|=\lim_{n}\|Ac_{n}\|\ge \lim_{n}d\|c_{n}\|=d\|x\|$. – DisintegratingByParts Jun 7 '14 at 20:10

Disclaimer: This prove is mostly due to T.A.E. - greatest regards to him!

Preparation:
(This is where real work has to be done.)

The distance of $\lambda$ to the numerical range gives us the estimate: $$d(\lambda,\mathcal{W})\leq|\langle \hat{x},(N-\lambda)\hat{x}\rangle|\leq\|(N-\lambda)\hat{x}\|\implies \|(N-\lambda)x\|\geq d(\lambda,\mathcal{W})\|x\|$$ Moreover since $N$ is normal and densely defined we have $N^*N=NN^*$ as a really nontrivial result (see proposition 4.17 b in [1]). Also since $N$ is normal it is closed and since it is densely defined $\mathcal{D}(NN^*)$ is a core for $N^*$ (see proposition 4.11 a in [1]). We thus have: $$d(\lambda,\mathcal{W})\leq|\langle \hat{c},(N-\lambda)\hat{c}\rangle|=|\langle (N^*-\overline{\lambda})\hat{c},\hat{c}\rangle|\leq\|(N^*-\overline{\lambda})\hat{c}\|\quad \hat{c}\in\mathcal{D}(N^*N)=\mathcal{D}(NN^*)\implies\|(N^*-\overline{\lambda})y\|=\lim_n\|(N^*-\overline{\lambda})c_n\|\geq\lim_n d(\lambda,\mathcal{W})\|c_n\|=d(\lambda,\mathcal{W})\|y\|$$

Main Work:
(The prove will be now an easy consequence.)

Assume $\lambda\notin\overline{\mathcal{W}}$.

Since $\lambda$ has positive distance to the numerical range $(N-\lambda)$ will be bounded below by the first estimate given in the preparation and due to this especially injective. Moreover since $N$ is normal it is closed so as well $(N-\lambda)$ and $(N-\lambda)^{-1}$. Thus by the closed graph theorem applied to $(N-\lambda)^{-1}$ the range of $(N-\lambda)$ is closed.

On the other hand we have that the $(N^*-\overline{\lambda})$ is bounded below as well due to the second estimate given in the preparation. But therefore the kernel is zero only and thus we have: $$\overline{\mathcal{R}(N-\lambda)}=\mathcal{N}(N^*-\overline{\lambda})^\perp=\{0\}^\perp=Y$$ That is the range of $(N-\lambda)$ is dense.

Collecting all together $(N-\lambda)$ is injective surjective and bounded below.
So $\lambda$ was in the resolvent set as was to be shown.

Reference: [1]: German version of Weidmann's 'Lineare Operatoren in Hilberträumen'