Spectrum of the Resolvent of a Self-Adjoint Operator

Question

Let $\mathcal{H}$ be a Hilbert space, and $A$ a self-adjoint operator with domain $D_{A} \subseteq \mathcal{H}$. Assume that $\lambda_0 \in \rho(A)$, where $\rho(A)$ is the resolvent set of $A$. For any $z \in \rho(A)$, let $R_{A}(z)=(A - z I)^{-1}$ be the resolvent of $A$.

Choose $\lambda \neq \lambda_0$. Then it is well known that $\lambda \in \rho(A)$ if and only if $(\lambda - \lambda_0)^{-1} \in \rho(R_{A}(\lambda_0))$ (see e.g. Schmudgen, Unbounded Self-adjoint operators on Hilbert Space, Proposition 2.10). So we have (note that by the spectral theorem $\sigma(A)$ is nonempty): \begin{equation} \sigma(R_{A}(\lambda_0)) \backslash \{0\} = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} If $A$ is a bounded operator on $\mathcal{H}$, then $0 \in \rho(R_{A}(\lambda_0))$, so that in this case, being $\sigma(A)$ closed, we have \begin{equation} \sigma(R_{A}(\lambda_0)) = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \} = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} Now suppose that $A$ is unbounded. In this case $0 \in \sigma(R_{A}(\lambda_0))$. If we could prove that $0$ is not an isolated point of $\sigma(R_{A}(\lambda_0))$ (which is the same to say that $\sigma(A)$ is not bounded), we could conclude also in this case that \begin{equation} \sigma(R_{A}(\lambda_0)) = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} So my question is the following: if $A$ is unbounded, can $\sigma(A)$ be bounded?

PS This question arouse from the answer given by TrialAndError in this post Norm of the Resolvent

Answer 1

Take $r > \max \sigma(A)$. Then $R_A(r)$ is self-adjoint and bounded. If $0$ is not in its spectrum, then $A = (R_A(r)^{-1}+rI$ is bounded. If $0$ is in its spectrum, it is an isolated point of the spectrum and therefore must be an eigenvalue: $R_A v = 0$ for some $v \in \mathcal H$. But that is impossible since $R_A(r) = (A-rI)^{-1}$, i.e. $R_A(r) v = u$ where $u \in D_A$ and $(A-rI) u = v$.

Answer 2

You mentioned you like Complex Analysis. So I thought I'd offer a proof using Complex Analysis applied to the resolvent. The proof comes down to evaluating the integral around all finite singularities of $(\lambda I-A)^{-1}x$ by determining the residue at $\infty$, which turns out to be $x$. This equivalence forces the completeness of spectral expansions for normal operators. It's a type of Complex Analysis conservation law that allows you to expand $x$ in terms of integrals in the finite plane. If the singularities are all discrete in the finite plane, you end up with an eigenfunction expansion of $x$. Continuous spectrum can lead to integral expansions, such as the classical Fourier integral expansions. More generally, the Spectral Theorem for sefadjoint operators can be proved using this conversation law; completeness is established by knowing that the resiude at infinity of $(\lambda I-A)^{-1}$ is $I$. So the technique is worth learning.

Suppose $A$ is a closed densely-defined normal operator on a Complex Hilbert space $\mathcal{H}$, and suppose that $\sigma(A)$ is a bounded set. By the previous problem you referenced, $$ \|(\lambda I -A)^{-1}\| \le \frac{1}{\mbox{dist}(\lambda,\sigma(A))}. $$ Therefore, $\lim_{\lambda\rightarrow\infty} (\lambda I-A)^{-1}=0$, and, for a fixed $x\in\mathcal{D}(A)$, the following limit is uniform in $\lambda$: $$ \lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}x=\lim_{\lambda\rightarrow\infty}x+(\lambda I-A)^{-1}Ax = x. $$ If $x\in\mathcal{D}(A)$, and if $R$ is large enough that $\sigma(A)\subseteq \{ \lambda : |\lambda| < R \}$, then $$ \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}xd\lambda = \lim_{R\rightarrow\infty}\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda(\lambda I-A)^{-1}x\frac{d\lambda}{\lambda}=x. $$ Because $\mathcal{D}(A)$ is dense and $\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda$ is a bounded operator, then $$ \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda =I. $$ For $x\in\mathcal{D}(A)$, \begin{align} Ax & = \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}Ax\,d\lambda \\ & = \frac{1}{2\pi i}\oint_{|\lambda|=R}-x+\lambda(\lambda I-A)^{-1}x\,d\lambda \\ & = \left(\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda (\lambda I-A)^{-1}d\lambda\right)x \end{align} So $A$ is bounded on $\mathcal{D}(A)$, which also forces $\mathcal{D}(A)=\mathcal{H}$ because $A$ is closed.