# Lecture 10: Spectrum and Resolvent ###### tags: 224a 2020 Topics: $M_x$ has no eigenvectors Definition of spectrum and resolvent set. Point, residual, and continuous spectrum. Compactness and nonemptiness of spectrum. Spectrum of Selfadjoint and Compact Operators See the class notes for full proofs. A few are available below. ### Nonemptiness of the Spectrum **Theorem.** If $A\in L(H)$ then $\sigma(A)\neq \varnothing.$ *Proof.* Assume $\sigma(A)$ is empty, i.e., $R_A(z)$ is an entire function. Then for any fixed $v,w\in H$ the scalar function $f(z):=v^*R_A(z)w$ is also entire. We saw in the previous lecture that $\|R_A(z)\|\rightarrow 0$ as $|z|\rightarrow\infty$, so the same is true of $f(z)$. Choose a closed disk $D$ such that $|f(z)|\le 1$ outside $D$; since $f$ is continuous on $D$, it is also bounded on $D$, and must be bounded everywhere. By Liouville's theorem $f$ must be constant, in fact zero by its behavior at infinity. Since this is true for every $v,w$, we conclude that $R_A(z)$ is identically zero, for every $z$, which is absurd since $\rho(A)\neq\varnothing$.$\square$ ### Spectrum of Selfadjoint Operators $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\ran}{\mathrm{Ran}}$ **Theorem.** If $A=A^*\in L(H)$ then $\sigma(A)\subset \R$ and $\sigma_r(A)=\varnothing$ *Proof.* By a $(v,Aw)=(Av,w)$ argument, $\sigma_p(A)\subset \R$. Noting that $\sigma(A)=\overline{\sigma(A)}$, we have $$\ker(\lambda-A)\neq\{0\}\iff \ker(\lambda^*-A)\neq\{0\}\iff \ran(\lambda-A)^\perp \neq \{0\},$$ so $\lambda\in\sigma_p(A)$ if and only if $\ran(\lambda-A)$ is not dense in $H$, implying that $\sigma_r(A)=\varnothing$. Now observe that for $\lambda=a+ib$ with $b\neq 0$, one has $$\|(a+ib-A)x\|^2 = \|(a-A)x\|^2 + |b|^2\|x\|^2 \ge |b|^2\|x\|^2.$$ This shows that $\lambda-A$ is injective, and that $\ran(\lambda-A)$ is closed, so $\lambda$ is not in $\sigma_p$ or $\sigma_c$. Since the residual spectrum is empty, it cannot be in the spectrum at all.$\square$