# Lecture 11: Spectral Radius, Cts Functional Calculus ###### tags: `224a 2020` ### Principle of Uniform Boundedness (i.e., Banach-Steinhaus Theorem) **Theorem.** If $T_n$ is a collection of operators in $L(X,Y)$ such that for every $v\in X$: $$ \sup_{n} \|T_nv\|<\infty$$ then $$\sup_n \|T_n\|<\infty.$$ *Proof.* We show the contrapositive. Choose a subsequence $T_n$ such that $\|T_n\|\ge 4^{n}$. Using the Lemma below, construct a sequence of vectors inductively by $x_0=0$ and $$\|x_n-x_{n-1}\|\le 3^{-n},\qquad \|T_n x_n\|\ge \|T_n\|3^{-n}.$$ The sequence $x_n$ is Cauchy so it converges to some $x$; moreover, we have the explicit error bound: $$\|x_n-x\|\le \sum_{j=n+1}^\infty \|x_j-x_{j-1}\|\le \sum_{j=n+1}^\infty 3^{-j}=3^{-n-1}(3/2)=3^{-n}/2.$$ Thus, we must have $$\|T_nx\|\ge \|T_nx_n\|-\|T_n(x-x_n)\|\ge (1/2)\|T_n\|(1/3)^n\rightarrow \infty,$$ as desired.$\square$ **Lemma.** For any vector $x\in H$ and $r\ge 0$, there is a vector $\|x'-x\|\le r$ with $\|Tx'\|\ge \|T\|r$. *Proof.* Choose a vector $e$ with $\|Te\|=\|T\|\|e\|=\|T\|r$ and observe that $$ 2(\|T(x+e)\|\lor\|T(x-e)\|)\ge \|T(x+e)\|+\|T(-x+e)\|\ge \|T(2e)\|=2\|T\|.$$ ### Gelfand Spectral Radius Formula $\newcommand{\C}{\mathbb{C}}$ We begin by recalling some facts from (scalar valued) complex analysis. **Fact.** If $f:D(z_0,r)\rightarrow \C$ is analytic on an open disk, then the power series $f(z)=\sum_k a_k (z-z_0)^k$ converges absolutely in the interior of the disk. **Corollary.** The radius of convergence of a power series of an analytic function $f:\Omega\rightarrow \C$ at at a point $z_0\in \Omega$ is equal to $\max\{r: D(z_0,r)\subset \Omega\}$. Let $r(A):=\lim_{n\rightarrow\infty}\|A^n\|^{1/n}$. This limit exists because $\log\|A^n\|$ is a subadditive sequence due to submultiplicativity of the norm (details are left to the homework). $\newcommand{\spr}{\mathsf{spr}}$ The power series $$ S(z):=\sum_{k=0}^\infty A^k/z^k$$ converges absolutely to $R_A(z):=(z-A)^{-1}$ whenever $|z|>r(A)$, so we conclude that $\{|z|>r(A)\}\subset \rho(A)$, whence $\spr(A):=\max_{\lambda\in \sigma(A)}|\lambda|\le r(A)$. We now show that this inequality is always tight. **Theorem. (Gelfand)** $r(A)=\spr(A)$. *Proof.* Since $\{|z|>\spr(A)\}\subset \rho(A)$, $$r_A(y):=(y^{-1}-A)^{-1}=y(I-yA)^{-1}$$ is analytic in ${D(0,1/\spr(A))}$. This implies that for every $v,w$: $$f_{v,w}(y):= v^*r_A(y)w$$ is a (scalar) analytic function in $D(0,1/\spr(A))$. Therefore the series obtained by Taylor expanding at zero: $$ s_{v,w}(y):=\sum_{k}(v^*A^kw) y^k$$ converges absolutely in $D(0,1/\spr(A))$. Thus, for any $1/t<1/\spr(A)$, the terms of the series must vanish and we have $$\sup_k |v^*A^kw|/t^k<\infty$$ whenever $t>\spr(A)$. By the principle of uniform boundedness, this implies $$\sup_k \|A^k\|/t^k<\infty$$ whenever $t>\spr(A)$. Taking $k^{th}$ roots shows that $r(A)<t$ whenever $t>\spr(A)$, so $r(A)\le \spr(A)$, as desired.$\square$ **Remark.** The above argument relied on the fact that a familiar fact from (scalar) complex analysis generalizes verbatim to "operator valued" analytic functions. The key device that allowed us to do this was the uniform boundedness principle. Though we we will not develop it here, this allows one to essentially generalize all of complex analysis to the operator valued setting. ## Continuous Functional Calculus Let $C(X)$ denote the Banach space of continuous functions on a compact set $X$ with the sup norm. **Theorem. (Continuous Functional Calculus)** If $A=A^*$ there is a linear map $\phi=\phi_A:C(\sigma(A))\rightarrow L(H)$ with the following properties: 1. $*$-homomorphism: $\phi(fg)=\phi(f)\phi(g), \phi(1)=I, \phi(\overline{f})=\phi(f)^*$. 4. Spectral Mapping: $\sigma(\phi_A(f))=f(\sigma(A))$ 2. Isometry: $\|\phi(f)\|=\|f\|_{\infty}$. 5. Eigenvector Mapping: If $A\psi=\lambda\psi$ then $\phi_A(f)\psi=f(\lambda)\psi$. 3. Positivity: If $f\ge 0$ then $\phi(f)\ge 0$. With the normalization that $\phi_A(x)=A$, the above is unique (in fact, assuming only (1) and continuity). We denote it by $\phi_A(f)=f(A)$. To prove the existence of such a mapping, we first show that it exists for the dense subspace $P\subset C(\sigma(A))$ of polynomial functions. **Lemma 1.(Spectral Mapping for Polynomials)** If $p\in \C[x]$, $p(\sigma(A))=\sigma(p(A))$. *Proof.* Reed and Simon VII.1$\square$ **Lemma 2.(Isometry for Polynomials)** If $p\in \C[x]\subset C(\sigma(A))$, $$\|p(A)\| = \|p\|_\infty.$$ The BLT theorem now implies that $\phi_A(\cdot)$ has a unique extension from $P$ to $C(\sigma(A))$. Note that self-adjointness was used in two places: (1) the polynomial functions are dense in $C(\sigma(A))$ (2) $r(A)=\|A\|$ for self-adjoint matrices. The theorem is simply not true without self-adjointness; consider the $2\times 2$ Jordan block with $\sigma(J)=\{0\}$ but $\|J\|=1$. The functional calculus gives a way to `probe' the spectrum of an operator using continuous functions. One example of this technique is the following. **Theorem.** If $A$ is self-adjoint $\lambda$ is an isolated point in $\sigma(A)$, then $\lambda\in \sigma_p(A)$. *Proof.* Since $\lambda$ is isolated, there is an $f\in C(\sigma(A))$ such that $f(\lambda)=1$ and $f(x)=0$ on $\sigma\setminus\{\lambda\}$. We also have $f^2=f$ and $\overline{f}=f$, so by the functional calculus, $f(A)$ must be a nonzero orthogonal projection $P$. Since $(x-\lambda)f(x)\equiv 0$, we have $(A-\lambda)P=0$, whence $\lambda\in\sigma_p(A)$.$\square$