Lecture 13: Spectral Radius Formula, Examples

tags: 224a

Nonemptiness of the Spectrum

Theorem. If

$A\in L(H)$ then

$\sigma (A)\ne \varnothing .$
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

$f(z):=v^{\ast }{R}_A(z)w$ is also entire. We saw in the previous lecture that

$\Vert R_A(z)\Vert \to 0$ as

$|z|\to \mathrm{\infty }$, 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

$\rho (A)\ne \varnothing$.

$◻$

Remark. This is the analogue of the fundamental theorem of algebra, which is also ultimately also a consequence of Liouville's theorem applied to the function

$1/p(z)$ for a polynomial

$p$. Note that boundedness is a crucial assumption; as we will see later, there are densely defined unbounded operators for which the spectrum is empty!

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

$v\in X$:

$$\underset{n}{sup}\Vert T_{n}v\Vert <\mathrm{\infty }$$
then

$$\underset{n}{sup}\Vert T_n\Vert <\mathrm{\infty }.$$
Proof. We show the contrapositive. Choose a subsequence

$T_n$ such that

$\Vert T_n\Vert \ge 4^n$. Using the Lemma below, construct a sequence of vectors inductively by

$x_0=0$ and

$$\Vert x_n-x_{n-1}\Vert \le 3^{-n},\Vert T_n{x}_n\Vert \ge \Vert T_n\Vert 3^{-n}.$$
The sequence

$x_n$ is Cauchy so it converges to some

$x$; moreover, we have the explicit error bound:

$$\Vert x_n-x\Vert \le \sum _{j=n+1}^{\mathrm{\infty }}\Vert x_j-x_{j-1}\Vert \le \sum _{j=n+1}^{\mathrm{\infty }}{3}^{-j}=3^{-n-1}(3/2)=3^{-n}/2.$$
Thus, we must have

$$\Vert T_{n}x\Vert \ge \Vert T_n{x}_n\Vert -\Vert T_n(x-x_n)\Vert \ge (1/2)\Vert T_n\Vert (1/3{)}^n\to \mathrm{\infty },$$
as desired.

$◻$

Lemma. For any vector

$x\in H$ and

$r\ge 0$, there is a vector

$\Vert x^{\prime }-x\Vert \le r$ with

$\Vert T{x}^{\prime }\Vert \ge \Vert T\Vert r$.
Proof. Choose a vector

$e$ with

$\Vert Te\Vert =\Vert T\Vert \Vert e\Vert =\Vert T\Vert r$ and observe that

$$2(\Vert T(x+e)\Vert \vee \Vert T(x-e)\Vert )\ge \Vert T(x+e)\Vert +\Vert T(-x+e)\Vert \ge \Vert T(2e)\Vert =2\Vert T\Vert .$$

Gelfand Spectral Radius Formula

$$
We begin by recalling some facts from (scalar valued) complex analysis.

Fact. If

$f:D(z_0,r)\to \mathbb{C}$ 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

$f:\mathrm{\Omega }\to \mathbb{C}$ at at a point

$z_0\in \mathrm{\Omega }$ is equal to

$max\{r:D(z_0,r)\subset \mathrm{\Omega }\}$.

Let

$r(A):=\underset{n\to \mathrm{\infty }}{lim}\Vert A^n{\Vert }^{1/n}$. This limit exists because

$\log \Vert A^n\Vert$ is a subadditive sequence due to submultiplicativity of the norm (details are left to the homework).

$$
The power series

$$S(z):=\sum _{k=0}^{\mathrm{\infty }}{A}^k/z^k$$
converges absolutely to

$R_A(z):=(z-A{)}^{-1}$ whenever

$|z|>r(A)$, so we conclude that

$\{|z|>r(A)\}\subset \rho (A)$, whence

$\mathsf{spr}(A):=\underset{\lambda \in \sigma (A)}{max}|\lambda |\le r(A)$. We now show that this inequality is always tight.

Theorem. (Gelfand)

$r(A)=\mathsf{spr}(A)$.
Proof. Since

$\{|z|>\mathsf{spr}(A)\}\subset \rho (A)$,

$$r_A(y):=(y^{-1}-A{)}^{-1}=y(I-yA{)}^{-1}$$ is analytic in

$D(0,1/\mathsf{spr}(A))$. This implies that for every

$v,w$:

$$f_{v,w}(y):=v^{\ast }{r}_A(y)w$$ is a (scalar) analytic function in

$D(0,1/\mathsf{spr}(A))$. Therefore the series obtained by Taylor expanding at zero:

$$s_{v,w}(y):=\sum _k(v^{\ast }{A}^{k}w)y^k$$
converges absolutely in

$D(0,1/\mathsf{spr}(A))$. Thus, for any

$1/t<1/\mathsf{spr}(A)$, the terms of the series must vanish and we have

$$\underset{k}{sup}|v^{\ast }{A}^{k}w|/t^k<\mathrm{\infty }$$
whenever

$t>\mathsf{spr}(A)$. By the principle of uniform boundedness, this implies

$$\underset{k}{sup}\Vert A^k\Vert /t^k<\mathrm{\infty }$$
whenever

$t>\mathsf{spr}(A)$. Taking

$k^{th}$ roots shows that

$r(A)<t$ whenever

$t>\mathsf{spr}(A)$, so

$r(A)\le \mathsf{spr}(A)$, as desired.

$◻$

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.

Measure Spaces

Spectrum of Selfadjoint Operators

$$

$$
Theorem. If

$A=A^{\ast }\in L(H)$ then

$\sigma (A)\subset \mathbb{R}$ and

${\sigma }_r(A)=\varnothing$
Proof. By a

$(v,Aw)=(Av,w)$ argument,

${\sigma }_p(A)\subset \mathbb{R}$. Noting that

$\sigma (A)=\overline{\sigma (A)}$, we have

$$\ker (\lambda -A)\ne \{0\}⟺\ker ({\lambda }^{\ast }-A)\ne \{0\}⟺\mathrm{Ran}(\lambda -A{)}^{\perp }\ne \{0\},$$
so

$\lambda \in {\sigma }_p(A)$ if and only if

$\mathrm{Ran}(\lambda -A)$ is not dense in

$H$, implying that

${\sigma }_r(A)=\varnothing$.

Now observe that for

$\lambda =a+ib$ with

$b\ne 0$, one has

$$\Vert (a+ib-A)x{\Vert }^2=\Vert (a-A)x{\Vert }^2+|b{|}^2\Vert x{\Vert }^2\ge |b{|}^2\Vert x{\Vert }^2.$$
This shows that

$\lambda -A$ is injective, and that

$\mathrm{Ran}(\lambda -A)$ 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.

$◻$

Defn. (Measure Space) The Borel sigma algebra of a topological space

$X$ is the collection of all sets obtainable from the open sets by countable union, countable intersection, and complements. A Borel measure is a countable additive measure defined on the Borel sets satisfying certain regularity properties (see RS I.4 for details) which will always be satisfied in this course. We will refer to the pair

$(X,\mu )$ as a measure space, and call it finite if

$\mu (X)<\mathrm{\infty }$.

Defn.(Lebesgue-Stieltjes Integral) Given a nonnegative measure

$\mu$ on

$X$, a function

$f:X\to \mathbb{C}$ is measurable if

$\mu (\{f(x)>t\})$ is measurable for every

$t$. We then define its Lebesgue-Stieltjes integral as:

$${\int }_{X}f(x)d\mu (x):={\int }_0^{\mathrm{\infty }}\mu (\{f>t\})dt.$$

Lebesgue measure is a special case of this, but there are other very different examples, e.g., the discrete measure

${\delta }_x$ supported on one point, and mixtures of such measures.

Basic Examples

1. Shift Operators.

   $\sigma (L)=\{|z|\le 1\}$, with point spectrum in the interior. Adjoint
   $R=L^{\ast }$ has no point spectrum. Residual and continuous spectrum on the homework.

2. Multiplication Operators.

   $g\in L^{\mathrm{\infty }}(X,\mu )$ for a finite measure space
   $\mu$ on
   $X\subset \mathbb{R}$. If
   $g$ is real-valued,
   $\sigma (M_g)=\{y:\mu (g^{-1}(y-ϵ,y+ϵ))>0\mathrm{\forall }ϵ>0\}$, the essential range of
   $g$.

3. Adjacency matrix of a path:

   $(Ax{)}_n=x_{n-1}+x_{n+1}$ on
   $H={\ell }^2(\mathbb{Z})$. Consider the isometry
   $U:H\to L^2(S^1)$ via Fourier series:
   
   $$Ux=\sum _{n\in \mathbb{Z}}{e}^{inx}{x}_n.$$
   In this basis, the operator is
   
   $$(UA{U}^{-1})f=e^{ix}f+e^{-ix}f=2\cos (x)f(x).$$
   Since conjugating by an isometry does not change the resolvent set or spectrum,
   $$\sigma (A)=\sigma (UA{U}^{-1})=\mathrm{ess}-\mathrm{range}(2\cos (x))=[-2,2].$$

In the next lecture we will show that the trick we used in example 3 always works.