# Basic spectral properties of banach algebra

Definition Let ${B}$ be a banach space, ${\mathcal{L}(B)}$ be the space of bounded linear operator from ${B \rightarrow B}$. If ${T \in \mathcal{L}(B)}$, then say a complex number ${\lambda}$ is in the resolvent set ${\rho(T)}$ of ${T}$ if ${\lambda I - T}$ has a bounded inverse. Then we define the resolvent of ${T}$ at ${\lambda}$ to be ${R_\lambda(T) = (\lambda I -T)^{-1}}$. If ${\lambda \not\in \rho(T)}$, then ${\lambda}$ is said to be in the spectrum ${\sigma(T)}$ of T.

We will learn some properties of the resolvent and the spectrum.

Lemma 1 ${\mathcal{L}(B)}$ is a banach algebra under composition.

Proof: Let ${T,P \in \mathcal{L}(B)}$, then

$\displaystyle \|TP\|_{\mathcal{L}(B)} = \sup_{\|x\|_B =1} \|TPx\|_B \le \sup_{\|x\|_B =1} \|T\|_{\mathcal{L}(B)} \|Px\|_B = \|T\|_{\mathcal{L}(B)} \|P\|_{\mathcal{L}(B)}$

$\Box$

Lemma 2 Let ${A}$ be a banach algebra with identity ${I}$. If ${x \in A}$ with ${\|x\|_A <1}$, then ${(I-x)^{-1}}$ exists.

Proof: Since ${A}$ is a banach algebra, we have ${\|x^n\| \le \|x\|^n}$. Then ${y:= \sum_0^\infty x^n}$ is absolutely convergent in ${A}$. Note that

$\displaystyle I - x^{n+1} = (I-x)(I +x +x^2 + \cdots + x^n) = (I +x +x^2 + \cdots + x^n) (I-x).$

Let ${n \rightarrow \infty}$ we will have ${I = (I-x)y = y(I-x)}$. Therefore ${y =(I-x)^{-1} }$. $\Box$

Theorem (First resolvent identity) If ${\lambda, \mu \in \rho(T)}$, then ${R_\lambda(T)}$ and ${R_\mu(T)}$ commute. Furthermore,

$\displaystyle R_\lambda(T) - R_\mu(T) = (\lambda - \mu) R_\lambda(T) R_\mu(T).$

Proof:

$\displaystyle R_\lambda(T) - R_\mu(T) = R_\lambda(T)(\mu I - T) R_\mu(T) - R_\lambda(T)(\lambda I - T) R_\mu(T)$

$\displaystyle = (\lambda - \mu) R_\lambda(T) R_\mu(T).$

Since we can interchange ${\mu}$ and ${\lambda}$, ${R_\lambda(T)}$ and ${R_\mu(T)}$ commute. $\Box$

Theorem ${\rho(T)}$ is an open subset of ${{\mathbb C}}$. Fixed ${T}$, ${\lambda \mapsto R_\lambda (T)}$ is an analytic ${\mathcal{L}(B)}$– valued function.

Proof: If ${\lambda_0 \in \rho(T)}$, first define ${\tilde{R}_\lambda(T) = R_{\lambda_0}(T) \sum_{n=0}^\infty (\lambda_0 - \lambda)^n (R_{\lambda_0}(T))^n}$. From the previous lemma, we know that if ${|\lambda_0 - \lambda| < \frac{1}{\|R_{\lambda_0}(T) \|}}$, the series defines the inverse of ${I - (\lambda_0 - \lambda)R_{\lambda_0}(T)}$. Therefore,

$\displaystyle \tilde{R}_\lambda(T) = R_{\lambda_0}(T) (I - (\lambda_0 - \lambda)R_{\lambda_0}(T))^{-1} = ((\lambda_0 I -T) - (\lambda_0 - \lambda))^{-1} = R_\lambda(T)$

if ${|\lambda_0 - \lambda| < \frac{1}{\|R_{\lambda_0}(T) \|}}$. So ${\rho(T)}$ is open. Since ${R_\lambda(T)}$ can be expressed as infinite series as ${\tilde{R}_\lambda(T)}$, it is analytic. $\Box$

Now we are going to show the spectrum of ${T \in \mathcal{L}(B)}$ is non-empty (actually works for any complex banach algebra), which gives a nice corollary by Gelfand and Mazur. Before proving that, we need a banach-valued version of Liouville theorem:

Theorem (Liouville theorem for banach-valued function) Let ${B}$ be a banach space. If ${f}$ is a function from ${{\mathbb C} \rightarrow B}$, which is entire and bounded, then ${f}$ is constant.

Proof: First note that if ${l \in B^*}$, then the function ${z \mapsto \langle l , f(z) \rangle_{B^* \times B}}$ is a complex-valued analytic function.

Pick a ${z_0 \in {\mathbb C}}$ and let ${g(z) = f(z) - f(z_0)}$. For any ${l \in B^*}$ , we have ${z \mapsto \langle l , g(z) \rangle_{B^* \times B}}$ is complex-valued analytic function and bounded in ${{\mathbb C}}$. Then by Liouville theorem ${\langle l , g(z) \rangle_{B^* \times B}}$ is a constant, and ${g(z_0)=0}$, therefore ${\langle l , g(z) \rangle_{B^* \times B} = 0}$ for all ${l \in B^*}$.

Assume ${g(z) \neq 0}$, by Hahn-Banach theorem there exists a nonzero linear bounded functional ${L \in B^*}$ such that ${\langle L , g(z) \rangle_{B^* \times B} \neq 0}$. By contraposition this implies ${g(z)=0}$, and hence ${f(z) = f(z_0)}$ is a constant. $\Box$

Remark: There might be a proof without using axiom of choice. I just don’t know it yet.

Theorem ${\sigma(T)}$ is a non-empty compact subset in ${{\mathbb C}}$.

Proof: First define ${\tilde{R}_\lambda(T) = \frac{1}{\lambda} \sum_0^\infty \left( \frac{T}{\lambda} \right)^n}$. By the previous lemma again, if ${|\lambda| > \| T\|}$, then ${\tilde{R}_\lambda(T)}$ defines the inverse of ${\lambda I -T}$. Therefore ${\sigma(T) \subset \{ \lambda\in {\mathbb C} : |\lambda| \le \|T\| \}}$ is a bounded subset. Since ${\rho(T)}$ is open, ${\sigma(T)}$ is closed. Hence ${\sigma(T)}$ is compact.

To see ${\sigma(T)}$ is non-empty. Assume it is, then ${R_\lambda(T)}$ is entire by the previous theorem. For ${|\lambda| > \|T\|}$, ${\| R_\lambda(T)\| =\| \tilde{R}_\lambda(T) \| \le \frac{C}{\lambda}}$, and goes to 0 as ${\lambda}$ goes to ${\infty}$. Then by Liouville theorem, ${R_\lambda(T) \equiv 0}$, which is not invertible. This gives a contradiction. $\Box$

Remark: As mentioned before, this argument works for any complex banach algebra.

Corollary (Gelfand-Mazur) Every complex banach division algebra ${A}$ is an isometric isomorphic to ${{\mathbb C}}$.

Proof: Since ${A}$ is a division algebra, the only non-invertible element is 0. By the previous theorem, we know that for any ${T \in A}$, there exists a ${\lambda \in {\mathbb C}}$ such that ${\lambda I - T =0}$. Then ${\|T\| = \| \lambda I\| = | \lambda |}$. So the map ${T \mapsto \lambda}$ gives an isometric isomorphism from ${A}$ to ${{\mathbb C}}$. $\Box$