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)}


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).


\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


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