Let (X,|| \cdot ||) be a Banach space over \mathbb{C} and A \in \mathcal{L}(X) be a continuous linear map from X to itself. The usual norm on \mathcal{L}(X) is M \mapsto \sup\limits_{||x||=1} ||Mx||, also denoted by || \cdot||. Setting \sigma(A)= \{ \lambda \in \mathbb{C} \mid A-\lambda \text{Id} \ \text{not invertible} \}, we want to show:

Theorem: \lim\limits_{n \to + \infty} ||A^n||^{1/n} = \max \{ |\lambda| \mid \lambda \in \sigma(A) \}.

The proof is interesting because it uses both functional analysis and complex analysis. The note finishes with some corollaries about complex Banach algebra and matrices.

Proof. For convenience, let \rho(A)= \mathbb{C} \backslash \sigma(A).

Claim 1: \sigma(A) is compact and included in the disk D(0,||A||).

The map \varphi : \left\{ \begin{array}{ccc} \mathbb{C} & \to & \mathcal{L}(X) \\ \lambda & \mapsto & A- \lambda \text{Id} \end{array} \right. is continuous and \rho(A)= \varphi^{-1}(\text{Isom}(X)) where \text{Isom}(X) \subset \mathcal{L}(X) is the set of isomorphisms. So we can show that \sigma(A) is closed just by noticing that \text{Isom}(X) is open in \mathcal{L}(X). Let u \in \text{Isom}(X) and h \in \mathcal{L}(X) such that ||h||<||u||. Then \sum\limits_{n \geq 0} (-u^{-1}h)^k is absolutely convergent, and a fortiori convergent since X is complete; moreover, it is the inverse of \text{Id}+u^{-1}h so u+h=u(\text{Id}+u^{-1}h) is invertible and \text{Isom}(X) is finally closed.

Then, if \lambda >||A|| then \sum\limits_{n \geq 0} \frac{1}{\lambda^n} A^n is absolutely convergent and it can be verifyed that (A-\lambda \text{Id})^{-1}=- \frac{1}{\lambda} \sum\limits_{n \geq 0} \frac{1}{\lambda^n} A^n. Therefore, \sigma(A) \subset D(0,||A||). In particulier, \sigma(A) is compact. \square

Claim 2: \sigma(A) is nonempty.

First, introduce R_A : \left\{ \begin{array}{ccc} \rho(A) & \to & \mathcal{L}(X) \\ \lambda & \mapsto & (A-\lambda \text{Id})^{-1} \end{array} \right. and notice that it is holomorphic. Indeed:

\begin{array}{ll} R_A(\lambda)- R_A(\lambda_0) & = (A-\lambda \text{Id})^{-1}(A-\lambda_0 \text{Id})(A- \lambda_0 \text{Id})^{-1}-(A- \lambda \text{Id})^{-1}(A-\lambda \text{Id})(A-\lambda_0 \text{Id})^{-1} \\ & =(A-\lambda \text{Id})^{-1} ((A-\lambda_0 \text{Id})-(A-\lambda \text{Id}))(A-\lambda_0 \text{Id})^{-1} \\ & = (\lambda-\lambda_0) (A-\lambda \text{Id})^{-1} (A-\lambda_0 \text{Id})^{-1} \end{array}

for \lambda, \lambda_0 \in \rho(A). By continuity, R_A is holomorphic with R_A'(\lambda)=(A-\lambda \text{Id})^{-2}.

For all \varphi \in \mathcal{L}(X)', let f_{\varphi} =\varphi \circ R_A : \rho(A) \to \mathbb{C}. If |\lambda| >||A||, then f_{\varphi}(\lambda)=- \frac{1}{\lambda} \sum\limits_{n \geq 0} \frac{1}{\lambda^n} \varphi(A^n); in particular, | f_{\varphi}(\lambda) | \leq \frac{||\varphi||}{|\lambda|-||A||} so \lim\limits_{|\lambda| \to + \infty} |f_{\varphi}(\lambda)|=0 and f_{\varphi} is bounded. According to Liouville’s theorem, if \rho(A)= \mathbb{C} then f_{\varphi} is constant (in fact identically zero because of the preceding limit). However, according to Hahn-Banach theorem, there exists \varphi \in \mathcal{L}(X)' such that f_{\sigma} is non identically zero. Therefore, \rho(A) \neq \mathbb{C} ie. \sigma(A) \neq \emptyset. \square

Thus, using claims 1 and 2, s_A=\max \{ |\lambda | \mid \lambda \in \sigma(A) \} is well defined.

Claim 3: For all \sigma > s_A, \limsup\limits_{n \to + \infty} ||A^n||^{1/n} \leq \sigma.

For any \varphi \in \mathcal{L}(X)', f_{\varphi} being holomorphic on \mathcal{C}\backslash (0,s_A), according uniqueness of power series expansion, f(\lambda)=- \frac{1}{\lambda} \sum\limits_{n \geq 0} \varphi(A^n) for every \lambda \in \mathbb{C} such that |\lambda|>s_A. In particuler, \sup\limits_{n \geq 0} |\varphi(\lambda^{-n}A^n) | <+ \infty for every \varphi \in \mathcal{L}(X)'. Therefore, using the lemma following the proof, there exists C(\lambda)>0 such that |\lambda|^{-n}||A^n|| \leq C(\lambda) for all n \geq 0. Therefore, \limsup\limits_{n \to + \infty} ||A^n||^{1/n} \leq \limsup\limits_{n \to + \infty} C(\lambda)^{1/n} |\lambda|= |\lambda|. \square

To conclude our proof, notice that the equality A^n-\lambda^n \text{Id}= (A-\lambda \text{Id}) \sum\limits_{k=0}^n \lambda^k A^{n-k-1} implies that if \lambda \in \sigma(A) then \lambda^n \in \sigma(A^n); using claim 1, we get |\lambda| \leq ||A^n||^{1/n}. Finally, s_A \leq \liminf\limits_{n \to + \infty} ||A^n||^{1/n} \leq \limsup\limits_{n \to + \infty} ||A^n||^{1/n} \leq s_A. \square

Lemma: Let B be a Banach space over \mathbb{C} and (B_n) a sequence in B. If for all \varphi \in B', \sup\limits_{n \geq 0} |\varphi(B_n)|<+ \infty, then \sup\limits_{n \geq 0} ||B_n||<+ \infty.

Proof. For any n \geq 0, there exists a linear functional \sigma_n : B' \to \mathbb{C} such that \sigma_n(\varphi)= \varphi(B_n) for all \varphi \in B'. Moreover ||\sigma_n||= \sup\limits_{||\varphi||=1} |\sigma_n(\varphi) | = \sup\limits_{||\varphi||=1} |\varphi(B_n)| \leq ||B_n|| so \sigma_n \in B''. According to Hahn-Banach theorem, there exists \varphi \in B' such that \varphi(B_n)=||B_n|| and ||\varphi||=1, so ||\sigma_n|| \geq |\sigma_n(\varphi)|= ||B_n||. Finally, ||\sigma_n||=||B_n||. By assumption, for any \varphi \in B', \sup\limits_{n \geq 0} | \sigma_n(\varphi) | <+ \infty; using Banach-Steinhaus theorem, we conclude that \sup\limits_{n \geq 0} ||B_n||= \sup\limits_{n \geq 0} ||\sigma_n||<+ \infty. \square

A simple corollary is the following (related to Frobenius theorem and Hurwitz theorem):

Theorem: (Gelfand-Mazur) Let X be a complex Banach algebra. If X is a division algebra then X is isometrically isomorphic to \mathbb{C}.

Proof. Let x \in X \backslash \{0\} such that ||x||=1. Suppose that there exists y \in X \backslash x\mathbb{C}. By assumption, y- \lambda x is invertible for all \lambda \in \mathbb{C} hence \sigma(yx^{-1})= \emptyset, contradicting claim 2. Therefore, X= x \mathbb{C}. Now, it is easy to verify that z \mapsto xz from \mathbb{C} to X is an isometric isomorphism. \square

Notice that if Y is a subspace of a complex Banach space X then the spectral radius formula still holds in Y (but with the spectral radius computed in Y); in particular, it can be useful to extend the formula to some real Banach spaces. For instance:

Property: Let A \in M_n(\mathbb{K}) be a matrix with \mathbb{K}= \mathbb{R} or \mathbb{C}. Then \lim\limits_{n \to + \infty} ||A^n||^{1/n} = \rho(A).

Corollary: For all A,B \in M_n(\mathbb{K}), \rho(AB) \leq \rho(A) \rho(B).

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