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 $A,B \in M_n(\mathbb{K})$, $\rho(AB) \leq \rho(A) \rho(B)$.