Spectral radius of $f(x)$ where $f$ is a Banach algebra map $f : A \to \mathbb{C^{n \times n}}$

Question

I have a question regarding the spectral radius. Let $A$ be a Banach algebra and $f : A \to \mathbb{C^{n \times n}}$ a Banach algebra map into the Banach algebra of square matrices. How can I show that the spectral radius of $f(x), \ x \in A$ is less or equal $||x||$? I know that the spectral radius of $f(x)$ must be smaller than the norm of $f(x)$ but I don't know how to proceed.

Answer 1

I suppose that by a "Banach algebra map" you mean that $f$ is linear and multiplikative.

By $e$ we denote the identity element in $A$. Hence $f(e)=I$.

Let us denote the spectral radius of $f(x)$ by r and suppose that $||x|| <r$. We find some $\mu \in \sigma(f(x))$ such that $|\mu|=r$. Then we have $|\mu|>||x||$ and derive that $z:=\mu e -x$ is invertible. It follows that

$I=f(z)f(z^{-1})=f(z^{-1})f(z)$. Hence $f(z)$ is invertible.

We have $f(z)=\mu I-f(x)$. Since $\mu \in \sigma(f(x))$, $f(z)$ is not invertible, a contradiction.