Banach algebra of $C^{1}([0,1])$

Question

I want to show that $\mathscr{A}=\{f \in C^{1}([0,1]) \}$ with norm $$\Vert f \Vert_{C^{1}} = \Vert f \Vert+ \Vert f^{'} \Vert,$$ is $\textbf{semi-simple}$.

I see a method as following:

To show that $r(f):= \sup\{ \vert \lambda \vert : \lambda e-f \notin G(\mathscr{A})\} =0 $, where $G(\mathscr{A})$ is the invertible elements of $\mathscr{A}$.

And $ r(f)=0 $ means $\mathscr{A}$ semi-simple.

But I do not understand the reason. Please help me to explain why does it work?

Answer 1

I'm guessing that instead of $r(f)$ you meant to put something along the lines of $\text{Rad}(\mathscr A)$, which denotes the intersection of all maximal ideals of $\mathscr A$. For if $\text{Rad}(\mathscr A)=\{0\}$, then by definition $\mathscr A$ is semi-simple.

To show $\text{Rad}(\mathscr A)=\{0\}$, just note for each $x\in[0,1]$, the evaluation $f\mapsto f(x)$ is a complex homomorphism of $\mathscr A$, whose kernel is a maximal ideal of $\mathscr A$. Thus if $f\in\text{Rad}(\mathscr A)$, $f(x)=0$ for all $x\in[0,1]$, so that $f=0$.