Spectrum of a linear function in the continous functions

Question

I am currently working through my course text and one of the examples for a unital, commutative Banach algebra is the space of continuous functions $C([a,b])$. Further, in the text, there is a theorem stating:

$\sigma(a) \neq 0$ $\forall a \in \mathcal{A}$, for $\mathcal{A}$ being a unital algebra.

This theorem is also closely related to the Gelfand Mazur Theorem.

Trying a few examples I came across the linear function $f(x) = x$, which is obviously in $C([0,1])$, and for which $\lambda id - f$ seems to be invertable for all $\lambda \in \mathbb{C}$. Thus f must have the resolvent $\rho(f) = \mathbb{C}$ and therefore $\sigma(f) = \mathbb{C} \setminus \mathbb{C} = \emptyset$ contradicting the theorem stated before. Does anyone see my mistake?

Answer 1

Note that the identity of the Banach algebra $C([0,1])$, which you denote by $\mathrm{id}$, is the constant function $\mathbf 1 \colon x \mapsto 1$. So, for $a \in C([0,1])$, $\lambda \in \mathbf C$, we have (as elements of $C([0,1])$ have inverses iff the do not have zeros: \begin{align*} \lambda \in \sigma(a) &\iff \lambda \mathbf 1 - a \text{ is not invertible}\\ &\iff \exists x \in [0,1]: \lambda - a(x) = 0 \\ &\iff \lambda \in a([0,1]) \end{align*} So, the spectrum of an element of $C([0,1])$ equals its range as a function, so for $f \colon x \to x$ we have $\sigma(f) = [0,1]$.

Answer 2

The reason for my misunderstanding was supposing that we look for the inverse regarding the composition of functions when it was rather the multiplicative inverse we were looking for.