# Spectrum of a linear function in the continous functions

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?

• $\lambda id-f$ is invertible for all $\lambda$ except one... Feb 15, 2022 at 22:53
• I still don't get it, could you enlighten me? I have tried various numbers. My only guess would be for $\lambda = f(x)$ but it has to be valid for all $x \in [0,1]$, no? Feb 15, 2022 at 23:23
• $f=id$, then if you take $\lambda=1$ you get the $0$ map. Feb 16, 2022 at 7:39

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 $$\sigma(f) = [0,1]$$.