Nonemptiness of the spectrum

Remarks

Proof idea: suppose $\sigma(a)$ is empty. Then the resolvent $R(\lambda) = (a - \lambda 1)^{-1}$ is defined on all of $\C$. For each continuous linear functional $\varphi$ on $A$, the function $\lambda \mapsto \varphi(R(\lambda))$ is entire and tends to $0$ as $\abs{\lambda} \to \infty$. By Liouville's theorem it is zero, and by the Hahn–Banach theorem this forces $R(0) = a^{-1} = 0$, which is impossible since $1 \ne 0$. Together with the compactness of the spectrum, the spectrum of every element is a nonempty compact subset of $\C$.

Example: in the algebra $C([0,1])$ of continuous complex functions with the sup norm, the spectrum of $f$ is its range $f([0,1])$, which is nonempty.

The complex scalars are needed. In the real Banach algebra of real $2 \times 2$ matrices, the rotation $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ minus $\lambda 1$ has determinant $\lambda^2 + 1 > 0$ for every real $\lambda$, so its real spectrum is empty. The condition $1 \ne 0$ is also needed: in the zero algebra every element is invertible.

Going deeper

For a student

This result is the first step of spectral theory in Banach algebras. It gives the Gelfand–Mazur theorem: a complex Banach division algebra is $\C$. It also makes the spectral radius $\max \set{\abs{\lambda} : \lambda \in \sigma(a)}$ well defined, which is the subject of Gelfand's spectral radius formula.