Compactness of the spectrum

Remarks

If $\abs{\lambda} > \norm{a}$, then $a - \lambda 1 = -\lambda (1 - a/\lambda)$, and the Neumann series $\sum_{n \ge 0} (a/\lambda)^n$ converges to an inverse of $1 - a/\lambda$. So $\sigma(a)$ lies in the closed disc of radius $\norm{a}$. If $u$ is invertible and $\norm{b} < 1/\norm{u^{-1}}$, then $u - b = u(1 - u^{-1} b)$ is invertible by the same series. Hence the invertible elements form an open set, the resolvent set is open, and $\sigma(a)$ is closed and bounded.

Completeness is necessary. In the normed algebra of complex polynomials with the sup norm on $[0,1]$, only nonzero constants are invertible. So the element $t$ has spectrum $\C$, which is not bounded.

For a complex Banach algebra the spectrum is also nonempty. For instance, for $a \in M_n(\C)$ with the operator norm, the theorem says that every eigenvalue has modulus at most $\norm{a}$.

Going deeper

For a student

This is the first basic fact of spectral theory in Banach algebras. Together with the holomorphy of the resolvent and Liouville's theorem, it gives the nonemptiness of the spectrum in the complex case. It is refined by Gelfand's spectral radius formula, $\max \set{\abs{\lambda} : \lambda \in \sigma(a)} = \lim_n \norm{a^n}^{1/n}$, which can be smaller than $\norm{a}$. The compactness allows the continuous functional calculus, which needs $\sigma(a)$ to be a compact space.