= Solution
For $x$ in a unital complex algebra $A$, the <spectrum of an element> is
$$
\sigma_A(x)=\{\lambda\in\mathbb C:x-\lambda1\text{ is not invertible}\}.
$$
For a nonunital algebra one uses its <unitization of an algebra>[unitization].
Now let $A$ be a <Banach algebra>. The invertible group is open, so the <resolvent of an element>[resolvent set] is open and the spectrum is closed. If $|\lambda|>\lVert x\rVert$, the <Neumann series>
$$
(\lambda1-x)^{-1}
=\lambda^{-1}\sum_{n=0}^\infty(x/\lambda)^n
$$
converges, so $\sigma_A(x)$ is contained in the closed disc of radius $\lVert x\rVert$ and is therefore compact.
If the spectrum were empty, $R(\lambda)=(\lambda1-x)^{-1}$ would be an entire $A$-valued function. For each $f\in A^*$, the scalar function $f(R(\lambda))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The <Liouville theorem> makes it identically zero. Since the <Hahn-Banach theorem> separates points, this would give $R(\lambda)=0$, contradicting its invertibility. Hence the spectrum is nonempty.
Back to article page
