Proposition 29.4.5.label Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_{A}(x) \ne \emptyset$.
Proof. Assume for contradiction that $\sigma_{A}(x) = \emptyset$, then by Definition 29.4.3, the resolvent
\[R_{x}: \complex \setminus \sigma_{A}(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}\]
is an entire function. Since
\[R_{x}(\lambda) = \frac{1}{\lambda - x}= \frac{\lambda^{-1}}{1 - \lambda^{-1}x}\]
which tends to $0$ as $|\lambda| \to \infty$, $R_{x} \in H(\complex; A) \cap C_{0}(\complex; A)$. By Liouville’s Theorem, $R_{x} = 0$, which is impossible.$\square$
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