Proposition 33.5.12.label Let $A$ be a unital Banach algebra, $B \subset A$ be a closed subalgebra containing $1$, and $x \in B$, then:
- (1)
$\sigma_{A}(x) \subset \sigma_{B}(x)$.
- (2)
$\partial \sigma_{B}(x) \subset \sigma_{A}(x)$.
- (3)
$\sigma_{B}(x)$ is the union of $\sigma_{A}(x)$ and some bounded components of $\complex \setminus \sigma_{A}(x)$.
Proof. (1): $G(B) \subset G(A)$.
(2): Let $\lambda \in \partial \sigma_{B}(x)$, then there exists $\seq{\lambda_n}\subset \complex \setminus \sigma_{B}(x)$ such that $\lambda_{n} - x \in G(B)$ for all $n \in \natp$, and $\lambda_{n} \to \lambda$ as $n \to \infty$. By Corollary 33.2.4, $\norm{(\lambda_n - x)^{-1}}_{A} \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_{n} - x)^{-1}\to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_{A} = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_{A}(x)$.$\square$
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