boundary of a spectrum proof

Question

Let $A$ be a closed unital subalgebra of banach algebra $B$. Prove that ${\delta}{\sigma}_{B}(x)$ is contained in ${\delta}{\sigma}_{A}(x)$ for every $x$ in $B$.

Answer 1

Solution: The problem appears to me to be false.

Let $B$ be the Banach algebra of continuous functions on the annulus $\mathscr{A}=\{ z \in \mathbb{C} : 1/2 \le |z| \le 1\}$ with $\|b\|=\sup_{z \in \mathscr{A}}|b(z)|$. Let $A$ be the subalgebra of $\mathscr{A}$ generated as the closure in $B$ of polynomials in $z$. Let $b(z)=z$. Then $\sigma_{B}(b) = \mathscr{A}$ and $\sigma_{A}(b)=\mathscr{D}$. Therefore $\delta\sigma_{B}(b)\not\subseteq\delta\sigma_{A}(b)$ because the boundary of the annulus $\mathscr{A}$ consists of two circles, while the boundary of $\mathscr{D}$ is a single circle.

However, the opposite inclusion is true. That is, $\delta\sigma_{A}(x)\subseteq\delta\sigma_{B}(x)$.

To see this, let $e$ be the unit in $B$. Let $x \in A$, and suppose $\mu$ is in the boundary of $\sigma_{A}(x)$. Because $\mu$ is in the boundary of $\sigma_{A}(x)$, then there exists a sequence $\{ \mu_{n}\}_{n=1}^{\infty} \subseteq \rho_{A}(x)$ which converges to $\mu$. Because $(x-\mu_{n}e)$ is invertible in $A$, then it is also invertible in $B$, and the inverses are the same. So $\{ \mu_{n}\}_{n=1}^{\infty} \subseteq \rho_{B}(x)$. If $\mu$ were to be in $\rho_{B}(x)$, then continuity of inversion would force $(x-\mu_{n}e)^{-1}\in A$ to converge to $(x-\mu e)^{-1}$ in $B$, which would also put $(x-\mu e)^{-1}\in A$, contrary to the assumption that $\mu \in \sigma_{A}(x)$. So every $\mu$ in the boundary of $\sigma_{A}(x)$ must also be in $\sigma_{B}(x)$. The boundary inclusion then follows because any $\mu \in \delta\sigma_{A}(x)$ is a cluster point of $\rho_{A}(x)\subseteq\rho_{B}(x)$ and is in $\sigma_{B}(x)$, which puts it in $\delta\sigma_{B}(x)$.