 Solution

Use ${\sigma }_A(x)$ and ${\rho }_A(x)$ for Banach algebra spectrum and algebra resolvent set. An inverse in the smaller unital algebra is also an inverse in the larger one, so ${\rho }_B(x)\subseteq {\rho }_A(x)$.

If $\lambda \in {\rho }_B(x)$ and $r=(\lambda 1-x{)}^{-1}\in B$, then for $|\mu -\lambda |\Vert r\Vert <1$ the Neumann series givesThus ${\rho }_B(x)$ is open. To prove relative closedness, suppose ${\lambda }_n\in {\rho }_B(x)$ and ${\lambda }_n\to \lambda \in {\rho }_A(x)$. Continuity of inversion in $A$ gives $({\lambda }_{n}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Since $B$ is norm closed, the limiting inverse belongs to $B$, so $\lambda \in {\rho }_B(x)$. Consequently ${\rho }_B(x)$ is relatively clopen in ${\rho }_A(x)$.