 Solution

Now $B=\overline{\mathbb{C}[1,x]}$. Let $U$ be a bounded component of ${\rho }_A(x)$ and $\lambda \in U$. Its boundary lies in ${\sigma }_A(x)$: an open disk about a boundary point in the algebra resolvent set would meet $U$ and would belong to the same component, contradicting boundary membership. The maximum modulus principle on the bounded domain $U$ therefore gives, for every complex polynomial $p$,Here spectral mapping theorem follows by factoring $p(z)-\mu$: its commuting factors are invertible exactly when every root avoids the Banach algebra spectrum.

The rule $p(x)\mapsto p(\lambda )$ is well-defined by the displayed bound, multiplicative and bounded on the dense polynomial subalgebra. It extends to a unital character ${\phi }_{\lambda }$ on $B$. Since ${\phi }_{\lambda }(\lambda 1-x)=0$, that element cannot have an inverse in $B$. Thus $U\subseteq {\sigma }_B(x)$. Together with part (iii), this proves that the Banach algebra spectrum of a polynomial-generated Banach subalgebra is exactly the original Banach algebra spectrum with all its bounded complementary components filled in.