Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 7 / 1

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 ${\rho }_A(x)$.

A connected component $U$ of ${\rho }_A(x)$ meets the relatively clopen subset ${\rho }_B(x)$ either everywhere or nowhere: its intersection and its complement would otherwise disconnect $U$. If it meets nowhere, $U\subseteq {\sigma }_B(x)$. If it meets everywhere, $U$ is a connected subset of ${\rho }_B(x)$, and any larger connected subset of ${\rho }_B(x)$ would be a connected subset of ${\rho }_A(x)$ larger than its component $U$. Thus $U$ is either a whole component of ${\rho }_B(x)$ or lies wholly in ${\sigma }_B(x)$.

For $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in $B$. The exterior of this disk is connected and belongs to both algebra resolvent sets. Since each Banach algebra spectrum is compact, each algebra resolvent set has exactly one unbounded component: any unbounded component meets that exterior, and two components meeting a common connected exterior must coincide. Part (ii) says that the component of ${\rho }_A(x)$ meeting the exterior is also a whole component of ${\rho }_B(x)$. Hence the unbounded components are identical.

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.

The hypothesis puts zero in the common unbounded resolvent component from part (iii). Hence $0\in {\rho }_B(x)$ for $B=\overline{\mathbb{C}[1,x]}$, and $(-x{)}^{-1}\in B$, or equivalently $x^{-1}\in B$. By the definition of this norm closure, for each $n$ choose a polynomial $p_n$ with $\Vert p_n(x)-x^{-1}\Vert <1/n$. ThereforeThis is an approximation in the algebra norm, not merely a scalar approximation on the Banach algebra spectrum.