= Solution
For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so $\sigma_B(x)\subseteq\sigma_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda1)^{-1}\in A$ is both open, by a local <Neumann series>, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus <spectrum in a closed unital subalgebra> says that $\sigma_A(x)$ is $\sigma_B(x)$ with some bounded complementary components filled in.
Now let $A$ be the <Banach subalgebra generated by one element> $x$ and put $K=\sigma_A(x)$. If $D$ were a bounded component of $\mathbb C\setminus K$, choose $\lambda\in D$. Since $(x-\lambda1)^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomials
$$
q_n(z)=(z-\lambda)p_n(z)
$$
satisfy $q_n(\lambda)=0$ and $q_n(x)\to1$. Applying the contractive <Gelfand transform> gives $q_n\to1$ uniformly on $K$, hence on $\partial D$. But the <maximum modulus principle> applied to $1-q_n$ gives
$$
1=|1-q_n(\lambda)|
\leq\sup_{z\in\partial D}|1-q_n(z)|\longrightarrow0,
$$
a contradiction. Therefore $\mathbb C\setminus K$ is connected.
The map
$$
\Phi_A\longrightarrow K,
\qquad
\varphi\longmapsto\varphi(x)
$$
is continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the <Banach-Alaoglu theorem>, while $K$ is Hausdorff, so this continuous bijection is a <homeomorphism>.
Under this identification, the Gelfand transform $\theta:A\to C(K)$ obeys
$$
\theta(f(x))(z)=f(z)
$$
by the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ gives
$$
\sup_{z\in K}|p_n(z)-f(z)|
\leq\|p_n(x)-f(x)\|\longrightarrow0.
$$
Thus every function holomorphic near $K$ is uniformly approximable there by polynomials.
Solved by gpt-5.6-sol high.
Back to article page
