The set $\mathbb{C} \setminus \sigma(x)$ is connected.

Question

Suppose that $A$ is a commutative Banach algebra with unit $e$ and the polynomials in $x \in A$ are dense in $A$. Show that the set $\mathbb{C} \setminus \sigma(x)$ is connected, where $\sigma(x)$ denotes the spectrum of $x$. I can show that $\lambda \notin \sigma(x)$ implies $P_n(x) \to (x-\lambda e)^{-1}$ for a sequence of polynomials $\{P_n(x)\}$ in $A$. However, I cannot prove that $P_n(z) \to (z-\lambda e)^{-1}$ uniformly for $z \in \sigma(x)$ without using the Gelfand transform. Any good suggestion?

Answer 1

$ \newcommand{\C}{{\mathbb{C}}} \newcommand{\N}{{\mathbb{N}}} \newcommand{\inv}{^{-1}} $

Suppose by contradiction that $\C\setminus \sigma (x)$ is disconnected, so it admits at least one bounded connected component, say $\Omega $.

Picking any point $\lambda \in \Omega $, we have that $x-\lambda $ is invertible and we may find a sequence $\{p_n\}_n$ of polynomials such that $$ \lim_{n\to \infty }\|p_n(x)-(x-\lambda )\inv\|=0. $$ Observing that the Gelfand transform $$ A\to C(\sigma (x)) $$ is contractive and sends $x$ to the identity function on $\sigma (x)$, we deduce that $$ \lim_{n\to \infty }\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|=0. $$ Notice that for every $n\in \N$ and every $z\in \sigma (x)$, one has that $$ |(z-\lambda )p_n(z)-1| = |z-\lambda ||p_n(z)-(z-\lambda )\inv|\leq \alpha \beta _n, \tag{1} $$ where $$ \alpha =\sup_{z\in \sigma (x)}|z-\lambda |, $$ and $$ \beta _n=\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|. $$ Observing that the boundary of $\Omega $ is contained in $\sigma (x)$, we deduce from the maximum principle that (1) also holds for every $z\in \Omega $, and hence $$ |p_n(z)-(z-\lambda )\inv|\leq {\alpha \beta _n\over |z-\lambda |}, \quad\forall z\in \Omega \setminus\{\lambda \}. $$ Letting $\gamma (t)=\lambda +re^{it}$, for $t\in [0,2\pi ]$, with $r$ sufficiently small so that $\gamma $ lies in $\Omega $, we than have that $$ 2\pi i= \left|\int_\gamma p_n(z)-(z-\lambda )\inv \,dz\right| \leq {\alpha \beta _n\over r}\text{Length}(\gamma ) =2\pi \alpha \beta _n. $$ Since $\beta _n\to 0$, as $n\to \infty $, this is a contradiction.