$\mathcal P(K)=\mathcal R(K)$ iff $\Bbb C\backslash K$ is connected

Question

Let $K$ be a compact set in $\Bbb C$. Let $\mathcal P(K)$ be the closed algebra generated by polynomials on $K$ and $\mathcal R(K)$ the closed algebra generated by rational functions without poles in $K$.

It is clear that $\mathcal P(K)\subset\mathcal R(K)$. If $\Bbb C\backslash K$ is a connected set, then Runge's approximation theorem implies that $\mathcal P(K)=\mathcal R(K)$. The converse of this which is

If $\Bbb C\backslash K$ is not connected, then $\mathcal P(K)\ne\mathcal R(K)$.

is apparently true, according to Graham Allan's Introduction to Banach Spaces and Algebras. How is it proved? Perhaps did I miss something obvious?

Answer 1

The following is not an entirely satisfactory answer but I thought I would put it down as something to be improved later. (I don't remember the context in which GRA stated this fact, so perhaps he had in mind some argument based on nearby parts of the book.)

It is a theorem that in a unital Banach algebra $A$ the spectrum of $a\in A$ relative to the closed unital subalgebra ${\rm alg}(a)$ generated by $a$ is the polynomial hull of $\sigma_A(a)$. Now if $A=R(K)$ and $a(z)=z$, then $\sigma_A(a)=K$; I have a vague recollection that this requires a little work to prove, by first showing that the character space of $R(K)$ is $K$.

Anyway, $P(K)={\rm alg}(a)$. So if $P(K)=R(K)$ then $K$ must equal its own polynomial hull, and this is known to imply ${\bf C}\setminus K$ is connected.

Answer 2

Perhaps a more elementary proof is the following :

Assume $\mathbb{C} \setminus K$ is disconnected. Let $V$ be a bounded component of $\mathbb{C} \setminus K$, and let $w \in V$. Let us show that the function $g(z):= 1/(z-w) \in \mathcal{R}(K)$ does not belong to $\mathcal{P}(K)$. Assume for a contradiction that there are polynomials $(p_n)$ such that $p_n \to g$ uniformly on $K$. Then in particular $(z-w)p_n \to 1$ uniformly on $\partial V \subset K$, so that $(z-w)p_n \to 1$ uniformly on $V$, by the maximum modulus principle. This implies that $p_n \to g$ uniformly on compact subsets of $V \setminus \{w\}$.

Now, let $\gamma$ be a circle in $V \setminus \{w\}$ around $w$. Uniform convergence and Cauchy's theorem gives $$0 = \int_\gamma p_n(z) dz \to \int_\gamma g(z) dz = 2\pi i,$$ a contradiction.