If $\Omega$ is a bounded region in $\mathbb{C}$ such that $\mathbb{C}\setminus\Omega$ is connected and $K$ is a compact subset of $\Omega$, show that $\hat{K}_{\Omega} = \hat{K}_{\mathbb{C}}$. ($\hat{K}_{\Omega} = \{z \in \Omega|\quad |f(z)| \leq \max_{\zeta \in K}{|f(\zeta)|, \forall f \in H(\Omega)} \}$ and $H(\Omega)$ denotes the space of analytic functions on $\Omega$.)
My thoughts:
Since $\mathbb{C}\setminus\Omega$ is connected, hence $\Omega$ is simply connected and bounded. WLOG, we can assume $\Omega$ is the open unit disk and $K$ is any compact subset $\{|z| \leq r, 0<r<1\}$. This means I need to show that for any point, $z_0$ outside the unit disk, $|f(z_0)|> \max_{\zeta \in K}|f(\zeta|$. Probably, I also need to use the fact that any analytic function in $K$ can be uniformly approximated by polynomials , according to Mergelyan's theorem. But I am not able to put all this together to form a coherent argument. Please help.
It is not clear that you can assume that $\Omega$ is the unit disk, since the biholomorphic mapping between $\Omega$ and $\mathbb{D}$ generally doesn't extend beyond the two domains, so composing an entire function with the biholomorphism need not lead to an entire function again, and hence it is not clear that the polynomially convex hulls correspond to each other.
However, Runge's approximation theorem makes things easy. Every $f \in H(\Omega)$ is the locally uniform limit of rational functions whose poles lie outside $\Omega$. Since $\Omega$ is simply connected, we have the even better
Since generally
$$\hat{K}_\Omega \subset \hat{K}_\mathbb{C}$$
is clear, all that remains is to show the reverse inclusion.
Consider a $z \in \Omega \setminus \hat{K}_\Omega$, and an $f \in H(\Omega)$ with $\lvert f(z)\rvert > \max \{ \lvert f(\zeta)\rvert : \zeta \in K\}$.
Throw Runge on it to see that $z \notin \hat{K}_\mathbb{C}$.