Let $K$ be a compact subset of an open set $G \subseteq \mathbb C$. I am trying to prove that the following are equivalent
- If $f$ is analytic in a neighborhood of $K$ and $\epsilon > 0$, then there is a $g \in H(G)$ with $\sup_{z \in K} |f(z) - g(z)|$
- If $D$ is a bounded component of $G\setminus K$, then $\overline{D} \cap \partial G \ne \emptyset$
- If $z$ is any point in $G\setminus K$, then there is a function $f \in H(G)$ with $|f(z)| > \sup_{w \in K} |f(w)|$
I managed to prove that 2, 3 are equivalent, and 2 implies 1. However, I am stuck with proving 1 implies 2.
So far, I showed that if $D$ is a bounded component of $G\setminus K$ with $\overline{D} \cap \partial G = \emptyset$, then $\partial D \subseteq K$.
My rough plan is to find $f \in H(G)$ so that there is some $z_0 \in D$ with $|f(z_0)| > \sup_{w \in K} |f(w)|$, and such $f$ will contradict the maximum modulus principle. But, it is not clear how to do this using 1.
I would appreciate any hint/reference.
take a point $w\in D$ and look at $f(z)=\frac{1}{z-w}$ which is analytic in a neighborhood of $K$; picking $g$ analytic on $G$ st $|g-f|_K \le \epsilon$ with $\epsilon$ to be determined, one has $|(z-w)g-1| \le \epsilon |z-w| \le \epsilon M$ on $K$ hence on $D$ by maximum modulus and the fact that $\partial D \subseteq K$, where $M=\max_{z \in K}|z-w|$
So if we choose $\epsilon$ small enough st $\epsilon M \le 1/2$ one gets $|(z-w)g-1| \le 1/2$ for all $z \in D$ hence for $z=w$, one gets $1 \le 1/2$ Contradiction!