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This question is exercise 8.2 in Conway's Functions of One Complex Variable I. It states:

Let $\mathbb{D}\subset\mathbb{C}$ be the open unit disk, and let $K=\{z\in\mathbb{D}: \frac{1}{4}\leq |z|\leq \frac{3}{4}\}\subset\mathbb{D}$ compact. Find a holomorphic function $f$ on $K$ (so on a neighborhood of $K$), which cannot be approximated on $K$ by holomorphic functions on $\mathbb{D}$.

After this, Conway makes the remark:

The next two problems are concerned with the following question. Given a compact set $K$ contained in $G$ open, can holomorphic functions on $K$ be approximated on $K$ (with the $\sup$ norm) by holomorphic functions on $G$? Exercise 2 says that for an arbitrary choice of $K$ and $G$, this is not true. Exercise 4 below gives criteria for a fixed $K$ and $G$ such that this can be done. Exercise 3 is a lemma which is useful in proving Exercise 4.

Exercise $3$ is a lemma, and Exercise 4 is the statement of Runge's theorem that was covered in my complex analysis course:

Let $G \subset \mathbb{C}$ be an open set and let $K\subset G$ be a compact subset. Then the following conditions are equivalent.

  1. Every holomorphic function on a neighborhood of $K$ can be approximated uniformly on $K$ by holomorphic functions on $G$.
  2. The open set $G\setminus K$ has no component whose closure in $G$ is compact.
  3. For every $z\in G\setminus K$ there exists a holomorphic function on $G$ such that: $$\sup_K |f| < |f(z)|$$

So, is there any suggestions for this problem? I couldn't think of exactly how to construct such a function. My guess would be something standard like $f(z)=1/z$, but I'm not sure how to show that this cannot be approximated on $K$.

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1 Answer 1

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You may indeed take $f(z)=\frac 1z$.
The function $|f|$ takes the value $4$ on the circle $C=\{|z|=\frac 14\}$ and the value $2$ on the circle $C'=\{|z|=\frac 12\}$.
But a function $g$ holomorphic on $\mathbb D$ satisfies $$\sup_{z\in C}|g(z)|\leq \sup_{z\in C'}|g(z)|$$ by the maximum principle and thus cannot approximate $f$ on $K$ well .

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  • $\begingroup$ Excellent, thank you. $\endgroup$
    – Moya
    May 29, 2015 at 22:15

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