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There is a proof of Mittag-Leffler's theorem with an explicit construction of a holomorphic function with the prescribed poles with prescribed order and residues, for a countable discrete set of points. I do not remember the reference; but my memory from my graduate course is that one defines a series sum and make certain adjustments. I was never quite good in this type of processes; so I am facing problem with the following exercise, which is nagging me for a long time. I thought of using Math Overflow with the hope that somebody can help me out.

Now I want to prove that every open set in the complex plane is now a domain of holomorphy. We take the boundary $\partial \Omega$ of the open set $\Omega$, and we take a countable dense sequence of points $z_i$ in $\partial \Omega$. If we are able to construct a series sum with poles at $z_i$, but so that it converges absolutely and uniformly on every compact set in the interior of $\Omega$, then we are done.

I would be most grateful if somebody can show me how to do the above.

flag	asked May 12 2010 at 23:25 Akela 1,773●6●19

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You have to assume the discrete set of points is CLOSED in Mittag-Leffler's theorem (Wikipedia erroneously forgets this condition ). For example it is impossible to find a meromorphic function on $\mathbb C$ with principal part equal to $1/(z−1/n)$ at $1/n$ , although the set of $1/n$'s in $\mathbb C$ is certainly discrete. By the way Mittag-Leffler's construction yields a meromorphic function, not a holomorphic one, and the prescription is principal parts, not residues. – Georges Elencwajg May 13 2010 at 8:25

You probably want to assume the open set is connected. Otherwise, a locally analytic function whose domain is not connected is by convention generally not thought of a a single analytic function. (Even when it is everywhere defined by a single series, such as in Josh Shadlen's elegant answer below.) Such a function can, for instance, violate the permanence principle. There are some fascinating examples of such functions in Hille's Analytic Function Theory, vol. II. – Daniel Asimov Jun 1 2010 at 20:38

Josh Shadlen · Accepted Answer · 2010-05-12 23:58:14Z UTC

Let $\zeta_k$ be a countable dense sequence of points in the boundary and consider $f(z) = \sum \frac{1}{2^k} \frac{1}{z-\zeta_k}$. The sum is plainly uniformly convergent on any subset of finite distance from the boundary, in particular on any compact subset of the interior.

Pete L. Clark · Answer 2 · 2010-06-01 18:41:05Z UTC

Rather than trying to put the poles on the boundary, choose a countable discrete subset $D = \{z_n\}$ of $\Omega$ whose closure contains $\partial \Omega$ (first convince yourself this is always possible) and then apply Mittag-Leffler's theorem to get a holomorphic function $f$ on $\Omega$ such that $\lim_{n \rightarrow \infty} |f(z_n)| = \infty$. Then show that this does what you want.

Addendum: I found a reference for the interpolation result I was using.

Theorem (Rudin, Real and Complex Analysis, Theorem 15.13): Let $\Omega$ be an open set in the complex plane and $A$ a closed, discrete subset of $\Omega$. To each $\alpha \in A$ we associate a non-negative integer $m(\alpha)$ and complex numbers $w_{\alpha,i}$ for $0 \leq i \leq m(\alpha)$. Then there exists a holomorphic function $f$ on $\Omega$ such that for all $\alpha \in A$ and all $0 \leq i \leq m(\alpha)$, $f^{(i)}(\alpha) = w_{\alpha,i}$.

This theorem -- and other variants involving meromorphic functions -- is indeed due to Gosta Mittag-Leffler and is often called the Anschmiegungssatz.

Domains of homolorphy in the complex plane

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