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.

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    $\begingroup$ 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. $\endgroup$ May 13, 2010 at 8:25
  • $\begingroup$ 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. $\endgroup$ Jun 1, 2010 at 20:38

2 Answers 2


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.


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.

  • $\begingroup$ Pete, I have upvoted this construction but I think you have to apply an interpolation theorem for functions defined on the $z_n$'s and not Mittag-Leffler's theorem stricto sensu if you want f to be holomorphic at the $z_n$'s (but maybe I'm missing some subtlety). $\endgroup$ May 13, 2010 at 8:48
  • $\begingroup$ @Georges -- you're right. I am using an interpolation theorem which is reminiscent of Weierstrass Factorization and Mittag-Leffler but not identical to either one. In fact the special case of $\mathbb{C}$ has been asked several times on MO, most recently by Kevin Buzzard. Offhand, I don't have a reference for the general case...sorry! $\endgroup$ May 13, 2010 at 17:34
  • $\begingroup$ @Georges -- the theorem is the following. You can construct an analytic function on a domain with prescribed isolated values on a countable set of isolated points. The proof goes along the same lines as that of Mittag-Leffler's theorem. $\endgroup$
    – Akela
    May 14, 2010 at 21:41

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