Domains of homolorphy in the complex plane - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:01:10Z http://mathoverflow.net/feeds/question/24448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24448/domains-of-homolorphy-in-the-complex-plane Domains of homolorphy in the complex plane Akela 2010-05-12T23:25:37Z 2010-06-01T18:41:05Z <p>There is a proof of <a href="http://en.wikipedia.org/wiki/Mittag-Leffler_theorem" rel="nofollow">Mittag-Leffler's theorem</a> 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.</p> <p>Now I want to prove that every open set in the complex plane is now a <a href="http://en.wikipedia.org/wiki/Mittag-Leffler_theorem" rel="nofollow">domain of holomorphy</a>. 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.</p> <p>I would be most grateful if somebody can show me how to do the above.</p> http://mathoverflow.net/questions/24448/domains-of-homolorphy-in-the-complex-plane/24452#24452 Answer by Pete L. Clark for Domains of homolorphy in the complex plane Pete L. Clark 2010-05-12T23:48:12Z 2010-06-01T18:41:05Z <p>Rather than trying to put the poles <em>on</em> the boundary, choose a countable discrete subset <code>$D = \{z_n\}$</code> 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. </p> <p><b>Addendum</b>: I found a reference for the interpolation result I was using.</p> <p>Theorem (Rudin, <em>Real and Complex Analysis</em>, 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 $0 \leq i \leq m(\alpha)$, $f^{(i)}(\alpha) = w_{\alpha,i}$. </p> <p>This theorem -- and other variants involving meromorphic functions -- is indeed due to Gosta Mittag-Leffler and is often called the <strong>Anschmiegungssatz</strong>.</p> http://mathoverflow.net/questions/24448/domains-of-homolorphy-in-the-complex-plane/24454#24454 Answer by Josh Shadlen for Domains of homolorphy in the complex plane Josh Shadlen 2010-05-12T23:58:14Z 2010-05-12T23:58:14Z <p>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. </p>