# Is Every Holomorphic Near an Entire?

Let $K\subset \mathbb C$ be a closed subset of the complex plane, not necessarily bounded.
Let $U$ be the interior of $K$.

Let $f:K\to \mathbb C$ be a continuous bounded function, whose restriction to $U$ is holomorphic.

Assume furthermore that for every closed curve $\gamma\subset K$, the integral $\int_\gamma f(z)dz$ vanishes. (this is only relevant if $K$ is not simply connected). If you prefer, you can assume that $K$ is simply connected.

Can $f$ be uniformly approximated by entire functions?

• Do you know Mergelyan's theorem? Jun 30 '15 at 16:26
• I have reworded the question, and I vote to keep it open. @Steven Gubkin: you seem to be knowledgeable in the area; if you have an answer to the question as currently worded, I would be curious to hear it. Jun 30 '15 at 16:48
• I think it is really just a literature search question, which I am not sure is appropriate, so I will just answer in the comments. In the case of an unbounded domain, one needs an extension of Mergelyan's theorem, namely the Keldysh approximation theorem. As long as the compliment of $U$ is connected as a subset of the Riemann sphere, this approximation goes through. (My thesis, just completed, was about an $L^2$ version of Mergelyan's theorem in several variables. So maybe I have just been swimming here so long I think nothing of it... ) Jun 30 '15 at 16:53
• @StevenGubkin One of the hardest things to realize at the end of your PhD is that you're really an expert now. :) Jun 30 '15 at 17:22
• In addition to the answers given, a literature reference: Dieter Gaier, "Lectures on Complex Approximation", includes Mergelyan's and Arakelyan's theorems, and much more! Jul 1 '15 at 16:59

As mentioned in the comments, this is true if $K$ is compact and the complement of $K$ in the Riemann sphere is connected : it is the content of Mergelyan's Theorem on uniform polynomial approximation of holomorphic functions.

EDIT If $K$ is only assumed to be closed, this is also true with the additional assumption that $K$ is locally connected at infinity : it is the content of Arakelian's theorem on the uniform approximation of holomorphic functions by entire functions.

It is false in general if the complement of $K$ is not connected : consider $K$ the annulus $\{1/2 \leq |z| \leq 2\}$ and $f(z):=1/z^2$. Then $f$ is continuous on $K$, holomorphic in the interior of $K$, and $\int_\gamma f(z)\, dz=0$ for all closed curve $\gamma$ contained in $K$. However, $f$ cannot be uniformly approximated on $K$ by entire functions $(f_n)$. Indeed, if it were the case, then the functions $g_n(z):=zf_n(z)$ would uniformly approximate $g(z):=zf(z)=1/z$ on $K$. This contradicts the fact that $\int_\mathbb{T} g_n(z)\, dz=0$ for all $n$, whereas $\int_\mathbb{T} g(z) \, dz = 2\pi i \neq 0$. Here $\mathbb{T}$ is the unit circle.

• A natural attempted fix would be to require "for any disc $D$ in $\mathbb{C}$ with $\partial D \subseteq K$ and any holomorphic $g$ on $D$, we have $\int_{\partial D} f(z) g(z) dz=0$." Any idea if the statement is true with that condition? Jun 30 '15 at 18:53
• I don't understand how this is just Mergelyan's Theorem on uniform polynomial approximation of holomorphic functions. Take the example of the function $f(z)=\sin(z)$, defined on some horizontal strip say $K=\{z\in\mathbb C : Im(z)\in[-1,1]\}$. This function CANNOT be uniformly approximated by polynomials. Jun 30 '15 at 18:54
• It should be noted that, even if the complement is not connected, we do get approximation by rational functions in much the same way (at least for bounded domains, I have not thought too much about unbounded ones). In fact, the proof of Mergelyans theorem consists of two steps: 1. prove that for each $\epsilon$ one can find an open set $U_\epsilon$ containing $K$, and $f_\epsilon$ holomorphic on $U_\epsilon$ such that $\operatorname{max}_{K}|f - f_\epsilon| < \epsilon$ 2. Use runge's theorem. Jun 30 '15 at 18:55
• @AndréHenriques your example cuts the Riemann sphere into two disconnected pieces Jun 30 '15 at 18:56
• @AndréHenriques The question does not ask for approximation by polynomials, instead by entire functions. In your case, the entire function $\sin(z)$ works perfectly! Jun 30 '15 at 19:09

This theorem does not follow in a straightforward way from Mergelyan's theorem, which on its face applies only to bounded domains.

There is, however, a related theorem called Arakelian's theorem which settles the matter. For completeness:

Definition: A "hole" of a closed subset $E$ of $\mathbb{C}$ is any bounded component of the complement of $E$.

Definition A set $E$ is Arakelian if $E$ has no holes, and if for every closed disk $D$, the union of all holes of $E \cup D$ is bounded.

Theorem Let $E$ be Arakelian. If $f$ is continuous on $E$ and holomorphic on the interior of $E$, then $f$ can be uniformly approximated by entire functions.

A proof deriving this result from Mergelyan's theorem can be found here.

I would also like to remark that Mergelyan's theorem (while usually stated for polynomial approximation) actually applies to domains with holes as well, if you just allow rational approximation with "poles in the holes".

I would also like to draw attention to how crazy Mergelyan's theorem is. One consequence is that one can approximate an antiholomorphic function by entire functions on any set without interior. For instance, one can approximate $\overline{z}$ on the "plus sign" consisting on the interval $[-1,1]$ on the real axis and "$[-i,i]$" on the imaginary axis. If anyone can explicitly construct such approximating functions, I would be very happy to see them!

• For your "plus sign" approximation, take $f(z) = z^3 g(z^4)$ where $g$ is a polynomial such that $|t^{3/4} (g(t) - t^{-1/2})| < \epsilon$ for $t \in [0,1]$. Such $g$ can be found using standard numerical approximation methods. For example, with $\epsilon \approx 0.1$, g(t) = 13.6602896164833+(-295.346171207092+(3061.90100782145+(-16221.4880068369+(48270.4896476068+(-84097.0658600084+(85091.1281127053+(-46285.3324386071+10463.1560731230*t)*t)*t)*t)*t)*t)*t)*t will work. Jun 30 '15 at 20:24
• @RobertIsrael That is very insightful. Jun 30 '15 at 20:40