If a holomorphic function $f$ is defined on some open set $\Omega$, then any connected open set $U$ containing $\Omega$ either has the property that there exists a holomorphic extension of $f$ on $U$ or does not. If there is a holomorphic extension, it's unique by analytic continuation. Thus, taking the union of all possible extensions, $f$ has a unique maximal connected, open domain on which it's holomorphic. (Right?)

What can we say about the behavior of $f$ at the boundary of this domain? Does it necessarily blow up? Or can it be bounded?


2 Answers 2


The "maximal" open domain is absolutely not unique (for example see a square root function) because of some bad global behaviour (depending on the path you walk to get to a new point, the values you find there may not agree).

As for natural boundaries (where the function cannot be extended even locally), anything can happen. With lacunary series it's very easy to make some functions that "look" really well-behaved, like $f(z) = \sum_{n \ge 1} n^{-2}z^{n^2}$ but still can't be extended past their boundary.

  • $\begingroup$ So for example we can extend $\sqrt z$ to either $\mathbb C$ minus the positive reals, or $\mathbb C$ minus the negative reals, yet we can't stitch those two extensions together? $\endgroup$
    – Jack M
    Sep 19, 2017 at 15:31
  • $\begingroup$ @mercio, with regards to the second part of your answer, with lacunary series the singularities are dense on the boundary domain of holomorphicity, so it does blow up there, or not? $\endgroup$ Sep 19, 2017 at 15:34
  • $\begingroup$ @Anaedonist my example is bounded on the unit disk by $\sum_{n \ge 1} n^{-2} = \pi^2/6$. However in this case it is the derivative that blows up. By integrating it a few times you can "hide" those blow ups more. Also, I don't think I know if there is an $f$ with a natural boundary where $f$ and all its derivative are bounded $\endgroup$
    – mercio
    Sep 19, 2017 at 15:39
  • $\begingroup$ @JackM yep! the two extensions will disagree on one quarter of the complex plane. $\endgroup$
    – mercio
    Sep 19, 2017 at 15:42
  • $\begingroup$ maybe $\sum_{n \ge 1} n^{- \sqrt n} z^n$ ? $\endgroup$
    – mercio
    Sep 19, 2017 at 15:52

Too long for a comment. I fear I am far competent enough on the matter, yet I found the question very interesting and will attempt an answer for learning purposes, just to see where my understanding fails.

I was in fact rather sure that there cannot exist a function $f$ holomorpic on an open set $\Omega$ (which is a maximal connected open domain as in the OP's question) and continuous on $\partial \Omega$. In other words, only singularities can prevent further analytical continuation.

I would argue so.

As $f$ is continuous on the boundary of $\Omega$, one can find an open set $K$ on which $f$ is continuous, such that both $ K \cap \Omega \neq \emptyset $ and $K \cap \bar {\Omega} \neq \emptyset$ hold.

Let then define as $M$ any open set contained in $K \cap \Omega$. The the Laplace equation on $M$ will be solved by a function $g$, coinciding with $f$ on $M$. As $f$ is continuous on $K$, I believe $g$ can be extended at least to $\partial \Omega$ as well.

Would not then $g$ constitute an analytical continuation of $f$ at least up to $\partial \Omega$, contrary to the assumed "maximal domain" hypothesis?


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