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Def: An open set $D \subset \mathbb C^n$ is called a domain of Holomorphy if there exists a holomorphic function $f$ on $D$ such that $f$ cannot be extended to a bigger set.

Is every non empty open set $D \subset \mathbb C$ is a domain of holomorphy?

I personally believe that this result is true but I'm unable to find a proof.Any ideas?

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  • $\begingroup$ Can you do the disk? $\endgroup$
    – Mariano Suárez-Álvarez
    Jun 26, 2015 at 21:34
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    $\begingroup$ @MarianoSuárez-Alvarez Yes, $\sum z^{n!}$. $\endgroup$
    – Zardo
    Jun 26, 2015 at 21:39
  • $\begingroup$ ANd why does that work? $\endgroup$
    – Mariano Suárez-Álvarez
    Jun 26, 2015 at 21:40
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    $\begingroup$ $D \neq \varnothing$ is necessary. Do you know the (general) Weierstraß product theorem? $\endgroup$
    – Daniel Fischer
    Jun 26, 2015 at 22:23
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    $\begingroup$ $D = \mathbb{C}$ is clear, so suppose $D\neq\mathbb{C}$. You take a sequence $(\zeta_n)$ in $D$ such that the set of accumulation points of the sequence is $\partial D$, and every $p\in\partial D$ is approached "from every side". You use Weierstraß' theorem to "find" a holomorphic function $f$ on $D$ with $f(\zeta_n) = 0$ for all $n$, and no other zeros. Then $f$ cannot be analytically continued across any boundary point, since each boundary point $p$ is an accumulation point of zeros of $f$ in any open subset $V\subset D$ such that $p\in \partial V$. $\endgroup$
    – Daniel Fischer
    Jun 27, 2015 at 9:32

2 Answers 2

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Yes, any nonempty open domain $ D\subset \mathbb C$ is a domain of holomorphy.
The proof is in two steps:

a) One proves that a holomorphically convex domain $D\subset \mathbb C^n$ is a domain of holomorphy .
This is not difficult: see for example Grauert-Fritzsche, Theorem 6.5, page 81

b) One proves that for $n=1$ any domain $D\subset \mathbb C$ is holomorphically convex.
This too is rather easy: given a compact subset $K\subset\subset D$ and a boundary point $a\in \partial D$, consideration of $\frac{1}{z-a}$ shows that the holomorphically convex hull $\hat {K}=\hat {K}_D$ cannot approach $\partial D$ and consideration of the holomorphic function $z$ shows that $\hat {K}$ is bounded.
Since $\hat {K}$ is closed in $D$ these considerations prove that $\hat {K}$ is compact.

NB Actually a) above is an equivalence: a domain $D\subset \mathbb C^n$ is holomorphically convex iff it is a domain of holomorphy.
This result was proved by Cartan-Thullen in 1932. Here are the authors reunited 55 years later.

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A nonrigorous but perhaps more accessible answer: the way one constructs a function holomorphic on $D$ but not any larger open set is to construct a meromorphic function on $\Bbb C$ that has poles surrounding the boundary of $D$, so closely that they cut off any analytic continuation. Such a meromorphic function can be chosen to be the reciprocal of a holomorphic function with prescribed zeros; the existence of such functions is, as Daniel Fischer suggested, guaranteed by the Weierstrass product theorem.

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  • $\begingroup$ Thank you for your response.Regards, $\endgroup$
    – Arpit Kansal
    Jun 27, 2015 at 11:04

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