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"Every bounded function that is holomorphic on $A$ is is constant."

For which $A\subseteq\mathbb{C}$ is this true?

Are there well-known examples of unbounded sets $A\subseteq\mathbb{C}$ on which there are non-constant bounded holomorphic functions?

Later edit: My striking through the second question was meant only to de-emphasize it. Feel free to post further on it if you wish. Some of the examples posted in response to it were already well known to me; I'd have thought of them if my attention had been on the second question rather than the first.

I'm envisioning a couple of possibilities: (1) Various other sorts of sets $A$ will be mentioned in answers; and (2) An answer will say that some nice theorem says this is true of a set $A$ if and only if whatever, where "whatever" is something non-trivially different from a tautologous "if and only if every bounded holomorphic function on $A$ is constant", and maybe "whatever" is somehow elegant or at least simple.

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You might want to assume something like connectedness because then you can have a "constant by parts" function on the different connected components. But this is a rather trivial remark. – Patrick Da Silva Aug 11 '12 at 22:41
What do you mean by holomorphic if $A$ isn't open? – Qiaochu Yuan Aug 11 '12 at 22:43
@QiaochuYuan : Possibly an interesting question. Maybe I should just ask for which open sets $A\subseteq\mathbb{C}$ this is true. But I see someone's posted an answer involving a set that is not open. I'm not sure what to make of that example yet. – Michael Hardy Aug 11 '12 at 23:12
I see plenty of answers to my second question. I'm not surprised; it seems like the easy part. The answer from Jose27 seems to attempt to address the first question, and I'm not sure I understand it yet. – Michael Hardy Aug 11 '12 at 23:14
$@$Michael: the only reasonable definition of a holomorphic function on an arbitrary subset $A$ of $\mathbb{C}$ that I've seen is: $f$ extends to a holomorphic function on some open subset $U \supset A$. Thus you may as well assume that $A$ is open. – Pete L. Clark Aug 11 '12 at 23:36

On the affirmative side you have that if $B$ is a discrete subset of $\mathbb{C}$ (meaning it has no accumulation points) then every bounded holomorphic function $f: \mathbb{C}\setminus B \to \mathbb{C}$ must be constant. This is because every "singularity" at $b\in B$ is removable (here you use that $B$ is discrete) and so $f$ extends to a bounded entire function.

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So if a function is defined near $b$ but not at $b$, and is bounded near $b$, then the singularity at $b$ is removable. I see one way of proving that, but I suspect there's a simpler way...... – Michael Hardy Aug 12 '12 at 22:28
@Michael Hardy: doesn't this follow almost immediately from Casorati-Weierstrass? – Niels Diepeveen Aug 12 '12 at 22:42
@NielsDiepeveen : So it does. So that's another way. I wonder if there's some identifiable simplest way. – Michael Hardy Aug 12 '12 at 22:50
.....but maybe that is the simplest one. – Michael Hardy Aug 12 '12 at 22:56
@Michael Hardy: The easiest way I can think of proving this is via Morera's theorem to the function $g(z)=(z-z_0)f(z)$. – Jose27 Aug 12 '12 at 23:40

For any $A$ whose complement contains a neighborhood of a point $z_o$, the holomorphic function $1/(z-z_o)$ is bounded on $A$.

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This is interesting; it shows that $\mathbb{C}$ minus a point is not biholomorphic to $\mathbb{C}$ minus a closed disk, which I don't think I knew before. – Qiaochu Yuan Aug 12 '12 at 15:57

Look at the exponential function on the left half-plane. If $a\le 0$, $$|\exp(a + ib)| = e^a \le 1.$$

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If $A$ is connected and open, by the uniformization theorem it is covered by either $\mathbb{C}$, the open disk, or the Riemann sphere. The third possibility does not occur in this case so we are reduced to the first two. In the first case any bounded entire function on $A$ pulls back to a bounded entire function on $\mathbb{C}$ so we reduce to the usual Liouville's theorem. In the second case the open disk clearly admits a non-constant bounded entire function to $\mathbb{C}$, namely the obvious inclusion. In particular, if $A$ is simply-connected and covered by the open disk then it is in fact biholomorphic to the open disk (so for example any open half-space has this property).

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Right, I was thinking along similar lines. This leaves open the case in which the universal cover is the open disk $\mathbb{D}$ and $U$ is not simply connected. Here I guess topological arguments will not be enough because there are compact Riemann surface quotients of $\mathbb{D}$ and on these every holomorphic function is constant. But there are certainly many cases at least where a planar $U$ carries a nonconstant bounded holomorphic function, e.g. whenever $U$ is not dense in $\mathbb{C}$, by Paul Garrett's answer. – Pete L. Clark Aug 11 '12 at 23:06
Did you omit the word "simply"? – Michael Hardy Aug 12 '12 at 15:52
OK, this takes care of the simply connected case. – Michael Hardy Aug 12 '12 at 15:54
@PeteL.Clark : You introduce the notation $U$. Exactly synonymous with "$A$" as used elsewhere in this thread? – Michael Hardy Aug 12 '12 at 15:55
@Michael: Sorry, that was the notation in my (deleted) answer. It is the same as $A$ except it's open! – Pete L. Clark Aug 12 '12 at 17:57

There are examples to your second question. Let $A$ to be the upper-half-plane, and consider the canonical conformal map from the upper-half-plane to the unit disc given by $$ F(z) = \frac{i-z}{i+z} $$

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