Let $V$ be a Riemann surface, $x\in V$, and $B:=B(x,r)$ some small ball (in a local chart). It is well known that there is a meromorphic function $f$ on $V$ with the only pole at $x$. What I’d like to ask of is if there is a meromorphic f on V that has a pole at x and additionally such that $|f|<1$ outside $B$?
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2$\begingroup$ Just multiply your function with a single pole on a sufficiently small number. $\endgroup$– Alexandre EremenkoSep 2, 2022 at 1:16
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$\begingroup$ My question was actually if it is possible to find a meromorphic function is bounded on $V\setminus B$ (look at $z+ 1/z$ on the plane --- what you want does not for this particular one) $\endgroup$– Lukasz KosinskiSep 2, 2022 at 6:50
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$\begingroup$ Sorry, I thought that the question is about compact Riemann surfaces. $\endgroup$– Alexandre EremenkoSep 2, 2022 at 13:55
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1 Answer
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For open surfaces, there are counterexamples. The first one was constructed by P. Myrberg:
Ueber die analytische Fortsetzung von beschrankten Funktionen, Ann. Acad. Sci. Fenn., Ser. A. I N:o 58 (1949)
Since this paper is difficult to obtain (and written in German), I refer to another paper
Heins, Maurice, Riemann surfaces of infinite genus. Ann. of Math. (2) 55 (1952), 296–317,
Which proves an even stronger result: there is an open Riemann surface, such that if you remove a disk from it, then on the remaining surface every non-constant meromorphic function takes all complex values, except at most two of them.
answered Sep 2, 2022 at 17:04
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$\begingroup$ Thanks a lot for your answer! Can I have one more question? I took a quick look at the paper and it seems that the surface constructed there is parabolic. Can I ask about the answer within the class of hyperbolic Riemann surfaces? $\endgroup$ Sep 2, 2022 at 18:32
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$\begingroup$ It is parabolic (before you remove the disk) in the sense that there is no Green function. So your new question is: suppose that a surface has a Green function. Does there exist a function with a single pole which is bounded outside a neighborhood of this pole? I suppose that the answer is still negative but do not have a ready example. $\endgroup$ Sep 2, 2022 at 18:59