Timeline for Simply-connected domain around a curve
Current License: CC BY-SA 2.5
8 events
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| Jul 29, 2010 at 12:41 | comment | added | Lasse Rempe | The function $f(z)=z+\sin(2\pi z)$ is perhaps the simplest example to illustrate the problem, and of course you are right that the issue is with several accesses to infinity in the same domain. I would be quite surprised if you could get enough control out of Arakelian's or Mergelian's theorems to avoid this. | |
| Jul 29, 2010 at 12:06 | comment | added | fedja | In general you are right: there is a possibility (which I missed) that we have escapes to infinity in the domain corresponding to finite points on the boundary, which will break my final argument about $0$ on the boundary. Still it seems salvageable because all we really need to show is that there is only one way to escape to infinity, i.e., that the boundary of $V$ is connected. This may, probably, be done by more careful approximation that kills off the unwanted branches somehow. Let me think a bit. | |
| Jul 29, 2010 at 11:50 | comment | added | Lasse Rempe | Dear fedja, There are many entire functions that omit zero and still do not have logarithmic singularities over infinity. As you note, here the tract will certainly be simply-connected, but the map is not a universal covering. See Bergweiler and Eremenko, "Direct singularities and completely invariant domains of entire functions", for an example of such a function. So I think your final claim cannot be correct. | |
| Jul 29, 2010 at 11:46 | comment | added | fedja | Needless to say, my $u$ is the real part of the entire function, not the logarithm of its absolute value :). | |
| Jul 29, 2010 at 11:43 | comment | added | fedja |
My $u$ is harmonic on the entire plane and your $\log|f|$ is most certainly not! The difference is that I can apply the minimum principle to the holes and get a contradiction an you cannot, so in my case the simple connectivity follows and in your case does not. Now, once we know simple connectivityof $V$, the composition of $u$ with the conformal map from the right half plane to $V$ is a positive harmonic function in the right half plane vanishing on the boundary. So, it must be $c\text{Re\,}z$ for some $c>0$.
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| Jul 29, 2010 at 9:01 | comment | added | Lasse Rempe | Indeed, the question was motivated precisely by the following: Which curves can be realized as asymptotic paths for logarithmic asymptotic values of entire functions? In other words, when is there an entire function $f$ as above such that $T$ is mapped as a universal covering, contains $\gamma$, and $f$ tends to $\infty$ along $\gamma$? | |
| Jul 29, 2010 at 8:59 | comment | added | Lasse Rempe | Dear fedja, Thank you for your answer. Indeed, any continuous function on $\gamma$ can be approximated by an entire function up to any continuous error; this follows from Arakelian's theorem (which of course is an extension of Mergelyan's theorem). However, I don't understand the final part of your argument. Take any entire function f and any component T of $f^{-1}(\{|z|>1\})$ (for some $R>0$). Then $T$ need not be simply-connected, and even if it is, $f$ need not map $T$ as a universal cover. However, $\log|f|$ is harmonic and 0 on the boundary. Am I missing something? | |
| Jul 29, 2010 at 5:14 | history | answered | fedja | CC BY-SA 2.5 |