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In a current project with a colleague, we have come across the following reasonably classical-sounding geometric question. While not vital to our work, it would be interesting if anyone has seen this type of issue discussed before, and if there are any references.

To keep things simple, I will formulate a special case, which appears to retain all issues present in the general case.

Question

Let $\gamma:[0,\infty)\to\mathbb{C}\setminus\{0\}$ be a continuous and injective curve in the punctured complex number plane, with $\gamma(t)\to\infty$ as $t\to\infty$.

Is there a simply-connected domain $V\subset\mathbb{C}\setminus\{0\}$ with $\gamma\subset V$ such that $\gamma$ tends to the boundary of $V$ horocyclically in $V$?

(The latter condition means the following: There is a conformal isomorphism $\phi$ that maps $V$ to the right half plane in such a way that $\operatorname{Re}(\phi(\gamma(t)))\to+\infty$ as $t\to\infty$.)

Remarks

  1. We may additionally suppose that a function $\delta(t)$ is given, and require that $V\subset\bigcup_{t\geq 0}B(\gamma(t),\delta(t))$, where $B(z,\delta)$ is the open disk of radius $\delta$ around $z$. This makes the statement of the question slightly more complicated, but makes the discussion of examples less cumbersome. (It is also part of the more general setup I mentioned above, which additionally replaces $\mathbb{C}\setminus\{0\}$ by an arbitrary open Riemann surface.)
  2. If the curve $\gamma$ is $C^1$, then it is easy to construct such a domain, by taking $V$ to be a (shrinking) "tubular" neighborhood of $\gamma$. In this case, it is even possible to ensure that the convergence is non-tangential, meaning that the argument of $\phi(\gamma(t))$ stays bounded as $t\to\infty$.
  3. It is not too difficult to see that non-tangential convergence cannot be obtained without some regularity assumption. However, perhaps a suitable uniform Lipschitz condition is sufficient.
  4. However, for horocyclic convergence, one can easily make do with much weaker conditions. For example, it seems enough to ask that the curve is $C^1$ on some sequence of intervals tending to infinity, while the behaviour in between can be as bad as you like.
  5. I tend to think that one can probably construct a counterexample for horocyclic convergence, but I may be wrong and in any case it is likely to involve some thought and effort. So I am hoping that someone can tell me that this or a similar question has been discussed in the literature.
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  • $\begingroup$ Note that the question can be restated, perhaps more concisely, as follows: Let be a Jordan arc in the complex plane, connecting 0 and 1. Is there a Jordan domain V, containing $\gamma\setminus\{0\}$, such that the convergence of gamma to 0 is horocyclic in V? $\endgroup$
    – Lasse Rempe
    Commented Jul 27, 2010 at 13:00
  • $\begingroup$ In view of your comment, I think that in the main question, you meant $\gamma(t)\to 0$ when $t\to\infty$. Am I right ? $\endgroup$
    – BS.
    Commented Jul 27, 2010 at 13:10
  • $\begingroup$ No, but it doesn't matter as you can apply the transformation $z\mapsto \frac{1}{z}$; hence my restatement. $\endgroup$
    – Lasse Rempe
    Commented Jul 28, 2010 at 22:24
  • $\begingroup$ Then, what takes the role of $0$ in the restatement ? $\endgroup$
    – BS.
    Commented Jul 29, 2010 at 14:08
  • $\begingroup$ 0 and $\infty$ are interchanged. In the restatement, the Jordan domain is supposed to be contained in $\mathbb{C}$ and have 0 on the boundary. If you transform this picture via 1/z, you get a Jordan domain that is contained in $\mathbb{C}\setminus\{0\}$ and contains $\infty$ on the boundary. $\endgroup$
    – Lasse Rempe
    Commented Jul 29, 2010 at 14:49

1 Answer 1

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First of all, notice that if $D$ is a compact and $\Gamma$ is a compact Jordan curve that starts on the boundary of $D$ but otherwise lies in the infinite component of the open complement of $D$, then you can find a polynomial that is approximately $0$ on $D$ and approximately $f$ on $\Gamma$ where $f$ is any continuous function on $\Gamma$ that is almost $0$ at the starting point of $\Gamma$. That is just Mergelyan (we can close all holes in $D$ losing nothing).

Using this, we can easily construct an entire function that approximates any continuous function $f$ we want on $\gamma$ with any precision. Indeed, choose $R_1>0$ and approximate $f$ on the piece $\Gamma_1$ of $\gamma$ that goes from the starting point until the last point of $\gamma$ on the circle of radius $R_1$ by $p_1$. Next choose $R_2$ and find a polynomial that approximates $0$ on $\Gamma_1\cup D(0,R_1)$ and $f-p_1$ on the part $\Gamma_2$ of $\gamma$ that runs from the end of $\Gamma_1$ to the last point on the circle $D(0,R_2)$, and so on. The series $\sum_j p_j$ will converge on the entire plane and be almost equal to $f$ on $\gamma$.

Now we conclude that we may have a harmonic on the entire plane function $u$ that is positive and tends to plus infinity along $\gamma$. It remains to take $V$ to be the connected component of the set $u>0$ containing $\gamma$ (then $u$ has to be the real part of the conformal mapping in question up to a positive constant factor just because there are very few positive harmonic functions in the right half plane that are $0$ on the boundary).

I leave the (harder) task to find this argument in the literature to somebody else :).

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  • $\begingroup$ 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? $\endgroup$
    – Lasse Rempe
    Commented Jul 29, 2010 at 8:59
  • $\begingroup$ 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 $\gamma$? $\endgroup$
    – Lasse Rempe
    Commented Jul 29, 2010 at 9:01
  • $\begingroup$ 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 $c>0$. $\endgroup$
    – fedja
    Commented Jul 29, 2010 at 11:43
  • $\begingroup$ Needless to say, my $u$ is the real part of the entire function, not the logarithm of its absolute value :). $\endgroup$
    – fedja
    Commented Jul 29, 2010 at 11:46
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    $\begingroup$ 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. $\endgroup$
    – fedja
    Commented Jul 29, 2010 at 12:06

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