Does there exist a rational function $f\in\Bbb{C}(z)$ whose Julia set coincides with $$ T:=\left\{\left(x,\sin\left(\frac{1}{x}\right)\right)\,\Big|\,x\in\left(0,\frac{1}{\pi}\right]\right\}\cup\big(\{0\}\times[-1,1]\big)\;? $$ What if we require the Julia set to only be homeomorphic with $T$?
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5$\begingroup$ The complement of $T$ is simply connected in the Riemann sphere, so the Fatou set of any map $f$ with $J(f) = T$ has a single component $U$, and the classification of Fatou components tells us what $U$ could be like. $U$ is neither a Hermann ring (because it's simply connected) nor a Siegel disk (because it's clearly not the image of the unit disk under an analytic map). So either $U$ is parabolic or it contains a single attracting fixed point. Have you investigated what either of these possibilities would imply about $f$? $\endgroup$– Sophie MCommented May 24 at 0:07
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2$\begingroup$ Slightly subtle aspect of the simple connectedness issue (because this caused me to doubt myself after I posted that comment): $U$ isn't a Hermann ring because it's simply connected in the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, and it's not a Siegel disk because it's not simply connected in the plane $\mathbb{C}$. $\endgroup$– Sophie MCommented May 24 at 6:48
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2$\begingroup$ @SophieM Thanks for your comments. I think the easiest way to argue that $f$ cannot admit a Siegel disk or Herman ring is to notice that $f$ should be injective on such a component. But there is only one Fatou component and a map of degree $d\geq 2$ restricts to a $d$-to-$1$ self-map on its Fatou set. $\endgroup$– KhashFCommented May 24 at 14:55
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The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.
Addressing the comment. Use the following Lemma. Let $g$ be a germ of an analytic function is a neighborhood of $z_0$, $w_0=g(z_0)$ and $E$ is an arbitrary set containing $w_0$. If the component of $g^{-1}(E)$ containing $x_0$ is a simple curve, then intersection of $E$ with a neighborhood of $w_0$ is either a simple curve, or a semi-closed simple curve with one end at $w_0$.
Proof. WLOG $z_0=w_0=0$, and $g(z)=z^m$. This makes the statement evident. In the first case, $m=1$, in the second case $m=2$.
It follows that the Julia set $J$ is a bordered closed 1-manifold, and thus $J$ is homeomorphic to a circle or to a segment.
Edit. There is also a reference on this result: Theorem A in this paper.
answered May 24 at 11:52
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$\begingroup$ I see why the Locally Eventually Onto property implies that the Julia set would have to be locally connected if it contains an arc with interior. But I cannot see why it would have to be a simple curve or Joran curve. Could you give a bit more explanation about this? $\endgroup$ Commented May 26 at 14:16
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$\begingroup$ @AlexandreEremenko Thanks for your answer. Indeed, once a small piece of the Julia set is a simple curve, the whole Julia set must be path connected by your argument because it is a surjective image of that piece. And we know that is not the case for the topologist's sine curve. $\endgroup$– KhashFCommented May 28 at 2:37
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1$\begingroup$ KhashF: yes. But I sketched a proof and gave a reference to a stronger result. $\endgroup$ Commented May 28 at 12:50