# Convex Julia sets

Consider the classical Julia set $$J_f$$ associated with $$f(z)=z^2+c$$. Since $$J_c$$ is completely invariant, we know that $$f^{-1}(J_f) \subseteq J_f$$. Now, let $$H_f$$ be the convex hull of $$J_f$$.

Is it true that $$f^{-1}(H_f) \subseteq H_f$$?

I have done some basic computer experiments, and it seem to hold for $$c \in [0,1]^2 \subset \mathbb{C}$$. Moreover, I suspect that the natural generalization of the statement above might hold for all polynomial maps. However, I have examples with rational maps where the statement is not true.

As an example, consider $$f(z)=z^3-iz + 0.2 + 0.4i$$. The blue points is the Julia set $$J_f$$ associated with $$f$$. The shaded region is the convex hull $$H_f$$ of the Julia set. Taking a uniform square grid $$G$$ on $$H_f$$, and plotting the points $$f^{-1}(G)$$ gives the black dots. As we can see, it is reasonable to guess that $$f^{-1}(H_f)\subset H_f$$.

• This is indeed true for certain values of $c$ e.g. $c<-2$ in which case $J_{f_c}$ is a Cantor subset of the real axis and $H_{f_c}$ is a compact interval. But I can't see how this could be true when the filled Julia set has non-empty interior (special cases with smooth Julia sets such as $c=0$ aside)? Have you verified it for the basilica Julia set (c=-1)? Apr 3, 2020 at 1:46
• @KhashF Indeed, the pictures I can make definitely supports the conjecture that it works for c=-1 also. Apr 3, 2020 at 9:53

Edited: The previously found sufficient condition is indeed necessary, but even better, it is satisfied by all polynomials of degree at least two. Thus the conjecture is true:

Theorem: Let $$p$$ be a complex polynomial of degree $$d \geq 2$$ and let $$H_p={\rm conv}J_p$$, the convex hull of the Julia set $$J_p$$ of $$p$$. Then $$p^{-1}(H_p) \subset H_p$$.

To prove this, I use several lemmata. First, there is the following version of Gauss-Lucas theorem, due to W. P. Thurston, which can be found in the preprint

A. Ch\'eritat, Y. Gao, Y. Ou, L. Tan: \emph{A refinement of the Gauss-Lucas theorem (after W. P. Thurston)}, 2015, preprint, hal-01157602

Lemma 1: Let $$p$$ be any polynomial of degree at least two. Denote by $$\mathcal{C}$$ the convex hull of the critical points of $$p$$. Then $$p: E \to \mathbb{C}$$ is surjective for any closed half-plane $$E$$ intersecting $$\mathcal{C}$$.

From this we have the following:

Lemma 2: Let $$p$$ be any polynomial of degree at least two. Then all zeros of $$p'$$ belong to $$H_p={\rm conv}J_p$$, the convex hull of the Julia set $$J_p$$ of $$p$$.

Proof: Suppose there is an $$x_0 \not \in H_p$$ such that $$p'(x_0)=0$$. By the hyperplane separation theorem , there exists a closed half-plane $$E$$ such that $$x_0 \in E$$ and $$E \cap H_p = \emptyset$$. In particular, $$E \cap J_p = \emptyset$$. By Lemma 1, $$p: E \to \mathbb{C}$$ is surjective. Take a $$z_0 \in J_p$$. Then on one hand $$p^{-1}(z_0) \subset J_p$$, while on the other hand $$p^{-1}(z_0) \cap E \neq \emptyset$$, a contradiction.

The next lemma is a modification of Exercise 2.1.15 in Lars Hörmander: Notions of Convexity Publisher Springer Science \& Business Media, 2007 (Modern Birkhäuser Classics) ISBN 0817645853, 9780817645854

Lemma 3: Let $$p(z)=\sum_{j=0}^d a_jz^j$$ be a polynomial in $$z \in \mathbb{C}$$ of degree $$d$$. Let $$B$$ be a closed convex subset of $$\mathbb{C}$$ containing all zeros of $$p'$$. Then the set $$C_B$$ of all $$w \in \mathbb{C}$$ such that all the zeros of $$p(\cdot)-w$$ are contained in $$B$$ is a convex set.

Proof of Lemma: Note that by continuity of roots $$C_B$$ is closed when $$B$$ is. Let $$w_1,w_2 \in C_B$$ and $$n_1,n_2 \in \mathbb{N}$$ and consider the polynomial (in one complex variable $$z$$) $$P(z):=(p(z)-w_1)^{n_1}(p(z)-w_2)^{n_2}$$. Then all zeros of $$P$$ lie in $$B$$ (by definition of $$C_B$$), so the convex hull of zeros of $$P$$ is contained in $$B$$. By Gauss-Lucas theorem (standard version), all zeros of $$P'$$ are contained in $$B$$. The zeros of $$P'$$ are respectively all the zeros of $$p(z)-w_1$$, all the zeros of $$p(z)-w_2$$ (if $$n_1, n_2 >1$$), all the zeros of $$p'$$ and all the zeros of $$p(\cdot)-\left (\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2\right)$$. By definition of $$C_B$$, $$\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2 \in C_B$$. Varying $$n_1,n_2$$ and using the property that $$C_B$$ is closed, we get that $$tw_1+(1-t)w_2 \in B$$ for all $$0 \leq t \leq 1$$.

Proof of Theorem: Applying the Lemma 3 to $$B=H_p={\rm conv}J_p$$, we get that the set $$C_p=\{w \in \mathbb{C}: \{z : p(z)-w=0\} \subset H_p\}$$ is convex. Furthermore, for $$w \in J_p$$ all solutions of $$p(z)-w$$ are in $$J_p \subset H_p$$, so $$J_p \subset C_p$$. Hence $$H_p \subset C_p$$, which implies that $$p^{-1}(H_p) \subset H_p$$.

For the quadratic family $$f_c(z)=z^2+c, \ c \in \mathbb{C}$$ it is straightforward (without appealing to Lemma 2) to check that the critical point $$0$$ is the center of symmetry of the Julia set $$J_c$$, so it is a convex combination of two points in $$J_c$$.