Newton basin fractals are visualizations of the Julia sets of functions of the form:

$$f_p(z) = z - p(z)/p'(z)$$

where $p$ is a complex polynomial. My question is:

When is the Julia set, $J(f_p)$, continuously determined by the roots of $p$, in the sense of the Hausdorff metric?

I have convinced myself that it is a necessary condition that the roots of $p$ be simple, and I'm happy to elaborate on my reasoning, but I suspect that the statement is either obvious (or obviously false) to experts. (Update: It turns out that it is obviously true.)

B. Krauskopf and H. Kriete proved that:

If a sequence of meromorphic functions $f_n$ converges uniformly on compact sets to a limit function $f$, and $F(f)$ is a union of basins of attracting periodic orbits, and $\infty\in J(f)$, then $J(f_n)$ converges to $J(f)$ in the Hausdorff metric.

And at first glance that theorem looks like a great starting point for showing that the simple roots condition is sufficient. Unfortunately it is not the case that $F(f_p)$ is a union of attracting basins whenever $p$ has simple roots.

Is anything more known about this? Perhaps there is some other condition (not merely simple roots) that makes $F(f_p)$ a union of attracting basins.

I expected for this to be a well studied problem. An answer to the related question of "When is convergence of Newton-Raphson iteration not sensitive to perturbation of roots?" seems like it would have useful applications. But I have not found much beyond results like Krauskopf and Kriete's theorem, above, which don't quite apply.

Edit: To give the question a little more context and motivation: Here is a program to demonstrate that in some sense, perhaps not exactly the way I've phrased it, a Newton basin fractal may transform continuously under changes to the roots of its polynomial. In the demonstration (which requires a WebGL enabled browser) the Newton basin fractal for the polynomial $z\to (z^2 - 1)(z - \lambda)$ is rendered in a 10x10 box centered at the origin. Clicking and dragging updates $\lambda$ and the resulting basins.

Edit 2: I've updated the demonstration in an attempt to get more information onto the screen. Now the usual Newton basin rendering is blended with a rendering of the parameters $\lambda$ such that the sequence of iterates starting at the critical point converges very slowly. These are the parameters for which the Fatou set might include components that are not basins of attracting periodic orbits, preventing application of Krauskopf and Kriete's theorem.

The roots are highlighted in yellow and the critical point is highlighted in cyan. It's easy to see that the Julia set experiences a discontinuous transformation when any two roots overlap, and that when $\lambda$ is in the "problem area" the critical point is sort of pressed towards the Julia set (which makes sense given how the problematic parameters are determined in the first place). But even as I move $\lambda$ through the problematic region I don't see the Julia set transforming in a discontinuous way, certainly nothing as obviously discontinuous as the transformations that occur when two roots overlap. I wonder if the Julia set actually is continuously determined by $\lambda$ in this region, even if the theorem above isn't quite enough to prove it.


1 Answer 1


The question of when the Julia set of a function depends continuously on the parameters is well-studied. For the case of polynomials, see Douady, Does a Julia set depend continuously on the Polynomial? Proceedings of Symposia in Applied Mathematics, 49, 91–135, 1994.

This implies that the answer for your question is negative. More precisely, consider the bifurcation locus of the family that you are looking at; i.e., the set of unstable parameter value. (It is well-known that such parameters exist; e.g. take points on the boundaries of little Mandelbrot set copies in the parameter plane.)

Now, in this locus, the set of parameters having a linearizable irrationally indifferent periodic point is dense. For these parameters, the periodic point is in the Fatou set, and hence has positive distance from the Julia set. On the other hand, arbitrarily small perturbations will lead to parameters with parabolic points, where the periodic point is in the Julia set. This proves that the Julia set does not depend continuously near such points.

(The answer is also negative near parabolic parameters.)

Of course, if you are in the complement of the bifurcation locus (say, with all critical points tending to attracting orbits), then the Julia set depends continuously, and even moves holomorphically.

There is always a semicontinuous dependence of the Julia set, due to the fact that repelling periodic orbits are dense, and that these are stable under perturbations. In other words, under small perturbations the Julia set can get larger in a discontinuous way, but not smaller.

  • $\begingroup$ Thank you for the answer, Lasse. I want to make sure I understand your comments, so please forgive this possibly naive clarification request: Does the failure of the (any) critical point to tend to an attracting cycle imply that the corresponding parameter belongs to the bifurcation locus, or is this merely a necessary condition? You mentioned the boundaries of the little Mandelbrot set copies, and indeed the set of "problematic parameters" from my example includes the boundaries of many scaled and rotated copies of the Mandelbrot set, but it is not composed entirely of such boundaries. $\endgroup$ Commented Aug 10, 2014 at 22:12
  • $\begingroup$ Even in the Mandelbrot set, it is not known whether the only stable parameters are given by those where the critical point tends to attracting cycles. This ("Density of Hyperbolicity") is a very famous conjecture, and would follow from local connectivity of the Mandelbrot set (which is perhaps the most famous open problem in the entire field). $\endgroup$ Commented Aug 10, 2014 at 23:09
  • $\begingroup$ Thank you for the extra information, Lasse. It sounds like the problem of characterizing the bifurcation loci of these families ($z\to z - p(z)/p'(z)$ for whatever polynomial $p$) is hard. I'm not sure how to make the idea precise, but the little Mandelbrot set copies make me think that characterizing the stability properties of even just the family, $z\to (z^2-1)(z-\lambda)$ should be "at least as hard" as resolving the Density of Hyperbolicity conjecture. $\endgroup$ Commented Aug 13, 2014 at 1:35

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