The Gauss--Lucas Theorem states that all zeros of derivative of a degree $n$ complex polynomial $p(z)$ are contained in the convex hull of the zeros of $p$. By iteration, this implies that the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$ are contained in the convex hull of the zeros of $p$.

The Riemann--Hurwitz Theorem (among others) implies that if a tract $D$ of $p$ (namely a component of the set $\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) contains all the zeros of $p$ in its bounded face, then all the critical points of $p$ are contained in $D$.

My conjecture is that in fact, if $D$ is a tract of $p$ and contains all the zeros of $p$, then $D$ also contains all the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$.

This certainly does not follow by straight-forward iteration, since in general there may not exist some tract $D'$ of $p'$ which contains all the zeros of $p'$ in $D$, and is itself contained in $D$. It seems that the tracts and level curves of $p'$ do not interact very nicely with the tracts and level curves of $p$ (even worse for $p'',p''',\ldots$).

This question was originally asked on M.SE, where it received no answers, however user Behavior did point out that this conjecture could be equivalently stated as:

Conjecture: Let $M=\max(|p(z)|:p'(z)=0)$, then if $w$ is a zero of any derivative of $p$, then $|p(w)|\leq M$.


2 Answers 2


Very interesting problem. At first I thought the statement might be true for real valued polynomials, but unfortunately I found the following,

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Unless I am misunderstanding the problem, I believe this is the counterexample you were looking for. Although, we note that the second derivative does look suspiciously still trapped inside of the tract. Maybe, your conjecture holds for only the first and second derivative.

  • 1
    $\begingroup$ Thanks very much for the counter-example. The conjecture definitely holds for the first deriv. If, $x_0$ is a crit. point of $p$, and $\lambda$ is the level curve of $p$ containing $x_0$, then the max. mod. thm. implies that each face of $\lambda$ contains a zero of $p$. Thus any level curve of $p$ containing the zeros of $p$ in its bounded face must also contain $\lambda$ in its bounded face, and thus contain $x_0$ in its bounded face. For the second derivative, it looks like if you tighten up the parameters on the counter-example, the second deriv. zeros might be outside the level curve. $\endgroup$ May 4, 2015 at 15:31
  • $\begingroup$ I did spend some time trying to put the second derivative's zeros outside the tract, but could not successfully. Need to spend some more time with it. $\endgroup$ May 4, 2015 at 16:08
  • $\begingroup$ I believe your conjecture is true for polynomials with only real zeros. Since any level set that contains all of the zeros must also contain the critical points, it follows that the interval $[x_0,x_n]$ must be contained in the level set as well (where $x_0$ is the smallest zero and $x_n$ is the largest zero). Since it is know that derivatives interlace, then your conjecture follows. $\endgroup$ May 4, 2015 at 17:01
  • $\begingroup$ I believe, I can prove your conjecture is true for all real valued cubic polynomials; i.e., the second derivative's zeros will be contained in the tract as well as the first derivative's zeros. Oh well, time for work. I will definitely give this some more thought. $\endgroup$ May 4, 2015 at 17:41
  • $\begingroup$ Nice example - how did you go about finding it? $\endgroup$ May 5, 2015 at 11:59

I believe that your conjecture is incorrect. Let me give you an indication why this should be true. (I am not claiming that this is a rigorous proof, although I believe one could be furnished with enough attention. However, knowing that the result should be false, it may be easier to find an example in some other way.)

Firstly observe that, if true, your statement would imply the same for entire functions with a finite set of singular values (critical or asymptotic values). Indeed, such a function f can be approximated, in the sense of locally uniform convergence, by polynomials with the same singular values. (The prototype of this is the convergence of unicritical polynomials (1+z/d)^d to the exponential map; in general, this can be shown using the concept of so-called "line complex"; see the final chapter of the book by Gol'dberg and Ostrovskii.)

Obviously, if a higher derivative of f had a zero that was outside of the disc in question, then the approximating polynomials would also.

I think the same is probably true for functions with bounded set of singular values, i.e. the so-called Eremenko-Lyubich class (where we approximate by polynomials whose set of critical values approaches the set of singular values of f).

Now, if you let M be the maximal modulus of a singular value, as in the final version of your conjecture, then every component of $V = \{z\in\mathbb{C}: |f(z)|>M\}$ is mapped to the punctured disc $W = \{|z|>M\}$ as a universal covering map. Your conjecture would imply that the higher derivatives of this map have no zeros, but you clearly cannot expect this. Indeed, essentially any such covering can be approximated in a suitable sense by a corresponding one coming from an entire function with bounded or even finite singular sets; see my paper on "Approximation by Eremenko-Lyubich functions" (Proc. LMS), and Chris Bishop's far more sophisticated work on "Models for the Speiser class". While there is some work on the details required, it seems pretty clear that you can make examples this way - although I doubt it is the easiest route ...

(See also Chris's previous proof that "true trees are dense"; Inventiones, 2014.)

I hope this makes sense, and apologies if I am missing something.

  • $\begingroup$ Thank you very much for the response, and the references. I follow your answer, and have been working through some examples. Several (naive) observations: An easy collection of examples are the functions $f(z)=p(z)e^{q(z)}$, where $p$ and $q$ are polynomials. For this case it appears that $V$ has a single component, and the simple examples I have already done (ie $ze^z$, $e^{z^2+z}$) work out fine. Alternatively, a simple example where $V$ has several (here $2$) components is $f(z)=\cos(z)$. Here also the higher derivatives of $f$ have no roots in $V$. (Question in the next commment.) $\endgroup$ Dec 12, 2014 at 2:39
  • $\begingroup$ Question 1) Is there an easy example you can think of where the higher derivs of $f$ have roots in $V$? (Obviously if the answer were immediate, you would have given it earlier, but perhaps my examples above my suggest something.) $\endgroup$ Dec 12, 2014 at 2:41
  • $\begingroup$ Question 2) I am not so sure now that I understand the application of your statement that if a higher deriv. of $f$ had a zero outside the disk in question, then the higher deriv of the approximating polys would as well. After all, while the function $f$ has a fixed tract containing all of its crit. pts., the tracts which contain the crit. points of the approximating polys may be growing very large, since they only have the same crit values as $f$, not necessarily the same crit. points as $f$, is this right? $\endgroup$ Dec 12, 2014 at 2:51
  • $\begingroup$ @TrevorRichards: The trouble with the "easy" examples is that they all essentially look like e^z, which does not have higher-order critical points ... One case of entire functions with a finite set of singular values that it's relatively easy to get your hands on, and that have quite different tracts from exponential maps, are Poincaré (linearising) functions of post-critically finite polynomials around repelling periodic points. $\endgroup$ Dec 12, 2014 at 9:47
  • $\begingroup$ @TrevorRichards Alternatively, if you are looking for explicit examples of polynomials, I suggest looking at some Shabat polynomials (= polynomial Belyi functions), i.e. polynomials with two critical values (or -1 and 1 may be the best normalisation for your question). Given any tree, with an embedding in the plane, there is a Shabat polynomial realising this tree. There are some programs for computing these (eg Don Marshall's "zipper", and Laurent Bartholdi also has a program). Not sure they're publicly available, but they exist. Just draw some "complicated" trees and experiment .. $\endgroup$ Dec 12, 2014 at 9:52

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