I recently found out (Piranian, "The Shape of Level Curves") that a polynomial tract (ie a connected component of a set of the form $G=\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) need not be starshaped with respect to the zeros of $p$ contained in $G$. This disappointed me bitterly, as that starshapeness was a pivotal step in a proposed "proof" I had of Smale's mean value conjecture.

The places where $G$ is not starshaped with respect to the zeros of $p$ in $G$ are near critical points of $p$ in $G$ or in $\partial G$, so I still hold out a tiny bit of hope for the starshapeness of $G$ with respect to the critical points of $p$ contained in $G$:

Conjecture: If $G$ is a tract of $p$ with smooth boundary containing more than one distinct zero of $p$, then $G$ is starshaped with respect to the critical points of $p$ contained in $G$.

Intuitions/proofs/disproofs/references are all welcome.

EDIT: Note that when I say that $G$ should be "starshaped with respect to the critical points", I mean that each point in $G$ can be seen by some one of the critical points of $p$ in $G$, not of course that some single critical point can see all points in $G$.

Note also that I added the assumption that $G$ contains more than one distinct zero of $p$ (since otherwise $G$ will not contain any critical points of $p$.

One reason I think this is plausible: If we consider the lemniscate of Bernoulli, and let $G$ be the interior of a level curve of $p$ which is a bit bigger, the critical point of $p$ is right in the center, so should be able to "see" both lobes. In the counter-example of Piranian to my desired conjecture (that tracts are star-shaped with respect to the zeros they contain), the points that killed the starshapeness were close to the boundary of $G$, so perhaps if we assume $\partial G$ is smooth, $G$ will contain enough critical points to see into all "corners".

  • $\begingroup$ Do you assume that $G$ is connected? $\endgroup$ Dec 4, 2014 at 20:06
  • $\begingroup$ Yes, a tract is a component of the set $\{z:|p(z)|<\epsilon\}$. I will add that to the definition. $\endgroup$ Dec 5, 2014 at 0:23

1 Answer 1


There is no hope: according to a theorem of Hilbert, every analytic Jordan curve $J$ can be approximated by a lemniscate $\{z:|P(z)|=\epsilon\}$. So the set does not have to be starlike with respect to any point. For this theorem of Hilbert, see, for example J. L. Walsh, ``Interpolation and Approximation by Rational Functions in the Complex Plane,'' 5th ed., Amer. Math. Society, Providence, RI, 1969.

  • $\begingroup$ There is also a simple proof using potential theory in the book "Potential theory in the complex plane" by Thomas Ransford, Theorem 5.5.8. I don't have access to Walsh's book right now, so I don't know if the proof is the same. $\endgroup$ Dec 3, 2014 at 0:18
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    $\begingroup$ Ah, but I do not need it to be starshaped with respect to just one point, I want it to be starshaped with respect the family of critical points. Thus, each point in the tract can be "seen" by some critical point. I will make that more clear in the question. $\endgroup$ Dec 3, 2014 at 15:06
  • $\begingroup$ Moreover, it seems that if the Jordan Curve $J$ that the lemniscate $\Lambda=\{z:|p(z)|=\epsilon\}$ approximates is very complicated, then the bounded face of $\Lambda$ may need to contain many zeros of $p$ for $\Lambda$ to approximate $J$ well. $\endgroup$ Dec 4, 2014 at 15:41
  • $\begingroup$ (continued) and thus of course the bounded face of $A$ must contain many zeros of $p'$ if it contains many zeros of $p$. $\endgroup$ Dec 6, 2014 at 22:50

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