The following problem arose in collaborative work with Subhro Ghosh:

Question: To any polynomial $P_n(z)=\sum_{i=0}^n a_i z^i =a_n \prod_{i=1}^n(z-z_i)$, attach the empirical measure of zeros $L_n=n^{-1}\sum \delta_{z_i}$ (a probability measure on the complex plane). Let ${\cal M}$ denote the collection of limit points (in the weak topology) of those empirical measures obtained when all $a_i$ are constrained to be non-negative reals (but otherwise, arbitrary $a_i$), with limit points being probability measures. Can ${\cal M}$ be characterized?

Remarks: of course, probability measures in $\cal M$ are symmetric about the real axis, so enough to look at restriction of measures to the (closed) upper half plane. Probability measures supported on the negative half plane belong to $\cal M$. On the negative side, by a theorem of Obrechkoff, any probability measure in $\cal M$ charges the symmetric cone of total width $\alpha$ around the real axis with mass at most $\alpha/\pi$. Finally, for random polynomials with $a_i$ iid with bounded moments of some order, the limit point is the uniform measure on the circle, and from this (or using products of polynomials $\sum_{i=0}^k z^i$) one easily shows that any radially symmetric probability measure belongs to $\cal M$. Results of Barnard et als on factoring polynomials with positive coefficients allow one to construct further examples of measures in $\cal M$. There are more examples, but we look for a characterization of $\cal M$.

  • $\begingroup$ Ofer, can you be more specific on "many examples"? Do you have an example which contradicts the conjecture that Obrechkoff's condition is necessary and sufficient? $\endgroup$ Jul 8, 2013 at 12:15
  • 1
    $\begingroup$ And what is the reference on Obrechkoff? $\endgroup$ Jul 8, 2013 at 12:56
  • 1
    $\begingroup$ Sorry for the delay in answering, I was travelling.The reference is CRAS (1923) gallica.bnf.fr/ark:/12148/bpt6k3130n/f112.image $\endgroup$ Jul 11, 2013 at 16:23
  • $\begingroup$ And I don't have an example contradicting the sufficiency of Abrechkoff's condition. I wrote ``more examples'' because you can use e.g. the above mentioned results of Barnard to sweep out all zeros from the positive sector, and create thus new examples from a given one. I will review your answer below, it looks very promising. Thanks. $\endgroup$ Jul 11, 2013 at 16:29
  • $\begingroup$ Ofer, thanks. What is the reference on Barnard? I am still confident that what I wrote is true, but the details are somewhat technical, and it will take some time to write. $\endgroup$ Jul 12, 2013 at 8:45

1 Answer 1


I don't have a complete proof yet, but I have a plausible conjecture. Let $\mu$ be a probability measure in the plane, define the potential $$u(z)=\int\log|1-z/t|d\mu(t).$$ Then I conjecture that $\mu\in M$ iff $u$ satisfies $u(z)\leq u(|z|)$ for all $z$.

It is evident that this condition is necessary. It seems that it is strictly stronger than the Obrechkoff condition. I don't think that this condition can be restated as a simple property of $\mu$ itself.

To prove the sufficiency, I am first going to restrict to a dense subclass of $\mu$ with convenient properties (it is clear that it is enough to prove sufficiency for a dense subclass). The convenient properties I have in mind is that $\mu$ does not charge some small angular sector $|\arg z|<\epsilon$ and that it behaves nicely near $0$ and $\infty$, say has some small atom at $-\epsilon$ and nothing else in the disc $|z|<100\epsilon$, and similarly at infinity. In addition, I want to require that $u(|z|)>u(z)$ for all $z$ except on the positive ray.

Then I am going to discretize the measure to obtain a polynomial, whose $(1/n)\log|P_n|$ approximates $u$ nicely near the positive ray, and apply the saddle point method to the integral $$\int_{|z|=r}\frac{P_n(z)}{z^k}\frac{dz}{z},$$ with $n\to\infty$, using the nice behavior near the positive ray, and obtain an asymptotic for the coefficients which will show that they are positive.

The difficulty is that the asymptotics must be uniform in $k$, but I hope to achieve this by the arrangement near $0$ and $\infty$ described above.

In fact, there is an (unpublished and unproved) conjecture of Alan Sokal that if a polynomial satisfies $|P(z)|<P(|z|)$ then some sufficiently high power has positive coefficients. This of course would imply sufficiency of my condition.

ADDED on July 19. The above outline is correct; we are writing a proof which will soon be posted on arxiv.

ADDED on August 23. Here is the precise statement. A probability measure $\mu$ is a limit measure if and only if it is symmetric with respect to complex conjugation, and $u(z)\leq u(|z|)$ where $$u(z)=\int_{|\zeta|\leq 1}\log|z-\zeta|d\mu(\zeta)+\int_{|\zeta|>1}\log|1-z/\zeta|d\mu(\zeta).$$ (The potential I wrote earlier may be divergent for some probability measures, so it has to be modified a little bit). A proof of this is now available: https://arxiv.org/abs/1409.4640.

UPDATE on September 10, 2014. What I called "Sokal's Conjecture" above (Theorem 1 in the preprint cited above) turned out to be known before. It was proved by V. de Angelis, MR1976089. This was found as a result of David Handelman's answer to another MO question: Stability of real polynomials with positive coefficients.

  • $\begingroup$ This is nice. I was confused about the "evidently necessary" condition at first. I mean, it is true when $\mu$ is given by a polynomial with positive coefficients and giving no weight to $z$. However, $\log\lvert 1-z/t\rvert$ is not a continuous bounded function of $t$, so passing to the weak closure doesn't seem obvious. However, if you write $v_z(t)=\log\lvert(1-\lvert z\rvert/t)/(1-z/t)\rvert$ then this is bounded as $t\to\infty$. If you write $v_\phi(t)=\int v_z(t)\phi(z)d^2z$ for smooth nonegative $\phi$ with compact support in $\mathbb{C}^*$ then this is continuous and bounded. $\endgroup$ Jul 13, 2013 at 14:24
  • $\begingroup$ The condition is then $\int v_\phi(t)d\mu(t)\ge0$ for all such $\phi$. That is, $\int(\log\lvert 1-\lvert z\rvert/t\rvert-\log\lvert 1-z/t\rvert)d\mu(t)\ge0$ holds for $z\in\mathbb{C}^*$ in the sense of distributions. I'm not sure if this is all obvious, or is how it was intended to be understood, but had me confused about the validity of the necessary condition for a while. $\endgroup$ Jul 13, 2013 at 14:28
  • $\begingroup$ George, have you already resolved your difficulty, or you need an explanation? The hint is that u(|z|) is a subharmonic function which depends only on |z|. When measures converge weakly such functions converge uniformly on every compact set. $\endgroup$ Jul 13, 2013 at 20:51
  • $\begingroup$ A good general reference for convergence of potentials is Hormander, Notions of Convexity, Birkhauser, 1994, especially Thm 3.2.13. $\endgroup$ Jul 14, 2013 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.