Encouraged by the progress made in a recently posted MO problem, here is a "conceptually related" problem originating from a 2003 joint paper of Sergei Konyagin and myself.

Suppose we are given $n$ points $z_1,...,z_n$ on the unit circle $U=\{z\colon |z|=1\}$ and $n$ weights $p_1,...,p_n\ge 0$ such that $p_1+....+p_n=n$, and we want to find yet another point $z\in U$ to maximize the product $$ \prod_{i=1}^n |z-z_i|^{p_i}. $$ How large can we make this product by the optimal choice of $z$?

Conjecture. For any given $z_1,...,z_n\in U$ and $p_1,...,p_n\ge 0$ with $p_1+...+p_n=n$, there exists $z\in U$ with $$ \prod_{i=1}^n |z-z_i|^{p_i} \ge 2. $$

Here are some comments.

  • If true, the estimate of the conjecture is best possible, as evidenced by the situation where the points are equally spaced on $U$ and all weights are equal to $1$.

  • We were able to resolve a number of particular cases; say, that where the points $z_i$ are equally spaced on $U$, and also that where all weights are equal to $1$.

  • The case $n=2$ is almost trivial, but already the case $n=3$ is wide open.

  • In the general case we have shown that the maximum is larger than some absolute constant exceeding $1$.

  • Although this is not obvious at first glance, this conjecture is actually about the maxima of polynomials on the unit circle.

I would be very interested to see any further progress!


Let's create a proof a la Koosis. All the techniques used below can be found in his book "The Logarithmic Integral".

Take $a>1$ and put $f(z)=a\prod_j(1+z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on the circle with slope $\frac 12 n$ (with respect to the usual circle length) and jumps at $z_j$. Let $I$ be the image of one of the arcs between two adjacent points $z_j$ and $z_{j+1}$ under the mapping $z\mapsto \operatorname{arg}f(z)$. We can transplant all functions defined on the circle arc $[z_j,z_{j+1}]$ to $I$ using this mapping. Note that the integral of any function over the arc with respect to the circle length is just $2/n$ times the integral of its transplant over $I$ with respect to the line length.

Let $\Phi$ be the transplant of $|f|$. Assume that $\Phi<2$ on $I$. The transplant of $f$ is then just $F(t)=\Phi(t)e^{it}$. The key observation is the following: $$ \int_I \log|2-F(t)|dt\ge \log 2(|I|-\pi). $$ Assuming that it is true, we conclude that the full integral of $\log|2-f|$ over the unit circle is at least $2/n$ times the sum of the right hand sides over the intervals corresponding to all arcs, which is $0$. On the other hand, if $a>1$, then $\log|2-f(0)|=\log|2-a|<0$, so $2-f$ must have a root inside the disk and the maximum principle finishes the story.

Now let us prove the observation claim. The only thing we really know about $\Phi$ is that it is log-concave and, thereby, unimodal. Fortunately, that's all we need. So, in what follows, $\Phi$ will be just any unimodal function on $I$ with values in $[0,2]$. Since we can always extend $\Phi$ by $0$ outside $I$, we can switch to any larger interval we want without making the inequality easier. So, WLOG, $I=[-2\pi n-\frac\pi 2,2\pi n+\frac\pi 2]$

Now, let us observe that for every fixed $t$, the integrand is minimized for $\Phi(t)=2\max(0,\cos t)$ (that is just the nearest point on the line) and that the farther we go away from this optimal value, the larger the integrand is. Therefore, to minimize the left hand side, we need to stay as close to the black regime on the picture (the graph of $2\cos_+ t$) as we can.

enter image description here

Suppose that the actual $\Phi$ is given by the blue line. Then, replacing $\Phi$ by the red line $\Psi$, we come closer to the optimum at every point. But the red line consists of several full periods (horizontal pieces) and several pieces that together constitute one full positive arc of $\cos t$. Now, each full period means running over some circle around the origin, so the average value of $\log|2-\Psi(t)e^{it}|$ over each full period is exactly $\log 2$. At last, the $2\cos t$ part gives $\int_0^\pi\log (2|\sin t|)dt=0$, which is exactly the loss of $\pi \log 2$ compared to $\log 2$ times its length $\pi$.

That's it. Feel free to comment and/or ask questions.

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    $\begingroup$ Took me a while but, having eventually read carefully -- an amazing argument! You should publish it. It would be interesting to analyze what configuration of points $z_i$ and weights $p_i$ the red line corresponds to. Are there any extremal configurations other than that where the points are equally spaced, and the weights are all equal to $1$? $\endgroup$ – Seva Jun 4 '11 at 20:40
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    $\begingroup$ No. The extremum is unique (up to rotation, of course). As to publishing, if there are other people interested in such things, I'll probably make a short writeup and put it on ArXiV. :) $\endgroup$ – fedja Jun 5 '11 at 0:59
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    $\begingroup$ @jeq Indeed. I'll see if the picture is still somewhere on my old laptop hard drive (probably not :-( ) $\endgroup$ – fedja Nov 6 '17 at 1:59
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    $\begingroup$ @shurtados Found it! Enjoy :-) $\endgroup$ – fedja Nov 10 '17 at 19:58
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    $\begingroup$ @shurtados How about now? I think I just forgot to save the edits last time... $\endgroup$ – fedja Nov 10 '17 at 22:07

Here is a second proof. Let $$H(z)=\prod_j|z-z_j|^{p_j}.$$ Let $$F(z)=\prod_j\left((z-z_j)(1/z-\bar{z}_j)\right)^{p_j/n},\quad |z|\geq 1.$$ It is easy to see that: $$|F(z)|=H^{2/n}(z),\quad |z|=1,$$ and that $F$ maps conformally the exterior of the unit disc on the exterior of a "star" consisting of $n$ straight segments $[0,a_j]$ with some complex $a_j$. Moreover, $F(z)\sim z$ as $z$ tends to infinity.

Now we see that the result is equivalent to the following theorem of Dubinin: Let $K$ be the union of some intervals of the form $[0,a_j]$, and suppose that capacity of $K$ is $1$. Then $\max_j|a_j|$ takes its minimal value when the star is symmetric: all |a_j| are equal and the angles between adjacent intervals are equal.

The reference is Dubinin, Uspekhi Mat. Nauk (=Russian Math. Surveys), 49 (1994). Fedja's proof is much simpler, so it is a new proof of Dubinin's theorem.

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  • $\begingroup$ Sounds interesting, but I do not quite get it yet. In the third displayed equality, the exponent should be $2/n$; correct? Could you explain why "$F$ maps conformally the exterior etc" (and what are the $a_i$s)? Exactly what do you mean by $F(z)\sim z$? What is the capacity in your context and why the assertion is equivalent to Dubinin's theorem? Thanks in advance for the explanations (and welcome to the MO)! $\endgroup$ – Seva Aug 4 '12 at 9:25
  • $\begingroup$ Seva, I corrected the misprint 2/n. $F(z)\sim z$ means as usual that $F(z)/z\to 1$ as $z\to\infty$. To see that it is a conformal map and to find the image, look how it behaves on the unit circle: the argument is constant on the arcs between any $z_j,z_{j+1}$. So these arcs are maped on some segments $[0,a_k]$. What are the $a_k$? Some complex numbers. $|a_k|$ is the maximum of $|F|$ on the corresponding arc. The angles between the segments are $2\pi p_j/n$. $\endgroup$ – Alexandre Eremenko Aug 4 '12 at 17:22
  • $\begingroup$ And what is the capacity of $K$? $\endgroup$ – Seva Aug 4 '12 at 18:11
  • $\begingroup$ You can look for various equivalent definitions of logarithmic capacity in any textbook, but for our purposes a connected compact has capacity $1$ if the conformal map of its exterior onto the exterior of the unit disc safisfies $F(z)/z\to 1$. Our map satisfies this. $\endgroup$ – Alexandre Eremenko Aug 4 '12 at 19:59

Not an answer, but maybe an useful reference:

Your statement reminds me of Schiefermayr's theorem:

If $p_n$ is a degree $n$ polynomial and $\Sigma\subseteq\mathbb{R}$. Then $$ \|p_n\|_{L^{\infty}(\Sigma)} \geq 2 Cap(\Sigma)^n $$ where $Cap(\Sigma)$ is the logarithmic capacity of $\Sigma$, e.g. $Cap([-2,2]) = 1$.

This is Corollary~5.7.7. in Barry Simon's book: Szego's Theorem and its decendents.

Unfortunately, this does not apply in your case, since it requires the work with subsets of the real line. However, you are close to this situation, since you require the zeros of your polynomial to be on the unit circle.

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  • $\begingroup$ Another problem seems to be that my product is not a polynomial in $z$. $\endgroup$ – Seva May 9 '11 at 5:24

Here is an observation that didn't get me very far. Let $p$ have degree $n$ and have $r$ distinct roots, all on the unit circle. Call these roots $e^{i \theta_j}$, where $\theta_k$ has multiplicity $c_j$.

Set $$q(u) = \frac{1}{n} \sum_{j=1}^r \left( c_j \left( u + e^{i \theta_j} \right) \prod_{\substack{1 \leq k \leq r \\ k \neq j}} \left( u - e^{i \theta_k} \right) \right).$$ Then $q$ is a monic polynomial of degree $r$, all of whose roots are on the unit circle. They occur at precisely the places where $|p(e^{i \phi})|$ has a local maximum. Your goal is to show that, at one of these local maxima, we have $|p| \geq 2^{n/r}$.

I had a detailed computation of this written out, but I couldn't find a way to make it useful, so I'm recording the formula in case it helps someone else. My plan was to show that $\prod_{\{\phi : q(e^{i \phi})=0\}} |p(e^{i \phi})| \geq 2^n$, but this inequality turned out to be false; if all the $\theta$'s are very close together, then one of the local maxima of $p$ is near $2^n$ but the other local maxima can be arbitrarily small.

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  • $\begingroup$ By the way, an interesting observation is that the product of the roots of $q$ is negative that of the roots of $p$. That also didn't help me, though. $\endgroup$ – David E Speyer May 23 '11 at 22:20

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