Dubinin's theorem

In today's post our aim is to prove the following:

For any given $z_1,\ldots,z_n \in U := \{z : |z| = 1\}$ and $p_1,\ldots,p_n \geq 0$ with $p_1 + \cdots + p_n = n$, there exists $z \in U$ with

\[\prod_{i=1}^n |z - z_i|^{p_i} \geq 2.\]

Equality holds precisely when the points are equally spaced on $U$ and all weights are equal to $1$.

I thought this seemed like a natural enough question, something that could've appeared on a university maths contest. We will see, however, that it is in fact equivalent to a theorem of Dubinin that took a lot of effort to prove! It is therefore all the more impressive that a short and clever proof of this result was found on MO, which we will follow.

Proof:  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 $\frac{1}{2n}$ (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 $z \mapsto \arg f(z)$. 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:

\[\int_I \log|2 - F(t)|\,dt \geq \log 2 (|I| - \pi).\]

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.

Recall that a  function $f : \mathbb{R}^n \to \mathbb{R}^+$ is logarithmically concave (or log-concave for short) if its domain is a convex set, and if it satisfies the inequality

${\displaystyle f(\theta x+(1-\theta )y)\geq f(x)^{\theta }f(y)^{1-\theta }}$

The only thing we really know about $\Phi$ is that it is log-concave and, thereby, unimodal, i.e. it attains its maximum at a unique value. 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, $I = [-\frac{2\pi}{n} - \frac{\pi}{2}, \frac{2\pi}{n} + \frac{\pi}{2}]$.

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.

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 $\pi$. 

Chasing through carefully and noting when equalities occur gives the equality case.$\blacksquare$

Definition: For a compact set $E$ in the complex plane we define the Robin's constant of $E$ \[V(E) = \inf_{\nu} \int_{E \times E} \ln \frac{1}{|u - v|} \,d\nu(u)\,d\nu(v), \]

and $\nu$ runs over each probability measure on $E$. 

Definition: The logarithmic capacity of $E$ is given by

\[\gamma(E) = e^{-V(E)}, \] 

and the set $E$ is said to be polar if $V(E) = +\infty$ or equivalently, $\gamma(E) = 0$.

Definition: Let \[\phi(z) = cz + c_0 + c_1 z^{-1} + c_2 z^{-2} + \cdots\]

be an analytic function, regular and univalent for $|z| > 1$, that maps $|z| > 1$ conformally onto the region $T$ preserving the point at infinity and its direction. Then the function $\phi(z)$ is uniquely determined and $c$ is called the transfinite diameter.

It turns out, though we won't prove it, that for a compact set $E$ the transfinite diameter and the logarithmic capacity are equal. We will use the transfinite diameter interpretation to show that the following theorem is equivalent to the above result:

Dubinin's theorem: 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.

We will use this theorem to sketch a proof of the Schinzel-Zassenhaus conjecture in the next post.

Proof of equivalence: Let

\[H(z) := \prod_j |z - z_j|^{p_j}.\]

and

\[F(z) := \prod_j \left((z - z_j)(1/z - \overline{z}_j)\right)^{p_j/n}, \quad |z| \geq 1.\]

 $F(z) \sim z$ as $z$ tends to infinity, so it has capacity/transfinite diameter 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 $a_j$. 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 $z_j,z_{j+1}$ (but this isn't necessarily 0 since there is ambiguity in the choice of logarithm, hence why we wrote $|F(z)$| instead of just $F(z)$). So these arcs are maped on some segments $[0,a_k]$. This gives a justification for why the roots are well-defined: almost all rays from the origin will give a valid branch cut for the logarithm.

The $a_k$ are some complex numbers such that $|a_k|$ is the maximum of $|F|$ on the corresponding arc. The angles between the segments are $2\pi p_j/n$. From this description one can also check that the $|a_k|$ are all equal when the $z_j$ are evenly spaced. Up to some scaling arguments in the $|a_k|$ one obtains the bounds on $F$ and also the corresponding equality cases. $\blacksquare$