Maximum of Polynomials in the Unit Circle

Question

Let $z_{1},z_{2},\ldots,z_{n}$ be i.i.d random points in the unit circle ($|z_i|=1$) with uniform distribution of their angles. Consider the random polynomial $P(z)$ given by $$ P(z)=\prod_{i=1}^{n}(z-z_i). $$

Let $m$ be the maximum absolute value of $P(z)$ on the unit circle $m=\max\{|P(z)|:|z|=1\}$.

How can I estimate $m$? More specifically, I would like to prove that there exist $\alpha>0$ such that the following holds almost surely as $n\to\infty$ $$ m\geq e^{\alpha\sqrt{n}}. $$

Any idea of what can be useful here?

Answer 1

The logarithm of your objective function is

$$\sum_{i=1}^n \ln\lvert z - z_i\rvert\;.$$

If you think of the points as continuously distributed for large $n$, this becomes $n$ times the average

$$\frac{1}{\pi}\int_{D_1} \ln\lvert x-x' \rvert \mathrm d^2x'\;.$$

This is the potential of a uniformly charged disk at distance $|x|$ from the centre, within the disk. It can be evaluated, as in the three-dimensional case of a uniformly charged sphere, by splitting the integral into a part for the rings interior to $x$, for which the potential has to be evaluated at radius $|x|$, and a part for the rings exterior to $x$, for which the potential has to be evaluated at the radius of the rings:

$$ \begin{eqnarray} \frac{1}{\pi}\int_{D_1} \ln\lvert x-x' \rvert \mathrm d^2x' &=& 2\left(\int_0^{|x|}\ln|x|r\mathrm dr+\int_{|x|}^1\ln r r\mathrm dr\right)\\ &=&\frac{1}{2}\left(|x|^2-1\right)\;. \end{eqnarray} $$

So the average contribution from a uniform density of points is zero at the boundary and negative everywhere else. Does that fit with your numerical experiments? It suggests that the maximum should typically be near the boundary for large $n$.

It also suggests that you can think of the square root term in your conjecture as arising from fluctuations around this uniform density, so the conjecture could be rephrased as something like "Almost surely, there will be local fluctuations in the density of the points to create a fluctuation of $\alpha\sqrt{n}$ in the sum of the logarithms at some point on the boundary." I'll give some more thought to how this might be made rigorous.

Answer 2

This seems $\textbf{somewhat related}$ to a Miklos Schweitzer $(1972)$ problem.

$\textbf{Problem 6.}$ Let $P(z)$ be a polynomial of degree $n$ with complex co-efficients, $$P(0)=1, \quad \ \text{and} \ |P(z)| \leq M \ \ \text{for} \ |z| \leq 1.$$ Prove that every root of $P(z)$ in the closed unit disc has multiplicity at most $c\sqrt{n}$ where $c=c(M) > 0$ is a constant depending only on $M$.

$\textbf{Solution 1.}$ (As given in "Contests in Higher Mathematics" by : Gabor J.Szekely). It is sufficient to examine the multiplicity of number $1$. Infact, if we prove something for $1$ then we may appy the result to the polynomial $p(z)=P(\alpha z)$ with $|\alpha| \leq 1$, and in this way, we obtain the same estimate for all roots lying in the unit disc. The idea of the solution is the following. We consider the integral $$F(P) = \int\limits_{0}^{2\pi} \log{|P(e^{i\phi})|} \ d\phi$$ and show that it exists and is non-negative. Then we estimate it from above, once in the neighborhood of $1$ with the aid of multiplicity of $1$ and the degree of $P$, and once at other points using the condition $|P(z)| \leq M$.

It is sufficient to prove the existence of the integral for polynomials of the form $z-z_{0}$. If $$P(z)= c \cdot\prod\limits_{i=1}^{n}(z-z_{i})$$ then $$\log{|P(z)|} = \log{|c|} + \sum\limits_{i=1}^{n} \log{|z-z_{i}|}$$

The existence of $$\int\limits_{0}^{2\pi} \log{|e^{i\phi}-z_{0}|}\ d\phi$$ is evident if $|z_{0}| \neq 1$. Next let $|z_0|=1$. Without loss of generality, we may assume that $z_{0}=1$ (a substitution $\phi=n+\phi_{0}$ takes them into each other). Then $$|e^{i\phi}-1| = \left\lvert\:2\sin\frac{\phi}{2}\right\rvert$$ and $$\int\limits_{0}^{2\pi} \log\:\left\lvert\:2\sin\frac{\phi}{2}\right\rvert \ d\phi$$ really exists and is equal to $0$.

Next, compute the integral $$f(\alpha)= \int\limits_{0}^{2\pi} \log\biggl|1-\frac{e^{i\phi}}{\alpha}\biggr| \ d\phi \qquad\qquad (\alpha\neq 0)$$ for $\alpha \neq 1$. Obviously, its value depends on the absolute value of $\alpha$ only (again, a subsitution as above may be applied), so it is the same for the numbers $\alpha\epsilon_{1},\alpha\epsilon_{2},\cdots, \alpha\epsilon_{n}$ where the $\epsilon_{j}$ are the $n$-th unit roots. Therefore, $$n f(\alpha) = \sum\limits_{j=1}^{n} f(\alpha\epsilon_{j})=\int\limits_{0}^{2\pi} \log\:\Biggl| \prod\limits_{j=1}^{n} \biggl(1-\frac{e^{i\phi}}{\alpha\epsilon_{j}}\:\biggr)\Biggr| \ d\phi$$

Now if $|\alpha|>1$, then for $n\to\infty$ the integral on the right hand side tends to zero since $1-e^{in\phi}/\alpha^{n} \to 1$ uniformly; thus $f(\alpha)=0$. On, the other hand if $|\alpha| <1$ then $$n(f(\alpha)+2\pi\log{|\alpha|})=\int\limits_{0}^{2\pi} \log{|\alpha^{n}-e^{in\phi}|} \ d\phi \to 0$$ since $1-|\alpha|^{n} < |\alpha^{n}-e^{in\phi}|<1+|\alpha|^{n}$; so in this case $f(\alpha)=-2\pi \log{|\alpha|}$ is only possible. In each of the three case, we have $f(\alpha) \geq 0$. In our case, the relation $P(0)=1$ implies, $$P(z)=\prod\limits_{j=1}^{n} \biggl(1-\frac{z}{z_j}\biggr)$$ whence $$F(P)= \sum\limits_{j=1}^{n} f(z_j) \geq 0 \qquad (1)$$

Now let $P(z)=(z-1)^{k} Q(z)=a_{0}+a_{1}z+\cdots +a_{n}z^{n}$, where $Q(1)\neq 0$. We estimate $F(P)$ with the help of $k$, $n$, and $M$. Let $$F(P)=\int\limits_{0}^{2\pi} = \int\limits_{-\epsilon}^{\epsilon} + \int\limits_{\epsilon}^{2\pi-\epsilon} = F_{1}+F_{2}$$

Then $$F_{2} \leq \int\limits_{0}^{2\pi} \log{M} \ d\phi = 2 \cdot \pi \cdot \log{M} \qquad (2)$$

We split $F_{1}$ again into two parts: $$F_{1} = \int\limits_{-\epsilon}^{\epsilon} \log{|(z-1)^{k}|} \ d\phi + \int\limits_{-\epsilon}^{\epsilon} \log|Q(e^{i\phi})| \ d\phi=F_{3}+F_{4} \qquad (3)$$

Clearly \begin{align*} F_{3} &= k \int\limits_{-\epsilon}^{\epsilon} \log\Bigl(2\cdot\sin\frac{|\phi|}{2}\Bigr) < 2k\int\limits_{0}^{\epsilon} \log\phi \ d\phi \\ &=2k\epsilon \cdot (\log\epsilon -1) \qquad (4) \end{align*}

For estimating $F_{4}$ we need an estimate of $Q$, which we obtain from the co-efficients of the expansion of $Q$ about $1$. Let $Q(1+z)= R(z)$. We calculate the co-efficients of $R$ from those of $P$ using the formula $R(z)= \frac{P(z+1)}{z^{k}}$

\begin{align*} P(z+1) &=\sum\limits_{j=0}^{n} a_{j}(z+1)^{j} =\sum\limits_{j=0}^{n}\sum\limits_{m=0}^{j}a_{j}\cdot {j \choose m} \cdot z^{m} \\ &=\sum\limits_{m=k}^{n} z^{m} \cdot \sum\limits_{j=m}^{n} a_{j} \cdot {j \choose m} \end{align*} since for $m < k$, the coefficient of $z^{m}$ is $0$ by our assumption. Thus $$R(z)=\sum\limits_{m=0}^{n-k}b_{m}z^{m}, \qquad b_{m}=\sum\limits_{j=m+k}a_{j} {j \choose m+k} \qquad (5)$$

Further, by the cauchy inequalities, $|a_{j}|=|P^{(j)}(0)|/j! \leq M$ Putting this into $(5)$, we find, $$|b_{m}| \leq M \sum\limits_{j=m+k}^{n} {j \choose m+k} = M {n+1 \choose m+k+1} \qquad (6)$$

If $|z|=\delta$ and $\delta(n-k)/(k+2) < 1$, then in view of $(6)$,

\begin{align*} \frac{|R(z)|}{M} & \leq \sum\limits_{m=0}^{\infty} \delta^{m} {n+1 \choose m+k+1} \\ &={n+1 \choose k+1} \cdot \Bigl(1+\delta \frac{n-k}{k+2} + \delta^{2}\frac{n-k}{k+2}\cdot\frac{n-k-1}{k+3} + \cdots \Bigr) \\ &\leq {n+1 \choose k+1} \sum\limits_{j=0}^{\infty} \Bigl(\delta \frac{n-2}{k+2}\Bigr)^{j} = {n+1 \choose k+1} \frac{1}{1-\delta \frac{n-k}{k+2}} \end{align*}

Since $|e^{i\phi}-1|=2|\sin\frac{\phi}{2}| \leq |\phi|$, therefore if $\epsilon < ((k+2)/2(n-k))$, then for $|\phi| \leq \epsilon$ we have $$|Q(e^{i\phi})|= |R(e^{i\phi}-1)| \leq M {n+1 \choose k+1}\frac{1}{1-\epsilon\frac{n-k}{k+2}} < 2M {n+1 \choose k+1}$$ If $k\geq 2$, $t! > t^{t}e^{-t}$ we obtain,

\begin{align*} {n+1 \choose k+1} &= \frac{(n+1) \cdot n \cdot (n-1) \cdots (n-k+1)}{(k+1)!} \\ &= \frac{(n^{2}-1)\cdot n \cdot (n-2) \cdot (n-k+1)}{(k+1)!} < \frac{n^{k+1}}{(k+1)!} < \Bigl(\frac{en}{k+1}\Bigr)^{k+1} \end{align*}

Thus $$\log{|Q(e^{i\phi})|} < \log{2M} + (k+1)\log\frac{en}{k+1}$$ and consequently $$F_{4} < 2\epsilon \biggl[ \log(2M) + (k+1)\log\frac{en}{k+1}\biggr]$$

Collecting everything, by $(1)$,$(2)$,$(3)$ and $(4)$ $$(\pi+\epsilon)\log{M} + \epsilon\log\frac{2en}{k+1} + \epsilon k \log\frac{en}{k+1} \geq 0 \qquad (7)$$

Now if $n=k^{2}/2c$, let $\epsilon = c/k$ (this fulfills the condition $\epsilon < ((k+2)/2(n-k))$). Then $(7)$ becomes, $$\Bigl(\pi+\frac{c}{k}\Bigr) \log{M} + \frac{c}{k} \log\frac{\epsilon k^{2}}{c(k+1)} + c \log\frac{k}{2(k+1)} \geq 0$$

We only make things worse if we also replace $k+1$ by $k$. Moreover, $c/k=k/2n \leq \frac{1}{2}$ gives $\pi+\frac{c}{k} < 4$, $c/k \log(ek/c) < (c/k) \cdot (ek/c)=e < 3$, So finally $$4\log{M} + 3 \geq c \log{2}, \qquad (8)$$ $$c < \log{M}+6$$ Since $k=\sqrt{2cn}$, relation $(8)$ means that we have proved the assertion of the problem with $$c(M)=\sqrt{16\log{M}+12}$$