# Minimize norm of a polynomial around a circle (count the solutions)

I already posted this question at MSE here, but as it received no significative feedback for a while I cross-post it here.

I also noticed a related question here on MO (which does not answer my present question).

Let $P=\sum_{k=0}^n a_kX^k$ ba a polynomial of degree $n \gt 0$, and let $r\gt 0$. Suppose that $P$ is not the monomial $a_nX^n$, in other words there is at least an $i<n$ such that $a_i\neq 0$. Denote by $C$ the circle $\big\lbrace z \ \big| \ |z|=r\big\rbrace$. Clearly, the continuous function $|P|$ attains a minimum value (let us denote it by $\mu$) on the compact set $C$. Let $M=\big\lbrace z\in C \ \big| \ |P(z)|=\mu\big\rbrace$. How many elements can we have in $M$ ? When $\mu=0$, $M$ contains only roots of $P$, so that the maximum cardinality is $n$.

For $l\in[-n,n]$, let us put $$b_l=\sum_{j=-n+{\sf max}(0,-l)}^{n+{\sf min}(0,-l)} \bar{a_j}a_{j+l}r^{2j+l}, B(X)=\sum_{l=-n}^{n}b_lX^{l+n}, D(X)=\frac{B(X)}{X^n}$$.

Then we have the identity $|P(re^{i\theta})|^2=D(e^{i\theta})$ for any $\theta\in{\mathbb R}$. Since $D'(X)=\frac{XB'(X)-B(X)}{X^{n+1}}$, we see that $|M| \leq |M'|$ where $M'=\big\lbrace z \in {\mathbb C} \ \big| \ zB’(z)-B(z)=0, |z|=1 \big\rbrace$. Remark that $D'$ is zero iff $B$ is a monomial in $X$, iff $P$ itself is a monomial in $X$. As the degree of $XB'(X)-B(X)$ is exactly $2n$, we deduce $|M| \leq 2n$.

Note however that we have only counted local extrema here, and the question is about the global extrema. Thus, one expects $|M|$ to be significatively lower than the upper bound $2n$. In fact, I conjecture the following :

Conjecture. $|M| \leq n$.

I have checked this conjecture on a few random numerical examples. Does anyone have an idea about how to prove or find a counterexample to this conjecture ?

-
Note that between any two succesive global minima of $|P|$ on the circle there must be at least one other local extremum. So it follows from your bound that there cannot be more than $n$ global minima. – WimC Mar 2 '14 at 20:14

Let me begin with a simple proof that $|M|\leq 2n$.
The curve $|P(z)|^2=\mu^2$ is a real algebraic curve of degree $2n$. At the points of $M$ this curve must be tangent to the unit circle. The unit circle is a curve of degree $2$, therefore by Bezout theorem these two curves have at most $4n$ intersections counting multiplicity. But the multiplicity is at least $2$ at each point.
Now suppose WLOG that $1$ is not in $M$. Consider the rational function $$f(z)=P\left(\frac{1}{1-z}\right).$$ The preimage of the unit circle under $1/(1-z)$ is a straight line $L$. And the set $|f(z)|^2=\mu^2$ is an algebraic curve $C$ of degree $2n$. The intersection $L\cap C$ may consist of at most $2n$ points, counting multiplicity. But the multiplicity of each point is at least $2$. This proves the statement.