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 ?

  • 7
    $\begingroup$ 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. $\endgroup$
    – WimC
    Mar 2, 2014 at 20:14

1 Answer 1


The conjecture is correct.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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