IMC 2009 · Day 1 · P4

The polynomials $p$ and $q$ are regular on the complex plane, so by the Maximum Principle, $\max_{|z| \le 1} |q(z)| = \max_{|z| = 1} |q(z)|$, and similarly for $p$. Let us denote $M_f = \max_{|z| = 1} |f(z)|$ for any regular function $f$. Thus it suffices to prove that $M_q \le M_p$.

First, note that we can assume $c_n = 0$. Indeed, for $c_n = 1$, we get $p = q$ and the statement is trivial; otherwise, $q(z) = c_n p(z) + (1 - c_n) r(z)$, where $r(z) = \sum_{j=0}^{n} \frac{c_j - c_n}{1 - c_n} a_j z^j$. The sequence $c_j' = \frac{c_j - c_n}{1 - c_n}$ also satisfies the prescribed conditions (it is a positive linear transform of the sequence $c_j$ with $c_0' = 1$), but $c_n' = 0$ too, so we get $M_r \le M_p$. This is enough: $M_q = |q(z_0)| \le c_n |p(z_0)| + (1 - c_n) |r(z_0)| \le c_n M_p + (1 - c_n) M_r \le M_p$.

Using the Cauchy formulas, we can express the coefficients $a_j$ of $p$ from its values taken over the positively oriented circle $S = \{|z| = 1\}$: $$a_j = \frac{1}{2\pi i} \int_S \frac{p(z)}{z^{j+1}}\,dz = \frac{1}{2\pi} \int_S \frac{p(z)}{z^j}\,|dz|$$ for $0 \le j \le n$, otherwise $$\int_S \frac{p(z)}{z^j}\,|dz| = 0.$$ Let us use these identities to get a new formula for $q$, using only the values of $p$ over $S$: $$2\pi \cdot q(w) = \sum_{j=0}^{n} c_j \left( \int_S p(z) z^{-j}\,|dz| \right) w^j.$$ We can exchange the order of the summation and the integration (sufficient conditions to do this obviously apply): $$2\pi \cdot q(w) = \int_S \left( \sum_{j=0}^{n} c_j (w/z)^j \right) p(z)\,|dz|.$$ It would be nice if the integration kernel (the sum between the brackets) was real. But this is easily arranged — for $-n \le j \le -1$, we can add the conjugate expressions, because by the above remarks, they are zero anyway: $$2\pi \cdot q(w) = \sum_{j=0}^{n} c_j \left( \int_S p(z) z^{-j}\,|dz| \right) w^j = \sum_{j=-n}^{n} c_{|j|} \left( \int_S p(z) z^{-j}\,|dz| \right) w^j,$$ $$2\pi \cdot q(w) = \int_S \left( \sum_{j=-n}^{n} c_{|j|} (w/z)^j \right) p(z)\,|dz| = \int_S K(w/z)\, p(z)\,|dz|,$$ where $$K(u) = \sum_{j=-n}^{n} c_{|j|} u^j = c_0 + 2 \sum_{j=1}^{n} c_j\, \Re(u^j)$$ for $u \in S$.

Let us examine $K(u)$. It is a real-valued function. Again from the Cauchy formulas, $\int_S K(u)\,|du| = 2\pi c_0 = 2\pi$.

If $\int_S |K(u)|\,|du| = 2\pi$ still holds (taking the absolute value does not increase the integral), then for every $w$: $$2\pi |q(w)| = \left| \int_S K(w/z)\, p(z)\,|dz| \right| \le \int_S |K(w/z)| \cdot |p(z)|\,|dz| \le M_p \int_S |K(u)|\,|du| = 2\pi M_p;$$ this would conclude the proof. So it suffices to prove that $\int_S |K(u)|\,|du| = \int_S K(u)\,|du|$, which is to say, $K$ is non-negative.

Now let us decompose $K$ into a sum using the given conditions for the numbers $c_j$ (including $c_n = 0$). Let $d_k = c_{k-1} - 2 c_k + c_{k+1}$ for $k = 1, \dots, n$ (setting $c_{n+1} = 0$); we know that $d_k \ge 0$. Let $F_k(u) = \sum\limits_{j=-k+1}^{k-1} (k - |j|) u^j$. Then $K(u) = \sum\limits_{k=1}^{n} d_k F_k(u)$ by easy induction (or see Figure for a graphical illustration).

So it suffices to prove that $F_k(u) \ge 0$ for $u \in S$. This is reasonably well-known (as $F_k$ is the Fejér kernel), and also very easy: $$\begin{align*} F_k(u) &= (1 + u + u^2 + \dots + u^{k-1}) (1 + u^{-1} + u^{-2} + \dots + u^{-(k-1)}) = \\ &= (1 + u + u^2 + \dots + u^{k-1}) \overline{(1 + u + u^2 + \dots + u^{k-1})} = \left| 1 + u + u^2 + \dots + u^{k-1} \right|^2 \ge 0 \end{align*}$$ This completes the proof.