IMC 2015 · Day 2 · P10

Let $$M = \max_{0 \le x \le 1} \bigl| p(x) \bigr|.$$ For every positive integer $k$, let $$J_k = \int_0^1 \bigl( p(x) \bigr)^{2k} dx.$$ Obviously $0 < J_k < M^{2k}$ is a rational number. If $(p(x))^{2k} = \sum\limits_{i=0}^{2kn} a_{k,i} x^i$ then $J_k = \sum\limits_{i=0}^{2kn} \frac{a_{k,i}}{i+1}$. Taking the least common denominator, we can see that $J_k \ge \dfrac{1}{\operatorname{lcm}(1, 2, \dots, 2kn+1)}$.

An equivalent form of the prime number theorem is that $\log \operatorname{lcm}(1, 2, \dots, N) \sim N$ if $N \to \infty$. Therefore, for every $\varepsilon > 0$ and sufficiently large $k$ we have $$\operatorname{lcm}(1, 2, \dots, 2kn+1) < e^{(1+\varepsilon)(2kn+1)}$$ and therefore $$\begin{gather*} M^{2k} > J_k \ge \frac{1}{\operatorname{lcm}(1, 2, \dots, 2kn+1)} > \frac{1}{e^{(1+\varepsilon)(2kn+1)}}, \\ M > \frac{1}{e^{(1+\varepsilon)\left(n + \frac{1}{2k}\right)}}. \end{gather*}$$ Taking $k \to \infty$ and then $\varepsilon \to +0$ we get $$M \ge \frac{1}{e^n}.$$ Since $e$ is transcendent, equality is impossible.

Remark. The constant $\frac{1}{e} \approx 0.3679$ is not sharp. It is known that the best constant is between $0.4213$ and $0.4232$. (See I. E. Pritsker, The Gelfond–Schnirelman method in prime number theory, Canad. J. Math. 57 (2005), 1080–1101.)