IMC 1995 · Day 2 · P11

a) Let us consider the integral $$\int_0^{2\pi} f(x) (1 \pm \cos x)\,dx = \pi (a_0 \pm 1).$$ The assumption that $f(x) \ge 0$ implies $a_0 \ge 1$. Similarly, if $f(x) \le 0$ then $a_0 \le -1$. In both cases we have a contradiction with the hypothesis of the problem.

b) We shall prove that for each integer $N$ and for each real number $h \ge 24$ and each real number $y$ the function $$F_N(x) = \sum_{n=1}^{N} \cos(x\, n^{\frac{3}{2}})$$ changes sign in the interval $(y, y+h)$. The assertion will follow immediately from here.

Consider the integrals $$I_1 = \int_y^{y+h} F_N(x)\,dx, \qquad I_2 = \int_y^{y+h} F_N(x) \cos x\,dx.$$ If $F_N(x)$ does not change sign in $(y, y+h)$ then we have $$|I_2| \le \int_y^{y+h} |F_N(x)|\,dx = \left| \int_y^{y+h} F_N(x)\,dx \right| = |I_1|.$$ Hence, it is enough to prove that $$|I_2| > |I_1|.$$ Obviously, for each $\alpha \ne 0$ we have $$\left| \int_y^{y+h} \cos(\alpha x)\,dx \right| \le \frac{2}{|\alpha|}.$$ Hence $$\tag{1} |I_1| = \left| \sum_{n=1}^{N} \int_y^{y+h} \cos(x\, n^{\frac{3}{2}})\,dx \right| \le 2 \sum_{n=1}^{N} \frac{1}{n^{\frac{3}{2}}} < 2 \left( 1 + \int_1^{\infty} \frac{dt}{t^{\frac{3}{2}}} \right) = 6.$$ On the other hand we have $$\begin{align*} I_2 &= \sum_{n=1}^{N} \int_y^{y+h} \cos x \cos(x\, n^{\frac{3}{2}})\,dx = \frac{1}{2} \int_y^{y+h} (1 + \cos(2x))\,dx \\ &\quad + \frac{1}{2} \sum_{n=2}^{N} \int_y^{y+h} \left( \cos\left( x (n^{\frac{3}{2}} - 1) \right) + \cos\left( x (n^{\frac{3}{2}} + 1) \right) \right) dx = \frac{1}{2} h + \Delta, \end{align*}$$ where $$|\Delta| \le \frac{1}{2} \left( 1 + 2 \sum_{n=2}^{N} \left( \frac{1}{n^{\frac{3}{2}} - 1} + \frac{1}{n^{\frac{3}{2}} + 1} \right) \right) \le \frac{1}{2} + 2 \sum_{n=2}^{N} \frac{1}{n^{\frac{3}{2}} - 1}.$$ We use that $n^{\frac{3}{2}} - 1 \ge \frac{2}{3}\, n^{\frac{3}{2}}$ for $n \ge 3$ and we get $$|\Delta| \le \frac{1}{2} + \frac{2}{2^{\frac{3}{2}} - 1} + 3 \sum_{n=3}^{N} \frac{1}{n^{\frac{3}{2}}} < \frac{1}{2} + \frac{2}{2\sqrt{2} - 1} + 3 \int_2^{\infty} \frac{dt}{t^{\frac{3}{2}}} < 6.$$ Hence $$\tag{2} |I_2| > \frac{1}{2} h - 6.$$ We use that $h \ge 24$ and inequalities (1), (2) and we obtain $|I_2| > |I_1|$. The proof is completed.