IMC 2005 · Day 1 · P3

Let $M = \max\limits_{0 \le x \le 1} |f'(x)|$. By the inequality $-M \le f'(x) \le M$, $x \in [0,1]$ it follows: $$-M f(x) \le f(x) f'(x) \le M f(x), \quad x \in [0,1].$$ By integration $$-M \int_0^x f(t)\,dt \le \frac{1}{2} f^2(x) - \frac{1}{2} f^2(0) \le M \int_0^x f(t)\,dt, \quad x \in [0,1]$$ $$-M f(x) \int_0^x f(t)\,dt \le f^3(x) - f^2(0) f(x) \le M f(x) \int_0^x f(t)\,dt, \quad x \in [0,1].$$ Integrating the last inequality on $[0,1]$ it follows that $$-M \left( \int_0^1 f(x)\,dx \right)^2 \le \int_0^1 f^3(x)\,dx - f^2(0) \int_0^1 f(x)\,dx \le M \left( \int_0^1 f(x)\,dx \right)^2 \Leftrightarrow$$ $$\left| \int_0^1 f^3(x)\,dx - f^2(0) \int_0^1 f(x)\,dx \right| \le M \left( \int_0^1 f(x)\,dx \right)^2.$$

Let $M = \max\limits_{0 \le x \le 1} |f'(x)|$ and $F(x) = -\int_x^1 f$; then $F' = f$, $F(0) = -\int_0^1 f$ and $F(1) = 0$. Integrating by parts, $$\int_0^1 f^3 = \int_0^1 f^2 \cdot F' = [f^2 F]_0^1 - \int_0^1 (f^2)' F = f^2(1) F(1) - f^2(0) F(0) - \int_0^1 2 F f f' = f^2(0) \int_0^1 f - \int_0^1 2 F f f'.$$ Then $$\left| \int_0^1 f^3(x)\,dx - f^2(0) \int_0^1 f(x)\,dx \right| = \left| \int_0^1 2 F f f' \right| \le \int_0^1 2 F f |f'| \le M \int_0^1 2 F f = M \cdot [F^2]_0^1 = M \left( \int_0^1 f \right)^2.$$