Putnam 2007, Problem B-2

Suppose that $f : [0,1] \to \mathbb{R}$ has a continuous derivative and that $\int_0^1 f(x) \, dx = 0$. Prove that for every $\alpha \in (0,1)$,
$$\left| \int_0^\alpha f(x) \, dx \right| \leq \frac{1}{8} \max_{0 \leq x \leq 1} |f'(x)|.$$

Solution:
Put $B = \max_{0 \leq x \leq 1} |f'(x)|$ and $g(x) = \int_0^x f(y) dy$. Since $g(0) = g(1) = 0$, the maximum value of $|g(x)|$ must occur at a critical point $y \in (0,1)$ satisfying $g'(y) = f(y) = 0$. We may thus take $\alpha = y$ hereafter.

Since $\int_0^\alpha f(x) dx = -\int_0^{1-\alpha} f(1-x) dx$, we may assume that $\alpha \leq 1/2$. By then substituting $-f(x)$ for $f(x)$ if needed, we may assume that $\int_0^\alpha f(x) dx \geq 0$. From the inequality $f'(x) \geq -B$, we deduce $f(x) \leq B(\alpha - x)$ for $0 \leq x \leq \alpha$, so
$$\int_0^\alpha f(x) dx \leq \int_0^\alpha B(\alpha - x) dx = -\frac{1}{2}B(\alpha - x)^2\bigg|_0^\alpha = \frac{\alpha^2}{2}B \leq \frac{1}{8}B$$
as desired.