SIB 2009, Problem #2
Using integration by parts (twice) we get:
$$\int_{0}^{2\pi} f(x) \cos x \text{ dx} = f'(2\pi) - f'(0) - \int_{0}^{2\pi} f''(x) \cos x \text{ dx}$$
Now $$\int_{0}^{2\pi} f''(x) \cos x \text{ dx} \le \int_{0}^{2\pi} |f''(x) \cos x| \text{ dx} \le \int_{0}^{2\pi} |f''(x)| \text{ dx}$$ $$ = \int_{0}^{2\pi} f''(x) \text{ dx} = f'(2\pi) - f'(0)$$