IMC 2017 · Day 1 · P2

Notice that $f'$ satisfies the Lipschitz-property, so $f'$ is continuous and therefore locally integrable.

Consider an arbitrary $x \in \mathbb{R}$ and let $d = f'(x)$. We need to prove $f(x) > \frac{d^2}{2L}$.

If $d = 0$ then the statement is trivial.

If $d > 0$ then the condition provides $f'(x - t) \ge d - Lt$; this estimate is positive for $0 \le t < \frac{d}{L}$. By integrating over that interval, $$f(x) > f(x) - f\bigl( x - \tfrac{d}{L} \bigr) = \int_0^{d/L} f'\bigl( x - t \bigr)\,dt \ge \int_0^{d/L} (d - Lt)\,dt = \frac{d^2}{2L}.$$ If $d < 0$ then apply $f'(x + t) \le d + Lt = -|d| + Lt$ and repeat the same argument as $$f(x) > f(x) - f\bigl( x + \tfrac{|d|}{L} \bigr) = \int_0^{|d|/L} \bigl( -f'(x + t) \bigr)\,dt \ge \int_0^{|d|/L} (|d| - Lt)\,dt = \frac{d^2}{2L}.$$