IMC 1994 · Day 2 · P7

Assume that there is $y \in (a,b]$ such that $f(y) \ne 0$. Without loss of generality we have $f(y) > 0$. In view of the continuity of $f$ there exists $c \in [a,y)$ such that $f(c) = 0$ and $f(x) > 0$ for $x \in (c,y]$. For $x \in (c,y]$ we have $|f'(x)| \le \lambda f(x)$. This implies that the function $g(x) = \ln f(x) - \lambda x$ is not increasing in $(c,y]$ because of $g'(x) = \frac{f'(x)}{f(x)} - \lambda \le 0$. Thus $\ln f(x) - \lambda x \ge \ln f(y) - \lambda y$ and $f(x) \ge e^{\lambda x - \lambda y} f(y)$ for $x \in (c,y]$. Thus $$0 = f(c) = f(c+0) \ge e^{\lambda c - \lambda y} f(y) > 0$$ — a contradiction. Hence one has $f(x) = 0$ for all $x \in [a,b]$.