TITLE: $(0 \leq f' \leq f$ on $\mathbb R$ and $f(a)=0)$ $\implies f=0$?
QUESTION [5 upvotes]: Given that $0 \leq f' \leq f$  on $\mathbb R$  and $f(a)=0\in\mathbb R$ for some $a\in\mathbb R$, how do I prove that $f$ is identically zero?  Using the mean value theorem naively didn't really get me anywhere.

REPLY [4 votes]: Hint: Suppose $a + 1 > x > a$. Then
$$f(x) = f(x) - f(a) = f'(c)(x-a)$$
For some $c \in (a,x)$. But
$$f'(c)(x-a) \leq f(c)(x-a) = (f(c) - f(a))(x-a) = f'(d)(c-a)(x-a)$$
For some $d \in (a,c)$. Suppose $f'$ is bounded by $M$ on $[a-1,a+1]$ (note that $f'$ is bounded on this set since $f$ is continuous so bounded on a compact set, and $f' \leq f$). Then
$$0 \leq f(x) \leq M (x-a)^{n}$$
for all $n$, and $0 \leq x - a < 1$. Let $n \to \infty$. Can you iterate this argument to show that $f$ is identically zero?