TITLE: Does $\varphi(x)\le\int_0^x\varphi(t)dt$ for all $x\in[0,\infty)$ imply $\varphi\equiv0$.
QUESTION [5 upvotes]: Let $\varphi$ be a nonnegative and continuous function on $[0,\infty)$ and such that $$\varphi(x)\le\int_0^x\varphi(t)dt$$ for all $x\in[0,\infty)$. Can we infer from here that $\varphi\equiv0$.
By taking limit on both sides, I got $\varphi(0)\le 0$. As $\varphi$ is nonnegative, I can say $\varphi(0)=0$. But what then. I could have progressed if I could show that $\varphi(x)\le \varphi'(x)$, but it is not quite evident from the given condition.

REPLY [11 votes]: The usual technique is to consider $$g(x) =e^{-x}\int_{0}^{x}\varphi(t)\,dt$$ and then $$g'(x) =e^{-x} \left(\varphi(x) - \int_{0}^{x}\varphi(t)\,dt\right)\leq 0$$ so that $g$ is decreasing and hence $g(x) \leq g(0)=0$. But $g$ is non-negative and hence $g(x) =0$ for all $x\geq 0$ and therefore $\int_{0}^{x}\varphi(t)\,dt=e^{x} g(x) $ and it's derivative $\varphi(x)$ vanish for all $x\geq 0$.