# Let $f:[0,a]\rightarrow[0,\infty)$ be a continuous function such that $f(t)\leq e^{\int_{0}^{t}f(s)ds}-1$ for all $t\in[0,a]$. Prove that $f\equiv0$

Let $$f:[0,a]\rightarrow[0,\infty)$$ be a continuous function such that $$f(t)\leq e^{\int_{0}^{t}f(s)ds}-1$$ for all $$t\in[0,a]$$. Prove that $$f\equiv0$$.

I have thought like this: Assume $$F(t)=\int_{0}^{t}f(s)ds\implies F'(t)=f(t)$$ . Then $$F'(t)\leq e^{F(t)}-1$$. Now how can proceed.

Since $$F'(t) \leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality $$e^{-F(t)-t} (F'(t)+1) \leq e^{-t}$$ $$\implies \frac {d}{dt} e^{-F(t)-t} \geq \frac {d}{dt} e^{-t}$$ $$\implies e^{-F(t)-t} \geq e^{-t}$$ $$\implies e^{F(t)} \leq 1$$ $$\implies F(t)\leq 0$$ But we know $$F$$ is non negative and hence $$F\equiv 0$$ So $$f\equiv 0$$