A function that is identically zero

Since we have $F(x) = \int_{0}^{x} f(t) dt \implies F'(x) = f(x)$. So, the given inequality becomes:

So, $g(x)= e^{-cx}F(x) $ is a decreasing function. $\implies g(x) \le g(0)$ for all $x \ge 0$. Now, $F(0)=0$. $g(x) \le 0$ for $x \ge 0 \implies e^{-cx}F(x) \le 0 $ for $x \ge 0. $

Since $f$ is non-negative, therefore $ F(x) =\int_{0}^{x} f(t) dt \ge 0 $ for all $x \ge 0 $. Therefore, we conclude that $F(x)= 0$ for all $x \ge 0$. Since $f$ is continuous and non-negative and $\int_{0}^{x}f(t)dt = 0$ for all $x \ge 0 \implies f(x)=0$ for all $x \ge 0$.