$\frac{1}{x}\int_0^x f(t)dt=cf(x),\forall{x\in(0,\infty)}$ $\Rightarrow$ $f=0$ almost everywhere? Suppose $f$ is a measurable real-valued function on $(0,\infty)$ such that $f\geq 0$ and
$$\frac{1}{x}\int_0^x f(t)dt=cf(x),\quad\forall{x\in(0,\infty)}$$
for some constant $c$. Does it follow that $f=0$ almost everywhere?
Since the left-hand side is the average of the function on $[0,x]$ it is intuitively true, but I am not sure how to prove it. Of course if $f=0$ almost everywhere we get this property immediately.

Consider $f(x)=\frac{x^{\frac{1}{c}-1}}{c}$.
