$\frac{1}{x}\int_0^x f(t)dt=cf(x),\forall{x\in(0,\infty)}$ $\Rightarrow$ $f=0$ 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.
 A: No. Counterexample: $c=1$ and $f(x)=k>0$ for all $x$. 
A: Consider $f(x)=\frac{x^{\frac{1}{c}-1}}{c}$.
A: @Eric, If your question is only : is it true that $f=0$ almost everywhere, you have
get the counterexample. But if your question is what can we said about $f,$
follow me:
From $\frac{1}{x}\int_{0}^{x}f(t)dt=cf(x),$ for all $x>0,$ we get $$
\int_{0}^{x}f(t)dt=cxf(x),\ x>0.
$$
Assume that $f$ is continuous. By the Fundamental Theorem the left hand side
is a differentiable function, so the right on should be too, that is $f$ is
necessary differentiable and then
\begin{eqnarray*}
\frac{d}{dx}\int_{0}^{x}f(t)dt &=&cf(x)+cxf^{\prime }(x),\ for\ x>0. \\
f(x) &=&cf(x)+cxf^{\prime }(x),\ for\ x>0 \\
(1-c)f(x) &=&cxf^{\prime }(x),\ \ \ \ \ for\ x>0.
\end{eqnarray*}
Discussion:
If $c=1$ then $f^{\prime }(x)=0,$ for all $x>0,$ so $f$ is (any) constant
including $f(x)=0$ everywhere.
If $c=0$ then $f(x)=0,$ for all $x>0.$ 
If $c\neq 0,1.$ Assume there exists an interval $I$ where $f(x)>0.$ So on
that interval one has
$$
\frac{f^{\prime }(x)}{f(x)}=\frac{(1-c)}{c}\frac{1}{x},\ for\ all\ x\in I.
$$
then%
$$
\ln \left\vert f(x)\right\vert =\frac{1-c}{c}\ln x+d,\ for\ all\ x\in
I\subset \left( 0,+\infty \right) 
$$
and so, let $(c-1)/c=k$
$$
\left\vert f(x)\right\vert =e^{k\ln x+d}=\alpha x^{k},\ for\ all\ x\in I,\
with\ \alpha >0.
$$
But in your hypothesis, you said $f\geq 0,$ so
$$
f(x)=\alpha x^{k},\ for\ all\ x\in I,\ with\ \alpha >0.
$$
Take $I$ be the maximal interval possible, so either it is $\left( 0,+\infty
\right) ,$ or $I_{\max }=\left( a,b\right) $ for some $b>0$ and some $a\geq
0.$ From continuity arguments, it follows that $f(b)=0.$ So from $(1-c)f(x)=cxf^{\prime }(x),\ for\ x>0.
$ it follows that $f^{\prime }(b)=0.$ But $f^{\prime }(x)=\alpha kx^{k-1}$
for all $x\in \left( a,b\right) $ then $\lim_{x\rightarrow b^{-}}f^{\prime
}(x)=\alpha kb^{k-1}=0$ implies $b=0$ which is impossible since $b>0.$ It
follows that $I_{\max }$ is unbounded from the right. We can show the same
way that $a=0$ necessarily and then $I_{\max }=\left( 0,+\infty \right) .$
It follows that either $f(x)=\alpha x^{k},\ for\ all\ x\in (0,+\infty ),\
with\ \alpha >0.$ or $f(x)=0$ for all $x\in (0,+\infty ).$
Conclusion:
If $c\neq 0,1$ then either $f(x)=0$ for all $x>0$ or $f(x)=f(1) x^{k}$ for
all $x>0$ with $k=\frac{c-1}{c}.$
