Is $f \equiv 1$ the only normalized solution of $f(x) = \frac 1x \int_0^x f(t) dt$? Let $f \in L^1(0,1)$ be non-negative a.e. and such that
$$
\int_0^1 f(x) dx = 1.
$$
Assume that
$$
f(x) = \frac 1x \int_0^x f(t), \quad x \in (0,1)
$$
a.e.
Is $f \equiv 1$ a unique (up to an equivalence class) function satisfying above assumptions? Or there are some other ones?
 A: Since $f \in L^1(0, 1)$ and 
\begin{align}
F(x) = \int^x_0 f(t)\ dt
\end{align}
then
$F$ is absolutely continuous and $F$ is differentiable a.e. where $F'=f$. Next, observe
\begin{align}
f'(x) = \frac{f(x)x- \int^x_0 f(t)\ dt}{x^2} = \frac{f(x)x-xf(x)}{x^2}=0 \ \ \text{ a.e.}
\end{align}
Combining everything, we have that $f\equiv$ const  since $f$is continuous. Lastly $f(1) = \int^1_0 f(t)\ dt =1$. 
Edit: It should be noted that $f$ is locally absolutely continuous on $(0, 1)$ since $F$ is absolutely continuous and $\frac{1}{x}$ is also absolutely continuous. Moreover, since we know $f'\equiv 0$ a.e. then by the Fundamental theorem of calculus, we have that
\begin{align}
f(1)-f(x) = \int^1_x f'(t)\ dt =0. 
\end{align}
A: Hint:
For $x\ne0$,$$xf(x)=\int_0^x f(t)dt,$$ so that by differentiation
$$xf'(x)+f(x)=f(x)$$ or  $$f'(x)=0.$$

If we assume $f$ piecewise constant, in a constant interval $[x_k,x_{k+1})$,
$$\frac1x\int_0^xf(x)dx=\frac1x\int_0^{x_k}f(x)dx+\frac1x\int_{x_k}^xf(x)dx=\frac{\sum_{j=0}^{k-1}f_j(x_{j+1}-x_j)+f_k(x-x_k)}x=f_k.$$
As you can check by induction, the only solution is $f_k=1$.
