Let $f_n$ be a$f_{n+1}\left(x\right)=\frac{1}{x}\int _0^xf_n\left(t\right)dt\:$ Let be $f_n $ be a sequence of functions and $f_0$ an continuous arbitrary function derivable in $0$ such that:
$$f_{n+1}\left(x\right)=\frac{1}{x}\int _0^xf_n\left(t\right)dt$$
for every $n$ positive integer.
The domain of $f_n$ is $[0,1]$
I was wondering if the following statement is true or not:

$f'_n $is uniformly convergent to the function:
$g(x)=0$ for every $x \in [0,1]$

I was thinking that I need to prove that:
$$sup_{x\in [0,1]}|f'_n(x)-f(x)|=0$$
that is to say for every $\epsilon>0$  there is a positive integer $N$ such that for every $n>N$ and ${x\in [0,1]}$ we have:
$$|f'_n(x)-f(x)|<\epsilon$$
How should I proceed?
 A: We can replace $f_n(x)$ by $f_n(x)-f_0(0)$ to reduce the proof to the case when  $f_0(0)=0$. In this case, differentiabilty of $f_0$ at $0$ implies that there exists a constant $M$ with $|f_0(x) |\leq Mx$ for all $x$.
Now  $xf_{n+1}'(x)+f_{n+1}(x)=(xf_{n+1}(x))'=f_n(x)$ so $|f_{n+1}'(x)|\leq  \frac  1 x (|f_n(x)+|f_{n+1} (x)|)$.  Use induction to show that $|f_n(x)| \leq \frac {Mx} {2^{n}}$ for all $n$. Hence, $f_{n+1}'(x) \to 0$ uniformly.
A: The operator $T(f)  =1/x\int_0^x f$ is a continuous contraction on the space $\mathcal{C}_0[0,1]=\{f \in \mathcal{C}[0,1]: f(0) = 0\}$.The easy estimate $\lVert T(f) \rVert \leq \lVert f \rVert$ shows T is continuous. Evaluating $T$ on a polynomial $f(x) = a_1x\dots +a_n x^n$ yields the estimate $\lVert T(f) \rVert \leq 1/2\lVert f \rVert$ for polynomials. We apply Stone-Weierstrass to get the estimate for all $f$, which proves $T$ is a contraction with fixed point $f\equiv 0$
A few details: $\lVert \rVert$ is sup norm.
Define $T(f)(0) = 0$ (if this wasn't clear). Then L'Hôpital's rule give $T(f) $ continuous on $[0,1]$.
Also, S-W insures $\exists \, p_n\rightarrow f$, $p_n$ polynomials in $\mathcal{C}[0,1]$. Then if $f \in \mathcal{C}_0[0,1]$, $(p_n - p_n(0)) \rightarrow (f - f(0)) = f$, in sup norm, so we can assume $p_n \in \mathcal{C}_0$ to apply the estimate.
