Series of Functions of integrals How do you solve this question? We had that in a test and I've been staring on it for around 30 minutes without any solution.
Given $g(x)$ a differentiable and bounded function over $\mathbb{R}$. Define, 
\begin{align}
f_0(x) & =g(x) \\
f_n(x) &= \int _0^xdt_{n-1}\int _0^{t_{n-1}}dt_{n-2}\int _0^{t_{n-2}}dt_{n-3}...\int _0^{t_3}dt_2\int _0^{t_2}dt_1\int _0^{t_1} g(t_0)\,dt_0
\end{align}
A) Prove that the series of functions $$\sum_{n=0}^{\infty }\:f_n(x)$$ converges uniformly in every closed interval $[0,a]$ for $a > 0$.
B) We define $G(x)= \displaystyle\sum_{n=0}^{\infty} f_n(x)$ for every $x \in [0,a]$. Prove that $G'(x)=\displaystyle\sum_{n=0}^{\infty } f_n'(x)$.
C) Show that for every $x \in [0,a]$ applies: $G(x)=\displaystyle\int _0^xe^{x-t}g'\left(t\right)dt\:$
Please, solve it if you may.
And something a little bit important as well, did anyone came across this question in the past? Where is the source of this question?

Show by induction that for $n \ge 1$,
$$f_n(x) =  \frac{1}{(n-1)!}\int_0^x (x - t)^{n-1} g(t)\, dt = \frac{1}{(n-1)!}\int_0^x t^{n-1}g(x - t)\, dt.$$
Let $a > 0$, and set $M = \max\{|g(x)|: 0\le x \le a\}$. For all $x\in [0,a]$, $|f_0(x)| \le M$ and for $n\ge 1$,
$$|f_n(x)| \le \frac{M}{(n-1)!}\int_0^a t^{n-1}\, dt = \frac{Ma^n}{n!}.$$ Since $\sum_{n = 0}^\infty Ma^n/n!$ converges (to $Me^a$, in fact), by the Weierstrass $M$-test, the series $\sum_{n = 0}^\infty f_n(x)$ converges uniformly on $[0,a]$. This proves A).
To prove B), note that since $f_n'(x) = f_{n-1}(x)$ and the series $\sum_{n = 0}^\infty f_n(x)$ converges uniformly on $[0,a]$, the series $\sum_{n = 0}^\infty f_n'(x)$ converges uniformly on $[0,a]$. Therefore, $G'(x) = \sum_{n = 0}^\infty f_n'(x)$. 
Part C) does not hold unless $g(0) = 0$. The formula for $G$ should be 
$$G(x) = g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt.$$
You can either use the formula for $f_n$ I displayed in the beginning to find $G$ directly (integrating term-wise then applying integration by parts), or by writing a first-order differential equation in $G$ and solving by method of integrating factors. If you choose the latter, then use the relation $f_n'(x) = f_{n-1}(x)$, $f_0(x) = g(x)$ and the value $G(0) = g(0)$ to develop the inital value problem
$$G' - G = g',\, G(0) = g(0).$$
The integrating factor is $e^{-x}$, so the general solution of the ODE is
$$G(x) = Ae^x + \int_0^x e^{x-t}g'(t)\, dt.$$
The initial condition $G(0) = g(0)$ yields $A = g(0)$. Hence
$$G(x) = g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt.$$
If you choose the former method, then write
\begin{align}
G(x) &= g(x) + \sum_{n = 1}^\infty \frac{1}{(n-1)!}\int_0^x (x - t)^{n-1}g(t)\, dt\\
&= g(x) + \int_0^x \sum_{n = 1}^\infty\frac{(x-t)^{n-1}}{(n-1)!}g(t)\, dt\\
&= g(x) + \int_0^x e^{x-t}g(t)\, dt\\
&= g(x) + \int_0^x e^{x-t}g'(t)\, dt - g(x) + g(0)e^x\\
&= g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt.
\end{align}
The interchange of sum and integral in second step is justified by the integrability of the terms $(x-t)^{n-1}/(n-1)!$ and the uniform convergence of the series $\sum_{n = 1}^\infty (x-t)^{n-1}/(n-1)!$ over $[0,a]$. The second to the last step is where integration by parts is used.
