TITLE: Uniformly convergence of $f_{n+1}(x)= \int^x_0 f_n(t)dt$
QUESTION [5 upvotes]: Let $f_0$ be integrable function on $[0,a]$ and a sequence of functions $f_{n+1}(x)= \int^x_0 f_n(t)dt$. 
   Show that $f_n$ is uniformly convergent on $[0,a]$ to $0$. 

I get so far is if $\int_0^xf_0(t)dt =x_0 $ so $\int_0^xx_0dt=xx_0 $ and $f_n(x)=x^nx_0  $
so  $f_n(x) \to 0$ for $x\in[0,1) $ but I don't know if it uniformly and what happen for $a>1$ ?
Thanks ahead.

REPLY [6 votes]: Show by induction that $|f_n(x)|\leq \frac{x^n}{n!}\|f_0\|_\infty$ for all $n\in\mathbb N, x\in[0,a]$
Then $\|f_n\|_\infty \leq \frac{a^n}{n!}\|f_0\|_\infty$. This sequence should look familiar.
Edit: This only works if $\|f_0\|_\infty$ is finite as pointed out in the comments. This can be obtained by using $f_1$ instead of $f_0$ for the argument and using $\|f_1\|_\infty \leq \|f_0\|_1$.