let $ b>0 $ and let $ f_{1}:[0,b]\to\mathbb{R} $ be an integrable functions.
Prove that the sequence of functions, that defined recursivly by:
$ f_{n+1}\left(x\right)=\int_{0}^{x}f_{n}\left(t\right)dt $
is uniformly convergent to the function $ f=0 $
Actually, I have no idea where to start, I cant really see anything that would justify that the values of the functions really approaches to $ 0 $ (I cant even see why $ f_n(x) $ is getting smaller for a constant $ x $ ).
What I did notice is that for $ n>3 $ the function $ f_n $ is the anti-deriviative of $ f_{n-1} $ (except for $ f_2 $ and $f_1 $ ), and ofcourse they are all continious functions (except for maybe $ f_1 $ ). Any hints would help, thanks in advance.
Since $f_1$ is Riemann integrable then by definition $f_1$ is bounded, i.e, $|f_1(x)|\leq M$.
Then $|f_2(x)|=|\int_{0}^x f_{1}(t)dt|\leq \int_{0}^x |f_{1}(t)|dt\leq Mx$.
So $|f_3(x)|=|\int_{0}^x f_{2}(t)dt |\leq \int_{0}^x |f_{2}(t)|dt\leq \int_{0}^x M tdt=M\frac{x^2}{2}.$
By induction we obtain $|f_n(x)|\leq M\frac{x^{n-1}}{(n-1)!}$.
Thus $|f_n(x)|\leq M\frac{x^{n-1}}{(n-1)!}\leq M\frac{b^{n-1}}{(n-1)!}\xrightarrow{n\rightarrow \infty}0$.
The result follows.