Consider the sequence $(f_{n})_{n=1}^{\infty}$ of continuous functions on $I = [0, \infty)$ defined recursively by
$f_{1}(x)=x, f_{n}(x)=x+\int_{0}^{x}f_{n-1}(t)\sin(x-t) dt, \forall n\geq 2$.
This sequence satisfies
$\vert f_{n+1}(x)-f_{n}(x)\vert\leq\frac{x^{n+1}}{(n+1)!}$
for all $n\geq 1$ and $x\geq 0.$
Prove that the sequence $(f_{n})_ {n=1}^{\infty}$ converges pointwise to a limit function $F$ on $[0, \infty)$, and that the convergence is uniform on $[0, A]$ for all $A > 0$.
In the previous problems I did, I used to begin by finding out the limit of the sequence of functions but I'm not really sure how to proceed with this one since it is recursively defined. Maybe someone knows of a plotter to help me visualize the sequence of functions?
My guess is that $\vert f_{n+1}(x)-f_{n}(x)\vert\leq\frac{x^{n+1}}{(n+1)!}$ means I need to use the Cauchy criterion? So if I'm correct, I think I need to prove that $\frac{x^{n+1}}{(n+1)!}$ is less than $\epsilon$.
Any help would be greatly appreciated, thanks!
You know that the series $\sum_n \frac{x^n}{n!}$ converges (it's the exponential function), so it is Cauchy. Fix $\varepsilon>0$ and $x\geq 0$, then there exists $N\in\mathbb{N}$ such that for all $p,q\geq N$, $q>p$, $$\sum_{n=p}^{q-1} \frac{x^{n+1}}{(n+1)!}\leq \varepsilon.$$ For these $p,q$, you have $$|f_q(x)-f_p(x)|=|\sum_{n=p}^{q-1} f_{n+1}(x)-f_n(x)| \leq \varepsilon $$ Hence, $(f_n)_n$ is Cauchy and thus converges.
For the uniform convergence, just make the same reasoning and deal with $\sup_{x\in [0,A]}e^x$.