Is uniform convergence required for a continuous limit function?

Question

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 $x\geq 0.$

I have completed the proof 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 $A > 0$. But the question next asks me to prove that $F$ is continuous on $[0, \infty)$. I am not sure how to proceed with this since I was under the impression that uniform convergence was required for the limit function to be continuous.

Any help would be greatly appreciated!

Answer 1

You have proved that for an arbitrary $A\gt0$ the succession of continuous functions $f_n$ restricted to $[0,A]$ converges uniformly to $F$ restricted to $[0,A]$, so we can conclude that $F$ restricted to $[0,A]$ is continuous...

The thesis then follows from the fact that $A$ is arbitrary

Uniform convergence of continuous function is a sufficient condition for the limit function to be continuous, while point wise convergence is not sufficient in general, as Jean-Claude Arbaut pointed out in a comment above.