I have this question from a real analysis assignment,
For $n\geq1$, let $f_n:[0,1]\to\mathbb{R}$ be a continuous function with $$|f_n(x)|\leq 1+\frac{n}{1+n^2x^2}$$
Define $F_n:[0,1]\to\mathbb{R}$ via $$F_n(x)=\int_0^xf_n(t)dt$$ Show that the sequence $\{F_n\}_{n\geq1}$ admits a subsequence that converges pointwise on $[0,1]$.
My first thought is that each $f_n$ is uniformly bounded (by 2, for instance), but this doesn't seem like a strong enough conclusion to reach from that first line, considering the bound is shrinking as $n$ grows.
My second thought is that for fixed $x$, $\lim_{n\to\infty}|f_n(x)|\leq 1$, so $\lim_{n\to\infty}|F_n(x)|\leq x$ , which seems like it would be useful, but I'm not sure if it's relevant or how to use it.
I'm also not sure how to approach actually obtaining a pointwise convergent subsequence. I've seen results from class that give a sequence with uniform convergence, but not pointwise. So, I'm assuming there isn't some theorem to invoke to reach our conclusion? Should I try and construct some sort of subsequence myself?
Any hints or tips would be much appreciated. (apologies if there's already a question like this, I spent quite a bit of time searching and couldn't find anything)
Best Answer
First of all, $\{F_n\}$ is uniformly bounded since
$$ |F_n(x)| \le \int_0^x \left(1+ \frac{n}{1+(nt)^2} \right) dt\le 1 + \int_0^1 \frac{n}{1+(nt)^2} dt= 1+ \arctan n \le 1+ \frac{\pi}{2}.$$
But $\{ F_n\}$ might not be equicontinuous (see the remark below).
To get around this, as you have pointed out in the comment, on each interval $[1/m, 1]$, $\{ F_n|_{[1/m, 1]|} \}$ is equicontinuous since $F'_n = f_n$ is uniformly bounded there. Using Arzela-Ascoli on each $[1/m,1]$ together with a diagonal sequence argument, there is a subsequence $\{F_{n_k}\}$ so that $\{F_{n_k}\}$ converges uniformly in $[1/m, 1]$ for all $m\in \mathbb N$. In particular, $F_{n_k}$ converges pointwisely in $(0,1]$. Lastly, since $F_n(0) = 0$ for all $n\in \mathbb N$, the subsequence $F_{n_k}$ converges pointwisely in $[0,1]$.
Remark Take for example $f_n(x) = 1+ \frac{n}{1+n^2x^2}$. Then \begin{align} F_n(x) &= \int^x_0 \left(1+ \frac{n}{1+(nt)^2} \right) dt \\ &= x + \arctan (nx) . \end{align}
In particular, $F_n(1/n) = 1/n + \arctan 1> \arctan 1$. Thus choosing $\epsilon_0 = \frac 12 \arctan 1$, there is no $\delta >0$ so that $$ |F_n(x) - F_n(0)| = F_n(x) < \epsilon_0$$ for all $x\in [0,\delta)$ and for all $n$. Thus $\{ F_n\}$ is not equicontinuous.