Show that there exists a subsequence $\{F_{n_{k}}\}$ which converges to uniformly on $[a,b]$.

Question

Let $\{f_n\}$ be uniformly bounded sequence of functions which are Riemann-integrable functions on $[a,b]$ and define for $a\leq x\leq b$. $$ F_n(x)= \int_a^x f_n(t)dt.$$

Show that there exists a subsequence $\{F_{n_{k}}\}$ which converges to uniformly on $[a,b]$.

Here actually the basic problem is to show that $\{F_{n}\}$ is equicontinuous right. So can anyone write explicitly the reason why it is equicontinuous?

Answer 1

We have the following estimate : without loss of generality, assume $x \le y$.
$$ \left| F_n(x) - F_n(y) \right| = \left| \int_x^y f_n(t) \, dt \right| \le \int_x^y |f_n(t)| \, dt \le M|y-x| $$ where $M$ is the uniform bound of the $f_n$'s. Therefore the sequence $F_n$ is equicontinuous ; the uniform boundedness is given by the bound $M(b-a)$. You can apply Arzelà-Ascoli.

Hope that helps,

Answer 2

This is a consequence of Arzelà–Ascoli theorem.

As the $f_n$'s are uniformly bounded, say by $M$, then $$ \lvert F_m(x)-F_m(y)\rvert\le\int_x^y \lvert\,f_m(t)\rvert\,dt\le M\lvert x-y\rvert, $$ and hence $\{F_m\}$ is a uniformly equicontinuous family, which is also uniformly bounded, as $$ \lvert F_m(x)\rvert\le M(b-a). $$ Thus according to Arzelà–Ascoli theorem it has a uniformly convergent subsequence.