2
$\begingroup$

Let ${f_n}$ be a uniformly bounded sequence of measurable functions on $[0,1]$. Define $$F_n(x)=\int_0^x f_n(t)dt.$$ Then there is a subsequence $\{F_{n_k}\}$ that converges uniformly on $[0,1].$

Attempt: We know $|f_n| \leq M$ for all $n \in \mathbb{N}$ with $M$ the least such upper bound. So $|F_n| \leq M(1-0)=M$ for all $n \in \mathbb{N}$ and $x \in [0,1]$. Since $\{|F_n|\}$ is a bounded and closed sequence of real numbers, we know there is a subsequence that converges to $\limsup |F_n|$.

We also know that, given $k \in \mathbb{N},$ there is a $j \in \mathbb{N}$ such that $||F_{n_l}|-\limsup |F_n|| \leq \frac{1}{k}$ for all $l \geq j$ due to a property of $\limsup.$ But this says exactly that there is a subsequence of $|F_n|$ that converges uniformly on $[0,1]$. $\square$

I know this answer is incomplete by a long shot (I need to look at the functions, and I'm assuming I'm taking the entire integral over $[0,1]$ in the proof), but I would like to know if the approach is hinting on anything, and regardless, what I need to do from here. Thanks!

$\endgroup$
1
  • 1
    $\begingroup$ You mean $f_n(t)$ as an integrand. $\endgroup$ Aug 9 '14 at 1:27
2
$\begingroup$

For a fixed bounded function $f(x)$, $\int_0^xf(t)dt$ is uniformly continuous in $x$. Show that $F_n(x)$ is equicontinuous and uniformly bounded on the unit interval. Now apply Arzela Ascoli.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.