1
$\begingroup$

Can someone please help me prove the following? I am having difficulties proving it.

Let $f_n(x) (n=1,2,\cdots)$ be increasing absolutely continuous functions on $[a,b].$ Assume $f(x) = \sum_{n=1}^\infty f_n(x)$ converges on $[a,b],$ prove that $f(x)$ is absolutely continuous on $[a,b].$

$\textbf{My idea:}$ For each $\epsilon > 0,$ there is a $\delta > 0$ such that for every finite disjoint collection $\{(a_k,b_k)\}_{k=1}^n$ of open intervals in $(a,b),$ $$\vert \sum_{k=1}^n [f(b_k) - f(a_k)]\vert < \epsilon, \text{ if } \sum_{k=1}^n [b_k - a_k] < \delta.$$

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

\begin{align}|\sum_k (f(b_k)-f(a_k))| &\le \sum_k |f(b_k) - f(a_k)| \\ &\le \sum_k |f(b_k) - f_m(b_k)| + \sum_k |f_m(b_k) - f_m(a_k)| + \sum_k |f_m(a_k) - f(a_k)| \end{align}

Choose $m$ large enough so the first and third terms are small. Choose $\delta$ small enough so that the middle term is small.

$\endgroup$
5
  • $\begingroup$ thank you so in the end you'll see that $\sum_k^n |f(b_k) - f(a_k)| < \epsilon$ if $\sum_k^n [b_k - a_k] < \min\{\delta^+, \delta^-\}$? $\endgroup$
    – rudinsimons12
    Oct 13, 2020 at 19:44
  • $\begingroup$ So $m$ depends on $a_k, b_k$. It's not clear how you choose $\delta >0$ so that the inequality holds for all choices of $(a_k, b_k)$. $\endgroup$
    – Arctic Char
    Oct 13, 2020 at 20:06
  • $\begingroup$ @ArcticChar Yes, it is ok for $m$ to depend on $a_k, b_k$. Once $m$ is fixed, applying the definition of absolute continuity of $f_m$ will allow you to make the middle term small no matter the choice of $a_k, b_k$. $\endgroup$
    – angryavian
    Oct 13, 2020 at 20:26
  • $\begingroup$ Can you write down precisely the definition of $\delta$? I really can't see how it is independent of $(a_k, b_k)$... $\endgroup$
    – Arctic Char
    Oct 13, 2020 at 20:32
  • $\begingroup$ @ArcticChar Suppose you want the middle term to be smaller than some number $\epsilon / 3$. The fact that $f_m$ is abs. cont. gives you a $\delta$ that guarantees this for any $a_k, b_k$ satisfying $\sum_k (b_k-a_k) < \delta$. $\endgroup$
    – angryavian
    Oct 13, 2020 at 20:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .