Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. If $f$ is absolutely continuous and $g$ satisfies a Lipschitz condition, then $g \circ f$ is absolutely continuous.

  2. If $f$ is absolutely continuous and strictly increasing and $g$ is absolutely continuous, then $g \circ f$ is absolutely continuous.

  3. There exist absolutely continuous functions $f$ and $g$ defined on $[0,1]$ such that $g\circ f$ is not absolutely continuous.

share|cite|improve this question

put on hold as too broad by Mice Elf, avid19, Solid Snake, John Ma, Claude Leibovici 24 secs ago

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer 1

The solutions to both (a) and (b) follow fairly directly from the definitions of absolutely continuous and Lipschitz continuous. Here's the solution to the first one:

(a) Since $g$ is Lipschitz continuous an interval $[a,b]$, there is some $M>0$ such that $|g(x)-g(y)|<M|x-y|$ for any $x, y \in [a,b]$. Now since $f$ is absolutely continuous on some other interval $[c,d]$, there is some $\delta >0$ such that, given $\epsilon > 0$,

$$ \displaystyle \sum_{k=1}^n |f(x_k)-f(y_k)|<\epsilon/M$$

whenever $\{[x_k, y_k] \mid k=1, \ldots, n \} $ is a (clearly, finite) collection of mutually disjoint subintervals of $[c,d]$ such that $\displaystyle \sum_{k=1}^n (y_k-x_k)<\delta$. Now it is straightforward to see that

$$\displaystyle \sum_{k=1}^n |(g \circ f)(x_k)-(g \circ f)(y_k)|= \sum_{k=1}^n|g(f(x_k))-g(f(y_k))|$$

$$\leq \sum_{k=1}^n M|f(x_k)-f(y_k)|<M(\epsilon / M) = \epsilon$$

Thus $g \circ f$ is absolutely continuous on $[c,d]$.

Is it clearer what to do for part (b)? For part (c) I suggest looking at this previous question.

EDIT: If you need further explanation, please let me know.

share|cite|improve this answer

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