Take the 2-minute tour ×
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|improve this question
1  
What have you tried? –  P.. Nov 14 '12 at 13:33

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|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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