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I am trying to show that if $f:[0,1]\to \mathbb{R}$ is absolutely continuous then $|f|^p$ is absolutely continuous for all $p>1$.

Since the product of absolutely continuous functions is absolutely continuous, the result is obvious if $p$ is an integer. My first thought in trying to prove the result was to try to generalize the proof that the product of absolutely continuous functions is absolutely continuous. Following that strategy and using the fact that $f$ is bounded, it's not difficult to reduce the problem to proving the result for the case that $1<p<2$. Once I get here though I am not sure how to proceed. Any help is appreciated!

Kyle Liss
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  • Not a full answer, but possibly helpful http://math.stackexchange.com/questions/622119/an-absolutely-continuous-function-that-isnt-one-when-raised-to-a-fractional-pow – TomGrubb Dec 28 '16 at 03:52
  • Thank you that's what I was looking for. Since $f$ is continuous on $[0,1]$, $|f(x)| \le M \forall x\in [0,1]$ for some $M > 0$. Since $|x|^p$ is lipschitz on $[-M,M]$ for $p>1$ we get the result by composition. – Kyle Liss Dec 28 '16 at 04:06

1 Answers1

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We know that for every $\delta>0$ there is an $S_\delta$ such that for any finite sequence of $n$ disjoint closed intervals $[a_i,b_i]$ with sum of lengths less than $\delta$ we have $\sum\limits_{i=1}^n|f(a_i)-f(b_i)|\leq S_\delta$.

Now notice that the function $f(x)=x^p$ defined on the interval $[0,a]$ is lipschitz continuous, so there exists a $\alpha >0$ such that $||x|^p-|y|^p|\leq \alpha|x-y|$.

Hence $\sum\limits_{i=1}^n|f(a_i)-f(b_i)|\leq S_\delta\alpha$.

Lets take a $\delta$ such that $S_\delta< \epsilon/\alpha$ and we are done.

Asinomás
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