# $f(x)$ and $h(x)$ are absolutely continuous functions. Is $e^{f(x)} |h(x)|$ as well?

Given that functions $f(x)$ and $h(x)$ are absolutely continuous on $[0,1]$, I want to show that $e^{f(x)} |h(x)|$ is absolutely continuous as well.

I know that (1) the product of two absolutely continuous function on $[0,1]$ is absolutely continuous. (2) the composition of a Lipschitz continuous function and an absolutely continuous function is absolutely continuous. So $|h|$ is absolutely continuous,

But the exponential function $e^x$ is not Lipschitz, so not absolutely continuous.

What's the key to solve the problem here?

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## 2 Answers

$f$ is continuous, hence bounded on it's (compact) domain of definition, and the exponential is locally Lipschitz.

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uhhmm - and 'not Lipschitz' does not imply 'not absolutely continuous'. –  user20266 Dec 7 '11 at 17:44
@mixedmath are you sure? As far as I know Lipschitz implies absolutely continuous, the reverse implication is new to me. Could you provide a reference, please? –  user20266 May 6 '12 at 19:54
Not only am I not sure, I'm absolutely wrong. Sorry for that! I stumbled across this post looking to see if a recent post was a duplicate. I then proceeded to give examples of AC but not Lip functions. Sorry! –  mixedmath May 6 '12 at 20:00

$e^x$ is not Lipschitz on $\mathbb{R}$, but is Lipschitz on any bounded interval (with a Lipschitz constant that depends on the interval). Since $f$ is continuous on $[0,1]$ it is bounded (I assume that you know this); let's say that $|f(x)|\le M$ for all $x\in[0,1]$. The exponential function is Lipschitz continuous on $[-M,M]$, so that $e^{f(x)}$ is absolutely continuous on $[0,1]$.

Forget about the following line. I was thinking in uniform continuity, not absolute continuity.

An alternative way: every continuous function on a bounded closed interval is absolutely uniformly continuous.

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absolute continuity implies differentiability Lebesgue almost everywhere in one dimension. This is by no means true for every continous function on a closed interval! –  user20266 Dec 7 '11 at 17:51
I read absolute, but thought uniform. –  Julián Aguirre Dec 7 '11 at 17:58
but uniform is not enough :-) –  user20266 Dec 7 '11 at 18:04