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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?

2 Answers 2

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

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    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 e$x$amples of AC but not Lip functions. Sorry!2012-05-06
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$e^x$ 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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    but uniform is not enough :-)2011-12-07