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If $f(x)$ is absolutely continuous (a.c.) on [a,b], is the function $e^{f(x)}$ also absolutely continuous on [a,b] ?

thanks

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I like that you introduce (a.c.) and then promptly don't use it. ;p –  mixedmath Aug 5 '12 at 3:06
    
See also this question: Absolutely Continuous Function composed with an Exponential Function and the question linked to it. –  Martin Sleziak Aug 5 '12 at 7:27

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It is true. Suppose $g$ is Lipschitz with rank $L$ on $[a,b]$, and $f$ is AC. Then $g \circ f$ is also AC.

To see this, suppose $f$ is AC, and let $\epsilon>0$. Choose $\delta>0$ such that if $(y_k,x_k)$ are a finite collection of pairwise disjoint intervals in $[a,b]$ with $\sum |y_k-x_k| < \delta$, then $\sum |f(y_k)-f(x_k)| < \frac{\epsilon}{L}$.

Now consider $\sum |g \circ f(y_k)-g \circ f(x_k)| = \sum |g(f(y_k))-g ( f(x_k))| \leq L \sum |f(y_k)-f(x_k)| < \epsilon$. Hence $g \circ f$ is AC.

Since $x \mapsto e^x$ is smooth, it is Lipschitz on any compact interval, hence the function $ x \mapsto e^{f(x)}$ is AC.

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I like that you introduce rank $L$ and then promptly don't use it. ;p –  Matt Aug 5 '12 at 3:13
    
I got stuck in the mathjax :-). –  copper.hat Aug 5 '12 at 3:15
    
Exercise 5:7.8 in Bruckner-Bruckner-Thomson gives some conditions when composition is absolutely continuous. –  Martin Sleziak Aug 5 '12 at 7:29
    
@MartinSleziak: Thanks for catching that, I removed the offending part. I've been down this overlapping road before, and forgot about it completely. The wonders of aging... –  copper.hat Aug 5 '12 at 7:41
    
And nice reference! –  copper.hat Aug 5 '12 at 7:42

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