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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. ;p2012-08-05
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    See also this question: [Absolutely Continuous Function composed with an Exponential Function](http://math.stackexchange.com/questions/89481/absolutely-continuous-function-composed-with-an-exponential-function) and the question linked to it.2012-08-05

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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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    [Exercise 5:7.8](http://books.google.com/books?id=1WY6u0C_jEsC&pg=PA219) in Bruckner-Bruckner-Thomson gives some conditions when composition is absolutely continuous.2012-08-05
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    @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...2012-08-05
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    A newer edition of that book is freely (and legally) available from here: http://classicalrealanalysis.info/com/FREE-PDF-DOWNLOADS.php2012-08-05