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I want to prove:

Let $f$ be Lipschitz on $\mathbb{R}$ and $g$ be absolutely continuous on $[a,b]$. Show that the composition $f\circ g$ is absolutely continuous on $[a,b]$.

Proof: For $f$ being Lipschitz, there exist a constant $c \geq 0$ such that $$|f(x')-f(x)| \leq c|x'-x| \ \forall x',x \in \mathbb{R}.$$

Now, let $g$ to be absolutely continuous and let $\epsilon>0$. Choose $\delta > 0$, such that for any disjoint collection $\{(a_k,b_k)\}_{k=1}^n$ of $(a,b)$, we have $$\sum_{k=1}^n[b_k-a_k] < \delta \Rightarrow \sum_{k=1}^n[g(b_k)-g(a_k)] < \frac{\epsilon}{c}.$$ In particular, for the composition $(f \circ g)(x) = f(g(x))$, we obtain $$|f(g(x'))-f(g(x))| \leq c|g(x')-g(x)| \ \forall x',x \in \mathbb{R}.$$ Thus, for each $k = \{1,2, \ldots,n\}$, we have $$|f(g(b_k))-f(g(a_k))| \leq c|g(b_k)-g(a_k)|,$$ it follows that $$\sum_{k=1}^n|f(g(b_k))-f(g(a_k))| \leq c\sum_{k=1}^n|g(b_k)-g(a_k)| < \epsilon,$$ and the proof is complete.

Michael Jesurum
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okie
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  • Yes, but not the way to write it up. Start with "Let $\epsilon>0$." Choose $\delta$ so that the inequality in the third line is "$ < \epsilon/c$" (Not "$<\epsilon$". [This will require that you figure out the $c$ in advance of using the absolute continuity definition.] Etc. – B. S. Thomson Dec 06 '15 at 20:42
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    I'm sure your proof now will be accepted. Some friendly advice: don't be content with just supplying your proof. Our job is not merely to be trained as good graduate students, but to move towards being mathematicians and thinking the way that mathematicians always think about such problems. So the natural question here is: Is the composition of two absolutely continuous functions necessarily absolutely continuous? Some research here will give you an answer to that, and you should consider appending that to your answer. – B. S. Thomson Dec 07 '15 at 03:39

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