Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been working on the following question:

If $F : \mathbb{R} \rightarrow \mathbb{R}$ is a Lipschitz function, then $F(x)=F(0)+\int_0^x F'(t) dt$.

I've already proved that Lipschitz implies $F'$ is exists a.e., and $F'$ is essentially bounded, but for whatever reason I've been stumped on this one. I looked at similar questions on here but couldn't seem to find too much that went into detail.

share|improve this question
You mean $F:\mathbb R\to \mathbb R$? –  Pedro Tamaroff Aug 18 '12 at 19:19
add comment

1 Answer

If $F$ is Lipschitz it is absolutely continuous. From Rudin's "Real & Complex Analysis" Theorem 7.20, we have that $F$ is differentiable a.e. and $F(x)=F(0)+\int_0^x F'(t) dt$.

share|improve this answer
I'm aware of that, and did not understand his proof. –  Frank White Aug 18 '12 at 19:40
@copper.hat: Just a nitpick, but F being absolutely continuous implies not only that F' exists a.e., but also that it's Lebesgue integrable. –  Bey Aug 18 '12 at 19:47
I know, I just included the relevant parts of the statement. –  copper.hat Aug 18 '12 at 19:48
But it is relevant to say F' is integrable, otherwise the expression involving the integral of F' doesn't necessarily make sense. In any case, like I said, it's a nitpicky little thing –  Bey Aug 18 '12 at 19:49
@Bey: Since $F$ is Lipschitz, we have $|F'(x)| \leq L$ where $L$ is the Lipschitz rank. Using the DCT shows that it is integrable, but again, that wasn't the focus, I think? –  copper.hat Aug 18 '12 at 19:53
show 6 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.