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 have a simple question on Sobolev space theory. Let $1\le p \le \infty. $How can one prove that $u\in W^{1,p}(1,0)$ is equal s.e. to an absolutely continuous function and that $u'$ exists a.e. and belongs $L^p(0,1)$?

Thank you for your assistance.

share|improve this question
What does equal s.e. mean? Also don't you mean $W^{1,p}(0,1)$? If I'm guessing right, it looks like a basic theorem in Sobolev space theory... –  tomasz Dec 2 '12 at 1:25
I mean the function can be represented by a function that is a.e. equal to an absolutely continuous function. –  Pooya Dec 9 '12 at 9:16

1 Answer 1

Consider the case $p=1$. Take $u\in W^{1,1}(0,1)$ and put $v(t)=u(0)+\int_0^tu'(s)ds$, then $v\in W^{1,1}(0,1)$ and is absolutely continuous. We have $v'=u'$ a.e. so $u=v+c$ a.e.

share|improve this answer
Thank you for the answer. The $p=1$ case can be proven by your argument. I still don't know how to prove in general case. –  Pooya Dec 9 '12 at 9:17
@Pooya: Just notice that for $p>1$ we have $W^{1,p}\subset W^{1,1}$ and $L^p\subset L^1$ since $(0,1)$ has finite measure. –  Jose27 Dec 9 '12 at 22:40

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.