# $u\in W^{1,p}(0,1)$ is equal a.e. to an absolutely continuous function?

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

Thank you for your assistance.

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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

## 2 Answers

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.

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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
@Jose27 Can you elaborate why $W^{1,p} \subset W^{1,1}$ is true for all $p > 1$? I apologize if this question is fundamental, but I too am studying Sobolev spaces but have little knowledge in measure theory. – Cookie Jan 10 at 23:09
@dragon: This is jsut Holder's inequality: $$\int_0^1 |f|dx \leq \left( \int_0^1 |f|^p dx \right)^{1/p}.$$ Now if $u\in W^{1,p}(0,1)$ then using $f=u$ and $f=u'$ we arrive at $u\in W^{1,1}(0,1)$. – Jose27 Jan 11 at 1:49
@Jose27 "is absolute continuous" I still cannot see what really make $u$ equal to an absolute continuous function $v$, what was used in obtaining the absolute continuity? I saw other notes, says "fundamental theorem of calculus" why? – math101 Jul 8 at 10:19

The answer posted by Jose27 is correct. For more details and related things, see the reference here: http://www.iadm.uni-stuttgart.de/LstAnaMPhy/Weidl/fa-ws04/Suslina_Sobolevraeume.pdf, especially Theorem 5.

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