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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.

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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
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1 Answer

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
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