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

Thank you for your assistance.

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  • $\begingroup$ 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... $\endgroup$
    – tomasz
    Dec 2, 2012 at 1:25
  • $\begingroup$ I mean the function can be represented by a function that is a.e. equal to an absolutely continuous function. $\endgroup$
    – Pooya
    Dec 9, 2012 at 9:16

3 Answers 3

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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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  • $\begingroup$ 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. $\endgroup$
    – Pooya
    Dec 9, 2012 at 9:17
  • $\begingroup$ @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. $\endgroup$
    – Jose27
    Dec 9, 2012 at 22:40
  • $\begingroup$ @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. $\endgroup$
    – Cookie
    Jan 10, 2015 at 23:09
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    $\begingroup$ @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 $u\in W^{1,1}(0,1)$. $\endgroup$
    – Jose27
    Jan 11, 2015 at 1:49
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    $\begingroup$ @Jose27 Hi Jose. I think we should just define $v$ by $v(t)=\int_0^tu'(s)ds$, i.e. without adding the $u(0)$ term. The reason is that, $u$ is a $L^1$ function and defined almost everywhere, so it makes no sense to say $u(0).$ However, though we define $v(t)$ in this slightly different way, it is still absolutely continuous. And so your reasoning still applies. $\endgroup$
    – Sam Wong
    Mar 17, 2020 at 15:05
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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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We want to show $u(x)-u(a) = \int_a^{x}u'(t)dt$ a.e., then consider the mollifier and convolve it with $u$, then the formal integration by parts yields: $u_{\epsilon}(x)-u_{\epsilon}(a) = \int_a^{x}u'_\epsilon(t)dt$ and by $L^1$ convergence of $u_\epsilon$ and $u'_\epsilon$: $\int_a^{x}u'_\epsilon(t)dt \to \int_a^{x}u'(t)dt$. By picking a subsequence of $u_\epsilon$, we may assume $u_\epsilon \to u$ a.e., then $u(x)-u(a) = \int_a^{x}u'(t)dt$ a.e., thus $u(x)$ can be identified as an absolutely continuous function $v$ such that $v(x) = u(a) + \int_a^{x}u'(t)dt$.

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