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$\newcommand{\R}{\mathbb R}$ Let $\Omega\subset \R^d$ be a smooth, bounded open set and fix $p\geq 1$.

  • Fact 1: the usual Lebesgue differentiation theorem says that, if $u\in L^p(\Omega)$, then $$ u(x)=\lim\limits_{r\to 0}\frac{1}{|B_r(x)|}\int_{B_r(x)}u(y)dy $$ for Lebesgue-a.e. $x\in \Omega$.

  • Fact 2: for $u\in W^{1,p}(\Omega)$ (the usual Sobolev space) it is well known that the boundary trace $v:= tr(u)$ is well-defined as an element of $W^{1-1/p,p}(\partial\Omega)$.

Question: is it true that $$ v(x)=\lim \limits_{r\to 0}\frac{1}{|B_r(x)\cap \Omega|}\int_{B_r(x)\cap \Omega}u(y)dy $$ for $\mathcal H^{d-1}$-a.e. $x$ in the boundary?

This seems a very natural question to ask and I hope that the answer is already written somewhere out there. I suspect that the notion of $p$-capacity should play a role here (being the precise representative of a $W^{1,p}$ function $p$-quasicontinuous), but I'm not too familiar with capacity theory so I'd rather avoid appealing to such raffinate machinery and reinventing the wheel, if possible. (I am also aware that the trace can be recovered as the continuous limit along any transversal curve, but somehow I did not manage to conclude from this.) Actually any reference to either a positive or negative statement would make me happy, I just need a black-box theorem that I could apply.

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  • $\begingroup$ I guess its $\cap \partial\Omega$ and not $\cap \Omega$ in your question. Have you tried checking the references of this post mathoverflow.net/questions/117233/… ? $\endgroup$
    – Ayman Moussa
    Commented Jan 26, 2023 at 13:52
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    $\begingroup$ no, I really mean $\cap \Omega$, that's the whole point! otherwise this would be a simple application of the Lebesgue differentiation theorem "along the boundary", i-e in $L^p(\partial\Omega)$ $\endgroup$
    – leo monsaingeon
    Commented Jan 26, 2023 at 13:55
  • $\begingroup$ Oh, I get it, sorry. $\endgroup$
    – Ayman Moussa
    Commented Jan 26, 2023 at 16:41

2 Answers 2

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It is true - this is Theorem 5.7 in Evans and Gariepy’s Measure Theory and Fine Properties of Functions (2015 version).

Note that the theorem is stated for BV functions, but Sobolev functions are BV, so it holds also for your case.

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    $\begingroup$ Thank you Nate, this is really what I needed. Much appreciated! (I had actually checked an older version of Evans-Gariepy, to no avail. I'll make sure to give a closer look at the 2015 version, I may learn something useful ;-) ) $\endgroup$
    – leo monsaingeon
    Commented Jan 27, 2023 at 10:52
  • $\begingroup$ @leomonsaingeon I also have an older version (from 1992) of Evans-Gariepy; in mine it's Theorem 2 (p. 181) in Chapter 5 'BV Functions and Sets of Finite Perimeter', Section 5.3 'Traces'. $\endgroup$
    – Leo Moos
    Commented Jan 27, 2023 at 10:57
  • $\begingroup$ @LeoMoos: ah, now I see it, for some reason I didn't bother checking the BV part. damn! $\endgroup$
    – leo monsaingeon
    Commented Jan 27, 2023 at 10:59
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See

Jonsson, A.; Wallin, Hans, A Whitney extension theorem in (L^p) and Besov spaces, Ann. Inst. Fourier 28, No. 1, 139-192 (1978). ZBL0369.46031.

Proposition 7.1 in Section 7.3 is exactly what you are looking for (and a bit more).

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    $\begingroup$ Thank you Willie for your input. However, before spending some time trying to understand and navigating the heavy notations therein: I don't see a proposition 7.3 there, only proposition 7.1 (which doesn't see to be related to my question). Is this really the statement I should try to understand? $\endgroup$
    – leo monsaingeon
    Commented Jan 26, 2023 at 15:37
  • $\begingroup$ Sorry, I mean proposition 7.1 in section 7.3. And yes, it is exactly what your question is asking. The notion of "strictly defined" is exactly "the formula in the Lebesgue differentiation theorem holds". $G_\alpha$ is the Bessel potensial, so $G_1*f$ given $f\in L^p$ is a $W^{1,p}$ function. In your case you should set $j = 0$ and $d = n-1$. $\endgroup$
    – Willie Wong
    Commented Jan 26, 2023 at 15:54
  • $\begingroup$ The statement becomes then a $W^{1,p}$ function has a representative that satisfies the Lebsegue differentiation theorem condition outside of a set of $(n-1)$-dimensional Hausdorff measure 0. $\endgroup$
    – Willie Wong
    Commented Jan 26, 2023 at 15:57
  • $\begingroup$ I would also check Evans and Gariepy "Measure Theory and Fine Properties of Functions"; but my copy is not in my office right now (on my desk at home). This seems like something that may be treated in it. $\endgroup$
    – Willie Wong
    Commented Jan 26, 2023 at 16:33
  • $\begingroup$ Huum, I'm not too sure this is really what I need: how does the boundary trace come into play here? of course I may extend $u\in W^{1,p}(\Omega)$ to $\bar u\in W^{1,p}(R^n)$, but then it is absolutely not clear to me that $tr(u)=\bar u|_{\partial\Omega}$ (where the restriction is well-defined thanks to prop. 7.1 in question). What's more, I need to compare the average of $\bar u$ over balls with the average of $u$ over partial balls (intersected with $\Omega$). And again, I don't see how to do this, unless some particular trick (reflexion?) is implemented. Am I missing something? $\endgroup$
    – leo monsaingeon
    Commented Jan 27, 2023 at 10:44

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