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It is well-known that we have the trace theorem for Sobolev spaces. Let $\Omega$ be an open domain with smooth boundary, we know that the map

$$ T: C^1(\bar\Omega) \to C^1(\partial\Omega) \subset L^p(\partial\Omega) $$

by $Tu(y) = u(y)$ for $y\in\partial\Omega)$ can be extended continuously to a linear map on Sobolev spaces for $p > 1$

$$ T: W^{1,p}(\Omega) \to L^p(\partial\Omega)$$

We also know that this map is not surjective, since the Trace Theorem (Sobolev embedding) tells us that when dropping 1 dimension, we have that the image of $T$ actually lives ([Edited May 10 2012] caveat: see my comment on the answer below) in a fractional Sobolev space,

$$ T: W^{1,p}(\Omega) \to W^{1-1/p, p}(\partial\Omega) \Subset L^p(\partial\Omega) $$

On the other hand, we know that this map $T$ has dense image in $L^p$, just using the density of $C^1$.

Question: Is there a known characterisation of precisely what the image set of $T$ is? A slightly weaker question is: consider‡ $w \in W^{s,q}(\partial\Omega)$ for $1 - 1/p \leq s \leq 1$ and $q \geq p$, does there necessarily exist some function $u\in W^{1,p}(\Omega)$ such that $Tu = w$?

For example, if we assume that $w$ is Lipschitz on $\partial\Omega$, then we can extend (almost trivially) $w$ to a Lipschitz function $C^{0,1}(\bar\Omega)\subset W^{1,p}$ for every $p$. So the case $s = 1, q = \infty$ has a positive answer. Whereas the Sobolev embedding theorem mentioned above tells us that it is impossible to go below $s < 1-1/p$ and $q < p$.

‡ The lower cut-off here is clearly not sharp. The trace theorem combined with Sobolev embedding can be used to trade differentiability with integrability. Out of sheer laziness I will not include the numerology here. One should interpret the conditions on $s,q$ to be that $s \leq 1$, $q \geq p$ plus the requirement that $(s,q)$ is at least as good as what can be guaranteed by Sobolev embedding and the trace theorem.

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    $\begingroup$ What is $C^1(\partial\Omega)$? Do you have a reference for the result $ T: W^{1,p}(\Omega) \to W^{1-1/p, p}(\partial\Omega) \Subset L^p(\partial\Omega) $? $\endgroup$
    – user14319
    Sep 11, 2016 at 13:02

1 Answer 1

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The image you are looking for equals the Besov space $B_{p,p}^{1-\frac1p} (\partial \Omega )$. See

H. Triebel. Interpolation theory, function spaces, differential operators. Leipzig, 1995 (in fact, I used the earlier Russian edition, Moscow, 1980).

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  • $\begingroup$ Thank you for the reference. I will look for it in the library tomorrow. $\endgroup$ Jul 19, 2011 at 19:03
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    $\begingroup$ incidentally this also means that the "folklore" version of the Sobolev trace theorem is not strictly correct. From Adams and Fournier Sobolev Spaces, paragraph 7.67 on p255, we see that $B_{p,p}^s = F_{p,p}^s$ (at least on $\mathbb{R}^n$) and thus if $p \geq 2$ $B_{p,p}^{1-1/p} \supseteq W^{1-1/p,p}$ with equality only when $p = 2$, and if $p\leq 2$ the inclusion is reversed. $\endgroup$ May 10, 2012 at 8:37
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    $\begingroup$ Willie Wong, your last comment has surprised me. What you are saying is that if $p< 2$, then $W^{1-1/p,p}(\partial \Omega)$ is strictly larger than $B^{1-1/p}_{p,p}(\partial\Omega)$? But then in view of Anatoly Kochubei's answer, there might not be possible to find a right inverse of the trace operator on the whole space $W^{1-1/p,p}(\partial \Omega)$ -- which however is a classical assertion, e.g. Thm. 1.5.1.3 in Grisvard's Elliptic problems in nonsmooth domains. $\endgroup$ Oct 30, 2013 at 14:34
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    $\begingroup$ It seems that Grisvard's not being careful in his exposition, or that he has a different definition of the fractional Sobolev spaces. The paper of Gagliardo that Grisvard cites for Theorem 1.5.1.3 is mentioned by Adams and Fournier in Remark 7.45 on page 241. In Item 3 they specifically indicate that the trace space as identified by Gagliardo is the Besov space that Anatoly mentioned above. See also Marschall, J. The trace of Sobolev-Slobodeckij spaces on Lipschitz domains, Manuscripta Math., 1987, 58, 47-65 $\endgroup$ Dec 2, 2013 at 8:57
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    $\begingroup$ I think this is "just" a problem of nomenclature, which stems from $B^s_{p,p}(R^n) \simeq W^{s,p}(R^n)$. The notion of such a space on $\partial\Omega$ is subject to the author's understanding of which came "first" (Triebel defines $W^{s,p}(R^n) = B^s_{p,p}(R^n)$ and shows later that the "standard" $W^{s,p}(R^n)$ norm is equiv.). In fact, Triebel (Ch.3.6.1), Grisvard (Def.1.3.3.2) and Marschall (p.50) all define the spaces on $\partial \Omega$ via pullback w.r.t- suitable charts of the spaces on $R^{n-1}$ which should imply $B^s_{p,p}(\partial\Omega) \simeq W^s_{p,p}(\partial\Omega)$.. $\endgroup$
    – Hannes
    Sep 12, 2016 at 8:02

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