# Image of the trace operator

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.

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

The image you are looking for equals the Besov space $B_{p,p}^{1-\frac1p} (\partial \Omega )$. See
• 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. May 10, 2012 at 8:37
• 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. Oct 30, 2013 at 14:34
• 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 $B^s_{p,p}(\partial\Omega) \simeq W^s_{p,p}(\partial\Omega)$.. Sep 12, 2016 at 8:02