# 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.

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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. – Willie Wong May 10 '12 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. – Delio Mugnolo Oct 30 '13 at 14:34