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.
