the inverse for the trace theorem

Question

The trace theorem says that the restriction of a $W^{1,p}(\Omega)$ function $u$, $Tu$ belongs to $W^{1-1/p,p}(\partial\Omega)$ if $\Omega$ satisfies some smooth condition, for example, $\Omega$ is convex. Now my question is the inverse of the Trace Theorem. Suppose $\Omega$ is convex, and $\phi\in W^{1-1/p,p}(\partial\Omega)$, is there exists a fucntion $\Phi\in W^{1,p}(\Omega)$ with $\|\Phi\|\leq C\|\phi\|?$

Dees the extension theorem is related to this question? But usually the extension Theorem talks about the extension from a domain to the whole space.

Answer 1

I think your question is answered in

Article (JonWal1978) Jonsson, A. & Wallin, H. A Whitney extension theorem in $L_p$ and Besov spaces Ann. Inst. Fourier (Grenoble), 1978, 28, vi, 139-192

and

Article (Marsch1987) Marschall, J. The trace of Sobolev-Slobodeckij spaces on Lipschitz domains Manuscripta Math., 1987, 58, 47-65

Theorem 2 of the latter paper states that if $\Omega$ is a Lipschitz domain, with $s,p$ satisfying some inequalities (as usual), then $W^{s,p}(\Omega)$ traces to (as a surjection) some Besov space on the boundary $\partial\Omega$ and that the trace operator has bounded linear right inverse if $s-1/p$ is not an integer.

Answer 2

At least if $p=2$ (Hilbert case), the answer is yes. Not only that, but there exist explicit right inverses, that is linear bounded operators $L:W^{1/2}(\partial\Omega)\rightarrow W^{1,2}(\Omega)$ such that $T\circ L$ is the identity. Since such operators are pseudo-differential, it is likely that they are bounded for $p\ne2$ as well, and thus satisfy again $T\circ L={\rm id}$.