# the inverse for the trace theorem

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.

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This is very closely related to my question here: mathoverflow.net/questions/70740/image-of-the-trace-operator – Willie Wong May 10 '12 at 8:01

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.

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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 $T\circ L={\rm id}$.

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Could you give a reference about these explicit right inverses? – user37238 Aug 3 '15 at 10:42
@37238. When $\Omega$ is a half-space, just make a Fourier transform in the variables tangent to the boundary. Then you build by hand the inverse. – Denis Serre Aug 3 '15 at 12:53
Thank you for your quick answer. In fact, I was thinking about explicit right inverses for a general domain (say $\Omega$ $\mathcal{C}^1$ bounded). – user37238 Aug 3 '15 at 12:55
One denomination is Babitch inverse – Denis Serre Aug 3 '15 at 12:57