# traces of sobolev spaces under additional assumptions

Let $p\in [1,\infty]$, $\Omega$ an open bounded domain with (smooth, if necessary) boundary $\partial \Omega$.

Is there a subspace $X\subset L^p(\Omega)$ - a simply describable space, ideally a Sobolev or Besov or Triebel-Lisorkin or somebody else's space - such that the trace operator is surjective from $\{f\in X:\Delta f\in L^p(\Omega)\}$ onto $L^p(\partial\Omega)$?

The answer is affirmative if $p=2$: One can take $X=H^\frac{1}{2}(\Omega)$. That the trace operator is surjective from $$\{f\in H^\frac{1}{2}(\Omega):\Delta f\in L^2(\Omega)\}\quad \hbox{onto}\quad L^2(\partial\Omega)$$ has been checked in [DOI 10.1007/s00020-002-1163-2], Lemma 3.1: The idea is to use a certain isomorphism that allows to construct a harmonic $H^\frac{1}{2}(\Omega)$-function with given initial data, but the proof is essentially an application of Lions-Magenes theory, which is restricted to the Hilbert case. (For me, the obvious place to look for a Banach space extension is the book of Grisvard, but I could not find anything there, either).

EDIT: My conjecture is of course that, more generally, $X=W^{\frac{1}{p},p}(\Omega)$ can be taken. This would yield the space $W^{\frac{1}{p},p}(\Omega)\cap D(\Delta;L^p(\Omega))$ with the notation of Grisvard. Such spaces do not seem to have been introduced in the books of Adams-Fourier or Demengel-Demengel.

EDIT #2: Perhaps it is convenient to explain my interest in this question. The point is that I would like to define not only a right inverse of the trace operator, as customary; but rather an operator that maps a given function on the boundary into the (unique) solution of an eigenvalue equation for $\Delta$ whose boundary values are prescribed by the given function - something that was considered e.g. in many articles of the Japanese school active on Dirichlet forms in the 1960s (Fukushima, Sato etc.). Now, I am willing to consider very weak notions of solution, but at the very least I want to make sure that both sides of the the eigenvalue equation are in $L^p$.

• Theorem 3.54, in Demengel & Demengel ? Oct 30 '13 at 11:59
• Taking $\sigma:={\rm grad }f$ for some $f\in W^{1,p}(\Omega)$? I am not quite sure this has to do with my question. It seems to me that the operator $S$ of Demengel-Demengel would then be a weakly defined normal derivative - more precisely, $S({\rm grad }f)=\frac{\partial f}{\partial \nu}$, and not the trace. Besides, $S$ is surjective onto $W^{1-\frac{1}{p'},p'}(\partial \Omega)$, not onto $L^p(\Omega)$, as I would like to have. Oct 30 '13 at 13:32
• Sorry, read too quickly. Maybe 3.67 can be pushed a bit? It is true that the trace of $W^{1,1}$ is $L^1(\partial\Omega)$ (same reference)...do you want $p>2$ or $p<2$? Oct 30 '13 at 14:35
• either case would be interesting for me. Oct 30 '13 at 14:53
• by the way, 3.67 surprises me quite a bit. is the space of all traces of $W^{1,p}(\Omega)$-functions not a Besov space? Oct 30 '13 at 15:06

Partial answer: according to Triebel (Theory of function spaces, 1983, Remark 2.7.5, p. 139), the trace of the Besov space $B^{1/p, p}_1 (\Omega)$ is $L^p (\partial \Omega)$, but the linear extension operator from $L^p (\partial \Omega)$ to $B^{1/p, p}_1 (\Omega)$ cannot be linear. In particular, the harmonic extension fails to give such an extension.
Edit 1: The proof can be found in the paper V.I. Burenkov, M.L. Gold'man, On extension of $L^p$ functions, Proceeding of the Steklov Institute of Mathematics, 1981, p. 33-54 (translated from a 1979 paper in Russian).
• According to Triebel "There does not exist a linear extension operator from $L^p (\partial \mathbb{R}^n_+)$ to $B^{1/p, p}_1 (\mathbb{R}^n_+)$", that is, there is no linear operator $\operatorname{ext}: L^p (\partial \Omega) \to B^{1/p, p}_1 (\Omega)$ such that $\operatorname{ext} \circ \operatorname{tr} = \operatorname{id}$. Nov 5 '13 at 16:23
• sorry, i do not have triebel's book at my disposal right now. does his result hold for all $p$? i am trying to make sense of how it relates to the assertion of lions-magenes i quote. i guess $B^{\frac12,2}_1(\mathbb R^n_+)$ does not easily compare to $H^{\frac12}(\mathbb R^n_+)$, right? Nov 5 '13 at 20:26
• $H^{1/2} (\mathbb{R}^n_+) = B^{1/2, 2}_2 (\R^n_+) \supsetne B^{1/2, 2}_1 (\R^n_+)$ Nov 6 '13 at 11:07