# Question on Sobolev extension onto boundary

Let $$U \subset \mathbb R^3$$ be an open, bounded and connected set with a $$C^2-$$regular boundary $$\partial U$$. I'm trying to understand the following implication:

If $$f\in W^{2-1/2,2}(U)$$ then $$f{\vert}_{\partial U} \in W^{1,2}(\partial U)(*)$$

So, I'm aware of this theorem:

General Trace Theorem: if $$f\in W^{1-1/p,p}(\partial \Omega)$$, then there exists a function $$f \in W^{1,p}(\Omega)$$ such that $$f{\vert}_{\partial \Omega}=f$$

QUESTION: Is the above theorem still valid if we replace $$\partial \Omega$$ with $$U$$ and $$\Omega$$ with $$\partial U$$ so that $$(*)$$ makes sense? If not, is there any other way to deduce $$(*)$$?

Any help is appreciated. Thanks in advance!

EDIT: regularity in $$(*)$$ fixed

• You're asking about whether you can extend a function from the boundary into the domain? – Ian Nov 13 '18 at 15:52
• I think that you have the regularities reversed. Anyway, see here. – Giuseppe Negro Nov 13 '18 at 15:55
• @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary... – kaithkolesidou Nov 13 '18 at 15:56
• The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning. – Ian Nov 13 '18 at 15:57
• @GiuseppeNegro What do you mean by reversed regularities? – kaithkolesidou Nov 13 '18 at 16:00

$$\newcommand{\R}{\mathbb{R}}$$ The implication (*) is invalid, and for obvious reasons. The trace $$f|_{\partial U}$$ is kind of a restriction of $$f$$, so you cannot expect it to have more derivatives than $$f$$!
To have an example, choose your favorite function $$g \in W^{1/2,2}(\R)$$ that doesn't belong to $$W^{1,2}(\R)$$. Then consider \begin{align*} U &= \{ (x,y,z) \in \R^3 : z > 0 \}, \\ f(x,y,z) & = g(x,y). \end{align*} Then $$f$$ is locally in $$W^{1/2,2}$$, but its restriction to the boundary $$\partial U = \R^2 \times \{0\}$$ is simply $$g$$, which is not in $$W^{1,2}$$.