Trace Operator on $L^2$ Functions

Question

in Why no trace operator in $L^2$? it is mentioned, that there exists a linear continuous trace operator from $L^2(\Omega)$ to $H^\frac12(\partial\Omega)$* for sufficiently smooth boundary. Can you give me any reference for this statement? I need something like this and can not find it anywhere else.

Answer 1

I recommend you have a look into Girault&Raviart's book on Finite element methods for Navier-Stokes equations. Chapter 1 provides a survey on basic concepts on Sobolev spaces including also the pretty involved definition of the trace spaces. The result you are out for is given in Thm. 1.5.

Personally I find the presentation very accessible.

Of course, as @Tomás says, you will find everything in the book by Lions&Magenes Non-Homogeneous Boundary Value Problems and Applications in a more general setting.

Answer 2

In Girault&Raviart's book, Thm 1.5 does not apply in this case. Indeed, the continuity of the trace operator from $W^{s,p}(\Omega)$ to $W^{s-\frac{1}{p},p}(\Omega)$ holds only if $s-\frac{1}{p}>0$ (this is implied by the assumptions).

Answer 3

@Michael: I think that this result is not true, even when the boundary is smooth. I've been looking for it (for a while now...) in the literature, but unfortunately I haven't been able to find anything..! It seems that the continuity of the trace operator $\gamma:H^s(\Omega)\rightarrow H^{s-\frac{1}{2}}(\partial\Omega)$ holds iff $s>1/2$ (for smooth enough domains). Even the limit case $H^{\frac{1}{2}}(\Omega)\rightarrow L^2(\partial\Omega)$ does not work (see Lions & Magenes, same reference as Tomás', theorem 9.5).

Actually, in order to have a trace exactly in $L^p(\partial\Omega)$, $p>1$, the function has to be in some Besov space, see Cornelia Schneider's article Trace operators in Besov and Triebel-Lizorkin spaces (Corollary 3.17).

@Tomás: In the reference you suggest (Lions & Magenes), I assume you are referring to Theorem 9.4. But, when considering the trace operator, i.e. $\mu=0$, the assumption $\mu<s-\frac{1}{2}$ implies that $s$ has to be strictly greater than $\frac{1}{2}$ for the result to apply.