4
$\begingroup$

Let $(M^m,g)$ be a compact Riemannian manifold with smooth nonempty boundary, and $N^n\subseteq \Bbb R^d$ a boundaryless isometrically embedded Riemannian manifold. For $1\le p<\infty$ we define as usual $$W^{1,p}(M,N):=\{u\in W^{1,p}(M,\Bbb R^d):u(x)\in N\text{ a.e.}\}.$$ Using the Euclidean trace theorem, we have a trace map $$T_N:W^{1,p}(M,N)\to L^p(\partial M,\Bbb R^d)$$ just by restricting the domain of $T:W^{1,p}(M,\Bbb R^d)\to L^p(\partial M,\Bbb R^d)$. Is it true that the image of $T_N$ is contained in $L^p(\partial M,N)$?

If $N$ and $p$ are such that $C^\infty(\overline M,N)$ is not dense in $W^{1,p}(M,N)$, then I don't see how this can be proved. If we approximate $u\in W^{1,p}(M,N)$ by smooth functions $u_j\in C^\infty(\overline M,\Bbb R^d)$, then the continuity of the trace map implies $$Tu_j\to T_Nu $$ in $L^p(M,\Bbb R^d)$, so $Tu_j\to T_Nu$ a.e. in $\partial M$. But the images of the $Tu_j$'s are not necessarily in $N$, so this doesn't say anything about $T_Nu$. Now $u_j\to u$ a.e., but since $\partial M$ has measure zero we run into the precise issue that forces us to have a trace theorem in the first place!

In this article by Bethuel and Demengel, they say this is easy to see (top of the second page).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.