Suppose that $\Omega$ is a bounded domain and Let $A\subseteq \Omega$ is a subset of measure zero. Is the set of smooth functions which are zero on $A$ dense in Sobolev space? For instance $W^{1,2}(\Omega)$?
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$\begingroup$ Did you want $A$ to be closed, or something like that? If $A$ is a dense set of measure zero then the set of smooth functions zero on $A$ is just $\{0\}$. $\endgroup$– Nate EldredgeNov 15, 2022 at 6:21
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$\begingroup$ I was wondering if the subspace of smooth functions which are zero on A is dense as a subspace of Sobolev space. It seems like that need not be the case, due to the answer below. $\endgroup$– Ryan VaughnNov 15, 2022 at 13:03
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$\begingroup$ Right, but the set consisting of only the zero function is obviously not dense in Sobolev space. So what I mean is that unless you have more conditions on the set $A$, there is a trivial counterexample. Jan Bohr shows that there are also nontrivial counterexamples. $\endgroup$– Nate EldredgeNov 15, 2022 at 15:10
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It is not always clear, what it means for a Sobolev function to vanish on a non-open subset $A\subset \Omega$. Suppose that $f\in H^s(\Omega)$, the $L^2$-bases Sobolev space of order $s\in \mathbb R$, and $A\subset \Omega$ is a nonempty submanifold of codimension $k$. Then $f\vert_A$ makes sense as Sobolev function only if $s>k/2$. However, the trace map $f\mapsto f\vert_A$ is then continuous from $H^s(\Omega)$ to $H^{s-k/2}(A)$ and its zero locus (that is, the set of $f$ with $f\vert_A=0$) must be closed and fails to be dense.
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$\begingroup$ Thank you! Do you know of a good reference for the fact that the trace map is continuous for codimension $k$ submanifold? $\endgroup$ Nov 14, 2022 at 17:04
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1$\begingroup$ Taylor's PDE book (vol 1) is a good reference, see e.g. Proposition 1.6 in Chapter 4. That's the local result for $k=1$. You can apply it several times to get higher $k$ and use the standard chart/cutoff procedure to get the result for, say, compact submanifolds. $\endgroup$– Jan BohrNov 14, 2022 at 23:29