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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 $W^{1,2}(\Omega)$?

Ryan Vaughn
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  • 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 ${0}$. – Nate Eldredge Nov 15 '22 at 06:21
  • 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. – Ryan Vaughn Nov 15 '22 at 13:03
  • 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. – Nate Eldredge Nov 15 '22 at 15:10

1 Answers1

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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.

Jan Bohr
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  • Thank you! Do you know of a good reference for the fact that the trace map is continuous for codimension $k$ submanifold? – Ryan Vaughn Nov 14 '22 at 17:04
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    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. – Jan Bohr Nov 14 '22 at 23:29