# Sobolev density of smooth functions which are zero on a measure zero subset

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

• 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 $\{0\}$. Nov 15, 2022 at 6:21
• 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. Nov 15, 2022 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. Nov 15, 2022 at 15:10

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.
• Thank you! Do you know of a good reference for the fact that the trace map is continuous for codimension $k$ submanifold? Nov 14, 2022 at 17:04
• 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. Nov 14, 2022 at 23:29