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Suppose $u\in W^{1,p}_0(\Omega)$, where $\Omega\subset\mathbb{R}^n$ is a bounded domain (open and connected) and $p\geq 1$. Let $a\in\mathbb{R}$ and suppose the set $\Omega_a=\{x\in\Omega:\ u(x)=a\}$ has positive measure. Is true that $\Omega_a$ must contains a open set?

Im stuck with this. I couldnt find a counter example. Any help is appreciated.


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What does "$u(x)=a$" even mean? You can't simply evaluate $L^p$-functions at points... (at least not without saying what that should mean) – Sh4pe Oct 25 '12 at 14:27
up vote 3 down vote accepted

No, the null set of such a function need not have non-empty interior. In fact, for any closed subset $C\subset \mathbb R$, there exists $u\in C^\infty(\mathbb R)$ such that $u^{-1}(0) = C$:

Proof: We can write the complement of $C$ as a countable, disjoint union of open intervals $\mathbb R \setminus C = \bigcup_n I_n$. Now choose cut-off functions $\psi_n\ge 0$ such that $\psi_n(x) > 0$ iff $x\in I_n$.

Next, choose coefficients $c_n>0$ such that $c_n |\psi_n |_{C^n(\mathbb R)}\le 2^{-n}$ for $n=1, 2, \dots$ Then the function $u(x) = \sum_{n=1}^\infty c_n\psi_n(x)$ satisfies $u(x)>0$ iff $x\in \bigcup_n I_n$ and $u(x) = 0$ otherwise. That is to say $u^{-1}(0) = C$.

Furthermore for any $k\in \mathbb N$, we have $$|u|_{C^k(\mathbb R)} \le \sum_{n\ge1} c_n|\psi_n|_{C^k(\mathbb R)} \le \sum_{n< k} c_n|\psi_n|_{C^k(\mathbb R)} + \sum_{n\ge k} 2^{-n} < \infty$$ i.e. $u\in C^\infty(\mathbb R^n)$. $\square$

Now you can take $C$ a fat Cantor set in $[0,1]$ and find a $C^\infty$ function $u$ satisfying $u^{-1}(0) = C$. This $u$ is then in $W^{1,p}_0((0,1))$, but $C$ is nowhere dense.

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Thank you @SamL. – Tomás Oct 25 '12 at 14:59

The points in the Sobolev space are not functions, they are equivalence classes of functions. In particular, you are allowed to alter your functions by sets of measure zero. So when you write $u \in W_0^{1,p}(\Omega)$, what you're actually doing is saying that $u$ is a representative of a class of functions in the Sobolev space. But I'm allowed to change $u$ on any set of measure 0 and not change the equivalence class. In particular, I can change the value of $u$ on any point of the form $(p,q)$ for rational $p,q$ and stay inside the equivalence class.

What does this mean for the problem? It means that even if the equivalence class of $u$ did happen to contain a function $v$ so that $\Omega_a(v)$ contains an open set and has positive measure, I could simply redefine $v(p,q) = a+1$ for every rational point $(p,q) \in \Omega_a(v)$ and this would be in the equivalence class of $u$. If $v'$ is this new, redefined function, then certainly $\Omega_a(v')$ has no open subsets... I've punched out all the rational points! But it also has positive measure, since I changed $v$ on a set of measure zero. So your statement does not hold for this function $v'$.

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Thank you @ZachL. – Tomás Oct 25 '12 at 14:58

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