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Given a non-negative function $\phi\colon\mathbb R^n\to\mathbb R$, which is in $C^k$, and define the set \begin{align} S:=\{x\in\mathbb R^k\colon \phi(x)>0\} \end{align} then can we say that $\partial S$ is of $C^k$, assuming that $S,S^c\ne\emptyset$?

The above has counterexamples, and I want to change the question as below.

The $\phi, S$ are as above and consider one connected component $\Omega$ of $S$, does $\Omega$ has $C^k$ boundary?

Yikun Qiao
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  • Not much, for instance, if $n=1$, $\partial S$ can be a Cantor set. Are you familiar with the implicit function theorem? – Moishe Kohan Feb 17 '18 at 18:14
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    Consider $\mathbb{R}^2 \to \mathbb{R}$, $(x, y) \mapsto x^2y^2$ (the squares only because you demanded that the function be non-negative). – hunter Feb 17 '18 at 18:15
  • @MoisheCohen I can not construct such an example, and I do not understand how to use the implicit function theorem here. – Yikun Qiao Feb 17 '18 at 18:18
  • What if I consider a connected component of $S$ and it seems like having $C^k$ boundary? – Yikun Qiao Feb 17 '18 at 18:20
  • The point of the example from hunter is that the derivative of the mapping does not have maximal rank (in several points). This is not about connected components. (That's also why Moishe Cohen asked about the implicit function theorem). – Thomas Feb 17 '18 at 18:33
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    The thing to know is that given any closed subset $E\subset R^n$ there exists a $C^\infty$ function $f: R^n\to [0,\infty)$ such that $E=f^{-1}(0)$. Restricting to a connected component will not help you. You pretty much have to assume that $0$ is a regular value of $f$, then the IFT will tell you the rest of the story. – Moishe Kohan Feb 17 '18 at 18:55

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Let me convert my comments into an answer. First of all, given any closed subset $E\subset R^n$ there exists a $C^\infty$ function $f: R^n\to [0,\infty)$ such that $E=f^{−1}(0)$. See for instance here. In particular, the boundary of your set $S$ can be as bad as you wish, for instance, it can be the Koch snowflake or a Cantor subset of $R^2$, or the Menger curve in $R^3$, etc. In particular, even if you $S$ is connected, its boundary need not even be a topological manifold.

The "correct" assumption to make is that $0$ is a regular value of $f$, i.e. for every $x\in f^{-1}(0)$, $\nabla f(x)\ne 0$. Then the Implicit Function Theorem will imply that $f^{-1}(0)$ is a smooth submanifold of dimension $n-1$. See for instance this wikipedia article. My favorite reference for this staff is Guillemin and Pollack "Differential Topology". You then can prove (quite easily) that $$ f^{-1}(0)= \partial S, $$ where $S=\{x: f(x)>0\}$.

Moishe Kohan
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