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Let $\Omega$ be an open smooth domain in $\mathbb{R}^n$, then does there exist a smooth function $u$ such that $\{u<0\}=\Omega$ and $\{u=0\}=\partial \Omega$?

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The answer is yes. When considering $\Omega \subset \mathbb{R}^n$, I assume you mean that $\Omega$ is a smooth manifold with boundary of some dimension $k \in \{1, \ldots, n\}$. I will roughly sketch the proof as it's given in Lee's Intro to Smooth Manifolds. In his text, this proposition is given on pages 118-119 and is a byproduct of the bigger subject of defining manifolds/submanifolds using immersions, embeddings, and submersions.

Proposition: Every smooth manifold with boundary $\Omega$ admits a boundary defining function: a smooth function $f:\Omega \to [0,\infty)$ such that $f^{-1}(0) = \partial \Omega$ and $df_p \neq 0$ for all $p \in \partial \Omega$.

Proof Sketch: Let $\{(U_\alpha, \varphi_\alpha)\}$ be a collection of smooth charts for $\Omega$ such that $\bigcup_\alpha U_\alpha = \Omega$. For each $\alpha$ define $f_\alpha:U_\alpha \to \mathbb{R}$ such that $f_\alpha \equiv 1$ if $U_\alpha \cap \partial \Omega = \emptyset$, otherwise if $U_\alpha$ intersects the boundary, let $f_\alpha(x^1, \ldots, x^n) = x^n$ be the $n$th coordinate (in local coordinates under $\varphi_\alpha$, $x^n = 0$ for boundary points). We see then that the $f_\alpha$ are positive on $Int \Omega$ and zero on $\partial \Omega$. We let $\{\psi_\alpha\}$ be a partition of unity subordinate to the cover $\{U_\alpha\}$, and define the function $f:\Omega \to [0, \infty)$ by

$$ f(p) \;\; =\;\; \sum_\alpha \psi_\alpha(p) f_\alpha(p). $$

This essentially covers the main details of the proof, but I would highly recommend looking into this text for a deeper explanation and the greater context in which this result operates. I'm sure there are other places you could look for this too. The key words I'd recommend looking up, if any of them were confusing for you, would be boundary-defining function, partition of unity, manifold with boundary, coordinate chart.

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    $\begingroup$ Thank you very much! This answer is really instructive. $\endgroup$ – student Jan 14 '17 at 20:27

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