# A question about partition of unity

If $Y\subset X$ are open sets in $\mathbb{R}^n$,then $Y$ is said to have a $C^1$ boundary in $X$ if for every boundary point $x_{0}\in X$ of $Y$ one can find a $C^1$ function $\rho$ in a neighborhood $X_{0}$ of $x_{0}$ such that $$\rho(x_0)=0,\quad d\rho(x_{0})\ne 0,\quad Y\cap X=\{x\in X_{0};\rho(x)<0\}$$ Then my question is how to prove that there exists a function$\rho\in C^{1}(X)$ such that $\rho=0$,$d\rho\ne 0$ on the boundary $\partial Y$ in $X$ and $Y\cap X=\{x\in X;\rho(x)<0\}$

It's very natural to think about using partition of unity to put the local result global,but I don't know how to make these properties still hold when choosing the partition of unity,and the boundary may be unbound and quite arbitrary.

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It depends on the set that you are considering. – Tomás Oct 10 '12 at 13:19

The fact that the boundary is unbounded and arbitrary is not really relevant. For the existence of the partition of unity $(\phi_k)$ you really only need paracompactness, and every open $X \subseteq \mathbb{R}^n$ satisfies this. If you have any point $x \in Y$, you get $\rho(x) = \sum_k \phi_k(x) \rho_k(x) <0$ since a convex combination of negative functions is negative. Similarly you get $\rho(x)>0$ for $x\notin Y$. For the condition on the gradient you have to use the chain rule: On $\partial Y$ (in $X$) you get $$\nabla \left(\sum_k \rho_k \phi_k\right) = \sum_k (\rho_k \nabla \phi_k + \phi_k \nabla \rho_k) = \sum_k \phi_k \nabla \rho_k.$$ This is (pointwise) a convex combination of outward pointing normal vectors, so you get an outward-pointing normal vector. In particular, you get something non-zero. The crucial point here is that for fixed $x\in \partial Y$ the direction of $\nabla \rho(x)$ is determined by the geometry of $\partial Y$.