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We say that $\Sigma \subset \mathbb{R}^n$ is a $C^\infty$ hypersurface if and only if, for all $p \in \Sigma$, there exists an open neighborhood $U_p \subset \mathbb{R}^n$ containing $p$ and a smooth function $f_p : U\to \mathbb{R}$ with $\nabla f_p \neq 0$ on $U_p$ such that

$$ U_p \cap \Sigma = \{x \in U_p | f_p(x) = 0\}.$$

I would like to know if the following proposition is true:

If $h$ is a smooth function such that $h(p) = 0$ for all $p \in \Sigma$, then $h = g f_p$ locally near each $p$, where $g$ is another smooth function defined near $p$ (and the $f_p$ used here is the same function described in the previous definition).

My background in differential geometry is not very strong. I would greatly appreciate any hints or nudges in the right direction and then hopefully I can post the correct argument (or counterexample) myself. Will Taylor's formula come into the picture here? Or perhaps the inverse/implicit function theorems?

JZS
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    This is true if $\Sigma$ is closed, because then there exists a global function $f\in C^\infty (\mathbb R^n )$ whose zero locus is $\Sigma$ and whose differential is non-zero at every $p$ : see here. Then any function $h$ vanishing on $\Sigma$ is a multiple $h=gf$ of $f$, with $g\in C^\infty (\mathbb R^n)$ . – Georges Elencwajg Aug 26 '16 at 19:04

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