Can a smooth map be extended to an open set?

Question

Let $X\subset\mathbb{R}^n$ and let $f:X\to\mathbb{R}^m$ be a smooth function i.e. for every $x\in X$, there exists an open set $U$ and a function $g:U\to\mathbb{R}^m$ which is smooth in the usual sense such that $g\vert_{U\cap X}=f\vert_{U\cap X}$. Does this imply that there exists an open set $U$ containing all of $X$ and a smooth function $g:U\to\mathbb{R}^m$ such that $g\vert_X=f$? For every example I could some up with, the answer was yes, but I do not see a clear way to prove this in general.

Answer 1

Yes, and the construction is by means of a partition of unity. That is, consider a cover of $X$ by open sets $U_\alpha$, and for each $\alpha$, find a function $g_\alpha:U_\alpha\to\mathbb R^m$ such that $g_\alpha|X = f|(U_\alpha\cap X)$. Then introduce a smooth partition of unity $\{\phi_\alpha\}$ subordinate to $\{U_\alpha\}$, and for $x\in \bigcup U_\alpha$, define $$ g(x) = \sum_{\alpha}\phi_\alpha(x)g_\alpha(x). $$ The function $g$ is well-defined and smooth, as part of the standard construction of a smooth partition of unity. (Crucially, the supports of the $\phi_\alpha$ are locally finite, meaning that every point has a neighborhood that intersects $\mathrm{supp}(\phi_\alpha)$ for only finitely many values of $\alpha$.) Moreover, for $x\in X$, $$ g(x) = (\sum_\alpha\phi_\alpha(x))f(x) = 1\cdot f(x) = f(x). $$ You can read more about the construction of a smooth partition of unity here. It is one of the standard and important constructions in analysis, with lots of uses.