Continuation of smooth functions on the bounded domain

Question

Given a bounded domain $\Omega\subset \mathbb{R}^n$ and a smooth function $f$ with bounded derivatives on $\Omega$, is it possible to extend $f$ to $\tilde{f} : \mathbb{R}^n \to \mathbb{R}$ such that it is smooth and compactly supported with given closed set $\Omega ' \supset \Omega$ ?

Answer 1

After the question has been edited, the answer is yes, if the boundary of $\Omega$ is sufficiently smooth and there is an open set $G$ with $\overline{\Omega}\subset G\subset\Omega´$. Note at first, that $f$ can be continuously extended to $\overline{\Omega}$ by a classical topological result for uniformly continuous functions. And as the same is true for all partial derivatives, this extension is infinitely continuously differentiable.

The smooth extension to all of $\mathbb{R}^n$ is basically Whitney's extension theorem, see e.g. this question. And multiplication with a suitable function gives the compact support.

Answer 2

No, consider $f: (0, 1) \to \mathbb{R}$, $f(x) = \frac{1}{x}$. You cannot extend $f$ over $0$. Generally, the problem of extending functions defined on open sets is hopeless.