Continuation of a smooth function Setting
Suppose I have two bounded open domains $\Omega' \subset \Omega \subset \mathbb{R}^n$ (I'm particularly interested in case n = 2 or n = 3). We assume that all boundaries of domains are $C^\infty$-smooth and that inner domain is lying properly (with it closure) inside the outer: $\bar{\Omega'} \subset \Omega$.
Suppose I'm given a smooth function $f \in C^\infty\left(\overline{\Omega \setminus \Omega'}; \mathbb{R}\right)$. We could assume that $L^\infty$ norm of all derivatives of $f$ is bounded.
Question
Is it possible to continue it smoothly inside $\Omega'$? What are the constructive ways to do it (or may be just with finite smoothness)?   
Example
If we seek for just a continuous continuation of $f$, then we could put a "rubber film" over inner domain, or, formally, continue $f$ with the solution of the following Poisson equation:
$\Delta u = 0$ in $\Omega'$ with Dirichlet boundary conditions: $u_{\partial \Omega'} = f_{\partial \Omega'}$.
Update By constructive I mean "numerically friendly", i.e. easy to code.
References and comments are appretiated! 
 A: This has basically already been answered, but because all your boundaries are smooth (or, more importantly, $C^1$), and your domains bounded, then if you want an explicit smooth continuation, you could do the following: for a sufficiently small $\epsilon > 0$, you can construct a subset $V \subset \Omega'$ such that for any $x \in \partial V$, dist($x,\partial \Omega'$) = $\epsilon$. Then, "radially" along these lines of length $\epsilon$ connecting the boundary of $V$ to the boundary of $\Omega'$, you can have your continuation of $f$ smoothly vanish to 0 using a scaling of the form $e^{-1/x}$. Then set $f = 0$ on $V$.
Therefore this continuation of $f$ will "mostly" vanish in $\Omega'$. You're basically just constructing a smooth mollification of $f \chi_{\Omega \setminus \Omega'}$.
The problem with this construction is that, from an application point of view, it's not very helpful. By making $f$ vanish on the interior set, you've basically lost all the information that was encoded in $f$ in the outer set.
A: This answer may not be practically useful to you, but I think it's nice from a conceptual point of view. The extension could be done in two steps. I'm presuming that you are defining a smooth function on the closed set $K=\overline{\Omega\setminus\Omega'}$ as one that is smooth on a small neighborhood $K'$ (which I'll also take as closed) $K$.
First, you could extend $f$ continuously from the closed set $K'$ to all of $\Omega$. Second, you could smooth the extension, preserving it on the interior of $K$. The continuous extension is a direct application of the Tietze extension theorem, while the smoothing is an application of the Steenrod approximation theorem.
