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.


Is it possible to continue it smoothly inside $\Omega'$? What are the constructive ways to do it (or may be just with finite smoothness)?


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!

  • 4
    $\begingroup$ Look in Stein, "Singular integrals and differentiability properties of functions" $\endgroup$
    – Deane Yang
    Nov 8, 2011 at 17:14
  • 3
    $\begingroup$ For finite order smoothness, the Babitch extension does the job. It is used to extend Sobolev functions $u\in H^s(\Omega)$ to $H^s({\mathbb R}^n)$. $\endgroup$ Nov 8, 2011 at 17:20
  • $\begingroup$ @Dean Yang Thanks, the parity of unity answers the question of existence. Although from computational point of view it doesn't look very helpful . $\endgroup$ Nov 8, 2011 at 17:52
  • $\begingroup$ If you know more about the domains (like explicit formulas or global co-ordinates), you don't need to use a partition of unity. You can probably do what Stein does locally but do it globally. $\endgroup$
    – Deane Yang
    Nov 8, 2011 at 18:10

2 Answers 2


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.

  • $\begingroup$ @Igor Not exactly: function is given to me as a result of numerical evaluation of Navier-Stokes. I don't know anything specific about it, so the continuation on a neighborhood is as hard as a total continuation. $\endgroup$ Nov 9, 2011 at 23:58
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    $\begingroup$ @Kirill, then yor hypotheses should say that $f$ is only smooth onthe interior of $\Omega\setminus\Omega'$. Also if you are not dealing with questions of principle, but only discretized approximations, you could just break up your domain info simplices and use splines. If you are then interested in the combinatorics of precisely how to do that, that's a different question. $\endgroup$ Nov 10, 2011 at 8:36
  • $\begingroup$ @Igor Ok, I mean let it has this smooth continuation. The question then is how to compute it inside (numerically)? As I understand splines give polynomial continuation, but may be there is still possibilities to do it either more elegant or computationally simple. $\endgroup$ Nov 10, 2011 at 17:53
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    $\begingroup$ @Kirill, you are now asking a very different question than originally posted. If you put any set of values on a finite grid, you can fit them with splines of sufficiently high polynomial order. That's what they are for: to construct finitely differentiable fits to arbitrary data. The construction of the splines will of course depend on the underlying grid. I realize that you probably won't find this answer satisfactory, but it's impossible to provide a better one without more constraints on the extension. Otherwise, as you can see, there are too many solutions to choose from. $\endgroup$ Nov 10, 2011 at 20:44
  • $\begingroup$ @Igor, I agree that the question in thew form I posted it useless and I need to change it. Anyway, thank you for comments and answering. $\endgroup$ Nov 13, 2011 at 20:54

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.


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