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Let $\Omega$ be a open, bounded set in $\mathbb{R}^n$.

Suppose $g$ is a continuous function defined on the boundary $\partial \Omega$.

Then, is it possible to show that there exists a function $f$ defined (and continuous) on $\Omega \cup \partial \Omega$ such that $f$ is smooth in $\Omega$ and $f$ agrees on $g$ on $\partial \Omega$?

Also, how much smoothness (e.g. $C^k$) can we get?

user74261
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    Intuitively, a solution of the Laplace equation with g as the boundary condition should be smooth (perhaps even if g has discontinuities). – Keith McClary Sep 28 '15 at 02:24
  • Somewhat repeating Keith: Check out Kakutani's formula and surrounding results. – Titus Sep 28 '15 at 04:19
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    You surely want $f$ to be continuous on $\overline \Omega$ as well. – zhw. Sep 28 '15 at 06:13
  • @zhw. $g$ is assumed to be continuous, so wouldn't it force $f$ to be continuous on the closure of $\Omega$? – user74261 Sep 28 '15 at 13:53
  • No, without this added requirement, you could just define $f=0$ on $\Omega,$ and $f=g$ on $\partial \Omega.$ – zhw. Sep 28 '15 at 16:06
  • @KeithMcClary It's not easy to show that the solution actually has the boundary value we need, and in general it doesn't: http://math.stackexchange.com/q/1016424 –  Sep 30 '15 at 03:14

1 Answers1

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Yes, there is a $C^\infty$ extension. The construction is due to Whitney, published in 1934. You can find it at the beginning of Singular integrals and differentiability properties of functions by Stein. It goes like this:

  1. Write $\Omega$ as the union of Whitney cubes $Q_j$.
  2. Let $\varphi_j$ be a $C^\infty$ partition of unity subordinate to the cover $\frac32 Q_j$ (dilated the cubes to create an open cover but still keep them away from $\partial\Omega$)
  3. Pick a point $x_j\in\partial\Omega$ that realizes the distance $\operatorname{dist}(Q_j,\partial\Omega)$.
  4. The function $f(x) = \sum_j g(x_j) \varphi_j(x)$ has the desired properties.

Indeed, $f$ is $C^\infty$ smooth, being a locally finite sum of $C^\infty $ functions. As $x\to \zeta\in\partial\Omega$, the values of $g$ used in the construction of $f(x)$ are taken from progressively smaller neighborhoods of $\zeta$; hence they are close to $g(\zeta)$. This and the fact that $\{\varphi_j\}$ is a partition of unity imply $f(x)\to g(\zeta)$.