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Let the open set $\Omega\subset{\mathbb R}^n$, and $k$ be a positive integer. $C^k(\Omega)$ will denote the space of functions possessing continuous derivatives up to order $k$ on $\Omega$, and $C^k(\overline{\Omega})$ will denote the space of all $u\in C^k(\Omega)$ such that $\partial^{\alpha}u$ extends continuously to the closure $\overline{\Omega}$ for $0\leq|\alpha|\leq k$.

As I understand, an extension means that there exists $\widetilde{\partial^{\alpha}u}$ which is defined on $\overline{\Omega}$ and $\widetilde{\partial^{\alpha}u}|_{\Omega}=\partial^{\alpha}u$. And "extends continuously" means $\widetilde{\partial^{\alpha}u}$ is continuous with respect to the relative topology on $\overline{\Omega}$.

Here are my questions:

  • How do people usually do such extension?

  • When does such extension exist and when does not?

  • Let $\Omega$ be an open subset of ${\mathbb R}$. Is there any non-trivial counterexample such that this kind of extension does not exist?

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1 Answer 1

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For simplicity I will assume that $\Omega$ is bounded. Consider first the case $k=0$. Then you are asking when a continuous function on $\Omega$ can be extended continuously to $\bar\Omega$. The answer (may be of not much practical use) is that this is possible if and only if $f$ is uniformly continuous. To do the extension, for any $x\in\partial\Omega$ choose a sequence $\{x_n\}\subset\Omega$ such that $\lim_{n\to\infty}x_n=x$ and define $f(x)=\lim_{n\to\infty}f(x_n)$. The uniform continuity of $f$ allows one to prove:

  1. $\{f(x_n)\}$ is a Cauchy sequence, and hence convergent;
  2. The value of $f(x)$ is independent of the sequence chosen to define it.

Examples of continuous functions on an open set that cannot be extended continuously to the boundary are $1/x$ and $\sin(1/x)$ on $(0,1)$. On $\mathbb{R}^n$, $n>1$, you can take as $f$ the solution of Dirichlet's problem on $\Omega$ with boundary value a discontinuous function on $\partial\Omega$.

For $k>0$, $f\in C^k(\Omega)$ can be extended to a $C^k(\bar\Omega)$ function if and only if its partial derivatives of order $k$ are uniformly continuous.

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