3
$\begingroup$

Consider the following definition.

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 $0\leq|\alpha|\leq k$.

Here is my question:

If $u\in C^k(\overline{\Omega})$, can $\partial^\alpha u$ extend continuously to $\mathbb{R}^n$ for $0\leq|\alpha|\leq k$?


This question is motivated by the following ones:

$\endgroup$
4
  • $\begingroup$ I think the answer is Yes, the proof is similar to the proof in Evans PDE 2nd page.269 (Extension theorem step 2). $\endgroup$
    – yoyo
    Sep 21, 2016 at 14:04
  • $\begingroup$ Evans proved the case for $k=1$, but the method are the same in general. $\endgroup$
    – yoyo
    Sep 21, 2016 at 14:06
  • $\begingroup$ @yoyo: I don't have any assumptions on $\partial\Omega$ for this question. $\endgroup$
    – user9464
    Sep 21, 2016 at 22:56
  • $\begingroup$ Oh!~~ You are right, the question is more complicated. Thanks for reminding. $\endgroup$
    – yoyo
    Sep 22, 2016 at 11:05

1 Answer 1

0
$\begingroup$

This post has been put as a more complete one in MO:
Two different kinds of definitions of $C^k(\overline{\Omega})$ — extension and restriction.

Thanks to a comment there, the answer to the question is yes. But the question in this post does not match the one in the title. See more details in the accepted MO answer.

$\endgroup$

You must log in to answer this question.