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This is cross-posted in MSE.

I have seen two different kinds of definitions of the notation $C^k(\overline{\Omega})$ — by "extension" of functions on $\Omega$ or by "restriction" of functions on $\mathbb{R}^n$. I'm not sure about how different these two kinds of definitions could be.

(I) 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 $0\leq|\alpha|\leq k$.

The above definition defines $C^k(\overline{\Omega})$ as a subset of $C^k(\Omega)$. See for instance Folland's Introduction to Partial Differential Equations.

On the other hand, one can also see in cited reference that $C^k(\overline{\Omega})$ is defined as restriction to $\overline{\Omega}$ of $C^k(\mathbb{R}^n)$ functions.

(II) For instance, in Hermann Sohr's The Navier-Stokes Equations — An Elementary Functional Analytic Approach (page.23), $C^k(\overline\Omega)$ means the space of all restrictions $u|_{\overline\Omega}$ to $\overline\Omega$ of functions $u\in C^k(\mathbb{R}^n)$ such that $$ \sup_{|\alpha|\leq k,x\in\mathbb{R}^n}|D^\alpha u(x)|<\infty. $$ Here $|\alpha|\leq k$ is replaced by $|\alpha|<\infty$ if $k=\infty$.

Here are my questions:

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

  • [Added thanks to Pietro Majer's comment below] If $u\in C^k(\overline{\Omega})$ as in (I), does there exist $w\in C^k(\mathbb{R}^n)$ such that $w|_\Omega=u$?

  • Are (I) and (II) essentially the same?
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    $\begingroup$ It should follow from Whitney's extension theorem. ams.org/journals/tran/1934-036-01/S0002-9947-1934-1501735-3/… $\endgroup$ – Willie Wong Sep 22 '16 at 2:29
  • $\begingroup$ @WillieWong: It seems that one needs no smooth assumption on $\partial\Omega$ according to the linked paper? It is quite misleading in the second part of the Wikipedia article on Whiney's extension theorem that one seems to need a smooth boundary of $\Omega$. $\endgroup$ – Jack Sep 22 '16 at 2:43
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    $\begingroup$ As you put the question, it's just a matter of continuous extension from closed sets. If $u\in C^k(\overline\Omega) $ as in (I) , by definition $\partial^\alpha u$ extends continuously to $\overline\Omega$, and by Tietze's theorem, yes, it has a continuous extension to $\mathbb{R}^n$. I suspect that was not what you meant... $\endgroup$ – Pietro Majer Sep 22 '16 at 7:04
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    $\begingroup$ @Jack: smoothness is enough to rule out the type of problems that user111 pointed out in the answer he gave below. But Whitney in his paper uses a somewhat weaker condition: his "Taylor-expansion" like condition is stronger than mere $C^k(\bar{\Omega})$ per definition (I) (the example user111 linked to fails the Taylor expansion condition at the origin) for irregular domains. $\endgroup$ – Willie Wong Sep 22 '16 at 13:09
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    $\begingroup$ For the purpose of PDE theory, however, the boundaries are usually assumed to be very nice, so for all intents and purposes of applications to PDEs the two definitions can be taken to be equal (provided you are aware of the situations that they are not.) $\endgroup$ – Willie Wong Sep 22 '16 at 13:11
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The standard reference for the extension problem is Whitney's extension theorem, which, has also many more modern presentations (a pretty good one is found in Stein's Singular Integrals and Differentiability Properties of Functions, Princeton University Press). It is worth paying attention, however, to the conditions that Whitney used in his extension theorem, which roughly states that

at every point in the domain (up to the closure) the function is compatible with having the 'correct order' Taylor expansion.

This condition is slightly stronger than mere $C^k(\bar{\Omega})$ in the sense of (I) of the question statement, as the example given in the link in this answer shows.

In the case where you have some regularity conditions on the boundary $\partial\Omega$, however, you can use the regularity to show that $C^k(\bar{\Omega})$ in the sense of (I) automatically implies the condition that Whitney requires. This is why in many sources you will find Whitney's extension theorem stated only for closed sets with smooth boundary.

In the context of PDE theory (such as Folland's book that you referred to for the definition (I)), domains usually have already nice boundaries (often you will need this for things besides just Whitney extension, such as boundary regularity estimates for elliptic equations), so in that setting the two definitions are essentially the same.

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The answer is no. See the example given at the end of this question:

Density of polynomials in $C^k(\overline\Omega)$

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