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Let $d\in\mathbb N$ and $\Lambda\subseteq\mathbb R^d$ be bounded and open. What's the definition of $C^k(\overline\Lambda)$ for some $k\in\mathbb N_0$? I've encountered this notation in a book that I'm reading. It's clear to me that $C^k(\Lambda)$ is the space of $k$-times continuously differentiable functions $\Lambda\to\mathbb R$. But since derivatives are usually defined on open sets only, I don't understand what $C^k(\overline\Lambda)$ could be. Maybe the space of those $f\in C^k(\Lambda)$ whose partial derivatives admit a continuous extension to $\overline\Lambda$?

0xbadf00d
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  • equivalently, in $\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ you require that $x \in \Lambda$ – reuns Sep 03 '16 at 11:03
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    @user1952009 "Equivalently" to what? – 0xbadf00d Sep 03 '16 at 11:31
  • try with $\Lambda = (a,b)$. start with $f^{(k)}(x)$ continuous on $[a,b]$, and integrate. – reuns Sep 03 '16 at 11:33
  • @user1952009 Maybe you could give me a clue what you want me to discover. One of the reasons why we usually want the set $M⊆ℝ^d$ in $C^k(M)$ to be open is that this is an easy way to ensure uniqueness of the derivative. But the case $d=1$ is different. $f(x):=x^2$ for $x∈[0,1]$, for example, is uniquely (onesidedly) differentiable even at the boundary points. However, I don't want to "try something" and then come up with my own definition of $C^k(\overline Λ)$. I thought there would be a well-known definition, cause the author of the referenced book use this notation without introducing it. – 0xbadf00d Sep 03 '16 at 12:18
  • I told you what is in my opinion the well-known definition : use $\lim_{x \to a, x \in \Lambda}$ in the definitions of the (partial) derivatives. Equivalently, it means $f$ has a $C^k$ Taylor expansion everywhere in $\overline{\Lambda}$, and it is also equivalent to what you wrote. – reuns Sep 03 '16 at 12:22
  • @user1952009 I found a definition for $C^k(\overline\Lambda)$ (Google Books link) in a book. They say that it's the space of "functions" $u$ for which ${\rm D}^\alpha u$ is bounded and uniformly continuous in $\Lambda$, for all $0\le\alpha\le k$. However, are these "functions" supposed to be defined on $\Lambda$, $\overline\Lambda$, $\mathbb R^d$ or something else? – 0xbadf00d Sep 03 '16 at 19:08

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There are a couple different ways one can try to handle this. Let $\Omega$ be an open bounded set.

The first is to say that $f \in C^k(\bar{\Omega})$ if there exists an open set $W$ with $\bar{\Omega} \subset W$ and a function $F \in C^k(W)$ such that $F = f$ on $\Omega$. This method basically says that there's a bigger open set on which there is a $C^k$ extension of the function, where the openness of $W$ means we use the usual understanding of differentiation in open sets.

However, there is another option. One can define $C^k(\bar{\Omega})$ to be the collection of functions in $C^k(\Omega)$ such that $\partial^\alpha u$ extends to a continuous function on all of $\bar{\Omega}$ for $|\alpha | \le k$.

The equivalence of these two is not so trivial. That the first implies the second is easy. For the converse, one must use the Whitney extension theorem, I believe, and it's possible that some regularity of the boundary is needed in order for Whitney to be applicable. If I recall correctly, there is a discussion of this in the Appendix of Leoni's A First Course in Sobolev Spaces.

The problem with just computing derivatives directly, as suggested in the comments, is that the boundary can be quite wild in general, and so it might not even make sense to form certain difference quotients. For instance, consider a domain with a cusp on the boundary. At the tip of the cusp there's really only one direction in which you can "enter" the domain, so how do you form the difference quotients to compute the partial derivatives in the other directions? A possible work-around is to drop the partials and work directly with the total derivative tensors, but the boundary regularity causes problems again, as one needs something like a tangent space in order to prove uniqueness for the derivatives.

Glitch
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