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I have just started learning about Sobolev spaces. So this might be trivial. I am working through the book "Partial Differential Equations" by Lawrence Evans, it came highly recommended.

Taking $\Omega \subset \mathbb{R}^{n}$ to be some open set. He defines the space $$W^{k,p}_{o}(\Omega) := \lbrace u \in W^{1,p}(\Omega): u|_{\partial \Omega} = 0 \rbrace$$ for $1 \leq p < \infty$.

He does not however define $W^{1,\infty}_{o}(\Omega)$. Does anyone know how this is generally defined? Is it simply $$W^{1,\infty}_{o}(\Omega) := \lbrace u \in W^{1,\infty}(\Omega): u|_{\partial \Omega} = 0 \rbrace$$ Why is it not dealt with in the same manner as for $1 \leq p < \infty$? I also checked Brezis book, he also does not deal with the case $p = \infty$.

Thanks.

1 Answers1

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The alternative definition is to define $W_0^{1,p}(\Omega)$ as the closure of $C_c^{\infty}(\Omega)$ in the sobolev norm. For $1\leq p < \infty$ these definitions are equivalent, but not for $ p = \infty $. He probably does not define it to avoid that ambiguity.

EDIT: Here is a more detailed answer. Let us define $W_0^{1,p}(\Omega) := \{f \in W^{1,p}(\Omega) \mid f|_{\partial \Omega} = 0\}$ and $W_c^{1,p}(\Omega) := \overline{C_c^{\infty}(\Omega)}$, where the closure is taken in the $W^{1,p}$ norm.

First note that Evans actually defines $W_0^{1,p}(\Omega)$ the way I have defined $W_c^{1,p}(\Omega)$ (see page 259 in my edition of the book).

Evans then shows (5.5 Theorem 2, page 273) that (in my notation) $W_0^{1,p} = W_c^{1,p}$ as long as $p< \infty$.

This is not the case for $p=\infty$, because $W_c^{1,\infty}(\Omega)$ is (by definition) the set of all limits $f = \lim_n f_n$, where $f_n \in C_c^{\infty}(\Omega)$ andthe limit is taken in $W^{1,\infty}$, which means $\Vert f - f_n \Vert_\infty \rightarrow 0$ and $\Vert \partial^{\alpha}f - \partial^{\alpha} f_n\Vert_\infty \rightarrow 0$. You can easily check that this implies $f \in C^1(\Omega)$ with $f|_{\partial \Omega} = 0$ and $\partial^{\alpha}f |_{\partial \Omega} = 0$.

But one can construct functions $f \in W_0^{1,\infty}(\Omega)$ (with my above definition), whose partial derivatives are discontinuous.

PhoemueX
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  • So are you saying that the definition of $W^{1,\infty}(\Omega)$ should be regarded as just the closure of $C^{\infty}_{c}(\Omega)$ in $W^{1,\infty}(\Omega)$? Also, do you know specifically why the two definitions are not equivalent for $p = \infty$, is it because of some property not shared by $L^{\infty}$ and $L^{p}$ spaces? –  May 24 '14 at 16:31
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    The difference is that limits of continuous functions w.r.t. the $L^\infty$ norm are again continuous, which is not true for the other $L^p$ norms. So the closure of $C_c^{\infty}$ for $p = \infty$ will be all $C^1$ functions which (together with the derivatives) vanish on the boundary. But there are $W^{1,\infty}_0$ functions (with the other definition) whose derivatives are not continuous. – PhoemueX May 24 '14 at 16:36
  • What limits of continuous functions are you referring to? –  May 24 '14 at 16:39
  • What definition of $W^{1,\infty}_{o}(\Omega)$ are you using? –  May 24 '14 at 16:42
  • Thanks. Just to confirm, are you saying that for $p = \infty$ we have $W^{1,\infty}{c}(\Omega) \subset W^{1,\infty}{o}(\Omega)$, where the inclusion is proper? If that is right then could I state that it is sufficient to take the definition as $W^{1,\infty}{o}(\Omega):= {f \in W^{1,\infty}(\Omega): f|{\partial \Omega} = 0 }$? Lastly would I be right in stating that if $v \in W^{1,\infty}(\Omega)$ and $u \in W^{1,\infty}{o}(\Omega)$ and if $v = u$ on $\partial O$ then $v \in W^{1,\infty}{o}(\Omega)$? –  May 29 '14 at 14:40
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    Yes, you can use that definition and your final statement is correct, BUT consider the following caveat: For $\partial \Omega$ sufficiently smooth, you (and Evans, ...) always write $u|_{\partial \Omega}$ for $u \in W^{1,p}$. It is important to remember here, that this notation can only be interpreted literally for $u \in W^{1,p}\cap C(\overline{\Omega})$ (cf. the definition of the trace operator). The reason for this is that $u \in W^{1,p}$ is only an equivalence class of a.e. identical functions. But note that for $p=\infty$ every(!) $u \in W^{1,p}$ has a continuous representative. – PhoemueX May 30 '14 at 10:06
  • Thanks again for all the assistance, very helpful. Firstly, can you give me a hint as to how to show that if $\Vert f -f_{n} \Vert_{\infty} \rightarrow 0$ and $\Vert \partial^{\alpha} f - \partial^{\alpha} f_{n} \Vert_{\infty} \rightarrow 0$ then it implies that $f \in C^{1}(\Omega)$? Secondly, are you saying that when considering the trace of $W^{1,\infty}(\Omega)$ we can always consider $u \in W^{1,\infty}(\Omega) \cap C(\bar{\Omega})$? Lastly, would I be right in stating that for $u \in W^{1,\infty}$ the continuous representative is not necessarily unique? –  Jun 03 '14 at 16:11
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    First, the continuous representative of $u \in W^{1,\infty}$ is unique. Generally, if $f,g$ are continuous, then ${x\mid f(x)\neq g(x)}$ is open, so is a null set iff it is empty. This is also the idea for your other question: if $\Vert f- f_n \Vert_\infty \rightarrow 0$ (as well as for derivatives) and $f_n$ is continuous, then $\Vert f_n - f_m \Vert_\sup = \Vert f_n - f_m \Vert_\infty \rightarrow 0$ (the first equality uses continuity). The same holds for derivatives. Then use a (higher dimensional analog of) http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability . – PhoemueX Jun 09 '14 at 09:09
  • So $W^{1,\infty}{c}(\Omega) \subset W^{1,\infty}{0}(\Omega)$.So it makes sense to choose $W^{1,\infty}{0}(\Omega)$ as the definition? Is that what you use? Then would I be right in stating that $f|{\Omega}$ is not defined in the same way was you define for $p < \infty$(which uses the trace operator $T:W^{1,p}(\Omega) \rightarrow L^{p}(\partial \Omega)$)? Since as you state it has a continuous representative. So we have a continuous representative say $f^{}$ on $\Omega$. Is $f|_{\partial \Omega}$ then the extension of $f^{}$ to $\overline \Omega$ restricted to $\partial \Omega$? –  Jun 23 '14 at 19:47