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Given that $\Omega \subset \mathbb{R}^{n}$ is an open, convex, Lipschitz bounded set. Let $O \subset \Omega$ be open bounded set then consider. $$u_{m} \rightharpoonup^{*} u \text{ in } W^{1,\infty}_{o}(O)$$ Assume that we can find $v_{m}$ such that $$v_{m} \rightharpoonup^{*} u \text{ in } W^{1,\infty}(O)$$$$ v_{m} = u \text{ on } \partial O$$ Assume that we apply the linear continuous operator $$P: W^{1,\infty}(O) \rightarrow W^{1,\infty}(\Omega)$$ We then extend $u$ from $O$ to $\Omega$ in such a way that the extension $\bar{u} \in W^{1,\infty}(\Omega)$ and similarly since $v_{m} = u$ on $\partial O$ we define ${\bar{v}}_{m} = v_{m}$ in $O$ and ${\bar{v}}_{m} = \bar{u}$ in $\Omega - O$.

We then have $${\bar{v}}_{m} \rightharpoonup^{*} \bar{u} \text{ in } W^{1,\infty}_{o}(\Omega)$$

  1. In Lawrence's book "Partial Differential Equations" the trace operator isn't defined for $p = \infty$. If the trace operator isn't defined how would you describe how '$v_{m} = u$ on $\partial O$' is defined?

  2. Can we immediately state that $v_{m} \in W^{1,\infty}_{o}(O)$ since $v_{m} = u$ on $\partial O$ and $u \in W^{1,\infty}_{o}(O)$?

  3. Do we get ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$ in $W^{1,\infty}_{o}(\Omega)$ since we have that from the extension operator it follows that $P_{2}: W^{1,\infty}_{o}(O) \rightarrow W^{1,\infty}_{o}(\Omega)$ is also a continuous linear extension operator? If this is true, how would you go about showing this?

Thanks for any assistance

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  • $\begingroup$ Do you have any regularity assumptions on the boundaries of $O$ and $\Omega$? The consideration of traces is rarely fruitful without them. If the domains are reasonable (i.e., quasiconvex), then $W^{1,\infty}$ is just the class of Lipschitz functions, for which the trace is naturally defined by pointwise limit. $\endgroup$ – user127096 Apr 21 '14 at 22:17
  • $\begingroup$ @user127096 Yes I omitted that by mistake. Let's take the domain to be Lipschitz and convex. What would it be a pointwise limit of? In Evans book he defines the trace for $1 \leq p < \infty$, where states that if $u \in C^{1}(\bar{U}) \cup W^{1,p}(O)$ then we have functions $u_{m} \in C^{\infty}(\bar{U})$ which converge uniformly to $u$ on $\bar{U}$, not sure if that can be applied to $p= \infty$. $\endgroup$ – Alex Apr 22 '14 at 8:13
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Expanding the comment by user127096.

The space $W^{1,\infty}({\cal{O}})$ is equal to the space of bounded locally Lipschitz functions. This means for every element $u\in W^{1,\infty}({\cal{O}})$ it is possible to find a function $v$ that is Lipschitz continuous by just rearranging a set of measure zero.

The trace then can be taken as you take the trace of continuous functions by pointwise convergence.

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  • $\begingroup$ Thanks for your response. There are a few things I don't get. Are you sure $W^{1,\infty}(O)$ is equivalent to $C^{0,1}$. Does this equivalence not only apply locally(for compact sets)? Secondly, what reference are you using to characterize $W^{1,\infty}_{0}(O)$? I don't quite get what you are saying there. $\endgroup$ – Alex May 9 '14 at 14:26
  • $\begingroup$ Hi,the theorem I am referring to is theorem 4. 1 of this lecture notes www.math.jyu.fi/research/reports/rep100.pdf Regarding the definition I am using Renardy & Rogers An Introduction to Partial Differential Equations theorem 6.110 $\endgroup$ – yess May 9 '14 at 16:26
  • $\begingroup$ What edition are you using for Renardy & Rogers? I can't find a theorem 6.110 in the second edition. As for the other notes, it says that $W^{1,\infty}(\Omega)$ is equivalent to bounded locally Lipschitz functions. So I'm guessing that the result that I was referring to does not require the functions to be bounded but instead states that $W^{1,\infty}_{loc}(\Omega)$ is equivalent to locally-lipschitz functions. As for the other question again, would I be right then in saying that $W^{1,\infty}(\Omega) := \{ v \in W^{1,p}(\Omega): v = 0 \text{ on } \partial \Omega \}$? $\endgroup$ – Alex May 12 '14 at 19:51
  • $\begingroup$ I made a mistake, the theorem I was using as stated in the book only applied to $W^{1,2}$. I will modified my answer. The characterization you are suggesting doesn't seem right. For example, the functions in $W^{1,\infty}$ must be Lipschitz continuous and this is not the case of a general element of a sobolev space with trace zero. Sorry, if I was a source of confusion. $\endgroup$ – yess May 13 '14 at 12:35
  • $\begingroup$ What characterization are you referring to? I made a mistake in my comment as well, I meant to write $W^{1,\infty}_{o}(\Omega) = \{ v \in W^{1,p}(\Omega): v = 0 \text{ on } \partial \Omega \}$, do you think that is fine? The result $W^{1,\infty}_{loc}(\Omega)$ equivalent to locally lipschitz functions on $\Omega$ is a result in Evans book, the proof is almost identical to your proof from the notes. $\endgroup$ – Alex May 13 '14 at 15:39
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I'm not sure about question 1. Interesting question though, since I note also that the trace is not defined for $p = \infty$ in two of Evans books (Partial Differential Equations and Measure Theory and Fine Properties of Functions) as well as Brezis book (Function Analysis, Sobolev Spaces and Partial Differential Equations).

For question 2: I would say yes. Since if the trace is defined for $p = \infty$, then by definition it follows that $u \in W^{1,p}_{o}(\Omega)$ iff $Tu = o$ on $\partial \Omega$. Where $T$ is the bounded linear trace operator $T: W^{1,p}(\Omega) \rightarrow L^{p}(\partial \Omega)$.

For question 3: I don't think you have to apply the extension theorem. You could just extend ${\bar{v}}_{m}$ and $\bar{u}$ by zero on $\Omega\setminus \bar{O}$. If we assume that $O$ is also lipschitz then we note that the $n$-dimensional measure of the boundary of $O$ is zero. So it does not affect our Lebesgue integrals. We can show ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}\text{ } \text{ in } W^{1,\infty}_{o}(\Omega)$ as follows:

Assuming we have the following extension:

$$\begin{align} \bar{u} = \begin{cases} u~~~ \text{ in } O \\ 0~~~~ \text{ in } \Omega\setminus O \end{cases} \end{align} $$ and

$${\bar{v}}_{m} = \begin{cases} v_{m}~~ \text{ in } O \\ \bar{u}~~~~~ \text{ in } \Omega \setminus O. \end{cases} $$

Take any $\phi \in L^{1}(\Omega)$, this yields $$\lim\limits_{m \rightarrow \infty} \int_{\Omega}({\bar{v}}_{m} - \bar{u})\phi dx = \int_{O}(v_{m}-u)\phi dx + \int_{\Omega \setminus O}(u-0)\phi dx.$$

The first integral converges to $0$ since you assumed that $v_{m} \rightharpoonup^{*} u$ in $W^{1,\infty}(O)$. The second integral is $\int_{\Omega \setminus O} 0 dx = 0$. This gives ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$ in $L^{\infty}(\Omega)$. Similary, you can show that $\nabla {\bar{v}}_{m} \rightharpoonup^{*} \nabla \bar{u}$ in $L^{\infty}(\Omega; \mathbb{R}^{n})$. Conclusion ${\bar{v}}_{m} \rightharpoonup^{*} \bar{u}$. $\square$

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