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Basically, I have the same question as in Extension of $W^{1,\infty}(\Omega)$: Given a bounded, open set $\Omega\subset\mathbb{R}^n$ with $\mathcal{C}^1$-boundary and another open bounded set $V\supset\supset\Omega$. I want to find an extension operator $E:W^{1,\infty}(\Omega)\rightarrow W^{1,\infty}(\mathbb{R}^n)$ which is linear, bounded and satisfies: $Eu=u$ $dx$-a.e on $\Omega$ as well as $\mathrm{supp}(Eu)\subset V$ for each $u\in W^{1,\infty}(\Omega)$. I do not know the result about the coincidence of $W^{1,\infty}(\Omega)$ with Lipschitz continuous functions as stated in the link above. Moreover, all approximation results I know for Sobolev functions are for $p<\infty$. I was wondering whether it is possible to construct a Cauchy sequence in $W^{1,\infty}(\mathbb{R}^n)$ by mollifying the zero-extension of $u\in W^{1,\infty}(\Omega)$. Maybe the limit has the desired properties. Unfortunately, I was not able to show the Cauchy-property. Does someone know an elementary proof of the desired result?

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  • $\begingroup$ This proposition is proved for $1\leq p\leq \infty$ in section 5.4 of L. Evans, *Partial Differential Equations.*Rather than mollifying, the idea in this reference is to use the local flatness of the boundary to reflect the function $u$ across it, and define the extension via this reflection. He also remarks that the proof gives an extension for $W^{2,p}$ as well, provided the boundary is $C^2$. $\endgroup$ – Gyu Eun Lee Oct 11 '15 at 12:46
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    $\begingroup$ @GyuEunLee I think, Evan's proof does not hold for $p=\infty$. His convergence results only hold for finite $p$ (approximations by $\mathcal{C}^\infty$-functions). In the proof (step 7) he makes use of these results. $\endgroup$ – JohnSmith Oct 11 '15 at 12:58
  • $\begingroup$ @JohnSmith Ah, yes, you're absolutely right. $\endgroup$ – Gyu Eun Lee Oct 11 '15 at 23:34
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    $\begingroup$ Mollification is hopeless; it has no reason to converge in the $W^{1,\infty}$ norm. $\endgroup$ – user147263 Oct 13 '15 at 6:33
  • $\begingroup$ Mollifying the zero-extension, you will get an uncontrollable jump at the boundary. $\endgroup$ – user90189 Oct 17 '15 at 5:09
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I will sketch two possible ways; tell me if you need more details at some point.

Choose a finite cover of $\partial\Omega$ made of open sets $U_i\Subset V$ for which a $C^1$-diffeomorphism $\psi_i:U_i\to B^{n-1}\times (-1,1)$ exists. Here we are thinking a point $x\in\mathbb{R}^n$ as a couple $(x',t)\in\mathbb{R}^{n-1}\times\mathbb{R}$, so the last set is $$B^{n-1}\times (-1,1)=\{x=(x',t):|x'|<1,\ |t|<1\}.$$ The diffeomorphism $\psi_i$ is required to "straighten the boundary", i.e. $\psi_i(U_i\cap\partial\Omega)=B^{n-1}\times\{0\}$; we also require that $\psi_i(U_i\cap\Omega)=B^{n-1}\times(0,1)$. In words, we are mapping $U_i\cap\Omega$ to the upper half of the cylinder $B^{n-1}\times(-1,1)$. (Why do such $U_i$'s and $\psi_i$'s exist?)

Using partitions of unity $\phi_i$ (subordinate to the cover $\{U_0\}\cup\{U_i\}$ of $\overline{\Omega}$, where $U_0$ covers the remaining part of the interior, i.e. $\Omega\setminus\cup U_i\subset U_0\Subset\Omega$) and looking at the functions $v_i:=(\phi_i u)\circ\psi_i^{-1}$, we reduce to the following problem: given $v\in W^{1,\infty}(B^{n-1}\times(0,1))$, we have to build an extension $Ev\in W^{1,\infty}(B^{n-1}\times(-1,1))$ in such a way that $E$ is a linear bounded operator and maps functions such as our $v_i$'s to some function lying also in $W^{1,\infty}(\mathbb{R}^n)$. Then it will suffice to sum all the functions $Ev_i\circ\psi_i$, together with $\phi_0u$, to obtain the desired extension.

This is simply done by reflection: put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=v(x',-t)$ if $t<0$. It is easy to see that $\frac{\partial (Ev)}{\partial x_i}$ exists and is given by the same formula for $i<n$: to check the definition of weak derivative against a $\phi\in C^1_c(B^{n-1}\times(-1,1))$ one can first consider $\phi\cdot\eta_k$, where $\eta\in C^\infty(\mathbb{R})$ satisfies $\eta(t)=0$ for $t<\frac{1}{2}$, $\eta(t)=1$ for $t>1$, and $\eta_k(t):=\eta(k t)$; then one lets $k\to\infty$.

We claim that $\frac{\partial (Ev)}{\partial x_n}$ also exists and equals $g$, where $g(x',t)=\frac{\partial v}{\partial x_n}(x',t)$ for $t>0$, while $g(x',t)=-\frac{\partial v}{\partial x_n}(x',t)$ for $t<0$. To see this, pick any test function $\phi\in C^1_c(B^{n-1}\times(-1,1))$ and observe that $$\int_{B^{n-1}\times (-1,1)}Ev\frac{\partial \phi}{\partial x_n}=\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$ where $\chi(x',t):=\phi(x',t)-\phi(x',-t)$. Since $\chi(x',0)=0$, we have the estimate $|\chi(x',t)|\le Mt$ for $t>0$, where $M:=\|\frac{\partial\chi}{\partial x_n}\|_\infty$. Now cutoff $\chi$ using $\eta_k$ as before, and apply the definition of weak derivative: $$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}(\eta_k\chi)=-\int_{B^{n-1}\times (0,1)}v\frac{\partial(\eta_k\chi)}{\partial x_n}$$ $$=-\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)-\int_{B^{n-1}\times (0,1)}\eta_k v\frac{\partial\chi}{\partial x_n}$$ (here we used sometimes $x_n$ and sometimes $t$ to denote the last component of $x$).

To conclude, it suffices to prove that $\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\to 0$ (as $k\to\infty)$: then in the limit we deduce $$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}\chi=-\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$ and the left hand side equals $\int_{B^{n-1}\times(-1,1)}g\phi$, while the right hand side is $-\int_{B^{n-1}\times(-1,1)}Ev\frac{\partial \phi}{\partial x_n}$, so we are done. But $$\left|\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\right|\le kCM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|t \le CM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|\to 0,$$ where we put $C:=\|\eta'\|_\infty$.

This extension operator has the defect of mapping $C^1(\overline{\Omega})$ outside $C^1(V)$. To circumvent this, another more clever construction is as follows: let's take for granted the extension theorem for $W^{1,1}$ (which comes from what we have done up to now). Put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=-3v(x',-t)+4v(x',-\frac{t}{2})$ if $t<0$. The key point now is that it suffices to show that $Ev\in W^{1,1}$ because then (restricting to the two halves) the weak gradient of $Ev$ is forced to be in $L^\infty$ (rather than only in $L^1$) as we have an explicit formula for it (write it down!). For any $W=B^{n-1}_{1-\epsilon}\times(-1+\epsilon,1-\epsilon)\Subset B^{n-1}\times (-1,1)$ we can find a sequence $(v_n)\subset W^{1,1}(W)\cap C^1$ converging to $v$ in $B^{n-1}_{1-\epsilon}\times(0,1-\epsilon)$ (use the extension for $W^{1,1}$ and mollify). Then $Ev_n\in C^1$ (check it), where of course we are now restricting to $W$, and $$\|Ev_n-Ev_m\|_{W^{1,1}}=\|E(v_n-v_m)\|_{W^{1,1}}\le C\|v_n-v_m\|_{W^{1,1}}.$$ Thus $(Ev_n)\subset W^{1,1}(W)$ is a Cauchy sequence. But it has to converge to $v$, so (at least locally) $v\in W^{1,1}$.

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  • $\begingroup$ Thank you very much for your detailed answer. I have one question to the end of the third paragraph: If you have constructed the "local" extension $Ev\in W^{1,\infty}(B^{n-1}\times (-1,1))$ and the $\mathcal{C}^1$-diffeomorphisms $\psi_i$, how do you ensure that $Ev_i\circ\psi_i$ is weakly differentiable? If I integrate this concatenation against a testfunction $\phi$ and try to get $\psi_i$ to the test function (using the transformation rule for integrals) I end up with a function like $\phi\circ\psi_i^{-1}$ which $\endgroup$ – McTuck Oct 19 '15 at 20:38
  • $\begingroup$ needs not to be a test function anymore. How to apply the weak differentiability of $Ev_i$? $\endgroup$ – McTuck Oct 19 '15 at 20:38
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    $\begingroup$ The function $w_i:=Ev_i\circ\psi_i$ belongs to $W^{1,\infty}(U_i)$, do you agree? Moreover, it vanishes outside some compact subset of $U_i$ (since the same happens for $Ev_i$), so $w_i=\rho w_i$ for some cutoff function $\rho\in C^\infty_c(U_i)$ (chosen s.t. $\rho\equiv 1$ on the compact set $\overline{\{w_i\neq 0\}}$). $\endgroup$ – Mizar Oct 19 '15 at 20:48
  • $\begingroup$ Unfortunately, I was too ambitious. Why is $w_i\in W^{1,\infty}(U_i)$? How do you prove weak differentiability? The $\psi_i$ is only $\mathcal{C}^1$, right? I've tried to reduce the issue to the weak differentiabilty of $Ev_i$. $\endgroup$ – McTuck Oct 19 '15 at 21:31
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    $\begingroup$ I see your problem (by the way, notice that you don't know the first equality yet since we are dealing with weak derivatives). The standard way to prove the chain rule, i.e. that $dv=du\cdot J\psi$, is to proceed by density by mollifying $u$ and passing to the limit. For the details you can see Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Prop. 9.6 $\endgroup$ – Mizar Oct 19 '15 at 23:30

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