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?
Functional Analysis – Extension Theorem for Sobolev Spaces W^{1,?}(?)
functional-analysissobolev-spaces
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I am not familiar with the Whitney extension theorem, etc. but maybe the following proof will do?
Fix $x^0\in\partial U$. Assume for now that for some ball $B=B(x^0, r)$ about $x^0$, the boundary is straightened, such that $B\cap U\subset \mathbb{R}_{+}^n$. Let $B^+=\{x\in B\ \mid \ x_n\geq 0\}$ and $B^-=\{x\in B\ \mid \ x_n\leq 0\}$. Define \begin{equation} \bar{u}(x)\equiv\left\{ \begin{array}{l l} u(x', x_n)&\quad x\in B^+\\ u(x', -x_n)&\quad x\in B^-. \end{array}\right. \end{equation} Let $u^+=\bar{u}|_{B^+}$ and $u^-=\bar{u}|_{B^-}$. On $\{x_n=0\}$ we observe that $u^+=u^-$. Since $u\in L_{\text{loc}}^1(U)$ we can deduce that $\bar{u}\in L_{\text{loc}}^1(B)$. We now claim that the weak derivative of $\bar{u}$ is \begin{equation}%\label{eq: dve} D\bar{u}(x)=\left\{ \begin{array}{l l} Du(x', x_n)&\quad x\in B^+\\ D_{x'}u(x', -x_n)&\quad x\in B^-\\ -D_{x_n}u(x', -x_n)&\quad x\in B^-. \end{array}\right. \end{equation} We prove it as follows. Let $\varphi\in C_{c}^{\infty}(B)$ and assume for now that $u\in C^1(\overline{B^+\setminus\mathbb{R}^{n-1}})$ then \begin{align*} \int_{B}\bar{u}\varphi_{x_n}\mathrm{d}x &=\int_{B^{+}}\bar{u}\varphi_{x_n}\mathrm{d}x+\int_{B^-}\bar{u}\varphi_{x_n}\mathrm{d}x\\ &=\int_{B^{+}}u\varphi_{x_n}\mathrm{d}x-\int_{B^+}u(y', y_n)\varphi_{y_n}(y', -y_n)\mathrm{d}y\\ &=-\int_{\{x_n=0\}}u\varphi\mathrm{d}S_x-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x+\int_{\{y_n=0\}}u\varphi\mathrm{d}S_y-\int_{B^+}-u_{y_n}(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\ &=-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^+}-u_{y_n}(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\ &=-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-u_{x_n}(x', -x_n)\varphi(x', x_n)\mathrm{d}x \end{align*} and using the same technique for $i\neq n$ we obtain for every multi index $\alpha$ such that $\vert\alpha\vert=1$ \begin{equation*} \int_{B}\bar{u}D^{\alpha}\varphi\mathrm{d}x=\left\{ \begin{array}{l} -\int_{B^+}D_{x'}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}D_{x'}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x\\ -\int_{B^+}D_{x_n}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-D_{x_n}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x. \end{array}\right. \end{equation*}
Let $C^+\equiv B^+\setminus\{x_n=0\}$. Since $u\in W^{1, \infty}(C^+)$ is locally integrable in $C^+$, we can define its $\varepsilon$ mollification in $C_{\varepsilon}^+$ and we also know that: \begin{equation*} D^{\alpha}u^{\varepsilon}(x)\equiv \eta_{\varepsilon}\ast D^{\alpha}u(x)\quad x\in C_{\varepsilon}^+, \ \vert\alpha\vert\leq1. \end{equation*} Here, we define the cutoff function $\eta:\mathbb{R}^n\rightarrow\mathbb{R}$ as \begin{equation*} \eta(x)\equiv\left\{ \begin{array}{l l} C\exp\left[\frac{-1}{1-\|x\|^2}\right] &\quad \|x\|<1\\ 0&\quad \|x\|\geq 1 \end{array}\right. \end{equation*} where $C$ is chosen such that $\int_{\mathbb{R}^n}\eta\mathrm{d}x=1$ and then we define $\eta_{\varepsilon}(x)\equiv\frac{1}{\varepsilon^n}\eta\left(\frac{x}{\varepsilon}\right)$.
Define $v_{\varepsilon}: C^{+}\rightarrow\mathbb{R}$ as \begin{equation*} v_{\varepsilon}(x)\equiv\left\{ \begin{array}{l l} u^{\varepsilon}(x) &\quad x\in C_{\varepsilon}^{+}\\ 0&\quad x\in C^{+}\setminus C_{\varepsilon}^{+}. \end{array}\right. \end{equation*} Now, because $u^{\varepsilon}\in C^{\infty}(D)$ for every open bounded set $D\subset C^+$, by repeated application of the mean value theorem, the derivatives of all orders are bounded on such $D$. Therefore $D^{\alpha}u$ is Lipschitz continuous on $D$ for all $\vert\alpha\vert\leq 1$. In particular, $D^{\alpha}u$ is uniformly continuous on open bounded subsets of $C^{+}$ for all $\vert\alpha\vert\leq 1$. So we can say that $v^{\varepsilon}\in C^1(\overline{C_{\varepsilon}^{+}})$. Now for $\vert\alpha\vert=1$ we claim that the weak derivative of $v_{\varepsilon}$ exists and is given by: \begin{equation} D^{\alpha}v_{\varepsilon}(x)\equiv\left\{ \begin{array}{l l} D^{\alpha}u^{\varepsilon}(x) &\quad x\in C_{\varepsilon}^{+}\\ 0&\quad x\in C^{+}\setminus C_{\varepsilon}^{+}. \end{array}\right. \end{equation} We check it as follows, let $\vert\alpha\vert=1$ and let $\varphi\in C_{c}^{\infty}(C^+)$, then \begin{align*} \int_{C^+}v_{\varepsilon}\varphi_{x_i}\mathrm{d}x &=\int_{C_{\varepsilon}^{+}}v_{\varepsilon}\varphi_{x_i}\mathrm{d}x\\ &=-\int_{C_{\varepsilon}^+}(v_{\varepsilon})_{x_i}\varphi\mathrm{d}x+0\\ &=-\int_{C_{\varepsilon}^+}u^{\varepsilon}_{x_i}\varphi\mathrm{d}x \end{align*} for $i=1, \dots, n$ whence we conclude that the weak derivative exists. Since $\|D^{\alpha}v_{\varepsilon}\|_{L^{\infty}(C^+)}\leq\|D^{\alpha}u\|_{L^{\infty}(C^+)}\infty$ for all $\vert\alpha\vert\leq 1$ we conclude that $v_{\varepsilon}\in W^{1, \infty}(C^+)$. Moreover as $\varepsilon\rightarrow0$ \begin{equation*} D^{\alpha}v^{\varepsilon}\rightarrow D^{\alpha}u\quad\text{a.e. in } C^+, \ \vert\alpha\vert\leq 1. \end{equation*} We can write \begin{equation*} C^+\equiv\bigcup_{i=1}^{\infty}C_{1/i}^{+} \end{equation*} and $D^{\alpha}v_{i}: C^{+}\rightarrow\mathbb{R}$ for each $\vert\alpha\vert\leq 1$ \begin{equation*} D^{\alpha}v_{i}(x)\equiv\left\{ \begin{array}{l l} D^{\alpha}v^{\frac{1}{i}}(x) &\quad x\in C_{1/i}^{+}\\ 0&\quad x\in C^+\setminus C_{1/i}^{+}. \end{array}\right. \end{equation*} Since $D^{\alpha}v_i$ is measurable and uniformly bounded almost everywhere in $C^+$ by $\|D^{\alpha}v\|_{\infty}$ for each $\vert\alpha\vert\leq 1$, we deduce by the dominated convergence theorem that: \begin{equation*} \lim_{i\rightarrow \infty}\int_{C^+}D^{\alpha}v_i\mathrm{d}x=\int_{C^+}D^{\alpha}u\mathrm{d}x\quad \vert\alpha\vert\leq 1. \end{equation*} Since $\mathcal{L}^{n}(\mathbb{R}^{n-1})=0$ we can define $D^{\alpha}v_i$ arbitrarily there and write \begin{equation*} \lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}v_i\mathrm{d}x=\int_{B^+}D^{\alpha}u\mathrm{d}x \quad\vert\alpha\vert\leq 1. \end{equation*} Now let $\bar{u}$ be as defined initially, $\alpha=(0, \dots, 1)$ and $\varphi\in C_{c}^{\infty}(B)$, then %replacing $D^{\alpha}v_i$ with $D^{\alpha}u_i$ we have \begin{align*} \int_{B}\bar{u}D^{\alpha}\varphi\mathrm{d}x &=\int_{B^{+}}\bar{u}D^{\alpha}\varphi\mathrm{d}x+\int_{B^-}\bar{u}D^{\alpha}\varphi\mathrm{d}x\\ &=\int_{B^{+}}uD^{\alpha}\varphi\mathrm{d}x-\int_{B^+}u(y', y_n)D^{\alpha}\varphi_{y_n}(y', -y_n)\mathrm{d}y\\ &=\lim_{i\rightarrow \infty}\int_{B^{+}}v_iD^{\alpha}\varphi\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}v_i(y', y_n)D^{\alpha}\varphi(y', -y_n)\mathrm{d}y\\ &=-\lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}v_i(x', x_n)\varphi(x', x_n)\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}-D^{\alpha}v_i(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\ &=-\int_{B^+}D^{\alpha}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^+}-D^{\alpha}u(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\ &=-\int_{B^+}D^{\alpha}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-D^{\alpha}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x. \end{align*} We obtain the corresponding result for any other multi index $\alpha$ such that $\vert\alpha\vert=1$. So we conclude that the weak derivative of $\bar{u}$ is as claimed.
Furthermore, we have that for $\vert \alpha\vert\leq 1$: \begin{equation*} \vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B)}\leq\vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B^+)}+\vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B^-)}=2\vert\vert D^{\alpha}u\vert\vert_{L^{\infty}(B^+)} \end{equation*} and hence \begin{equation*} \vert\vert \bar{u}\vert\vert_{W^{1, \infty}(B)}\leq 2\vert\vert u\vert\vert_{W^{1, \infty}(B^+)}. \end{equation*} If the boundary is not already straightened out we can do so under the map $\Phi: W\rightarrow B$ and unstraighten it under $\Psi: B\rightarrow W$ as defined in Evans, where $\Psi (B)=W$. Writing $\Phi(x)=y$ and $\Psi(y)=x$ we define $u'(y)\equiv u\circ\Psi (y)$ . Then we have: \begin{equation*} \vert\vert \bar{u'}\vert\vert_{W^{1, \infty}(B)}\leq 2\vert\vert u'\vert\vert_{W^{1, \infty}(B^+)} \end{equation*} and converting back to original coordinates we have \begin{equation} \vert\vert \bar{u}\vert\vert_{W^{1, \infty}(W)}\leq 2\vert\vert u\vert\vert_{W^{1, \infty}(U)} \end{equation} We know that $\overline{U}\subset\mathbb{R}^n$ is paracompact so it admits locally finite partitions of unity subordinate to any countable cover of open sets. Cover the boundary $\partial U$ with the cover $\{W_x\}_{x\in\partial U}$ such that each open set $W_x$ is a ball about $x$ and we obtain inequality as above for each $W_x$. The boundary is compact so we can pass to a finite subcover $\{W_i\}_{i=1}^{N}$ and have the corresponding extensions $\bar{u}_i$. Choose any $W_0\subset\subset U$ such that $U\subset\bigcup_{i=0}^{N} W_i$ and define $\bar{u}_0\equiv u$ in $W_0$.
Let $\{\zeta_i\}_{i=1}^{N}$ be a locally finite smooth partition of unity subordinate to the cover $\{W_i\}_{i=0}^{N}$. Define $\bar{u}\equiv\sum_{i=0}^{N}\zeta_i\bar{u}_i$. Then we obtain the bound: \begin{align*} \vert\vert \bar{u}\vert\vert_{W^{1, \infty}(\mathbb{R}^n)}&=\left\| \sum_{i=0}^{N}\zeta_i\bar{u}_i\right\|_{W^{1, \infty}(\mathbb{R}^n)}\\ &\leq\sum_{i=0}^{N}\|\zeta_i\bar{u}_i\|_{W^{1, \infty}(W_i)}\\ &\leq\sum_{i=0}^{N}\|\bar{u}_i\|_{W^{1, \infty}(W_i)}\\ &\leq \sum_{i=0}^{N}2\|u\|_{W^{1, \infty}(U)}\\ &\leq C\|u\|_{W^{1, \infty}(U)}. \end{align*}
Furthermore we can arrange for the support of $\bar{u}$ to lie within some $V\supset\supset U$. Define $Eu\equiv\bar{u}$. Then we can see that the map $E: W^{1, \infty}(U)\rightarrow W^{1, \infty}(\mathbb{R}^n)$ is a linear operator such that
- $Eu=u$ almost everywhere in $U$.
- $Eu$ has support within $V\supset\supset U$
- $\vert\vert Eu\vert\vert_{W^{1, \infty}(\mathbb{R}^n)}\leq C\|u\|_{W^{1, \infty}(U)}$ for $C=2N$.
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.
Best Answer
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) $v\in W^{1,1}$.