# Extension Thoerem for the Sobolev Space $W^{1, \infty}(U)$

I am trying to find a way of extending functions in the Sobolev Space $W^{1, \infty}(U)$ to $W^{1, \infty}(\mathbb{R}^n)$ where $U\subset\mathbb{R}^{n}$ is open such that $U\subset\subset V$ for $V\subset\mathbb{R}^{n}$ also open and bounded. Furthermore, assume $\partial U$ is $C^1$.

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Your space $W^{1,\infty}(U)$ is in fact the space of Lipschitz continuous functions defined on $U$. It looks like you need something like this: en.wikipedia.org/wiki/Kirszbraun_theorem – Beni Bogosel Jun 15 '12 at 18:11
Kirszbraun's theorem works, but for scalar-valued functions. McShane-Whitney formula $U(y)=\inf_{x\in U}u(x)+L|x-y|$ is all one needs. Either way, one first needs to know that $W^{1,\infty}(U)=\mathrm{Lip}(U)$ for the indicated class of domains $U$. If this nontrivial fact cannot be taken for granted, it will be the main part of the proof. – user31373 Jun 15 '12 at 18:22
@LeonidKovalev : What should one take care of in this case compared to extension of $W^{1,p}$ ? why does it not work for $p=\infty$ – Theorem Jun 15 '12 at 18:46
@Theorem The Whitney-type extension, as described, e.g., in Stein's 1970 book, works fine for all $1\le p\le \infty$. But when $p=\infty$, one is tempted to give a shorter proof using the identification of $W^{1,\infty}$ with the Lipschitz space, see the comment by Beni Bogosel. I made a remark to the effect that the identification is not entirely trivial. ... Side remark: it's hard to answer question like this without knowing the tools that the questioner has in disposal. I can't tell if Nirav knows about Whitney decomposition etc. – user31373 Jun 15 '12 at 18:55

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 $$\bar{u}(x)\equiv\left\{ \begin{array}{l l} u(x', x_n)&\quad x\in B^+\\ u(x', -x_n)&\quad x\in B^-. \end{array}\right.$$ 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 $$%\label{eq: dve} D\bar{u}(x)=\left\{ \begin{array}{l l} Du(x', x_n)&\quad x\in B^+\\ D_{x'}u(x', -x_n)&\quad x\in B^-\\ -D_{x_n}u(x', -x_n)&\quad x\in B^-. \end{array}\right.$$ 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*}

If we let $C^+\equiv B^+\setminus\{x_n=0\}$ then we know that if $v: C^+\rightarrow\mathbb{R}$ is locally integrable in $C^+$ then we can define its $\varepsilon$ mollification in $C_{\varepsilon}^+$ and if moreover $v\in W^{1, \infty}(C^+)$ then we also know that: \begin{equation*} D^{\alpha}v^{\varepsilon}(x)\equiv \eta_{\varepsilon}\ast D^{\alpha}v(x)\quad x\in C_{\varepsilon}^+\subset C^+, \ \vert\alpha\vert\leq1 \end{equation*} and as $\varepsilon\rightarrow0$ \begin{equation*} D^{\alpha}v^{\varepsilon}\rightarrow D^{\alpha}v\quad\text{a.e. in } C_{\varepsilon}^+, \ \vert\alpha\vert\leq 1. \end{equation*} We can write \begin{equation*} C^+\equiv\bigcup_{i=1}^{\infty}C_{1/i}^{+} \end{equation*} and because $D^{\alpha}v^{1/i}\rightarrow D^{\alpha}v\quad\text{a.e. in } C_{1/i}^+$ for each $i$ then \begin{equation*} D^{\alpha}v^{1/i}\rightarrow D^{\alpha}v\quad\text{a.e. in } C^+,\ \vert\alpha\vert\leq 1. \end{equation*} Since $D^{\alpha}v^{1/i}$ is measurable and uniformly bounded almost everywhere in $C^+$ by $\|D^{\alpha}v\|_{\infty}$, we deduce by the dominated convergence theorem that: \begin{equation*} \lim_{i\rightarrow \infty}\int_{C^+}D^{\alpha}v^{1/i}\mathrm{d}x=\int_{C^+}D^{\alpha}v\mathrm{d}x. \end{equation*} Since $\mathcal{L}^{n}(\mathbb{R}^{n-1})=0$ we can write \begin{equation*} \lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}v^{1/i}\mathrm{d}x=\int_{B^+}D^{\alpha}v\mathrm{d}x. \end{equation*} Now let $\bar{u}$ be as defined originally, $\alpha=(0, \dots, 1)$ and $\varphi\in C_{c}^{\infty}(B)$, then 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\\ &=\lim_{i\rightarrow \infty}\int_{B^{+}}u^{1/i}D^{\alpha}\varphi\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}u^{1/i}(y', y_n)D^{\alpha}\varphi(y', -y_n)\mathrm{d}y\\ &=-\lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}u^{1/i}(x', x_n)\varphi(x', x_n)\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}-D^{\alpha}u^{1/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 $$\vert\vert \bar{u}\vert\vert_{W^{1, \infty}(W)}\leq 2\vert\vert u\vert\vert_{W^{1, \infty}(U)}$$ 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

1. $Eu=u$ almost everywhere in $U$.
2. $Eu$ has support within $V\supset\supset U$
3. $\vert\vert Eu\vert\vert_{W^{1, \infty}(\mathbb{R}^n)}\leq C\|u\|_{W^{1, \infty}(U)}$ for $C=2N$.
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