Sobolev extension problems of $W^1_\infty(\Omega)$

Question

Recently I have read the paper Whitney's problem on extendability of functions and an intrinsic metric written by Nahum Zobin and published by Advances in Mathematics in 1998. I am confused about one proposition of this paper. Now let me state the background of this paper.

Let $\Omega$ be a bounded connected open set in $\mathbb{R}^n$. Consider the following Sobolev function space \begin{equation}\tag{#}\label{W1infty} W^1_\infty(\Omega)=\{f\in C(\Omega):\forall \alpha\in \mathbb{Z}^n_+,|\alpha|=1,f^{(\alpha)}\in L^\infty(\Omega)\}. \end{equation} (This definition follows from E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, 1970, Chap. V, Section 6.2.) Here $|\alpha|=\sum_{i=1}^n \alpha_i$ for $\alpha=(\alpha_1,\cdots,\alpha_n)\in \mathbb{Z}^n_+$, $f^{(\alpha)}$ denotes the corresponding distributional partial derivative, $C(\Omega)$ denotes the space of continuous functions, $L^\infty(\Omega)$ denotes the space of essentially bounded functions on $\Omega$.

Let $W^1_\infty(\mathbb{R}^n)|_\Omega$ denote the space of restrictions to $\Omega$ of functions from $W^1_\infty(\mathbb{R}^n)$.

For $x,y\in\Omega$, let \begin{equation*} d_\Omega(x,y)=\text{infimum of lengths of polygonal paths in $\Omega$ joining $x$ and $y$}. \end{equation*} $d_\Omega(x,y)$ is called the intrinsic metric in $\Omega$.

Now fix $x\in \Omega$, consider the function $f(y)=d_\Omega(x,y)$. Then $f\in W^1_\infty(\Omega)$ and \begin{equation}\tag{$*$}\label{1} \sup_{\substack{z\in \Omega\\ |\alpha|=1}} |f^{(\alpha)}(z)|=1. \end{equation} If $W^1_\infty(\Omega)\subset W^1_\infty(\mathbb{R}^n)|_\Omega$, then $f$ is extendible to $\tilde{f}\in W^1_\infty(\mathbb{R}^n)$ and the open mapping theorem guarantees that on can choose $\tilde{f}$ such that \begin{equation}\tag{$**$}\label{2} \sup_{\substack{z\in \mathbb{R}^n\\ |\alpha|=1}} |f^{(\alpha)}(z)| \leqslant C\sup_{\substack{z\in \Omega\\ |\alpha|=1}} |f^{(\alpha)}(z)|=C. \end{equation} So, \begin{align} d_\Omega(x,y)&=|f(y)-f(x)|=|\tilde{f}(y)-\tilde{f}(x)|\notag\\ &\leqslant \sup_{\substack{z\in \mathbb{R}^n\\ |\alpha|=1}}|\tilde{f}^{(\alpha)}(z)|\cdot |x-y| \leqslant C|x-y|.\tag{$***$}\label{3} \end{align}

My questions are:

1. Why $f\in W^1_\infty(\Omega)$ and \eqref{1} holds?

2. How to use the open mapping theorem and why \eqref{2} holds?

3. Why the first inequality of \eqref{3} holds?

About Question 1, I know $f$ is Lipschitz with constant $1$ under the intrinsic metric, that is, \begin{equation*} |f(z_1)-f(z_2)|\leqslant d_\Omega(z_1,z_2),\quad \forall z_1,z_2\in\Omega. \end{equation*} But I don't know whether this property can help us deal with this question.

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Here I provide more materials of those questions.

1. This paper defined the Sobolev space as follows: \begin{equation*} W^{k+1}_\infty(\Omega)=\{f\in C^k(\Omega): \forall \alpha\in \mathbb{Z}^n_+,|\alpha|=k+1,f^{(\alpha)}\in L^\infty(\Omega)\}, \end{equation*} where $C^k(\Omega)$ denotes the space of $k$ times continuously differentiable functions. Others are same as before. Previously, I was also confused of $C^k(\Omega)$, but unfortunately, this paper does not provide additional information about it. Since I only need $k=0$, I wrote it as \eqref{W1infty}.

2. The definition of Sobolev space in E. M. Stein's book (pp. 121-122, Chap. V, Sect. 2): for any nonnegative integer $k$ and $1\leqslant p\leqslant \infty$, the Sobolev space $L^p_k(\mathbb{R}^n)$ is defined as the space of functions $f$, with $f\in L^p(\mathbb{R}^n)$ and where all $f^{(\alpha)}$ exist and $f^{(\alpha)}\in L^p(\mathbb{R}^n)$ in the distributional derivative sense, whenever $|\alpha|\leqslant k$. The space of functions can be normed by the expression \begin{equation*} \|f\|_{L^p_k(\mathbb{R}^n)}=\sum_{|\alpha|\leqslant k}\|f^{(\alpha)}\|_{L^p(\mathbb{R}^n)}. \end{equation*}

The Stein's book also writes the following (pp. 159, Chap. V. Sect. 6.2). $f$ belongs to $L^\infty_k(\mathbb{R}^n), k\geqslant 1$ if and only if $f$ can be modified on a set of measure zero so that $f$ has continuous partial derivatives of total order $\leqslant k-1$. Moreover, whenever $g=f^{(\alpha)}$, $|\alpha|\leqslant k-1$, then \begin{equation*} \sup_x |g(x)|<\infty,\quad \text{and}\quad \sup_{x,x'}\dfrac{|g(x)-g(x')|}{|x-x'|}<\infty. \end{equation*}

Answer 1

Point $(**)$ with the open mapping theorem is meaningless if the space "$W^1_\infty$" is not arranged to be normable. So I assume that by "$C(\Omega)$" it is meant the space $C_b(\Omega)$ of bounded continuous functions. Then $f\in W^1_\infty(\Omega)$ need not hold as one sees by considering the two-dimensional set $\Omega=\{\,(t,s):0<t<1\text{ and }\sin(t^{-1})<s<\sin(t^{-1})+1\,\}\,$.

So one needs to take as an additional assumption that $f$ be bounded. Under this assumption the Lipschitz property of $f$ gives $(*)$ by noting that by Rademacher's theorem a (locally) Lipschitzian function is almost everywhere pointwise differentiable with the (a.e.) pointwise derivative also being the distributional one, by absolute continuity along lines. One also uses that $|u-y|=d_\Omega(u,y)$ holds when $u,y$ are in a ball included in $\Omega\,$. For convenience, in view of $(*\!*\!*)$ I take $|z|=|z|_{\ell^{\,1}}\,$, i.e. sum of absolute values of coordinates. Note that here all norms are equivalent.

For $(**)$ one needs to assume that $W^1_\infty(\Omega)\subseteq W^1_\infty(\mathbb{R}^n)|_\Omega$ and hence $W^1_\infty(\Omega)= W^1_\infty(\mathbb{R}^n)|_\Omega$holds. Then the restriction map $R:g\mapsto g\,|\,\Omega$ is a continous linear surjection $E=W^1_\infty(\mathbb{R}^n)\to W^1_\infty(\Omega)=F\,$, and hence open by the Open Mapping Theorem. So, letting $B_X(r)$ be the open ball in $X$ centered at zero with radius $r\,$, there is $\varepsilon\in\mathbb R^+$ with $B_F(\varepsilon)\subseteq R\,[\,B_E(1)\,]\,$. Since $\varepsilon\,(1+\,\|\,f\,\|_F)^{-1\,}f\in B_F(\varepsilon)$ holds, there is $f_1\in B_E(1)$ with $\varepsilon\,(1+\,\|\,f\,\|_F)^{-1\,}f=R\,f_1$ and for $\tilde f=\varepsilon^{-1\,}(1+\,\|\,f\,\|_F)\,f_1$ and $M=\varepsilon^{-1\,}(1+\,\|\,f\,\|_F{}^{-1}\big)$ then $\|\,\tilde f\,\|_E < \varepsilon^{-1\,}(1+\,\|\,f\,\|_F) = M\,\|\,f\,\|_F$ holds. So we have $f=R\,\tilde f$ and $\|\,\tilde f\,\|_{W^1_\infty(\mathbb R^n)} < M\,\|\,f\,\|_{W^1_\infty(\Omega)}$ whence \begin{align} \sup_{\substack{z\in \mathbb{R}^n\\ |\alpha|=1}} |\tilde f{}^{(\alpha)}(z)| & \le\|\,\tilde f\,\|_{W^1_\infty(\mathbb R^n)} < M\,\|\,f\,\|_{W^1_\infty(\Omega)} \\ & = M\,\big(\,\|\,f\,\|_{C_b(\Omega)} + \sup_{\substack{z\in \Omega\\ |\alpha|=1}} |f^{(\alpha)}(z)|\,\big)\le M\,\big(\,\|\,f\,\|_{C_b(\Omega)} + 1) = C\,, \end{align} noting that $\|\,f\,\|_{C_b(\Omega)} < +\infty$ holds as $f$ is assumed to be a bounded function.

Point $(*\!*\!*)$ is a direct consequence of $(**)$ by noting that $C$ is a Lipschitz constant for $\tilde f\,$. If one wants a proof of this, one takes a mollified sequence $\langle\,f_i\,\rangle$ of smooth function converging to $\tilde f$ and applies the mean value theorem to the functions $f_i\,$.

Note that if one uses some other norm like $|\,z\,|_{\ell^{\,2}}\,$ then one gets for $C$ a different value.

Remark. The above leaves it open what is a possible relation between the required assumptions that $f$ be bounded and $W^1_\infty(\Omega)\subseteq W^1_\infty(\mathbb{R}^n)|_\Omega\,$, i.e. does possibly one imply the other. Saying that $\Omega$ is intrinsically bounded iff $f$ is bounded, and saying that $\Omega$ allows $W^{1,+\infty}\,$−extension iff the latter holds, by suitably modifying Example 3 on page 3 here we see that at least the implication "$\,\Omega$ intrinsically bounded $\Rightarrow\Omega$ allows $W^{1,+\infty}\,$−extension" does not hold. So it is only left open whether the assumption that $f$ be bounded is really needed, or whether it follows from the extension property.

Added 27.6.2024: The Theorem below shows that the explicit assumption that $f$ be bounded is superfluous.

Theorem. If $\Omega$ allows $W^{\,1,+\infty\,}$−extension, then $\Omega$ is intrinsically bounded.

Proof. For the indirect verification we assume that $f$ is not bounded and that $\Omega$ allows $W^{\,1,+\infty\,}$−extension. Since $\Omega$ is bounded, there is a sequence $\langle\,x_i\,\rangle$ in $\Omega$ that converges to some $z_9\in\mathbb R^n$ and satisfies $f(x_i) + 2 < f(x_{i+1})$ for all $i\in\mathbb N_0\,$. Letting $B_i$ be the open ball centered at $x_i$ with radius $1\,$, let $U_i$ be the connected component of $\Omega\cap B_i$ that contains $x_i\,$. By the triangle inequality for $d_{\,\Omega}$ then $U_i\cap U_j = \emptyset$ for different $i,j\in\mathbb N_0\,$. Then we let $g$ be defined by $U_i\owns z\mapsto(-1)^{\,i\,}\sup\,\{\,0,1-|\,z-x_i\,|\,\big\}$ for $i\in\mathbb N_0\,$, and $z\mapsto 0$ for $z\in\Omega\setminus\bigcup_{\,i\in\mathbb N_0}U_i\,$. Now $g$ is locally $1\,$−Lipschitz for the norm metric, hence in $W^1_\infty(\Omega)$ and so there is $\bar g$ in $W^1_\infty(\mathbb R^n\big)$ with $g\subseteq\bar g\,$. Then $\bar g$ is Lipschitz (for the norm metric) and so there is $M\in\mathbb R^+$ with $|\,\bar g\,y - \bar g\,z\,|\le M\,|\,y - z\,|$ for all $y,z\in\mathbb R^n$. In particular, we have \begin{align} 2 & = |\,g\,(x_i)-g\,(x_{i+1})\,| = |\,\bar g\,(x_i) - \bar g\,(x_{i+1})\,| \\ & \le M\,|\,x_i - x_{i+1}\,|\le M\,\big(\,|\,x_i - z_9\,| + |\,z_9 - x_{i+1}\,|\,\big)\to 0 \end{align} as $i\to\infty\,$, a contradiction.

Answer 2

This is only a too long comment trying to make sense of some obscurities/omissions in Zobin's article. On page 99 there he writes $\bigcap_{\,l=1}^{\,k}W^l_\infty(\Omega)\not\subset L^\infty(\Omega)$ which suggest that he does not intend $W^l_\infty(\Omega)$ to be the usual Banach space $W^{l,+\infty}(\Omega)$ but instead a Fréchet space in whose definition $C^k(\Omega)$ indeed is intended to be the space of not necessarily bounded functions with topology of uniform convergence of derivatines on compact sets. Then his application of the Open Mapping Theorem may be the following:

For any fixed $x\in\Omega$ define $f_x$ by $\Omega\owns y\mapsto d_{\,\Omega}(x,y)\,$. Then let $U$ in the space $E=W^1_\infty(\mathbb R^n)$ be the set of all $f_1$ with first order (a.e. defined weak) partial derivatives having absolute value less than $1$ on $\mathbb R^n$. Now $U$ is an open zero neighbourhood in $E$ whence by the Open Mapping Theorem there is a zero neighbourhood $V$ in $F=W^1_\infty(\Omega)$ with $V\subseteq R\,[\,U\,]\,$. Then there is a compact set $K\subseteq\Omega$ and $\varepsilon\in\mathbb R^+$ such that for $g$ in $F$ with $|\,g\,z\,|\le\varepsilon$ for $z\in K$ and first order partials having absolute value at most $\varepsilon$ on $\Omega$ we have $g\in V$.

Fixing some $x_0\in\Omega\,$, for any $x\,,z\in\Omega$ we have $f_x\,z\le f_{x_0}\,z+f_{x_0}\,x\,$. Since $f_{x_0}$ can be extended to a continuous function on $\mathbb R^n$, there is $M_9\in\mathbb R^+$ such that $f_x\,z\le M_9$ holds for all $x\,,z\in\Omega\,$. Taking $C=\varepsilon^{-1\,}(1+M_9)\,$, for $g=C^{-1}f_x$ we see that $g\in V$ holds, and so there is $f_1\in U$ with $g=R\,f_1\,$. For $\tilde f=C\,f_1$ then $f_x\subseteq\tilde f$ with $|\,\partial^{\,\alpha}\tilde f\,z\,|< C$ for $z\in\mathbb R^n$ and $|\,\alpha\,|=1\,$.

Note that, if my interpretation above of Zobin's quite weird definition of the space $W^1_\infty(\Omega)$ as a topological linear supremum of the Fréchet space $C\,(\Omega)$ embedded in $\mathscr D'(\Omega)$ and the inverse image of the weak gradient from $L^{+\infty}(\Omega,\mathbb R^n)$ is correct, then the condition $W^1_\infty(\Omega)\subset W^1_\infty(\mathbb R^n)\,|_\Omega$ is, at least apriori, much more restrictive than it would be for the usual definition of the Sobolev space $W^{1\,,+\infty}(\Omega)$ as a Banach space.