(continuously) extend a function from a $C^1$-boundary to the whole space

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Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$ be bounded and open. Assume $\partial\Omega$ is a $(d-1)$-dimensional embedded $C^\alpha$-submanifold of $\mathbb R^d$ for some $\alpha\in\mathbb N$.$^1$

If $f:\partial\Omega\to\mathbb R$, can we extend $f$ to $\overline\Omega$ or even $\mathbb R^d$? such that

  1. this extension is continuous, if $f$ is continuous?
  2. this extension is differentiable, if $f$ is the restriction of a $C^1(U,\mathbb R)$-function, where $U\subseteq\mathbb R^d$ is open and $\partial\Omega\subseteq U$?

And are those extensions, if they exist, unique?

EDIT: The following is clear to me: Let $k\in\{1,\ldots,d\}$ and $M$ be a compact $k-$dimensional embedded $C^\alpha$-submanifold of $\mathbb R^d$. Since $M$ is compact, there is a $k$-dimensional $C^\alpha$-atlas $((M_i,\phi_i))_{i\in I}$ of $M$ for some finite set $I$. By definition, $M_i$ is an open subset of $M$ and $\phi_i$ is a $C^\alpha$-diffeomorphism from $M_i$ onto an open subset of $\mathbb R^k$ for all $i\in I$ and $$M\subseteq\bigcup_{i\in I}M_i\tag1.$$ Now, $$T_xM={\rm D}\phi_i^{-1}\left(\phi_i(x)\right)\mathbb R^k\;\;\;\text{for all }x\in M_i\text{ and }i\in I\tag2.$$ Let $(e_1,\ldots,e_k)$ denote the standard basis of $\mathbb R^k$, $$\sigma_j(x):={\rm D}\phi_i^{-1}\left(\phi_i(x)\right)e_j\;\;\;\text{for }j\in\{1,\ldots,k\}\text{ and }x\in M_i\text{ for some }i\in I$$ and $\left(\tau_1(x),\ldots,\tau_k(x)\right)$ denote the orthonormal basis of $\mathbb R^k$ obtained from $\left(\sigma_1(x),\ldots,\sigma_k(x)\right)$ by the Gram-Schmidt orthogonalization process. If I'm not missing something, $\sigma_j$ (and hence $\tau_j$) is in $C^{\alpha-1}(M,\mathbb R^k)$.

Now, in the situation of the question, $M=\partial\Omega$, $N_x\partial\Omega$ is $1$-dimensional and $$\nu_{\partial\Omega}(x):=\frac{\left({\rm D}\phi(x)\right)^\ast e_k}{\left\|\left({\rm D}\phi(x)\right)^\ast e_d\right\|},$$ where $\phi$ is a $C^\alpha$-diffeomorphism of an open neighborhood of $x$ onto an open subset of $\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$, is an orthonormal basis of $N_x\partial\Omega$ for all $x\in\partial\Omega$. As before, $\nu_{\partial\Omega}$ should belong to $C^{\alpha-1}(\partial\Omega,\mathbb R^d)$, but please correct me, if I'm wrong.

The open questions are: (a) How can we use the orthonormal bases $\left(\tau_1(x),\ldots,\tau_{d-1}(x),\nu_{\partial\Omega}(x)\right)$, $x\in\partial\Omega$, of $\mathbb R^d$ to construct the desired extension? (b) Which argument do we need to show that $\sigma_j(x)$ and $\nu_{\partial\Omega}(x)$ are well-defined, i.e. independent of the choice of the chart (for $\sigma_j(x)$: $x$ could belong to $M_i\cap M_j$ for some $i\ne j$. Why is this not a problem?)


$^1$ i.e. $\partial\Omega$ is locally $C^\alpha$-diffeomorphic to $\mathbb R^{d-1}$.

2

There are 2 best solutions below

6
On

It is not completely clear what is given and what can be chosen here. I describe the situation where $1\leq k<d$ and we have a compact $k$-dimensional $C^\alpha$ ($\alpha\geq 1$) submanifold $M$ (without boundaies) of ${\Bbb R}^d$ and a function $f\in C^\beta(M)$, $0\leq \beta\leq \alpha$.

For each $a\in M$ you may choose an open neighborhood $U_a \subset {\Bbb R}^d$ of $a$ and a $C^\alpha$ diffeo $\phi_a : U_a \rightarrow V_a=\phi_a(U_a)\subset {\Bbb R}^d$ so that $\phi_a(a)=0_d$ and $$ \phi_a(U_a \cap M) = V_a\cap({\Bbb R}^k \times \{0_{d-k}\}).$$

(This is one definition of a smooth submanifold and equivalent to any other as far as I know). Let $\pi:{\Bbb R}^k \times {\Bbb R}^{d-k} \rightarrow{\Bbb R}^k \times \{0_{d-k}\}$ be the natural projection. Possibly shrinking neighborhoods we may assume that $\pi (V_a)\subset V_a$.

Use compactness to construct a finite cover $U_1,\ldots,U_N$ with an associated smooth partition of unity (see wiki) for $M$ subordinate to the cover. In other words, we obtain $C^\infty$ functions $\chi_i\in C^\infty_c(U_i)$ such that $\sum_i \chi_i(x)=1$ for each $x\in M$. Given the function $f\in C^\beta(M)$ we obtain a function $f_i$ with compact support and of the same regularity by declaring for each $i$: $$f_i(x) = f(\phi_i^{-1} \circ \pi \circ \phi_i(x)) \chi_i(x), \ x\in U_i,$$ and extending by zero to the rest of ${\Bbb R}^d$. Note that for $x\in M\cap U_i$ we have by construction $f_i(x)=f(x)\chi_i(x)$. Therefore, $F(x) = \sum_i f_i(x)\in C^\beta_c({\Bbb R}^d)$ provides an extension of $f$ as required.

Remarks: There is no need for tubular neighborhoods (created e.g. using normal flows) to make such an extension unless you want to study formulae for co-area, Laplacians or similar stuff. It makes life more complicated than necessary The above extension is obviously far from unique.

0
On

I think we don't need to assume compactness as in H. H. Rugh's answer. In fact, I guess we only need to ensure that there is a suitable partition of unity.

For clarity, remember the following: If $E_i$ is a $\mathbb R$-Banach space, $f:\Omega_1\to E_2$ and $\alpha\in\mathbb N_0\cup\{\infty\}$, then $f$ is $C^\alpha$-differentiable at $x_1\in\Omega_1$ if $$\left.f\right|_{O_1\:\cap\:\Omega_1}=\left.\tilde f\right|_{O_1\:\cap\:\Omega_1}\tag3$$ for some $\tilde f\in C^\alpha(O_1,E_2)$ for some $E_1$-open neighborhood $O_1$ of $x_1$. In that case $(O_1,\tilde f)$ is called $C^\alpha$-extension of $f$ at $x_1$. $f$ is $C^\alpha$-differentiable if $f$ is $C^\alpha$-differentiable at $x_1$ for all $x_1\in\Omega_1$.

Assume $E_1=\mathbb R^d$ for some $d\in\mathbb N$. We've got the following result:

Theorem Let $\Omega_1\subseteq\mathbb R^d$ be closed, $\alpha\in\mathbb N_0\cup\{\infty\}$, $f:\Omega_1\to E_2$ be $C^\alpha$-differentiable and $O_1$ be an $\mathbb R^d$-open neighborhood of $\Omega_1$. Then there is a $\tilde f\in C^\alpha(\mathbb R^d,E_2)$ (i.e. $\tilde f$ is a $\alpha$-times continuously Fréchet differentiable function from $\mathbb R^d$ to $E_2$) with $$\operatorname{supp}\tilde f\subseteq O_1\tag4$$ and $$\left.\tilde f\right|_{\Omega_1}=f\tag5.$$

Proof: Let $\left(O_{x_1},f_{x_1}\right)$ be a $C^\alpha$-extension of $f$ at $x_1$ for $x_1\in\Omega_1$. Since $O_1$ is an $\mathbb R^d$-open neighborhood of $\Omega_1$, $O_{x_1}\cap O_1$ is an $\mathbb R^d$-open neighborhood of $x_1$ for all $x_1\in\Omega_1$ and hence, by replacing $O_{x_1}$ with $O_{x_1}\cap O_1$, we may assume that $$O_{x_1}\subseteq O_1\tag6$$ for all $x_1\in\Omega_1$. Since $\mathbb R^d\setminus\Omega_1$ is open, $$\mathcal C:=\left\{O_{x_1}:x_1\in\Omega_1\right\}\cup\{\mathbb R^d\setminus\Omega_1\}$$ is an open cover of $\mathbb R^d$. Thus, by Theorem 4 below (applied for the $C^\infty$-manifold $\mathbb R^d$), there is a $C^\infty$-partition of unity $\left\{\rho_{x_1}:x_1\in\Omega_1\right\}\cup\{\rho_0\}$ subordinate to $\mathcal C$. Note that $$\rho_{x_1}\tilde f_{x_1}\in C^\alpha\left(O_{x_1},E_2\right)\;\;\;\text{for all }x_1\in\Omega\tag7.$$ Let $$g_{x_1}:=\left.\begin{cases}\rho_{x_1}\tilde f_{x_1}&\text{on }\operatorname{supp}\rho_{x_1}\\0&\text{on }\mathbb R^d\setminus\operatorname{supp}\rho_{x_1}\end{cases}\right\}\;\;\;\text{for }x_1\in\Omega_1.$$ Note that $O_{x_1}\setminus\operatorname{supp}\rho_{x_1}$ is open and $$\left.g_{x_1}\right|_{O_{x_1}\setminus\operatorname{supp}\rho_{x_1}}=0=\left.\rho_{x_1}\tilde f_{x_1}\right|_{O_{x_1}\setminus\operatorname{supp}\rho_{x_1}}\tag8$$ and hence $$g_{x_1}\in C^\alpha\left(\mathbb R^d,E_2\right)\tag9$$ for all $x_1\in\Omega_1$. Since $\left(\operatorname{supp}\rho_{x_1}\right)_{x_1\in\Omega_1}$ is locally finite, $$\tilde f:=\sum_{x_1\in\Omega_1}g_{x_1}\in C^\alpha\left(\mathbb R^d,E_2\right)$$ is well-defined. Note that $$\operatorname{supp}\tilde f\subseteq\bigcup_{x_1\in\Omega_1}\operatorname{supp}\rho_{x_1}\subseteq\bigcup_{x_1\in\Omega_1}O_{x_1}\subseteq O_1\tag{10}.$$ Now let $y_1\in\Omega_1$. Then, $$\tilde f_{x_1}(y_1)=f(y_1)\;\;\;\text{for all }x_1\in\Omega_1\text{ with }y_1\in O_{x_1}\tag{11}$$ and hence (since $\operatorname{supp}\rho_{x_1}\subseteq O_{x_1}$ for all $x_1\in\Omega_1$) $$\tilde f_{x_1}(y_1)=f(y_1)\;\;\;\text{for all }x_1\in\Omega_1\text{ with }\rho_{x_1}(y_1)\ne 0\tag{12}.$$ Since $\rho_0(y_1)=0$, \begin{equation}\begin{split}\tilde f(y_1)&=\sum_{x_1\in\Omega_1}g_{x_1}(y_1)=\sum_{\substack{x_1\in\Omega_1\\\rho_{x_1}(y_1)\ne0}}\rho_{x_1}(y_1)\tilde f_{x_1}(y_1)=\sum_{x_1\in\Omega_1}\rho_{x_1}(y_1)f(y_1)\\&=\underbrace{\left(\sum_{x_1\in\Omega_1}\rho_{x_1}(y_1)+\rho_0(y_1)\right)}_{=\:1}f(y_1)=f(y_1).\end{split}\tag{13}\end{equation}

Now, in the context of the question, this result can be applied to the manifold boundary $\partial M$ of a $d$-dimensional properly embedded $C^\alpha$-submanifold $M$ of $\mathbb R^d$ with boundary, since it can be shown that the $\partial M$ is actually equal to the topological boundary and hence is a closed subset of $\mathbb R^d$.

If $\Omega$ is as in the question, then $\overline\Omega$ is a particular instance of $M$. Conversly, if $M$ is bounded, then $M^\circ$ (the manifold interior of $M$ which can be shown to coincide with the topological interior of $M$) is a particular instance of $\Omega$. This was shown by Jack Lee here.


Let $E$ be a topological space and $I$ be a nonempty set.

Definition 1 $(B_i)_{i\in I}\subseteq M$ is called locally finite if $$\forall x\in E:\exists N\subseteq E:x\in N\wedge\left|\left\{i\in I:B_i\cap N\ne\emptyset\right\}\right|\in\mathbb N_0\tag{14}.$$

Let $(\Omega_i)_{i\in I}$ be an open cover of $E$.

Definition 2 $(\rho_i)_{i\in I}$ is called partition of unity subordinate to $(\Omega_i)_{i\in I}$ if

  1. $\rho_i:M\to\mathbb R$ is continuous and nonnegative for all $i\in I$;
  2. $\operatorname{supp}\rho_i\subseteq\Omega_i$ for all $i\in I$;
  3. $\left(\operatorname{supp}\rho_i\right)_{i\in I}$ is locally finite;
  4. $\sum_{i\in I}\rho_i=1$.

Assume $E$ is a $C^\infty$-manifold with boundary.

Definition 3 $(\rho_i)_{i\in I}$ is called $C^\infty$-partition of unity subordinate to $(\Omega_i)_{i\in I}$ if $(\rho_i)_{i\in I}$ is a partition of unity subordinate to $(\Omega_i)_{i\in I}$ and $\rho_i$ is $C^\infty$-differentiable for all $i\in I$.

Theorem 4 There is a $C^\infty$-partition of unity subordinate to $(\Omega_i)_{i\in I}$.

Proof: Introduction to Smooth Manifolds, Lemma 2.23.