2
$\begingroup$

Knowing that $\omega\Subset\Omega\subset\mathbb{R}^2$ (compactly included) are two open and bounded sets with $C^2$ boundary, is it true that for any function $\phi_0:\overline{\omega}\to\mathbb{R},\ \phi_0\in C^1(\overline{\omega})$ ($\overline{\omega}$ is the closure of $\omega$) we can find an extension $\phi:\Omega\to\mathbb{R}$ with $\phi\in C^1_{c}(\Omega)$ (compactly supported in $\Omega$)?

Motivation

If this type of result is true than we can obtain simple formulas for perimeter of implicitly defined curves in $\mathbb{R}^2$, putting $\phi_0$ the unit outer normal vector to a regular curve (which is defined in a neighborhood of the boundary of $\omega$). See here: Perimeter continuity of $BV$ sets on any sequence from $W^{1,1}$

What did I do?

I proved by standard methods (using cut-off functions and convolution) that we can obtain a mollifying sequence $\phi_n,\ n\in\mathbb{N}^*$ compactly supported in $\Omega$ that tends to $\phi_0$ in $L^1(\Omega)$, but I cannot prove that we can indeed have an extension.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

$\newcommand\de\delta\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$The answer is yes. Indeed, by Whitney's theorem, there is a function $f\in C^1(\mathbb R^2)$ whose restriction to $\overline\omega$ is $\phi_0$. Now take any open set $\Omega_0$ such that $\omega\Subset\Omega_0\Subset\Omega$ and then any function $g\in C^1(\mathbb R^2)$ such that $g=1$ on $\overline\omega$ and $g=0$ on $\mathbb R^2\setminus\Omega_0$. It remains to let $\phi$ be the restriction of $gf$ to $\Om$. (The conditions that $\om$ and $\Om$ have $C^2$ boundaries and that $\Om$ be bounded are not needed. The condition that $\om$ be bounded follows from $\om\Subset\Om$.)

Detail: The set $\Om_0$ and the function $g$ can be constructed as follows. Since $\om\Subset\Om$, there is some real $\de>0$ such that the closure $\overline{\om_{2\de}}$ of the $(2\de)$-neighborhood $\om_{2\de}$ of $\om$ is contained in $\Om$. Let then $\Om_0:=\om_{2\de}$ and $g:=1_{\om_\de}*\psi_\de$, where $\psi_\de$ is any nonnegative function in $C^1(\R^2)$ supported on the ball of radius $\de$ in $\R^2$ centered at $0$ and such that $\int_{\R^2}\psi_\de=1$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Is it true that the Whitney Extension of $\phi_0$ satisfies the inequality $|f|\leq |\phi_0|$ on $\overline{\omega}$? $\endgroup$
    – Bogdan
    Commented Jul 22, 2021 at 15:14
  • $\begingroup$ @Bogdan : Since $f$ and $\phi_0$ are continuous and $f=\phi_0$ on $\omega$, we have $f=\phi_0$ on $\overline\omega$. $\endgroup$
    – Iosif Pinelis
    Commented Jul 22, 2021 at 15:42
  • $\begingroup$ Sorry. I mean in $|f|\leq sup_{x\in\overline{\omega}} |\phi_0|$ in $\Omega$. I found an article which says something like this but I do not understand exactly. Here it is: core.ac.uk/download/pdf/82133103.pdf (the proposition at page 326). Is it indeed true that $f$ can be chosen that way? $\endgroup$
    – Bogdan
    Commented Jul 22, 2021 at 17:09
  • 1
    $\begingroup$ @Bogdan : I think this is impossible in general. Consider e.g. $\phi_0(x_1,x_2):=x_1$ for $(x_1,x_2)\in\overline\omega$, where $\omega$ is the open unit disk. Then $\sup_{\overline\omega}|\phi_0|=1$, but $f(x_1,0)>1$ for all $x_1>1$ close enough to $1$. Perhaps, you misunderstood the proposition. $\endgroup$
    – Iosif Pinelis
    Commented Jul 22, 2021 at 17:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .