# $C^1$ extension with compact support

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.

## 1 Answer

$$\newcommand\de\delta\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$$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 $$\int_{\R^2}\psi_\de=1$$.

• Is it true that the Whitney Extension of $\phi_0$ satisfies the inequality $|f|\leq |\phi_0|$ on $\overline{\omega}$? Commented Jul 22, 2021 at 15:14
• @Bogdan : Since $f$ and $\phi_0$ are continuous and $f=\phi_0$ on $\omega$, we have $f=\phi_0$ on $\overline\omega$. Commented Jul 22, 2021 at 15:42
• 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? Commented Jul 22, 2021 at 17:09
• @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. Commented Jul 22, 2021 at 17:34