Let $\Omega\_1$, $\Omega\_2$ be bounded,convex, open domains with smooth boundary in $\mathbb{R}^2$ and $\overline\Omega\_1\subset\Omega\_2$. Suppose we are given a $C^1$ function $f:\overline\Omega\_1\cup(\mathbb{R}^2\setminus\Omega\_2)\rightarrow\mathbb{R}$ satisfies the following properties:


(1)$f\equiv 1$ on $\partial\Omega\_1$ and $f\equiv 0$ on $\partial\Omega\_2$.


(2)$\nabla f\cdot \nu\_k< 0$ on $\partial\Omega\_k$, where $\nu\_k$ is the unit outward normal vector to $\Omega\_k$, ($k=1,2$).


Now the question is that can we extend $f$ to be $C^1$ in $\mathbb{R}^2$ such that $0\leq f\leq 1$ and $|\nabla f|\neq 0$ in $\Omega\_2\setminus\overline\Omega\_1$?


I think the answer must be yes because I can imagine its figure as a frustum of a cone. I have read some references about the Whitney's extension theorem but they do not match. I would be very appreciate if anyone can provide the proof or references.



Looking for a reference for an extension problem of function
The boundary of a convex domain is a Jordan curve (this is the only property of
the boundaries that will be used). Then $\Omega\_2\backslash\Omega\_1$ is a topological ring, and by the well known theorem, there is a conformal map
$\phi:\Omega\_2\backslash\Omega\_1\to A$, where $A=\{ z:r<|z|<1\}$, for some $r\in(0,1)$. This map is continuous on the boundary. Now take
$f=-\log|\phi|/\log r$ as your extesion. Since $\phi'(z)\neq 0$, you have $\nabla f(z)\neq 0$. You only have to be careful when you use the normal vector, since you did not state any smoothness condition of $\partial\Omega\_j$.


Remark. Your condition 2 is redundant, since you did not say that you want the extended $f$ to be $C^1$ in $\Omega\_2$. But this also can be achieved, if desired, by the same method, which reduce the question
to the round ring. You only have to be careful when you mention the normal vector since you did not state any smoothness conditions of $\partial\Omega\_j$.


