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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.

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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 $\partial\Omega_j$.

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  • $\begingroup$ Sorry I have edited the question. What I want is $f$ to be $C^1$. And in this case it seems that $f$ cannot simplily be taken as a function of $\phi$. Could you give me more hints? Thank you. $\endgroup$
    – Jeff J
    Commented Sep 4, 2021 at 12:55
  • $\begingroup$ As I wrote you need some condition on smoothness of $\partial\Omega_j$ for your condition 2 to make sense. $\endgroup$ Commented Sep 4, 2021 at 13:01
  • $\begingroup$ I have supposed that $\Omega_k$ has smooth boundary in the beginning. $\endgroup$
    – Jeff J
    Commented Sep 4, 2021 at 13:32
  • $\begingroup$ If the boundaries are smooth, the same method works, since in this case $\phi'$ extends to the boundaries and is different from $0$. You just have to work a bit to construct $h$ in the round ring so that $f=h(\phi)$ solves the problem. $\endgroup$ Commented Sep 4, 2021 at 13:34
  • $\begingroup$ So the question can be reduced to $\Omega_1=B_r(0)$ and $\Omega_2=B_1(0)$. The key is even in the reduced case such an extension can be done? It sounds like obvious but I do not know how to prove it. $\endgroup$
    – Jeff J
    Commented Sep 4, 2021 at 14:15

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