A lemma about extension of function

Suppose that $f(M)$ is a $\mathcal C^n$-function whose domain is $\mathcal X$. If $f^*(M)$ is a $\mathcal C^n$-function whose domain is $\mathcal X^*$, and $f(M)=f^*(M)$ whenever $M\in\mathcal X\cap\mathcal X^*$, we call that $f^*$ is an ($\mathcal C^n$-)extension of $f$, and $f$ is ($\mathcal C^n$-)extended into $\mathcal X^*$.

Given that $f(x,y)$ is $\mathcal C^n$ ($n\ge1$) function on some bounded open set $\mathcal M\subset\Bbb R^2$, whose boundary is $\mathcal L$, and for each point on $\mathcal L$, there’s a neighborhood into which $f$ could be $\mathcal C^n$-extended. We conclude that $f$ could be extended into $\Bbb R^2$.

Григорий Михайлович Фихтенгольц

I found that the proof on the book was so complicated for me to understand, so I’m looking for some explanation, as intuitive as possible. Can anyone help me? Thanks a lot!

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As $\bar M=M\cup L$ is compact I suggest working with a so called partition of unity:

For each $p\in \bar M$ there is an $\epsilon>0$ (depending on $p$) such that $f$ can be smoothly extended to $U_{3\epsilon}(p)$. The family $\bigl(U_\epsilon(p)\bigr)_{p\in L}$ is an open covering of $\bar M$; therefore there exist points $p_k\in \bar M$ $\ (1\leq k\leq N)$ such that $$\bar M\subset\Omega:=\bigcup_{k=1}^N U_{\epsilon_k}(p_k)\ .$$
For each $k\in[N]$ there is a $C^n$- (even a $C^\infty$-) function $\phi_k: {\mathbb R}^2\to [0,1]$ with $\phi_k(z)\equiv1$ for $z\in U_{\epsilon_k}(p_k)$ and $\phi_k(z)\equiv0$ outside $U_{2\epsilon_k}(p_k)$. (First one has to construct once and for all a $C^n$-function $\chi$ which is $\equiv1$ for $0\leq t\leq1$ and $\equiv0$ for $t\geq2$. Then put $\phi_k(z):=\chi\bigl({|z-p_k|\over\epsilon_k}\bigr)$. )

For technical reasons we need the auxiliary function
$$\phi_*(z):=\prod_{k=1}^N \bigl(1-\phi_k(z)\bigr)\in[0,1]\ .$$
It is $\equiv0$ on $\Omega\supset \bar M$ and $=1$ in all points $z$ where the $\phi_k$ simultaneously vanish. Now put
$$\psi_k(z):={\phi_k(z)\over \phi_*(z)+\sum_{k=1}^N\phi_k(z)}\qquad(1\leq k\leq N)\ .$$
Each $\psi_k:{\mathbb R}^2\to[0,1]$ is smooth and $\equiv0$ outside $U_{2\epsilon_k}(p_k)$, and what is essential: The $\psi_k$ sum to $1$ on $\Omega$.

For each $k$ choose an extension $f_k$ of $f$ to $U_{3\epsilon_k}(p_k)$ and put
$$g_k(z):=\cases{\psi_k(z)f_k(z) &$\bigl (z\in U_{3\epsilon_k}(p_k)\bigr)$\cr 0 & (otherwise)\cr}\ .$$
The $g_k$ are smooth in all of ${\mathbb R}^2$, and one easily checks that
$$f_*(z):= \sum_{k=1}^N g_k(z)$$ is a smooth extension of $f$ to all of ${\mathbb R}^2$.


The hypothesis that given a point $x$ on the boundary $\mathcal L$ there exists a neighborhood of that point where you have your extension helps a lot, I believe. So, for each $x$ on the boundary take the union of all the neighborhood of $x$. This will be an open set, and intuitively some “open band” around your domain, that also enters the domain itself a bit.

Good, so what do we do now? Take for each $x$ on the boundary of your domain, the minimal distance $d_x = d(x, \complement E)$ to the complement of your domain including the extension with the open band. So, you might have some trouble now. For example, does this distance go to $0$? It might if you do it this way. However, what you often can do is some compactness argument and take a finite subcover and you will be great. In that case you will have a positive distance. The idea now is that you make an indicator function (a function that is either $1$ or $0$) which is $1$ up to the boundary of your original function and extends a bit into your “open band” (which has to be reduced to make things still positive).

Right, the idea then is that you have a nice function which is not yet smooth at all but it does nothing to the function domain itself (after multiplication) but it does cut-off the extended domain slightly. A nice trick now is convolution. You take something as smooth as a babys butt and then you convolve! The result will have similar properties as your original function but will be smooth. However, be careful.

I am fully aware that this is not a full proof (or not at all) but I wanted to give a possible idea.

Some possible details:
Note that the boundary of an open set is closed. This is the set difference of the closure of the set itself and its complement. Hence, we have a closed set.

Good, now our set including the boundary is compact because it is both closed and bounded. This means that for any open cover we can select finite subcover. This is something we would like to have as this means we can cover our set with finitely many balls.

Then we invoke our condition of having balls on the boundary. Let us take all these balls and make an open cover for our set. We must be careful as although the boundary is closed it does not have to be bounded in general I believe. I’m not very familiar with these topological quirks, so I would have to think about it. Perhaps the boundary can be some space-filling curve and that would be quite annoying, to say the least 8-).

In any way, you have to be careful, you would only like to reduce the cover of the boundary to a finite subcover, if you just naively add this to your original cover it might just happen that at some boundary points there is “no space left”. Just consider say a circle with a cover of your boundary. If your cover has the circle itself this is what will remain.

All I want to do is make sure there is some slightly larger set than your original domain where I am allowed to change the values of the function. That’s the idea. The rest are the technicalities 8-).

It’s just an outline of improved Blatter’s proof.

Conclusion 0 Given $f$ is a $\mathcal C^n$-function whose domain is the open set $\mathcal X$, and $f$ could be $\mathcal C^n$-extended into the open set $\mathcal X^*$, then $f$ could be $\mathcal C^n$-extended into $\mathcal X\cup\mathcal X^*$.

Conclusion 1 If $f$ is $\mathcal C^n$ on open set $\mathcal M_1$ and $\mathcal M_2$, then $f$ is $\mathcal C^n$ on $\mathcal M_1\cup\mathcal M_2$.

Conclusion 2 Let $\overline{\mathcal M}=\mathcal M\cup\mathcal L$, we have $\overline{\mathcal M}$ is an closed set.

Conclusion 3 For each point in $\overline{\mathcal M}$, there’s a neighborhood into which $f$ could be $\mathcal C^n$-extended.
(Hint: for each $p\in\mathcal M$, there’s a neighborhood completely in $\mathcal M$)

Conclusion 4 For each point $P$ in $\Bbb R^2$ and $r>0$, there’s a $\mathcal C^n$-function $\phi:\Bbb R^2\to\Bbb R$, where $\phi(M)=1$ whenever $|MP|\le r$, and $\phi(M)=0$ whenever $|MP|\ge2r$.

Now let’s start the proof. As in conclusion 3, we can find a open neighborhood for each point in $\overline{\mathcal M}$. These neighborhoods cover the closed set $\overline{\mathcal M}$, so there’re finity many neighborhoods, say $U_{3r_1}(P_1),\ldots,U_{3r_m}(P_m)$ (where 3 in $3r_k$ is important), covering $\overline{\mathcal M}$. As in conclusion 4, we can define a $\mathcal C^n$-function $\phi_k$ for each point $P_k$ and radius $r_k$.

Suppose that $\phi_*(M)=(1-\phi_1(M))\cdots(1-\phi_m(M))$, $\psi_k(M)=\phi_k(M)/(\phi_*(M)+\sum_{j=1}^m\phi_j(M))$, we have $\phi_*$, $\psi_k$ is $\mathcal C^n$, and $\psi_k(M)=0$ when $M$ is outside $U_{2r_k}(P_k)$.

For each $k$ choose an extension $f_k$ of $f$ to $\mathcal M\cup U_{3r_k}(P_k)$ (for conclusion 0) and let
$$g_k(M)=\begin{cases}\psi_k(M)f_k(M),&\qquad M\textrm{ is in $\mathcal M\cup U_{3r_k}(P_k)$}\\0&\qquad M\textrm{ is outside $U_{2r_k}(P_k)$}\end{cases}$$

Hint: You’re confused about the disjoint of two cases until you realize that $\psi_k(M)=0$ whenever $M$ is outside $U_{2r_k}(P_k)$.

$g_k$ is $\mathcal C^n$ because of conclusion 1. Now we let $f^*(M)=g_1(M)+\cdots+g_m(M)$, and we have $f^*$ is $\mathcal C^n$.

For each $M\in\mathcal M$, we have