# A lemma about extension of function

Definition 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 $\mathcal X^*$.

Lemma 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 $\Bbb R^2$.

Source 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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My idea is: Let $P$ denote the set of open sets $U\supseteq \mathcal M$ into which $f$ can be extended, partially ordered by inclusion. By hypothesis $P$ is nonempty. Let $\{U_i\}_{i\in I}$ be a chain in $P$ and consider $V=\bigcup\limits_{i\in I} U_i$. If we show that $f$ can be extended into $V$, then every chain in $P$ has an upper bound and so by Zorn's lemma $P$ has some maximal element. But any maximal element must have empty boundary, as otherwise we could extend $f$ to a larger open set. Since $\mathbb R^2$ is connected, the only nonempty set with empty boundary is $\mathbb R^2$. – Alex Becker Jul 1 '12 at 11:09
@Alex: That is quite a cool proof. But, what is actually going on there? – Jonas Teuwen Jul 1 '12 at 11:20
@JonasTeuwen Well, it's not a proof yet because I don't know how to show that $f$ can be extended into $V$. But the idea is to show that $f$ can be extended onto a set so large it can't be extended any more (using Zorn's) and using the criteria for $f$ and connectedness of $\mathbb R^2$ show that this set must be $\mathbb R^2$. – Alex Becker Jul 1 '12 at 11:23
@Alex Yes, I agree. But I was wondering, that if you can do that, what would it look like... AC is quite non-constructive. Perhaps if it works, it will work for a very large class of domains, but maybe in general, the domains are not so bad and you can have a (reasonably) explicit construction. – Jonas Teuwen Jul 1 '12 at 11:25
It just occurred to me that I might be misinterpreting the statement. Can any extension of $f$ be extended on the boundary of its domain? – Alex Becker Jul 1 '12 at 11:26

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 ${\mathbb R}^2$.

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It seems good, and I'll try to understand tomorrow. First, I should know some concepts, for example, smooth exactly means that $\mathcal C^n$? It seems that $\mathcal C^n$ is smooth enough. – Frank Science Jul 1 '12 at 14:14
How can you show that $g_k(z)$ is smooth, especially when $z$ tends to the boundary of $U_{2\epsilon_k}(p_k)$? The derivatives from the outer space of $U_{2\epsilon_k}$ is zero, but from the inner space, I don't know. – Frank Science Jul 2 '12 at 2:37
Pretty good, but some small dilemma. I'll post an answer, improved from yours, and you can read my answer, but I'll accept your answer. – Frank Science Jul 2 '12 at 5:08
@Frank Science: I think I've fixed the "small dilemma". At the same time I extended the POU to all of $\bar M$ to make the well-definedness of $f_*$ more obvious. – Christian Blatter Jul 2 '12 at 7:53

Intuitively:

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-).

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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 $$f^*(M)=\sum_{k=1}^mg_k(M)=\sum_{k=1}^m\psi_k(M)f(M)=\sum_{k=1}^m\psi_k(M)=1$$ Q.E.D

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