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I have a function $f\in C^\infty: \mathbb{R}^n \to \mathbb{R}$ with the properties :

1) $f(x) = 0$, when $\Vert{x}\Vert \le 1$,

2) $0 < f(x) < 1$, when $1 < \Vert{x}\Vert < 2$,

3) $f(x) = 1$, when $\Vert{x}\Vert \ge 2$.

I want to generalize my $f$ to $g\in C^\infty: \mathbb{R}^n \to \mathbb{R}$, so that for some closed subset $C$ and some open subset $U$ ($C \subset U$), one has

1) $g(x) = 0$, when $x \in C$,

2) $0 < g(x) < 1$ when $x \in U / C$,

3) $g(x) = 1$ when $x \not\in U$.

What is the simplest way to do this?

Ximik
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1 Answers1

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For a set $E$ in $\mathbb{R}^n$ and for any point $x$ in $\mathbb{R}^n$ define $$ d(x,E)=\inf_{y\in E}d(x,y) $$ where $d(x,y)$ is the usual Euclidean distance between $x$ and $y$.

Easy things to show about this:

1.) $d(x,E)=d(x,\overline{E})$, where $\overline{E}$ is the closure of $E$.

2.) $d(x,E)=0$ if and only if $x\in E$.

3.) $x\to d(x,E)$ is a continuous function for any set $E$.

Now given any two disjoint closed sets $A$ and $B$ in $\mathbb{R}^n$, define $f:\mathbb{R}^n\to\mathbb{R}$ by $$ f(x)=\frac{d(x,A)}{d(x,A)+d(x,B)} $$ We have $f^{-1}(0)=A$ and $f^{-1}(1)=B$.

Edit: Note this gives a continuous function, but I don't think it will necessarily always be $C^1$ or whatever.

Edit 2: I realize that this no longer meets all the desired properties (as the function is not $C^\infty$), but I think this construction gives a simple function that is at least continuous. It also ties in with Urysohn functions. So hopefully with this in mind I can keep this answer up for viewing.

TomGrubb
  • 13,177
  • Your hint is right. I've found very close solution here http://math.stackexchange.com/questions/198748/c-infty-version-of-urysohn-lemma-in-bbb-rn – Ximik May 27 '15 at 20:59