$C^\infty$ version of Urysohn Lemma in $\Bbb R^n$

I'm trying to solve an exercise which its conclusion seems to be the title of this post. The exercise is:

1. Show that the function $h:\Bbb R\to [0,1[$ given by $$h(t)=\begin{cases} e^{-1/t^2} &\text{if } t\neq 0\\ 0 &\text{otherwise} \end{cases}$$ is $C^\infty$.
2. Show that the functions $$h_+(t)=\begin{cases} e^{-1/t^2} &\text{if } t\gt 0\\ 0 &\text{otherwise} \end{cases}\quad\text{and}\quad h_{-}(t)=\begin{cases} e^{-1/t^2} &\text{if } t\lt 0\\ 0 &\text{otherwise} \end{cases}$$ are $C^\infty$.
3. Show that the function $k:\Bbb R\to [0,1[$ given by $k(t)=h_-(t-b)h_+(t-a)$ is $C^\infty$ and positive for $t\in ]a,b[$.
4. Let $R$ the rentangle $]a_1,b_1[\times\cdots\times]a_n,b_n[$. Show that there is a $C^\infty$ function $g:\Bbb R^n\to [0,1[$ strictly positive on $R$.
5. Conclude that if $K$ is a compact subset of $\Bbb R^n$ and $U$ is an open neighborhood of $K$, there is a $C^\infty$ function $f:\Bbb R^n\to [0,1]$ such that $f_{|K}\equiv 1$ and its support is contained in $U$.

From 1.-4. I can prove that for any open and bounded set $O\subset \Bbb R^n$, there is a $C^\infty$ function with its support contained in $O$. So my first attempt was apply this to the open $U\setminus K$. Then I get a $C^\infty$ function $f$ that is $0$ (in particular) over $K$. If I just consider $\chi_K+f$ that function can fail to be $C^\infty$.

In a discussion on the chat, robjohn suggest this. It works fine, but then my question is:

Can 5. be proved by using 1.-4.? If yes, how?

• Where are you reaching a snag? That would be helpful in informing people as to where to begin advising. – Christopher A. Wong Sep 18 '12 at 21:46
• Note that 2. implies 1 by adding $h=h_++h_-$. Also 3. follows immediately from 2. and the chain rule. So all you have to show is 2. – nullUser Sep 18 '12 at 22:09

Since $K$ is compact and $U^C$ is closed, there is a positive distance, $\Delta$, from $K$ to $U^C$.

We can cover each point $k\in K$ with an open cube $Q_k(\Delta/\sqrt{n})$ centered at $k$ and side $\Delta/\sqrt{n}$. Note that the entire cube is within $\frac\Delta2$ of $k$. Since $K$ is compact, choose a finite subcover of these cubes $\{Q_{k_j}(\Delta/\sqrt{n}):1\le j\le N\}$. For each cube in this subcover, define the function $f_j$ mentioned in step 4 above on $Q_{k_j}(2\Delta/\sqrt{n})$ but divide it by its minimum on $\overline{Q}_{k_j}(\Delta/\sqrt{n})$. Thus, $f_j$ is supported in $Q_{k_j}(2\Delta/\sqrt{n})$ and is $\ge1$ on $Q_{k_j}(\Delta/\sqrt{n})$.

$\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $\ge1$ on $K$. Then $\phi\circ\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $1$ on $K$, where

$$\phi(x)=\frac{h_+(x)}{h_+(x)+h_+(1-x)}$$

Note that $\phi\in C^\infty$, and $\phi(x)=0$ for $x\le0$ and $\phi(x)=1$ for $x\ge1$.

• I just found what I've asked is just the Section 2.6 "Constructions of Smooth Functions" of the book Differentiable Manifolds by Lawrence Conlon. What do you mean by "Note that the entire cube is within $\frac{\Delta}{2}.$" – leo Sep 20 '12 at 4:21
• @leo: If the side of the cube centered at $k$ is $\Delta/\sqrt{n}$, then the distance from $k$ to the corner of the cube is $\Delta/2$. – robjohn Sep 20 '12 at 4:30
• @leo: you want the overlap! The important part is that $\sum f_j=0$ outside of $U$ and $\sum f_j\ge1$ on $K$. That way, when you apply $\phi$ it becomes $0$ outside $U$ and $1$ on $K$. – robjohn Sep 20 '12 at 19:28
• Ok, I see now. thanks – leo Sep 20 '12 at 21:09
• It's long time ago, but... Seems like your prove is incomplete. It's possible $x \in U \setminus K: F(x)=1$ ($F$ is your result function). Am i wrong? It would be much better to have something like "$=1$ if and only if $x \in K$". – Ximik May 28 '15 at 21:24

$\newcommand{\d}{\mathrm d}\newcommand{\supp}{\operatorname{supp}}$ This is just the end of the Section 2.6: Constructions of Smooth Functions, in Differentiable Manifolds by Lawrence Conlon.

Let $a\lt b$, and consider the function $k:\Bbb R\to [0,1[$ as in 3. Now, let $l:\Bbb R\to[0,1[$ defined by $$l(x)=\frac{\int_a^x k(t)\d t}{\int_a^b k(t)\d t}.$$ Notice that $l$ is no decreasing, $l\equiv 0$ in $]-\infty,a]$, $l\equiv 1$ in $[b,\infty[$, and $0\lt l(x)\lt 1$ for $x\in ]a,b[$.

Now, as in @robjohn's answer, it's possible to cover $K$ by a finite number of open intervals, say $\{I_j\}_1^N$, so that $\bar{I_j}\subset U$ for each $j\in\{1,\ldots,N\}$.

For each $j\in\{1,\ldots,N\}$, pick a soft function $g_j:\Bbb R^n\to [0,1[$ as in 4. Then the function $g:\Bbb R^n\to [0,1[$ given by $$g=g_1+\cdots+g_N$$ is $C^\infty$, positive over $K$ and with $$\supp(g)\subseteq U.$$ Now, $g$ attains its minimum over $K$, say in $x_\ast\in K$, thus $g(x_\ast)\gt 0$. Let $\delta=g(x_\ast)$.

Take an $l$ as the one discussed at the beginning with $a=0$ and $b=\delta$. The function $f=l\circ g$ have the desired properties.