# An stronger form of the existence of a smooth Urysohn function on $\mathbb R^n$

I proved the following form of the existence of a smooth Urysohn function::

proposition: For any compact set $K\subset\mathbb R^n$ and any open set $U\subset\mathbb R^n$ where $K\subset U$, there is a smooth($C^{\infty}$) function $f:\mathbb R^n\rightarrow [0,1]$ such that $f_{|_{K}}\equiv1$ and $\ \text{supp}(f)\subset U$.

For the its proof I used compactivity of $K$ strictly for defining $f$ .

Now, I wish to show that if $K$ be only a closed set then the above proposition still hold in the following form:

The problem: For any closed set $C\subset\mathbb R^n$ and any open set $U\subset\mathbb R^n$ where $C\subset U$, there is a smooth($C^{\infty}$) non negative function $f:\mathbb R^n\rightarrow [0,\infty)$ such that $f_{|_{C}}>0$ and $\text{supp}(f)\subset U$.

How?

• Partition of unity plus the compact case. – Daniel Fischer Aug 5 '15 at 8:07
• @ Daniel Fischer:I don't want to use partition of unity. I want to show it directly like to the first proposition. – bigli Aug 5 '15 at 8:14
• Why don't you want to use partitions of unity? – Daniel Fischer Aug 5 '15 at 8:18
• @ Daniel Fischer: Because of that in the book am reading, the problem is introdused befor than partition of unity. – bigli Aug 5 '15 at 8:43

Since we don't require that $f \equiv 1$ on $C$, we can get by without partitions of unity, although the construction takes us a few steps towards (smooth) partitions of unity.

For $k \in \mathbb{N}\setminus \{0\}$, define

$$C_k := \{ x \in C : k-1 \leqslant \lVert x\rVert \leqslant k\}.$$

Then each $C_k$ is compact (maybe empty), and

$$C = \bigcup_{k = 1}^\infty C_k.$$

Further define small open neighbourhoods

$$U_k := \bigl\{ x \in U : \operatorname{dist}(x,C_k) < \tfrac{1}{3}\bigr\}$$

of $C_k$.

Apply the proposition to all pairs $C_k \subset U_k$ to have a smooth $f_k \colon \mathbb{R}^n \to [0,1]$ with $f_k \equiv 1$ on $C_k$ and $\operatorname{supp} f_k \subset U_k$.

Since $U_k \subset \bigl\{ x \in \mathbb{R}^n : k-\frac{4}{3} < \lVert x\rVert < k + \frac{1}{3}\bigr\}$, every ball with radius $\leqslant \frac{1}{6}$ meets at most two of the $U_k$, hence

$$f(x) := \sum_{k = 1}^\infty f_k(x)$$

converges locally uniformly, and since every point has a neighbourhood on which all but finitely many $f_k$ vanish identically, $f$ is smooth, and $\operatorname{supp} f \subset \bigcup\limits_{k = 1}^\infty U_k \subset U$. And we have $f(x) \geqslant 1$ for $x\in C$.

• 1. Why $C = \bigcup_{k = 1}^\infty C_k$? 2. Why is convergent the series ? – bigli Aug 5 '15 at 9:22
• 1. For every $x \in C$, we have $\lVert x\rVert \geqslant 0$. Let $k = \lfloor \lVert x\rVert\rfloor + 1$, then $x \in C_k$. 2. I would like that you find the answer yourself. How does $f_k$ behave on $C_m$ when $m \neq k$? – Daniel Fischer Aug 5 '15 at 9:24
• My idea is that all of $U_k:=U$. – bigli Aug 5 '15 at 10:00
• Maybe, maybe not. Anyway $\bigcup U_k \subset U$, which is enough. The support of $f$ is contained in $\bigcup U_k$. – Daniel Fischer Aug 5 '15 at 10:02
• In fact, the family $\{ U_k : k \in \mathbb{N}\setminus \{0\}\}$ has a property that makes questions of convergence a non-issue. Try to make a sketch to see what. – Daniel Fischer Aug 5 '15 at 10:23