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 $\text{supp}(f)\subset U$.


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

  • $\begingroup$ 1. Why $C = \bigcup_{k = 1}^\infty C_k$? 2. Why is convergent the series ? $\endgroup$ – bigli Aug 5 '15 at 9:22
  • $\begingroup$ 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 $m \neq k$? $\endgroup$ – Daniel Fischer Aug 5 '15 at 9:24
  • $\begingroup$ My idea is that all of $U_k:=U$. $\endgroup$ – bigli Aug 5 '15 at 10:00
  • $\begingroup$ Maybe, maybe not. Anyway $\bigcup U_k \subset U$, which is enough. The support of $f$ is contained in $\bigcup U_k$. $\endgroup$ – Daniel Fischer Aug 5 '15 at 10:02
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Aug 5 '15 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.