A question on smooth version of Urysohn lemma

Question

Hi everyone: Let $ B(x_{0},R) $ be an arbitrary open ball in $ \mathbb{R}^{N} $ with $ N\geq2 $. We know that for every $ \varepsilon>0, $ there exists a test function $ \phi_{\varepsilon} $ with compact support in $ B(x_{0},R+\varepsilon)$ that is equal to 1 on $ B(x_{0},R)$. My question is: are the family $ \phi_{\varepsilon} $ uniformly bounded?

Answer 1

The $\varphi_\varepsilon $ may be chosen to be uniformly bounded, namely you can arrange that $0\leq \varphi_\varepsilon \leq 1$. Much more generally, given two disjoint closed (but not necessarily compact) sets $A,B\subset \mathbb{R}^{N} $ there exists $ \varphi \in C^\infty (\mathbb R^N)$ with $0\leq \varphi \leq 1, \varphi \mid A=0$ and $\varphi\mid B=1$, exactly as in the topological Urysohn theorem.

Answer 2

You can certainly make it have bound $1$. Just convolve the indicator function of $B(x_0, R+\epsilon/2)$ with a smooth nonnegative function supported in $B(0,\epsilon/2)$ whose integral is $1$.