Let $K$ be a compact subset of $\mathbb{R}^n$ and $\mathcal{U}$ be a finite collection of open subsets covering $K$ satisfying the minimality property: *for every $U\in \mathcal{U}$, the sub-collection $\mathcal{U}-\{U\}$ does not cover $K$.*


*Notation:* Given any subset $A\subseteq K$ denote the relative complement $K-A:=\{x\in K:\, x\not\in A\}$. Given a point $x\in K$ and a subset $A\subseteq K$ let $\|x-A\|:=\inf\_{a\in A} \|x-a\|$.


Is there a condition on the open cover $\mathcal{U}$ such that the partition of unity $\{\psi\_U\}\_{U \in \mathcal{U}}$ is such that each
$$
\psi\_U(x):= \frac{\|x-(K-U)\|^2}{\sum\_{V \in \mathcal{U}}\,\|x-(K-V)\|^2}
$$
is Lipschitz?


If so, what are explicit bounds on the Lipschitz constant of the functions $\{\psi\_U\}\_{U\in \mathcal{U}}$?



Lipschitz-regularity of partition of unity
For convenience write


$$ \phi\_U(x) = \| x - (K-U)\|. $$


Let $N = |\mathcal{U}|$ the number of open sets in the cover.
Define
$$ f\_1(x) = \frac{1}{N} \sum \phi\_U(x), \qquad f\_2(x) = \left( \frac1N \sum \phi\_U^2(x) \right)^{1/2} $$
Note that since $\mathcal{U}$ is a cover we have
$$ 0 < f\_1(x) \leq f\_2(x) \leq \sqrt{N} f\_1(x) $$
Let
$$ \delta\_1 = \inf f\_1, \quad \delta\_2 = \inf f\_2 $$
and we also have $0 < \delta\_1 \leq \delta\_2 \leq \sqrt{N} \delta\_1 $. Note also that $\delta\_1$ is a Lebesgue number of the covering $\mathcal{U}$.


Now, you've defined
$$ \psi\_U(x) = \frac{\phi\_U^2(x)}{N f\_2^2(x)} $$
so
$$ N(\psi\_U(x) - \psi\_U(y)) = \frac{\phi\_U^2(x) f\_2^2(y) - \phi\_U^2(y) f\_2^2(x) }{f\_2(x) f\_2(y)} $$
$$ = \frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)} + \frac{\phi\_U^2(y) (f\_2^2(y) - f\_2^2(x))}{f\_2(x)^2 f\_2(y)^2}. $$
So we have the following estimate
$$ N |\psi\_U(x) - \psi\_U(y)| \leq \left|\frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)}\right| + N \left| \frac{f\_2^2(y) - f\_2^2(x)}{f\_2(x)^2} \right| $$


Now, it suffices to consider only those $x,y$ with $\|x-y\| \leq \delta \leq \delta\_1$ for a fixed $\delta$. This is because


1. We know that $0 \leq \psi\_U(x) \leq 1$ and hence if $\|x-y\| > \delta$, then the corresponding difference quotient is $\leq \frac{1}{\delta}$.
2. On the other hand, if we define
$$ \epsilon = \inf\_{U\in \mathcal{U}} d( U \setminus \cup (\mathcal{U} \setminus \{U\}), K \setminus U ) $$
then essentially the [same argument of Pietro Majer](https://mathoverflow.net/questions/424603/lipschitz-regularity-of-partition-of-unity#comment1091500_424603) shows that the Lipschitz constant of the family is at least $1/\epsilon$. (Note: minimality of the covering is used to get that $U\setminus \cup(\mathcal{U}\setminus \{U\}) \neq\emptyset$.) Note that $\epsilon \geq \delta\_1$. So we don't expect a much better upper bound estimate of the Lipschitz constant than $1/\delta$.


Now, $\phi\_U(x)$ is well-known to be $1$-Lipschitz. So we have that $$\phi\_U(x) + \phi\_U(y) \leq 2 \phi\_U(x) + \delta \leq (2N+1) f\_1(x)$$
So
$$ \left|\frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)}\right| \leq \frac{2N+1}{\delta\_2} \|x - y\| $$
Similarly, we have
$$ |f^2\_2(x) - f^2\_2(y)| \leq \frac1N \sum |\phi\_U(x) - \phi\_U(y)| |2\phi\_U(x) + \delta| \leq \|x-y\| \cdot (\delta + 2 f\_1(x)) $$
so
$$ N \left| \frac{f\_2^2(y) - f\_2^2(x)}{f\_2(x)^2} \right| \leq \frac{3N}{\delta\_2} \|x-y\|$$


So putting everything together we find that the Lipschitz constant of the family $\psi\_U$ are uniformly bounded by $6 / \delta\_2$.


### Summary


1. For the finite collection $\mathcal{U}$, each function in the corresponding partition of unity is necessarily Lipschitz.
2. The uniform Lipschitz constant can be controlled by a quantity similar to the Lebesgue number of the covering.


