# Lipschitz-regularity of partition of unity

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 $$\{\psi_U\}_{U\in \mathcal{U}}$$?

• Note that if an open set $U$ can't be removed from the covering $\mathcal U$, i.e. $K\cap U\setminus \bigcup_{U\neq V\in\mathcal U}V\neq\emptyset$, then the $\psi_U$ supported in $U$ in any partition of unity subordinated to $\mathcal U$ must have Lipschitz constant at least 1/diam(U). Jun 13, 2022 at 11:21
• @PietroMajer This I am having a lot of trouble seeing; I was also suspecting that $L=1/r$ where $r>0$ is the Lebesgue number of $\mathscr{U}$. (Ideally, do you have a reference to this or your claim?) Jun 13, 2022 at 11:30
• Well, what I wrote is just this: if, for a given $x\in K$, $U$ is the only open set in $\mathcal U$ such that $x\in U$, then $\psi_V(x)=0$ for all $V\neq U$, so $\psi_U(x)=1$; since $\psi_U(y)=0$ on $\partial U$ we need the Lipschitz constant of $\psi_U$ at least 1/diam(U). Jun 13, 2022 at 13:49
• @PietroMajer Ah true, but then can we upper-bound the Lipschitz constant in a similar fashion? Jun 13, 2022 at 13:57
• @LSpice Thanks, its all fixed now; sorry about the typos. Jun 13, 2022 at 22:56

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 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.
• This has kept me busy for a day. The Lipschitz constant can also be written $6 \sqrt{N} / \delta_1$, which looks like a much better upper bound than what the published literature. Then it finally dawned on me that $\delta_1$ is quite a conservative lower bound for the Lebesgue number. It may generally only be the $N$-th part of the best possible Lebesgue number $\delta_\ast$. The best known bounds actually scale like $N / \delta_\ast$. Aug 24 at 22:50
• Is there a specific reason why one would use the squares of the distance functions rather than the distance functions themselves? Aug 29 at 15:38
• I used the square because that's what is used in the original question. @shuhalo Your question is probably better directed to the OP. Aug 30 at 1:01