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

  • 1
    $\begingroup$ 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). $\endgroup$ Jun 13, 2022 at 11:21
  • $\begingroup$ @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?) $\endgroup$ Jun 13, 2022 at 11:30
  • $\begingroup$ 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). $\endgroup$ Jun 13, 2022 at 13:49
  • $\begingroup$ @PietroMajer Ah true, but then can we upper-bound the Lipschitz constant in a similar fashion? $\endgroup$ Jun 13, 2022 at 13:57
  • 1
    $\begingroup$ @LSpice Thanks, its all fixed now; sorry about the typos. $\endgroup$ Jun 13, 2022 at 22:56

1 Answer 1


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 $6 / \delta_2$.


  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.
  • $\begingroup$ 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 $N / \delta_\ast$. $\endgroup$
    – shuhalo
    Aug 24 at 22:50
  • $\begingroup$ Is there a specific reason why one would use the squares of the distance functions rather than the distance functions themselves? $\endgroup$
    – shuhalo
    Aug 29 at 15:38
  • $\begingroup$ I used the square because that's what is used in the original question. @shuhalo Your question is probably better directed to the OP. $\endgroup$ Aug 30 at 1:01

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