0
$\begingroup$

This is mainly a reference request. Consider a finite collection of (let's say, for simplicity) of open balls $B_i, i = 1, 2, ..., m$ in (again, for simplicity) $\mathbb{R}^n$. I am looking for results that state the following: from the given collection, we can select a subcollection $B_1, B_2, ....,B_k, k \leq m$ (relabelled) such that $\cup_{i = 1}^mB_i \subset \cup_{j = 1}^k rB_j$, with $r > 1$ being very close to $1$, and such that the new subcollection has as little intersection between $B_1, B_2,...B_m$ as possible. Of course, phrases "very close to $1$" and "as little intersection" are very vague. I am looking for references which have quantitative results of this type.

$\endgroup$
1
  • $\begingroup$ Is the corresponding version of the Besikovich covering lemma good enough for your purposes? (the corresponding nearly disjointness property is that every point in the space is contained in at most $N(r,n)$ balls $rB_j$ or, alternatively, the covering family can be partitioned into $N(r,n)$ subfamilies consisting of pairwise disjoint balls) $\endgroup$
    – fedja
    Nov 8, 2015 at 21:56

1 Answer 1

1
$\begingroup$

A possible answer in the spirit of your question is the following result of Chanillo and Muckenhoupt: given a number $\delta \in (0, 1/2)$ and a finite collection of balls $B_\alpha$ in $\mathbb{R}^n$, we can find a subcollection $B_1, ..., B_N$ such that $$\bigcup_{\alpha} B_\alpha \subset \bigcup^N_{i = 1}(1 + \delta)B_i,$$ and $$\Sigma_{i = 1}^N \chi_{B_i}(x) \leq 4^n\delta^{-n}.$$ As you can see, the second condition captures "minimal intersection" quantitatively. The proof is pretty nice and elementary, see Lemma 3 of this paper.

Additional comment: The main aim of the paper cited is to improve the famous Donnelly-Fefferman estimates of the BMO norm of $\text{log}(|u|)$, where $u$ is an eigenfunction of the Laplacian $-\Delta$ on a compact Riemannian manifold. Covering lemmas of this type are somewhat common in harmonic analysis though.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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