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Suppose I have a compact set $K \subset B_1(0) \subset \mathbb{R}^n$. Can I always find a family of open balls $\{B_{r_j}(x_j)\}$ such that

  1. $x_j \in K$ and $B_{r_j}(x_j) \subset B_1(0)$ for each $j$;

  2. $K \subset \bigcup_j B_{r_j}(x_j)$; and

  3. The collection $\{B_{2r_j}(x_j)\}$ has bounded overlap, that is to say that there is some number $N = N(n)$ such that each point of $\bigcup_j B_{2r_j}(x_j)$ lies in at most $N$ balls from the collection. In other words, $$ \sum_j \mathbf{1}_{B_{2r_j}(x_j)} \leq N $$


The motivation behind a question like this is a fairly common situation where at each point $y \in K$ I can prove an estimate like $$ \int_{B_{r/2}(y)} f \leq C\int_{B_r(y)} g, $$ but what I really want is an estimate of the form $$ \int_{\{x\ :\ \mathrm{dist} (x,K) < r/10\}} f \leq C \int_{\{x\ :\ \mathrm{dist} (x,K) < 10r\}} g $$ The numbers 1/2 and 10 are not important, but to do this it is natural to try to find a collection of balls as described. In specific cases I have constructed this collection myself but now I wonder if there is a general lemma that I happen not to have heard of.

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    $\begingroup$ If B is the ball centered at x of radius r, I would usually use 2B to mean the ball centered at 2x of radius 2r, but to make your question and motivation make sense it seems you're using 2B to mean the ball centered at x of radius 2r. Correct? $\endgroup$ – Yoav Kallus Oct 17 at 12:57
  • $\begingroup$ yes; perhaps I should edit as such...we must come from different "schools" $\endgroup$ – T_M Oct 17 at 12:59
  • $\begingroup$ Unless I am misreading the question, it seems pretty clear there is such N(n). Take some lattice covering of R^n by equal sized balls for each n. Doubling the balls gives you another covering where each point is covered by no more than some number of balls particular to your chosen original covering. $\endgroup$ – Yoav Kallus Oct 17 at 13:01
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    $\begingroup$ Ah. Seems I left off the crucial ingredient! The centres must be on K as hinted at in the motivation; my apologies $\endgroup$ – T_M Oct 17 at 13:05
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    $\begingroup$ I think this still leaves open the trivial solution of covering K with a single large ball, no? Do you want to specify that the radii need to be arbitrarily small? $\endgroup$ – Yoav Kallus Oct 17 at 13:13
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Let $r$ be less than half the distance from $K$ to the complement of $B_1(0)$. Start with a ball of radius $r$ centered at each point of $K$. By https://en.wikipedia.org/wiki/Besicovitch_covering_theorem there is a subcover of $K$ by balls that is a union of $C_n$ collections $A_i$, where each $A_i$ consists of pairwise disjoint balls of radius $r$. For each collection $A_i$ of disjoint balls, the balls of twice the radius can overlap at most with multiplicity $3^n$ because a ball of radius $3r$ around the point of overlap can contain at most $3^n$ disjoint balls of radius $r$. So overall we conclude that $N(n) \le 3^n C_n$.

PS Here is a self-contained argument that avoids using the Besicovitch covering lemma:

Given $r$, let $Z=\{z_j\}_{j=1}^m$ be a maximal subset of $K$ such that the open balls $\{B(z,r/2): z \in Z\}$ are pairwise disjoint. Then the open balls $\{B(z,r): z \in Z\}$ form a cover of $K$ (If $y \in K$ was not covered, that would contradict the maximality of $Z$). Finally, the collection of balls $S=\{B(z,2r): z \in Z\}$ has overlap multiplicity at most $5^n$: Indeed, if a point $w$ is in the intersection of $L$ balls from $S$, then $B(w,5r/2)$ contains $L$ disjoint balls of radius $r/2$; comparing volumes gives $L \le 5^n$.

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  • $\begingroup$ OK I think I can see this! Thank you. While the constant is not important to me: When I consider this point of overlap in the enlarged collection, should I maybe consider the ball of radius $3r$ about it? (i.e. the centres are < 2r away so to contain the r balls about them..?) $\endgroup$ – T_M Oct 17 at 13:24
  • $\begingroup$ Right. Corrected. $\endgroup$ – Yuval Peres Oct 17 at 13:29

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