# Stronger version of Besicovitch covering theorem

I'm wondering if the following strengthening of the Besicovitch covering theorem holds: Suppose $$A\subset\mathbb R^n$$ is a bounded subset and suppose $$x\mapsto r_x$$ is a function $$A\to(0,\infty)$$. Is it possible to choose constants $$0<\lambda<1$$ and $$N\in\mathbb N$$ (both depending only on the dimension $$n$$) so that there exists a countable subset $$C\subset A$$ such that the following two statements hold?

(1) the collection of balls $$\{B(x,\lambda r_x)\}_{x\in C}$$ covers $$A$$ and,

(2) we have the pointwise inequality $$\sum_{x\in C} \chi_{B(x,r_x)}\le N$$ (where $$\chi_E$$ denotes the indicator function of a set $$E$$). In words, no point of $$\mathbb R^n$$ is contained in more than $$N$$ of the balls $$B(x,r_x)$$ (with $$x\in C$$).

If we replaced $$\lambda$$ by $$1$$, this statement would be true as a consequence of the usual Besicovitch covering theorem. I'm somewhat stuck trying to modify that proof to get this stronger statement (I would just like $$\lambda$$ to be strictly smaller than $$1$$, it can be as close to $$1$$ as we like).

Is this stronger version with some $$0<\lambda<1$$ true? If not, is there an easy counterexample? If it is true, a proof (or sketch/hint/reference) would be appreciated!

If it helps to answer the question, I don't mind adding the additional assumption that $$A$$ is actually a compact set (i.e., its closed in addition to being bounded) and $$x\mapsto r_x$$ is a continuous (or even Lipschitz) function.

I think the statement is false. Consider $$n = 1$$. Let $$A = \{x_i\}_{i \in \mathbb{N}}$$ where $$x_i = 2^{-i}$$.

Define $$r_i$$ as follows: if $$k^2 \leq i < (k+1)^2$$, let $$r_i = 2^{-i} - 2^{-(k+1)^2}$$.

Notice that $$x_{k^2}$$ only belongs to $$B(x_{k^2}, r_{k^2})$$ and no other ball. So any cover of $$A$$ must include all of $$B(x_{k^2}, \lambda r_{k^2})$$.

Let $$\eta = \lceil-\log_2(1 - \lambda)\rceil$$, for $$\lambda < 1$$. Note that $$\eta \in \mathbb{N}$$.

Observe that $$B(x_i, \lambda r_i) \subset B(x_i, \lambda x_i)$$, which means that $$x_{i + \eta} \not\in B(x_i, \lambda r_i)$$.

Now, suppose we have a cover of $$A$$. It must include the point $$x_{k^2}$$. The argument above implies that $$x_{k^2+\eta}$$ is not included. By construction if $$x_{k^2 + \eta}$$ belong to $$B(x_i, \lambda r_i)$$ for some $$i$$, then $$i \leq k^2 + \eta$$. This implies that $$x_{k^2 + 2\eta}$$ cannot belong to whichever ball that covers $$x_{k^2+\eta}$$.

This argument can be iterated $$\sigma$$ times, where $$\sigma$$ is the biggest integer such that $$k^2 + \sigma \eta < (k+1)^2$$, or that $$\sigma = \lfloor (2k+1) / \eta\rfloor$$.

Therefore for any fixed $$\lambda < 1$$, choosing a starting $$k$$ sufficiently large we see that the cover for $$A$$ must include at least $$\lfloor (2k+1) / \eta \rfloor$$ many $$x_i$$ with $$i$$ between $$k^2$$ and $$k^2 + 1$$. The corresponding balls for all these $$x_i$$ have non-empty total intersection, on which set the sum of the characteristic function is at least $$\sigma$$, which can be made arbitrarily large.

Compactness of $$A$$ will not help. I can add to the set $$A$$ the origin which we label as $$x_\infty$$. For any fixed $$\lambda, N$$ I can choose a sufficiently large $$k$$ such that the corresponding $$\sigma > N$$. Then if I take the corresponding $$r_\infty \ll 2^{-(k+2)^2}$$ we would have a counterexample with a compact $$A$$.

As $$A$$ is discrete, obviously $$x_i \mapsto r_i$$ is continuous; even in the compact case we can modify the definitions of $$r_i$$ for $$i > (k+2)^2$$ where $$k$$ is the critical value above so that the radius function is continuous at $$x_\infty$$.

• Very nice! It seems to that you're able to make this construction (and get continuity, even Lipschitzness, of the radius function) because $A$ is "almost discrete" in your example. Do you think an example with $A$ being a compact interval (or ball for $n>1$) can be constructed with the radius function being continuous/Lipschitz? – Mohan Swaminathan Jun 20 '19 at 14:09
• @MohanSwaminathan: I am not sure. I think one should be able to fill in the "gaps" between the discrete points, All the gaps are pretty straightforward EXCEPT the ones between $(2^{-k^2}, 2^{- k^2 + 1})$. The naive extension will of course cause the $r_x$ to be no longer continuous. I think in that interval we can probably linearly change $r_x$ from (on the right) $2^{-k^2}$ to (on the left) $2^{-k^2} - 2^{-(k+1)^2}$. Then up to some (pretty annoying looking) detail checking, the argument should still work, maybe with some small additional changes. – Willie Wong Jun 21 '19 at 16:28