Let $x_0$ be a point on the plane and $2^K$ be a large number. Given a family $\mathcal C$ of circles with equal radius $R$ satisfying the following property:
- The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.
- The circles have bounded overlap, that is, there is an absoulute constant $M$ such that for all $x$ $$ \sum_{c\in \mathcal C}1_{c}(x)\leq M. $$ Then what can we say about the cardinality of $\mathcal C$, in terms of $K$?
Maybe it is easier to think of the case $M=1$ first. I came across this question when I am reading A study guide for the l 2 decoupling theorem by Bourgain and Demeter. It is one step for an inequality which they say it is easy to verify, but I found it looks like much deeper.
A very rough upper bound on the cardinality of $\mathcal{C}$ comes from the fact that each circle $c\in\mathcal{C}$ covers an area of at least $\frac{\pi}{2}R^2-\varepsilon $ of the annulus $$A:=\{x\in\Bbb{R}^2: 2^K R<x<2^{K+1}R\},$$ that its center must be inside, where $\varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2\pi$ and the circles cover the annulus at most $M$ times, so there are at most $$\frac{M2^{2K}R^2\pi}{\frac{\pi}{2}R^2-\varepsilon}\approx M2^{2K+1},$$ circles in $\mathcal{C}$, where I just waved the $\varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $\mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.