I was studying this paper by J. Duoandikoetxea et al. The paper concerns some inequalities related to the $k$-plane transform and a maximal operator related to it. In one of the results, the authors want to prove the equivalence of the usual Hardy-Littlewood maximal function with an annular maximal function. The idea behind this is to get an estimate of a covering of an annular region by balls having the same radius as that of the annular region.
To make notations precise, I will follow the authors' symbols. Given $0 \leq a < b < \infty$, $A_{a, b} = \left\lbrace x \in \mathbb{R}^n | a < \| x \| < b \right\rbrace$. Also, $B \left( x, r \right) = \left\lbrace y \in \mathbb{R}^n | \| x - y \| < r \right\rbrace$. Given a ball $B = B \left( x, R \right)$, we consider $a = \max \left\lbrace 0, \| x \| - R \right\rbrace$, and $b = \| x \| + R$. So, we have the situation in the following figure.
We will assume that $\| x \| \geq 2R$. Next, the authors say that if $N_1$ is the number of balls obtained by a rotation of $B$ which are disjoint and inside $A_{a, b}$, then $N_1 \geq C \left( \frac{\| x \|}{R} \right)^{n - 1}$. While in two dimensions it is clear to me how this inequality comes, the higher dimensional picture is not so clear. To present my idea, I will first give the two-dimensional arguments that I believe lead us to the conclusion.
Let $N_1$ be the number of disjoint balls that fill up the entire annular region (and ar obtained by rotating $B$). Therefore, they must be at least as many to fill up half of it. By elementary trigonometry (see figure below), one sees that the cone formed at the origin due to the ball $B$ has angle $2 \sin^{-1} \left( \frac{R}{\| x \|} \right) \leq C \frac{R}{\| x \|}$. Then, we can choose $N_1 \in \mathbb{N}$ such that $C \frac{R}{\| x \|} N_1 \geq \pi$, or what is the same, $N_1 \geq C \frac{R}{\| x \|}$. The relation $\sin^{-1} t \leq C t$ follows near $0$ since $\lim\limits_{t \rightarrow 0} \frac{\sin^{-1} t}{t} = 1$.
This gives a clear picture of the two-dimensional case. Now to proceed to three-dimensions, we start by doing this on the equator of the sphere that contains $x$ (as shown in the next figure). Now, the remaining annular region would be filled up if we rotate the string of balls we obtained by above process in a direction perpendicular to the equator considered. However, this time, instead of rotating it by $\pi$ radians, we only rotate it by $\frac{\pi}{2}$ to get a "spherical" square on the surface of the sphere. This "square" will be filled up by $C \left( \frac{\| x \|}{R} \right)^{2}$ many balls identical to the one we started with.
For higher dimensions, we continue this process in every perpendicular direction until we run out of linearly independent directions. This would give us a "cube" on the surface of the sphere which can be filled up by $C \left( \frac{\| x \|}{R} \right)^{n - 1}$ many balls.
Is this idea correct? Or am I missing something in this?
PS: We only need a lower estimate on the number $N_1$ of balls that can be chosen, and not the exact number.