# Specialized covering lemma for a Hardy-Littlewood maximal inequality

In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:

Suppose $$K$$ is a compact set, and for every $$x \in K$$, we are given an open ball $$B(x,r_x)$$ that is centered at $$x$$ and of radius $$r_x$$. Assume that $$R:= \sup_{x \in K} r_x < \infty.$$ Let $$\mathcal{B}$$ be this collection of balls, i.e. $$\mathcal{B} = \{B(x,r_x) \colon x \in K\}.$$ Then given any $$\varepsilon > 0$$, there exists a finite subcollection $$\mathcal{C}$$ of balls from $$\mathcal{B}$$, so that the balls in $$\mathcal{C}$$ are pairwise disjoint, and so that the (concentric) dilates of balls in $$\mathcal{C}$$ by $$(2+\varepsilon)$$ times would cover $$K$$.

The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.

Note: This is not homework.

Well, I am almost 2 years late, but I figured I could show my answer. I didn't quite follow the selection procedure in @ydx 's answer.

We need to consider $$2+\epsilon$$ dilation because we will need $$2$$ radii to get to a centre and an $$\epsilon$$ more to cover the ball around said centre. Fix $$0<\epsilon<1$$.

Around each $$x\in K$$, there's a ball of radius $$r_x$$. Cover $$K$$ by $$B(x, \epsilon r_x)$$ and obtain a finite subcover $$B(x_1, \epsilon r_{x_1}), \dots, B(x_n, \epsilon r_{x_n})$$.

Now, from the collection $$B(x_i, r_{x_i}), 1\leq i\leq n$$ obtain a disjoint subcollection by choosing the largest radius at each step of the selection (similar to the construction in the finite Vitali covering). Let this subcollection be $$B(x_1, r_{x_1}), \dots, B(x_m, r_{x_m})$$ with $$r_{x_1}\geq r_{x_2}\geq\dots\geq r_{x_m}$$ (by construction).

Now, for $$j>m, B(x_j, r_{x_j})$$ intersects some ball in the subcollection, say $$B(x_1, r_{x_1})$$. By construction $$r_{x_1}\geq r_{x_j}$$. Therefore, $$B(x_1, 2r_{x_1})$$ contains the centre $$x_j$$ and going an $$\epsilon$$ further will contain $$B(x_j, \epsilon r_{x_j})$$, i.e., $$B(x_1, (2+\epsilon)r_{x_1})\supseteq B(x_j, \epsilon r_{x_j}).$$

Since $$\epsilon<1$$, the collection $$B(x_i, (2+\epsilon)r_i)$$ covers the union of $$B(x_i, \epsilon r_i)$$, hence covers $$K$$.

Consider the open cover $$\{B(x,\varepsilon r_x):\ x\in K\}$$ and get a finite subcover, say $$\{B(x_j,\varepsilon r_j):\ j=1,\dots,N\}$$. WLOG, assume $$r_1\geq r_2\geq\dots\geq r_N$$. Then, for each $$j=1,\dots,N$$, select the open ball $$B(x_j,r_j)$$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set $$\{B(x_{j_k},r_{j_k}):\ k=1,\dots,M\}$$ where $$M\leq N$$ and $$j_{1}\leq j_2\leq\dots\leq j_M$$.

For each $$j$$, if $$B(x_j,\varepsilon r_j)\not\subset\bigcup_{k=1}^MB(x_{j_k},(2+\varepsilon)r_{j_k})$$, then $$|x_j-x_{j_k}|+\varepsilon r_j>(2+\varepsilon)r_{j_k}$$ for each $$k$$. Since $$B(x_j,r_j)$$ is not selected, we can also find some $$k$$ such that $$B(x_j,r_j)\cap B(x_{j_k},r_{j_k})\neq \varnothing$$, so $$|x_j-x_{j_k}|< r_j+r_{j_k}$$. Note that we may choose the smallest $$k$$ with such a property, so we may assume $$B(x_j,r_j)\cap B(x_{j_l},r_{j_l})= \varnothing$$ if $$l. But this implies that $$(2+\varepsilon)r_{j_k}-\varepsilon r_j, i.e. $$r_{j_k}. Here is a contradiction, since we should have chosen $$B(x_j,r_j)$$ instead of $$B(r_{j_k},r_{j_k})$$.

• Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center. Commented Dec 16, 2018 at 2:20
• Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<\infty, so I guess there should be a proof relying on this assumption. Commented Dec 16, 2018 at 2:49