I am trying to understand the following fact:

Suppose $\{B_i\}_i$ are disjoint balls in $\mathbb R^n$, and $A_i \subset 100 B_i$ is a subset with $|A_i| \geq c |B_i|$. Then for any nonnegative $f$, we have $\sum_i |B_i| \inf_{A_i} f \lesssim \int_{\cup_i A_i} f$, where the implied constant depends only on $c$ and the dimension $n$.

(Here, $|\cdot|$ denotes Lebesgue measure, and $100B$ denotes the ball with the same center as $B$ and $100$ times the radius.)

Is there a way to prove this with (some combination of) covering lemmas, maximal function estimates, or weighted inequalities? I couldn't see an easy way to prove this.

Some background (which isn't needed for my question): The statement above is taken from Chapter 13 of David and Semmes's *Singular integrals and rectifiable sets in $\mathbb R^n$*. (It appears in the middle of a proof. They do not state this as a separate lemma.)

Here is a sketch of the proof in the book:

Let $p \in (1, \infty)$ and for each $i$, let $w_i$ be a function on $A_i$ (all TBD). By Holder,

\begin{align*} \inf_{A_i} f \leq \left(\frac{1}{|A_i|}\int_{A_i} f^{1/p} \right)^{p} \leq \left(\frac{1}{|A_i|}\int_{A_i} f w_i \right) \left(\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} \end{align*}

so

\begin{align} \sum_i |B_i| \inf_{A_i} f &\lesssim \sum_i \left(\int_{A_i} f w_i \right) \left(\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} \\ &\leq \left(\int f \textstyle\sum_i 1_{A_i} w_i \right) \left(\sup_i\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} \end{align}

To complete the proof, we just need to choose $p$ and $w_i$ so that (i) $\sum_i 1_{A_i} w_i \lesssim 1$ and (ii) $\sup_i\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \lesssim 1$. This can be accomplished as follows:

Let $p = 3$. Introduce an ordering on the indices so that $i \prec j$ if $|B_i| < |B_j|$ (and break ties arbitarily). Set $w_i(x)^{-1/2} = \sum_{j \preceq i} 1_{A_j}(x) = \# \{ j : x \in A_j \text{ and } j \preceq i\}$.

Note that if $j \preceq i$ and $A_j \cap A_i \neq \emptyset$, then $B_j \subset 300B_i$. This, with the disjointness of the $B_j$, implies $$\int_{A_i} w_i^{-1/2} \leq \sum_{j \preceq i, A_j \cap A_i \neq \emptyset} |A_j| \approx \sum_{j \preceq i, A_j \cap A_i \neq \emptyset} |B_j| \leq |300B_i| \approx |A_i|.$$

This proves (ii). (Also, this implies $w_i(x) > 0$ for almost every $x \in A_i$.)

Finally, for any fixed $x$, if $w_i(x) = w_j(x) \neq 0$, then $i=j$. Since $w_i$ takes values in $\{m^{-2} : m \in \mathbb N\} \cup \{0\}$, we have the pointwise bound $\sum_i 1_{A_i} w_i \leq \frac{\pi^2}{6}$, which shows (i) holds and completes the proof.

I don't really have a good intuition for this proof, especially how to motivate the choice of $p$ and $w_i$ (other than "because it works"). In particular, I am mystified (and amazed) at how the authors use $\sum_{m=1}^\infty m^{-2} < \infty$ to control the overlap of the $\{A_i\}_i$. This is why I'd be interested to see if there was another proof.

Singular integrals and rectifiable sets in $\mathbb R^n$. I edited the question to add this information. I also elaborated on the $\pi^2/6$ step (which works exactly in the way you said). $\endgroup$ – Alan C Nov 26 '20 at 18:013more comments