# Can this inequality be proved using weighted maximal function estimates?

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.

• I did not have to think about your question (it is 2am), but is this post useful here? mathoverflow.net/a/320747/121665 Nov 26, 2020 at 7:06
• Why $\pi^2/6$? Could you say from which textbook is the proof taken? Nov 26, 2020 at 9:20
• Well, I see. I think the proof has been invented since the overlapping of the 100 balls cannot be controlled having no information on radii. Then it works since the natural numbers $w_i(x)^{-1/2}$ are all different. No idea if a simpler proof is available. Nov 26, 2020 at 11:26
• another general idea behind this proof is that the estimate $\inf_A f\leqslant M_p(f):=(\frac1{|A|}\int_A f^p)^{1/p}$ becomes worse when $p$ (that is called power-mean inequality, a partial case of Holder). Since $p=1$ does not work, it is natural to get smaller $p$. After that we get something not-linear in $p$, so it is natural to get a further linear upper bound which itself again follows from Holder. Nov 26, 2020 at 16:29
• @GiorgioMetafune The proof is from Chapter 13 of David and Semmes's 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). Nov 26, 2020 at 18:01

It suffices to show that $$\sum_i |B_i| 1_{\inf_{A_i} f > t} \lesssim \int_{\bigcup A_i} 1_{f>t}$$ for any $$t>0$$, since the claim follows by integrating in $$t$$ and using the Fubini-Tonelli theorem (i.e., use the layer cake decomposition). (Equivalently: to prove the claim, it suffices to do so in the special case when $$f$$ is an indicator function.) But one has $$M (1_{\bigcup A_i} 1_{f>t})(x) \gtrsim 1$$ whenever $$x \in B_i$$ and $$\inf_{A_i} f>t$$, so the claim follows from the Hardy-Littlewood maximal inequality.