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This is Excercise 19 in Terry's notes on Differentiation theorem. I find it interesting but stumbled to get what he meant in his hint.

The Hardy-Littlewood maximal inequality for an absolutely integrable function $f$ reads:

$$ m(\{Mf(x)\geq \lambda\}) \leq \frac{C_d}{\lambda}\int_\mathbb{R^{d}}|f(t)|dt $$

where $m$ is the Lebesgue measure, $\lambda>0$, and $Mf(x)$ is the Hardy-Littlewood maximal function. Using a Vitali type of covering lemma one can show the constant can be $3^d$. The exercise asks to improve this to $2^d$, by noticing that $2$-scaled balls cover the centers. Terry hinted that one need to do some $\epsilon$ adjustment. However, I failed to see why and how can small adjustments to $2$-scaled balls (or perhaps something else) can still lead to a sufficient cover, given that choice of covering balls seems to be fixed.

What am I missing here?

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  • $\begingroup$ I don't understand completely the last part of Tao's hint (maybe "centers" has to be replaced by "balls of twice the radius"?). Anyway: given the balls $B'_j$ from the Vitali covering (which are fixed once and for all) observe that $(2+\epsilon)B'_j$ cover the desired set $K$, which coincides with the set of all the centers. From here you obtain the inequality with $(2+\epsilon)^d$, and then it suffices to take the limit as $\epsilon\to 0$ $\endgroup$ – Del Oct 26 '16 at 13:07
  • $\begingroup$ @Del I don't see why $K$ coincides with the set of all the centers. Maybe you mean the Vitali cover defined as this? But this doesn't appear in Tao's construction. Could you explain a little more here? Thanks!:) $\endgroup$ – Syl.Qiu Oct 26 '16 at 15:49
  • $\begingroup$ yes you're right, that's the point I was misinterpreting. I apologize, I don't know what I had in mind...I'll think about it. Moreover we could use the Vitali lemma that you linked only if there were a $\limsup_{r\to 0}$ in the definition of maximal function, or something like that, which is not the case. $\endgroup$ – Del Oct 26 '16 at 16:29
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    $\begingroup$ @Del Actually now I think your idea is right (except the Vitali covering part, but in fact it's not needed). After some discussion with others I can be sure that the finite covering balls cannot be the original one, which is what I was missing. Thanks! :) $\endgroup$ – Syl.Qiu Oct 28 '16 at 3:05
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Thanks to discussion with Del and Po-lam, I realised that Tao did't mean to use the original finite covering balls (those to be dilated by a magnitude of $3$) in the Vitali covering lemma.

Once realising this, one tries to realize the following (credit to Yung Po-lam and Ding Cong ), see also the comments below for some further hint: 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 $K$.

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  • $\begingroup$ Glad you found out! I still don't understand though, did you prove a different version of Vitali's theorem? $\endgroup$ – Del Oct 28 '16 at 9:44
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    $\begingroup$ @Del not quite so. Here we have to choose sufficient amount of covering balls (i.e. every epsilon neighborhood of $p \in K$ has to contain at least one center). Here is the very point why we consider "an epsilon of room";) $\endgroup$ – Syl.Qiu Oct 31 '16 at 16:56

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