Does the Lebesgue version of Hardy-Littlewood really require a multiple of 3?

Question

I'm reading Axler's proof of the following which he calls the Hardy-Littlewood maximal inequality.

Before writing the proof I'll flag that my question is about the use of the Vitali Covering Lemma which produces a multiple of 3.

Theorem: Suppose $h\in\mathcal L^1(\mathbf R)$. Then

$$ |\{b\in\mathbf R:h^*(b)>c\}|\le\frac 3 c \| h\|_1 $$

for every $c>0$.

Proof: Suppose F is a closed bounded subset of $\{b\in\mathbf R:h^*(b)>c\}$. [...]

For each $b\in F$ there exists $t_b>0$ such that

$$ \frac{1}{2t_b}\int_{b-t_b}^{b+t_b}|h|>c. $$

Clearly

$$ F\subset \bigcup_{b\in F}(b-t_b,b+t_b). $$

The Heine-Borel Theorem tells us that this open cover of a closed bounded set has a finite subcover. In other words, there exist $b_1,...,b_n\in F$ such that

$$F\subset (b_1-t_{b_1},b_1+t_{b_1})\cup\cdots\cup (b_n-t_{b_n},b_n+t_{b_n}).$$

To make the notation cleaner, relabel the open intervals above as $I_1,\dots,I_n$.

Now apply the Vitali Covering Lemma to the list $I_1,\dots,I_n$ producing a disjoint sublist $I_{k_1},\dots,I_{k_n}$ such that

$$I_1\cup\cdots\cup I_n\subset (3 * I_{k_1})\cup\cdots\cup (3*I_{k_m}) $$

Thus

$$\begin{aligned} |F|&\le|I_1\cup\cdots\cup I_n|\\ &\le|(3*I_{k_1})\cup\cdots\cup(3I_{k_m})|\\ &\le |3*I_{k_1}|+\cdots+|3*I_{k_m}|\\ &<\frac 3 c \left(\int_{I_{k_1}}|h|+\cdots+\int_{I_{k_m}}|h| \right)\\ &\le\frac 3 c \int_{-\infty}^\infty|h|. \end{aligned}$$

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The proof goes on from there but this already contains what I want to ask about, which are the final two inequalities.

It seems to me that we used the Vitali Covering Lemma in order to obtain disjoint intervals. Disjointness is what allows for the final inequality.

However, because the finite subcover $I_1,\dots,I_n$ was just any subcover, could we not have merely merged overlapping intervals into a single interval? Doing so recursively, finite many times, we can be guaranteed to always get a collection of disjoint intervals.

Wouldn't that be a simpler proof and remove the need for the odd-looking 3 in the result?

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[Edit for clarification]

Perhaps what I was suggesting was unclear, so let me say more: Could we not just replace $I_1,\dots,I_n$ with $J_1,\dots,J_m$ in the following way? Take the least index $I_k$ which overlaps any other interval, $I_{k'}$ and merge these into a new interval $J_1$ so that $J_1=I_k\cup I_{k'}$. Repeat as before, possibly replacing any $J_l$ with a new $J_l$ that results from merging a $J_l$ with some $I_k$. Repeat until none of the intervals overlap. If there are intervals which did not merge, still labeled by some $I_k$, then label them with some $J_l$.

(There are more precise ways to formalize this but the notation gets distracting. But just to give an example, if you start from the intervals $(0,2),(1,3), (3,4)$ you end up with the intervals $(0,3),(3,4)$.)

Anyway, the result is now a collection of disjoint open intervals $J_1,\dots,J_m$ such that, rather than having any subsets or inequalities, we get a precise equality,

$$ \bigcup_{k=1}^n I_k = \bigcup_{k=1}^m J_m $$

Then we never use the Vitali Covering Lemma and get no multiple of 3. The upper bound becomes

$$ |\{b\in\mathbf R:h^*(b)>c\}|\le \frac 1 c \|h\|_1 $$

Answer 1

As noted on page 107 of my book Measure, Integration & Real Analysis (whose electronic version is legally available for free at https://measure.axler.net), the smallest constant that can replace $3$ in the one-dimensional Hardy-Littlewood inequality is $$ \frac{11 + \sqrt{61}}{12} \approx 1.56752, $$ as was proved by Antonios Melas.

However, as noted in a comment above, the important consequences of the Hardy-Littlewood inequality require only that some constant works.

Answer 2

the finite subcover $I_1,\cdots,I_n$ was just any subcover

That doesn't seem right however one interprets the phrase "any subcover". The proof applied Heine-Borel theorem by first selecting a particular open cover. Then the Heine-Borel theorem says that this open cover has a finite subcover. It doesn't say that any subcover of that open cover is finite.

could we not have merely merged overlapping intervals into a single interval?

This is exactly what the Vitali Covering Lemma does. At least, the version as presented in Stein/Shakarachi, Princeton Lectures in Analysis Vol.$3$, where the Hardy-Littlewood maximal inequality is discussed.

If two distinct intervals $I_m=(b_m-t_{b_m},b_m+t_{b_m})$ and $I_k=(b_k-t_{b_k},b_k+t_{b_k})$ overlap, and say $t_{b_m}\ge t_{b_k}$, then an interval concentric with $I_m$ would need at least $3$ times the radius $t_{b_m}$ to cover both intervals. (Think of what happens when the intervals $I_m,I_k$ are "almost" disjoint!)

Why don't we take another point in $I_m\cup I_k$ as our new center, i.e. the center of the interval which covers both $I_m$ and $I_k$? Well, of course we can, and we may get a different constant. For example in this question on Terry's notes we see how the constant is $2$ instead of $3$. Also note that the version of Hardy-Littlewood maximal inequality is $d-$dimensional, not $1-$dimensional, which is where the exponent $d$ comes from, so theyh ave $3^d$ or $2^d$ instead of $3$ or $2$.

Finally, it should be clear that no disjoint sublist of the subcover can have union containing the union of all elements in the subcover, unless all the intervals in that subcover is already pairwise disjoint. In that case you can just take the entire subcover as the sublist.

There has to a constant there. That's kind of the point of Hardy-Littlewood maximal inequality. Besides, the expression wouldn't make sense if you don't have something there: what is $A\le \dfrac{}{x}$ even supposed to mean?

Answer 3

I finally found the mistake in my suggested alternate proof. Although it is possible to "merge" intervals together as I suggested, it then causes the identity $|I_i|=2t_i$ to no longer hold.