Below is an outline of the notes I wrote up on the basic theory Hardy—Littlewood maximal funtion and its variants for a seminar. The notes assume familiarity with measure theory.

Let $$f:[a,b] \to \mathbb{R}$$ be continuous. The fundamental theorem of calculus states that

$F(x) = \int_a^x f(y) \, dy$

is differentiable in $$(a,b)$$ and $$F'(x) = f(x)$$ for all $$x \in (a,b)$$.

Could we generalize this result to a larger class of functions? Note that the above result is equivalent to

$(1) \hspace{2em} \lim_{r \to 0} \frac{1}{2r} \int_{x-r}^{x+r} f(y) \, dy = f(x)$

for all continuous functions $$f$$. In this form, the fundamental theorem of calculus is a statement about the behavior of the integral mean value

$(2) \hspace{2em} (\mathcal{A}_{r} f)(x) = \frac{1}{2r} \int_{x-r}^{x+r} f(y) \, dy$

as the length of the interval $$(x-r,x+r)$$ centered at $$x$$ decreases to 0. Since $$(\mathcal{A}_rf)(x)$$ is well-defined for all $$f \in L^1([x-r,x+r])$$, it makes sense to try and prove (1.1) for all $$f \in L^1_{\mbox{loc}}(\mathbb{R})$$, the space of measurable functions on $$\mathbb{R}$$ whose integral is finite on each compact subset of $$\mathbb{R}$$.

We also consider $$d$$-dimensional generalizations of (1). To this end, we must determine what we wish to take as an $$n$$-dimensional generalization of intervals. We abstract three properties of intervals: compactness, convexity, and symmetry.

Definition. A nonempty set $$B \subseteq \mathbb{R}^d$$ is convex if, for each pair of points $$x_1$$ and $$x_2$$ in $$B$$, the convex combination $$(1-\lambda) x_1 + \lambda x_2$$ is in $$B$$ for all $$0 \leq \lambda \leq 1$$.

Definition. A nonempty set $$B \subseteq \mathbb{R}^d$$ is centrally symmetric with respect to $$p \in B$$ if $$B$$ is invariant under the affine transform $$x \mapsto 2p - x$$. This is equivalent to saying that $$p + h \in B$$ if and only if $$p - h \in B$$.

We write centrally symmetric convex body to refer to a subset of $$\mathbb{R}^d$$ that is compact, convex, and centrally symmetric with its center $$p$$ at the origin.

Since we can rewrite (2) as

$(\mathcal{A}_r f)(x) = \frac{1}{2r} \int_{-r}^r f(x+y) \, dy,$

it suffices to consider centrally symmetric convex bodies with respect to the origin in asking the following question:

Question. Given a centrally symmetric convex body $$B \subseteq \mathbb{R}^d$$, does the integral mean value

$(\mathcal{A}_{rB}f)(x) = \frac{1}{m(rB)} \int_{rB} f(x+y) \, dy$

of $$f \in L^1_{\mbox{loc}}(\mathbb{R}^d)$$ converge pointwise to $$f(x)$$ as $$r \to 0$$? Here we have used $$rB$$ to denote the scaled set $$rB = \{ry : y \in B\}$$.

Whenever $$f$$ is continuous, the argument for the one-dimensional fundamental theorem of calculus can be applied with minor modification to answer Question 1.5 in the affirmative. If $$f$$ is merely $$L^1(\mathbb{R})$$, then, for each $$\varepsilon > 0$$, we can find a $$g \in \mathscr{C}_c(\mathbb{R}^d)$$ such that $$\| f-g \|_1 < \varepsilon$$. We can then rewrite $$(\mathcal{A}_{rB}f)(x) - f(x)$$ as

$\mathcal{A}_{rB}(f-g)(x) + (\mathcal{A}_{rB}g)(x) - g(x) + g(x) - f(x).$

By continuity, we have $$(\mathcal{A}_{rB}g)(x) \to g(x)$$ as $$r \to 0$$, whence we have the estimate

\begin{align*} &\limsup_{r \to 0} \vert (\mathcal{A}_{rb}f)(x) - f(x) \vert \\ \leq& \limsup_{r \to 0} \vert \mathcal{A}_{rb}(f-g)(x) \vert| \\ &+ \limsup_{r \to 0} \vert(\mathcal{A}_{rB}g)(x) - g(x) \vert \\ &+ \vert f(x) - g(x) \vert \\ =& \limsup_{r \to 0} \mathcal{A}_{rB}(\vert f - g \vert)(x) + \vert f(x) + g(x) \vert.\end{align*}

Therefore, the study of the integral mean values of $$f$$ at $$x \in \mathbb{R}^d$$ depends crucially on the quantity

$\limsup_{r \to 0} \mathcal{A}_{rB}(\vert f - g \vert)(x),$

which is bounded from above by the maximal function

$\sup_{r > 0} \mathcal{A}_{rB}(\vert f - g \vert)(x).$

This motivates us to introduce our main object of study:

Definition. The Hardy—Littlewood maximal function of $$h \in L^1_{\mbox{loc}}(\mathbb{R}^d)$$ over a centrally symmetric convex body $$B \subseteq \mathbb{R}^d$$ is

\begin{align*}(\mathcal{M}_Bf)(x) &= \sup_{r > 0}(\mathcal{A}_{rB}\vert f \vert)(x) \\ &= \sup_{r > 0} \frac{1}{(m(rB)} \int_{rB} \vert f(x+y) \vert \, dy.\end{align*}

What we have discussed so far assures us that if $$f \in L^1(\mathbb{R}^d)$$, then, for each $$\varepsilon > 0$$, there exists a $$g \in L^1(\mathbb{R}^d)$$ that yields the estimate

$(3) \hspace{3em} \limsup_{r \to 0} \vert (\mathcal{A}_{rB}f)(x) - f(x) \vert \leq \mathcal{M}_B(f-g)(x) + \vert f(x) - g(x)\vert.$

Ideally, we would have liked to show that

$\mathcal{M}_B(f-g)(x) + \vert f(x) - g(x) \vert C \varepsilon$

for some constant $$C$$ independent of $$\varepsilon$$, thus proving that $$\mathcal{A}_{rB} f \to f$$ pointwise everywhere. But this is too much to hope for, as

$(\mathcal{A}_{(-r,r)}\chi_{(0,1)}(0) = \frac{1}{2r} \int_0^r \, dy = \frac{1}{2}$

for all $$0 < r < 1$$, which does not converge to $$\chi_{(0,1)}(0) = 0$$ as $$r \to 0$$.

So then, if we hope to obtain an affirmative answer to the above Question, then we must settle for an almost-everywhere statement. This equivalent to the statement that the set

$\left\{x \in \mathbb{R}^d : \limsup_{r \to 0} \vert (\mathcal{A}_{rB}f)(x) - f(x) \vert > 0 \right\}$

is of Lebesgue measure zero. Since the above set is the intersection of the sets

$(4) \hspace{3em} E_k = \left\{x \in \mathbb{R}^d : \limsup_{r \to 0} \vert (\mathcal{A}_{rB}f)(x) - f(x)\vert > \frac{1}{k}\right\},$

it suffices to show that $$m(E_k) = 0$$ for all $$k \in \mathbb{N}$$.

Now, Estimate (3) implies that

(5) \hspace{3em} \begin{align*} m(E_k) \leq& m \left( \left\{ x : \mathcal{M}_B(f-g)(x) > \frac{1}{2k}\right\}\right) \\ &+ m \left( \left\{ x : \vert f(x) - g(x) \vert > \frac{1}{2k} \right\}\right) \end{align*}

Observe that

\begin{align*} m\left(\left\{ x : \vert f(x) - g(x) \vert > \frac{1}{2k} \right\}\right) &= \int_{\{x : \vert f(x) - g(x) \vert > \frac{1}{2k}\}} 1 \, dy \\ &\leq \int_{\{x : \vert f(x) - g(x) \vert > \frac{1}{2k}\}} \frac{\vert f(y) - g(y)\vert}{1/2k} \, dy \\ &\leq \int_{\mathbb{R}^d} \frac{\vert f(y) - g(y)\vert}{1/2k} \, dy \\ &= 2k \|f-g\|_1. \end{align*}

It is therefore natural to hope for a bound of the form

$(6) \hspace{3em} m \left(\{x : \mathcal{M}_B(f-g)(x) > \frac{1}{2k}\}\right) \leq 2kA \|f-g\|_1$

for some constant $$A$$ independent of $$(f-g)$$, so that (5) can be written as

$m(E_k) \leq 2k(A+1) \|f-g\|_1 < 2k(A+1)\varepsilon.$

Since $$\varepsilon$$ was arbitrary, we can then conclude that $$m(E_k) = 0$$.

(6) is established by the following foundational result in maximal function theory:

Theorem (Weak-type $$(1,1)$$-bound on the maximal function). If $$B \subseteq \mathbb{R}^d$$ is a centrally symmetric convex body, then there exists a constant $$A_{d,1,B}$$ such that

$m(\{x \in \mathbb{R}^d : \mathcal{M}_Bf(x) > \alpha\}) \leq \frac{A_{d,1,B}}{\alpha}\|f\|_1$

for each $$\alpha > 0$$ and every $$f \in L^1(\mathbb{R}^d)$$. The constant $$A_{d,1,B}$$ depends only on the dimension $$d$$ and the body $$B$$.

The weak-type bound now implies the below pointwise (almost everywhere) convergence result for the $$L^1$$ case. The general case of the theorem follows easily from the $$L^1$$ case by considering the compact cutoff function $$f\chi_{B(0;k)}$$ with respect to closed balls $$B(0;k)$$ of radius $$k \in \mathbb{N}$$ centered at the origin.

Theorem (Lebesgue differentiation theorem). If $$f \in L^1_{\textrm{loc}}(\mathbb{R}^d)$$, and if $$B \subseteq \mathbb{R}^d$$ is a centrally symmetric convex body, then

$\lim_{r \to 0}(\mathcal{A}_{rB}f)(x) = f(x)$

for almost every $$x \in \mathbb{R}^d$$.

The method of establishing a weak-type bound to prove a pointwise convergence result turns out to be extremely powerful. In fact, we cannot do better:

Theorem (Stein's $$L^1$$ maximal principle). Let $$G$$ be a compact, Hausdorff, abelian topological group equipped with the Haar measure $$\mu$$. If $$(\varphi_n)_{n=1}^\infty$$ is a sequence of operators in $$L^\infty(G)$$ such that, for each $$f \in L^1(G)$$, we have the "pointwise convergence criterion"

$\limsup_{n \to \infty} \vert (f \ast \varphi_n)(x) \vert < \infty$

on a set $$E_f$$ of positive measure, then the maximal operator

$Mf(x) = \sup_{n \in \mathbb{N}} \vert (f \ast \varphi_n)(x) \vert$

satisfies the weak-type $$(1,1)$$-bound.

The weak-type bound of a general maximal operator is typically established through a judicious use of the Hardy–Littlewood maximal operator. It is thus of interest to tighten the bound

$m(\{x \in \mathbb{R}^d : \mathcal{M}_Bf(x) > \alpha\}) \leq \frac{A_{d,1,B}}{\alpha}\|f\|_1$

by reducing the size of the constant $$A_{d, 1, B}$$ as much as possible.

The classical proof yields $$A_{d,1,B} = O(5^d)$$ when $$B$$ is the Euclidean ball. This is a consequence of the following lemma:

Lemma (Infinitary Vitali covering lemma). If $$\{B(x_\beta,r_\beta)\}_{\beta}$$ is a collection of Euclidean balls in $$\mathbb{R}^d$$ whose radii are uniformly bounded, then there exists a pairwise-disjoint countable subcollection $$\{B(x_n,r_n)\}_n$$ such that

$\bigcup_{\beta} B(x_\beta,r_\beta) \subseteq \bigcup_n B(x_n,5r_n).$

An 1988 result of Stein and Strömberg reduces the constant to $$O(d)$$ for the Euclidean ball and $$O(d \log d)$$ for an arbitrary centrally symmetric convex body. Naor and Tao showed in 2011 that $$d \log d$$ is, in fact, optimal for a broad class of metric measure spaces.

The question of finding a tight bound for the Euclidean-space case remains open. In 2003, Melas showed that $$\frac{11+\sqrt{61}}{21}$$ is the optimal constant for the weak-type bound of the Hardy–Littlewood maximal function on the one-dimensional Euclidean ball. This remains the only known tight bound.

Stein and Strömberg conjectured that the bound may, in fact, be $$O(1)$$. While the conjecture has not been settled for the Euclidean ball case, Aldaz showed in 2011 that the constant grows without bound on the $$l^\infty$$ ball as the dimension increase.

Much work has been done on establishing dimension-independent bound for $$L^p$$ when $$p > 1$$. A classical proof using real interpolation methods gives us the bound

$\|\mathcal{M}_Bf\|_p \leq A_{d, p, B} \|f\|_p$

for all $$p > 1$$ with constant

$A_{d, p, B} = 2^{\frac{p-1}{p}} A^{\frac{1}{p}}_{d, 1, B} \left( \frac{p}{p-1} \right)^{1/p}.$

Stein showed in 1982 that, given a fixed centrally symmetric convex body $$B$$, the constant can be made $$O(1)$$ with respect to the dimension $$d$$. Moreover, Bourgain's far-reaching generalization in 1986 shows that, for each $$p > 3/2$$,

$\sup_{d, p, B} A_{d, p, B} < \infty$

where the supremum is taken over all $$d \geq 1$$ and centrally symetric convex bodies $$B \subseteq \mathbb{R}^d$$. In 2014, Bourgain brought $$p$$ down to $$p > 1$$ for the $$l^\infty$$ ball. The question of establishing the above uniform bound for a large class of centrally symmetric convex bodies in the range $$p > 1$$ remains open.