Maximal function theory

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

Download the notes here: MaximalFunctionTheory.pdf (29 pages)

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

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

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

as the length of the interval $(x-r,x+r)$ centered at $x$ decreases to 0. Since $(\mathcal{A_{\mathnormal{r}}}f)(x)$ is well-defined for all $f \in L^1([x-r,x+r])$, it makes sense to try and prove (1.1) for all $f \in L^1_{\text{loc}}(\mathbb{R})$, 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

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

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

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

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

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 $g \in \mathscr{C_{\mathnormal{c}}}(\mathbb{R}^d)$ such that $\lVert f-g \rVert_1 < \varepsilon$. We can then rewrite $(\mathcal{A_{\mathnormal{rB}}}f)(x) - f(x)$ as

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

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

$$\begin{align*} &\limsup_{r \to 0} \vert (\mathcal{A_{\mathnormal{rB}}}f)(x) - f(x) \vert \\ \leq& \limsup_{r \to 0} \vert \mathcal{A_{\mathnormal{rB}}}(f-g)(x) \vert \\ +& \limsup_{r \to 0} \vert(\mathcal{A_{\mathnormal{rB}}}g)(x) - g(x) \vert \\ +& \vert f(x) - g(x) \vert \\ =& \limsup_{r \to 0} \mathcal{A_{\mathnormal{rB}}}(\vert f - g \vert)(x) + \vert f(x) + g(x) \vert. \end{align*}$$

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

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

which is bounded from above by the maximal function

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

This motivates us to introduce our main object of study:

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

$$\begin{align*}(\mathcal{M_{\mathnormal{B}}}f)(x) &= \sup_{r > 0}(\mathcal{A_{\mathnormal{rB}}}\vert f \vert)(x) \\ &= \sup_{r > 0} \frac{1}{(m(rB)} \int_{rB} \vert f(x+y) \vert \, dy.\end{align*}$$

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 $\mathcal{A_{\mathnormal{rB}}} f \to f$ pointwise everywhere. But this is too much to hope for, as

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

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

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

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

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

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

Now, Estimate (3) implies that

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

Observe that

$$\begin{align*} m\left(\left\lbrace x : \vert f(x) - g(x) \vert > \frac{1}{2k} \right\rbrace\right) &= \int_{\lbrace x : \vert f(x) - g(x) \vert > \frac{1}{2k}\rbrace} 1 \, dy \\ &\leq \int_{\lbrace x : \vert f(x) - g(x) \vert > \frac{1}{2k}\rbrace} \frac{\vert f(y) - g(y)\vert}{1/2k} \, dy \\ &\leq \int_{\mathbb{R}^d} \frac{\vert f(y) - g(y)\vert}{1/2k} \, dy \\ &= 2k \lVert f-g\rVert_1. \end{align*}$$

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

$$(6) \hspace{3em} m \left(\lbrace x : \mathcal{M_{\mathnormal{B}}}(f-g)(x) > \frac{1}{2k}\rbrace\right) \leq 2kA \lVert f-g\rVert_1$$

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

$$m(E_k) \leq 2k(A+1) \lVert f-g \rVert_1 < 2k(A+1)\varepsilon.$$

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

$$m(\lbrace x \in \mathbb{R}^d : \mathcal{M_{\mathnormal{B}}}f(x) > \alpha\rbrace) \leq \frac{A_{d,1,B}}{\alpha}\lVert f\rVert_1$$

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

$$\lim_{r \to 0}(\mathcal{A_{\mathnormal{rB}}}f)(x) = f(x)$$

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

$$m(\lbrace x \in \mathbb{R}^d : \mathcal{M_{\mathnormal{B}}}f(x) > \alpha\rbrace) \leq \frac{A_{d,1,B}}{\alpha}\lVert f\rVert_1$$

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

$$\lVert \mathcal{M_{\mathnormal{B}}}f \rVert_p \leq A_{d, p, B} \lVert f \rVert_p$$

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 symmetric 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.