---
tag: 744826NNN9
kind: definition
name: Hardy–Littlewood maximal function
msc: [42B25, 28A15]
defines: VX1A56P5KQ
uses: [2Z2QVWZ262, X7GTF3D98F, 0YE5868HZB]
sources:
  - url: https://en.wikipedia.org/wiki/Hardy–Littlewood_maximal_function
---

## Definition

Let $n \ge 1$, let $\lambda$ be [[Lebesgue measure|2Z2QVWZ262]] on $\R^n$, and let $f \colon \R^n \to \R$ be locally [[integrable|X7GTF3D98F]].
For $x \in \R^n$ and $r > 0$, let $B(x, r)$ be the [[open ball|0YE5868HZB]] of centre $x$ and radius $r$.
The *Hardy–Littlewood maximal function* of $f$ is the function $Mf \colon \R^n \to [0, \infty]$ given by
$$Mf(x) = \sup_{r > 0} \frac{1}{\lambda(B(x, r))} \int_{B(x, r)} \abs{f(y)} \, d\lambda(y).$$

## Remarks

Example: let $n = 1$ and $f = \mathbf{1}_{[0, 1]}$, and let $x > 1$.
For $r \ge x$ the average is $1/(2r) \le 1/(2x)$.
For $x - 1 < r < x$ the average is $(r - x + 1)/(2r)$, which increases with $r$ and tends to $1/(2x)$.
For $r \le x - 1$ the average is $0$.
So $Mf(x) = 1/(2x)$.

Non-example of integrability: this shows that $Mf$ need not be integrable even when $f$ is, since $\int_1^\infty dx/(2x) = \infty$.
In fact $Mf$ is integrable on $\R^n$ only when $f = 0$ almost everywhere.

Variants use balls containing $x$ instead of balls centred at $x$ (the uncentred maximal function), or cubes instead of balls.
They are comparable to $Mf$ up to constants depending on $n$.