The Centered Hardy-Littlewood Maximal Function

Definition by Centered Ball Averages

At a point $x$, the centered Hardy–Littlewood maximal function compares the average size of a function over every ball centered at $x$. The radii vary, but the centers do not. Even when each individual average is finite, the supremum over all radii can be infinite. Keeping those two facts separate is essential when using the definition.

Throughout, let $n\ge1$, and use Lebesgue measure $m$ on $\mathbb{R}^n$ with its Lebesgue sigma-algebra $\mathcal{L}$. The functions $f:\mathbb{R}^n\to\mathbb{R}$ or $f:\mathbb{R}^n\to\mathbb{C}$ considered below are measurable and locally integrable: $\int_K |f|\,dm<\infty$ for every compact set $K\subset\mathbb{R}^n$. In particular, $\int_{B(x,r)}|f|\,dm<\infty$ for every $x\in\mathbb{R}^n$ and every $r>0$. Write $B(x,r)=\{y:|y-x|<r\}$, and let $\omega_n=m(B(0,1))$, so $m(B(x,r))=\omega_n r^n$.

Definition: For a measurable, locally integrable real- or complex-valued function $f$, its centered Hardy–Littlewood maximal function is the extended nonnegative-valued function $$ Mf(x)=\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|\,dm(y) =\sup_{r>0}\frac{1}{\omega_n r^n}\int_{B(x,r)}|f(y)|\,dm(y). $$ The supremum is taken over all positive real radii. Thus $Mf(x)\in[0,\infty]$: it may be finite or $+\infty$. For each fixed $r>0$, however, the average in the supremum is finite by local integrability.

The absolute value is part of the definition, so the maximal function is nonnegative even when $f$ is signed or complex-valued. The definition uses averages over balls whose center is exactly $x$; a ball that contains $x$ but has a different center is not one of the balls in this supremum.

Basic Properties and Their Proofs

Proposition: Let $f,g:\mathbb{R}^n\to\mathbb{R}$ or $\mathbb{C}$ be measurable and locally integrable. If $|f|\le |g|$ almost everywhere, then $Mf(x)\le Mg(x)$ for every $x$. Also $M(f+g)(x)\le Mf(x)+Mg(x)$ for every $x$. For every scalar $c\ne0$, $M(cf)(x)=|c|Mf(x)$ for every $x$; separately, $M(0f)(x)=0$ for every $x$.

Proof. Fix $x\in\mathbb{R}^n$ and $r>0$. If $|f|\le|g|$ almost everywhere, integrating this inequality over $B(x,r)$ gives $\int_{B(x,r)}|f|\,dm\le\int_{B(x,r)}|g|\,dm$. Division by the positive number $\omega_n r^n$ preserves the inequality. Taking the supremum over the same set of radii proves $Mf(x)\le Mg(x)$. For the sum, the triangle inequality gives $|f(y)+g(y)|\le |f(y)|+|g(y)|$ almost everywhere. Integrating and dividing by $\omega_n r^n$ shows that the average of $|f+g|$ over $B(x,r)$ is at most the sum of the corresponding averages for $f$ and $g$. Each of those averages is at most its supremum, so the average for $f+g$ is at most $Mf(x)+Mg(x)$. Taking the supremum over $r>0$ proves the claim, with the inequality interpreted in $[0,\infty]$. If $c\ne0$, then $|cf|=|c||f|$, so every average for $cf$ is $|c|$ times the corresponding finite average for $f$. Taking suprema gives $M(cf)(x)=|c|Mf(x)$, also when the supremum is infinite. If $c=0$, every average of $|0f|$ is zero, and its supremum is zero. This proves the separate zero-scalar statement without multiplying zero by $+\infty$. $\square$

These properties are pointwise statements. In particular, if two locally integrable functions agree almost everywhere, then their integrals of absolute value over every ball agree, and their maximal functions agree at every point. The proposition does not assert that the maximal function is finite everywhere: its value is defined as an extended value precisely to allow an unbounded supremum.

Worked Example: A Constant Function

Let $f(y)=c$ for every $y\in\mathbb{R}^n$, where $c$ is a real or complex constant. For any center $x$ and radius $r>0$, the integral of $|f|$ is the constant $|c|$ times the volume of the ball:

\int_{B(x,r)}|f(y)|\,dm(y) =|c|\,m(B(x,r)) =|c|\omega_n r^n.

Dividing by $\omega_n r^n$ gives the average $|c|$ for every radius. Its supremum is therefore $Mf(x)=|c|$ at every $x$. If $c=0$, this calculation says directly that every average is zero. Check: substituting any $r>0$ back into the defining average gives $|c|\omega_n r^n/(\omega_n r^n)=|c|$, so the value is independent of both the center and radius.

Radial Calculations at the Origin

For radial functions, centered averages at the origin can be computed using polar coordinates. If $g(y)=G(|y|)$ is nonnegative and radial, the polar-coordinate formula is $\int_{B(0,r)}g(y)\,dm(y)=n\omega_n\int_0^r G(\rho)\rho^{n-1}\,d\rho$, whenever the integral exists. The ball volume in the denominator is $\omega_n r^n$. We use this formula to compute the averages for each radius, then take their supremum; the supremum need not occur at a radius where the profile is largest pointwise.

Worked Example: A Radial Quadratic Profile Supported in the Unit Ball

Define $f(y)=|y|^2$ when $|y|<1$, and $f(y)=0$ when $|y|\ge1$. This is a nonnegative measurable, locally integrable function. We compute $Mf(0)$ by considering separately $0<r\le1$ and $r\ge1$.

If $0<r\le1$, the whole ball $B(0,r)$ lies in the region where $f(y)=|y|^2$. Therefore

\int_{B(0,r)}f\,dm =n\omega_n\int_0^r \rho^2\rho^{n-1}\,d\rho =n\omega_n\frac{r^{n+2}}{n+2}, \qquad \frac{1}{\omega_n r^n}\int_{B(0,r)}f\,dm =\frac{n}{n+2}r^2.

This average is at most $n/(n+2)$, since $r^2\le1$, and equals $n/(n+2)$ when $r=1$. If $r\ge1$, the integral includes the entire support of $f$, so

\int_{B(0,r)}f\,dm =n\omega_n\int_0^1 \rho^{n+1}\,d\rho =\frac{n\omega_n}{n+2}, \qquad \frac{1}{\omega_n r^n}\int_{B(0,r)}f\,dm =\frac{n}{n+2}r^{-n}.

Because $r^{-n}\le1$ for $r\ge1$, this average is also at most $n/(n+2)$, with equality at $r=1$. Thus $Mf(0)=n/(n+2)$. Check: the two formulas for the average both give $n/(n+2)$ at $r=1$, and each is bounded above by that value on its stated range. The supremum is attained by the unit ball.

Worked Example: A Radial Annulus

Let $f(y)=3$ when $1<|y|<2$, and $f(y)=0$ otherwise. The strict inequalities specify the annulus; including or excluding its boundary spheres does not alter any integral because those spheres have Lebesgue measure zero. At the origin, first take $0<r\le1$. The ball contains no points of the annulus except possibly boundary points, so the integral is zero and the average is zero.

For $1<r\le2$, the part of the annulus inside $B(0,r)$ has measure $\omega_n(r^n-1)$. The integral and average are therefore

\int_{B(0,r)}f\,dm=3\omega_n(r^n-1), \qquad \frac{1}{\omega_n r^n}\int_{B(0,r)}f\,dm =3(1-r^{-n}).

Since $r^{-n}$ decreases as $r$ increases, this average increases on $1<r\le2$ and reaches $3(1-2^{-n})$ at $r=2$. For $r\ge2$, the ball contains the full annulus up to boundary sets of measure zero, so

\int_{B(0,r)}f\,dm=3\omega_n(2^n-1), \qquad \frac{1}{\omega_n r^n}\int_{B(0,r)}f\,dm =3(2^n-1)r^{-n}.

This last expression decreases for $r\ge2$ and at $r=2$ equals $3(2^n-1)/2^n=3(1-2^{-n})$. Comparing the three radius ranges gives $Mf(0)=3(1-2^{-n})$. Check: the middle and outer formulas agree at $r=2$, and the average is zero for $r\le1$; hence no radius has a larger average.

Finite Averages Can Have an Infinite Supremum

Worked Example: A Locally Integrable Radial Singularity

For $y\ne0$, define $f(y)=|y|^{-n/2}$, and set $f(0)=0$. This is a nonnegative measurable function. It is locally integrable: for any $R>0$, polar coordinates give

\int_{B(0,R)}f\,dm =n\omega_n\int_0^R \rho^{n-1-n/2}\,d\rho =n\omega_n\int_0^R \rho^{n/2-1}\,d\rho =2\omega_n R^{n/2}<\infty.

For each radius $r>0$, dividing this integral by $\omega_n r^n$ gives the centered average $2r^{-n/2}$, which is finite. But the averages have no finite upper bound as $r$ decreases toward zero. For example, the radii $r=2^{-k}$, $k=1,2,\ldots$, give averages $2(2^{-k})^{-n/2}=2^{1+kn/2}$, which tend to $+\infty$. Consequently $Mf(0)=+\infty$, even though each individual ball average is finite. Check: at $r=1$ the average is $2$, while at $r=2^{-k}$ it is $2^{1+kn/2}$; these values are unbounded as $k\to\infty$. No finite radius has an infinite average: the displayed formula is finite for every $r>0$.

A Pitfall About Infinite Values

A tempting but invalid shortcut is to apply the homogeneity identity $M(cf)=|c|Mf$ with $c=0$ without checking what the right-hand side means. In the singularity example, $Mf(0)=+\infty$. Substituting $c=0$ into the expression $|c|Mf(0)$ produces $0\cdot(+\infty)$, which is undefined in extended-real arithmetic. The definition itself resolves the case: every ball average of $0f$ is zero, so $M(0f)(0)=0$. The homogeneity identity is stated for $c\ne0$, and the zero scalar is handled separately.

A second useful check in radial calculations is to distinguish the integral over the part of the support inside a ball from the integral over the full support. In the annulus example, a ball of radius $r$ with $1<r<2$ contains only the annular portion with radii between $1$ and $r$, whose volume is $\omega_n(r^n-1)$. Using the full annulus volume $\omega_n(2^n-1)$ at such a radius would incorrectly count points outside the ball. The radius-by-radius cases above prevent that error and also verify where the supremum occurs.

The central distinction is that local integrability guarantees a finite average for each fixed ball, not a uniform bound over all radii at a point. The maximal function records that supremum, allowing the value $+\infty$. Its monotonicity, sublinearity, and homogeneity for nonzero scalars follow directly from the corresponding properties of absolute value and integration, with the zero-scalar case requiring its own direct calculation.

Check Your Understanding

Use the definition and verify the range of radii used in every radial calculation.

For $f(y)=c$ on $\mathbb{R}^n$, compute every centered ball average and determine $Mf(x)$ for all $x$.
For $f(y)=|y|^2$ on $B(0,1)$ and zero outside, derive the average at the origin separately for $0<r\le1$ and $r\ge1$.
For the annulus function with value $3$ on $1<|y|<2$, explain why the average increases for $1<r\le2$ and decreases for $r\ge2$.
For $f(y)=|y|^{-n/2}$ away from the origin, verify local integrability near zero and explain why $Mf(0)=+\infty$ although each fixed-radius average is finite.
Why must the homogeneity statement for $M(cf)$ treat $c=0$ separately when $Mf(x)$ may equal $+\infty$?