Uncentered Maximal Functions and Geometric Comparison

Two Choices of Balls

A ball average can be tested using only balls centered at the point in question, or using every ball that contains that point. The second choice gives more candidates, so its maximal function is at least as large as the centered one. The key geometric fact is that any ball containing a point fits inside a ball centered at that point with twice the radius. Comparing the volumes then gives a dimension-dependent bound in the other direction.

Throughout, the ambient measure space is $(\mathbb{R}^n,\mathcal{L},m)$, where $\mathcal{L}$ is the Lebesgue sigma-algebra and $m$ is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. Write $B(a,r)=\{y\in\mathbb{R}^n:|y-a|<r\}$ for the open Euclidean ball with center $a$ and radius $r>0$. Its measure is $m(B(a,r))=\omega_n r^n$, where $\omega_n=m(B(0,1))$ is positive and finite.

Definition: Let $f:\mathbb{R}^n\to\mathbb{R}$ or $f:\mathbb{R}^n\to\mathbb{C}$ be Lebesgue measurable and locally integrable, meaning $\int_K |f|\,dm<\infty$ for every compact set $K\subset\mathbb{R}^n$. For $x\in\mathbb{R}^n$, define the centered and uncentered ball maximal functions by $$ M_{\mathrm{c}}f(x)=\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f|\,dm,\qquad M_{\mathrm{u}}f(x)=\sup_{\substack{a\in\mathbb{R}^n,\ r>0\\x\in B(a,r)}}\frac{1}{m(B(a,r))}\int_{B(a,r)}|f|\,dm. $$ The values are in $[0,\infty]$. These are pointwise definitions; this tutorial does not assert their measurability. Changing values of $f$ on a null set does not change these maximal functions, since all ball integrals are unchanged.

Local integrability ensures that every displayed ball integral is finite: the closure of a ball is compact, and the ball lies inside that closure. Each denominator is positive and finite. Absolute values are used even when $f$ is signed or complex-valued, so the averages in the definitions are nonnegative. The supremum may nevertheless be infinite.

The Geometric Comparison

Theorem: For every locally integrable real- or complex-valued $f$ on $(\mathbb{R}^n,\mathcal{L},m)$, the following inequalities hold at every $x\in\mathbb{R}^n$, with extended values allowed: $$ M_{\mathrm{c}}f(x)\le M_{\mathrm{u}}f(x)\le 2^n M_{\mathrm{c}}f(x). $$

Proof. Fix $x\in\mathbb{R}^n$. Every centered ball $B(x,r)$ contains $x$, so it is one of the balls used in the supremum defining $M_{\mathrm{u}}f(x)$. Taking suprema gives $M_{\mathrm{c}}f(x)\le M_{\mathrm{u}}f(x)$.

For the reverse bound, take any ball $B(a,r)$ with $x\in B(a,r)$. Thus $|a-x|<r$. For each $y\in B(a,r)$, the triangle inequality gives $$ |y-x|\le |y-a|+|a-x|<r+r=2r. $$ Consequently $B(a,r)\subset B(x,2r)$. Since $|f|\ge0$, monotonicity of the Lebesgue integral gives $$ \int_{B(a,r)}|f|\,dm\le \int_{B(x,2r)}|f|\,dm. $$ The volume formula yields $m(B(x,2r))=\omega_n(2r)^n=2^n\omega_n r^n=2^n m(B(a,r))$. Dividing by the positive measure $m(B(a,r))$, we obtain $$ \frac{1}{m(B(a,r))}\int_{B(a,r)}|f|\,dm \le 2^n\frac{1}{m(B(x,2r))}\int_{B(x,2r)}|f|\,dm \le 2^n M_{\mathrm{c}}f(x). $$ This holds for every ball in the uncentered supremum. Taking that supremum proves $M_{\mathrm{u}}f(x)\le2^nM_{\mathrm{c}}f(x)$. If $M_{\mathrm{c}}f(x)=\infty$, the last inequality is automatic in the extended order; otherwise the preceding bound applies to every candidate average and hence to their supremum. $\square$

The factor comes from the ratio of the volumes of the containing ball and the original ball, not from a comparison of their integrals alone. In particular, the argument uses both the containment and the fact that Lebesgue measure scales by the $n$-th power of the radius.

Worked Example: A Constant Function

Let $f(y)=c$ for a fixed real or complex number $c$, on $\mathbb{R}^n$. This function is locally integrable, including when $c=0$. For any ball $B(a,r)$, its average of $|f|$ is $$ \frac{1}{\omega_n r^n}\int_{B(a,r)}|c|\,dm =\frac{|c|\,m(B(a,r))}{m(B(a,r))} =\frac{|c|\omega_n r^n}{\omega_n r^n} =|c|. $$ Thus every candidate average in either definition equals $|c|$, and $M_{\mathrm{c}}f(x)=M_{\mathrm{u}}f(x)=|c|$ for every $x$. For $c=0$, both functions are identically zero; for any $c$, the comparison reads $|c|\le |c|\le 2^n|c|$, as required.

Worked Example: The Comparison Factor in One Dimension

On $(\mathbb{R},\mathcal{L},m)$, let $f=\chi_{[0,1]}$ and evaluate the two maximal functions at $x=2$. This is a measurable, locally integrable function, and endpoint values do not affect any of its Lebesgue integrals.

A centered ball is the interval $B(2,r)=(2-r,2+r)$, of length $2r$. If $0<r\le1$, its intersection with $[0,1]$ has measure zero. If $1<r<2$, that intersection has length $r-1$, so the average is $(r-1)/(2r)=1/2-1/(2r)$. This increases as $r$ increases to $2$. If $r\ge2$, the intersection has measure $1$, and the average is $1/(2r)$, which decreases as $r$ increases from $2$. Therefore the largest centered average is attained at $r=2$, and $$ M_{\mathrm{c}}f(2)=\frac{1}{2\cdot2}=\frac14. $$

For an uncentered interval $I=(a,b)$ containing $2$, one has $a<2<b$. If it has positive overlap with $[0,1]$, then $a<1$. When $0\le a<1$, the overlap has measure $1-a$, while $b-a>2-a$, giving $$ \frac{m(I\cap[0,1])}{m(I)}<\frac{1-a}{2-a}\le\frac12. $$ When $a<0$, the overlap has measure at most $1$ and the interval has length $b-a>2-a>2$, so its average is less than $1/2$. If $a\ge1$, the overlap has measure zero. Conversely, intervals $I=(0,2+\delta)$, for $\delta>0$, contain $2$ and have average $1/(2+\delta)$, which tends to $1/2$ as $\delta$ decreases to zero. Hence $M_{\mathrm{u}}f(2)=1/2$, and the comparison is exact: $$ M_{\mathrm{u}}f(2)=\frac12=2M_{\mathrm{c}}f(2). $$ This shows that in dimension one the factor $2$ cannot be replaced by a smaller universal factor.

Worked Example: Ball Volumes in Two Dimensions

On $(\mathbb{R}^2,\mathcal{L},m)$, let $f=\chi_{B(0,1)}$ and $x=(3,0)$. The support of this indicator is the closed unit disk $\overline{B(0,1)}$; using the open ball in the definition makes no difference to its integrals. The ball $B(0,4)$ contains $x$, since $|x|=3<4$, and contains the closed unit disk. Its area is $\pi\cdot4^2=16\pi$. Thus its uncentered average is $$ \frac{1}{16\pi}\int_{B(0,4)}\chi_{B(0,1)}\,dm =\frac{\pi}{16\pi} =\frac1{16}. $$ The centered ball $B(x,8)$ also contains the closed unit disk: if $|y|\le1$, then $|y-x|\le|y|+|x|\le4<8$. Its area is $\pi\cdot8^2=64\pi$, and its average is $$ \frac{1}{64\pi}\int_{B(x,8)}\chi_{B(0,1)}\,dm =\frac{\pi}{64\pi} =\frac1{64}. $$ These are particular candidate averages, so they imply $M_{\mathrm{u}}f(x)\ge1/16$ and $M_{\mathrm{c}}f(x)\ge1/64$; they are not asserted to be the exact maximal values. Their ratio is $4=2^2$, illustrating the volume ratio in the comparison proof.

Null Sets and Pointwise Values

Proposition: If $f$ and $g$ are locally integrable real- or complex-valued functions on $(\mathbb{R}^n,\mathcal{L},m)$ and $f=g$ $m$-a.e., then $M_{\mathrm{c}}f(x)=M_{\mathrm{c}}g(x)$ and $M_{\mathrm{u}}f(x)=M_{\mathrm{u}}g(x)$ for every $x\in\mathbb{R}^n$.

Proof. Since $f=g$ outside a Lebesgue-null set, $|f|=|g|$ outside that same set. For any ball $B$, the nonnegative measurable functions $|f|\chi_B$ and $|g|\chi_B$ therefore agree $m$-a.e., so their Lebesgue integrals are equal. Their averages over $B$ are equal because the denominator $m(B)$ is the same and is positive. This applies to every centered ball and every ball admissible in the uncentered definition. Taking the corresponding suprema proves both equalities at each $x$. $\square$

Worked Example: Changing a Function at One Point

On $\mathbb{R}$, set $g(y)=7$ if $y=0$ and $g(y)=0$ otherwise. This function is measurable and locally integrable. Every ball, whether or not it contains $0$, intersects the set $\{0\}$ in a set of Lebesgue measure zero. Consequently, for every ball $B$, $$ \int_B |g|\,dm=\int_B 7\chi_{\{0\}}\,dm =7m(B\cap\{0\})=0. $$ Every centered and uncentered average is therefore zero, so $M_{\mathrm{c}}g(x)=M_{\mathrm{u}}g(x)=0$ for every $x$. The pointwise value $g(0)=7$ does not create a positive ball average.

A Common Mistake and Why the Comparison Matters

A tempting but incorrect argument is: “The uncentered maximal function is no larger than the centered one, because a ball in either definition is just a ball average.” This reverses the inclusion of the families over which the suprema are taken. Centered balls are among the uncentered candidates, so the correct first inequality is $M_{\mathrm{c}}f\le M_{\mathrm{u}}f$. In the one-dimensional example, the incorrect claim would say $1/2\le1/4$, contradicting the computed values. The correction is to use containment: any ball containing $x$ lies in a centered ball at $x$ of twice its radius. The ratio of lengths in one dimension is $2$, giving $M_{\mathrm{u}}f\le2M_{\mathrm{c}}f$; in $\mathbb{R}^n$, the volume ratio is $2^n$.

These pointwise inequalities allow estimates for one maximal function to be transferred to the other, at the cost of the explicit factor $2^n$. The argument concerns pointwise values and ball geometry only; it does not require a claim about measurability of either maximal function. It also does not require the ambient measure space to be finite: only the finite, positive volumes of the Euclidean balls and local integrability of $f$ enter the proof.

Check Your Understanding

Use the definitions and the containment argument to answer each question.

Why does $M_{\mathrm{c}}f(x)\le M_{\mathrm{u}}f(x)$ hold for every $x$, and which inclusion of candidate families does it use?
For $x\in B(a,r)$, prove directly that $B(a,r)\subset B(x,2r)$.
In $\mathbb{R}^3$, what volume ratio replaces the one-dimensional factor $2$, and which radius scaling gives it?
For the indicator of $[0,1]$ at $x=2$, why is the uncentered value a supremum rather than an average attained by the intervals $(0,2+\delta)$ with $\delta>0$?
If two locally integrable functions agree $m$-a.e., explain why their maximal functions agree at every point, not merely almost everywhere.