Maximal Functions of Interval Indicators

Centered Averages of an Interval

At a point outside a set, a centered interval can still capture part of that set, and its average depends on both the amount captured and the radius. For an interval indicator, this competition can be computed exactly. For a finite union, one centered interval may capture several components at once, so the separate maximal functions of the components do not in general combine by taking their maximum.

All measures and integrals below are Lebesgue measure $m$ and Lebesgue integrals on $\mathbb{R}$. The function $f=\chi_E$ is real-valued and measurable. For a locally integrable function $f$, its centered Hardy–Littlewood maximal function is

Mf(x)=\sup_{r>0}\frac{1}{2r}\int_{x-r}^{x+r}|f(y)|\,dy.

For an indicator, the numerator is $m(E\cap(x-r,x+r))$. Whether interval endpoints are included in $E$, or in the averaging interval, does not change this measure. The supremum is over all positive radii; it need not be attained.

Proposition: Let $I\subset\mathbb{R}$ be an interval with finite positive length $L=b-a$, where $a<b$. For $f=\chi_I$, the centered maximal function is given everywhere by $$ M\chi_I(x)= \begin{cases} 1, & a<x<b,\\[2pt] \frac12, & x=a\text{ or }x=b,\\[2pt] \frac{L}{2(L+d)}, & x\notin[a,b], \end{cases} $$ where $d=\operatorname{dist}(x,[a,b])$. These values are independent of whether either endpoint belongs to $I$.

Proof. If $a<x<b$, then for every sufficiently small $r>0$, the interval $(x-r,x+r)$ lies in $I$ up to endpoints. Its average of $\chi_I$ is $1$. No average can exceed $1$, so $M\chi_I(x)=1$.

Suppose $x<a$, and write $d=a-x>0$. For $r\le d$, the overlap has measure zero. For $r>d$, its measure is $\min(r-d,L)$. Thus the average is zero for $r\le d$, is $(r-d)/(2r)$ when $d<r\le d+L$, and is $L/(2r)$ when $r\ge d+L$. On the middle range, $(r-d)/(2r)=\tfrac12(1-d/r)$ is increasing in $r$; on the final range, $L/(2r)$ is decreasing. The maximum is therefore at $r=d+L$, with value $L/[2(d+L)]$. If $x>b$, reflection about the midpoint of $I$ gives the same calculation with $d=x-b$.

At $x=a$, every centered interval of radius $r\le L$ overlaps $I$ in length $r$, giving average $1/2$. For $r\ge L$, the overlap length is at most $L$, so the average is at most $L/(2r)\le1/2$. Hence the supremum is $1/2$. At $x=b$, the reflected argument gives $1/2$. This proves the formula in every case. $\square$

Worked Example: The Superlevel Sets for One Interval

Let $I=[2,5]$, so $L=3$. Compute $M\chi_I$ and the strict superlevel set $\{x:M\chi_I(x)>\lambda\}$ for $\lambda=1/4,\ 1/2,\ 3/4,\ 1$.

For $x<2$, the distance is $d=2-x$, and the formula gives $M\chi_I(x)=3/[2(3+2-x)]=3/[2(5-x)]$. For $x>5$, it gives $3/[2(3+x-5)]=3/[2(x-2)]$. In the interior the value is $1$, and at $2$ and $5$ it is $1/2$.

For $0<\lambda<1/2$, the left tail satisfies $3/[2(3+d)]>\lambda$ exactly when $d<3(1-2\lambda)/(2\lambda)$; the right tail has the same condition. The interior and both endpoints also belong to the strict superlevel set. Therefore the set is $(2-R,5+R)$, where $R=3(1-2\lambda)/(2\lambda)$. At $\lambda=1/4$, $R=3$, so the set is $(-1,8)$. At $\lambda=1/2$, the endpoints have value exactly $1/2$ and are excluded, leaving $(2,5)$. At $\lambda=3/4$, the tails and endpoints are below the threshold, while every interior point has value $1$, so the set is again $(2,5)$. At $\lambda=1$, no point has value strictly greater than $1$, so the set is empty.

Check: At $x=-1$, the distance to $[2,5]$ is $3$, and $3/[2(3+3)]=1/4$, exactly the threshold, so this boundary point is correctly excluded for $\lambda=1/4$. The endpoint values at $2$ and $5$ are exactly $1/2$, as required by the formula.

Finite Unions: Overlap Lengths and Superlevel Sets

Write a finite union of bounded intervals, after merging overlaps and intervals that meet at endpoints, as $E=\bigcup_{j=1}^N[a_j,b_j]$, with positive-length, pairwise separated components ordered from left to right. Open or half-open endpoint choices make no difference to $M\chi_E$, since they do not change any overlap measure. Set $L_j=b_j-a_j$. For $x\in\mathbb{R}$ and $r>0$, the overlap length is

q_E(x,r)=m(E\cap(x-r,x+r)) =\sum_{j=1}^N\max\!\left(0,\min(b_j,x+r)-\max(a_j,x-r)\right).

Theorem: For this finite union, at every $x\in\mathbb{R}$, $$ M\chi_E(x)=\sup_{r>0}\frac{q_E(x,r)}{2r}. $$ For every $\lambda>0$, its strict superlevel set is exactly $$ \{x:M\chi_E(x)>\lambda\} =\bigcup_{r>0}\{x:q_E(x,r)>2\lambda r\}. $$ For fixed $x$, the supremum can be found by checking the finitely many positive breakpoints $r=|x-a_j|$, $r=|x-b_j|$, and the limit as $r\downarrow0$; the limiting value as $r\to\infty$ is $0$.

Proof. The overlap formula follows by intersecting each component $[a_j,b_j]$ with $(x-r,x+r)$; the components are disjoint up to endpoints, so their overlap measures add. Substituting this formula in the definition of the centered maximal function proves the first equality. A supremum of numbers is strictly greater than $\lambda$ if and only if at least one of those numbers is strictly greater than $\lambda$. Applying this fact to the averages gives the superlevel-set equality, including its strict inequality.

For the finite-radius assertion, fix $x$. Each component’s overlap length is a continuous piecewise-linear function of $r$; its formula can change only when $x-r$ or $x+r$ reaches one of that component’s endpoints. Thus $q_E(x,r)=Ar+B$ on each interval between consecutive positive breakpoints, for constants $A,B$ on that interval. There the average is $A/2+B/(2r)$. If $B>0$, this is decreasing in $r$; if $B<0$, it is increasing; if $B=0$, it is constant. Its supremum on each such interval is therefore given by an endpoint value or a one-sided endpoint limit. The limits at positive breakpoints are included by continuity. The limit as $r\to\infty$ is $0$, because $q_E(x,r)\le m(E)<\infty$. The remaining possible endpoint limit is $r\downarrow0$. There are finitely many breakpoints, so these checks determine the supremum. If a listed breakpoint is $0$, it is accounted for by that limit rather than treated as a positive radius. $\square$

The theorem also gives a direct way to identify finite-union superlevel sets: for each radius, use the displayed overlap formula and solve the strict, piecewise-linear inequality $q_E(x,r)>2\lambda r$; then take the union over positive radii. In particular, the superlevel set is open when $\lambda>0$: for fixed $r$, the overlap is continuous as a function of $x$, so $\{x:q_E(x,r)>2\lambda r\}$ is open, and a union of open sets is open. The radius description is useful even when the resulting endpoints have no short single-interval formula.

Worked Example: Two Separated Intervals at the Midpoint

Take $E=[0,1]\cup[2,3]$ and $x=3/2$. Compute $M\chi_E(3/2)$, and compare it with the two separate component maximal functions at that point.

For $r\le1/2$, the centered interval has zero overlap with $E$. For $1/2<r\le3/2$, it overlaps each component in length $r-1/2$, so $q_E(3/2,r)=2r-1$ and the average is $1-1/(2r)$, increasing in $r$. For $r\ge3/2$, both components are covered and $q_E(3/2,r)=2$, so the average $1/r$ decreases. The maximum occurs at $r=3/2$: it equals $2/(2\cdot3/2)=2/3$. In particular, $3/2\in\{M\chi_E>1/2\}$.

For either component separately, the distance from $3/2$ to that component is $1/2$, and its length is $1$. The single-interval formula gives $1/[2(1+1/2)]=1/3$ for each. Thus the maximum of the two separate values is $1/3$, not the value $2/3$ for their union. Check: at radius $3/2$, the averaging interval is $(0,3)$; its overlap with the union has measure $2$, and its total length is $3$, giving $2/3$.

Worked Example: A Component Endpoint Near Another Interval

Let $E=[0,1]\cup[1.1,2.1]$, and evaluate $M\chi_E(1)$, which is an endpoint of the first component.

For $0<r\le0.1$, only the first component is met, in length $r$, and the average is $1/2$. For $0.1<r\le1$, the overlaps have lengths $r$ and $r-0.1$, respectively. Thus the total overlap is $2r-0.1$, and the average is $1-0.05/r$, increasing with $r$. For $1\le r\le1.1$, the first component is fully covered and the second contributes $r-0.1$; the average is $(r+0.9)/(2r)=1/2+0.45/r$, decreasing with $r$. For $r\ge1.1$, the total overlap is $2$, and $1/r$ is decreasing. The largest value is therefore at $r=1$, and equals $1.9/2=0.95$.

Check: at radius $1$, the averaging interval is $(0,2)$. Its overlap with $[0,1]$ has length $1$, and its overlap with $[1.1,2.1]$ has length $0.9$. The total is $1.9$, divided by the interval length $2$, giving $0.95$. Thus, unlike the endpoint of a single interval, this component endpoint has maximal value $0.95$.

Endpoint Behavior and a Common Pitfall

For one interval, each endpoint has maximal value exactly $1/2$. For a finite union, that conclusion need not hold at the endpoint of a component: a larger centered interval may also capture a nearby component, as the preceding computation shows. The overlap formula, rather than a componentwise endpoint rule, determines the correct value. It also ensures that strict superlevel sets use the correct endpoint convention: a point with value exactly $\lambda$ is not in $\{M\chi_E>\lambda\}$.

Common mistake. For $E=[0,1]\cup[2,3]$, one might calculate each component’s maximal function at $x=3/2$ and take the larger value as the maximal function of the union. Each component gives $1/3$, since its length is $1$ and its distance from $3/2$ is $1/2$. But a single radius-$3/2$ interval captures both components and gives average $2/3$. The operation of taking a centered average is applied to the whole set before taking the supremum over radii; maximizing component averages separately loses the benefit of capturing several components at once.

This explicit computation is useful when examining maximal-function superlevel sets: even for simple sets, the geometry of a centered interval can combine separated pieces. For a finite union, the overlap expression records that geometry exactly, gives a finite pointwise calculation, and keeps endpoint decisions tied to the actual strict threshold rather than to an assumed single-component rule.

Check Your Understanding

Use the overlap formula and pay attention to strict inequalities and endpoint values.

For $I=[-1,2]$, compute $M\chi_I(4)$, and determine whether $4$ belongs to $\{M\chi_I>1/4\}$.
For $I=[0,2]$, find the strict superlevel set $\{M\chi_I>1/2\}$, including the status of the endpoints.
For $E=[0,1]\cup[2,3]$, compute the average at $x=3/2$ for radius $r=1$. Is that the maximal value?
Explain why $q_E(x,r)$ is piecewise linear as a function of $r$ for fixed $x$, and name all possible positive breakpoints.
Can the endpoint of a component of a finite union have maximal value greater than $1/2$? Verify your answer using $E=[0,1]\cup[1.1,2.1]$ at $x=1$.

Maximal Functions of Interval Indicators

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