advanced38 min studyLesson 378 in Mathematics

Maximal functions and weak-type bounds

Measure the largest local average, derive an exact interval example, and distinguish weak L¹ control from strong norm bounds.

On this page

What you’ll understand

  • Define the centered Hardy–Littlewood maximal function.
  • Interpret weak-type bounds as estimates for exceptional-set measure.
  • Explain why maximal averaging does not generally preserve L¹.

One local average can miss a feature because its radius is poorly chosen. The Hardy–Littlewood maximal function takes the largest average over every radius at each location, exposing a function's possible concentration across scales. This is a nonlinear operation even though each individual average is linear before taking absolute values. Its natural endpoint control is a bound on how large the exceptional set can be, rather than a finite integral of the output. That distinction makes maximal estimates useful in differentiation and convergence arguments without promising an L¹ norm bound that is false.

A supremum of local absolute averages

For locally integrable ff on Rd\mathbb R^d, define Mf(x)Mf(x) as the supremum over positive radii of the average of ∣f∣|f| on the Euclidean ball centered at xx. The output may initially be infinite at some points. Taking absolute values before averaging prevents cancellation, and taking a supremum makes the operator sublinear rather than linear. Centered balls are part of this definition; allowing every ball containing the point yields a related uncentered maximal operator with comparable bounds, but different exact values.

A bounded input has maximal function bounded by its essential supremum, because each average has that bound. For integrable inputs a more subtle estimate uses a covering argument: points with large maximal value have witnessing balls with large average, and a suitably selected disjoint collection controls the measure of their union after enlargement. Constants depend on dimension and the geometry of the balls. Simply summing every witnessing ball would overcount without control, which is why the covering step is essential.

Mf(x)=sup⁡r>01∣B(x,r)∣∫B(x,r)∣f(y)∣ dyMf(x)=\sup_{r>0}\frac1{|B(x,r)|}\int_{B(x,r)}|f(y)|\,dy
The definition is centered and uses Euclidean Lebesgue volume. Local integrability makes each finite-radius average meaningful.↔ Scroll wide equations horizontally.

Weak control is not an integral bound

For f∈L1(Rd)f\in L^1(\mathbb R^d) and every λ>0\lambda>0, the set where MfMf exceeds λ\lambda has measure at most Cd∥f∥1/λC_d\|f\|_1/\lambda. This is weak type (1,1), also described as a weak-L¹ estimate. It implies that an integrable input has finite maximal function almost everywhere, but it does not imply that the maximal function itself is integrable. A tail proportional to the reciprocal of distance in one dimension already shows the difference.

Combining the weak endpoint estimate with the L∞ averaging bound through an appropriate sublinear interpolation argument gives strong Lᵖ bounds for every exponent strictly greater than one and finite; the infinity bound is separate and direct. Constants need not remain bounded as the exponent approaches one. Do not substitute p=1 into the strong estimate or infer a uniform constant across exponents. Endpoint spaces record information that ordinary strong-norm statements lose, and their exceptional-set formulation is often exactly what an almost-everywhere argument needs.

∣{x:Mf(x)>λ}∣≤Cdλ∥f∥1,∥Mf∥p≤Cd,p∥f∥p(1<p<∞)|\{x:Mf(x)>\lambda\}|\le\frac{C_d}{\lambda}\|f\|_1,\qquad\|Mf\|_p\le C_{d,p}\|f\|_p\quad(1<p<\infty)
The first estimate assumes f in L¹ and λ>0; the second assumes f in Lᵖ and does not extend as a strong L¹ bound.↔ Scroll wide equations horizontally.

Connect estimates to local recovery

The Lebesgue differentiation theorem states that for a locally integrable function, averages of its absolute difference from its value tend to zero over shrinking balls at almost every point. Such points are Lebesgue points. One proof approximates by continuous functions, where local recovery is straightforward, then uses maximal weak bounds to control the set where approximation errors can spoil the conclusion. The estimate controls the exceptional set before passing to the approximation limit; norm density alone would not give the needed pointwise statement.

Differentiation remains an almost-everywhere theorem. A single jump can have a local average different from any arbitrarily assigned value, and changing the input on a null set changes no ball integral. The maximal function also takes a supremum over large as well as small balls, so it is not the same as the shrinking-radius limit. In applications separate the operator that measures worst-case averages from the limiting process it helps control. Numerical inspection of finitely many radii can illustrate the idea but cannot replace the supremum or its measure estimate.

A centered maximal function has a nonintegrable tail

Let f be the indicator of [0,1] on the line. Compute Mf(x) for x>1, evaluate it at x=2, and determine whether Mf belongs to L¹.

  1. At x>1 a centered interval of radius r≤x−1 misses the support. For x−1<r<x its overlap length is r−(x−1), so the average is [r−(x−1)]/(2r), increasing with r.

  2. For r≥x the interval covers all of [0,1], giving average 1/(2r), decreasing with r. The supremum occurs at r=x, so Mf(x)=1/(2x). In particular Mf(2)=1/4.

  3. Although the input has L¹ norm one, integrating its maximal tail from x=1 to infinity gives the divergent integral of 1/(2x). Thus Mf is not in L¹. This is consistent with weak type (1,1), which controls level-set measures rather than the output integral.

The result

The centered maximal function of a compactly supported integrable input can have an infinite L¹ norm. The exact tail also distinguishes centered from uncentered averaging conventions.

Explore the model

Interactive lab

Interactive lab · function

Compute local averages from an area readout

Pause and Reset, then set both controls to a=1 and x=1. For the local test f(y)=y², divide the displayed area from 0 to 1 by its length to predict the centered average over [−1,1], namely 1/3 by symmetry. Reset and set a=1 and x=2; area 8/3 gives centered average 4/3 over [−2,2]. The polynomial is only locally integrable, not globally L¹; this illustrates radius-dependent averages and does not test the global weak-type bound.

Compute local averages from an area readoutThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTaking a maximum over radii is an extra operation beyond evaluating one average; area, average, and point value are distinct.

Check your understanding

What does the weak type (1,1) estimate for the maximal operator control?

Try it yourself

For the same centered maximal function of the indicator of [0,1], find the measure of {Mf>1/5}. Use symmetry and the formula from the worked example.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Inside (0,1) the maximal value is one. Reflect the right-hand tail about x=1/2 to obtain the left-hand tail.

Compare with the solution

For x>1 the inequality 1/(2x)>1/5 gives x<5/2. For x<0 symmetry gives Mf(x)=1/[2(1−x)], and the inequality gives x>−3/2. The support interior also lies in the superlevel set, and endpoint conventions do not affect its measure. The set is the interval (−3/2,5/2) up to boundary points, with measure four. The input norm divided by the threshold is five, so this example is compatible with a weak estimate but says nothing about the sharp universal constant.

Take it with you

  • The centered maximal function takes a supremum of absolute local averages.
  • Weak L¹ control concerns exceptional-set size, not output integrability.
  • Maximal estimates connect norm approximation with almost-everywhere local recovery.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.