Weak Control Is Not Strong Control
The weak-type estimate for the Hardy–Littlewood maximal function controls the measure of the set where the maximal function exceeds a threshold. It does not say that the maximal function is integrable. Interval indicators make the distinction visible: their maximal functions have tails proportional to the reciprocal of distance. A truncated reciprocal profile gives a second, quantitative view of the same obstruction, with large lower bounds even when the output is integrated only over a bounded interval.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This space is complete, sigma-finite, and not finite. The functions \(f\) below are real-valued and Lebesgue measurable, and the maximal function may take values in \([0,\infty]\). For \(f\in L^1(m)\), define the centered maximal function by
The superlevel set will always mean the strict set \(\{x\in\mathbb{R}:Mf(x)>\lambda\}\), with \(\lambda>0\). Measurability of \(Mf\) and the weak-type estimate below are established by the earlier material on maximal-function measurability and interval covering. We use the one-dimensional estimate in the form
Here \(L^1(m)\) consists of equivalence classes of measurable functions with finite integral of their absolute value; the displayed norm is \(\|f\|_1=\int_{\mathbb{R}}|f|\,dm\). The estimate bounds the measure of a superlevel set, not the integral of \(Mf\). We first compute an exact superlevel set to see how much of the threshold dependence is unavoidable.
Worked Example: An Exact Superlevel Set for an Interval Indicator
Let \(f=2\chi_{[1,3]}\), so \(f\) is real-valued, measurable, and \(\|f\|_1=2\cdot 2=4\). Compute \(\{Mf>1/2\}\).
For \(1<x<3\), sufficiently small centered intervals about \(x\) lie in \([1,3]\), so their averages equal \(2\), and \(Mf(x)=2\). At either endpoint, centered intervals with radius tending to zero have half their length inside \([1,3]\); their averages equal \(1\), so \(Mf(1)\ge1\) and \(Mf(3)\ge1\). Thus every point of \([1,3]\) belongs to the strict superlevel set.
If \(x<1\), a centered interval of radius \(r\) has no intersection of positive measure with \([1,3]\) when \(r\le1-x\). For \(1-x<r<3-x\), its average is $$ \frac{2(x+r-1)}{2r}=1-\frac{1-x}{r}, $$ which increases as \(r\) increases. For \(r\ge3-x\), the interval contains all of \([1,3]\), and its average is \(4/(2r)\), which decreases as \(r\) increases. Therefore \(Mf(x)=4/[2(3-x)]\) for \(x<1\). The right-hand side is attained at \(r=3-x\).
For \(x>3\), the same calculation on the right gives \(Mf(x)=4/[2(x-1)]\): while the centered interval meets only part of \([1,3]\), its average increases up to radius \(x-1\), when it first contains the whole interval; beyond that radius the average decreases. Consequently the left and right strict threshold conditions are, respectively,
with \(x<1\) in the first calculation and \(x>3\) in the second. The boundary points \(-1\) and \(5\) give equality, so they are excluded. Combining both tails with \([1,3]\) gives \(\{Mf>1/2\}=(-1,5)\), whose measure is \(6\). Check: at \(x=0\), the left-tail value is \(4/[2(3)]=2/3>1/2\); at \(x=4\), the right-tail value is \(4/[2(3)]=2/3>1/2\). The weak-type ratio is \((1/2)\cdot6/4=3/4\), consistent with the bound \(3\).
The Threshold Dependence Cannot Be Improved
The weak-type estimate has a factor \(1/\lambda\). The next calculation shows that this order cannot be replaced by a smaller order as \(\lambda\) decreases to zero, even for one fixed integrable function. This is a statement about the dependence on \(\lambda\), not a claim that the numerical constant \(3\) is optimal.
Worked Example: A Fixed Indicator Tests the Weak-Type Order
Take \(f=\chi_{[0,1]}\) on \((\mathbb{R},\mathcal{L},m)\). Then \(\|f\|_1=1\). For \(0<\lambda<1/2\), compute its strict superlevel set.
For \(x<0\), the centered interval first reaches the support at radius \(-x\). As its radius grows to \(1-x\), the average increases; after radius \(1-x\), the whole support is included and the average decreases. Hence \(Mf(x)=1/[2(1-x)]\). For \(x>1\), the corresponding maximum occurs at radius \(x\), giving \(Mf(x)=1/(2x)\). For \(0<x<1\), sufficiently small intervals give \(Mf(x)=1\). At \(x=0\) and \(x=1\), the supremum is \(1/2\), which is greater than the assumed \(\lambda\).
Solving the strict inequalities in the two exterior regions gives $$ \frac{1}{2(1-x)}>\lambda\ \Longleftrightarrow\ x>1-\frac{1}{2\lambda}, \qquad \frac{1}{2x}>\lambda\ \Longleftrightarrow\ x<\frac{1}{2\lambda}. $$ Because \(\lambda<1/2\), these bounds extend past \(0\) and \(1\), respectively. Equality at either outer endpoint is excluded by the strict threshold. Thus $$ \{Mf>\lambda\}=\left(1-\frac{1}{2\lambda},\,\frac{1}{2\lambda}\right), \qquad m(\{Mf>\lambda\})=\frac{1}{\lambda}-1. $$ It follows that \(\lambda m(\{Mf>\lambda\})=1-\lambda\), which tends to \(1\) as \(\lambda\) decreases to zero.
Check: at \(\lambda=1/4\), the endpoints are \(1-2=-1\) and \(2\), so the set has measure \(3\); the formula gives \(1/(1/4)-1=3\), and \(\lambda m(\{Mf>\lambda\})=3/4=1-\lambda\).
Proof. Use \(f=\chi_{[0,1]}\), for which the calculation above gives \(\lambda m(\{Mf>\lambda\})=1-\lambda\) whenever \(0<\lambda<1/2\). The proposed bound would imply \(1-\lambda\le\eta(\lambda)\), which is impossible as \(\lambda\downarrow0\), since the left side tends to \(1\) and the right side tends to \(0\). For a proposed bound \(m(\{Mf>\lambda\})\le C\lambda^{-\alpha}\|f\|_1\) with \(\alpha<1\), the same example would require \(\lambda^{-1}-1\le C\lambda^{-\alpha}\). Multiplying by \(\lambda\) gives \(1-\lambda\le C\lambda^{1-\alpha}\), whose right side tends to zero while its left side tends to \(1\). This is a contradiction. \(\square\)
Why Strong \(L^1\) Boundedness Fails
Proof. Let \(f=\chi_{[0,1]}\), so \(\|f\|_1=1\). For \(x>1\), the centered interval \((x-x,x+x)=(0,2x)\) contains \([0,1]\). Its average of \(f\) is \(1/(2x)\), and therefore \(Mf(x)\ge1/(2x)\). For every \(R>1\), monotonicity of the Lebesgue integral gives $$ \int_{\mathbb{R}}Mf\,dm\ge\int_1^R\frac{1}{2x}\,dm(x)=\frac12\ln R. $$ As \(R\to\infty\), the right side tends to infinity, so \(\int_{\mathbb{R}}Mf\,dm=\infty\). Thus this integrable input has a maximal function that is not integrable. In particular, there cannot be a finite constant \(C\) such that \(\|Mf\|_1\le C\|f\|_1\) for every \(f\in L^1(m)\). \(\square\)
This example also identifies why the weak-type estimate is compatible with failure of strong \(L^1\) control. The bound \(Mf(x)\ge1/(2x)\) yields a logarithmically divergent integral, while its superlevel sets have finite measure at every positive threshold. A bound on the size of each superlevel set is not, by itself, an integrable majorant for \(Mf\).
Worked Example: A Truncated Reciprocal Profile
For an integer \(N\ge16\), define \(f_N(t)=1/t\) on \([1/N,1]\) and \(f_N(t)=0\) elsewhere. This is a nonnegative, real-valued Lebesgue measurable function in \(L^1(m)\), and direct integration gives $$ \|f_N\|_1=\int_{1/N}^{1}\frac{dt}{t}=\ln N. $$ We obtain a lower bound for \(Mf_N\) on the bounded interval \([N^{-1/2},1/2]\).
Fix \(x\) in that interval and choose the centered radius \(r=x\), so the averaging interval is \((0,2x)\) and has length \(2x\). Since \(x\ge N^{-1/2}\ge1/N\) and \(2x\le1\), its intersection with the support of \(f_N\) is \([1/N,2x]\). Therefore $$ Mf_N(x)\ge \frac{1}{2x}\int_{1/N}^{2x}\frac{dt}{t} =\frac{\ln(2Nx)}{2x}. $$ Integrating this lower bound over the stated interval yields $$ \int_{N^{-1/2}}^{1/2}Mf_N(x)\,dm(x) \ge \frac12\int_{N^{-1/2}}^{1/2}\frac{\ln(2Nx)}{x}\,dx =\frac14\left((\ln N)^2-\bigl(\ln(2\sqrt N)\bigr)^2\right). $$ The last quantity is positive for \(N>4\), since \(\ln N>\ln(2\sqrt N)\) is equivalent to \(N>4\).
Writing \(L=\ln N\), the lower bound divided by the input norm is $$ \frac{L^2-(L/2+\ln2)^2}{4L} =\frac{3L}{16}-\frac{\ln2}{4}-\frac{(\ln2)^2}{4L}. $$ As \(N\to\infty\), \(L\to\infty\), so this ratio tends to infinity. Thus even these lower bounds on bounded regions cannot be controlled by a fixed multiple of \(\|f_N\|_1\). Check: for \(N=16\), \(\|f_{16}\|_1=\ln16=4\ln2\), and the lower bound is \(\bigl((4\ln2)^2-(3\ln2)^2\bigr)/4=7(\ln2)^2/4\), in agreement with the displayed formula. The full integral of \(Mf_N\) is also infinite: for \(x>1\), the centered interval \((0,2x)\) contains the support and gives \(Mf_N(x)\ge(\ln N)/(2x)\), whose integral over \((1,\infty)\) diverges.
A Common Error About Sharpness
A common calculation error is to use the length of an interval in the numerator but then forget that the function has height \(c\). For \(f=c\chi_{[b,b+a]}\), the integral over its support is \(ca\), not \(a\). For example, with \(c=2\), \(a=2\), and \([b,b+a]=[1,3]\), the whole-interval average on a centered interval of length \(2(3-x)\) to the left is \(4/[2(3-x)]\), not \(2/[2(3-x)]\). At \(x=0\), these expressions are \(2/3\) and \(1/3\), respectively; only the first exceeds \(1/2\). The right tail likewise has numerator \(4\). Omitting the height changes the superlevel set and invalidates the claimed endpoint.
A second interpretive pitfall is to call the weak-type estimate “sharp” without specifying what is sharp. The interval calculation proves that the order \(1/\lambda\) cannot be improved to a smaller order uniformly. It does not prove that the constant \(3\) in the stated weak-type estimate is the best possible constant. The truncated reciprocal calculation serves a different purpose: it makes the failure of strong \(L^1\) control quantitative, while the interval indicator already proves that an integrable input can have a nonintegrable maximal function.
Check Your Understanding
In each question, use the strict superlevel convention \(\{Mf>\lambda\}\) and state the range of thresholds for which your calculation applies.
- For \(f=c\chi_{[b,b+a]}\), with \(a,c>0\), compute the maximal-function values outside the interval and derive the superlevel set when \(0<\lambda<c/2\).
- For \(f=\chi_{[0,1]}\), calculate \(\lambda m(\{Mf>\lambda\})\) for \(0<\lambda<1/2\), and explain what this shows about replacing the weak-type factor \(1/\lambda\).
- Use a centered interval to prove that \(M\chi_{[0,1]}(x)\ge1/(2x)\) for \(x>1\), and conclude that \(M\chi_{[0,1]}\notin L^1(m)\).
- For the truncated reciprocal function \(f_N\), verify that \([1/N,2x]\) is the relevant integration interval when \(N^{-1/2}\le x\le1/2\).
- Explain why an exact lower bound on the weak-type constant does not follow merely from showing that the dependence \(1/\lambda\) cannot be improved.