Local Maximal Estimates for Locally Integrable Functions

Why Local Integrability Requires a Local Estimate

The global weak-type estimate for the Hardy–Littlewood maximal function assumes that the function is integrable over all of $\mathbb{R}^n$. A function can be integrable on every bounded set and still have arbitrarily large averages over sufficiently large balls. To obtain a useful estimate from local integrability, we therefore restrict the radii of the balls and examine the maximal function only on a bounded region. Then every ball relevant to the estimate lies inside a larger bounded set, where the function is integrable.

Throughout, $m$ denotes Lebesgue measure on $\mathbb{R}^n$. Functions may be real- or complex-valued, are Lebesgue measurable, and are locally integrable when $\int_C |f|\,dm<\infty$ for every compact set $C\subset\mathbb{R}^n$. The maximal functions below use $|f|$, so they are nonnegative and may take the value $+\infty$. All level sets use the strict inequality $> \lambda$.

Definition: For $R>0$, the centered maximal function restricted to radii at most $R$ is $$ M_R f(x)=\sup_{0<r\le R}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|\,dm(y), $$ where $B(x,r)$ is the open Euclidean ball with center $x$ and radius $r$. The unrestricted centered maximal function is $Mf(x)=\sup_{r>0}m(B(x,r))^{-1}\int_{B(x,r)}|f|\,dm$. The restricted function is measurable when $f$ is locally integrable, by the measurability result for maximal functions.

We use the global weak-type estimate for the centered maximal function, established in the earlier material on maximal functions and covering lemmas: there is a dimensional constant $C_n<\infty$ such that, for every $g\in L^1(\mathbb{R}^n)$ and every $\lambda>0$, $$ m(\{x\in\mathbb{R}^n:Mg(x)>\lambda\})\le \frac{C_n}{\lambda}\int_{\mathbb{R}^n}|g|\,dm. $$ Only this global result requires global integrability. The argument below shows how to apply it when $f$ is merely locally integrable.

The Truncation That Captures Every Relevant Ball

Let $K\subset\mathbb{R}^n$ be bounded and measurable, and let $R>0$. Define its closed $R$-neighbourhood by $$ D_R(K)=\{y\in\mathbb{R}^n:\operatorname{dist}(y,K)\le R\}. $$ This set is closed and bounded, hence compact. If $x\in K$ and $0<r\le R$, every $y\in B(x,r)$ satisfies $\operatorname{dist}(y,K)\le |y-x|<r\le R$. Thus $B(x,r)\subset D_R(K)$. This inclusion is the reason the truncation works.

Theorem (Local weak-type estimate): Let $f$ be locally integrable on $\mathbb{R}^n$, let $K\subset\mathbb{R}^n$ be bounded and measurable, let $R>0$, and let $\lambda>0$. Then $$ m(\{x\in K:M_R f(x)>\lambda\}) \le \frac{C_n}{\lambda}\int_{D_R(K)}|f|\,dm. $$ The constant $C_n$ is the same dimensional constant as in the global weak-type estimate. No assumption that $f\in L^1(\mathbb{R}^n)$ is needed.

Proof. If $K$ is empty, the level set on the left is empty and the claimed inequality holds. Otherwise, $D_R(K)$ is compact. Local integrability therefore gives $\int_{D_R(K)}|f|\,dm<\infty$. Define $g=|f|\chi_{D_R(K)}$. Then $g$ is measurable and belongs to $L^1(\mathbb{R}^n)$, with $\int_{\mathbb{R}^n}g\,dm=\int_{D_R(K)}|f|\,dm$. For every $x\in K$ and every $0<r\le R$, the ball $B(x,r)$ lies in $D_R(K)$, so $$ \int_{B(x,r)}g\,dm=\int_{B(x,r)}|f|\,dm. $$ Taking the supremum over $0<r\le R$ gives $M_R f(x)\le Mg(x)$ (in fact, the restricted supremum of these averages of $g$ equals $M_R f(x)$). Consequently, $$ \{x\in K:M_R f(x)>\lambda\} \subseteq \{x\in\mathbb{R}^n:Mg(x)>\lambda\}. $$ Apply the global weak-type estimate to $g$, which is integrable as verified above. Monotonicity of measure and the identity for $\int g$ yield the asserted bound. $\square$

Corollary: If $K$ is compact and $U$ is an open set containing $K$, choose $R>0$ small enough that $D_R(K)\subset U$. Then the local estimate can use only the integral over $U$: $$ m(\{x\in K:M_R f(x)>\lambda\}) \le \frac{C_n}{\lambda}\int_U|f|\,dm. $$

Proof. Since $K$ is compact and $U$ is open with $K\subset U$, the distance from $K$ to the closed set $U^c$ is positive when $U^c$ is nonempty. Choose $R$ smaller than that distance. Then $D_R(K)\subset U$, and nonnegativity of $|f|$ gives $\int_{D_R(K)}|f|\,dm\le\int_U|f|\,dm$. The theorem proves the result. If $U=\mathbb{R}^n$, the inclusion holds for every $R>0$, and the same integral comparison applies. $\square$

Worked Local Bounds

Worked Example: An Indicator on a Bounded Interval

Take $n=1$, $f=\chi_{[0,1]}$, $K=[-1,1]$, $R=1$, and $\lambda=\tfrac12$. The function is locally integrable, and the relevant neighbourhood is $D_1(K)=[-2,2]$. The local theorem gives $$ m(\{x\in[-1,1]:M_1 f(x)>\tfrac12\}) \le \frac{C_1}{1/2}\int_{-2}^{2}\chi_{[0,1]}(y)\,dy =2C_1\cdot 1=2C_1. $$ Here the integral is $1$, since the indicator equals $1$ on the interval $[0,1]$ of length $1$ and vanishes elsewhere, up to endpoints of measure zero.

Check: the averaging balls centered in $[-1,1]$ with radii at most $1$ are all contained in $[-2,2]$, exactly the region on which the integral was taken. Also $M_1 f(x)\le1$, because every average of an indicator is at most $1$; thus the level set is a subset of $[-1,1]$ and has measure at most $2$. This is consistent with the stated weak bound.

Worked Example: A Locally Integrable Function That Grows

In $\mathbb{R}$, let $f(x)=2+|x|$, $K=[0,1]$, $R=1$, and $\lambda=3$. Then $D_1(K)=[-1,2]$, and $f$ is locally integrable. Compute the integral by splitting at $0$: $$ \int_{-1}^{2}(2+|x|)\,dx =\int_{-1}^{0}(2-x)\,dx+\int_0^2(2+x)\,dx =\frac52+6=\frac{17}{2}. $$ Therefore, $$ m(\{x\in[0,1]:M_1 f(x)>3\}) \le \frac{C_1}{3}\cdot\frac{17}{2} =\frac{17C_1}{6}. $$ The function need not be bounded on all of $\mathbb{R}$; only its integral on the bounded neighbourhood is used.

Check: the first integral is $[2x-x^2/2]_{-1}^{0}=5/2$, and the second is $[2x+x^2/2]_0^2=4+2=6$. Their sum is $5/2+12/2=17/2$, as used in the estimate.

Worked Example: An Unbounded Locally Integrable Function

Define $f(0)=0$ and $f(x)=|x|^{-1/2}$ for $x\ne0$. This function is unbounded near $0$, but it is locally integrable because its singularity has a finite integral. Set $K=[-1,1]$, $R=1$, and $\lambda=2$. Then $D_1(K)=[-2,2]$, and $$ \int_{-2}^{2}|f(x)|\,dx =2\int_0^2 x^{-1/2}\,dx =2[2\sqrt{x}]_0^2 =4\sqrt{2}. $$ The local estimate is consequently $$ m(\{x\in[-1,1]:M_1 f(x)>2\}) \le \frac{C_1}{2}\cdot4\sqrt{2} =2\sqrt{2}\,C_1. $$ The value assigned at $0$ does not change the integral, since the singleton $\{0\}$ has Lebesgue measure zero.

Check: integrating on the positive and negative halves gives $2\sqrt2$ on each half, hence $4\sqrt2$ in total. The bound uses a finite integral even though $f$ is unbounded.

Why the Radius Restriction Cannot Be Dropped

The local estimate concerns $M_R f$, not automatically the unrestricted $Mf$. In a ball of radius greater than $R$, points can lie outside $D_R(K)$; truncating $f$ there would change the average. More fundamentally, local integrability alone places no control on the integral over arbitrarily large balls. A common error is to apply the global weak-type estimate directly to a locally integrable function, or to truncate it on $K$ alone and then claim that all balls centered in $K$ are unaffected. The balls of radius at most $R$ require the larger neighbourhood $D_R(K)$; unrestricted balls require additional global information.

For a concrete obstruction, take $f(x)=e^{x^2}$ on $\mathbb{R}$. It is real-valued, continuous, and locally integrable, but it is not globally integrable. Fix any $x\in\mathbb{R}$, and take $r$ sufficiently large that $2r\ge1$ and $x+r-1\ge0$. The interval $[x+r-1,x+r]$ lies inside $B(x,r)=(x-r,x+r)$, up to an endpoint, and $e^{y^2}\ge e^{(x+r-1)^2}$ on that interval. Therefore $$ \frac{1}{m(B(x,r))}\int_{B(x,r)}e^{y^2}\,dy \ge \frac{e^{(x+r-1)^2}}{2r}\longrightarrow\infty \quad\text{as }r\to\infty. $$ Thus $Mf(x)=\infty$ for every $x$. In particular, its unrestricted superlevel set is all of $\mathbb{R}$ for every finite $\lambda>0$, even though the integral of $f$ on any fixed bounded neighbourhood is finite.

When $f\in L^1(\mathbb{R}^n)$, the global weak-type estimate does apply directly to $Mf$. When $f$ is only locally integrable, the proven local estimate instead supplies the control needed for averages over bounded radii and bounded regions. Keeping those hypotheses distinct is essential: local integrability gives finite integrals on compact sets, not a bound for averages over all scales.

Check Your Understanding

In each question, keep track of the bounded region containing the averaging balls and of whether the maximal function is restricted in radius.

For $K=[2,3]\subset\mathbb{R}$ and $R=\tfrac12$, write $D_R(K)$ explicitly.
State the local weak-type bound for a locally integrable $f$, a bounded measurable $K$, $R>0$, and $\lambda>0$.
For $f=\chi_{[0,1]}$, $K=[-1,1]$, and $R=1$, compute $\int_{D_R(K)}|f|\,dm$.
Why does truncation to $D_R(K)$ preserve every average used in $M_R f(x)$ for $x\in K$?
Give the missing hypothesis that permits applying the global weak-type estimate directly to the unrestricted maximal function $Mf$.

Local Maximal Estimates for Locally Integrable Functions

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