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Fall 2012 - Problem 6

Hardy-Littlewood maximal inequality, Lebesgue differentiation theorem

Let $f \in L^1\p{\R}$ and let $\mathcal{M}f$ denote its maximal function, that is,

\p{\mathcal{M}f}\p{x} = \sup_{0<r<\infty} \frac{1}{2r} \int_{-r}^r \abs{f\p{x - y}} \,\diff{y}.

By the Hardy-Littlewood maximal function theorem,

m\p{\set{x \in \R \mid \p{\mathcal{M}f}\p{x} > \lambda}} \leq 3\lambda^{-1}\norm{f}_{L^1} \quad\text{for all } \lambda > 0.

Using this, show that

\limsup_{r\to0} \frac{1}{2r} \int_{x-r}^{x+r} \abs{f\p{y} - f\p{x}} \,\diff{y} = 0 \quad\text{ for almost every } x \in \R.

Solution.

Let

\begin{aligned} \p{A_rf}\p{x} &= \frac{1}{2r} \int_{x-r}^{x+r} \abs{f\p{x} - f\p{y}} \,\diff{y} \\ \p{Af}\p{x} &= \limsup_{r\to0}\,\p{A_rf}\p{x}, \end{aligned}

so we need to show that $Af = 0$ for almost every $x \in \R$ .

Let $\epsilon > 0$ . Since compactly supported continuous functions are dense in $L^1\p{\R}$ , there exists $g \in C_c\p{\R}$ such that $\norm{f - g}_{L^1} < \epsilon$ . Thus,

\p{A_rf}\p{x} \leq \p{A_r\p{f - g}}\p{x} + \p{A_rg}\p{x}.

It's clear by continuity that the Lebesgue differentiation theorem holds at points of continuity, so $Ag = 0$ everywhere. Hence, if we set $h = f - g$ , $\p{Af}\p{x} \leq \p{Ah}\p{x}$ .

For $\lambda > 0$ , consider the set $E_\lambda = \set{x \in \R \st \p{Af} \geq \lambda}\p{x}$ . Notice that by the triangle inequality,

\begin{aligned} \p{A_rh}\p{x} &= \frac{1}{2r} \int_{x-r}^{x+r} \abs{h\p{x} - h\p{y}} \,\diff{y} \\ &\leq \frac{1}{2r} \int_{x-r}^{x+r} \abs{h\p{x}} \,\diff{y} + \frac{1}{2r} \int_{x-r}^{x+r} \abs{h\p{x}} \,\diff{y} \\ &\leq \p{\mathcal{M}h}\p{x} + \abs{h\p{x}}. \end{aligned}

Since this is independent of $r$ , we have $\p{Af}\p{x} \leq \p{Ah}\p{x} \leq \p{\mathcal{M}h}\p{x} + \abs{h\p{x}}$ . Thus,

E_\lambda \subseteq \set{x \in \R \st \p{\mathcal{M}h}\p{x} \geq \frac{\lambda}{2}} \cup \set{x \in \R \st \abs{h\p{x}} \geq \frac{\lambda}{2}}.

By the Hardy-Littlewood maximal theorem and Chebyshev's inequality, we get

m\p{E_\lambda} \leq \frac{6}{\lambda}\norm{h}_{L^1} + \frac{2}{\lambda}\norm{h}_{L^1} \leq \frac{8\epsilon}{\lambda} \xrightarrow{\epsilon\to0} 0.

Thus,

\set{x \in \R \st \p{Af} > 0} = \bigcup_{n=1}^\infty E_{1/n} \implies m\p{\set{x \in \R \st \p{Af} > 0}} \leq \sum_{n=1}^\infty m\p{E_{1/n}} = 0.

Hence, $Af = 0$ for almost every $x \in \R$ , which completes the proof.