Spring 2018 - Problem 4

Solution.

1. Let

   $$Hf\p{x} = \sup_{r > 0} \frac{1}{2r} \int_{x-r}^{x+r} \abs{f\p{y}} \,\diff{y}$$

   be the Hardy-Littlewood maximal function. Observe that $\norm{Hf}_{L^\infty} \leq \norm{f}_{L^\infty}$, so by sublinearity,

   $$\begin{aligned} m\p{\set{Hf > \alpha}} &\leq m\p{\set{Hf\chi_{\set{\abs{f} > \frac{\alpha}{2}}} > \frac{\alpha}{2}}} + m\p{\set{Hf\chi_{\set{\abs{f} \leq \frac{\alpha}{2}}} > \frac{\alpha}{2}}} \\ &= m\p{\set{Hf\chi_{\set{\abs{f} > \frac{\alpha}{2}}} > \frac{\alpha}{2}}}. \end{aligned}$$

   Then by the Hardy-Littlewood maximal inequality,

   $$\begin{aligned} \norm{Hf}_{L^p}^p &= \int_0^\infty p\alpha^{p-1} m\p{\set{Hf > \alpha}} \,\diff\alpha \\ &\leq \int_0^\infty p\alpha^{p-1} m\p{\set{Hf\chi_{\set{\abs{f} > \frac{\alpha}{2}}} > \frac{\alpha}{2}}} \,\diff\alpha \\ &\leq \int_0^\infty Cp\alpha^{p-2} \int_{\set{\abs{f\p{x}} > \frac{\alpha}{2}}} \abs{f\p{x}} \,\diff{x} \,\diff\alpha \\ &= Cp \int_\R \abs{f\p{x}} \int_0^{2\abs{f\p{x}}} \alpha^{p-2} \,\diff\alpha \,\diff{x} && \p{\text{Fubini-Tonelli}} \\ &= \frac{C2^{p-1}p}{p - 1} \int_\R \abs{f\p{x}}^p \,\diff{x} \\ &= \frac{C2^{p-1}p}{p - 1} \norm{f}_{L^p}^p. \end{aligned}$$

   for some constant $C > 0$. We now apply it twice: observe that by Fubini-Tonelli,

   $$\norm{y \mapsto \norm{f_y}}_{L^p\p{\R}}^p = \int_\R \norm{f_y}_{L^p}^p \,\diff{y} = \int_\R \p{\int_\R \abs{f\p{x, y}}^p \,\diff{x}} \,\diff{y} = \norm{f}_{L^p\p{\R^2}}^p,$$

   and so

   $$\begin{aligned} \abs{\frac{1}{4r\rho} \int_{-r}^r \int_{-\rho}^\rho f\p{x + h, y + \ell} \,\diff{h} \,\diff{\ell}} &\leq \frac{1}{2r} \int_{-r}^r \p{\frac{1}{2\rho} \int_{-\rho}^\rho \abs{f\p{x + h, y + \ell}} \,\diff{h}} \,\diff\ell \\ &\leq \frac{1}{2r} \int_{-r}^r Hf_{y+\ell}\p{x} \,\diff\ell \\ &\leq H\p{Hf_y\p{x}}\p{y} \\ \implies \abs{\br{Mf}\p{x, y}} &\leq H\p{Hf_y\p{x}}\p{y}. \end{aligned}$$

   Hence, if we replace $C$ with $\frac{C2^{p-1}p}{p - 1}$, we have

   $$\begin{aligned} \norm{Mf}_{L^p\p{\R^2}}^p &\leq \norm{\p{x, y} \mapsto H\p{Hf_y\p{x}}\p{y}}_{L^p\p{\R^2}} \\ &= \norm{x \mapsto \norm{y \mapsto H\p{Hf_y\p{x}}\p{y}}_{L^p\p{\R}}}_{L^p\p{\R}} \\ &\leq C\norm{x \mapsto \norm{y \mapsto Hf_y\p{x}}_{L^p\p{\R}}}_{L^p\p{\R}} \\ &\leq C^2 \norm{x \mapsto \norm{y \mapsto f_y\p{x}}_{L^p\p{\R}}}_{L^p\p{\R}} \\ &= C^2 \norm{f}_{L^p\p{\R^2}}. \end{aligned}$$

2. Let

   $$\br{H_rf}\p{x, y} = \frac{1}{4r^3} \int_{-r}^r \int_{-r^2}^{r^2} \abs{f\p{x + h, y + \ell}} \,\diff{h} \,\diff{\ell},$$

   which we showed via the previous calculation that $H_r$ is strong type $\p{p, p}$. As in the proof of the Lebesgue differentiation theorem, let $\alpha > 0$. First, observe that by Chebyshev's inequality,

   $$m\p{\set{\p{x, y} \in \R^2 \st \br{H_rf}\p{x, y} > \alpha}} \leq \frac{C}{\alpha} \norm{H_rf}_{L^p} \leq \frac{C}{\alpha} \norm{f}_{L^p}$$

   for any $f \in L^p\p{\R}$. Notice that the result is certainly true for continuous functions, so let $\epsilon > 0$ and pick $g$ continuous and compactly supported such that $\norm{f - g}_{L^p} < \epsilon$. Then

   $$\begin{gathered} \abs{\br{A_rf}\p{x, y} - f\p{x, y}} \leq \abs{\br{A_r\p{f - g}}\p{x, y}} + \abs{\br{A_rg}\p{x, y} - g\p{x, y}} + \abs{f\p{x, y} - g\p{x, y}} \\ \implies \limsup_{r\to0} \,\abs{\br{A_rf}\p{x, y} - f\p{x, y}} \leq \limsup_{r\to0} \,\abs{\br{A_r\p{f - g}}\p{x, y}} + \abs{f\p{x, y} - g\p{x, y}}. \end{gathered}$$

   Hence,

   $$\begin{aligned} m\p{\set{\limsup_{r\to0} \,\abs{\br{A_rf}\p{x, y} - f\p{x, y}} > \alpha}} &\leq m\p{\set{\limsup_{r\to0} \,\abs{\br{A_r\p{f - g}}\p{x, y}} > \frac{\alpha}{2}}} + m\p{\set{\abs{f\p{x, y} - g\p{x, y}} > \frac{\alpha}{2}}} \\ &\leq m\p{\set{\br{H_r\p{f - g}\p{x, y} > \frac{\alpha}{2}}}} + m\p{\set{\abs{f\p{x, y} - g\p{x, y}} > \frac{\alpha}{2}}} \\ &\leq \frac{2C}{\alpha} \norm{f - g}_{L^p} + \frac{2}{\alpha} \norm{f - g}_{L^p} \\ &\leq \p{\frac{2C + 2}{\alpha}} \epsilon. \end{aligned}$$

   Thus, the set where the $m\p{\set{\limsup_{r\to0} \,\abs{\br{A_rf}\p{x, y} - f\p{x, y}} > \alpha}} = 0$ for all $\alpha > 0$, so $\br{A_rf} \to f$ for almost every $\p{x, y} \in \R^2$.