Let $\mathcal{R}$ denote the set of all open rectangles in $\mathbb{R}^2$ with sides parallel to the coordinate axis. Given a function on $\mathbb{R}^2$, consider the maximal function: $$f_{\mathcal{R}}^*(x)=\sup_{x\in R\in \mathcal{R}}\frac{1}{|R|}\int_R|f(y)|dy,\enspace x\in \mathbb{R}^2 ,$$ where the supremum is taken over all rectangles in $\mathcal{R}$ that contain the point $x$.
Problem Let $B=B_1(0)$ denote the unit ball in $\mathbb{R}^2$ and let $\varphi(x)=\frac{1}{\pi}\chi_B(x)$. For $\delta>0$, set $$\varphi_{\delta}(x)=\delta^{-2}\varphi(x/\delta)=\frac{1}{|B_{\delta}(0)|}\chi_{B_{\delta}(0)}(x). $$ Prove for that $x=(x_1,x_2)\in \mathbb{R}^2$ such that $|x_1x_2|>0$, $$\liminf_{\delta\to 0^+}(\varphi_\delta)_{\mathcal{R}}^*(x)\geq \frac{1}{16|x_1||x_2|}. $$
Discussion For $\delta>0$, let $R$ be an open rectangle containing $x$ such that $\delta B:=B_{\delta}(0) \subseteq R$. Then: \begin{align*} (\varphi_\delta)_{\mathcal{R}}^*(x)&\geq \frac{1}{|R|}\int_R \frac{1}{|\delta B|}\chi_{\delta B}(y) dy \\ &=\frac{1}{|R||\delta B|}\int_{\delta B} dy=\frac{1}{|R|}\\ \end{align*}
I'm not too sure how to proceed my brain is mush, I've been working on the assignment for a few days and any help or ideas is appreciated!
Continuing from the discussion above, we have: $$(\varphi_{\delta})_{\mathcal{R}}^*(x)\geq \frac{1}{R}$$ where $R$ is such that $x=(x_1,x_2)\in R$ and $\delta B\subset R$. We can specifically find such a rectangle: Consider $R:=(-2|x_1|-\delta,2|x_1|+\delta)\times (-2|x_2|-\delta,2|x_2|+\delta )$. Then it is easy to see that $\delta B\subset R$, and $x=(x_1,x_2)\in R$.
$$(\varphi_{\delta})_{\mathcal{R}}^*(x)\geq \frac{1}{R}=\frac{1}{(4|x_1|+2\delta)(4|x_2|+2\delta)}=\frac{1}{16|x_1||x_2|+8\delta(|x_1|+|x_2|)+4\delta^2}.$$ Taking $\liminf$ as $\delta\to 0^+$, we the desired result: $$ \liminf_{\delta\to 0+}(\varphi_{\delta})_{\mathcal{R}}^*(x)\geq \frac{1}{16|x_1||x_2|}$$ and of course our work also requires $|x_1x_2|>0$.