Hardy-Littlewood maximal function $f_{\epsilon}^*(x)$ tends to $\frac{1}{|x_1||x_2|}$ as $\epsilon \to 0^+$ Consider the family of rescaled functions $f_{\epsilon}(x)=\frac{1}{\epsilon^{2}}f(\frac{x}{\epsilon})$, for $f(x)=\frac{1}{m(B)}\chi_B(x)$, where $\epsilon >0 $ and $B$ is the unit ball in $\mathbb{R}^2$.
Now consider the Hardy-Littlewood maximal function $$f_{\epsilon}^*(x)=\text{sup}_\mathfrak{R}\big\{ \frac{1}{m(R)}\int_R |\frac{1}{\epsilon^2 m(B)}\chi_B(x/\epsilon)|\big\},$$ where the sup ranges over all $R\in \mathfrak{R}(x)$, and $\mathfrak{R}(x)$ is the family of all rectangles containing $x$ with sides parallel to the axes. 

I want to show that $f_{\epsilon}^*(x)$ tends to $\frac{1}{|x_1||x_2|}$ as $\epsilon \to 0^+$, where $x=(x_1,x_2)$.

Some thoughts: 
First off, since $f(x)$ is non-negative, bounded, supported on the unit ball and $\int_{\mathbb{R}^2}f(x)=1$, we have that the family $f_\epsilon(x)$ is an approximation of the identity, as defined in Real Analysis book of Stein-Shakarchi, that is: $|f_\epsilon(x)|\leq \frac{A}{\epsilon^2}$ and $|f_\epsilon(x)|\leq \frac{A\epsilon}{|x|^{3}}$, for all $\epsilon>0$ and for all $x$ and the constant $A$ does not depend on $\epsilon$ (I guess it suffices to take $A=\frac{1}{m(B)}$ ?).\
I have:
$\frac{1}{|x_1||x_2|}=\text{sup}_\mathfrak{R}\big\{ \frac{1}{m(R)}\int_R |\frac{1}{ m(B)}\chi_B(x)|\big\}=f^*(x)=\text{lim}_{\epsilon\to 0}\int_{\mathbb{R}^2} f^*(x-y)f_\epsilon(y) dy=\text{lim}_{\epsilon\to 0}f^*(x)*f_\epsilon(x)=\text{lim}_{\epsilon\to 0}f^*_\epsilon(x)$ 
for $x=(x_1,x_2)$.
Does this make sense?
 A: Your $f_\epsilon$ represents a mass on $1$ uniformly distributed on a ball of radius $\epsilon$.  
Let's try to approximate the $\sup$ for small $\epsilon$.  Suppose $x$ is in the open first quadrant, i.e. $x_1, x_2 > 0$.  If $\epsilon$ small enough, the ball will be entirely down and to the left of $x$, so we can assume without loss of generality that any rectangle $R$ has $x$ as the top-right corner.  
Now, on the one hand,
$$f^*_\epsilon(x) \geq \frac{1}{m([-\epsilon, x_1] \times [-\epsilon, x_2])} \int_{[-\epsilon, x_1] \times [-\epsilon, x_2]} f_\epsilon(y) \;dy = \frac{1}{(x_1 + \epsilon)(x_2 + \epsilon)}$$
since the term on the right is a value of 
$$\frac{1}{m(R)}\int_R \frac{1}{\epsilon^2 m(B)}\chi_B(x/\epsilon)\;dx$$
for $R \in \mathcal{R}(x)$.
On the other hand, if the rectangle is $R = [y_1, x_1] \times [y_2, x_2]$, then $\int_R f_\epsilon(y)\;dy = 0$ if $y_1 \geq \epsilon$ or $y_2 \geq \epsilon$, and $\int_R f_\epsilon(y)\;dy \leq 1$.  Thus, certainly
$$f_\epsilon^*(x) \leq \frac{1}{m([\epsilon, x_1] \times [\epsilon, x_2])} = \frac{1}{(x_1 - \epsilon)(x_2 - \epsilon)}$$
So, taking $\epsilon \to 0$, we conclude by the squeeze theorem that the limit is $\frac{1}{x_1x_2}$ for $x_1, x_2 > 0$.
The situation for other values of $x$ in the plane is similar, and I leave it to you.
A: This isn't a full answer but too long for a comment. Your definition of the Hardy-Littlewood function doesn't make sense.  If $g$ is a nonnegative measurable function, then for any $x \in \mathbb R^2$ you have
$$g^*(x) = \sup_{R \in {\cal R}(x)} \frac{1}{m(R)} \int_R g(y) \, dy.$$ Thus
\begin{align*}f_\epsilon^*(x) = \sup_{R \in {\cal R}(x)} \frac{1}{m(R)} \int_R f_\epsilon(y) \, dy &= \sup_{R \in {\cal R}(x)}  \frac{1}{m(R)} \int_R \frac{1}{\epsilon^2m(B)} \chi_B\left( \frac y\epsilon \right) \, dy\\
&= \sup_{R \in {\cal R}(x)}  \frac{1}{m(R)m(B)} \int \frac{1}{\epsilon^2} \chi_R(y)\chi_B\left( \frac y\epsilon \right) \, dy\end{align*}
Make the substitution $z = \dfrac y\epsilon$ to rewrite the inner integral as
$$ \int \chi_R(\epsilon z) \chi_B(z) \, dz = \int \chi_{R/\epsilon}(z) \chi_B(z) \, dz= m(R/\epsilon \cap B).$$ This shows what you are really looking at is
$$f_\epsilon^*(x) = \frac{1}{m(B)}\sup_{R \in {\cal R}(x)} \frac{m(R/\epsilon \cap B)}{m(R)}$$ and its limit as $\epsilon \to 0^+$.
