Real Analysis, Folland problem 3.3.23 Differentiation on Euclidean Space Background information: If $f\in L^{1}_{loc}$, we define its Hardy-Littlewood maximal function $Hf$ by $$Hf(x) = \sup_{r > 0}A_r|f|(x) = \sup_{r > 0}\frac{1}{m(B(r,x))}\int_{B(r,x))}|f(y)|dy$$ where $L^{1}_{loc}$ is the space of locally integrable functions. Also, $Hf$ is measurable.
A useful variant of the Hardy-Littlewood maximal function is $$H^{*}f(x) = \sup\{\frac{1}{m(B)}\int_B |f(y)|dy: B \ \text{is a ball and} \ x\in B\}$$
Show that $H f\leq H^{*}f\leq 2^n Hf$.
Attempted proof: First, we will show that $Hf\leq H^{*}f$. I believe this is true since the way Hardy-Littlewood maximal function $Hf$ is defined where $r$ is strictly positive. I am not sure how to show this in a rigorous fashion though. Any suggestions is greatly appreciated. 

First of all, the supremum in the definition of $H^*f(x)$ is taken over a set which contains the set in the definition of $Hf(x)$. Since suprema get larger as the sets get larger, this shows that $Hf(x)\leq H^*f(x)$.
For the other inequality, let $B$ be any ball containing $x$, and suppose that $B$ has radius $r$. Then $B\subset B(2r,x)$, hence
$$\frac{1}{m(B)}\int_B|f(y)|\;dy\leq \frac{m(B(2r,x))}{m(B)}\frac{1}{m(B(2r,x))}\int_{B(2r,x)}|f(y)|\;dy\leq 2^n Hf(x)$$
Since $B$ was any ball containing $x$, taking the supremum over all such balls shows that $H^*f(x)\leq 2^nHf(x)$.
