An inequality for a maximal function on an $n$-ball. We have $Mf(x) = \sup_{r>0} \frac{c_n}{r^n} \int_{|y|\le r} |f(x-y)| dy$ the maximal function, where $r^n/c_n$ is the volume of the n-dimensional ball of radius $r$, $|y|\le r$.
I want to show that for $f_j=\chi_{(2^{j-1},2^j)}$, where $\chi$ is the charactersitic function, and $j=1,2,3,\ldots$, that: $Mf_j(x) \geq 1/8$ if $|x|\le 2^j$.
Here's, what I have done:
$2^{j-1} \le |x-y| \le 2^j$, so we get that $|y|<2^{j+1}$ and $|y|\le r$.
So we have that $Mf_j(x) \ge c_n/r^n \cdot r^n/(c_n\cdot 2^{(j+1)n})$ 
I don't see how to get the eighth there, anyone can help me with this?
It appears in Stein's real variable methods, orthogonality, and oscillatory integrals, 1993 edition pages 50-51.
 A: This estimate is a consequence of the explicit computation of the maximal function of a nonempty open interval $(a,b)$. By considering cases on $x$ and $r$, it is not hard to verify the following result.

Lemma For $-\infty\leq a<b\leq\infty$,
  $$M\chi_{(a,b)}(x)=\begin{cases} 1 & {a<x<b}\\ \\
2^{-1}\dfrac{b-a}{x-a} & {x\geq b}\\ \\ 2^{-1}\dfrac{b-a}{b-x} & 
{x\leq a}\end{cases}$$

Proof: Fix $-\infty\leq a<b\leq\infty$. Let $\mathcal{A}_{r}(x)$ denote the symmetric average at $x$ of radius $r>0$. We consider three cases for $x$: $a<x<b$, $x\leq a$ and $x\geq b$.


*

*Suppose $a<x<b$. Observe $\left|x-y\right|<r$ and $a<y<b$ imply that
$$\mathcal{A}_{r}(x)=\dfrac{1}{2r}\int_{x-r}^{x+r}\chi_{(a,b)}(y)dy=\dfrac{\min\left\{x+r,b\right\}-\max\left\{x-r,a\right\}}{2r}$$
If $r\leq\min\left\{\left|x-a\right|,\left|x-b\right|\right\}$, $\mathcal{A}_{r}(x)=1$. If $\left|x-a\right|\geq r>\left|b-x\right|$, then
$$\mathcal{A}_{r}(x)=\dfrac{b-(x-r)}{2r}=\underbrace{\dfrac{b-x}{2r}}_{<1/2}+\dfrac{1}{2}<1$$
The argument for $\left|x-b\right|\geq r>\left|x-a\right|$ is completely analogous, and if $r>\max\left\{\left|b-x\right|,\left|a-x\right|\right\}$, then
$$\mathcal{A}_{r}(x)=\dfrac{b-a}{2r}<\dfrac{b-a}{\left|x-b\right|+\left|x-a\right|}=1$$
We conclude that $M\chi_{a,b}(x)=1$, if $a<x<b$.

*Now suppose $x\geq b$. If $r\leq\left|x-b\right|$, then $\mathcal{A}_{r}(x)=0$. If $r>\left|x-b\right|$, then
$$\mathcal{A}_{r}(x)=\dfrac{b-\max\left\{x-r,a\right\}}{2r}$$
If $r\leq\left|a-x\right|$, then
$$\mathcal{A}_{r}(x)=\dfrac{b-(x-r)}{2r}=-\dfrac{\left|x-b\right|}{2r}+\dfrac{1}{2},$$
which is maximal when $r=\left|x-a\right|$. As $\mathcal{A}_{r}(x)$ is decreasing on $r\geq\left|x-a\right|$, we conclude that $M\chi_{(a,b)}(x)$ is as stated.

*Lastly, suppose that $x\leq a$. If $r\leq\left|x-a\right|$, then $\mathcal{A}_{r}(x)=0$. If $r>\left|x-a\right|$, then
$$\mathcal{A}_{r}(x)=\dfrac{\min\left\{x+r,b\right\}-a}{2r}$$
If $r\leq\left|b-x\right|$, then
$$\mathcal{A}_{r}(x)=\dfrac{(x+r)-a}{2r}=-\dfrac{\left|x-a\right|}{2r}+\dfrac{1}{2},$$
which is maximal for $r=\left|x-b\right|$. As $\mathcal{A}_{r}(x)$ is decreasing on $r\geq\left|x-b\right|$, we conclude that $M\chi_{(a,b)}(x)$ is as stated.
$\Box$
Restricting now to the case $(a,b)=(2^{j-1},2^{j})$ and $\left|x\right|\leq 2^{j}$, we have that 
$$Mf_{j}(x)=\begin{cases} 1 & {2^{j-1}<x<2^{j}}\\ \\ \dfrac{1}{2} & {x=2^{j}} \\ \\ 2^{-1}\dfrac{2^{j-1}}{2^{j}-x}\geq \dfrac{2^{j-2}}{2^{j}-(-2^{j})}=\dfrac{1}{8} & {-2^{j}\leq x\leq 2^{j-1}}\end{cases}$$
