Maximal function is not bounded on $L^1(\Bbb R^n)$ 
Prove: There exists a constant $c_p,p>1$, with $c_p>{c}/{(p-1)}$, so that if $f$ in $L^p$ then $$\|Mf\|_P \ge c_p \|f\|_P$$.

This is a problem in E.M stein's Harmonic analysis(8.14(b)). I can prove 8.14(a) that if $f$ is supported in a set of finite measure, then $Mf$ in $L_1$ iff $|f|log(1+|f|)$ in $L_1$. I prove 8.14(a) by using Calderon-Zygmund decomposition.
I am not really sure where to begin with 8.14(b). Any help would be greatly appreciated.
 A: From Lebesgue differentiation Theorem, we have that for every $f\in L^p$
$$|f(x)|=\left|\lim_{r\to 0}\frac{1}{|B(x,r)|}\int_{B(x,r)} f(y)dy\right| \le Mf(x)~~~a.e$$ that is
$$|f(x)|\le Mf(x)~~~ae \implies \|f\|_p\le \|Mf\|_p$$
To see that M is not bounded on $L^1$ fixed $f\neq 0$ in $L^1$ and 
chose $f_j =f\mathbb{1}_{B(0,j)}$ then $f_j\to f$ in $L^1$ and therefore we can find ${j_0}$ such that, $f_{j_0}\not \equiv 0$ 
taking $r = |x| +j_0$ yields that $B(0,j_0) \subset B(x,|x|+j_0)$
Therefore
$$ Mf(x)\ge Mf_{j_0}(x) = \sup_{r>0}\frac{1}{|B(x,r)|}\int_{B(x,r)} \left|f(y)\right|dy \ge \frac{1}{|B(x,|x|+j_0)|}\int_{B(x,|x|+j_0)} \left|f_{j_0}(y)\right|dy = \frac{1}{|B(x,|x|+j_0)|}\int_{B(x,|x|+j_0)\cap B(0,j_0)} \left|f(y)\right|dy = c_n(|x|+j_0)^{-n} \int_{ B(0,j_0)} \left|f(y)\right|dy $$
That is $$Mf_{j_0}(x)\ge c_n(|x|+j_0)^{-n} \int_{ B(0,j_0)} \left|f(y)\right|dy$$
this implies that 
$$\int_{\Bbb R^n} Mf (x)dx \ge c_n\|f_{j_0}\|_1 \int_{\Bbb R^n}(|x|+j_0)^{-n}dx =\infty $$
Hence $$Mf\not\in L^1$$
