Stein - maximal function and $L \log L$ spaces Stein makes the claim that if a function $f$ is supported on a ball $B$, then the maximal function $\mathcal{M}f \in L^1(B)$ if and only if $f \in L \log L(B)$. Can someone direct me to a proof for this?
 A: The idea is based on a classic paper of Calderon-Zygmund. The $1$-d result also appears in Zygmund's Trigonometric Series, Vol I book (pp. 32). Here is a sketch of a solution, which is valid for any $f\in L^1_{loc}$ and any bounded set $B$. This is from notes I prepared for a  Harmonic Analysis course a few years ago:
Denote by $m$ the Lebesgue Measure on $\mathbb{R}^d$.
For any bounds set $B$
$$\begin{align}
\int_BMf&= \int^\infty_0 m\big(x\in B: Mf> t\big)\,dt\\
&=2\int^\infty_0 m\big(x\in B: Mf> 2t\big)\,dt\\
&\leq 2m(B)+2\int^\infty_1 m\big(x\in B: Mf> 2t\big)\,dt
\end{align}$$
For any $t>0$, $f=f\mathbb{1}_{\{|f|>t\}}+f\mathbb{1}_{\{|f|\leq t\}}$. Let $f_1=f\mathbb{1}_{\{|f|>t\}}$, and $f_2=f-f_1$. 
As $Mf_2\leq t$, $\{x\in B:Mf>2t\}\subset\{x\in B: Mf_1>t\}$. Hence, by Hardy-Littlewood's maximal inequality
$$\begin{align}
\int^\infty_1 m\big(x\in B: Mf> 2t\big)\,dt&\leq \int^\infty_1 m\big(x\in B: Mf_1> t\big)\,dt\\
&\leq \int^\infty_1\frac{C_d}{t}\Big(\int_{|f|>t}|f|\,dm\Big)\,dt\\
&=C_d\int_{\mathbb{R}^d}|f|\Big(\int^{\max\{|f|,1\}}_1\frac{dt}{t}\Big)\,dm\\
&=C_d\int_{\mathbb{R}^d}|f|\log_+|f|\,dm
\end{align}$$
where $\log_+(x):=\max(0,\log x)$, and $C_d$ is a constant depending only on the dimension $d$.
The converse is a little more difficult and was proved in E.M. Stein, Note on the class $L\,\log\,L$, Studia Math. 32 (1969)) for $L_1(\mathbb{R}^d,m)$. The main idea is to apply a well known results by Calderón-Zygmund that states that for any $t>0$, there are $d$-dimensional cubes $Q_k$ with pairwise disjoint interiors and such that
(1) $f\leq t$ $\mu$--almost surely on $F=\bigcap_k(\mathbb{R}^d\setminus Q_k)$; (2)
$$ \begin{align}
t<\frac{1}{m(Q_k)}\int_{Q_k}|f|\,dm\leq 2^d t\tag{1}\label{one}
\end{align}$$
It follows that there is a constant $c_d$ such that for all $x\in Q_k$,
$$Mf (x)>(c_d)^{-1} t$$
Take for example the ball $B(x;r_k)$ centered at $x$ of radius $r_k\operatorname{diam}(Q_k)$. Then
$$\begin{align}
Mf(x)&\geq\frac{1}{m(B(x;r_k))}\int_{B(x;r_k)}|f|\,dm\geq \frac{m(Q_k)}{ m(B(x;r_k))}\frac{1}{m(Q_k)}\int_{Q_k}|f|\,dm>\frac{t}{d^{d/2}\omega_d}
\end{align}$$
As a consequence $\bigcup_kQ_k\subset \{Mf>(c_d)^{-1}t\}$. By \eqref{one}
$$m(Mf>(c_d)^{-1}t)\geq \sum_km(Q_k)\geq\frac{1}{2^dt}\int_{\bigcup_kQ_k}|f|\geq \frac{1}{2^dt}\int_{\{|f|>t\}}|f|\,dm$$
If $Mf\in L_1(B,m)$ then one can argue that $Mf\in L^{loc}_1(\mathbb{R}^d)$. Also, if $f$ is supported in $B$, then it is easy to check that $\lim_{|x|\rightarrow\infty}Mf(x)=0$. In particular, $\{Mf>1\}$ will be contained in a bounded set. Putting things together, we obtain
$$\begin{align}
\infty>\int_{\{Mf>1\}}Mf\,dm&\geq \int^\infty_1 m(Mf>t)\,dt\\
&\geq \int^\infty_1\frac{1}{2^dc_d t}\int_{\{f>c_dt\}}|f|\,dm\\
&=\frac{1}{2^dc_d}\int_B|f|\Big(\int^{\max(|f|/c_d,1)}_1\frac{dt}{t}\Big)\,dm\\
&=\frac{1}{2^dc_d}\int_B|f|\log_+(|f|/c_d)\,dm
\end{align}
$$
