The Hardy-Littlewood maximal function of a function $f$ is defined by $$ M f(x):=\sup_{0<r<\infty}\frac{1}{|B_r|}\int_{B_r}|f(x+y)|dy, $$ where $|B_r|$ denotes the Lebesgue measure of the ball $B_r$. It is well-known that for $p\in(1,\infty]$, there exists a constant $C_{d,p}>0$ such that \begin{align} \|M f\|_p\leq C_{d,p}\|f\|_p. \label{mf} \end{align} That is, $M$ is bounded in $L^p$-spaces.

The zero order Besov space can be defined as follows. Recall the Littlewood-Paley operators is deifned as $$ \Lambda_jf:=h_j*f \quad (convolution), $$ where $h_j\in C_0^\infty(R^n)$. We have $$ f=\sum_{j\geq -1}\Lambda_jf. $$ Then, $B^0_{p,\infty}(R^n)$ is defined as $f$ satisfying $$ \sup_{j}\|\Lambda_jf\|_p<\infty. $$

Is the operator $M$ bounded in $B^0_{p,\infty}(R^n)$? (the problem I met is the change of order for $M$ and convolution. Moreprecisely, is it true that for a $C>0$ $$ \Lambda_jMf\leq CM\Lambda_jf?? $$ If this is true, then $\|\Lambda_jMf\|_p \leq C\|M\Lambda_jf\|_p\leq C\|\Lambda_jf\|_p$, this means that $\|Mf\|_{B^0_{p,\infty}}\leq C_d\|f\|_{B^0_{p,\infty}}$.)