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 $\|Mf\|\_{B^0\_{p,\infty}}\leq C\_d\|f\|\_{B^0\_{p,\infty}}$.)



Is the maximal function bounded on the Besov space?
* The original inequality in the question was $\Lambda\_jMf\leq M\Lambda\_j f$.
**This inequality is not true in general** even if $f\geq 0$ since the opposite inequality is true.


Assume that $h\_j$ and $f$ are non-negative. Let $M\_r=|B(0,r)|^{-1}\chi\_{B(0,r)}$. Then $Mf=\sup\_{r>0} M\_r\*f$. Hence
$$
h\_j\*M\_r\*f=M\_r\*h\_j\*f,
\qquad
h\_j\*Mf\geq M\_r\*h\_j\*f,
$$
because $M\_rf\leq Mf$ and taking the supremum over $r$ yields
$$
\Lambda\_jMf= h\_j\*Mf\geq M(h\_j\*f)=M\Lambda\_j f.
$$
If your inequality were true $\Lambda\_jMf\leq M\Lambda\_j f$, then we would have equality: $\Lambda\_jMf= M\Lambda\_j f$ which holds rarely. I do not have an example on top of my head, but such an example is easy to find.


* If we drop the assumption that $f\geq 0$, then it is easy to show that even a **stronger inequality from the current formulation of the problem: $\Lambda\_jMf\leq CM\Lambda\_j f$ is not true**. Here is an example.


Let $R>0$ be large and let $\lambda\_j\approx \chi\_{B(0,R)}$ be a $C\_0^\infty$ approximation of the characteristic function of the ball. Let $f\in C\_0^\infty(B(0,1))$, $\int f=0$, $\int |f|=1$. Then 
$|\Lambda\_j f|=|\lambda\_j\*f|\leq 1$
and $\Lambda\_j f\neq 0$ in a neighborhood of $\partial B(0,R)$ of thickness $1$. 


For $|x|$ much larger than $R$ we have:
$$
M\Lambda\_j f\approx \frac{R^{n-1}}{|x|^n}.
$$
However,
$$
M f\approx \frac{1}{|x|^n}
\qquad
\text{so}
\qquad
\Lambda\_jMf(x)\approx \frac{R^n}{|x|^n}
$$
and the inequality $\Lambda\_jMf\leq CM\Lambda\_j f$ would require $C$ to be of magnitute $R$.


* One easily can generalize the previous example to the case in which $\lambda\_j$ is supported in an annulus which, I believe, is the case you are interested in.


