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

  • $\begingroup$ I added another example that disproves your stronger inequality. I also changed the title to a more appealing one. $\endgroup$ – Piotr Hajlasz Apr 27 '18 at 22:22
  • 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.
  • $\begingroup$ Then, the answer you posed are right. $\endgroup$ – Wenguang Zhao Apr 27 '18 at 21:32

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