For a given graph $G$ and a function $f\in L^1(G)$. then we define the Hardy-Littlewood maximal operator as follow

$$M_Gf(v)=\sup_{r\in N}\frac{1}{|B(v,r)|}\sum_{w\in B(v,r)}|f(w)|.$$

Where $B(v,r)$ is a ball centred at $v$ with radius $r$. While the spherical maximal operator defined as $$M_G^*f(v)=\sup_{r\in N}\frac{1}{|S(v,r)|}\sum_{w\in S(v,r)}|f(w)|.$$ Where $S(v,r)$ is a sphere centred at $v$ with radius $r$.

Using the fact that any ball can be written as a disjoint union of spheres, how can we prove the following inequality.

$$M_Gf(v)\le M_G^*f(v).$$


closed as off-topic by Shaun, Strants, José Carlos Santos, Vladhagen, Namaste Aug 27 '18 at 22:29

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This is honestly trivial from the definitions. Ok:

If $A$ is a finite set define $$f_A=\frac1{|A|}\sum_{x\in A}f(x),$$the average of $f$ over $A$. If $A$ and $B$ are disjoint then $$f_{A\cup B}=\frac{|A|}{|A\cup B|}f_A+\frac{|B|}{|A\cup B|}f_B.$$ Since the coefficients $|A|/|A\cup B|$ and $|B|/|A\cup B|$ are non-negative and add up to $1$ it follows that $$f_{A\cup B}\le\max(f_A,f_B).$$

Details: $$\begin{align}\frac{|A|}{|A\cup B|}f_A+\frac{|B|}{|A\cup B|}f_B &\le \frac{|A|}{|A\cup B|}\max(f_A,f_B)+\frac{|B|}{|A\cup B|}\max(f_A,f_B)\\&=\max(f_A,f_B).\end{align}$$

  • $\begingroup$ @majduleenzeyadeh I don't know what "Even it less than the maximum, not each one?" means, sorry. $\endgroup$ – David C. Ullrich Aug 27 '18 at 20:00

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