We know that Hardy-Littlewood maximal function is $(p,p)$ for any $p>1$. But one proves first that it is weak type $(1,1)$ and then use interpolation. I am just curious to know: is there a way of proving the $(p,p)$ property directly, i.e. without using interpolation? 



Hardy-Littlewood maximal function
Here is an old trick. Use the Poisson kernels instead of balls to define the central maximal function. Suppose that $p=2$. Then $Mf(x)=\int P\_{a(x)}(x-y)f(y)dy$ for some function $a(x)>0$. 
Consider the adjoint operator $Vg(y)=\int P\_{a(x)}(x-y)g(x)dx$ with some positive $g\in L^2$. We have $Vg(y)^2=\iint P\_{a(x)}(x-y)g(x)P\_{a(z)}(z-y) g(z)dx dz$, so, integrating with respect to $y$, we get
$$
\int Vg(y)^2dy=\iint P\_{a(x)+a(z)}(x-z)g(x)g(z)dxdz
$$ 
Rewrite the last integral as 
$$
\begin{aligned}
&2\iint\_{\{a(x)\ge a(z)\}} P\_{a(x)+a(z)}(x-z)g(x)g(z)dxdz
\cr
&\le C\iint P\_{a(x)}(x-z)g(x)g(z)dxdz
=C\int (Vg)g
\end{aligned}
$$
Thus $\|Vg\|\_{L^2}^2\le C\langle Vg,g \rangle $, so $\|Vg\|\_{L^2}\le C\|g\|\_{L^2}$. The duality finishes the story.


I leave it to you to exploit what one can squeeze from this trick for $p\ne 2$.


