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?
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$\begingroup$ Is the idea for the proof of interpolation allowed(splitting into two functions)? Or you expect a totally new way of proving? $\endgroup$– Shaoming GuoCommented Jan 15, 2012 at 13:19
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$\begingroup$ I am talking about a proof where you do not boil down the question to proving that it is weak $(1,1)$ and $(\infty, \infty)$. For the central H-L maximal function on the upper half space one can prove $(p,p)$ directly. $\endgroup$– sprCommented Jan 16, 2012 at 8:28
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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 $p\ne 2$.
