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The question is for 1 $\leq$ p,q $\leq$ $\infty$

I can prove that HL Maximal function is weak (1,1) using the method of finite collection of bounded balls. I can also prove that it is strong (p,p) as it is a standard proof. However, what can be said about other cases?

Terence Tao's lecture notes make this comment without proof:

Dimensional analysis (analysing how $f$ and $Mf$ react under dilation of the domain $\mathbb R^d$ by a scaling parameter λ) shows that no weak or strong type (p, q) estimates are available in the off-diagonal case p $\neq$ q.

Could this be elaborated upon, perhaps in a mathematically rigorous way? Exactly how to do this dimensional analysis and how does it prove there are no estimates for off-diagonal case?

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    $\begingroup$ Using Riesz-Thorin interpolation theorem, weak type $(1,1)$ and strong type $(p,p)$ gives all the strong type $(q,q)$ for $1 < q < p$. According to my understanding, Tao means that: assume that some strong type $(p,q)$ holds for $p \neq q$, we should consider $\frac{1}{\lambda} f$ and $M ( \frac{1}{\lambda} f)$. Then some non-homogeneity shows up, and we could pass $\lambda \to 0$ or $\lambda \to \infty$ to derive a contradiction. $\endgroup$ – mathdoge Jun 22 at 12:17
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    $\begingroup$ Can't you just use that $|f|\leq M f$, so if you would have the boundedness (at least strong type), you would get $\|f\|_q \leq \|M f\|_q \lesssim \|f\|_p$, which is clearly false in general. Probably something similar can be done for the weak type estimate. $\endgroup$ – PhoemueX Jun 22 at 15:30
  • $\begingroup$ Sorry, I made a typo above: should be Marcinkiewicz interpolation instead of Riesz-Thorin. $\endgroup$ – mathdoge Jun 22 at 15:47
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The definition is $$Mf(x)=\sup_{r>0} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f(t)|dt$$ where $B(x,r)=\{y \in \Bbb R^n: ||x-y||<r\}$

Then dilate your $f$ by some $\delta>0$, $f^{\delta}(x):=f(\delta x)$

Let's assume that the Hardy-Littlewood Maximal function is strong $(p,q)$. We'll basically show that $p=q$

Now take $L^p$ norms, $||f^{\delta}||_p=(\int_{\Bbb R^n}|f^{\delta}(t)|^pdt)^{1/p}=(\int_{\Bbb R^n}|f(\delta t)|^pdt)^{1/p}$

Now a change of variable $\delta t=y$ gives, $||f^{\delta}||_p=(\int_{\Bbb R^n}|f(\delta t)|^pdt)^{1/p}=({\delta}^{-n}\int_{\Bbb R^n}|f(y)|^pdy)^{1/p}={\delta}^{-n/p}||f||_p \dots (*)$

Now $$Mf^{\delta}(x)=\sup_{r>0} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f^{\delta}(t)|dt=\sup_{r>0} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f(\delta t)|dt$$ $$=\sup_{r>0} \frac{{\delta}^{-n}}{|B(x,r)|}\int_{B(\delta x,\delta r)}|f(y)|dy \text{ ( same substitution as above)}$$ $$=\sup_{r>0} \frac{1}{|B(\delta x,\delta r)|}\int_{B(\delta x,\delta r)}|f(y)|dy$$ $$=\sup_{\lambda>0} \frac{1}{|B(\delta x,\lambda)|}\int_{B(\delta x,\lambda)}|f(y)|dy$$ $$=Mf(\delta x)=(Mf)^{\delta}(x)$$

Then $||Mf^{\delta}||_q=(\int_{\Bbb R^n} |Mf^{\delta}(x)|^q dx)^{1/q}= (\int_{\Bbb R^n} |Mf(\delta x)|^q dx)^{1/q}= {\delta}^{-n/q}||Mf||_q \le C \delta^{-n/p}||f||_p$

But then, ${\delta}^{-n(\frac{1}{q}-\frac{1}{p})}=1 \implies p=q$

Now for the weak off-diagonal case, it's a simple application of the Off diagonal version of Marcinkewicz interpolation combining with the fact that the Maximal function is not strong in the off-diagonal setup, which we just proved!

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