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Let $\mathcal{M}$ be the Hardy-Littlewood maximal operator in $\mathbb{R}^{n}$ defined by

$$\mathcal{M}(f)(x):=\sup_{r>0}\frac{1}{|B(x,r)|}\int_{B(x,r)}|f|.$$

It is known that for $p>1$ and for any $f\in L^{p}(\mathbb{R}^{n})$, one has $\mathcal{M}(f)\in L^{p}(\mathbb{R}^{n})$ and $\mathcal{M}:L^{p}(\mathbb{R}^{n})\rightarrow L^{p}(\mathbb{R}^{n})$ is a bounded operator.

Let $C_{p,n}$ be the norm of $\mathcal{M}$ as an operator in $L^{p}(\mathbb{R}^{n})$. Thus, as a consequence $$\|\mathcal{M}(f)\|_{L^{p}}\leq C_{p,n}\|f\|_{L^{p}},\forall f\in L^{p}(\mathbb{R}^{n}).$$

It is unknown what is the best bound of $C_{p,n}$, see.

My question is: is it true that for $p\geq 2$, there is an absolute constant $C_{0}$ (for example $C_{0}=10$) such that $$C_{p,n}<C_{0},\forall p\geq 2,\forall n?$$

Thanks so much for any suggestions.

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    $\begingroup$ What is $C_{p,n}$? Any constant that works?? Those could be arbitrarily large. $\endgroup$ – mathworker21 Aug 6 at 9:36
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    $\begingroup$ Once you make the question make sense, the answer is "no". $C_{p,n}$ blows up for any fixed $n$ as $p \downarrow 1$ $\endgroup$ – mathworker21 Aug 6 at 9:37
  • $\begingroup$ @mathworker Thanks for your comment, I have editted my question. $\endgroup$ – John Hana Aug 6 at 9:41
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    $\begingroup$ did you read my first comment? let's say $C_{p,n} = 10$ works for each $p,n$. Then $C_{p,n} = 10+p+n$ works as well, and those blow up. What do you mean by $C_{p,n}$? $\endgroup$ – mathworker21 Aug 6 at 9:42
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    $\begingroup$ @JohnHana (1) Wikipedia says that for each $p$, there is some $C_p$ such that for all $n$, $||Mf||_p \le C_p ||f||_p$ for each $f \in L^p(\mathbb{R}^n)$. In particular, there are $C_2,C_\infty$. (2) Therefore, by Marcinkiewicz interpolation, $||Mf||_p \le C||f||_p$ for some absolute constant $C$ (solely based on $C_2,C_\infty$, which are absolute) for $2 \le p \le \infty$. $\endgroup$ – mathworker21 Aug 6 at 9:56
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I see that your question was already solved in the comments. Nevertheless, I would like to comment about a related (and more interesting) result in that direction. Consider the Maximal Hardy-Littlewood function associated to centered cubes, that is, the MHL function defined by $$ \mathcal{M}f(x):=\sup_{r>0}\,\dfrac{1}{\vert \mathcal{Q}_{x,r}\vert}\int_{\mathcal{Q}_{x,r}}\vert f(y)\vert dy, $$ where $\mathcal{Q}_{x,r}$ is the cube centered at $x$ with side-length $r$. A basic result shows that this operator (as the one associated to centered balls) is unbounded for $p=1$. However, it is a bounded operator in the $L^{1,\infty}$ sense, which means that there exist a constant $c>0$ such that for all $\alpha\geq0$ it holds $$ \vert\{x\in\mathbb{R}^d: \ \mathcal{M}f(x)\geq \alpha \}\vert\leq \dfrac{c}{\alpha}\Vert f\Vert_{L^1}, $$ where $\vert\cdot\vert$ stands for the Lebesgue measure of the set and note that $c$ depends on the dimension $d$. A longstanding open question due to Elias Stein and J. Strömberg was about the behavior of the optimal constant $c_d$ depending on the dimension. This question was solved by J. M. Aldaz in a really short paper (published on annals) with a beautiful proof (here), where he showed that the optimal constant $$ c_d\to\infty \quad \hbox{as} \quad d\to\infty. $$ Thus, proving that the analogous to your question in the case $p=1$ is false.

Even when this is not exactly an answer to your original question, I think this can help you to have a better understanding of the Maximal Hardy-Littlewood operator in a (much) harder case.

PS: Note also that this is for the operator associated to cubes.

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