# Is the norm of the Hardy-Littlewood maximal operator bounded?

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}

Thanks so much for any suggestions.

• What is $C_{p,n}$? Any constant that works?? Those could be arbitrarily large. – mathworker21 Aug 6 at 9:36
• Once you make the question make sense, the answer is "no". $C_{p,n}$ blows up for any fixed $n$ as $p \downarrow 1$ – mathworker21 Aug 6 at 9:37
• @mathworker Thanks for your comment, I have editted my question. – John Hana Aug 6 at 9:41
• 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 $C_{p,n}$? – mathworker21 Aug 6 at 9:42
• @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 $2 \le p \le \infty$. – mathworker21 Aug 6 at 9:56

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.