What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

2011-05-25T08:52:55Z

The (uncentered) Hardy-Littlewood maximal function $M(f)$ of (a locally integrable) function $f$ on $\mathbb{R}^{n}$ is defined by the rule $M(f)(x)=\sup_{\delta>0,\left|y-x\right|<\delta} \text{Avg}_{B(y,\delta)} \left|f\right|$, where $\text{Avg}_{B(y,\delta)} \left|f\right| = \int_{\left|z\right|<\delta} f(y-z) dz$.

The following results regarding the (uncentered) Hardy-Littlewood maximal function are well-known and can be found in many basic texts on analysis (e.g. Loukas Grafakos' "Classical Fourier Analysis", Chapter 2, pages 78-81):

The Hardy-Littlewood maximal function is a bounded operator from $L^1(\mathbb{R}^n)$ to $L^{1,\infty}(\mathbb{R}^n)$ (i.e., weak $L^1$) of norm at most $3^n$ ($n$ is the dimension of the Euclidean space).
Since the Hardy-Littlewood maximal function is also a bounded operator from $L^{\infty}(\mathbb{R}^n)$ to itself with norm at most $1$ (this is clear), we can apply the Marcinkiewicz interpolation theorem to conclude that for all $1 < p < \infty$, the operator norm of the Hardy-Littlewood maximal function is at most $2\left(\frac{p}{p-1}\right)^{\frac{1}{p}}3^{\frac{n}{p}}$. In fact, there is a slightly better bound: $\frac{p}{p-1}3^{\frac{n}{p}}$.
The bound given above grows exponentially with the dimension $n$ (if $p$ is fixed). It is a fact that it cannot be improved to a bound that does not grow exponentially with the dimension $n$.

My questions:

Is an exact value for the norm of the (uncentered) Hardy-Littlewood maximal function, viewed as a bounded operator from $L^p$ to itself ( $1<p<\infty$ ), known? If so, what is it?

Also, what is the norm of the Hardy-Littlewood maximal function when it is viewed as an operator from $L^1$ to weak $L^1$ (if it is known)?

Are the answers to the analogous questions regarding the centered Hardy-Littlewood maximal function known?

I apologize if this question is too basic. It seems like a fairly simple question but it is not clear (at least to me) how to solve it.

Answer by Shaoming Guo for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

2011-05-25T09:18:47Z

I just searched in google, "best constants for uncentered maximal functions", by Grafakos and Smith. "the best constants for the centered H-L maximal inequality", by AD Melas.(1-D, weak type (1,1)). maybe you can find more results following this two.

Answer by Alex I for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

2013-02-22T15:58:20Z

Uncenteed maximal operator is proven to be unbounded as $L^p \to L^p$ for general norm (see L. Grafakos and J. Kinnunen, "Sharp inequalities for maximal functions associated with general measures") Actually the same result is for $L^1 \to L^{1,\infty}$ (ADAM OSEKOWSKI "BEST CONSTANTS IN THE WEAK-TYPE ESTIMATES FOR UNCENTERED MAXIMAL OPERATORS"). In those papers one can find some results regarding strong maximal function.

What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function? - MathOverflow

What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

Answer by Shaoming Guo for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

Answer by Alex I for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?