What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:12:08Z http://mathoverflow.net/feeds/question/65935 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65935/what-is-the-lp-norm-of-the-uncentered-hardy-littlewood-maximal-function What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function? Amitesh Datta 2011-05-25T08:52:55Z 2013-05-06T17:22:00Z <p>The (uncentered) Hardy-Littlewood maximal function $M(f)$ of (a locally integrable) function $f$ on $\mathbb{R}^{n}$ is defined by the rule <code>$M(f)(x)=\sup_{\delta&gt;0,\left|y-x\right|&lt;\delta} \text{Avg}_{B(y,\delta)} \left|f\right|$, where $\text{Avg}_{B(y,\delta)} \left|f\right| = \int_{\left|z\right|&lt;\delta} f(y-z) dz$</code>. </p> <p>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):</p> <ul> <li><p>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).</p></li> <li><p>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 $1 &lt; p &lt; \infty$, 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: $\frac{p}{p-1}3^{\frac{n}{p}}$.</p></li> <li><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 $n$.</p></li> </ul> <p>My questions:</p> <p><strong>Is an exact value for the norm of the (uncentered) Hardy-Littlewood maximal function, viewed as a bounded operator from $L^p$ to itself (<code>$1&lt;p&lt;\infty$</code>), known? If so, what is it?</strong> </p> <p><strong>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)?</strong></p> <p><strong>Are the answers to the analogous questions regarding the <em>centered</em> Hardy-Littlewood maximal function known?</strong></p> <p>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.</p> http://mathoverflow.net/questions/65935/what-is-the-lp-norm-of-the-uncentered-hardy-littlewood-maximal-function/65937#65937 Answer by Shaoming Guo for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function? Shaoming Guo 2011-05-25T09:18:47Z 2011-05-25T09:18:47Z <p>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.</p> http://mathoverflow.net/questions/65935/what-is-the-lp-norm-of-the-uncentered-hardy-littlewood-maximal-function/122642#122642 Answer by Alex I for What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function? Alex I 2013-02-22T15:58:20Z 2013-02-25T13:31:34Z <p>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.</p>