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(This is essentially a continuation of my previous question, here.)

Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. Further assume (though you can remove this assumption if you like) that each ball has positive finite volume. Let $M$ be the uncentred Hardy-Littlewood maximal operator, given by $$Mf(x) := \sup_{B \ni x} \frac{1}{\mu(B)} \int_B |f(y)| \; d\mu(y)$$ with the supremum taken over balls containing $x$.

In my previous question, I asked when $M$ is not of weak type $(1,1)$, and was presented with an example ($\mathbb{R}^2$ with Gaussian measure) which isn't of weak type $(1,1)$, but which is of strong type $(p,p)$ for all $p>1$. (See Sjögren ('83), and Forzani, Scotto, Sjögren, Urbina ('02).)

Now I'm interested in the following question: are there any metric measure spaces $(X,d,\mu)$ for which $M$ is of strong type $(p,p)$ if and only if $p > p_0$ (or $p \geq p_0$) for some $p_0 > 1$ (note the strict inequality here)? In other words, do we ever have some but not all midpoint strong type estimates?

I know that there are maximal operators out there for which this is true, but I'm specifically interested in the uncentred Hardy-Littlewood maximal operator (and to some extent the centred version, though we have to assume some regularity of the measure here to ensure measurability).

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  • $\begingroup$ Do you have any "geometric doubling" condition on your metric space? Probably you already know that but I would start by scanning Heinonen's "lectures on Analysis on metric spaces". I doubt you will find the answer to your exact question there, but maybe you will find more info on when the maximal function is (1,1), in order to exclude these spaces from your studies. I don't know of a specific example of a maximal operator in $\mathbb R^n$ that satisfies this property. The closest I can think of is the universal maximal function on radial functions is bounded on $L^p$ only when $p>n$. $\endgroup$
    – ioannis.parissis
    Commented May 1, 2013 at 21:43
  • $\begingroup$ For the purpose of this question the metric space is arbitrary. Nevertheless, there exist Borel measures on $\mathbb{R}^n$ for which the associated uncentred maximal operator is only bounded on $L^\infty$ (see Infante ('11): ams.org/journals/proc/2011-139-08/S0002-9939-2011-10727-5). So not only does geometric doubling not always imply the weak (1,1) estimate, it isn't strong enough to ensure any nontrivial (p,p) estimates! (comment continued below) $\endgroup$
    – user14166
    Commented May 1, 2013 at 22:01
  • $\begingroup$ I've seen the results on the strong maximal operator (looking at averages over arbitrary rectangles), but this can't be realised as a Hardy-Littlewood maximal operator. Of course, if there's some way of linking this to the HL case, I'd be very interested! $\endgroup$
    – user14166
    Commented May 1, 2013 at 22:04
  • $\begingroup$ A small correction: I guess you meant to write $\frac{1}{\mu(B)}\int_B |f(y)| \color{blue}{ d \mu(y)}$ above. $\endgroup$
    – ioannis.parissis
    Commented May 14, 2013 at 13:15
  • $\begingroup$ of course, thanks for letting me know. I'll edit the post and fix it $\endgroup$
    – user14166
    Commented May 14, 2013 at 14:30

2 Answers 2

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For the uncentred maximal function, one can cook up examples by using the star graph with a suitable weight on the vertices. In more detail, if one takes a star graph (with the graph metric) with $n$ spokes, with the root vertex having measure one and the leaves having measure $n^{-1/p_0}$, then the uncentred maximal function has bounded $L^p$ operator norm for $p \geq p_0$ but unbounded norm for $p > p_0$, by testing this function on the indicator function of the root vertex. If one then takes a disjoint union of such star graphs with the $n$ parameter going to infinity, one gets an example of the form you wanted.

For the centred maximal function, the star graph construction is no longer useful, but there are other, more complicated, constructions (based on metrics in vector spaces over finite fields) in Section 6 of my paper with Assaf Naor which can probably be adapted to give a suitable counterexample, though this would require a bit of tinkering (the examples there were designed to have weak (1,1) fail but strong (p,p) for all $p>1$).

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  • $\begingroup$ Very nice, thanks! This leads to another question: can we find such a space with uniform control on ball volume? In this example the volume of the closed ball of radius 1 centred at the $n$th root vertex is $n^{1+1/p_0}$. I'm curious to see if having such a 'critical exponent' $p_0$ relates to nonuniformity in any way. $\endgroup$
    – user14166
    Commented May 15, 2013 at 16:38
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It was proved by Alex Ionescu that the uncentered Hardy Littlewood maximal function is restricted weak type $(2,2)$ and not strong type $(p,p)$ for $p<2$ for hyperbolic spaces in http://arxiv.org/pdf/math/0007200v1.pdf

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