# For which metric measure spaces is the Hardy-Littlewood maximal operator not of weak type (1,1)?

Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. I'll denote the Hardy-Littlewood maximal operator - either centred or uncentred, I don't mind which - by $M$. More precisely, for a function $f$ on $X$, define $$Mf(x) := \sup_{B \ni x} \frac{1}{\mu(B)} \int_B f(y) \; dy,$$ with the supremum taken over either all balls centred at $x$ or over all balls containing $x$.

There are plenty of theorems of the form 'suppose $(X,d,\mu)$ has property A. Then $M$ is of weak-type (1,1)', (for example, property $A$ could be '$(X,d,\mu)$ is doubling') thus implying that $M$ is bounded on $L^p(X)$ for $1 < p \leq \infty$ (since $L^\infty$ boundedness of $M$ is always obvious).

I'm interested in the converse. What is an example of a space $(X,d,\mu)$ is not of weak-type (1,1)? And how would one go about proving such a result?

Further, if there are any spaces with $M$ not of weak-type (1,1), do we still have $L^p$ boundedness for any $p > 1$?

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Such counterexamples are known. For instance, P. Sjögren has shown that the associated uncentered maximal to the $2$-dimensional Gaussian measure on $\mathbb{R}^2$ isn't weak type $(1,1)$. In a later paper, Forzani, Scotto, Sjögren and Urbina showed that this operator is strong type $(p,p)$ for $p>1$.
A well-known problem in this area is to determine if the ball (with Lebesgue measure) in $\mathbb{R}^n$ satisfies a weak type $(1,1)$ bound with a constant independent of the dimension. Related to this, there are a number of constructions of infinite families of metric spaces with nice properties, but for which the weak type $(1,1)$ constant can't be taken independent of the space in the family. See, for instance, the section Lower bounds" (page 8) of this paper of Naor and Tao for some such constructions.
Very interesting, thanks! Do you know of any counterexamples given by the geodesic distance and Riemannian volume on a Riemannian manifold? (i.e. an unweighted manifold, unlike Gaussian $\mathbb{R}^n$ – Alex Amenta Feb 27 '13 at 1:44