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 $p > 1$?


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.

  • $\begingroup$ 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 $\mathbb{R}^n$ $\endgroup$ – user14166 Feb 27 '13 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy