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Fix $c>1$. Let $(X,d)$ be a separable compact metric space, does there necessarily exist a Borel probability measure $\nu$ on $(X,d)$ such that

  • $\operatorname{sup}_{x \in X,r>0}\frac{\nu(\mathrm{Ball}(x,cr))}{\nu(\mathrm{Ball}(x,r))}<\infty$,
  • $\nu(\mathrm{Ball}(x,r))>0$ for every $x \in X$ and $r>0$.
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    $\begingroup$ What is $c$? Also, why doesn't the second point imply the first? Or do you want some uniform boundedness? $\endgroup$ – Steve Jun 26 '19 at 15:56
  • $\begingroup$ I clarified c and point 1. It isn't implied by the second one anymore (excuse my typo) $\endgroup$ – AIM_BLB Jun 26 '19 at 19:27
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Are you asking if there is a $c>1$ and a measure $\nu$ with this property? And is the first condition supposed to hold with a uniform upper bound? (If not, any probability measure satisfying the second condition would work, as Steve points out.)

If so, you are asking for the existence of a non-trivial "doubling measure". A necessary and sufficient condition for a complete metric space to carry a non-trivial doubling measure is that it is a "doubling metric space".

A metric space is a "doubling metric space" if every ball $B(x,r)$ can be covered by $N$ balls of radius $r/2$, where $N$ is a fixed constant. See wikipedia: https://en.wikipedia.org/wiki/Doubling_space

If a complete metric space carries a non-trivial doubling measure, it must be a doubling metric space, by a fairly standard covering argument. (See, for example, Heinonen's book Lectures on Analysis on Metric Spaces.)

The reverse implication, that every complete doubling metric space carries a non-trivial doubling measure, is more difficult. It was proven by Vol'berg-Konyagin in the compact case and Luukkainen-Saksman in the general case. See Luukkainen-Saksman (https://www.ams.org/journals/proc/1998-126-02/S0002-9939-98-04201-4/S0002-9939-98-04201-4.pdf) for the details and references.

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  • $\begingroup$ Doubling property is definitely sufficient but I'm not sure it's necessary. Here c is given and fixed but I'm not requiring that the balls all be compared by the same doubling constant. For example, I think this should hold for the unit ball in $\ell^2$ which isn't doubling $\endgroup$ – AIM_BLB Jun 26 '19 at 19:26
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    $\begingroup$ @AIM_BLB The metric doubling condition is necessary. See Proposition 59 in Measure Theory that you can find here: sites.google.com/view/piotr-hajasz/teaching/… $\endgroup$ – Piotr Hajlasz Jun 26 '19 at 20:19
  • $\begingroup$ Hi Piotr, thanks I worked it out on the Tram home and indeed it is necessary. I just found it surprising since the paper of Ambrosio which it comes from treated the two cases separately. $\endgroup$ – AIM_BLB Jun 26 '19 at 20:32

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