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Given a metric measure space $(X,d,m)$ (i.e. a metric space $(X,d)$ equipped with a $\sigma$-finite Borel measure $m$), we can consider the space $L^2(X,m)$ of "square-integrable functions":

$$L^2(X, m) = \left\{f: X\to \mathbb R, f \text{ measureable and }\int_X |f|^2 dm <\infty\right\}.$$

Further, $L^2(X.m)$ admits a norm $|| f ||_{L^p}:=\left(\int_{X}|f|^2dm\right)^{1/2}$, which can be used to turn it into a metric space.

My questions are the following:

  • Is there any measure on $L^2(X,m)$ which turns it into a metric measure space?
  • if there is, then, is there a "nice relation" between $(X,d,m)$ and $L^2(X,m)$? (Unfortunately I don't have yet decided what this should mean but any interactions would be welcome).

EDIT: For the sake of obtaining something non-trivial, I would like to also require that the measures involved have non-empty support. I am aware that Dirac measures work, however, if possible I would like to know about a measure that has more to do with the original measure $m$ on $X$.

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  • $\begingroup$ Don't you want further relations between the metric and the measure? You could take a Dirac measure in any element of $L^2$. By the way, in order to have really a metric on $L^2$ you need to pass to the quotient w.r.t. to the subspace of a.e. equal functions. $\endgroup$ – Jochen Aug 26 '16 at 8:17
  • $\begingroup$ For your first question, you could just take the zero measure. For your second question, I'll just point out that $L^2$ is in general an infinite dimensional vector space, and there is no translation invariant $\sigma$-finite Borel measure on an infinite dimensional vector space. $\endgroup$ – Justthisguy Aug 26 '16 at 13:53
  • $\begingroup$ @Jochen Thank you for your answer. I am aware that the Dirac measures work and that one has to identify functions that coincide almost everywhere. $\endgroup$ – Sak Aug 26 '16 at 17:54
  • $\begingroup$ @Justthisguy thank you for your answer. Wouldn't the Hausdorff measure be translation invariant? Am I missing something? Also, even though the zero-measure works I was hoping for something a little less trivial. I will update my question to reflect this. Thank you again. $\endgroup$ – Sak Aug 26 '16 at 17:55
  • $\begingroup$ The $d$ dimensional Hausdorff measure will not be $\sigma$-finite unless $L^2(X)$ is isomorphic to $\mathbb R^d$, i.e. if $X$ has $d$ points. $\endgroup$ – Justthisguy Aug 26 '16 at 18:07

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