Let $(X,\mu, d)$ be a metric measure space where $\mu$ is a doubling measure. For a relatively compact set $U\in X$ consider the following quantity $$I(U,\mu,d)=\int_U \int_U \log^2(d(x,y)) d\mu(x) d\mu(y). $$ Are there examples of metric measure spaces with relatively compact subsets for which $I$ isnt finite?

1$\begingroup$ If $U$ is compact, then in particular it's bounded and so $d(x,y)$ is upperbounded by some $M<\infty$. Doesn't that make the integral trivially finite? $\endgroup$ – Aryeh Kontorovich Aug 30 '18 at 20:07

1$\begingroup$ No.it doesnt.d(x,y) can be very small , consequently log^2 can be very large. In many cases integal becomes bounded but even in R^n it's not obvious, at least to me, why for an arbitrary measure the double integral should be bounded. it's trivial that its bounded bellow though. :) $\endgroup$ – BigM Aug 30 '18 at 20:16

1$\begingroup$ Ah ok thanks. I somehow only thought the large values of $d(x,y)$ could be problematic  ignored the small ones. $\endgroup$ – Aryeh Kontorovich Aug 30 '18 at 20:41

$\begingroup$ In fact, for some nice Riemannian manifold our double integral is bounded. My limited personal experience suggests its bounded for general care. $\endgroup$ – BigM Aug 30 '18 at 20:52

3$\begingroup$ $X=\{ x\}$ gives such examples. $\endgroup$ – Christian Remling Aug 31 '18 at 2:09
(What follows does not claim to be a complete answer, but rather a possible reasoning line for a proof that such integrals converge in the general case.)
In the Euclidean $\mathbb{R}^n$ case this should be true for any relatively compact subset $U$, because the following integrals converge for all $R \in \mathbb{R}^+$ and all $\alpha \geq 0$ (and in particular for the nonnegative integers $\alpha := n1$, with $n\geq 1$): \begin{equation} \int_0^R \log^2(r) r^{\alpha} dr = \frac{R^{α + 1} \big((α + 1)^2 \log^2(R)  2 (α + 1) \log(R) + 2\big)}{(α + 1)^3} \end{equation} On a general doubling space $X$ we can invoke Assouad's Theorem to see that there exist constants $n$ and $\epsilon \in (0,1)$ such that the $\epsilon$snowflaked version of $X$, with metric $(d_X)^\epsilon$, can be embedded into Euclidean $\mathbb{R}^n$ by a biLipschitz function $f:X \rightarrow \mathbb{R}^n$. In particular, a relatively compact $U \subset X$ would be mapped homeomorphically onto a relatively compact $f(U) \subset \mathbb{R}^n$.
At this point we could perform the integration on $f(U)$ (going through the biLipschitz homeomorphism $f^{1}:f(U) \rightarrow U$), use biLipschitzianity and observe that the following integrals also converge, for all $n\geq 1$ and all $\epsilon \in (0,1)$: \begin{equation} \int_0^R \log^2(r^\epsilon) r^{\epsilon(n1)} dr = \epsilon^2\int_0^R \log^2(r) r^{\epsilon(n1)} dr \end{equation}

$\begingroup$ For the sake of simplicity, let's stick to R. Why does the double integral behave as you stated? $\endgroup$ – BigM Aug 31 '18 at 15:59

$\begingroup$ This is roughly what I was thinking. Fix $R$, and write $U_R$ for the union of $x \in U$ such that the open ball of radius $R$ around $x$ is completely contained in $U$. For fixed $x \in U_R$, break the integral over $y \in U$ into a short range part (radius $r \in [0,R]$) and a long range part (radius $r>R$). The long range part is bounded because of relative compactness. The short range part is bounded because of the integral expression above. You then integrate over x, and get again something bounded by relative compactness. You then let $R$ go to zero. $\endgroup$ – Stefano Gogioso Aug 31 '18 at 16:38

$\begingroup$ It should work. But I'm going to sit detail and fill in the details..btw thanks for refering to Assouad's embedding theorem. I wasnt aware of that. Seems if we can prove it for R^n then we have it for arbitrary doubling measure spaces. $\endgroup$ – BigM Aug 31 '18 at 16:52


$\begingroup$ I wrote a sketch. I can do it with a few rather strong assumptions added. The ambient space should be Euclidean. Measure has to be a.c with repstect to Lebeague measure with bounded derivative . As C. Remling pointed out atomic measures and discrete spaces give cheap counterexamples $\endgroup$ – BigM Sep 4 '18 at 22:15