2
$\begingroup$

Let $b:(0,\infty)\to (0,\infty)$ be monotonically increasing. Call $b$ limit-tight, if $$ \lim_{\varepsilon\to 0}\ \limsup_{T\to\infty}\frac{b(T-\varepsilon)}{b(T)} =\lim_{\varepsilon\to 0}\ \limsup_{T\to\infty} \frac{b(T+\varepsilon)}{b(T)}, $$ where $\varepsilon $ is meant to approach zero from above. I invented the name, so there might be a better one out there.

Let $G$ be a group acting isometrically and transitively on a complete, locally compact metric space $X$, which also carries a $G$-invariant Radon measure $\mu$.

My question is this: Is the function $$ b(T)=\mu(B(T)) $$ limit-tight? Here $B(T)$ is the open ball of radius $T$ around any point of $X$.

Improve this question
$\endgroup$
2
  • $\begingroup$ Hm. On the definition: Am I right that the left limit is always $\leq 1$, while the right one is always $\geq 1$, so both should equal 1? $\endgroup$
    – Ilya Bogdanov
    Commented Nov 19, 2015 at 13:32
  • $\begingroup$ @Ilya Bogdanov: Indeed. $\endgroup$
    – user1688
    Commented Nov 20, 2015 at 5:21

1 Answer 1

Reset to default
3
$\begingroup$

Let $G$ be a free group on $p>1$ generators and $X=G$. It is a locally compact complete metric space with its invariant distance defined by length of words. The invariant measure given by cardinality has $\mu B(T)= 1 + \frac p{p-1}\big((2p-1)^{\lceil T\rceil}-1\big)$ so for any $\epsilon>0$ one has $\displaystyle \limsup_{T\to+\infty} \frac{\mu B(T+\epsilon)}{\mu B(T)}= 2p-1$.

Improve this answer
$\endgroup$

You must log in to answer this question.