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Recall that a probability $\mu$ measure on a metric space $(X,d)$ is called Ahlfors $q$-regular if there are $0<c\le C$ such that: for $\mu$-a.e.\ $x\in X$ one has $$ cr^q \le \mu(B(x,r)) \le Cr^q, $$ for any $0\le r\le \operatorname{diam}(X,d)$.

If $(X,d)$ is a finite metric space, then can it support an Ahlfors regular measure? If so, what are concrete examples of Ahlors $q$-regular measures on discrete metric spaces; for $q>1$?

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Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0,\infty)$ and $B(x,r)=\{x\}$ for $r\in(0,R)$.

Letting now $r\downarrow0$, from $cr^q \le \mu(B(x,r)) \le Cr^q$ we get $m\le 0$ if $q>0$ and $\infty\le m$ if $q<0$. So, we get a contradiction with $m\in(0,\infty)$ unless $q=0$.

So, $\mu$ can be Ahlfors $q$-regular only if $q=0$.

On the other hand, any probability measure over a finite set is clearly Ahlfors $0$-regular (assuming the convention that $r^q=1$ if $r=0=q$).

Thus, any probability measure over a finite set is Ahlfors $q$-regular iff $q=0$.

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  • $\begingroup$ Ah but the issues is largely coming from the lower bound then? So there can be "upper Ahlfors $q$-regular measures" on $X$ with $q>0$ (where by upper Ahlors regular I mean on the RHS $\mu(B(x,r))\le Cr^q$ is to hold)? $\endgroup$
    – Math_Newbie
    Commented Jan 31, 2023 at 23:27
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    $\begingroup$ @PiotrK : No, for $q>0$ the obstacle comes from the inequality $\mu(B(x,r)) \le Cr^q$, which implies $0<m\le0$, a contradiction. $\endgroup$
    – Iosif Pinelis
    Commented Feb 1, 2023 at 0:49
  • $\begingroup$ Oh whoops I read that backwards; yes it is clear now. Thanks. $\endgroup$
    – Math_Newbie
    Commented Feb 1, 2023 at 1:56

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