Let $(X,d)$ be a metric space and let $\mathcal{H}^s$ be the $s$-dimensional Hausdorff measure on $X$.

For a measure $\mu$ on $X$, we say that $\mu$ is a *Frostman measure* (sometimes referred as *upper Ahlfors regular measure*) if for some $\alpha >0$, $\mu(B(x,r))\leq C r^\alpha$ for every $x \in X$ and $r>0$.

Under what conditions on $s$ and $d$ is $\mathcal{H}^s$ a Frostman measure?

Some (silly) partial answers. For $X=\mathbb{R}^d$, $\mathcal{H}^d$ is clearly a Frostman measure, as it is a scalar multiple of Lebesgue measure. For $s>d$, $\mathcal{H}^s$ satisfies, trivially, $0=\mathcal{H}^s(B(x,r)) \leq C r^s$, so the answer is positive also in this case. For $s<d$, we have $\mathcal{H}^s(B(x,r)) = \infty$ so the inequality can't hold.

Obviously, if the metric space equipped with the measure $\mathcal{H}^s$ is $s$-Ahlfors regular, the measure is Frostman.

I've tried to think about the more general case, but I fail to have any intuition on a metric space $X$, so apologies if the answer turns out to be easy.