This question is related to [this one](https://mathoverflow.net/questions/401910/effect-of-snowflaking-on-metric-capacity). Let $(X,d)$ be a metric space, let $\epsilon\in [0,1)$ and consider the snowflake $(X,d^{1-\epsilon})$. Suppose that $(X,d)$ has a finite doubling constant, where the doubling constant $\lambda\_{(X,d)}$ of $(X,d)$ is defined by:



> 
> There exists $\lambda \in \mathbb{N}$ such that every ball of radius 2r can be covered
> with at most $\lambda$ balls of radius r. The least such constant is $\lambda\_{(X,d)}$, the
> doubling constant of X.
> 
> 
> 


In [Theorem 10.18 of Heinonen's Book](https://link.springer.com/content/pdf/10.1007%2F978-1-4613-0131-8_10.pdf), it is stated that quasi symmetries of doubling metric spaces must be doubling, so by [the discussion circa (10.2)](https://link.springer.com/content/pdf/10.1007%2F978-1-4613-0131-8_10.pdf) $(X,d^{1-\epsilon})$ must also be a doubling metric space. My question is, how do the doubling constants of $(X,d)$ and of $(X,d^{1-\epsilon})$ (defined as above) relate to one another explicitly?  

For instance, is it true that:
$$
 \lambda\_{(X,d^{1-\epsilon})}^{\frac1{1-\epsilon}}
\leq C\lambda\_{(X,d)} ,
 \qquad \forall \epsilon \in (0,1)
$$
for some absolute constant $C>0$?



Effect of snowflaking on doubling constants
Suppose $(X,d)$ has doubling constant $\lambda$. This means that for every $r$, every $d$-ball of radius $2r$ can be covered by $\lambda$ many $d$-balls of radius $r$. With respect to the metric $d^\alpha$, this means that every $d^\alpha$-ball of radius $(2r)^\alpha$ can be covered by $\lambda$ many $d^\alpha$-balls of radius $r^\alpha$. Letting $s = r^\alpha$, that means that every $d^\alpha$-ball of radius $2^\alpha s$ can be covered by $\lambda$ many balls of radius $s$.


Now putting $t = 2^{\alpha - 1}s$, we get that every $d^\alpha$-ball of radius $2t$ can be covered by $\lambda$ many $d^\alpha$-balls of radius $2^{1 - \alpha}t$. And each of those smaller balls can be covered by $\lambda$ many $d^\alpha$ balls of radius $2^{1 - 2\alpha}t$, and so on. After $k$ steps, with $k \geq 1/\alpha$, we get down to $d^\alpha$-balls of radius $2^{1 - k\alpha}t \leq t$. Conclusion: the original radius $2t$ ball can be covered by $\lambda^k$ radius $t$ balls, i.e., the doubling constant for the snowflaked space is at most $\lambda^{\lceil 1/\alpha\rceil}$.


