This question is related to this one. 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, it is stated that quasi symmetries of doubling metric spaces must be doubling, so by the discussion circa (10.2) $(X,d^{1-\epsilon})$ 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 $C>0$?
Best Answer
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 $\lambda^{\lceil 1/\alpha\rceil}$.