Raising metric to a power/equivalent metrics Suppose we have a locally compact separable bounded metric space $(X,d)$. It is well known that for $\epsilon\in(0,1)$, $d_1(x,y)=d(x,y)^\epsilon$ is a metric. My question is, can we relate $d_1$ and $d$. For example, are $d_1$ and $d$ (strongly) equivalent?
Thanks
 A: The  metrics $d_1,D$ are equivalent.
Let $a \in (0,1)$ and $d_1(x,y)=d(x,y)^a$
To prove that two metrics are equivalent you can prove that the function $I:(X,d) \rightarrow (X,d_1)$ such that $I(x)=x$ is a  $homeomorphism$ ,in other words, is a continuous bijection with a continuous inverse.
Clearly $I$ is a bijection thus invertible.
We will prove that it is continuous and its inverse is also continous.
Let $x_0 \in X$ and $U \subseteq X$ an $d_1$-open neighbourhood of $x_0$,thus exists an open ball $B_1(x_0, \epsilon) \subseteq U$.From the definition of continuity,we must find a $d-$open ball $B$ with radius  $\delta>0$ ,which contains $x_0$ such that $I(B) \subseteq B_1(x_0, \epsilon)$. 
Take $\delta=\sqrt[a]{\epsilon}$ and $B=B(x_0,\delta)$ and you have that $I(B) \subseteq B_1(x_0, \epsilon) \subseteq U$ thus $I$ is continuous in any point fo $X$ 
Now take the inverse $I^{-1}:(X,d_1) \rightarrow (X,d)$.
Let $x_0 \in X$ and $V \subseteq X$ a $d-$open neighbourhood  of $x_0$.
Exists an $d-$open ball $B(x_0, \epsilon) \subseteq V$
We must find a $d_1-$ open  ball $B_1 \subseteq X$ with a radius $\delta>0$ which contains $x_0$ such that $I^{-1}(B_1) \subseteq B$
For this take $\delta= \epsilon^a$ and $B_1=B_1(x_0, \delta)$ and you have that $I^{-1}(B_1) \subseteq B(x_o, \epsilon) \subseteq V$
Thus $I^{-1}$ is continuous, therofore $I$ is a homeomorphism.
$$Second$$ $$way$$

$Theorem:$Two metrics $d_1,d_2$ are equivalent  $iff$  $\forall x_n \in X$ such that $x_n \longrightarrow^{d_1} x_0$ then $x_n \rightarrow^{d_2} x_0$ and vice versa.

This is an easier way to prove that two metris are equivalent.You can try it.
