Here is a quotation of book "C*-algebras and Finite-Dimensional Approximations" by Nate and Taka.
In the proof of Proposition 4.4.5. (In P132), the author says:
The assertion $(1)\Longleftrightarrow (2)$ follows from the fact that $\mu\rightarrow\mu^{1/2}$ is a uniform homeomorphism from $\mathrm{Prob}(\Gamma)$ into $l^{2}(\Gamma)$.
Here, the $\mathrm{Prob}(\Gamma)$ be the space of all probability measures on $\Gamma$:
$$\mathrm{Prob}(\Gamma)=\{\mu\in l^{1}(\Gamma): \mu\geq0\mbox{ and }\sum\limits_{t\in \Gamma}\mu(t)=1\}.$$
My question is what does the uniform homeomorphism mean here? And how to verify it?
I assume it is "uniform" as in "uniform continuity". What happens is that $$\|\mu^{1/2}-\eta^{1/2}\|_2=\sum (\mu (t)^{1/2}-\eta (t)^{1/2})^2\leq\sum|\mu(t)-\eta(t)|=\|\mu-\eta\|_1. $$ The inequality is, as the book says, a consequence of the inequality $(a-b)^2\leq|a^2-b^2|$ for all positive numbers $a,b$. The proof of this inequality is very simple. First assume that $a=1$ and $b\leq 1$. Then, as $b^2\leq b$, $$ (1-b)^2=1+b^2-2b\leq1+b^2-2b^2=1-b^2. $$ Now assume that $a\geq b$ (otherwise exchange roles). Then $a/b\leq1$ and $$ (a-b)^2=a^2(1-\frac ba)^2\leq a^2\left(1-\frac{b^2}{a^2}\right)=a^2-b^2=|a^2-b^2|. $$