Here is a statement from Michel Ledoux's book(let) on Logarithmic Sobolev Inequalities.
Let $\nu,\mu$ be respectively two-sided exponential, and Gaussian measures on the Borel $\sigma-$algebra of $\mathbb{R}$, (namely, they have the following densities with respect to Lebesgue measure: $\frac{d\nu}{dx} = \frac{1}{2}e^{-|x|/2}$, and $\frac{d\mu}{dx}=Ce^{-x^2/2}$).
Let $\Phi:\mathbb{R}\to\mathbb{R}$ be an increasing map, which takes $\nu$ to $\mu$ (namely, for any Borel subset $A$, $\mu(A)=\nu(\Phi^{-1}(A))$.
Ledoux comments that ``It is a simple matter to check that'' for some $C>0$, $$ |\Phi(x)-\Phi(y)|\leq C\min\{|x-y|,|x-y|^{1/2}\},\forall x,y\in\mathbb{R}. $$
I tried to argue via considering the interval $(\Phi(a),\Phi(b))$, and $\Phi^{-1}((\Phi(a),\Phi(b)))=(a,b)$ and got something like, $$ \int_{a}^b \frac{1}{2}e^{-|x|/2}\;dx = \int_{(\Phi(a),\Phi(b))}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\;dx, $$ but couldn't obtain a bound (left hand side is upper bounded by |b-a| type of a term). Can anyone help me? Thanks!