Finding range and inverse of $f(x,y) := (x\sqrt{y},y\sqrt{x})$

Question

Consider the function $f: (0,\infty)\times (0,\infty) \to \mathbb{R}^2$ defined by $$f(x,y) := (x\sqrt{y},y\sqrt{x})$$ I know that in general it is hard to tell if $f$ is injective or not and to determine the image of $f$. So I started calculating a possible inverse function, where the domain is to be determined later. If $f(x,y) = (u,v)$ I got $$(x,y) = (u^{4/3}v^{-2/3},v^{4/3}u^{-2/3})$$ How do I now find $f((0,\infty)\times(0,\infty))$ and where that $f$ is invertible?

Answer 1

You've done most of the work.

If $f(x,y)=(u,v)$, then $$ \begin{cases} u=x\sqrt{y} \\[4px] v=y\sqrt{x} \end{cases} $$ so $y=u^2/x^2$ and $$ v=u^2x^{-3/2} $$ Thus $x^{3/2}=u^2v^{-1}$ and, by symmetry, $y^{3/2}=u^{-1}v^2$. Finally, $$ x=\sqrt[3]{u^4v^{-2}}\qquad y=\sqrt[3]{u^{-2}v^4} $$ and therefore $f$ has an inverse, so it is injective. There is no restriction on $u$ and $v$ so long as they're positive. The range is $(0,\infty)\times(0,\infty)$.

If you want to consider $(0,1)\times(0,1)$ as the domain, injectivity poses no problem, because the extended function is injective. Now you need to impose, always with $u>0$ and $v>0$, \begin{cases} u^4v^{-2}<1 \\[4px] u^{-2}v^4<1 \end{cases} that simplifies to \begin{cases} u^2<v \\[4px] v^2<u \end{cases} The region is the intersection of the interior of two parabolas in the $(u,v)$ plane.