$f(x_1,x_2)=\frac{x_1^2}{x_2}$ quasiconvex and/or quasiconcave or nothing on $\mathcal R\times \mathcal R$? Related to the 3.16e question in Boyd's book. It asks what is $f$ in $\mathcal R\times R_{++}$. I am not interested in it but related thing when the domain is larger. So $f(x_1,x_2)=\frac{x_1^2}{x_2}$ is quasiconvex on $\mathcal R\times \mathcal R_{++}$ and quasiconcave on $\mathcal R\times \mathcal R_{--}$ where $\mathcal R_{--}$ and $\mathcal R_{++}$ do not contain zeros. $\mathcal R$ is the real numbers. The function is undetermined when $x_2=0$. Now I started wondering: 

what is $f$ when $\mathcal R \times \mathcal R$? If I get a question like this, should I define it with intervals separately or say nothing over the whole domain?


 A: On the whole domain $\mathbb{R}\times\mathbb{R}$ it's certainly not quasiconvex or quasiconcave:
Consider the preimage of $(-\infty,\epsilon)$ for small $\epsilon>0$.  The preimage comprises the half-plane $x_2<0$ and a region in $x_2>0$ with boundary given by  $$x_1^2/x_2=\epsilon\Longrightarrow x_2=\epsilon x_1^2.$$
Any point $(x_1,\frac{\epsilon}{2}x_1^2)$ lies on a segment between two points of the preimage, e.g., $(x_1,-1)$ and $(x_1,2\epsilon x_1^2)$, but it is not in the preimage.
Hence the preimage is not convex, and the function is not quasiconvex.
Similarly the preimage of $(-\epsilon,\infty)$ is not convex, hence the function is not quasiconcave.
A: Say nothing over the whole domain. 
Think of a function of one variable that have  negative second derivative at all but finite points and at these points the function is not differentiable (it has kinks), the first derivative changes sign. This function is not quasi-concave nor quasi-convex but you can divide its domain in regions where it is concave. But we don't have a name for this type of function. 
