I have a quartic polynomial of shape $$(a\cdot x+b+c\cdot xy-d\cdot y+y^2\big)^2$$ where $a,b,c,d\in\mathbb Z_{>0}$ are coefficients.

The polynomial is not convex.

  1. However is it quasiconvex for various values of $(a,b,c,d)$?

  2. My problem is to find the regions of $(a,b,c,d)\in\mathbb Z_{>0}^4$ on which the polynomial is quasiconvex?

Note this polynomial is always non-negative and expressible as sum of squares. If so may be explicit representation will help. Note $0$ level is always the lower bound. It might be easy to handle.


Let $g(x,y) = ax+b+cxy-dy+y^2$ so that $f(x,y) = (ax+b+cxy-dy+y^2)^2 = g(x,y)^2$. If $g$ has a non-convex zero set, then $f$ won't be quasi-convex - for small enough $\epsilon >0$, $f^{-1}((-\infty,\epsilon))$ will be a neighborhood of $g^{-1}(0)$ which will again be nonconvex. So our only hope of $f$ being quasiconvex is when $g=0$ is convex.

Our goals are for this conic to be a single line, a single point, or empty, as these are the only conics which are convex. We can read off when $ax+b+cxy-dy+y^2=0$ is of the appropriate form by examining the matrix form of our conic, $$G =\begin{pmatrix} 0 & c/2 & a/2 \\ c/2 & 1 & -d/2 \\ a/2 & -d/2 & b \end{pmatrix}$$

Let $U$ be the upper left $2\times 2$ minor.

In order to get a single line, a single point, or an empty conic, we must have $\det G =0$ and $\det U \geq 0$. On the other hand, $\det U = \frac{-c^2}{4}$, which as $c\neq 0$ by the problem statement, is always negative, so $f$ is never quasiconvex.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.