Does there exist a bivariate polynomial $p \in \Bbb R[x,y]$ that is positive iff $x, y > 0$?
My motivation was originally to state multiple positivity conditions with one expression but now I'm just curious and unable to find the answer via search.
$\begingroup$
$\endgroup$
0
Add a comment
|
$\begingroup$
$\endgroup$
3
No.
Consider such a polynomial $P$.
$P$ must be $0$ on the positive real $x$ and $y$ axes, as it must cross from being positive to negative there. As a result, both $x$ and $y$ divide such a polynomial. Write $P=xyQ$.
Now, $Q$ is positive everywhere except the third quadrant, and so $R(x,y)=-Q(-x,-y)$ is positive only on the first quadrant. As a result, we have a polynomial $R$ that satisfies the same property, but has lower degree than $P$. This process cannot continue infinitely, giving a contradiction.
answered Aug 11 '20 at 0:17
-
$\begingroup$ I think you have to be a bit more careful. $Q$ is certainly positive in the first quadrant, but in the second, third and fourth quadrants, it might be zero. Importantly, I don't see how we can assert that $Q$ is nonzero everywhere in the third quadrant? $\endgroup$– VincentAug 11 '20 at 9:35
-
$\begingroup$ @Vincent You're right that there are some issues with $0$, but the proof works verbatim to show that "there is no nonzero polynomial that is nonnegative on the first quadrant and nonpositive on the other three quadrants." $\endgroup$ Aug 11 '20 at 21:02
-
1