13
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

22
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Vincent
    Commented Aug 11, 2020 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." $\endgroup$ Commented Aug 11, 2020 at 21:02
  • 1
    $\begingroup$ You're right, thanks :) $\endgroup$
    – Vincent
    Commented Aug 12, 2020 at 2:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .