Non-negativity of symmetric bivariate polynomials of arbitrary degree over unit square

Question

What conditions need to be imposed on the coefficients of an $m$-degree bivariate polynomial $f(x,y)= f(y,x)$ over the unit square $x, y \in [0,1] \times [0,1]$, so that the polynomial (essentially proportional to a probability distribution) be non-negative over the domain?

Answer 1

It's unlikely that you'll get a nice, general necessary and sufficient condition that isn't basically a restatement of the problem.

One thing you can do, given a particular candidate polynomial $f$, is look at minimizing $f$ over the unit square. The minimum will occur either at an interior point where the gradient is $0$, a point on one of the edges where the tangential component of the gradient is $0$ and the outward normal component is non-positive, or a corner of the square.

EDIT: Some sufficient conditions might involve domination of the negative terms by some collection of positive terms. If you assume positivity of some particular set of coefficients, you might then find suitable conditions on the other coefficients.

For example, one sufficient condition is that $$c_{0,0} \ge \sum_i \sum_j \max(0, -c_{i,j})$$ where the polynomial is $\sum_i \sum_j c_{i,j} x^i y^j$.