Given the following set of polynomials that are non-negative on the interval $[0,1]^2$
$$ B = \{x, 1-x, y,1-y,xy,x-xy,y-xy,1-x-y+xy\} = \{b_1,b_2,...,b_8\} $$
Can we definitively say that all polynomials with linear terms (I have no idea what the term is for this, I mean that there are no $x^2$ or $y^2$) that are non-negative on $[0,1]^2$ can be written as a linear combination of non-negative coefficients and these polynomials?
Or in math notation
Given $p(x,y)\mid p(x,y)\geq0 \forall(x,y)\in[0,1]^2$
Does there exist a $c_1,c_2,...,c_8\in \mathbb{R}_{\geq 0} \mid c_1b_1+c_2b_2+ ... +c_8b_8=p(x,y)$
My gut instinct says yes because the 1-dimensional case is true:
Given $b=\{x, 1-x\} = \{b_1, b_2\}$, and $p(x)\in \mathbb{P}_1\mid p(x)\geq 0 \forall x\in[0,1]$
Prove $\exists c_1, c_2 \in \mathbb{R}_{\geq0} \mid c_1b_1 + c_2b_2 = p(x)$
Proof: Since $p(x)\in \mathbb{P}_2$, $p(x) = mx+a$. From the theorem, $p(x) = c_1b_1+c_2b_2 = c_1x+c_2(1-x) = (c_1-c_2)x+c_2$. Since $p(x)$ is non-negative at $p(0)$, $a\geq0$, and we can see that $a=c_2\therefore c_2\geq 0$. We apply the same process to see that $p(1) = m+a\geq0$. We also see that $m = c_1-c_2 \implies m+c_2 = c_1 \implies m+b = c_1 \therefore c_1\geq0$. This shows that we can write every polynomial as a positive combination of our basis Q.E.D.
Obviously this process is very tedious, and I wanted to see if there was an easier way to prove it. Thank you!
These polynomials have 4 degrees of freedom: their values in the four corners of the square. Each such polynomial $p$ can be written as $$ p(x,y) = p(1,1)xy + p(1,0)x(1-y) + p(0,1)(1-x)y + p(0,0)(1-x)(1-y), $$ which should help your case.