Assume that $P^d(\mathbb{R})$ is the cone of polynomials $f\in\mathbb{R}[x_1,...,x_n]$ such that $\deg f\leq d$ and $f(x)\geq 0$ for all $x\in\mathbb{R}^n$. Is it true that the boundary of this cone is the polynomials with at least one root?
At the first glance, I thought the answer was yes but then I saw polynomials like $f=(xy-1)^2+x^2$ which is strictly positive on $\mathbb{R}^2$ but there exists no $a\in\mathbb{R}$ such that $a>0$ and $f(x)\geq a$ for all $x\in\mathbb{R}^2$. If the answer is yes, what is the situation of this polynomial?
Suppose $f$ is a limit of positive polynomials, ie there exists a continuous map $p:(0,1)\to P^d(\Bbb R)$ so that $\lim_{t\to 0^+} p(t)=f$. Then $f$ cannot be negative at any point $x_0$, as then $\lim_{t\to0^+} p(t)(x_0) = f(x_0)$, but the LHS is strictly positive. Furthermore, I claim that if $f$ is a polynomial which is nonnegative, I can find it as a limit of strictly positive polynomials: take $f+t$, let $t\to0$. It follows that all limit points of $P^d(\Bbb R)$ are polynomials which are non-negative, and the assertion should be clear.
As for the second option, here is a heuristic way to think about it. Up to scaling by a constant (and perhaps one or two more slight omissions), positive polynomials of degree at most d parameterize subvarieties of $\Bbb A^n_{\Bbb R}$ of degree at most $d$ which do not intersect the real points of $\Bbb A^n_{\Bbb R}$. It shouldn't be so strange to think of moving the points which solve $(xy-1)^2+x^2$ in any direction just a little without introducing any real solutions, which should correspond to being able to wiggle the polynomial and stay within the cone. Of course, this could be made much more rigorous, but I did say that it was a heuristic.