Positivity of real functions in two variables

Question

Assume that $f_0,f_1,f_2$ are polynomial functions of degree two in two variables. This means that the $f_i$ are linear combinations with real coefficients of $x^2,xy,x,y^2,y,1$.

Consider the function $f = f_1^2-af_0f_2:\mathbb{R}^2\rightarrow\mathbb{R}$ where $a\in \mathbb{R}_{>0}$. Is it true that for a "random" choice (whit a suitable definition of random) of the $f_i$ there exists $(x_0,y_0)\in\mathbb{R}^2$ such that $f(x_0,y_0) \geq 0$.

Clearly, this does not hold for any choice of the $f_i$. Take for instance $f_1\equiv 0$, $f_0 = x^2+1$, $f_2 = y^2+1$. Then $f(x,y) = -a(x^2y^2+x^2+y^2+1) < 0 $ for all $(x,y)\in\mathbb{R}^2$.

Write $$f_0 = a_1 x^2 + a_2xy+ a_3 x+ a_4 y^2+ a_5 y+ a_6;$$ $$f_1 = b_1 x^2 + b_2xy+ b_3 x+ b_4 y^2+ b_5 y+ b_6;$$ $$f_2 = c_1 x^2 + c_2xy+ c_3 x+ c_4 y^2+ c_5 y+ c_6;$$ Then $f$ corresponds to the point $(a_1,\dots,c_6)\in\mathbb{R}^{18}$.

Taking $$f_0 = \epsilon_0(x^2 + y^2 + 1);$$ $$f_1 = \epsilon_1(x^2 + y^2 + 1);$$ $$f_2 = \epsilon_2(x^2 + y^2 + 1);$$ we have $f = (\epsilon_0^2-a\epsilon_1\epsilon_2)(x^2 + y^2 + 1)$ which is always negative when $\epsilon_0^2-a\epsilon_1\epsilon_2 < 0$.

But the point $(f_0,f_1,f_2)$ lies inside the linear subspace $a_2 = a_3 = a_5 = b_2 = b_3 = b_5 = c_2 = c_3 = c_5 = 0$.

Thank you very much.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\c{\mathsf c}\newcommand\ep{\varepsilon}$The answer is no. Indeed, after clarifications given by the OP in comments and in the original post, the question can be stated as follows:

For $i=0,1,2$, let $$f_i(x,y)=a_{i,0}+a_{i,1}x+a_{i,2}y+a_{i,3}xy+a_{i,4}x^2+a_{i,5}y^2,$$ where the $a_{i,j}$ are real numbers. For a real $a>0$, let $M_a$ be the set of all matrices $(a_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $f(x,y):=f_1(x,y)^2-af_0(x,y)f_2(x,y)\ge0$ for some $(x,y)\in\R^2$. Is it true that the dimension (in whatever appropriate sense) of the complement $M_a^\c$ of $M_a$ to $\R^{3\times6}$ is strictly less that $18$ (the dimension of $\R^{3\times6}$)?

Let $g(x,y):=1+x^2+y^2$. For real $\ep>0$, let $N_\ep$ denote the set of all matrices $(a_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $|a_{i,j}-b_j|<\ep$ for all $i=0,1,2,\,j=0,\dots,5$, where $(b_0,\dots,b_5)=(1,0,0,0,1,1)$. Then for $i=0,1,2$ and all real $x,y$ we have $$|f_i(x,y)-g(x,y)|<\ep(1+|x|+|y|+|xy|+x^2+y^2) \le2\ep g(x,y),$$ since $|x|\le(1+x^2)/2$, $|y|\le(1+y^2)/2$, and $|xy|\le(x^2+y^2)/2$.

So, taking now any $a\in(1,2)$ and any $\ep\in(0,\frac{\sqrt a-1}{2(\sqrt a+1)})$, we get ($\ep<1/2$ and) for all real $x,y$ $$f_1(x,y)^2-af_0(x,y)f_2(x,y)\le g(x,y)^2[(1+2\ep)^2-a(1-2\ep)^2]<0.$$ So, $N_\ep\subseteq M_a^\c$ and the dimension of $N_\ep$ is $18$. Thus, the dimension of $M_a^\c$ is $18$, not $<18$.