I've got two polynomials $p, \hat{p}:\mathbb{R}^2\rightarrow \mathbb{R}$ of degree $2\times2\ $ which are close together around $0$: $$|p(\mathbf{x})-\hat{p}(\mathbf{x})|<\varepsilon \quad \forall \mathbf{x}\in\mathbb{R}^2 \ni||\mathbf{x}||<K$$ $p$'s coefficients are denoted: $\{c_{0,0}, c_{0,1}, c_{0,2}, \dots c_{2,2}\}$ and $\hat{p}$'s: $\{\hat{c}_{0,0}, \hat{c}_{0,1}, \hat{c}_{0,2}, \dots \hat{c}_{2,2}\}$.
How do you get a bound on how close each pair of coefficients must be to one-another? In other words, what is $w_{i,j,\varepsilon} \ni |c_{i,j}-\hat{c}_{i,j}|<w_{i,j,\varepsilon}$?
Here was my train of thought, though I am not sure it leads anywhere. The difference of the two polynomials is another polynomial, so it is enough to find bounds for the coefficients of a polynomial which stays within $(-\varepsilon,\varepsilon)$ when $||\mathbf{x}||<K$
If $p(x,y) = \sum_{i=0}^2 \sum_{j=0}^2 c_{ij} x^i y^j$, take $9$ points $(x_k, y_k)$ with $\|(x_k,y_k)\| \le K$ and solve the system of equations $p(x_k,y_k) = a_k$ for the coefficients $c_{ij}$ (this can be done for "generic" choice of points). From the resulting equations you can get estimates: if all $|a_k| \le \epsilon$, then $|c_{ij}| \le C_{ij} \epsilon$ for some positive constant $C_{ij}$.