Given $P_1,P_2\in \mathbb R^2$, integer $n>1$, and $f\in \mathbb R[X,Y]$, $f=g_1+g_2$, partial derivatives of order $<n$ of $g_i$ vanishing at $P_i$

Question

Let $P_1,P_2\in \mathbb R^2$ and let $n\ge 1$ be an integer. Given $f(X,Y)\in \mathbb R[X,Y]$, how to show that there exists $g_1(X,Y),g_2(X,Y)\in \mathbb R[X,Y]$ such that $f(X,Y)=g_1(X,Y)+g_2(X,Y)$ where all partial derivatives of $g_i(X,Y)$, of order $\le n$, vanishes at $P_i$ ?

Answer 1

First, we must assume $P_1\neq P_2$ (otherwise, this is false).

Observe that an affine transformation $T:P\mapsto M(P)+Q$, where $M$ is a invertible linear map and $Q \in \mathbb{R}^2$, maps a polynomial $g$ whose partial derivatives of order $\leq n$ vanishes at $P$, to a polynomial $g\circ T^{-1}$ whose partial derivatives of order $\leq n$ vanishes at $T(P)$. This is because partial derivatives of order $\leq n$ of $g\circ T^{-1}$ at $T(P)$ are linear combinations of partial derivatives of order $\leq n$ of $g$ at $P$, by the chain rule.

Observe also that the group of affine transformation of $\mathbb{R}^2$ acts transitively on the set of distincts pairs of points.

These two observations means that we may (and will) assume that $P_1=(0,0)$ and $P_2=(1,0)$.

Consider now the polynomials $X^{n+1}$ and $(X-1)^{n+1}$. These are coprime in the principal ring $\mathbb{R}[X]$, so by Bezout, there exists polynomials $h,k \in \mathbb{R}[X]$ such that $$1=h(X)X^{n+1}+k(X)(X-1)^{n+1}.$$ Thus, multiplying by $f$ gives $$f(X,Y)=h(x)X^{n+1}f(X,Y)+k(X)(X-1)^{n+1}f(X,Y).$$ Call $g_1=h(x)X^{n+1}f(X,Y)$, $g_2=k(X)(X-1)^{n+1}f(X,Y)$.

You can check that any partial derivative of $g_1$ of order $\leq n$ vanishes at $(0,0)$ (in fact, on the whole line $X=0$), and any partial derivative of $g_2$ of order $\leq n$ vanishes at $(1,0)$ (in fact, on the whole line $X=1$).