Direct sum factorization of polynomials

Question

• I have been recently reading the paper "Mixed finite elements for second order elliptic problems in three variables" by Brezzi et. al.
• I noticed the claim in the proof of Lemma $2.1$, which basically boils down to given a polynomial $\mathrm{p}\left(x,y\right) \in P_{k}$, we can write it as $$ \mathrm{p}\left(x,y\right) = c\left(1 - x - y\right) + x\,\mathrm{p}_{1}\left(x,y\right) + y\,\mathrm{p}_{2}\left(x,y\right) $$ for some $c$ constant and $\mathrm{p}_{1}\left(x,y\right),\ \mathrm{p}_{2}\left(x,y\right) \in P_{k - 1}$. This is equivalent to the claim that $$ P_{k} = xP_{k - 1}\oplus yP_{k - 1} \oplus \text{Span}\left\{1 - x - y\right\} $$ as far as I understand.
• It sounds plausible but I am not sure if it is true. Does anyone know more about this ?.

Answer 1

Here, few days after I came up with this answer.

Let's consider the basis of the $P_k$, e.g. $\{x^iy^j | i,j\geq 0 \text{ and } i+j\leq k \}$. Observe that we can write each function in this basis as given above

$ 1 = 1\times (1-x-y) + x\times 1 + y\times 1 $

$ x^i = 0\times (1-x-y) + x\times x^{i-1} + y\times 0$, for $i>0$

$ y^j = 0\times (1-x-y) + x\times 0 + y\times y^{j-1}$, for $j>0$

$ x^iy^j = 0\times (1-x-y) + x\times \tfrac{1}{2}x^{i-1}y^j + y\times \tfrac{1}{2}x^{i-1}y^{j-1}$, for $j>0$ and $i>0$. Since sum of $k$-th degree polynomials is a $k$-th degree polynomial and $1$, $x^{i-1}$, $y^{j-1}$, $\tfrac{1}{2}x^{i-1}y^j$ and $\tfrac{1}{2}x^{i-1}y^{j-1}$ are $(k-1)$-st degree polynomials, we have the answer.

I am going to leave this here in case someone else (or future me) needs it.