If $K\subseteqℝ^2$ is a closed triangle spanned by $z^1,z^2,z^3$ and $P:K\toℝ$ is a polynomial of degree at most $1$ with $P(z^i)=0$, then $P\equiv0$

Question

Let

• $K\subseteq\mathbb R^2$ be a closed triangle spanned by $z^1,z^2,z^3$ (and assume that $z^i\ne z^j$)
• $P:K\to\mathbb R$ be a polynomial of degree at most $1$ with $$P(z^i)=0\;\;\;\text{for all }i\in\left\{1,2,3\right\}\tag1$$

How can we show that $P\equiv 0$?

My idea is the following:

• Let $L_1\subseteq K$ be the edge between $z^2$ and $z^3$
• Assume, for the moment, that $z_1^2\ne z_1^3$
• Let $$I_1:=[\min(z_1^2,z_1^3),\max(z_1^2,z_1^3)]$$
• It's clear that$^1$ there is a linear function $f_1:\mathbb R\to\mathbb R$ with $$f_1(I_1)=L_1\tag2$$
• Let $$P_1(x):=P(x,f_1(x))\;\;\;\text{for }x\in I_1$$
• It's easy to see that $P_1$ is a polynomial of degree at most $1$
• Using the result of my other question, we obtain the existence of some $C\in\mathbb R$ with $$P_1=C_1\left.f_1\right|_{I_1}\tag3$$
• Now, we know that $$C_1f_1(z_1^2)=P(z^2)=0=P(z^3)=C_1f_1(z_1^3)\tag4\;,$$ but is this sufficient to conclude?

So, the question is: How can we conclude and what do we do, if $z_1^2=z_1^3$?

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$^1$ Since $z_1^2\ne z_1^3$, we can choose $$f_1(x):=z_2^2+\frac{x-z_1^2}{z_1^3-z_1^2}\left(z_2^3-z_2^2\right)\;\;\;\text{for }x\in\mathbb R\;.$$

Answer 1

The key observation is that any point of $K$ can be written as $\ z\ =\ t\cdot z_r+(1-t)\cdot z_s\ $ for $t\in [0,1]$ and $r,s\in\{1,2,3\}$.

Now write a general polynomial $p$ of degree $1$ as $\ p(x,y)=ax+by+c\ $ (with $a,b,c\in\Bbb R$), then express $p(z)$ by means of $p(z_r)$ and $p(z_s)$.

Answer 2

Every point $x$ in the triangle $T$ can be written as $$ x = \sum_{i=1}^3 \lambda_i x_i, \ \sum_{i=1}^3 \lambda_i=1, \ \lambda_i\ge 0\forall i. $$ Take an affine linear function $f(x)=a^Tx+b$ with $f(x_i)=0$ for all $i$. Then with the characterization of $x\in T$ above $$ f(x) = a^Tx + b = (\sum_{i=1}^3 \lambda_i a^Tx_i )+ b =\sum_{i=1}^3\lambda_i (a^Tx_i+b) = \sum_{i=1}^3\lambda_if(x_i)=0. $$ Hence $f$ is zero on the triangle. If the triangle is not degenerate (collapsing to line or point), then the expansion $$ x = \sum_{i=1}^3 \lambda_i x_i, \ \sum_{i=1}^3 \lambda_i=1 $$ is valid for all $x\in \mathbb R^2$, which gives $f\equiv0$.