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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:

triangle

  • 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 $z_1^2=z_1^3$?

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

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    What is a polynomial of a point in $\Bbb R^2$? – 2017-02-13
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    what is, e.g. $f((2,1))$ for $f(x)=x+1$? By the usage of letter $z$, I assume it is rather about *complex numbers*, via the identification $\Bbb R^2\to\Bbb C,\quad (a,b)\mapsto a+bi$. – 2017-02-13
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    @AdamHughes A polynomial $Q:\mathbb R^d\to\mathbb R$ is a function of the form $$Q(x)=\sum_{i_1,\;\ldots\;,i_n}\alpha_{i_1,\;\ldots\;,i_d}x^{i_1}_1\cdots x^{i_d}_d\;\;\;\text{for all }x\in\mathbb R^d\;.$$ If its degree is at most $r$, then $i_1+\cdots+i_d\le r$ for each $\alpha_{i_1,\;\ldots\;,i_d}\ne 0$. – 2017-02-13
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    @pepa.dvorak So, $$P(x)=\alpha+\beta x_1+\gamma x_2\;\;\;\text{for all }x\in K$$ for some $\alpha,\beta,\gamma\in\mathbb R$. – 2017-02-13
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    @Berci Nothing is complex here. The $z^i$'s are just the vertices of the triangle as shown in the picture. – 2017-02-13
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    Right! Too fast.. – 2017-02-13

2 Answers 2

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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 $f\equiv0$.

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    I don't understand the last part of your answer. First of all, the triangle should be nondegenerate, if the $z^i$'s are distinct (which is the case by assumption in the question). Now, what do you mean by "the expansion [...] is valid for all $x\in\mathbb R^2$"? The expansion should only be valid for all $x\in K$ (with $K$ being the triangle in the question; you called it $T$) and, most importantly, this should be all we need (or is your polynomial $f$ supposed to be defined on all of $ℝ^2$?) to conclude $f\equiv0$ (since from the perspective of the question, $f$ should be a map $K\toℝ$) . – 2017-02-18
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    If the $z_i$ are on one line, then the triangle is degenerate, and $f=0$ does not follow. 'Expansion.. valid': every point in $\mathbb R^2$ can be written as such a linear combination (without the condition $\lambda_i\ge0$). – 2017-02-18
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    "nondegenerate": Yes, you're right. I didn't thought about that case. "expansion": Still doesn't make sense. You still have the condition $\sum_{i=1}^3\lambda_i=1$ which forces $x$ be in $K$. Or do you intend to drop that condition too? (In that case, your last sentence in the answer is pretty misleading.) Otherwise, I don't see why each $x\in\mathbb R^2$ has such an expansion. What's the argument? – 2017-02-18
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    The key is 'affine independent' points: Take $z_1=(0,0)$, $z_2=(1,0)$, $z_3=(0,1)$. Then every point $x$ can be written as $\sum \lambda_i z_i$ with $\sum\lambda_i=1$. – 2017-02-19
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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 $p(z_s)$.

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    No, every point in the interior of $K$ is not the convex combination of only two corner points. – 2017-02-16
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    @daw Maybe he meant the following: Let $$u:=z^2-z^1$$ and $$v:=z^3-z^1\;.$$ Then, if $x\in K$, $$x=z^1+su+tv$$ for some $s,t\ge0$ with $s+t\le1$. – 2017-02-18
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    @0xbadf00d which is exactly the expansionof my answer ;) – 2017-02-18