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Let $P_k$ be a degree-$k$ real polynomial of $m$ variables $x_1,\cdots,x_m$. $P_k$ can be written as $$P_k(x_1,\cdots,x_m) = \sum_{\alpha_1+\cdots+\alpha_m\leq k} \xi_{\alpha_1,\cdots,\alpha_m}x_1^{\alpha_1}\cdots x_m^{\alpha_m}.$$

My question is to prove the following:

$P_k(x_1,\cdots,x_k)=0$ almost everywhere on $0\leq x_1,\cdots,x_k\leq 1$ if and only if $\xi_{\alpha_1,\cdots,\alpha_m}=0$ for all monomials $\alpha_1,\cdots,\alpha_m$.

The "if" part is trivially true. Also, in univariate case ($m=1$) one can extract the non-zero components and use the fundamental theorem of algebra. However, I cannot think of a simple proof that handls $m>1$.

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  • $\begingroup$ You should probably mention what field(?) the coefficients $\xi$ belong to. $\endgroup$
    – hardmath
    Commented Jan 2, 2018 at 13:57
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    $\begingroup$ Good point. In my problem $\xi\in\mathbb R$, but I think the claim is also true even if $\xi\in\mathbb C$. $\endgroup$ Commented Jan 2, 2018 at 13:58
  • $\begingroup$ I would try induction on the number of variables, since you already have the case $m=1$ in hand. $\endgroup$
    – hardmath
    Commented Jan 2, 2018 at 14:04

1 Answer 1

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Over $\mathbb R$: take $Q_k=P_k^2$ and then $Q_k(x_1,\ldots,x_k)=0$ a.e. on $[0,1]^k$. This means that

$$\int_{[0,1]^k} Q_k(x_1,\ldots,x_k)dx_1\cdots dx_n=0$$

Having in mind that $Q_k$ is continuous and nonnegative, it follows that $Q_k(x_1,\ldots,x_k)=0$ on $[0,1]^k$, so also $P_k(x_1,\ldots,x_k)=0$.

All we need to prove now is that all coefficients are zero, but notice that, up to a constant factor, they are (or, at least, the degree-$k$ ones are) partial derivations of $P_k$ of some sort, thus they must also all be 0.

To see that on an example, look at e.g. $P_3(x,y)=ax^2y+\ldots$, then e.g. $a=\frac12\frac{\partial^3P}{\partial x^2\partial y}$ The general statement can be proven by induction on $k$.

Note: over $\mathbb C$, you would take $Q_k=P_k\overline{P_k}$ and the rest of the proof is the same.

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  • $\begingroup$ Thank you! very elegant answer. $\endgroup$ Commented Jan 2, 2018 at 14:14

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