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Let $\tilde{\mathbb{E}}$ be a degree-$r$ pseudo-distribution. If $\tilde{\mathbb{E}}P^2 = 0$, how to prove that $\tilde{\mathbb{E}}PQ=0$ for all polynomials $Q$ such that $\mathrm{deg}(PQ)\le r$?

We can prove the case $\mathrm{deg}(PQ) < r$ by induction, but it fails when $\mathrm{deg}(PQ) = r$.

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That implication is not actually true. Consider $P$, $Q$, and the pseudodistribution given by:

$$P = x, \quad Q = x^3,$$ $$r = 4,$$ $$\mathbb{\tilde E} 1 = 1, \quad \mathbb{\tilde E} x = 0,$$ $$\mathbb{\tilde E} x^2 = 0, \quad \mathbb{\tilde E} x^3 = 0,$$ $$\mathbb{\tilde E} x^4 = 1$$

Then $\mathbb{\tilde E}P^2 = 0$ but $\mathbb{\tilde E} PQ = 1$.

This is a valid degree-4 pseudodistribution since its moment matrix is positive-semidefinite.

$$ \left[\begin{array}{ccc} \mathbb{\tilde E}1 & \mathbb{\tilde E}x & \mathbb{\tilde E}x^2 \\ \mathbb{\tilde E}x & \mathbb{\tilde E}x^2 & \mathbb{\tilde E}x^3 \\ \mathbb{\tilde E}x^2 & \mathbb{\tilde E}x^3 & \mathbb{\tilde E}x^4 \\\end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\\end{array}\right] $$

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