How to prove $\tilde{\mathbb{E}}PQ=0$ when $\tilde{\mathbb{E}}P^2 = 0$? 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 $\mathrm{deg}(PQ) &lt; r$ by induction, but it fails when $\mathrm{deg}(PQ) = r$.
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.

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