Positive definite 2-sequence from positive definite 1-subsequences?

Question

If $p = (p_{n,m})_{n,m \geq 0}$ is a 2-sequence for which it is known that all the 1-subsequences $s^{(ab)} = (p_{ka,kb})_{k \geq 0}$ are positive semidefinite, for arbitrary $a,b \in \mathbb{N}_0$, can we conclude that $p$ is a positive semidefinite 2-sequence?

Answer 1

No. A counter-example is the 2-sequence $p$ with $p_{2k,k} = 1$ for all $k \geq 0$ and $p_{n,m} = 0$ elsewhere. The 1-subsequences $s^{(ab)}$ are all positive semidefinite (since they are either $(1,0,0,\cdots)$ or $(1,1,\cdots)$) but $p$ is not, because for the three pairs of indices $(n_1,m_1) = (0,0), (n_2,m_2) = (1,0)$ and $(n_3,m_3) = (1,1)$ and the three-vector $c = (1,1,-1)$ we have $$\sum_{i,j=1}^3 c_i c_j p_{n_i+n_j,m_i+m_j} = -1 < 0.$$ This violates the requirement for the positive semidefiniteness of a sequence (e.g. Nussbaum, Ark. Mat. 6, 10 (1965)).