Suppose that $f(t)$ is a polynomial of degree d with real roots $a_1 \leq a_2 \leq \dotsb \leq a_d$, and $g(t)$ is a polynomial of degree $d-1$ with real roots $b_1 \leq b_2 \leq \dotsb \leq b_{d-1}$. Let’s say that $g(t)$ "weakly interlaces" $f(t)$ if we have
$$a_1 \leq b_1 \leq a_2 \leq b_2 \leq \dotsb \leq b_{d-1} \leq a_d,$$
and that $g(t)$ "strongly interlaces" $f(t)$ if the inequalities are all strict.
For arbitrary $f(t)$ and $g(t)$ (not necessarily real rooted), let $B(f,g)$ be the Bézout matrix, whose $(i,j)$ entry is equal to the coefficient of $x^{i-1} y^{j-1}$ in the polynomial $$\frac{f(x)g(y) - f(y)g(x)}{x-y}.$$ It seems to be a well known fact that both polynomials are real rooted and $g(t)$ strongly interlaces $f(t)$ if and only if $B(f,g)$ is positive definite. For example, see Corollary 9.145 here: Fisk - Polynomials, roots, and interlacing.
It follows that, if both polynomials are real rooted and $g(t)$ weakly interlaces $f(t)$, $B(f,g)$ must be positive semidefinite. (You can just perturb $f(t)$ and $g(t)$ so that they strongly interlace, and the limit of positive definite matrices is positive semidefinite.) I would like to know if the converse is true. That is, if $B(f,g)$ is positive semidefinite, does it follow that both polynomials are real rooted and $g(t)$ weakly interlaces $f(t)$?
This assertion is made in Theorem 2.13 of this paper: Kummer, Naldi, and Plaumann - Spectrahedral representations of plane hyperbolic curves. However, the source that they cite (Krein and Naimark) only appears to treat the positive definite case. It may indeed be true that the positive semidefinite case follows from the positive definite case, but I would like to understand why.