My question is:
If $I$ is a homogenous ideal of $S=K[x_1,\dots,x_n]$ and $\mathrm{in}_{<}(I)$ is the initial ideal of $I$, with respect to a term order $<$ on $S$, then $S/I$ is Gorenstein if and only if $S/\mathrm{in}_{<}(I)$ is Gorenstein.
It looks so from here but I'm not sure at 100%.
It is not true in general that if S/I is Gorenstein, then S/in(I) is also Gorenstein. For example, consider $S = K[x,y,z]/(x^2,y^2,z^2,xy-xz,xz-yz)$. This is an Artinian Gorenstein ring and with respect to just about any monomial order, the initial ideal is $J = (x^2,y^2,z^2,xy,xz)$; however, $S/J$ is not Gorenstein. You can see this either by looking at the graded Betti numbers in Macaulay2; it looks something like this: $\begin{matrix} & 0 & 1 & 2 & 3\\ \text{total:} & 1 & 5 & 6 & 2\\ 0: & 1 & . & . & .\\ 1: & . & 5 & 5 & 1\\ 2: & . & . & 1 & 1 \end{matrix}$
The lack of symmetric means it is not Gorenstein.
Alternatively, you could note that the socle of $S/J$ is $[J:(x,y,z)]/J = (x,yz)/J$, which is not one-dimensional.
So the "squarefree initial ideal" hypothesis is doing a lot in the Conca-Varbaro paper, which is still quite a nice result. (Note that our initial ideal is not at all squarefree.)
