3.32 of Boyd's Convex Optimization book says:
Products and ratios of convex functions. In general the product or ratio of two convex functions is not convex. However, there are some results that apply to functions on $\Bbb R$.
If $f$ and $g$ are convex, both nondecreasing (or nonincreasing), and positive functions on an interval, then $fg$ is convex.
If $f$, $g$ are concave, positive, with one nondecreasing and the other nonincreasing, then $fg$ is concave.
If $f$ is convex, nondecreasing, and positive, and $g$ is concave, nonincreasing, and positive, then $f /g$ is convex.
I found a reference to 3.32 in a comment in Submodularity of the product of two non-negative, monotone increasing submodular functions as well.
That said, this is my question:
What are the required conditions along with proof for the product of two submodular functions to be submodular?
Note: The comment in the link above extends 3.32 to the submodular case without proof, while the 3.32 pasted above from the book doesn't mention the property of submodularity of $f$ or $g$ anywhere.
There are results for set functions directly analagous to results 1 and 2. These are not required conditions, but sufficient.
If $f$ and $g$ are supermodular, both nondecreasing (or nonincreasing), and nonnegative functions, then $fg$ is supermodular.
If $f$, $g$ are submodular, nonnegative, with one nondecreasing and the other nonincreasing, then $fg$ is submodular.
These statements can be proven with the "product rule" for set functions: $$f(A+s)g(A+s)-f(A)g(A) = (f(A+s)-f(A))g(A+s) + f(A)(g(A+s)-g(A))$$ In a more compact notation, writing $D_s f = f(A+s)-f(A)$ and $S_s f = f(A+s)$: $$ D_s fg = D_s f \ S_s g + f D_s g $$ Apply the product rule twice, we get an expression of second order differences that can be used to prove submodularity or supermodularity of the product $fg$: $$ D_t D_s fg = D_t D_s f \ S_t S_s g + D_s f \ D_t S_s g + D_t f \ S_t D_s g + f \ D_t D_s g $$