For what functions $f(x)$ is $f(x)f(y)$ convex? For which functions $f\colon [0,1] \to [0,1]$ is the function $g(x,y)=f(x)f(y)$ convex over $(x,y) \in [0,1]\times [0,1]$ ? Is there a nice characterization of such functions $f$?
The obvious examples are exponentials of the form $e^{ax+b}$ and their convex combinations. Anything else?
EDIT: This is a simple observation summarizing the status of this question so far. The class of such functions f includes all log-convex functions, and is included in the class of convex functions. So now, the question becomes: are there any functions $f$ that are not log-convex yet $g(x,y)=f(x)f(y)$ is convex?
EDIT: Jonas Meyer observed that, by setting $x=y$, the determinant of the hessian of $g(x,y)$ is positive if and only if $f$ is a log-convex. This resolves the problem for twice continuously differentiable $f$. Namely: if $f$ is $C^2$, then $g(x,y)$ is convex if and only if $f$ is log-convex.
 A: Suppose $f$ is $C^2$.  First of all, because $g$ is convex in each variable, it follows that $f$ is convex, and hence $f''\geq0$.  I did not initially have Slowsolver's insight that log convexity would be a criterion to look for, but naïvely checking for positivity of the Hessian of $g$ leads to the inequalities
$$f''(x)f(y)+f(x)f''(y)\geq0$$
and
$$f'(x)^2f'(y)^2\leq f''(x)f(x)f''(y)f(y)$$
for all $x$ and $y$, coming from the fact that a real symmetric $2$-by-$2$ matrix is positive semidefinite if and only if its trace and determinant are nonnegative.  The first inequality follows from nonnegativity of $f$ and $f''$.  The second inequality is equivalent to $f'^2\leq f''f$.  To see the equivalence in one direction, just set $x=y$ and take square roots; in the other direction, multiply the inequalities at $x$ and $y$.  Since $\log(f)''=\frac{f''f-f'^2}{f^2}$, this condition is equivalent to $\log(f)''\geq0$, meaning that $\log(f)$ is convex.
A: This may just be a silly observation.
But, let $\alpha=(a_1,a_2)$ and $\beta=(b_1,b_2)$. If $f$ is convex and the sum of $g$ at the four corners of the square formed by $\alpha$ and $\beta$ is non-positive, then $g(t\alpha + (1-t)\beta) \leq tg(\alpha)+(1-t)g(\beta)$, where $t \in [0,1]$
