Monotone log-supermodular function is supermodular.

Question

Let $X$ and $Y$ be lattices. Let $f: X \times Y \rightarrow \Re$. Function $f$ is log-supermodular if for all $x'>x$ and $y'> y$

\begin{equation} f\left(x', y'\right)f\left(x, y\right) \geq f\left(x, y'\right)f\left(x', y\right). \end{equation}

Function $f$ is supermodular if

\begin{equation} f\left(x', y'\right) + f\left(x, y\right) \geq f\left(x, y'\right) + f\left(x', y\right). \end{equation}

Finally, function $f$ is monotne in $x$ if $f\left(x', y\right) \geq f\left(x, y\right)$ for all $y$.

My question is this:

If I know that $f$ is log-supermodular and monotone in $x$, does this imply that $f$ is supermodular? I have tried to prove this formally without success. However, I think this should be correct.

Thank you so much for help.

Answer 1

I will assume $f$ is differentiable so the answer is straightforward.

Remember that $f\in C^1$ is super-modular iff $\partial^2_{xy}f\ge 0$. We also have that: $$\partial^2_{xy}\log(f)=\dfrac{\partial^2_{xy}f\cdot f - \partial_x f\cdot \partial_y f}{f^2}$$

So it is possible to have $\partial^2_{xy}\log(f)>0$ (log-supermodular), $\partial_x f>0$ (increasing in $x$) and $\partial^2_{xy}f<0$ (sub-modular).

Consider $f(x,y)=2+x-y-10^{-100}xy$ for $(x,y)\in[0,1]\times[0,1]$.

Answer 2

Given we already have the smooth case, let's do the general. Let $x, x' \in X$. Then by log-supermodularity:

$$ f(x \vee x') \cdot f(x \wedge x') \ge f(x) \cdot f(x') $$ so, re-arranging and subtracting one from both sides: $$ \frac{f(x \vee x')}{f(x)} - 1 \ge \frac{f(x')}{f(x \wedge x')} - 1. $$ But by monotonicity: $$ \frac{f(x)}{f(x \wedge x')} \bigg[\frac{f(x \vee x')}{f(x)} - 1\bigg]\ge \frac{f(x')}{f(x \wedge x')} - 1. $$ Simplifying and re-arranging yields: $$ f(x \vee x') + f(x \wedge x') \ge f(x) +f(x') $$ as required.