# Proving that twice-differentiability implies supermodularity

Letting $$\theta: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}$$, we say $$\theta(x;q)$$ satisfies the single crossing property if , for all $$x,y\in\mathbb{R}^n$$ and $$q,r\in\mathbb{R}^m$$ such that $$x\geq y$$ and $$q\geq r$$:

$$\theta(x;r) \geq \theta(y;r) \to \theta(x;q) \geq \theta(y;q)$$

and

$$\theta(x;r) > \theta(y;r) \to \theta(x;q) > \theta(y;q).$$

If $$\theta$$ is twice-differentiable with $$\frac{\partial \theta}{\partial x_i \partial q_k}>0$$ for all $$i$$ and $$k$$, then $$\theta$$ satisfies the single crossing property.

I am not quite sure about this, my first attempt would make use of the definition of second derivative assuming dimension of $$x$$ and $$q$$ are both one.

$$\frac{\partial^2 \theta}{\partial x \partial q} = \lim_{h\to 0}\lim_{\Delta \to 0} \frac{\theta(x+h,q+\Delta)-\theta(x,q+\Delta)-\theta(x+h,q)+\theta(x,q)}{\Delta h}$$

If as we reach the limit the sign is maintained

$$\theta(x+h,q+\Delta)-\theta(x,q+\Delta) >\theta(x+h,q)-\theta(x,q) >0$$ with the last step following from dimensions being one.

Actually given the sign on the cross partial derivatives, the stronger inequality $$\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r),$$ holds for all $$x \ge y$$ and $$q \ge r$$.
To see this, one can use the multivariate mean value theorem (assuming that $$\theta$$ is $$C^2$$).
First we have that there exists some $$s$$ for which $$r \le s \le q$$ and: $$\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum_i \left[\frac{\partial \theta(x;s)}{\partial q_i} - \frac{\partial \theta(y;s)}{\partial q_i}\right](q_i - r_i).$$
For the term between square brackets, we can once more apply the multivariate mean value theorem: For all $$i$$, there exists a $$z^i$$ such that $$y \le z^i \le x$$ and (I index $$z^i$$ by $$i$$ as the value of the vector might be different for different $$j$$): $$\frac{\partial \theta(x;s)}{\partial q_i} - \frac{\partial \theta(y;s)}{\partial q_i} = \sum_j \frac{\partial^2 \theta(z^i;s)}{\partial q_i \partial x_j}(x_j - y_j)$$
Putting the two together gives: $$\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum_i \sum_j \left[\frac{\partial^2 \theta(z^i;s)}{\partial q_i \partial x_j}\right](x_j - y_j)(q_i - r_i).$$ If all cross partial derivatives are positive, then the right hand side is also positive if $$x \ge y$$ and $$q \ge r$$. As such, $$\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r)$$