How to prove this fact about the argmax of a function given properties of partial derivatives I need to prove that assuming differentiability etc that the following holds. If we have a function $f(x,y)$ and we know that $\frac{\partial ^ 2 f}{\partial x \partial y} < 0$, then this implies that $g(y) = \arg \max_x f(x,y)$ is a decreasing function of $y$. i.e. as $y$ increases, $g(y)$ decreases.
My intuition is that this is because $\frac{\partial ^ 2 f}{\partial x \partial y} = \frac{\partial}{ \partial y} \left( \frac{\partial f}{\partial x} \right) < 0$ and thus the function $h(x,y) = \frac{\partial f}{\partial x}(x,y)$ is decreasing with respect to y. I am not sure this makes a conclusive argument, however.
Any help will be appriecated!
 A: I assume that $f$ is defined such that for all $y$, a unique $\arg \max_x f(x,y)$ exists, and that the domain over which $f$ is defined is convex.
Here is an argument you could use.  Consider any $y_0 < y_1$. Let $x_0 = g(y_0), x_1 = g(y_1)$. Suppose for the purpose of contradiction that $x_0 \leq x_1$.
By the definition of $g$, we see that $f(x_0,y_0) > f(x_1,y_0)$, and $f(x_1,y_1) > f(x_0,y_1)$. Now, define $h(y) = f(x_1,y) - f(x_0,y)$. We see that
$$
\begin{align}
h(y_0) &= f(x_1,y_0) - f(x_0,y_0) < f(x_1,y_1) - f(x_0,y_0)
\\ & < f(x_1,y_1) - f(x_0,y_1) = h(y_1).
\end{align}
$$
That is, $h$ increases from $y_0$ to $y_1$. It follows from the mean value theorem that there must exist a $y_*$ with $y_0 < y_* < y_1$ for which $\frac{dh}{dy}(y_*) > 0$, which is to say that
$$
f_y(x_1,y_*) > f_y(x_0,y_*).
$$
Now, by the mean value theorem, there must exist an $x_*$ with $x_0 < x_* < x_1$ for which
$$
f_{xy}(x_*,y_*) = \frac{\partial }{\partial x} f_y|_{(x,y) = (x_*,y_*)} = \frac{f_y(x_1,y_*) - f_y(x_0,y_*)}{x_1 - x_0} > 0,
$$
which contradicts our premise that $f_{xy} < 0$ over the domain of $f$.
