What I want to show is that if $f$ satisfies $f(x,y) > f(w,z)$ IFF $\dfrac{x}{w} > \dfrac{y}{z}$ then $f$ is homogenous of degree $0$. (i.e. $f(\alpha x,\alpha y ) = f(x,y)$)
I saw a proof of this somewhere along the following lines, but I do not quite understand it. So I will outline it, state my confusion and will be very happy if someone can clarify.
My two questions are below in yellow like this
Suppose $f(x,y) > f(w,z)$ IFF $\dfrac{x}{w} > \dfrac{y}{z}$,
- then $f(x,y) > f(w,z)$ IFF $\dfrac{x}{y} > \dfrac{w}{z}$.
- Now consider any $c,d$ then the condition that $f(x,y)=f(c,d)$ is invariant under scaling $x,y$.
- This follows because the condition, $f(x,y)=f(c,d)$, depends only on the ratio $\dfrac{x}{y}$ (by assumption)
- (I am confused here.) As a result, $f(x,y)$ must depend only on the ratio $\dfrac{x}{y}$ and must be proportional to $f\left(\dfrac{x}{y},1\right)$.
By assumption doesn't $f\left(\dfrac{x}{y},1\right) = f(c,d)$ have to be true when $f(x,y)=f(c,d)$? Which would imply $f\left(\dfrac{x}{y},1\right)=f(x,y,)$?
- (More generally) Why must it be proportional to $f\left(\dfrac{x}{y},1\right)$. (I am also not quite clear why it must depend only on the ratio, so bonus points if that can be explained. (But I think that follows from changing the ratio must make the function equal a different value)
- (And I am most confused here) Setting $x=y$ then gives that the proportionality constant is $1$
I have no idea why this is.
It seems that this proof takes unnecessary detour. Here is a proof based on the same idea but is somewhat more direct.
Because $f(x_1, y_1) > f(x_2, y_2) \Leftrightarrow \dfrac{x_1}{x_2} > \dfrac{y_1}{y_2}$ and $\dfrac{x_1}{x_2} > \dfrac{y_1}{y_2} \Leftrightarrow \dfrac{x_1}{y_1} > \dfrac{x_2}{y_2}$, then$$ f(x_1, y_1) > f(x_2, y_2) \Longleftrightarrow \frac{x_1}{y_1} > \frac{x_2}{y_2}, \tag{1} $$ which also implies$$ f(x_1, y_1) < f(x_2, y_2) \Longleftrightarrow \frac{x_1}{y_1} < \frac{x_2}{y_2}. \tag{2} $$ Combining (1) and (2) yields$$ f(x_1, y_1) = f(x_2, y_2) \Longleftrightarrow \frac{x_1}{y_1} = \frac{x_2}{y_2}, \tag{3} $$ which implies $f$ is homogeneous of degree $0$.