show that xf(x)=c if xyf(xy)=(1-x)yf((1-x)y)

Question

In an article about statistics, Leonard J. Savage makes the following claim:

Let $f:(0,1)\rightarrow\mathbb{R}$ differentiable. Then

$$ xyf(xy)=(1-x)yf((1-x)y)\mbox{ for all }x,y\in(0,1) $$

implies

$$ xf(x)=k\mbox{ for some }k\in\mathbb{R}\mbox{ and all }x\in(0,1) $$

I can't quite see why that is the case. Along which lines would I prove this?

Answer 1

The given equation can be written as $$ \frac{f(xy)}{f((1-x)y)} = \frac{1-x}{x}$$ Now for any $s,t \in (0,1)$ with $s+t < 1$, we can take $s = xy$, $t=(1-x)y$ where $y = s+t$ and $x = s/(s+t)$, and we get $$ \frac{f(s)}{f(t)} = \frac{1-x}{x} = \frac{t}{s}$$ But that means $s f(s)$ must be constant.