Can $f(a-b)=\frac{1}{|a-b|}$ for vectors two $a,\ b\in\mathbb{R}^2$ be separated into functions $g(|a|)$ and $h(|b|)$ such that $f(a,b)=g(|a|)h(|b|)$?

Question

I would like to separate a function $f(a,b) = f(a-b)=\frac{1}{|a-b|}$ where $a$ and $b$, with $a \neq b$, are vectors in $\mathbb{R}^2$, into two functions $g(|a|)$ and $h(|b|)$ such that $f(a,b) = f(a-b)=g(|a|)h(|b|)$. Can this be done?

Answer 1

Absolutely not. For picking $b = (0,0)$, the function $g(\|a\|)$ is just some constant nonzero multiple (namely $h(0)$) times $1 / \| a \|$.

Now consider the case $a = 2b$. You'd have $$ f(a-b) = f(2b-b) = f(b) = \frac{1}{\|b\|} = g(2b) h(b) = h(0) \frac{1}{\|2b\|} h(b) $$ for every $b$. THus $$ 1 = h(0) h(b)/2 $$ and $h$ is constant. That's not possible.

Answer 2

No it cannot… assume there exist such $g$ and $h$ then $$f(a-b) = g(|a|)h(|b|) = g(|a|)h(|-b|) = f(a+b)$$ but in general $$f(a-b) \not= f(a+b)$$