# doubt with absolute value

Assume a function

$$f(x,y) = 2\vert x \vert + 2\vert y\vert - \vert x-y \vert - \vert x+y \vert$$ If $$x>y$$ then $$f(x,y) = 2y$$, If $$x then $$f(x,y) = 2\vert x \vert$$

How to prove this? If $$x>y$$ then for $$x>0$$, $$f(x,y) = 2x + 2y -x + y -x - y = 2y$$ for $$x<0$$, we have $$f(x,y) = -2x + 2y + x + y +x - y = 2y$$

Now for $$y>x$$, can I write $$f(x,y) = 2\vert x \vert + 2\vert y\vert - \vert y-x \vert - \vert x+y \vert$$, such that for $$y>0$$, we have $$f(x,y) = 2x + 2y - y + x -x - y = 2x$$ and for $$y<0$$, we have $$f(x,y) = 2x - 2y + y + x - x + y = - 2x$$

Is my logic of proof sound?

It is a bit easier to consider this equality first:

$$\max(|x|,|y|)=\frac{|x+y|+|x-y|}2$$

Indeed let $$y=cx$$ ($$x\neq 0$$, it is trivially verified for $$x=0$$), then let consider $$|x+y|+|x-y|=|x|\,(|1+c|+|1-c|)$$

This transformation $$y=cx$$ allows us to deal only with one variable $$c$$ instead of two $$x,y$$ and to reduce the number of subcases to study.

• $$c\ge 1\implies |x|(1+c+c-1)=2c|x|=2|y|$$
• $$c\le -1\implies |x|(-1-c+1-c)=-2c|x|=2|y|$$
• $$-1\le c\le 1\implies |x|(1+c+1-c)=2|x|$$

The first two cases correspond to $$|c|\ge 1\iff |y|\ge|x|$$ and the third one to $$|y|\le |x|$$, hence the result.

As a consequence the original problem's expression becomes:

$$2|x|+2|y|-|x+y|-|x-y|=2|x|+2|y|-2\max(|x|,|y|)=2\min(|x|,|y|)$$