$\max\{x,y\}$ unique solution to functional equation?

Question

I know that the function $f:\mathbb R^2\mapsto \mathbb R,\,\,f(x,y):=\max\{x,y\}$ satisfies the equation $$f(x,y)+f(-x,-y)=\lvert x-y\rvert.$$ I want to prove/disprove if this is the only continuous solution. An other solution could be composed of two continous functions $f_1$ and $f_2$ defined on the two pairs of opposite quadrants $$Q_1:=\{(x,y):xy\geq0\},$$$$Q_2:=\{(x,y):xy\leq0\}$$such that $$f_1(x,0)=f_2(x,0),$$$$f_1(0,y)=f_2(0,y)$$ for all $x,y\in\mathbb R$ and $$f(x)=\begin{cases}f_1(x,y)\text{ if }(x,y)\in Q_1,\\f_2(x,y)\text{ if }(x,y)\in Q_2.\end{cases}$$ How can I continue?

Answer 1

Hint: Check if $f(x,y)=\frac{1}{2}|x-y|$ satisfies $f(x,y)+f(-x,-y)=|x-y|$.