Give an example of a function $f:\mathbb{R}^2 \mapsto \mathbb{R}^2$ such that
Solution 1:
take $f(x,y) = ( x^3 + y^3)^{1/3} $
$$ f(ax, ay) = ((ax)^3 + (ay)^3)^{1/3} = a f(x, y) $$
But,
$f(1,0) = 1 = f(0,1) $ and $f(1,1) = (2)^{1/3} $
It is then easy to extend this function to a new one $F: R^2 \to R^2 $ such that $F( \underline{x} ) = ( f( \underline{x}), 0 ) $ where $\underline{x} = (x,y) $
Solution 2:
You can let $f((r\cos\theta, r\sin\theta))$ be equal to $r g(\theta)$, where $g$ is any $\pi$-antiperiodic function you like. For instance, let $g(\theta)=\sin^3\theta e_x$ to obtain $$f(v)=r(v)\sin^3\theta(v)e_x=\frac{(v\cdot e_y)^3 e_x}{|v|^2}.$$