A non-linear map $f: \mathbb{R}^2 \to \mathbb{R}$ s.t. $f(av) = af(v)$ for $a \in \mathbb{R}, ~ v \in \mathbb{R}^2$ Is there a non-linear map $f: \mathbb{R}^2 \to \mathbb{R}$ s.t. $f(av) = af(v)$ for $a \in \mathbb{R}, ~ v \in \mathbb{R}^2$? I thought taking $f(x,y) = \sqrt{y^2}$ but it doesn't cover all of $\mathbb{R}$, nor satisfies $f(av) = af(v)$ for all $a \in \mathbb{R}, ~ v \in \mathbb{R}^2$. 
 A: Try considering the magnitude of elements in $\mathbb{R}^2$ as a starting point.
$$f(x,y) = \sqrt{x^2 + y^2}$$
It's clearly linear with respect to multiplication for positive $a$, but what happens if you consider any two points that aren't mutually collinear with the origin? 
The problem here is that it doesn't have the power to distinguish the sign of $a$ (and it seems you want a surjective map).  
You could update it without much effort:  let instead $f(x,y) = \text{sgn}(x)\sqrt{x^2 + y^2}$ or anything along those lines.  
Here I mean $\text{sgn(x)}$ to be the signum function which evaluates to $0$ when $x = 0$ and is otherwise $x/|x|$
A: Hint. This hint assumes you are comfortable with identifing $\mathbb R^2$ with $\mathbb C$. Consider the map $f(re^{i\theta})=r\sin\theta e^{i\theta}$. Is $f(av)=af(v)$ true? However, is $f(1+i)=f(1)+f(i)$ true?
A: Let $S=\{(\cos A, \sin A): A\in [0,\pi)\}.$ For each $v\in S$ choose an arbitrary $f(v)\in \Bbb R.$ And for  $a\in \Bbb R$ and $v\in S$ let $f(av)=af(v).$
For example let $f(1,0)=f(0,1)=1$ and $f(1/\sqrt 2\;,1/\sqrt 2\;)=1000000.$ Obviously $f(u)+f(v)\ne f(u+v)$ if $u=(1/\sqrt 2\;,0)$ and $v=(0,1/\sqrt 2\;).$
