Homogeneity does not suffice for a map between vector spaces to be linear

Question

The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:

Problem: Give an example of a function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f(av) = af(v)$ for all $a \in \mathbb{R}$ and all $v \in \mathbb{R}^2$ but $f$ is not linear.

I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.

Do you know such an example? Thank you in advance!

Answer 1

consider the map: $$f:\mathbb{R}^2 \to \mathbb{R}\\ (x,y) \mapsto \begin{cases} x \text{ if } y=0 \\ 0 \text{ else }\end{cases} $$ then this is homogenous $f(a(x,y))(f(ax,ay))=\begin{cases}ax \text{ if } y=0 \\ a0\text{ else }\end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 \neq f(2,0) =f((1,1)+(1,-1)).$$

Answer 2

I just finished working on this problem and this is what I got:

Let

$$ f:\mathbb{R}^2 \to \mathbb{R}\\ (x,y) \mapsto \begin{cases} \frac{x^2}{y} \ , \ y \neq 0 \\ 0 \ , \ y=0 \end{cases} $$ This is a homogeneous map since $$f\Bigl(a(x,y)\Bigr) = f\Bigl((ax,ay)\Bigr) = a\ \frac{x^2}{y} = af\Bigl((x,y)\Bigr)$$ whenever $a \neq 0$ and in the case of $a=0$ we get: $$f\Bigl(\ 0(x,y)\ \Bigr) = f\Bigl((0,0)\Bigr) = 0 = 0f\Bigl((x,y)\Bigr)$$ However, it is not additive, in general, since $$f\Bigl((x_1,y_1)+(x_2,y_2)\Bigr) = \frac{(x_1 + x_2)^2}{y_1+y_2} \neq \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2} = f\Bigl((x_1,y_1)\Bigr)+f\Bigl((x_2,y_2)\Bigr)$$ hence not linear.