Suppose $f$ is a function defined on a vector space which satisfies one of the requirements for a linear function, the “homogeneous” condition: $$\alpha \cdot f(v) = f(\alpha v).$$
This doesn't imply that $f$ must satisfy the other condition of a linear function, the “additivity” condition: $$f(u+v) = f(u) + f(v).$$ To easily see this, consider any arbitrary mapping of numbers to the hemisphere, then as long as these are scaled appropriately, $f$ will be homogeneous.
On the flip side, additivity doesn’t imply homogeneity either. However, additivity implies homogeneity over the rationals. Furthermore, additivity + continuity does imply homogeneity. See the question asked here: [https://math.stackexchange.com/questions/1648504/additivity-implies-homogeneity-of-rational-scalars].
This inspires the following question:
Let $v \in \mathbb{R}^n.$ Suppose $f(v)$ is homogeneous. Are there some additional assumptions on $f,$ weaker than additivity itself, that imply $f$ is additive?
Consider also functions like $$f(x,y,z) = f_x(x,y,z)\cdot x + f_y(x,y,z)\cdot y + f_z(x,y,z)\cdot z$$ Which satisfy homogeneity as long as the $f_i$ have the property that $f_i (v) = f_i (\alpha v)$ (but not necessarily additivity/linearity).
Let's call this a "weakened" linearity. What are the minimal assumptions needed to show that homogeneity implies "weakened" linearity?
