# What, along with homogeneity, implies additivity?

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?

If you assume additivity for orthonormal vectors i.e., $$f(u+v) = f(u) + f(v)$$ for $$u,v$$ orthonormal (or additivity for a chosen set of (scaled) orthonormal basis) and homogenity then additivity extends to all vectors $$u,v$$. Not sure if there is any other way. If you can eloborate on what you expect it will help others. Thanks man.
Note that additivity + homogenity + finite dimensionality => matrix representation of $$f$$. So lets pick standard basis. So if you assume: $$f([x,y,z]) = x + f([0,y,z])$$. and similarly for y and z. then this also works but by the same condition on orthonormal vectors as i said. Interesting question though !