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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?

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  • $\begingroup$ You should clarify your subject line, as you are not asking what your subject line asks. $\endgroup$ Nov 15, 2022 at 4:36
  • $\begingroup$ It is not quite clear what you are asking, actually. $\endgroup$ Nov 15, 2022 at 6:36

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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 !

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    $\begingroup$ Please stop writing "u" for "you". You are not texting your friends here. $\endgroup$ Nov 15, 2022 at 4:47

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