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I would like to prove the following statement. Let $n \in \mathbb{N}$ be arbitrary and $\emptyset \neq S \subseteq \mathbb{R}^n$. Then $\phi : S \rightarrow \mathbb{R}$ is affine iff there exists $\alpha_1, ..., \alpha_n, \beta$, each of them $\in \mathbb{R}$, such that $\phi(x) = \sum_{i=1}^n\alpha_ix_i \ + \beta$, for all $x \in S.$

I am familiar with the fact that, provided that $0 \in S$, $\phi \in \mathbb{R}^S$ is affine iff $\phi - \phi(0)$ is a linear map on $S$ and I believe this will be useful.

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  • $\begingroup$ Is $S$ just an arbitrary subset? Or should it be an affine subspace? $\endgroup$
    – Dustan Levenstein
    Feb 27, 2018 at 14:48
  • $\begingroup$ It is an arbitrary subset of $\mathbb{R}^n$ $\endgroup$
    – Gabriel Jardanovski
    Feb 27, 2018 at 17:11
  • $\begingroup$ certainly $S$ has to be affine set. $\endgroup$
    – Red shoes
    Feb 27, 2018 at 22:44
  • $\begingroup$ @Redshoes not really... There's a perfectly good notion of an affine function on $\mathbb R^n$ restricted to an arbitrary subset. If you really wanted to, you could even define this notion "intrinsically" using, say, the metric on $S$. As to whether this is a useful notion, I don't know. But it's certainly not a meaningless one. $\endgroup$
    – Dustan Levenstein
    Feb 27, 2018 at 23:35
  • $\begingroup$ If $S$ is not an affine set then $\phi$ can be naturally and uniquely extended to an affine function on entire $aff (S)$. So there is no points and no advantages of considering affine functions on non affine sets. @DustanLevenstein $\endgroup$
    – Red shoes
    Feb 27, 2018 at 23:47

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I don't know what definition of affine you're given, but I'm guessing it'll be helpful for you to know this fact:

Fix a choice of $v_0 \in S$. Then $\phi$ is affine if and only if the function $$S-v_0 \to \mathbb R$$ given by $$w \mapsto \phi(w+v_0)$$ is affine. Here $S-v_0$ denotes the set $\{v-v_0\mid v \in S\}$.

This reduces your problem to understanding that transformation between $S$ and $S-v_0$ and handling the case $0 \in S$. See if you can prove that and then use it together with the fact you quoted.

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