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Does anyone know whether the following is an elementary result in linear algebra (to be found in a undergrad book for example)?

Let $M: \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a mapping. Then:

$M$ is affine ($M(x)=Ax+\gamma$, for some fixed matrix $A$ and vector $\gamma$) if and only if $M$ satisfies property (1) below

(1): for all $a,b,c,d$ in $\mathbb{R}^d$: $a-b = c-d\,\,$ implies $\,\,M(a)-M(b)=M(c)-M(d)$.

I'm not so much interested in a proof (unless the theorem would be false), but more in whether this is a generally known elementary result.

Edit: The first part of Property (1) is also called "arithmetic proportion" or "difference proportion". https://encyclopediaofmath.org/wiki/Arithmetic_proportion

Edit 2: I think this is related to the fact that affine transformations preserve parallelism, but it would go in both directions (i.e., preservation of parallelism -> affine)

user144248
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  • In this form the result seems wrong to me even in case $d=1$ as there are discontinuous solutions of Cauchy's functional equation, see https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – Gerd Feb 12 '22 at 12:49
  • @gerd I'm not sure I follow what you want to say. Do you have a counter-example? (one part of the implication is obvious, btw.) – user144248 Feb 12 '22 at 13:02
  • I elaborate in an answer. – Gerd Feb 12 '22 at 14:05

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It is known that Cauchy's functional equation $f(x+y)=f(x)+f(y)$ has discontinuous solutions $f:\mathbb{R} \to \mathbb{R}$. Such a function is clearly not affine, as affine functions are continuous. Now if $a-b=c-d$ then $a+d=b+c$ hence $$ f(a)+f(d)=f(a+d)=f(b+c)=f(b)+f(c) \Rightarrow f(a)-f(b)=f(c)-f(d). $$ For the existence of such functions see non-continuous function satisfies $f(x+y)=f(x)+f(y)$

I think there is a better chance for this equivalence if you assume that the functions under consideration have a given quality such as 'continuous' or 'measurable'.

Gerd
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  • Thanks @Gerd. I am asking this question, because I saw the "proof" in a paper. See here (https://arxiv.org/abs/2004.01079), Section 3.2. What do you think is wrong in the proof? – user144248 Feb 12 '22 at 14:26
  • The author of this paper proves that $M$ is $\mathbb{Q}$-linear. To get $\mathbb{R}$-linearity from that you need something in addition, for example that $M$ is continous. If the proofs in the paper are correct (I didn't check everything) and $M$ is continuous then it is affine. – Gerd Feb 12 '22 at 14:41
  • The authors are aware of this problem (read footnote 2 on page 4). – Gerd Feb 12 '22 at 15:13
  • Alright, thanks – user144248 Feb 12 '22 at 16:51