What extra assumption makes this transformation affine?

Asked Mar 28 '14 at 18:48

Active Mar 28 '14 at 19:30

Viewed 115 times

Let a vector space $V$ be given. Let $f:V\to V$ have the property that for all $x,y,a\in V$, $$ f(x+a)-f(y+a) = f(x) - f(y) \tag{$\star$} $$ Q1. I'd like to know how weak one can make additional assumptions to guarantee that $f$ is affine, namely that there exists a linear transformation $L:V\to V$, and an element $v\in V$ such that \begin{align} f(x) = L(x) + v \end{align} for all $x\in V$.

I think, for example, that if one assumes that $V = \mathbb R^n$, and if $f$ is differentiable, then $(\star)$ implies that $f$ is affine. This makes me think that perhaps one needs a notion of smoothness or continuity in general and therefore possibly a norm? Also

Q2. Is there a common name for the property $(\star)$?

edited Mar 28 '14 at 19:30

asked Mar 28 '14 at 18:48

joshphysics

1,821

Is $f$ just a map or a linear map? – BIS HD Mar 28 '14 at 18:49
@BISHD Just a map. – joshphysics Mar 28 '14 at 18:50
If $f$ were a linear map, you see that the assumption is trivially true and every linear map is a linear transformation for $v=0$. – BIS HD Mar 28 '14 at 18:51
@BISHD Yes I understand that. I am not assuming that $f$ is linear. – joshphysics Mar 28 '14 at 18:52
Ok, just to be clear. You've asked for extra assumptions ;-) Linearity is one of them. But wait for the analysis experts to help you with analytic assumptions. – BIS HD Mar 28 '14 at 18:54
@BISHD lol ok. I had hoped what I was asking was clear from the context, but I agree it's not entirely clear, so I changed the wording to "how weak one can make additional assumptions." How's that? – joshphysics Mar 28 '14 at 18:57
Sounds great to me, I'm curious about the answer. ;-) – BIS HD Mar 28 '14 at 18:58
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@BISHD Great! One might think that such curiosity would warrant an upvote ;) ? Thanks for the edits btw. – joshphysics Mar 28 '14 at 19:00
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Already done that! Let's hope for some experts. – BIS HD Mar 28 '14 at 19:03
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Let $y=0$ and you get $f(x+a)-f(a)=f(x)-f(0)$. Define $g(x)=f(x)-f(0)$ and you have $g(x+a)=g(x)+g(a)$. So $g$ is additive, and you want to show that it is linear. I would expect that continuity should be sufficient, though I haven't checked that the same proof goes through. – Mar 28 '14 at 19:10
@Rahul Ah thanks! Does one need to restrict the field over which $V$ is defined? – joshphysics Mar 28 '14 at 19:16
Also, @BISHD Linearity is obviously a stronger condition than affineness! – Mar 28 '14 at 19:17
@joshphysics: I have no idea. As I said, I haven't looked at the proof closely enough to see how well it generalizes. – Mar 28 '14 at 19:18
@Rahul Ok thanks for the insight. – joshphysics Mar 28 '14 at 19:18
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Your question is related to this question on Cauchy's functional equation. – Christoph Mar 28 '14 at 19:23

What extra assumption makes this transformation affine?

0 Answers0