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We know that if $f(x+y)=f(x)+f(y)$ and $f$ meets some "reasonable" conditions, then $f$ is linear.

I've been considering the following extension: consider the reals under some unknown group operation $\oplus$ which is isomorphic to the reals under standard addition, i.e. $f(x\oplus y)=f(x)+f(y)$. Under what conditions must we conclude that $f(x)=cx$ where $f$ is the isomorphism?

I think over the rationals we could use the same argument as for Cauchy, but I'm not sure about over the reals.

Update: In another question Joy gives an example where $f:(\mathbb{R},\oplus)\to (\mathbb{R},+)$ with $f$ continuous yet $f(x)\not=cx$. So the answer to this generalization is not the same as the answer to the normal Cauchy.

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Xodarap
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  • what argument do you have to show the necessary condition for cauchy? of course, in this case as well continuity is sufficient. – suissidle Mar 29 '13 at 14:31
  • @suissidle: are you asking for a more exact definition of what I called "reasonable" above? Monotonicity is my preferred condition, but I'd be interested in the case where f is continuous is well. – Xodarap Mar 29 '13 at 14:47
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    under both conditions the only automorphisms of $(\mathbb{R},+)$ are multiplications with nonzero scalars:

    (1) http://math.stackexchange.com/questions/115486/what-is-operatornameaut-mathbbr

    (2) http://math.stackexchange.com/questions/60810/the-only-group-automorphisms-of-the-additive-group-of-real-numbers-that-are-also

     – suissidle Mar 29 '13 at 15:06
  • @suissidle: Sorry, I meant to say it is homomorphic. (So I guess the question reduces to "which homomorphisms are isomorphisms"...) – Xodarap Mar 30 '13 at 00:44
  • i don't know exactly what you mean by that. 'isomorphic' made sense unequivocally. the content of my comment to diego's answer is that any choice of 'group operation' is equivalent to a choice of isomorphism from $\mathbb{R}$ to itself (automorphism) - to justify talking about automorphisms $f$ instead of group operations $\oplus$ – suissidle Mar 30 '13 at 00:59

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If you conclude $f(x)=cx$, then $c(x\oplus y)=f(x\oplus y)=f(x)+f(y)=cx+cy=c(x+y)$, that is, $\oplus=+$. Now for an automprhism of $f$ of $(\mathbb{R},+)$, to conclude that $f(x)=cx$ then $f$ must be continuous, or equivalently, $\lim_{x\rightarrow 0}f(x)=0$.

Diego
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  • note that group automorphisms $g$ correspond to isomorphic structures on the group. if you define the new operation by $x\oplus y=g(x+y)$, then $g^{-1}(x\oplus y)=x+y$ and $g^{-1}$ gives you the isomorphism. and vice versa... – suissidle Mar 29 '13 at 21:07
  • Why is $g^{-1}$ an isomorphism? $g^{-1}(x\oplus y)\not = g^{-1}(x) + g^{-1}(y)$ unless $g(x)=x$ it seems. – Xodarap Mar 30 '13 at 21:43
  • I guess more generally I don't understand why we're talking about the automorphisms of $(\mathbb{R},+)$ since we don't know that $+=\oplus$. – Xodarap Mar 30 '13 at 22:53
  • You are asking the following: What condition do I have to add to an isomorphism $f:(\mathbb{R},\oplus)\rightarrow (\mathbb{R},+)$ in order to conclude that $f(x)=cx$. I am saying that no matter the conditions are, if you conclude that $f(x)=cx$, then necessarily $+=\oplus$. – Diego Mar 31 '13 at 07:01
  • Fair enough, but if we define $+=\oplus$ that kind of begs the question. I'm wondering if, similar to the $Aut(\mathbb{R},+)$ case there are non-obvious restrictions which end up forcing $+=\oplus$. – Xodarap Mar 31 '13 at 13:57