Does there exist nonlinear functions $f$ and $g$ such that $f(x+y)=g(x)+g(y)$ for every $x$ and $y$?
Maybe there is a trivial solution but I can't see it. Thank you.
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4Assuming the axiom of choice, this is even possible with $f=g$ (assuming linear means $\mathbb R$-linear, not $\mathbb Q$-linear). So, please specify what you mean with linear here (it's probably affine linear anyway). – Hagen von Eitzen Jan 12 '14 at 12:46
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Sorry I'm not a mathematician. I want to exclude the simple (affine) linear solution : f(x) = ax+b, x belongs to ℝ. – Patrick Jan 12 '14 at 13:32
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Easy to see $f(0)=2g(0)$.
Let $y=0$, we get $f(x)=f(x+0)=g(x)+g(0)$, i.e., $g(x)=f(x)-g(0)$.
Hence $f(x+y)=g(x)+g(y)=[f(x)-g(0)]+[f(y)-g(0)]=f(x)+f(y)-f(0)$. Moreover, $f(x+y)-f(0)=[f(x)-f(0)]+[f(y)-f(0)]$.
Def $h(x)=f(x)-f(0)$, one has $h(x+y)=h(x)+h(y)$. $\forall x,y \in \mathbb{R}$.
From this we can show that $h(r)=h(1)r$ for $\forall r \in \mathbb{Q}$. Further assume that $f(x)$ is continuous, so is $h(x)$. Then $h(x)=h(1)x$ for $\forall x \in \mathbb{R}$. That is, $f(x)-f(0)=[f(1)-f(0)]x$, $f(x)=[f(1)-f(0)]x+f(0)$, $f$ is therefore linear.
Xucheng Zhang
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@janmarqz Maybe we have different understanding on linear, you may regard it as homo-genius. But in fact the solution is correct. – Xucheng Zhang Jan 12 '14 at 14:45
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this kind of map is called affine-transformation and it isn't linear in the sense of linear algebra, cf. [https://en.wikipedia.org/wiki/Affine_transformation] – janmarqz Jan 12 '14 at 14:51
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being so, your answer fits perfectly the OP, because Patrick is asking for non-linear $f$ and $g$. – janmarqz Jan 12 '14 at 14:53
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Because...I think solution not the form of solution that matters. – Xucheng Zhang Jan 12 '14 at 14:58
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So guys, is it correct to say that there are no functions, other than linear or affine, that satisfy my requirements? – Patrick Jan 12 '14 at 15:04
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