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.

  • 4
    $\begingroup$ Assuming 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). $\endgroup$ – Hagen von Eitzen Jan 12 '14 at 12:46
  • $\begingroup$ Sorry I'm not a mathematician. I want to exclude the simple (affine) linear solution : f(x) = ax+b, x belongs to ℝ. $\endgroup$ – Patrick Jan 12 '14 at 13:32

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.

  • $\begingroup$ this expression for $f(x)$ isn't linear unless $f(0)=0$, sorry. $\endgroup$ – janmarqz Jan 12 '14 at 14:41
  • $\begingroup$ @janmarqz Maybe we have different understanding on linear, you may regard it as homo-genius. But in fact the solution is correct. $\endgroup$ – Xucheng Zhang Jan 12 '14 at 14:45
  • $\begingroup$ So, what is your definition of linear?? $\endgroup$ – janmarqz Jan 12 '14 at 14:47
  • $\begingroup$ f(x)=ax+b, like this. b=0 is just a special case. $\endgroup$ – Xucheng Zhang Jan 12 '14 at 14:48
  • $\begingroup$ this kind of map is called affine-transformation and it isn't linear in the sense of linear algebra, cf. [en.wikipedia.org/wiki/Affine_transformation] $\endgroup$ – janmarqz Jan 12 '14 at 14:51

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