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Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.

Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.

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You might look at the Wikipedia article on the Cauchy functional equation. –  André Nicolas May 8 '13 at 15:11
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If you assume axiom of choice, these function is exists - you can construct this function to use Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$. - But if you does not assume axiom of choice, you can't show that nonlinear additive function - If you assume axiom of determinacy, every real function is measurable. And every measurable additive real function is linear. –  tetori May 8 '13 at 15:18
    
This has got to be a dupe. –  Aryabhata May 8 '13 at 17:01
    
relevant mathoverflow.net/questions/57426/… –  clark May 8 '13 at 17:03

2 Answers 2

Yes continuity is enough: You can quickly show that $f(x)=x\cdot f(1)$ for $x\in\mathbb N$, then for $x\in\mathbb Z$ and then for $x\in\mathbb Q$; assuming continuity, this implies validity for all $x\in\mathbb R$.

Any other functions only exist per Axiom of Choice: View $\mathbb R$ as a vector space over $\mathbb Q$ and take any $\mathbb Q$-linear map (which need not be $\mathbb R$-linear).

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Also, it is well known that graph $\{(x, f(x):x\in \Bbb R\}$ of every solution $f$ except $f(x)=ax$ is dense in the plane $\Bbb R^2$.

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