Does local additivity imply local linearity? Suppose $D$ is a finite subset of $\mathbb{R}$ such that $0\in D$ and $-a\in D$ whenever $a\in D$.
Suppose $f:D\rightarrow\mathbb{R}$ satisfies $f(0)=0$, $f(-a)=-f(a)$ and $f(a)>f(a')$ whenever $a>a'$.
Suppose also that there exist $x,y\in D$ such that $x>y>0$, $x+y\in D$ and  $f(x+y)=f(x)+f(y)$. 
Do the above imply $f(x)=\alpha x$ and $f(y)=\alpha y$ for some $\alpha>0$?
 A: Keen's answer is excellent but here is an example that shows that $f$ need not really be 'fine-tuned' to $D$:
Just take $f$ the function defined by $f(d) = d^3$ for all $d \in D$ and let $D$ be any set containg $y = 0$ and $x = 1$.
The point is: your restrictions dictate that $f$ is a monotonously increasing, odd function but not much more. There are a lot of such functions, $d \mapsto d^3$ is just the simplest non-linear one. 
Now once you have the function the condition of there just being ONE pair $(x, y)$ satisfying $f(x + y) = f(x) + f(y)$ is really putting very little restrictions. Of course, the more such pairs $(x, y)$ you want to exist, the harder it becomes to find an $f$. 
In an extreme case: when you demand $f(x + y) = f(x) + f(y)$ for ALL $x, y$ in $D$ we do need that $f$ is linear when $D$ consists only of rational numbers. An interesting intermediate question is what happens in the case where you demand that for every $x \in D$ there is an $y \neq 0$ in $D$ such that $f(x +y) = f(x) + f(y)$. 
A: I don't think it does.
Take $D=\{(-1-e),-e,-1,0,1,e,1+e\}$ 
f(1)=1 , f(e)=2e and f(1+e)=1+2e and such that f(-x)=-f(x). 
It should satisfy all your conditions, but not the conclusion. (By the way it's not important that I took an irrational number: I just think it makes things a little bit cleaner.)
