A weak version of Cauchy's inequality on a finite set of integers

Question

Let $A:=\{-k,\ldots,-2,-1,0,1,2,\ldots,k\}$, $k<\infty$, $k\in \mathbb{N}$.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be such that $f(-x)=-f(x)$ for all $x\in A$ and $f(x+y)=f(x)+f(y)$ for all $x,y\in A$ such that $x+y\in A$.

Does it follow that there is $\alpha\neq 0$ such that $f(x)=\alpha x$ for all $x\in A$?

Answer 1

Sort of, and $\alpha=f(1)$ (whic may however be $0$).

From $f(-0)=-f(0)$ we find $f(0)=0$. Assume for some $m$, $0\le m<k$ we have shown that $f(x)=xf(1)$ for all $x\in A$ with $|x|\le m$. (And as just seen, this does hold true for $m=0$). Then $f(m+1)=f(m)+f(1)=mf(1)+f(1)=(m+1)f(1)$ and $f(-m-1)=-f(m+1)=(-m-1)f(1)$ and so we also have $f(x)=xf(1)$ for all $x\in A$ with $|x|\le m+1$. By induction, $f(x)=xf(1)$ for all $x\in A$ with $|x|\le k$, i.e., for all $x\in A$.

Answer 2

$f$ identically zero satisfies your request. If you suppose f is not identically zero you can observe that $f(0)=0$ and $f(n)=nf(1)$ for all $n\in A$. So your constant $\alpha$ is $f(1)$ that can even be zero if you don't require an extra condition on it.