# Cauchy's functional equation $f (x+y)=f (x)+f(y)$ in subdomains

Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is additive ($f(x+y)=f(x)+f(y)$) and monotonic on a set $D\subset\mathbb{R}$ such that $|D|>1$, $0\in D$ and $-a\in D$ whenever $a\in D$. Assume nothing about the behavior of $f$ in $\mathbb{R}\setminus D$.

Is it true that, for all $x\in D$, $f(x)=\alpha x$ for some $\alpha\neq 0$?

• @Moshe, in this case for all $x\in D$ you have $f(x)=\alpha x$ – Yanko Aug 21 '17 at 16:47
• You should include the functional equation, both in title and body. Most will know it, but it may improve searchability. If $D$ isn't closed under addition, the answer is trivially negative (but the question doesn't make a lot of sense, either, then, sorry). – Professor Vector Aug 21 '17 at 17:13
• thank you. and what if $D$ is closed under addition but is still finite? – user_xyz Aug 21 '17 at 17:24
• @user_xyz Finite and closed under addition? That's practically a synonym for "a subset of $\{0\}$"... which clearly contradicts your assumption that $|D|>1$. – Erick Wong Aug 21 '17 at 18:08
• @user_xyz Are you asking whether $D=\mathbb R$ is the only additively closed domain for which the original question holds for all $f$? Certainly not: even without the monotonicity constraint, the claim is true for $D =\mathbb Q$ and $D =\mathbb Z$. – Erick Wong Aug 21 '17 at 21:50

The answer is No!" considering axiom of choice which is sufficient to prove that there is a Hamel basis.
Let $H$ be a Hamel basis containing, say $1$ and $\sqrt 2$. Define $f$ such that $f ( 1 ) = 1$, $f ( \sqrt 2 ) = 2$ and $f ( a ) = 0$ for each $a \in H \backslash \{ 1 , \sqrt 2 \}$. Then taking $D = \{ - \sqrt 2 , -1 , 0 , 1 , \sqrt 2 \}$, $f$ is increasing on $D$ but there is no $\alpha \in \mathbb R$ such that $f ( x ) = \alpha x$ for all $x \in D$.