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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 $\alpha\neq 0$?

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  • $\begingroup$ @Moshe, in this case for all $x\in D$ you have $f(x)=\alpha x$ $\endgroup$ – Yanko Aug 21 '17 at 16:47
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    $\begingroup$ 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). $\endgroup$ – Professor Vector Aug 21 '17 at 17:13
  • $\begingroup$ thank you. and what if $D$ is closed under addition but is still finite? $\endgroup$ – user_xyz Aug 21 '17 at 17:24
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    $\begingroup$ @user_xyz Finite and closed under addition? That's practically a synonym for "a subset of $\{0\}$"... which clearly contradicts your assumption that $|D|>1$. $\endgroup$ – Erick Wong Aug 21 '17 at 18:08
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    $\begingroup$ @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 $D =\mathbb Z$. $\endgroup$ – Erick Wong Aug 21 '17 at 21:50
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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 $.

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