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It is known that, for real increasing function $f:[0,1]\to\mathbb R$, if $f(ax)=af(x)$, then $f$ is linear.

Now consider $a=n$ where $n$ is natural.

Is it possible that $f(x)$ is not linear on a positive measure of points?

I stuck here: say $f(x)=cx$ on the rationals, then, how to prove that $f(x)=cx$ on all reals?

daw
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dodo
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    The condition enforces $\mathbb{Q}$-multiplicativity. That $f$ is increasing then shows it is $\mathbb{R}$-linear. – David Gao Jun 08 '24 at 10:06
  • @DavidGao How do I prove the second step? – dodo Jun 08 '24 at 10:07
  • @DavidGao Might be related: https://math.stackexchange.com/questions/3008548/increasing-function-on-r-that-is-discontinuous-on-the-rationals – dodo Jun 08 '24 at 10:14
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    By $\mathbb{Q}$-multiplicativity, I meant $f(qx) = qf(x)$ for all $q \in \mathbb{Q}$, $x \in \mathbb{R}$. Now, if $c \in \mathbb{R}$, then take a sequence of rationals $q_n \searrow c$ and a sequence of rationals $p_n \nearrow c$. Assume $x \geq 0$. Then $cf(x) = \lim_n p_nf(x) = \lim_n f(p_nx) \leq f(cx) \leq \lim_n f(q_nx) = \lim_n q_nf(x) = cf(x)$. So $f(cx) = cf(x)$. The proof is similar when $x < 0$. From there additivity is obvious. – David Gao Jun 08 '24 at 10:56
  • I don't understand your question. First you recall we know $f$ is linear (with no constraint on $a$) and then you ask if when $a\in\Bbb N$, $f$ is not linear? – Anne Bauval Jun 08 '24 at 17:16

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