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Suppose that $h:(-\frac{1}{3}, \frac{1}{3})\to \mathbb R$, $h(x+y)=h(x)+h(y)$ for all $x,y\in (-\frac{1}{6},\frac{1}{6})$ and the function is bounded. Does it follow that $h(x)=x\cdot c$? I know that this is true if $h$ would have been defined over the real line and the additive equation would have been true for all the reals (by induction you prove that $h(nx)=nh(x)$ for all positive integers $n$ and then on rationals and then use density to get the statement for the irrationals too). However, I don't get it. This can't be true (in my opinion) for a closed interval, as the classic proof for the real line doesn't work the same way anymore.

My struggle began while watching this video (watch at 16:40):

USAMO 2018 #2

Also on AoPS in the thread for the USAMO 2018 #2 there exists a solution where the same thing is used. I am very confuse as I don't know how to prove $h$ is linear. Please help me, but provide a solution if the statement is true. Thank you!

furfur
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The same proof should work. That is, if we assume $h:(-1/3, 1/3) \to \mathbb{R}$ is continuous and that $h(x+y) = h(x)+h(y)$ whenever $x+y \in (-1/3, 1/3)$ then we have that $h(nx) = nh(x)$ whenever $nx \in (-1/3, 1/3)$. Then you also get that $h(x/n) = \frac{1}{n}h(x)$, so for any rational number q we have $h(qx)=qh(x)$ whenever $qx \in (-1/3, 1/3)$. Then if $h$ is continuous we can extend to all real numbers using the density of the rationals.

Jacob FG
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  • I managed to prove that $h(x)=cx$ for all rationals $x$ in the interval, but I can't prove that it's also true for irrationals. Remember that the OP is about bounded functions, not continuous. I tried by letting the function $u(x)=h(x)-cx$ and then supposing function $u$ is not 0, there exists a number t such that $u(t)\neq 0$. Using $u(qx)=qu(x)$ for all rationals we should break the boundaries. However, I don't know how to do that. Could you please help? – furfur Apr 30 '20 at 17:07
  • Do you mean bounded in the sense that $h(x) < M|x|$ for some fixed $M$? If so this is equivalent to $h$ being continuous at 0. Then by additivity $h$ is continuous everywhere. There is nothing you do differently here from when the domain is $\mathbb{R}$. – Jacob FG Apr 30 '20 at 17:25
  • I mean bounded in the sense that $-\frac{1}{3}<h(x)<\frac{2}{3}$ for all $x$ in $(-\frac13,\frac23)$. – furfur Apr 30 '20 at 17:35
  • This is equivalent. Assume $|h(x)| < \frac{2}{3}$, then I claim $|h(x)| < 2|x|$. Assume to the contrary that there is an $x$ such that $|h(x)|>2|x|$. Specifically let's say $|h(x)| = 2|x| + \varepsilon$. Then by density of the rationals we can find a rational number $q>1$ such that $\frac{1}{3} - \varepsilon < q|x| < \frac{1}{3}$, but then $|h(qx)| = q(2|x|+\varepsilon) > \frac{2}{3}$. – Jacob FG Apr 30 '20 at 18:32
  • But can't there be any solution without using continuity? I don't know that concept. Thank you so much! – furfur Apr 30 '20 at 19:40
  • I doubt it. The fact that the rationals are dense in the reals is crucial, and is entirely a fact about continuity. I would instead recommend you do try to understand what continuity is. – Jacob FG Apr 30 '20 at 19:51
  • Thank you for helping me. :) – furfur Apr 30 '20 at 19:55