Q:

If a function is additive on the interval $(-1/3, 1/3)$ does it follow it is linear? 

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! 

A:

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. 