Let $g(x)$ be continuous for all $x$ and let $g(a+b) = g(a) + g(b)$ for all $a$,$b$...

Question

Show $g(x) = g(1) \cdot x$ for $x = c, \frac{1}{c}$ and $\frac{d}{c}$, where $c, d \in \mathbb{Z}, c \neq 0$.

For $x=c$, I was thinking of doing an induction proof, as $g(c) = g(\frac{c}{2}) + g(\frac{c}{2})$. However, I'm not sure how to handle the latter two cases.

Answer 1

Guide:

For the second one note that $$\sum_{i=1}^cg\left( \frac1c \right)=g(1)$$

and also, we have

$$\sum_{i=1}^cg\left( \frac{d}c \right)=g(d)$$

now, use the first result to conclude.