Continuous Functions such that $f(0) = 1$ and $f(3x) - f(x) = x$ Just had a midterm with the following problem:
Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.
I was just curious how this would end up.
During the exam I tried setting $f(1) = c$, and end up getting $f(3) = 2c, f(9) = 4c, \ldots$ and from that $c = 3/2$ with the $f(0) = 1$ condition, but I was unable to make more progress on it.

We have: $f(x) - f\left(\frac{x}{3}\right) = \dfrac{x}{3}$
$f\left(\frac{x}{3}\right) - f\left(\frac{x}{9}\right) = \dfrac{x}{9}$
.....
$f\left(\frac{x}{3^{n-1}}\right) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3^n}$.
Add the above equations:
$f(x) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3}\cdot \left(1 + \frac{1}{3} + ... + \frac{1}{3^{n-1}}\right)$.
Letting $n \to \infty$ in the above equation and using continuity of $f$ at $x = 0$ we have:
$f(x) - f(0) = \dfrac{x}{3}\cdot \dfrac{1}{1-\frac{1}{3}} = \dfrac{x}{2}$. Thus:
$f(x) - 1 = \dfrac{x}{2}$, and $f(x) = 1 + \dfrac{x}{2}$
