If $f(3x)=f(x)$ and $f$ is continuous, show that $f(x)$ is a constant function. If $f(x)$ is a continuous function such that $f(3x)=f(x)$ and the domain of $f$ is all non-negative real numbers. Prove that $f$ is a constant function.
What I did:
$$f(3x)=f(x)=f\left(\frac{x}{3}\right)=\cdots= f\left(\frac{x}{3^n}\right)$$
Now as $n$ tends to infinity, $f(\frac{x}{3^n})$ tends to $f(0)$ and hence $f(x)=f(0)$ for all $x$.
However, I think the second step is a bit dodgy. I can't quite tell how, since I lack sufficient mathematical maturity, but it just doesn't seem right. I'd be glad if someone could provide a more rigorous proof of the problem. Thanks in advance.
 A: To make this argument a bit more formal, try something along these lines:
Suppose (for contradiction) that $f(c) \neq f(0)$ for some $c \in \mathbb{R}$. 
By the continuity of $f$ at $0$, there exists a $\delta > 0$ such that whenever $|x| < \delta$, $|f(x) - f(0)|< |f(c) - f(0)|$.  Noting that $f(c) = f\left( \frac c{3^n}\right)$, we may derive a contradiction.
That is, we may select an $n \in \mathbb{N}$ such that $|c/3^n| < \delta$.  The fact that $|c/3^n| < \delta$ and $|f(c/3^n)-f(0)| = |f(c)- f(0)|$ is a contradiction of our definition of $\delta$.
By this contradiction, we are forced to conclude that $f(x) = f(0)$ for all $x \in \mathbb{R}$.  That is, $f$ is constant.
A: Your argument is not dodgy at all. $f(x)$ being continuous at $x = 0$ is equivalent to the statement that for all $a_1,a_2,...$ converging to $0$ one has $f(a_1), f(a_2),...$ converges to $f(0)$. So take $a_n = {x \over 3^n}$ in your example.
A: Sorry I misread the question.
Your argument would work perfectly. For any $x\in \mathbb{R}$, we have 
$$f(x) = \lim_{n\rightarrow \infty} f(x) = \lim_{n\rightarrow \infty}  f(x3^{-n}) = f(\lim_{n\rightarrow \infty} x3^{-n}) = f(0)$$
