TITLE: Solve the functional equation $f(x)=f\left({x\over 3}\right)+f\left({2x\over 3}\right)$ with $f : [0,\infty) \to \mathbb R$ continuous
QUESTION [9 upvotes]: Solve the functional equation 

$$f(x)=f\left({x\over 3}\right)+f\left({2x\over 3}\right)\qquad \forall x\geq 0$$ 
  with $f : [0,\infty) \to \mathbb R$ continuous.

I can't manage to get this one to the form of Cauchy's functional equation, but I imagine that's how it's maybe done.

REPLY [8 votes]: To my surprise, there are non-linear solutions to this functional equation.
Consider functions of the form:
$$F(x;\alpha,k) = x^\alpha \sin( k\log x )$$
If one can choose $\alpha \in (0,1)$ and $k \in (0,\infty)$ such that:
$$\begin{align}
\left(\frac13\right)^\alpha \cos\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \cos\left(k\log\frac23\right) &= 1\\
\left(\frac13\right)^\alpha \sin\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \sin\left(k\log\frac23\right) &= 0\tag{*}
\end{align}$$
then $F$ is continuous over $[0,\infty)$ and satisfies:
$$ F(x;\alpha,k) =  F(\frac{x}{3};\alpha,k) + F(\frac{2x}{3};\alpha,k)$$
By trial and error, I find at least one solution of $(*)$ with $(\alpha,k) \sim ( 0.7586093, 16.4941542 )$ and the corresponding $f(x)$ looks like this: