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 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.
 A: 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:

A: We surely know that $f(0)=0$ as 
$$f(0)=f(0)+f(0)$$
An obvious solution is $f(x)=x$, and multiple of that $f(x)=\alpha x$ with $\alpha \in \mathbb{R}$. 
Do you need all solutions, or a proof that this are all functions? Or do you just need to find a single solution?
I am not sure if it is the unique solution, I give you some of my thoughts.
(In fact this reminds me  a bit  at devils staircase (the cantor function)  an so I think the solution will not be unique).
We have the equation
\[ f(x)=f(\tfrac{x}{3}) + f(\tfrac{2x}{3})\]
This means that 
$$f(x)-f(\tfrac{x}{3})=f(\tfrac{2x}{3})$$ 
Using the equation above again we have
\[ f(x)=f(\tfrac{x}{9}) + f(\tfrac{2x}{9}) + f(\tfrac{2x}{9}) + f(\tfrac{4x}{9})\]
This is obvious equal to
\[ f(x) = f(\tfrac{x}{9}) +2 f(\tfrac{2x}{9}) + f(\tfrac{4x}{9})\]
When we iterate this one again we get
\[ f(x) = f(\tfrac{x}{27}) + 3 f(\tfrac{2x}{27}) +3f(\tfrac{4x}{27})+ f(\tfrac{8x}{27})\]
More general we have
\[ f(x) = \sum_{i=0}^n \binom{n}{i} f\left(\frac{2^i x}{3^i}\right)\]
A: Observe that : $$f(x)=f\left(\frac{x}{3}+\frac{2x}{3}\right)=f\left(\frac{x}{3}\right)+f\left(\frac{2x}{3}\right)$$ 
you can make a substitution : 
$$\frac{x}{3}=y;$$
$$\frac{2x}{3}=z.$$
$$f(y+z)=f(y)+f(z)$$ which is a Cauchy functional equation with  the solution: 
$$f(y)=Ky, K \in \mathbb{R}.$$
or $$f\left(\frac{x}{3}\right)=\frac{x}{3}$$ 
$$\frac{x}{3} \mapsto X $$
so $$f(X)=kX$$ with $k \in \mathbb{R}$.
