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.
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:

$$\begin{align}
\left(\frac13\right)^\alpha \cos\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \cos\left(k\log\frac23\right) &amp;= 1\\
\left(\frac13\right)^\alpha \sin\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \sin\left(k\log\frac23\right) &amp;= 0\tag{*}
\end{align}$$

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:
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 $k \in \mathbb{R}$.