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

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.

Answer 1

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 \[ f(x) = \sum_{i=0}^n \binom{n}{i} f\left(\frac{2^i x}{3^i}\right)\]

Answer 2

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 $k \in \mathbb{R}$.