Solve Functional Equation $3f(x)=f\left(\frac{x}{3}\right)+f\left(\frac{1+x}{3}\right)$

Question

Is there any general strategy for the (non-trivial) solutions to equations similar to $$ 3f(x)=f\left(\frac{x}{3}\right)+f\left(\frac{1+x}{3}\right)? $$ with $x \in \mathbb{R}$. I can solve $$ 2f(x)=f\left(\frac{x}{2}\right)+f\left(\frac{1+x}{2}\right) $$ to be $f(x) = \pi \cot(\pi x)$ by using the fact that its Mellin transform satisfies $$ \frac{1}{1-x} = (1+x)\frac{1}{1-x^2} \\ f(x) = (1+x)f(x^2) $$ but this doesn't seem to generalise. Thanks for any help or ideas you can offer.

Answer 1

Partial answer: if you assume that $f$ is an integrable function on $\mathbb R$ you can solve the equation by taking Fourier transforms. After some simple manipulations we get the simple equation $\hat {f} (t)=(1+e^{it}) \hat {f} (3t)$. By iteration this gives $\hat {f} (t)=\frac {\hat {f} (t3^{-n})} {(1+e^{it})(1+e^{it/3})...(1+e^{it/3^{n-1})}}$ and we can let $n \to \infty$. Since the numerator tends to $\hat {f} (0)$ we see that $f$ is uniquely determined up to a constant factor. PS It turns out that the infinite product in the denominator does not converge, so there is no integrable function other than the zero function which solves the given equation! I hope this information is of some interest.