Solving $f(x) = f(x/2) + f(x/3) + x$

Question

How would one proceed to proving that the solution to the functional equation $$f(x) = f(x/2) + f(x/3) + x$$ is $f(x)=6x$ which is also unique?

To clarify, I am not aware of neither the proof of the solution, neither the uniqueness or not.

Answer 1

Unfortunately, $f(x)=6x$ is not the unique solution to your functional equation. Let us define the constant $a\approx 0.787885$ as the unique real solution to the equation $$\frac{1}{2^a}+\frac{1}{3^a}=1$$ Then we may see that the function $$f(x)=|x|^a+6x$$ also satisfies the given functional equation, since $$f(x)=|x|^a+6x=\frac{|x|^a}{2^a}+\frac{6x}{2}+\frac{|x|^a}{3^a}+\frac{6x}{3}+x=f(x/2)+f(x/3)+x$$