I know that of functional identity f(a+b)=f(a)+f(b), y=kx is definitely a solution. Other solutions may be constructed by treating the real numbers as a vector field over the rational numbers, which are pathological.
How to construct all pathological functions satisfying the identity f(x)=f(x2)? I know that f must be constant if it's continuous, but what are the other cases?
Answer
Edited 4 Jan to incorporate details I skipped pointed out by user517969.
Let ϕ(x) and ψ(x) be periodic with period log2, let a,b∈R. Then f(x)={aif x=0ϕ(log(−log|x|))0<|x|<1b|x|=1ψ(log(log(|x|))|x|>1
satisfies your functional equation , and can be as pathological (or as smooth) as ϕ and ψ are. Because f(−x)=f(x) there is no loss of generality in using the absolute value signs in the innermost logarithm. I think this is the most general form of solution.
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