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 $f(x)=f(x^2)$? 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 $\phi(x)$ and $\psi(x)$ be periodic with period $\log 2$, let $a,b\in\mathbb R$. Then $$f(x)=\begin{cases}a & \text{if }x=0\\\phi(\log(-\log|x|))&0<|x|<1\\b&|x|=1\\\psi(\log(\log(|x|))&|x|>1\end{cases}$$
satisfies your functional equation , and can be as pathological (or as smooth) as $\phi$ and $\psi$ 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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