Let $\mu(n)$ the Möbius function, we define $F:[0,1]\to[0,1]$ as $$F(x)=\sum\_{n=1}^\infty\frac{\mu(n)}{n}x^n.\tag{1}$$


For a function of this kind (I presume that this continuous function has image $[0,1]$) was defined, for example in last paragraph of page 986, what is a periodic point, and its corresponding order of periodicity.



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> **Question.** Can you find examples of periodic points of $(1)$? Is it possible or feasible to prove or refute that our function has periodic points of order $3$? **Many thanks.**
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Since I think that these questions are very difficult feel free to add comments about your thoughts and reasonings, heuristics or conjectures. I believe that this question concerning the function $(1)$ isn't in the literature.


References:
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[1] Tien-Yien Li, James A. Yorke, *Period Three Implies Chaos*, The American Mathematical Monthly, Vol. 82, No. 10 (1975).



What about of periodic points of $\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n$, $0<x<1$, where $\mu(n)$ is the Möbius function?
I claim $F(x)<x$ for all $0<x<1$. From there it follows that the only periodic point is the fixed point $0$.


Let $m(x)=\sum\_{n\leq x}\frac{\mu(n)}{n}$. We have (see e.g. equation (5) [here](https://terrytao.wordpress.com/2009/08/30/an-elementary-inequality-involving-the-mobius-function/)) $m(x)\leq 1$ for all $x$, with strict inequality e.g. for $n=2$ (indeed, for all $n>1$, but it's not needed). Using [summation by parts](https://en.wikipedia.org/wiki/Summation_by_parts), we get
$$\sum\_{n=1}^\infty\frac{\mu(n)}{n}x^n=\sum\_{n=1}^\infty(m(n)-m(n-1))x^n\\=\sum\_{n=1}^\infty m(n)(x^n-x^{n+1})<\sum\_{n=1}^\infty(x^n-x^{n+1})=x$$


