TITLE: Find $f$ such as $f(x) = \sum_{n=1}^\infty \frac{f(x^n)}{2^n}$
QUESTION [9 upvotes]: Find $f \in C^0([0,1] , \mathbb{R})$ such as $$f(x) = \sum_{n=1}^\infty \frac{f(x^n)}{2^n}$$
My try :
Constant functions work fine.
We can notice :
$$f(x) = \frac{f(x)}{2}+\sum_{n=2}^\infty \frac{f(x^n)}{2^n}$$ so
$$f(x) = \sum_{n=1}^\infty \frac{f(x^{n+1})}{2^n}$$
which make me believe constant functions are the only one to please the problem.
Any help will be appreciated.

REPLY [5 votes]: $f$ is continuous on the compact $[0,1]$ and $$\left\vert \frac{f(x^n)}{2^n}\right\vert\leq \frac{M}{2^n}$$
which ensures the normal convergence and then the pointwise convergence.
Moreover,
$\sum_{n=1}^{+\infty}\frac{1}{2^n}=1$ then constant functions are solutions.
and rewrite the equation :
$$
\sum_{n=1}^{+\infty}\frac{f(x)-f(x^n)}{2^n}=0
$$

Define for $x<1$ : $m(x)$ is any point from $[0,x]$ where $f$ attains its minimum on $[0,x]$ and $M(x)$ is any point from $[0,x]$ where $f$ attains its maximum on $[0,x]$

By definition of $m(x): f\bigl(m(x)^n\bigr)\ge f(m(x))$, for every $n$. So,
applying the equation for $m(x)$, with a negative general term we can deduce that 
$$
\forall n \quad f\bigl(m(x)^n\bigr)= f(m(x))
$$
As $m(x)<1$ we can pass to the limit as $n$ tends to $+\infty$ and 
$$
f(m(x))=f(0)
$$
Applying the same reasoning to $M(x)$ :
$$
f(M(x))=f(0)
$$

Therefore on the interval $[0,x],  f$ is constant : $f$ is contant on [0,1) and by continuity on [0,1]