Let $f$ be a continuous real valued function from $[0,1]$ such that $$f(x)+f(x^2)=x$$ for all $x\in [0,1]$. Does there exist such a function?
Plugging $x=0$ and $x=1$ respectively in the given equation we obtain $f(0)=0$ and $f(1)=\frac{1}{2}$. By the intermediate value theorem, $f$ attains any value between $0$ and $1$. Moreover, the range of $f$ is $[m,M]$ where $m$ (resp. $M$) is the minimum (resp. maximum) value of the function over $[0,1]$. How to use these facts to decide whether such function exists or not? Please give some hint to proceed. Thank you.
No, there is no such function. Suppose that $F(x)+F(x^2)=x$ and $F$ is continuous on $[0,1]$. We use the function $f(x), \ g(x)$ in my answer to this post where it is proved that $f(x)-g(x)$ is not a constant function. The functions $f(x), \ g(x)$ originate from Hardy's 'Divergent Series'. To recall the definition, $$ f(x)=\sum_{n=0}^{\infty} (-1)^n x^{2^n}, $$ $$ g(x)=\sum_{n=0}^{\infty} \frac{(\log x)^n}{(2^n+1)n!}. $$ It is easy to check that $f(x)+f(x^2)=x$ and $g(x)+g(x^2)=x$.
Now, since $F(x)$ on $[0,1]$ is continuous and $F(x)+F(x^2)=x$, we have $\Phi(x)=F(x)-g(x)$ satisfying $$ \Phi(x)=-\Phi(x^2)=\Phi(x^4). $$ Note that $g(x)$ is continuous on $(0,1]$. Thus, $\lim_{x\rightarrow 1-} \Phi(x)$ exists. Then such $\Phi$ satisfying $\Phi(x)=\Phi(x^4)$ must be a constant function. Let $\Phi(x)=c$. We have $$ F(x)-g(x)=c \ \ \textrm{ if } x\in (0,1]. \ \ (1) $$ Similarly, let $\Psi(x)=f(x)-F(x)$. Then $\Psi(x)=\Psi(x^4)$. As $f$ is a power series at $0$, we have $\lim_{x\rightarrow 0+} f(x) = 0$. Then we must have $\lim_{x\rightarrow 0+} \Psi(x)=-F(0)$. Such $\Psi$ satisfying $\Psi(x)=\Psi(x^4)$ must be a constant function. Thus, $$ f(x)-F(x)=-F(0) \ \ \textrm{ if } x\in [0,1). \ \ (2) $$ Summing up (1) and (2), we have $$ f(x)-g(x)=c-F(0) \ \ \textrm{ if } x\in (0,1). $$ Therefore, we have a contradiction since $f(x)-g(x)$ cannot be constant on $(0,1)$ as proved in my MO answer.
Remark The phenomenon we are observing here is that $f$ is oscillatory around $1$, and $g$ is oscillatory around $0$.