Continuous function satisfying $f^{k}(x)=f(x^k)$
Solution 1:
Case $k=1$ is trivial. So, suppose $k>1$.
Suppose k is odd. For even k one should correct this solution a bit. First note, that $a^k=a$ is equivalent to $a\in{-1,0,1}$. We have this equation for a=f(-1), a=f(0), a=f(1).
Note that for $x\in (-1,1)$ sequence $x,x^k,(x^k)^k,\dots$ goes to 0. Denote this sequence by $b_n$: $b_n = x^{(k^n)}$. From continuity of f we know, that $f(b_n)\rightarrow f(0)$ and $(f(x))^{(k^n)}\rightarrow f(0)$. Therefore for $x\in(-1,1)$ we know, that $f(x) \in (-1,1)$ or f is constant on this interval and equal to 1 or -1.
Now suppose x is in $(0,\infty)$. By the same reason the sequence $(f(x))^{(k^{-n})}$ goes to f(1). It is true iff (f(x)=0 and f(1)=0) or (f(x) is in $(-\infty,0)$ and f(1)=-1) or (f(x) is in $(0,\infty)$ and f(1)=1). The same is for x in $(-\infty,0)$ and f(-1).
So we have the following cases
- f(0)=1, then f(x)=1 for $x \in
[-1,1]$. Other values of f can be
uniquely determined by values on
$[-2^k,-2]\cup[2,2^k]$. This values can be chosen arbitrary to form continuous function
to $(0,\infty)$ such that $f(-2^k)=(f(-2))^k$ and $f(2^k)=(f(2))^k$.
Now I will just list all other cases without going to details, which are similar to shown in the first case.
f(0)=-1
f(0)=0,f(-1)=1, f(1)=1
f(0)=0,f(-1)=1, f(1)=-1
f(0)=0,f(-1)=1, f(1)=0
f(0)=0,f(-1)=-1, f(1)=1
f(0)=0,f(-1)=-1, f(1)=-1
f(0)=0,f(-1)=-1, f(1)=0
f(0)=0,f(-1)=1, f(1)=1
f(0)=0,f(-1)=1, f(1)=-1
f(0)=0,f(-1)=1, f(1)=0
Solution 2:
For a continuous $f:[0,1] \rightarrow (0, \infty)$ it is easy to show that $f(x)=1$ for $\forall x \in [0,1]$.
Solution 3:
Define $g$ on $[2,2^{k}]$ by $(g(2))^{k}=g(2^k)$. Then using the functional equation you can extend to an $f$ on all numbers $\geq 1$. We can then use the similar trick for $(0,1)$ and for the negative numbers as well.