How does one set out to find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ which satisfy $f^{k}(x)=f(x^k)$ , where $k \in \mathbb{N}$?
Motivation: Is $\sin(n^k) ≠ (\sin n)^k$ in general?
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12$\begingroup$ $f^k(x)$ usually means $\underbrace{f(f(...f}_k(x)...)$. Better write $(f(x))^k$. $\endgroup$– kennytmAug 13, 2010 at 14:11
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6$\begingroup$ @Chandru1: Heed the diamond. $\endgroup$– Tom StephensAug 13, 2010 at 14:22
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11$\begingroup$ @Chandru: It's a convention, nothing related to maturity. Moreover, $f(f(...f(x)...) = f(x^k)$ is also a valid functional equation, so it could really cause confusion written like this. $\endgroup$– kennytmAug 13, 2010 at 14:30
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2$\begingroup$ I'm not sure that I agree with KennyTM's statement about what $f^k(x)$ usually means (I'm thinking about $\sin^k(x)$), which means that regardless of what it usually means, it should be clarified. $\endgroup$– IsaacAug 13, 2010 at 16:11
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6$\begingroup$ I'm sorry. Are you fixing a k and asking for all functions satisfying that equation? Or do you want functions that satisfy that equation for all k simultaneously? $\endgroup$– Jason DeVito - on hiatusAug 13, 2010 at 17:03
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3 Answers
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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
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For a continuous $f:[0,1] \rightarrow (0, \infty)$ it is easy to show that $f(x)=1$ for $\forall x \in [0,1]$.
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1$\begingroup$ f(x)=0 seems to work too. What's your `easy' proof that f(x)=1 is the only solution? $\endgroup$– rgrigAug 13, 2010 at 13:56
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$\begingroup$ @grig: If i was to have a proof, i shall be posting it as soon as possible. $\endgroup$– anonymousAug 13, 2010 at 14:00
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$\begingroup$ 0 is not in the co-domain. So clearly f(1)=1 and f(0)=1. Let $x \in (0,1)$. Then since $f$ is continuous we have $lim_{k \to \infty} f^{k}(x)=lim_{k \to \infty} f(x^k) =f(0)=1. So f(x)=1. $\endgroup$– MykieAug 13, 2010 at 14:07
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$\begingroup$ @jennifer: Your reasoning does not hold, since $k$ in the problem is not arbitrary but given and fixed. $\endgroup$– HansNov 15, 2013 at 16:42
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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.