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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?

  • 12
    $f^k(x)$ usually means $\underbrace{f(f(...f}_k(x)...)$. Better write $(f(x))^k$.2010-08-13
  • 0
    @Kenny TM: I think people who visit this site are mature enough to understand what $f^{k}(x)$ means.2010-08-13
  • 6
    @Chandru1: Heed the diamond.2010-08-13
  • 11
    @Chandru: It's a [convention](http://en.wikipedia.org/wiki/Exponentiation#Exponential_notation_for_function_names), 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.2010-08-13
  • 0
    @Kenny TM: OK. No problem.2010-08-13
  • 2
    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.2010-08-13
  • 6
    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?2010-08-13
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    @Isaac: $sin^k$ and other trigonometric functions seem to be the exception, rather than the norm2010-08-14
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    @Casebash: that's quite possible. I can't recall a textbook using superscript numerals on a function at all, except for as exponents on trig functions and in parentheses as higher-order derivatives, which makes me think that it's a notation that is generally avoided because it does not have a single, clear meaning.2010-08-14
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    Chandru, will you please answer Jason's question?2011-06-17
  • 0
    @Jonas: I was not fixing $k$.2011-06-17

3 Answers 3

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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

  1. 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.

  1. f(0)=-1

  2. f(0)=0,f(-1)=1, f(1)=1

  3. f(0)=0,f(-1)=1, f(1)=-1

  4. f(0)=0,f(-1)=1, f(1)=0

  5. f(0)=0,f(-1)=-1, f(1)=1

  6. f(0)=0,f(-1)=-1, f(1)=-1

  7. f(0)=0,f(-1)=-1, f(1)=0

  8. f(0)=0,f(-1)=1, f(1)=1

  9. f(0)=0,f(-1)=1, f(1)=-1

  10. 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 $\forall x \in [0,1]$.

  • 1
    f(x)=0 seems to work too. What's your `easy' proof that f(x)=1 is the only solution?2010-08-13
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    @grig: If i was to have a proof, i shall be posting it as soon as possible.2010-08-13
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    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)=1.2010-08-13
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    @jennifer: Your reasoning does not hold, since $k$ in the problem is not arbitrary but given and fixed.2013-11-15
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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.