I'm thinking of $f$ as a function either from the reals to the reals or from the naturals to the naturals.
Edit: Okay, what if $f$ is from the naturals to the naturals?
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6$f=\frac{1}{x}$ – W. mu Jan 24 '18 at 03:22
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The question from the naturals to the naturals is much more interesting. – Fimpellizzeri Jan 24 '18 at 03:26
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Are there any where $f$ is continuous on $\mathbb{R}$? – Carl Schildkraut Jan 24 '18 at 03:30
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1@CarlSchildkraut Take $f(x) = -x$ for $x < 0$ and $f(x) = 0$ for $x \ge 0$ – Trevor Gunn Jan 24 '18 at 04:23
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@TrevorGunn That's a nice example; thanks. – Carl Schildkraut Jan 24 '18 at 06:39
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@W.mu Another one of those is $f(x)=\frac{x}{Cx-1}$ for any real $C$. – Alexander Burstein Jan 30 '18 at 09:17
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@AlexanderBurstein If you take $f(f(x))=x$, then $f(x)=f^{-1}(x)$, which means the figure $y=f(x)$ is symmetric about the line $y=x$, and these functions are infinite. – W. mu Jan 31 '18 at 01:28
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@W.mu Right, but I was thinking about rational functions with that property, like $C/x$ or $C-x$. In fact, the example I gave is the conjugate of $C-x$ by $1/x$. – Alexander Burstein Jan 31 '18 at 09:04
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Another family of examples for $\mathbb N$: Let $p$ be prime, $q$ coprime to $p$. $f(n) = p n$ if the $p$-adic order of $n$ is even (i.e. $n = p^e m$ where $e$ is even and $m$ is coprime to $p$), otherwise $f(n) = q n/p$. Then $f(f(x)) = q x$.
Robert Israel
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If you want one from the naturals to the naturals, take $f(x)$ to be the function $f(1)=2,f(2)=1$, and $f(n)=n$ for $n\geq 3.$
TomGrubb
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2Both examples exploit the fact that the identity is linear, but I wonder if these can be adapted to a case when, say, $f(f(x))=2x$. My intuition says the answer is no on the naturals. – Fimpellizzeri Jan 24 '18 at 03:31
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@Fimpellizieri That would be interesting. I agree with your intuition but I don't see a quick reason why it'd be impossible, especially when you can define $f$ to be some nasty piecewise defined thing – TomGrubb Jan 24 '18 at 03:39
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1@Fimpellizieri Here's a question similar to yours. It gives a solution to $f(f(x))=2x$ on the naturals. – Carl Schildkraut Jan 24 '18 at 03:59
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No. For example,$$ f(x) = \begin{cases} \displaystyle \frac{1}{x}; & x \neq 0\\ 0; & x = 0 \end{cases}. $$
Ѕᴀᴀᴅ
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As for $\Bbb N$, let $f(n)=n+1$ for odd and $=n-1$ for even $n$. One can twist this to fancier constructions by fancier partitions of $\Bbb N$.
Torsten Schoeneberg
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This doesn't directly answer the question, but if $f$ is continuous from $\mathbb{R}\to\mathbb{R}$, it still can be nonlinear. For a concrete example, see the comments on this answer to one of my previous questions in a similar vein to this one; essentially, one can rotate the graph of
$$y=\epsilon \cos(x)$$
for small enough $\epsilon$ exactly $45^{\circ}$ counterclockwise, and it satisfies the property.
Carl Schildkraut
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