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I am looking for a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$, or a proof that no such function exists.

Replacing $x$ by $1/x$ shows we must have $f(x) = f(1/x)$.

I am most interested in (non-)existence of smooth non-constant $f$.

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  • $\begingroup$ Are you sure about the $f(x)=f(1/x)$ part? $\endgroup$ Dec 9 '20 at 23:04
  • $\begingroup$ @alex.jordan yes, on the one hand $f(x+1/x)$ equals $f(x)$ by the original functional equation, and on the other hand it equals $f(1/x)$, by replacing $x$ by $1/x$ in the functional equation (that substitution leaves $x+\frac{1}{x}$ invariant). $\endgroup$ Dec 9 '20 at 23:19
  • $\begingroup$ Thanks, I was missing the bridge. $\endgroup$ Dec 9 '20 at 23:20
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There should be infinitely many continuous solutions, one for each continuous function $g:[1,2]\to \mathbb{R}$ with $g(1)=g(2)$. After imposing appropriate boundary and differentiability conditions on $g$, we can make the function smooth.

Let $x_1=1$ and $x_{n+1}=x_n+\frac{1}{x_n}$. Then $1\le x_n\le n$ and by the divergence of the harmonic series, $x_n\to\infty$ as $n\to \infty$. Since $h:t\mapsto t+\frac{1}{t}$ is strictly increasing on $[1,\infty)$, each $x\in[1,\infty)$ belongs to exactly one $[x_{n+1},x_{n+2})$ and $x=h^n(y)$ for exactly one $y\in[1,2)$. Then we define $f(x)=g(y)$. Using the relation $f(x)=f(1/x)$, this extends to $(0,\infty)$. It is continuous since it is continuous on each $[x_n,x_{n+1}]$ and agrees at end points.

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  • $\begingroup$ For example, the boundary conditions on $g$ could be that all derivatives are $0$, like $\exp\mathopen{}\left(\frac{1}{(x-1)(x-2)}\right)\mathclose{}$. $\endgroup$ Dec 10 '20 at 2:10

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