# Is there a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$?

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 $$f$$.

• Are you sure about the $f(x)=f(1/x)$ part? Dec 9 '20 at 23:04
• @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). Dec 9 '20 at 23:19
• Thanks, I was missing the bridge. Dec 9 '20 at 23:20

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