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I am trying to solve the functional equation $$f(x)=f\left(\frac{1}{x}\right),\quad x>0$$ for a real-valued function $f$, with $f(1)=2$. We may assume $f$ is continuous or infinitely differentiable. There is, obviously, the solution $f(x)=2$ for all $x>0$, but I have been unable to show that this is the only solution. By differentiating the original equation, one easily that $f^{(2n+1)}(1)=0$, where $f^{(k)}$ denotes the $k$th derivative of $f$, and $n$ is a non-negative integer. I cannot see this helping, though. By evaluating $f$ on $f(x)$ and applying the equation which we want it to satisfy, one gets that $f$ must satisfy $$f\left(\frac{1}{f(x)}\right)=f\left(f\left(\frac{1}{x}\right)\right).$$ This doesn't seem to help either. Substituting values on the original equation yields nothing. Another observation is that $$\lim_{x\to0+}f(x)=\lim_{x\to0+}f\left(\frac{1}{x}\right)=\lim_{x\to+\infty}f(x),$$ if the limit even exists. I know these are very superficial observations, but I am inexperienced with functional equations. Any hints?

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    $\begingroup$ Basically $f(x)=S(x,\frac 1x)$ where $S$ is a symmetric function in its arguments. Though this can be a little hidden like with $|\ln x |$. $\endgroup$
    – zwim
    Oct 22 '20 at 10:13
  • $\begingroup$ Just one specific example would be $f(x) = x^n + \left( \frac{1}{x} \right)^n$ for any $n$. $\endgroup$
    – A.J.
    Oct 22 '20 at 10:16
  • $\begingroup$ Probably the simplest solution is $f(x)=x+\frac{1}{x}$.EDIT: just noticed @A.J 's solution. $\endgroup$
    – K.defaoite
    Oct 22 '20 at 10:17
  • $\begingroup$ @zwim Absolutely right, I cannot believe I did not think about that $\endgroup$
    – Ray Bern
    Oct 22 '20 at 10:19
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Consider any function $g : [1,+\infty) \rightarrow \mathbb{R}$ such that $g(1)=2$.

Then the function $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ defined by $$f(x)=g(1/x) \quad \text{if } 0<x<1 \quad \quad \quad \text{and} \quad \quad f(x)=g(x) \quad\text{if } x\geq 1 $$

is a solution to your problem. (So there are many solutions)

(If you want continuous solutions, just ask that $g$ is continuous, it will give a continuous $f$. If you want differentiable solutions, then try to see at which condition on $g$ the constructed function will be differentiable - hint : you will have a condition about the derivative of $g$ at $1$)

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Since $x\gt0,$ there is $t\in\mathbb{R}$ such that $x=e^t$ which transform your equation to $f(e^t)=f(e^{-t}).$ Hence the functional equation only tells us that $f\circ\exp$ is even. So, take any even function $g$ and let $$f(x)=g(\ln(x)).$$

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I think, there are infinitely many such functions.

For example, $$f(x)=\frac{(x^{4040}-1)\ln^{2021}x}{x^{2020}}.$$

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    $\begingroup$ There is a bijection between the set of solutions of this problem and the set of functions $g$ defined on $[1,+\infty)$ such that $g(1)=2$ (see my answer). So of course, there are infinitely many solutions ! $\endgroup$ Oct 22 '20 at 10:25
  • $\begingroup$ @TheSilverDoe Yes, of course. It's exactly that I also wanted to say. $\endgroup$ Oct 22 '20 at 10:28

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