Solving the functional equation $f(x)=f(1/x)$ for $x>0$ I am trying to solve the functional equation
$$f(x)=f\left(\frac{1}{x}\right),\quad x&gt;0$$
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 $x&gt;0$, 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?
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
$$f(x)=g(1/x) \quad \text{if } 0&lt;x&lt;1 \quad \quad \quad \text{and} \quad \quad f(x)=g(x) \quad\text{if } x\geq 1  $$
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 $1$)
I think, there are infinitely many such functions.
For example, $$f(x)=\frac{(x^{4040}-1)\ln^{2021}x}{x^{2020}}.$$