# Solving the functional equation $f(x)=f(1/x)$ for $x>0$

I am trying to solve the functional equation $$f(x)=f\left(\frac{1}{x}\right),\quad x>0$$ 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?

• 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 $|\ln x |$.
– zwim
Commented Oct 22, 2020 at 10:13
• Just one specific example would be $f(x) = x^n + \left( \frac{1}{x} \right)^n$ for any $n$.
– A.J.
Commented Oct 22, 2020 at 10:16
• Probably the simplest solution is $f(x)=x+\frac{1}{x}$.EDIT: just noticed @A.J 's solution. Commented Oct 22, 2020 at 10:17
• @zwim Absolutely right, I cannot believe I did not think about that Commented Oct 22, 2020 at 10:19

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

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 $$1$$)

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 $$f(x)=g(\ln(x)).$$

I think, there are infinitely many such functions.

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

• 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 ! Commented Oct 22, 2020 at 10:25
• @TheSilverDoe Yes, of course. It's exactly that I also wanted to say. Commented Oct 22, 2020 at 10:28