All solutions of $f(x)f(-x)=1$

What are all the solutions of the functional equation $$f(x)f(-x)=1\,?$$

This one is trivial: $$f(x)=e^{cx},$$ as it is implied (for example) by the fundamental property of exponentials, namely $e^a e^b=e^{a+b}$. But there is another solution: $$f(x)=\frac{c+x}{c-x}.$$ Are there any more solutions? How can I be sure?

Solution 1:

Presumably, you want $f:\mathbb{R}\to\mathbb{R}$. If you want to use a different domain or codomain, the answer is probably not going to change much.

You can simply pick any $h:\mathbb{R}_{>0}\to\mathbb{R}_{\neq 0}$ and $\epsilon\in\{-1,+1\}$. Then, define the function $f:\mathbb{R}\to\mathbb{R}$ by $$f(x):=\left\{\begin{array}{ll} h(x)&\text{if }x>0\,,\\ \epsilon&\text{if }x=0\,,\\ \dfrac{1}{h(-x)}&\text{if }x<0\,. \end{array}\right.$$ Then, $f$ satisfies the required functional equation. Note that any such function $f$ takes the form above.

If you demand that $f$ is continuous, then $h$ has to be continuous and $\lim\limits_{t\to 0^+}\,h(t)=\epsilon$. This is all you need. It is a much more interesting problem to characterize all smooth or analytic functions $f$ that satisfy your functional equation. It turns out that the solutions are $f(x)=\epsilon\,\exp\big(g(x)\big)$, where $\epsilon\in\{-1,+1\}$ and $g:\mathbb{R}\to\mathbb{R}$ is a smooth or analytic, odd function. If you want $f$ to be just $k$-time differentiable, then $g$ is $k$-time differentiable.

Solution 2:

Here is the most general solution: Let $g: [0,\infty) \to \mathbb R\setminus \{0\}$ be any function such that $g(0)=\pm 1$ and define $f(x)=g(x)$ if $x \geq 0$ and $f(x) =\frac 1 {g(-x)}$ if $x \leq 0$.

Solution 3:

Take any function that is odd, that means $g(-x)=-g(x)$. For example $\sin x, x^3+2x$ and so on. Then $\exp(g(x))$ works.

Solution 4:

This is the case if and only if $x\mapsto\ln \lvert f(x)\rvert$ is an odd function, i.e. if and only if there are some odd function $g$ and some function $\theta:\Bbb R\to\{-1,1\}$ such that $f(x)=\theta(x)e^{g(x)}$

It is clear that the correspondence is bijective because $\theta=\frac f{\lvert f\rvert}$ and $g=\ln\lvert f\rvert$.