What is the solution of $f(x)\cdot f(-x) = 1$

Question

What is the general solution of the equation?

$$f(x) \cdot f(-x) = 1$$

I know that $f(x) = A^{k \cdot x}$ is a solution, and I am feeling this is the general solution, but I don't have any proof.

EDIT: After I read Didier's answer, I realised I might not have clarified the question. Didier's answer is correct, but I was mainly looking of other than the exponential functions. Vhailor's seem to have given the correct answer, because basically I can get any function, limit it to only x >= 0, then define g(-x) as 1/f(x). So I will mark this question as answered, but if you can think of some examples satisfying the conditions, please share them.

Answer 1

This equation has an awful lot of solutions. To wit, every function $f$ such that $f(x)=s(x)\mathrm e^{g(x)}$ for every $x$, where $s(x)=+1$ or $s(x)=-1$ and where $s$ is an even function and $g$ is an odd function, is a solution. And a moment of thought shows that every solution can be written like that. For some basics about (even functions and) odd functions, see there.

Edit Since it seems that the moment of thought alluded to above should have been described more explicitly, here it goes. Assume that $f$ is a solution. Then $f(x)\ne0$ for every $x$, hence $$ f(x)=s(x)\mathrm e^{g(x)}, $$ for some real numbers $g(x)$ and $s(x)$, where $s(x)=+1$ or $-1$ is the sign of $f(x)$, and where $g(x)=\log|f(x)|$. Thus, the condition on $f$ reads as $s(x)s(-x)=+1$, which means that $s(x)=s(-x)$, and as $\mathrm e^{g(x)+g(-x)}=1$, which means that $g(x)+g(-x)=0$, or equivalently, that $g$ is an odd function. One sees in particular that, as is the case for every odd function, $g(0)=0$, a fact which was obvious from the condition that $f(0)^2=1$.

A simple example outside of the scope of the formula given by the OP is $f(x)=\mathrm e^{\sin(x)}$.

Answer 2

Well, if you impose no additional restrictions on the function (continuity, etc.) then you can extend any injection $f$ defined on the positive reals to $\mathbb{R}$ so that it satisfies this condition, just define

$g(x) = f(x)\quad$ if $x \geq 0$

$g(x) = 1/f(-x)\quad$ if $x<0$.

Edit : If you want your function to be continuous just add the requirements that $f$ be continuous, $f(0)=±1$ and $f(x)\neq 0$.

Answer 3

According to http://eqworld.ipmnet.ru/en/solutions/fe/fe2104.pdf, the general solution of $f(x)f(-x)=1$ should be $f(x)=\pm e^{\Phi(x,-x)}$, where $\Phi(x,-x)$ is any antisymmetric function of $x$ and $-x$.