$f(x)$ is an analytic function in $\mathbb{R}$ such that $f(-x)f(x)=1$. What else can we find out about $f(x)$? Well, I know that there are some easy things we can say immediately:


*

*$f(0)= \pm 1$, follows immediately

*$f(x)=\pm 1$ is the obvious solution, so let's look for other solutions. Moreover, let's consider only the case $f(0)=1$ for now

*The obvious identities, such as
$$f(x)^2=\frac{f(x)}{f(-x)}$$

And now for the series:
$$f(x)=a_0+a_1 x+ a_2 x^2+a_3 x^3+\dots$$
We can immediately see by multiplying the series for $f(x)$ and $f(-x)$ that (for the case $f(0)>0$):
$$a_0=1$$
$$a_2=\frac{a_1^2}{2}$$
$$a_4=a_1 a_3-\frac{a_1^4}{8}$$
$$a_6=a_1 a_5+\frac{a_3(a_3-a_1^3)}{2}+\frac{a_1^6}{16}$$
And so on. The coefficients for the even powers will be related to the ones for the odd powers. 
But that's the extent of what we can really say, or so I think.

What else can we say about $f(x)$ based on these two restrictions only? And what is the weakest restriction we need to get $f(x)=c^x$?

 A: You have that $f(x)f(-x)=1$ is equivalent on saying that $f(x)= \pm \exp(g(x))$, for some analytic odd function $g$.
In fact, if $g$ is any analytic odd function, then
$$e^{g(x)}e^{g(-x)}=e^0=1$$
On the other hand, if $f(x)f(-x)=1$ (WLOG $f(0)=1$), then $f$ never vanishes. So we can think that $f$ is positive everywhere, and taking logarithms
$$\log f(x) + \log f(-x) = 0$$
i.e. $\log f$ is an odd function. If we assume $f(0)=-1$, then $\log (-f)$ is odd.
A: $f$ hasn't a root, because if exists $a\in \mathbb{R}/f(a)=0\to 0=1$, abs. Because $f$ is continuous, then $f$ is all positive with $f(0)=1$ or all negative with $f(0)=-1$. All function differentiable satisfying this, is an answer.
A: Take any analytical function in $R^{+}$ from $0$ to infinity which satisfies only $f(0)=-1$ or $f(0)=1$ and $f(x) \neq 0, x>0$. 
Create $g(x) = \begin{cases}
f(x) & \text{ if } x \geq 0 \\ 
\frac{1}{f(x)} & \text{ if } x < 0 
\end{cases}$
This is a very large family of functions. Any definition of $f(x)=c^x$ in $R^{+}$ would contain $f(-x)f(x)=1$ already because $f(-x)=\frac{1}{f(x)}=f(x)^{-1}$, just like $f(ax)=f(x)^{a}$ would be in $R^{+}$ for $f(x)=c^x$. Basically we would do no more than extend $f(x)=c^x$ over negative values.
