Any map $f \colon \mathbb{R} \to \mathbb{R}$ induces a "composition map"
$$f^\circ\colon \mathbb{R} \times \mathbb{N} \to \mathbb{R},$$


where 
$$f^{\circ n}(x) = \underbrace{f \circ \dotsb \circ f}\_{n \textrm{ times}} (x).$$


If $f$ happens to have an inverse, the domain of $f^\circ$ can be extended to $\mathbb{R} \times \mathbb{Z}$. Here's my question: is there a "nice way" to further extend $f^\circ$ to have domain $\mathbb{R} \times [0, \infty)$ in general, or $\mathbb{R} \times \mathbb{R}$ if $f$ is invertible?


I want to leave the "nice way" intentionally vague to allow for a variety of answers. But two constraints on this. Obviously one can stitch together something contrived, so "a nice way" should have an element of naturality to it. And in the best case scenario the answer would be a function $f^\circ \colon \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ that, under certain reasonable restrictions (e.g. logarithmic convexity as used in the [Bohr-Mollerup Theorem](https://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem "Bohr-Mollerup Theorem")), uniquely satisfies


1. $f^{\circ 0} = \textrm{id}$, and
2. $f^{\circ (t+1)} = f \circ f^{\circ t}$ for all $t \in \mathbb{R}$.


I'm curious if there's a general answer for arbitrary smooth / analytic $f$, but specifically would like to know the answer for $f(x) = e^x$.



Elegant / Canonical way to Extend Integer Iterates of a Function to a Real Parameter
Notice that $e^x$ does not have an inverse on the whole real line.


Extension of iterates is possible if $f$ has a fixed point $x\_0$. Suppose for example, that this fixed point is repelling that is $f(x\_0)=x\_0$ and $\lambda=f'(x\_0)>1.$


I assume that $f$ is analytic, strictly increasing on $R$ and maps $R$ onto
itself. The Poincare equation
$$F(\lambda y)=f(F(y))$$
has a unique analytic solution subject to initial conditions $F(0)=x\_0$ and $F'(0)=1$. This is a theorem of Poincare. You first find a convergent power series
near $0$ which satisfies it, and then extend $F$ to the whole real line by
the functional equation, using that $\lambda>1$.
This solution is also strictly monotone, and you can define
$$f^{\circ t}(x)=F(\lambda^t F^{-1}(x)).$$
Similar construction will work with $0<\lambda<1$ if you know that $f$ is surjective (just apply the construction to $f^{-1}$).


If $f$ is not surjective, then $F$ is not surjective, the fractional
iterates can still be defined but their domain will not be $R$.


