Generalising "composition $n$ times" to the reals over differentiable functions - can it be done?

Proving the statement below is sufficient, but any proof will be enjoyed!

Statement:
For any $f\colon\mathbb R \to\mathbb R$ differentiable over all reals, one can find a function $G : [0,1] \times \mathbb{C} \to \mathbb{C}$ defined such that:

• $G$ is differentiable over $x$


• $G(a,G(b,x)) = G(a+b,x)$ for $0 <=a+b <=1$


• $G(0,x) = x;$


• $G(1,x) = f(x)$

Given the $G$ above:

• Is $G$ unique? If not, how many such functions exist for a given $f$?



• Is $G$ differentiable over $a$?

If you can relax the condition to continuous $f$ and $G$ that would be even cooler.

EDIT:
In response to a proof that there exists no $G$ for monotonously shrinking $f$, such as $f(x) = -x$ I decided to relax the demand for G from $G : [0,1] \times \mathbb{R} \to \mathbb{R}$ to $G : [0,1] \times \mathbb{C} \to \mathbb{C}$, which allows you to make an infinite number of functions matching my requests simply by saying $G(a,x) = x*(f(x)/x)^a$ for any $k \in \mathbb N$ which makes the whole thing a bit trivial. My original question has been answered and the remaining one doesn't seem to be too interesting. What's the policy here? Should I delete the page?

EDIT 2: D'oh! $G(a,x) = x*(f(x)/x)^a$ only works for linear functions. Disregard edit 1. My statement has still been disproved though, the answer will be accepted as soon as I figure out how.

Answer

You seem to be looking for a fractional iterate of any $f\colon\mathbb R\to\mathbb R$; that is, to extend the notion or construct of $f^{\circ n}$ to reals $n$ for which $n\notin\mathbb N$, but still keeping within the domain of continuous functions. For trivial reasons, not possible for any monotone decreasing function $f$. Like $f(x)=-x$. So there can not be such a $G$.

Since I think cow-slowly, I may have misunderstood your question completely. If so, pardon it, I’ll remove.

EDIT: Probably my “trivial reasons” weren’t clear enough. First, your fractional iterate, in particular $f^{\circ 1/2}$, the “half-fold iterate”, must be one-to-one and onto $\mathbb R$. If continuous, it must turn intervals into intervals (continuous image of a connected is connected) and bounded closed intervals into b.c. intervals (continuous image of a compact is compact). It results that any continuous one-to-one onto map of $\mathbb R$ must be monotone increasing or monotone decreasing, and of course strictly so. But the composition of two decreasing functions is increasing. Thus $f(x)=-1$ can not be the “composition-square” of any continuous function.