# 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.

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I've added some formatting, you should click the edit button and take a look. The best way to learn is by example. – Jim Mar 15 '13 at 18:13
There seems to be a big problem here. You want $G(a,G(b,x)) = G(a+b,x)$ for all $a,b \in [0,1]$. However, $[0,1]$ is not closed under addition! For example, what if $a=1$ and $b=1$? You want $G(1,G(1,x)) = G(2,x)$, but $a=2$ is not in the domain of $G$. – Fly by Night Mar 15 '13 at 18:43
@FlybyNight and I suggest exchange $[0,1]$ by the group $\mathbb{S}^1$ – MathOverview Mar 15 '13 at 18:47
@אליהוצלע that could be dangerous. I'd want to know the fundamental group of the image before I started using circles. – Fly by Night Mar 15 '13 at 18:50
With such a G defined, the idea is that I can then extend G(a,x) to any a in the whole set of positive reals simply by repeatedly applying G – Curious Programmer Mar 15 '13 at 20:36

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 $G$.
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.
Thanks, I get it now ^^. I've found a $G$ for $f(x)=-x$ : $G(a,x) = x*i^a$ in the complex functions, so I'll loosen the restriction for $G$'s type to $[0,1] \times \mathbb{C} \to \mathbb{C}$. Hopefully there might be a generalisation out there still! – Curious Programmer Mar 15 '13 at 21:30
er, I mean $x * e^((2\pi)*a)$, not $x*i^a$ – Curious Programmer Mar 15 '13 at 21:40
Believe me, there are serious problems in the case of more elaborate functions than just $f(x)=cx$. Generally you can’t get fractional iterates except perhaps locally. A continuing question showing up here at SE is to get a “half-fold iterate” of $f(x)=\cos x$ or $f(x)=e^x$. Mostly they don’t exist. It’s a great problem to exercise your mind on, though! – Lubin Mar 16 '13 at 2:05