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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), 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 $f(x) = e^x$.

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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 $R$.

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  • $\begingroup$ What about functions that lack a fixed point? Again I'm thinking of $e^x$. Restricted domain of fractional iterates is no problem. $\endgroup$
    – Twiffy
    Commented Nov 21, 2017 at 22:22
  • $\begingroup$ For $e^x$ it is answered in the answer of bo198214 to the question mentioned in the comment of thel above. $\endgroup$ Commented Nov 22, 2017 at 0:27

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