Finding the “root” of a monotone function (in the sense of composition) Let $f:[0,\infty)\rightarrow [0,\infty) $ be a smooth and monotone function s.t $f(0)=0$. Let $N\in\mathbb{N}$. Can we find a function $g: [0,\infty) \rightarrow [0,\infty) $ s.t $g\circ\cdots\circ g$ ($g$ composed with itself $N$ times) equals $f$?
Can we say something about $g$‘s monotonicity? Its smoothness? I cannot come up with any basic answers. Thanks in advance to the helpers.
 A: I think I came in 5 minutes too late,  but since my answer is already typed up,  here it goes:
As it is clear by  the interesting  comments,  this question is not a walk in the park,  so let me offer my two cents by
addressing the simpler problem in which $f$ is assumed to be strictly increasing and onto $[0,\infty )$.
I will  ignore
the smoothness of $f$, hence also giving up on the question of the smoothness of $g$, and concentrate instead on an
increasing homeomorphism
$$
  f: [0,\infty )\to   [0,\infty ).
  $$
Hopefully the argument below can be refined by bringing the smoothness aspects back in to the picture.
Considering any  increasing homeomorphism  $h:{\mathbb  R}\to (0,\infty )$, we may replace $f$ by $h^{-1}fh$, and hence think of $f$ as
an increasing homeomorphism
$$
  f:{\mathbb  R}\to{\mathbb  R}.
  $$
Case 1: If $f(x)>x$, for all $x\in  {\mathbb  R}$, then there exists a homeomorphism $\varphi :{\mathbb  R}\to {\mathbb  R}$, such that
$$
  \varphi f\varphi ^{-1}(x) = x+1, \quad\forall x\in  R,
  $$
and therefore  $f$ has an increasing  $n^{\text{th}}$ root $g$, given by
$$
  g(x) = \varphi ^{-1}\big (\varphi (x)+1/n\big ).
  $$
Proof:  Choose any $x_0$ in ${\mathbb  R}$, and observe that for every $x$ in ${\mathbb  R}$, there exists a unique $n=n_x\in {\mathbb  Z}$, such that
$$
  f^n(x)\in \big[x_0,f(x_0)\big).
  $$
Letting $L$ denote the length of the interval $[x_0,f(x_0)\big)$,
we may then write $f^n(x)=x_0+aL$, for a unique $a=a_x\in [0,1)$.  The desired homeomorphism can then be taken to be
$\varphi (x) = n_x+a_x$.  QED
Case 2: $f(x)<x$, for all $x\in  {\mathbb  R}$.  This follows as an easy generalization of  case (1) with the obvious modifications (e.g.,
replacing $x+1$ by $x-1$).
To treat the general case,  let
$$
  F=\{x\in {\mathbb  R}: f(x)=x\},
  $$
and let the complement of $F$ be written as the disjoint union of open intervals
$$
  {\mathbb  R}\setminus F = \bigcup_{k=1}^\infty  J_k.
  $$
Clearly each $J_k$ is invariant under $f$, and since $J_k$ is order-homeomrphic to ${\mathbb  R}$, we may apply either case (1)
or (2) to produce an $n^{\text{th}}$ root $g_k$, on each open interval,  and then patch them together to obtain a global $n^{\text{th}}$ root $g$.
A: Thank you everyone, @Hanno posted a collection of references with a paper which addresses this problem:
http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.desklight-3881187f-67b6-4d8e-a977-31e52dd4e414/c/apm22_2_09.pdf
