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.
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 $g$.
Thank you everyone, @Hanno posted a collection of references with a paper which addresses this problem: