TITLE: square root of diffeomorphism of R: does it always exist?
QUESTION [23 upvotes]: Let $f:\mathbb R\to\mathbb R$ be a smooth, orientation preserving diffeomorphism of the real line.
Is it the case that there always exist another diffeomorphism $g:\mathbb R\to\mathbb R$
such that $g\circ g = f$?
Note: it is relatively easy to show that a continuous $g$ exists, but I am not managing to find a smooth (i.e. $\mathcal C^\infty$) solution of the above equation.

REPLY [2 votes]: Here this problem (and some more general) are studied for the case $f$ has only one fixed point. In general, one can get $C^1$ square roots, but for higher smoothness there are obstructions when the derivative at the origin is the identity which are studied there.