Schwarzian derivative of a diffeomorphism is zero iff Linear-fractionals? I have found the following derivation of the Schwarzian derivatives in the book of Ovsienko and Tabachnikov:
For a diffeomorphism $\gamma$ which acts on 4 points $t_1,t_2,t_3,t_4 \in \mathbb{RP}_1$,  We assume that the 4 points are spread so that $t_2,t_3,t_4$ can be defined by their distances to $t_1$ as a function of $\epsilon\in\mathbb{R}$:  $t_2 = t_1 + \epsilon$, $t_3 = t_1 + 2\epsilon$ and $t_4 = t_1 + 3\epsilon$. So, the 4 points become $t_1, t_1 + \epsilon, t_1 + 2\epsilon,  t_1 + 3\epsilon$ and they are related by the variable $\epsilon$. 
The Schwarzian derivative measures the effect of $\gamma$ on the cross-ratio as $\epsilon$ tends to zero. In other words, the Schwarzian derivative measures the cross-ratio of the points when they are infinitesimally close. To obtain the Schwarzian derivative one forms the Taylor expansion of $\Phi$ when $\epsilon$ goes to zero and keeps the first non-zero term of the expansion:
$$\Phi(\gamma(t_1),\gamma(t_2),\gamma(t_3),\gamma(t_4))= \Phi(t_1,t_2,t_3,t_4)
- \epsilon^2 S[\gamma](t_1) + O(\epsilon^3) \tag 1$$
In the above equation $S[\gamma]$ is the Schwarzian derivative for $\mathbb{RP}_1$, defined by:
$  S[\gamma]=\frac{\gamma'''}{\gamma'}-\frac{3}{2} \left (\frac{\gamma''}{\gamma'}\right )^2$
By construction when $\gamma$ is a linear-fractional function, $S[\gamma]$ and all the higher order terms are zero as linear-fractional exactly preserves the cross-ratio. 
According to the book, the converse is also true i.e. if $S[\gamma]=0$, $\gamma$ is a linear-fractional. But I cannot see why, because in equation (1), there are higher order terms of $\epsilon$. Could anyone please tell know what I am missing here? 
 A: $S[f]=0$ is a third order ordinary differential equation, and all fractional-linear functions
are solutions. This is a $3$-parametric family.
You can match any initial conditions with a fractional-linear function, then
uniqueness theorem says that all solutions are fractional-linear.
A: The Schwatzian derivative measures the deviation of $\phi$ (which I take here as an orientation preserving diffeomorphism of $\mathbb R$, for simplicity's sake) from being a Moebius transformation; here is a direct computational proof. Of course, Alexandre Eremenko's proof is more elegant.
$$
S(\phi)=0 \iff \phi(x)=\frac{ax+b}{cx+d}\text{  for }
\begin{pmatrix} a & b \\ c & d\end{pmatrix} \in SL(2,\mathbb R).
$$
Indeed, $S(\phi)=0$ if and only if $g=\log(\phi')'=\frac{\phi''}{\phi'}$
satisfies the differential equation $g'= g^2/2$, so that
$\frac{2\,dg}{g^2}=dx$ or $\frac{-2}{g}=x+\frac dc$
which means
$\log(\phi')'(x)=g(x)=\frac{-2}{x+d/c}$ or again
$$\log(\phi'(x))=\int\frac{-2dx}{x+d/c}=-2\log(x+d/c)-2\log(c)=\log(\frac 1{(cx+d)^{2}}).$$
Therefore, $\phi'(x)= \frac 1{(cx+d)^{2}}=\partial_x\frac{ax+b}{cx+d}$.
