Question in Schwarzian Derivative

Question

In Devaney's book An Introduction to Chaotic Dynamical Systems, an observation regarding Schwarzian Derivative is given: The fact that each $F^n_4$ has negative Schwarzian derivative allows us to observe that if $K\subset\mathbb{R}$ is any interval on which $(F^n_4)'\neq0$, then the minimum value of $(F^n_4)'(x)$ occurs at one of the endpoints of K.

Why is this true in general?

Here $F_4(x)=4x(1-x)$. But I think the author mean the statement is true for any $f$, not just for $F^n_4$.

Answer 1

It is not true at the Schwarzian is negative for all $f : \mathbb{R} \to \mathbb{R}$. For example, let $f(x)=x+x^3$. Then $Sf(0)=1>0$. However, it is true that

Claim 1 $S(F_4^n)(x)<0$ for all $n>0$.

This follow from the following claims:

A. $F_4(x)$ has negative Schwarzian.
B. If $f, g: \mathbb{R} \to \mathbb{R}$ satisfy $Sf(x)<0$ and $Sg(x)<0$, then $S(f \circ g)(x)<0$.

Using these two facts, we can prove Claim 1.

Proof of Claim 1 By induction. Base case $n=1$ follows by point 1. Induction step: $S(F_4^{n-1})(x)<0$ and $S(F_4)(x)<0$ by induction hypothesis. Thus, by point B above, $S(F_4^{n})(x)=(S(F_4^{n-1})\circ S(F_4))(x)<0$. Q.E.D.

Now we prove the intermediate claims A and B.

Proof of A. To ease notation, let $p=F_4$. Then $p'(x)=4(1-2x)$, $p''(x)=-8$. Then a calculation shows $Sp(x)=\frac{-6}{(2x-1)^2}<0$.

Proof of B. We have the following composition rule for the Schwarzian: $S(g \circ f)(x)= f'(z)^2(S(g) \circ f)(x) + S(f)(x)$. From this rule, B follows immediately.

Thus, the original claim 1 follows. For completeness, let us prove the composition rule for the Schwarzian, which is a straightforward computation.

Proof of compostion rule for Schwarzian We first compute the following derivatives: \begin{equation*} \begin{aligned} (g \circ f)'(x) &= g'(f(x))f'(x)\\ (g \circ f)''(x) &= g''(f(x))f'(x)^2+g'(f(x))f''(x)\\ (g \circ f)'''(x) &= g'''(f(x))f'(x)^3+3g''(f(x))f'(x)f''(x) + g'(f(x))f'''(x) \end{aligned} \end{equation*} Further, we have that \begin{equation*} \begin{aligned} S_{g \circ f}(x) &= \dfrac{(g\circ f)'''}{(g \circ f)'} - \dfrac{3}{2}\Bigg(\dfrac{(g \circ f)''}{(g \circ f)'} \Bigg)^2 \\ &= \dfrac{g'''(f(x))}{g'(f(x))}f'(x)^2 + 3\dfrac{g''(f(x))}{g'(f(x))}f''(x)+ \dfrac{f'''(x)}{f'(x)} - \dfrac{3}{2}\Bigg(\dfrac{g''(f(x))}{g'(f(x))}f'(x)+\dfrac{f''(x)}{f'(x)}\Bigg)^2 \\ &= \Bigg[ \dfrac{g'''(f(x))}{g'(f(x))}-\dfrac{3}{2}\Bigg( \dfrac{g''(f(x)}{g'(f(x))} \Bigg)^2 \Bigg] f'(x)^2 + \Bigg[ \dfrac{f'''(x)}{f'(x)} - \dfrac{3}{2} \Bigg( \dfrac{f''(x)}{f'(x)} \Bigg)^2 \Bigg] \\ &= S_g(f(x))f'(x)^2 + S_f(x) \end{aligned} \end{equation*}