# Exact quadratic growth for the derivatives of iterates of interval
diffeomorphisms with only parabolic fixed points

Leonardo Dinamarca Opazo & Andrés Navas

Abstract. We consider $C^{2}$ diffeomorphisms of a closed interval with only
parabolic fixed points. We show that the maximal growth of the derivatives of
the iterates of such a diffeomorphism is exactly quadratic provided it has a
non quadratically tangency to the identity at a fixed point that is
topologically repelling on one side. Moreover, in absence of such fixed
points, the maximal growth of the derivatives of the iterates is subquadratic.

Keywords: Interval, diffeomorphism, normal form, derivative growth,
distortion.

MCS 2020: 37C05 37C10, 37C15, 37E05.

In the beautiful work [PS04], Polterovich and Sodin prove the following
somewhat surprising result: for every $C^{2}$ diffeomorphim of a closed
interval, the growth of the derivatives of the iterates is at most quadratic.
More precisely, if we let
$\Gamma^{n}(f):=\sup_{x}\max\\{Df^{n}(x),Df^{-n}(x)\\},$ then

$\limsup_{n\to\infty}\frac{\Gamma^{n}(f)}{n^{2}}<+\infty.$

In this article we prove that the quotient above actually converges. To do
this, we obtain finer information for positive iterates. We hence let
$\Gamma^{n}_{+}(f):=\sup_{x}Df^{n}(x)$, and we study the asymptotic of this
sequence.

In all what follows, we will implicitly assume that the map $f$ we are dealing
with is orientation preserving and nontrivial. We call a component of $f$ any
subinterval fixed by $f$ that contains no fixed point in its interior. We let
$\mathcal{C}(f)$ be the family of components of $f$. Note that, for each
$I\in\mathcal{C}(f)$, one (and exactly one) of the endpoints is topologically
repelling for $f$. We denote it by $r_{I}$.

Recall that, by the work of Szekeres [Sz58] and Kopell [Ko68], for every
$C^{2}$ diffeomorphism of a half-closed interval with no fixed point inside,
there exists a unique $C^{1}$ vector field whose time-1 map coincides with the
diffeomorphism. If $I$ is a component of $f$, we denote by $X_{I}$ the unique
$C^{1}$ vector field defined on the half-closed interval
$I^{+}:=I^{\mathrm{o}}\cup\\{r_{I}\\}$ whose time-1 map coincides with the
corresponding restriction of $f$.

Main Theorem. For every $C^{2}$ diffeomorphism $f$ of a closed interval, the
expression $\Gamma_{n}^{+}(f)/n^{2}$ converges to a finite limit as $n$ goes
to infinite, and

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}=\sup_{I\in\mathcal{C}(f)}\left[\frac{1}{2}D^{2}f(r_{I})\cdot\sup_{y\in
I}X_{I}(y)\right].$ (1)

Since $\Gamma^{n}(f)=\max\\{\Gamma^{n}_{+}(f),\Gamma_{+}^{n}(f^{-1})\\}$, the
result above implies Polterovich-Sodin’s theorem (by an application to both
$f$ and $f^{-1}$ simultaneously). Also, it implies that, if $f$ has a
quadratic tangency to the identity at each of its fixed points, then
$\Gamma^{n}(f)/n^{2}$ converges to zero. Let us point out, however, that this
was first established by Watanabe in [Wa04] via a clever modification of
Polterovich-Sodin’s arguments. Nevertheless, our proof in this case (which is
essentially that of Watanabe up to a slight modification in the end) gives
extra information. For instance, for $f$ with no fixed point in the interior,
we prove the vanishing of the limit of $\Gamma^{n}_{+}(f)/n^{2}$ if the
tangency to the identity is quadratic at the topologically repelling fixed
point, but not necessarily at the other one. This means that topologically
contracting fixed points can only yield subquadratic growth of the derivative
along iterates. Actually, if such a point is not flat, then it yields uniform
decay of the derivatives along iterates. Rather surprisingly, this is not the
case of flat fixed points. We will illustrate this phenomenon with an explicit
example in Section 4.

The new key ingredient to establish the exact quadratic rate growth in the
Main Theorem are the Szekeres vector fields. It is worth mentioning that this
was somehow suggested in [PS04], where it is explicitly proposed to use normal
forms around fixed points in the analysis. Of course, this is unavailable for
infinitely flat fixed points, but fortunately, subquadratic growth in this
framework has been already established by Borichev in class $C^{\infty}$
[Bo04] and by Watanabe in class $C^{2}$ [Wa04]. In the non flat case, however,
there are still technical problems. In particular, normal forms are available
only in high differentiability (see [Ta73] for a classical result in class
$C^{\infty}$ and [EN23] for recent sharp results in finite regularity).
Moreover, the use of normal forms allows establishing the order of growth of
the derivatives along iterates, but does not detect its precise asymptotic.

The crucial observation is that dealing with the exact growth of derivatives
along iterates of an interval diffeomorphism $f$ naturally leads to consider
the 1-parameter groups of $C^{1}$ diffeomorphisms that realize $f$ as its
time-1 map. When normal forms are available, these flows are easily detected.
However, these exist even for flat fixed points. In particular, if $f$ has no
fixed point in the interior, classical work of Szekeres and Kopell shows that
there are (at most) two such flows, each of them arising by looking $f$ as a
diffeomorphism of a half-closed interval (these flows may happen to be
different). In our discussion, only one of them becomes relevant, namely the
one that is associated to the topologically repelling fixed point. This flow
is encoded by a $C^{1}$ (generating) vector field $X$ which hence satisfies
$X\circ f=X\cdot Df$. This relation obviously implies $Df^{n}=X\circ f^{n}/X.$
Studying the growth of $Df^{n}$ then reduces to study the decay of $X$ near
the topologically repelling fixed point as well as the behavior of the orbits
(of the inverse map $f^{-1}$) around it. Fortunately, the precise asymptotic
in the case we need to settle is already available, so after some preparation
work all fits nicely, and the exact formula (1) arises. Most of this is
carried out in full detail in Section 2.2. Quite remarkably, none of the
arguments require more regularity than $C^{2}$.

##  1 On Polterovich-Sodin’s growth lemma

A great deal of the analysis done in [PS04] is based on the following crucial
lemma. Although it is less relevant for our approach, we will still use it at
a key step; see Remark 2.4 on this.

Lemma [on quasiconvex sequences] Let $(a_{n})_{n\geq 0}$ be a sequence of real
numbers with $a_{0}=0$ and such that, for all $n\geq 1$, the inequality

$2a_{n}-a_{n-1}-a_{n+1}\leq Ce^{-a_{n}}$ (2)

holds for a certain (fixed) $C\geq 0$. Assume that
$\liminf_{n\to\infty}\frac{a_{n}}{n}=0$. Then, for all $n\geq 1$,

$e^{a_{n}}\leq\left(n\,\sqrt{\frac{C}{2}}+1\right)^{2}.$

Following Polterovich and Sodin, below we establish a general quantitative
lemma that incorporates some of Watanabe’s remarks [Wa04]. Although our
analysis mostly relies on the generating vector fields, this will simplify the
exposition at several parts.

######  Lemma 1.1.

Let $f$ be a $C^{2}$ diffeomorphism of a closed interval $L$ that has no fixed
point in the interior. If both endpoints are parabolic then, for all $n\geq
1$, it holds

$\Gamma^{n}_{+}(f)\leq\left(n\,\sqrt{\frac{C}{2}}+1\right)^{2},$

where $C:=C^{\prime}e^{C^{\prime}}$ , with
$C^{\prime}:=|L|\cdot\left\|\frac{D^{2}f}{Df}\right\|_{\infty}.$

###### Proof.

Assume that $f$ pushes all points to the right, otherwise just conjugate it by
the reflexion of $L$. For simplicity, denote $a_{n}:=\log\Gamma_{+}^{n}(f)$.
Let $x_{n}$ be a point that realizes $\Gamma^{n}_{+}(f)$, that is,
$Df^{n}(x_{n})=\Gamma_{+}^{n}(f)$. Then, as in [PS04], we have

$a_{n}=\log Df^{n}(x_{n})=\sum_{i=0}^{n-1}\log Df(f^{i}(x_{n})),$
$a_{n-1}\geq\log Df^{n-1}(f(x_{n}))=\sum_{i=1}^{n-1}\log Df(f^{i}(x_{n})),$
$a_{n+1}\geq\log Df^{n+1}(f^{-1}(x_{n}))=\sum_{i=-1}^{n-1}\log
Df(f^{i}(x_{n})).$

Therefore,

$2a_{n}-a_{n-1}-a_{n+1}\leq\log Df(x_{n})-\log Df(f^{-1}(x_{n})).$

Since $f$ is a $C^{2}$ diffeomorphism, there exists a point
$\xi_{n}\in[f^{-1}(x_{n}),x_{n}]$ such that

$\log Df(x_{n})-\log Df(f^{-1}(x_{n}))=D\log
Df(\xi_{n})\cdot[x_{n}-f^{-1}(x_{n})]=\frac{D^{2}f}{Df}(\xi_{n})\cdot[x_{n}-f^{-1}(x_{n})],$

hence

$2a_{n}-a_{n-1}-a_{n+1}\leq\frac{D^{2}f}{Df}(\xi_{n})\cdot[x_{n}-f^{-1}(x_{n})]=\frac{D^{2}f}{Df}(\xi_{n})\cdot\frac{[x_{n}-f^{-1}(x_{n})]}{[f^{n}(x_{n})-f^{n-1}(x_{n})]}\cdot[f^{n}(x_{n})-f^{n-1}(x_{n})].$

Also, there is $x_{n}^{\prime}\in[f^{-1}(x_{n}),x_{n}]$ such that

$Df^{n}(x_{n}^{\prime})=\frac{[f^{n}(x_{n})-f^{n-1}(x_{n})]}{[x_{n}-f^{-1}(x_{n})]},$

hence

$2a_{n}-a_{n-1}-a_{n+1}\leq\frac{D^{2}f}{Df}(\xi_{n})\cdot\frac{[f^{n}(x_{n})-f^{n-1}(x_{n})]}{Df^{n}(x_{n}^{\prime})}.$

Finally, the classical Denjoy’s control-of-distortion argument gives

$\frac{Df^{n}(x_{n})}{Df^{n}(x_{n}^{\prime})}\leq e^{V(f)},$

where $V(f):=\mathrm{var}(\log Df)$. (See [Na11] for a full treatment of this
argument.) Therefore,

$2a_{n}-a_{n-1}-a_{n+1}\leq
e^{V(f)}\cdot\frac{D^{2}f}{Df}(\xi_{n})\cdot[f^{n}(x_{n})-f^{n-1}(x_{n})]\cdot
e^{-a_{n}}.$ (3)

Since $|f^{n}(x_{n})-f^{n-1}(x_{n})|\leq|L|$ and

$V(f)=\int_{L}\left|\frac{D^{2}f}{Df}\right|\leq C^{\prime},$ (4)

this gives the announced quasiconvexity of $(a_{n})$. Since the endpoints are
parabolic, one has

$\lim_{n\to\infty}\frac{a_{n}}{n}=0$ (5)

(this is an elementary exercise using the continuity of the derivative; see
also [PS04, (1.2)] for a more sophisticated -though clarifying- proof).
Polterovich-Sodin’s lemma on quasiconvex sequences then allows concluding the
proof. ∎

##  2 A proof of the Main Theorem: no interior fixed point

We start by establishing the Main Theorem for diffeomorphisms of a closed
interval $L:=[a,b]$ with no interior fixed point. With no loss of generality,
we can suppose that $f$ pushes all interior points to the right, otherwise we
can conjugate it by the reflexion of $L$.

###  2.1 The case of a quadratic tangency at the repelling fixed point

Assuming that $D^{2}f(a)=0$, our goal is to prove that

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}=0.$ (6)

To do this, we closely follow Watanabe’s arguments from [Wa04], with an
important modification in the end.

Fix $\varepsilon>0$. For each positive $\mu<1$, let

$\delta:=\max\left\\{\frac{\mu\,|L|}{1-\mu},\,\mu\cdot\left\|\frac{D^{2}f}{Df}\right\|_{\infty}\right\\}.$

Also define

$C^{\prime}:=|L|\cdot\left\|\frac{D^{2}f}{Df}\right\|_{\infty}.$

Fix $\mu>0$ small enough such that $\delta e^{C^{\prime}}\leq 2\varepsilon$.
Let $\mu^{\prime}>0$ be such that

$\left|D^{2}f(\xi)\right|\leq\mu\quad\mbox{for all }\xi\in[a,a+\mu^{\prime}],$
(7)

and let $N$ be large enough so that $f^{N-1}(a+\mu^{\prime})>b-\mu$. We claim
that the sequence

$b_{n}:=\log\left(\frac{\Gamma_{+}^{n+N}(f)}{\Gamma^{N}_{+}(f)}\right)=a_{n+N}-a_{N},$

satisfies the quasiconvexity inequality

$2b_{n}-b_{n-1}-b_{n+1}\leq 2\varepsilon e^{-a_{N}}e^{-b_{n}}.$ (8)

Before checking this note that, since $b_{0}=0$, by (5) and the lemma on
quasiconvex sequences this implies that, for all $n\geq 1$,

$\Gamma^{n+N}_{+}(f)=e^{a_{n+N}}=e^{a_{N}}e^{b_{n}}\leq
e^{a_{N}}(n\sqrt{\varepsilon
e^{-a_{N}}}+1)^{2}=(n\sqrt{\varepsilon}+\sqrt{e^{a_{N}}})^{2}.$

This obviously gives

$\limsup_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}\leq\varepsilon,$

and since this is true for every $\varepsilon>0$, this implies (6).

Now, to prove (8), we refer to (3). There are two cases to consider:

  * •

If $x_{n+N}\leq a+\mu^{\prime}$ then $\xi_{n+N}\leq\mu^{\prime}$, which in
virtue of (7) gives $|D^{2}f(\xi_{n+N})|\leq\mu$. Since $Df(a)=1$, this
implies that $Df(y)\geq 1-\mu$. By (3) and (4), this gives

$2a_{n+N}-a_{n+N-1}-a_{n+N+1}\leq
e^{C^{\prime}}\frac{\mu\,|L|}{1-\mu}e^{-a_{n+N}}\leq\delta
e^{C^{\prime}}e^{-a_{n+N}}\leq 2\varepsilon e^{-a_{n+N}},$

which immediately yields (8).

  * •

If $x_{n+N}\geq a+\mu^{\prime}$ then, by the choice of $N$, we have
$f^{N-1}(x_{n+N})\geq b-\mu$. This implies that
$f^{n+N}(x_{n+N})-f^{n+N-1}(x_{n+N})<\mu$ which, by (3) and (4), gives

$2a_{n+N}-a_{n+N-1}-a_{n+N+1}\leq
e^{C^{\prime}}\left\|\frac{D^{2}f}{Df}\right\|\mu\,e^{-a_{n+N}}\leq\delta
e^{C^{\prime}}e^{-a_{n+N}}\leq 2\varepsilon e^{-a_{n+N}},$

which again shows (8).

###  2.2 The case of a nonquadratic tangency at the repelling fixed point

Recall that we are supposing that our $C^{2}$ diffeomorphism $f$ of the closed
interval $L:=[a,b]$ pushes all interior points to the right. Here, we also
assume that $f$ is not quadratically tangent to the identity at $a$, which
implies that $D^{2}f(a)>0$. Under these assumptions, our goal is to prove that

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}=\frac{1}{2}D^{2}f(a)\cdot\sup_{y\in
L}X_{L}(y).$ (9)

To do this, we first state a lemma that allows localizing the points $x_{n}$
that realize the maximal derivative for $f^{n}$. By reasons that will become
clear later, we will first establish only a half of it. The (much harder)
second half will be proved after some further developments.

######  Lemma 2.1.

There exists a compact subinterval $[A,B]\subset(a,b)$ such that if $(x_{n})$
is any sequence satisfying $Df^{n}(x_{n})=\Gamma_{+}^{n}(f)$, then
$f^{n}(x_{n})$ belongs to $[A,B]$.

Proof of the first half of Lemma 2.1 (existence of $A>a$). Since
$D^{2}f(a)>0$, the derivative $Df$ is strictly increasing on a certain
interval $[a,A]$. We claim that, for all $n\geq 1$, one necessarily has
$f^{n}(x_{n})\geq A$. Assume otherwise. Then $[x_{n},f^{n}(x_{n})]$ is
contained in $[a,A]$, where $Df$ is increasing. This implies that
$Df(f^{n}(x_{n}))>Df(x_{n})$, hence

$Df^{n}(f(x_{n}))=Df^{n}(x_{n})\cdot\frac{Df(f^{n}(x_{n}))}{Df(x_{n})}>Df^{n}(x_{n}).$

However, this contradicts the fact that $x_{n}$ realizes the maximum of the
derivative of $f^{n}$. $\hfill\square$

Before passing to the proof of the second half of the lemma, we first need to
control de orbits of the inverse map $f^{-1}$. This is the content of the next
folklore result (which seems to go back to Fatou), for which we include a
sketch of proof since most (all?) of the proofs in the literature use higher
regularity (see for instance [Re13] and [EN23, Proposition 3.2] for finer
results in class $C^{3}$). We denote by $L^{\mathrm{o}}$ the interior of the
interval $L$.

######  Lemma 2.2.

For every point $y\in L^{\mathrm{o}}$, one has
$\lim_{n\to\infty}n\,(f^{-n}(y)-a)=\frac{2}{D^{2}f(a)}$. Moreover, the
convergence is uniform on compact subintervals of $L^{\mathrm{o}}$.

###### Proof.

Changing $f$ by its inverse and conjugating by a translation, we are reduced
to show the following: for a $C^{2}$ diffeomorphism $g$ of an interval $[0,c]$
with no interior fixed point and which is of the form $g(x)=x-sx^{2}+o(x^{2})$
around the origin (with $s>0$), one has

$\lim_{n\to\infty}n\,g^{n}(y)=\frac{1}{s}$

for each $y\in(0,c)$, and the limit is uniform on compact subintervals. To
prove this first note that, since all compact subintervals eventually enter
into any prescribed neighborhood of the origin, we may argue in such a
neighborhood. Now consider the conjugate map $G:=I\circ g\circ I$, where $I$
denotes the inversion $I(z):=1/z$. We are thus reduced to show that
$G^{n}(z)/sn$ converges to 1 as $n$ goes to infinite, and that this
convergence is uniform on compact intervals. To do this, we compute:

$G(z)=\frac{1}{g(\frac{1}{z})}=\frac{1}{\frac{1}{z}-\frac{s}{z^{2}}+o\big{(}\frac{1}{z^{2}}\big{)}}=\frac{z}{1-\frac{s}{z}+o\big{(}\frac{1}{z}\big{)}}=z+s+o(1).$

Using this, one easily convinces that $G^{n}(z)=z+ns+o(n)$, which allows
establishing the announced limit. Moreover, the convergence is easily seen to
be uniform on compact subsets. We leave the details to the reader. ∎

The preceding lemma ensures that the growth of $\Gamma^{n}_{+}(f)$ is bounded
from below by a quadratic sequence, as we next state.

######  Lemma 2.3.

One has

$\liminf_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}\geq\frac{[A-f^{-1}(A)]\cdot
D^{2}f(0)}{2}.$

###### Proof.

For each $n\geq 1$, denote $A_{n}:=f^{-n}(A)$. There exists
$A_{n}^{\prime}\in[f^{-1}(A_{n}),A_{n}]$ such that

$Df^{n}(A_{n}^{\prime})=\frac{A-f^{-1}(A)}{f^{-n}(A_{n})-f^{-n-1}(A_{n})},$

hence

$\Gamma_{+}^{n}(f)\geq\frac{A-f^{-1}(A)}{f^{-n}(A_{n})-f^{-n-1}(A_{n})}.$ (10)

By the previous lemma,
$f^{-n-1}(A_{n})=a+\frac{2}{n\,D^{2}f(a)}+o(\frac{1}{n})$. Since close to $a$
we have

$f(x)=f(a)+\frac{D^{2}f(a)}{2}(x-a)^{2}+o(|x-a|^{2}),$

we thus conclude

$\displaystyle f^{-n}(A_{n})-f^{-n-1}(A_{n})$ $\displaystyle=$ $\displaystyle
f(f^{-n-1}(A_{n}))-f^{-n-1}(A_{n})$ $\displaystyle=$
$\displaystyle\frac{D^{2}f(a)}{2}\big{|}f^{-n-1}(A_{n})-a\big{|}^{2}+o\,\big{(}\big{|}f^{-n-1}(A_{n})-a\big{|}^{2}\big{)}$
$\displaystyle=$
$\displaystyle\frac{D^{2}f(a)}{2}\left(\frac{2}{n\,D^{2}f(a)}\right)^{2}+o\,\Big{(}\frac{1}{n^{2}}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{2}{n^{2}\,D^{2}f(a)}+o\Big{(}\frac{1}{n^{2}}\Big{)}.$

Introducing this equality into (10) and passing to the limit finally proves
the lemma. ∎

We can now proceed to show the second part of Lemma 2.1.

Proof of the second half of Lemma 2.1 (existence of $B>b$). By the previous
lemma, there exists $\varepsilon>0$ such that

$\Gamma_{+}^{n}(f)\geq 2\varepsilon\,n^{2}\quad\mbox{for all }n\geq 1.$ (11)

Fix $B^{*}>A$ close enough to $b$ so that

$\left\|\frac{D^{2}f}{Df}\right\|_{\infty}\cdot[b-B^{*}]\cdot\exp\left(\left\|\frac{D^{2}f}{Df}\right\|_{\infty}\cdot|L|\right)\leq
2\varepsilon.$ (12)

Recall that we are dealing with a sequence $(x_{n})$ such that
$Df^{n}(x_{n})=\Gamma_{+}^{n}(f)$, for which we let $y_{n}:=f^{n}(x_{n})$. The
proof will be completed after showing the next two claims.

Claim 1: There exists $B^{\prime}>B^{*}$ such that the inequalities $x_{n}\leq
A$ and $y_{n}\geq B^{\prime}$ cannot simultaneously hold.

Let $X=X_{L}$ be the Szekeres vector field associated to $f$ on the interval
$L^{+}=[a,b)$. We have $X\circ f=X\cdot Df$, which implies $X\circ
f^{n}=X\cdot Df^{n}$ for all $n\geq 1$. Now, let $(f^{t})$ be the flow of $X$
(so that $f^{1}=f$). Note that $X\circ f^{t}=X\cdot Df^{t}$ holds for all $t$.
Following Sergeraert [Se77], we compute

$\frac{f(x)-x}{X(x)}=\frac{1}{X(x)}\int_{0}^{1}\frac{d}{dt}f^{t}(x)\,dt=\int_{0}^{1}\frac{X(f^{t}(x))}{X(x)}\,dt=\int_{0}^{1}Df^{t}(x)\,dt.$

Since the right-side expression converges to $1$ as $x$ goes to the origin, we
conclude that, close to $a$,

$X(x)\sim f(x)-x\sim\frac{1}{2}D^{2}f(a)\cdot(x-a)^{2}.$

We hence let $A^{\prime}$ be such that
$a<A^{\prime}<\min\\{A,f^{-1}(B^{*})\\}$ and

$X(x)\geq\frac{1}{3}D^{2}f(a)\cdot(x-a)^{2}\quad\mbox{for all
}x\in[a,A^{\prime}].$ (13)

Similarly, let $Y$ be the Szekeres vector field associated to $f$ on the
interval $(a,b]$. Again, we have $Y\circ f^{n}=Y\cdot Df^{n}$. Also, close to
$b$, we have $Y(x)\sim f(x)-x.$ Thus, letting

$C^{\prime}:=\frac{\max\\{1,\max_{x\in[A^{\prime},B^{*}]}X(x)\\}}{\min_{x\in[A^{\prime},B^{*}]}Y(x)},$

we may choose $B^{\prime}>B^{*}$ such that

$Y(y)\leq\min\left\\{\frac{\varepsilon}{C^{\prime}},\frac{\varepsilon}{3C^{\prime}\cdot
D^{2}f(a)}\right\\}\quad\mbox{for all }y\in[B^{\prime},b].$ (14)

Recall from Lemma 2.2 that $n\,(f^{-n}(A)-a)$ converges to $2/D^{2}f(0)$. We
may hence choose $N$ such that

$f^{-n}(A)-a\geq\frac{1}{n\cdot D^{2}f(0)}\quad\mbox{for all }n\geq N.$ (15)

Finally, we may slightly increase $B^{\prime}$ if necessary in order to
satisfy $B^{\prime}>f^{N}(A^{\prime})$ (as well as (14)).

Now assume for a contradiction that $x_{n}\leq A$ and $y_{n}\geq B^{\prime}$.
There are two cases to consider:

  * •

If $x_{n}\geq A^{\prime}$ then, by the definition of $C^{\prime}$, we have
$1/Y(x_{n})\leq C^{\prime}$, and therefore

$Df^{n}(x_{n})=\frac{Y(y_{n})}{Y(x_{n})}\leq\varepsilon,$

which strongly contradicts (11).

  * •

If $x_{n}<A^{\prime}$, we let $i=i_{n}\geq 1$ be the (unique) integer such
that $f^{i}(x_{n})\in[A^{\prime},f(A^{\prime}))$. Then we compute

$Df^{n}(x_{n})=Df^{i}(x_{n})\cdot
Df^{n-i}(f^{i}(x_{n}))=\frac{X(f^{i}(x_{n}))}{X(x_{n})}\cdot\frac{Y(y_{n})}{Y(f^{i}(x_{n}))}.$

Using (13) and (14) and later using the definition of $C^{\prime}$, we obtain

$Df^{n}(x_{n})\leq\frac{X(f^{i}(x_{n}))}{Y(f^{i}(x_{n}))}\cdot\frac{\varepsilon}{3C^{\prime}\cdot
D^{2}f(a)}\cdot\frac{3}{D^{2}f(a)\cdot(x_{n}-a)^{2}}\leq\frac{\varepsilon}{(D^{2}f(a))^{2}\cdot(x_{n}-a)^{2}}.$

Finally, since $B^{\prime}>f^{N}(A)$, we must have $n\geq N$, which by (15)
implies

$x_{n}-a\geq f^{-n}(A)-a\geq\frac{1}{n\cdot D^{2}f(a)}.$

However, introducing this in the previous inequality gives

$Df^{n}(x_{n})\leq\varepsilon\,n^{2},$

which still contradicts (11).

Claim 2: The inequality $x_{n}\geq A$ can hold only for finitely many integers
$n$.

Define the sequence $(a_{n}^{*})$ by

$a_{n}^{*}:=\max_{x\in[A,b]}\max\big{\\{}\log Df^{i}(x):0\leq i\leq
n\big{\\}}.$

There are two cases to consider:

  * •

Assume first that this sequence is bounded, say by a constant $C$. Fix an
integer $N$ strictly larger than $\sqrt{\frac{C}{2\varepsilon}}$. By (11), if
$n\geq N$ then one cannot have $x_{n}\geq A$, which gives the desired
conclusion in this case.

  * •

Assume now that the sequence $(a_{n}^{*})$ diverges, and let $x_{n}^{*}$ be a
sequence of points in $[A,b]$ realizing $a_{n}^{*}$, that is, such that $\log
Df^{i_{n}}(x_{n}^{*})=a_{n}^{*}$ for some $i_{n}\leq n$. We first claim that
there exists $N$ such that $x_{n}^{*}\geq f(A)$ for all $n\geq N$. To show
this, let $\Delta:=\min_{x\in[A,f(A)]}(f(x)-x)$, and fix $N^{\prime}$ such
that $f^{N^{\prime}}(A)>b-e^{-V(f)}\cdot\Delta$. If $n\geq N^{\prime}$ then,
for any $x^{*}\in[A,f(A)]$,

$f(f^{n}(x^{*}))-f^{n}(x^{*})<b-f^{N^{\prime}}(A)<e^{-V(f)}\cdot\Delta\quad\mbox{
and }\quad f(x^{*})-x^{*}\geq\Delta.$

Taking $\xi^{*}\in[x^{*},f(x^{*})]$ such that

$Df^{n}(\xi^{*})=\frac{f^{n}(f(x^{*}))-f^{n}(x^{*})}{f(x^{*})-x^{*}},$

this implies

$Df^{n}(\xi^{*})\leq\frac{e^{-V(f)}\cdot\Delta}{\Delta}=e^{-V(f)}.$

Moreover, by Denjoy’s control of distortion argument, one has
$Df^{n}(x^{*})/Df^{n}(\xi^{*})\leq e^{V(f)}$. Therefore, for all $n\geq
N^{\prime}$ and all $x^{*}\in[A,f(A)]$,

$Df^{n}(x^{*})<1.$

Since $Df^{n}(x^{*})\leq(\|Df\|_{\infty})^{N^{\prime}}$ obviously holds for
all $n\leq N^{\prime}$ and all $x^{*}\in[A,f(A)]$, the divergence of
$(a_{n}^{*})$ shows that $x_{n}^{*}$ cannot belong to $[A,f(A)]$ for $n$
larger than a certain integer $N$.

Next we claim that, for all $n\geq N$, the following inequality holds:

$2a_{n}^{*}-a_{n-1}^{*}-a_{n+1}^{*}\leq 2\varepsilon e^{-a_{n}^{*}}.$ (16)

Indeed, let $i_{n}\leq n$ be such that

$a_{n}^{*}=\log Df^{i_{n}}(x_{n}^{*}).$

Obviously,

$a_{n-1}^{*}\geq\log Df^{i_{n}-1}(f(x_{n}^{*})).$

Moreover, since $x_{n}^{*}\geq f(A)$, we have $f^{-1}(x_{n}^{*})\geq A$, hence

$a_{n+1}^{*}\geq\log Df^{i_{n}+1}(f^{-1}(x_{n}^{*})).$

Having this at hand, to prove (16) one may proceed as in the proof of Lemma
1.1 using (12) in the end.

Now define the sequence $b_{n}^{*}:=a_{n+N}^{*}-a_{N}^{*}$. Note that
$b_{0}=0$ and that inequality (16) gives

$2b_{n}^{*}-b_{n-1}^{*}-b_{n+1}^{*}\leq 2\varepsilon
e^{-a_{N}^{*}}e^{-b_{n}^{*}}.$

Moreover, (5) easily implies that $a_{n}^{*}/n$ converges to 0, which is hence
also the case of $b_{n}^{*}/n$. The lemma on quasiconvex sequences then
implies that, for all $n\geq 1$,

$e^{b_{n}^{*}}\leq\left(n\sqrt{\frac{2\varepsilon
e^{-a_{N}^{*}}}{2}}+1\right)^{2}$

that is,

$e^{a_{n+N}^{*}}\leq\left(n\sqrt{\varepsilon}+\sqrt{e^{a_{N}^{*}}}\right)^{2},$

which obviously gives

$\limsup_{n\to\infty}\frac{e^{a_{n}^{*}}}{n^{2}}\leq\varepsilon.$

In virtue of (11), this implies the claim in this case.

We are finally in position to conclude the proof of (the second half of) Lemma
2.1. Indeed, by Claim 1 above, if $y_{n}\geq B^{\prime}$ then $x_{n}$ must lie
in $[A,b]$. However, according to Claim 2, this may only happen finitely many
times, say for the powers $n_{1}<n_{2}<\ldots<n_{k}$. For each of these
$n_{i}$, the derivative $Df^{n_{i}}(x)$ converges to $1$ as $x$ goes to $b$.
Therefore, there exists $\delta$ such that any $x$ realizing the maximum of
$Df^{n_{i}}$ satisfies $x\leq b-\delta$. The number
$B:=\max\\{B^{\prime},f^{n_{k}}(b-\delta)\\}$ then satisfies all of our
requirements. $\hfill\square$

######  Remark 2.4.

The proof of Section 2.1 as well as that of the second case in Claim 2 above
are the only parts of the proof of our Main Theorem where we do not know how
to proceed without using Polterovich-Sodin’s lemma on quasiconvex sequences.
The problem is that we may be dealing with an infinitely flat fixed point $b$
and, in this situation, there seems to be no control of the generating vector
field $Y$ leading to a direct proof of subquadratic growth of derivatives
along iterates about it, despite the fact that it is topologically
contracting. See the example in Section 4 that illustrates this phenomenon.
Needless to say, this issue remains somewhat mysterious to us.

We are finally in position to finish the proof of the Main Theorem in the
present case. To do this, we consider again the the Szekeres vector field
$X=X_{L}$ associated to $f$ on the interval $L^{+}=[a,b)$. We have

$\Gamma^{n}_{+}(f)=Df^{n}(x_{n})=\frac{X(y_{n})}{X(x_{n})}.$ (17)

Also, we know that, for $x$ close to $a$,

$X(x)\sim\frac{1}{2}D^{2}f(a)\cdot(x-a)^{2}.$

Since $y_{n}$ belongs to $[A,B]$ for all $n$, Lemma 2.2 implies that

$x_{n}=f^{-n}(y_{n})\sim a+\frac{2}{n\,D^{2}f(a)}.$

Therefore,

$X(x_{n})\sim\frac{1}{2}D^{2}f(a)\left(\frac{2}{n\,D^{2}f(a)}\right)^{2}=\frac{2}{n^{2}\,D^{2}f(a)}.$

Introducing this into (17) and passing to the limit gives

$\limsup_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}\leq\lim_{n\to\infty}\frac{\sup_{y\in
L}X(y)}{n^{2}\,X(f^{-n}(y_{n}))}\leq\frac{1}{2}D^{2}f(a)\cdot\sup_{y\in
L}X(y).$ (18)

Now recall that the vector field $X$ continuously extends to the closure of
$L$ by letting $X(b)=0$ (see [Yo95, Chapter IV]). In particular, there exists
a point $y^{+}\in L^{\mathrm{o}}$ such that $X(y^{+})=\sup_{y\in L}X(y)$. If
we let $x_{n}^{+}:=f^{-n}(y^{+})$, then

$\Gamma_{+}^{n}(f)\geq
Df^{n}(x_{n}^{+})=\frac{X(f^{n}(x_{n}^{+}))}{X(x^{+}_{n})}=\frac{X(y^{+})}{X(x_{n}^{+})}=\frac{\sup_{y\in
L}X(y)}{X(x_{n}^{+})}.$

Again,
$x_{n}^{+}=f^{-n}(y^{+})=a+\frac{2}{n\,D^{2}f(a)}+o\big{(}\frac{1}{n}\big{)}$
and $X(x_{n}^{+})\sim\frac{2}{n^{2}\,D^{2}f(a)}$, which implies that

$\limsup_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}\geq\frac{1}{2}D^{2}f(a)\cdot\sup_{y\in
L}X(y).$ (19)

Finally, putting (18) and (19) together yields (9).

##  3 A proof of the Main Theorem: interior fixed points

Having proved that the Main Theorem is valid for diffeomorphisms with no
interior fixed points, we next proceed to prove it in general. Again, we will
systematically use Polterovich-Sodin’s lemma on quasiconvex sequences, though
one can avoid this using generating vector fields and Yoccoz’ continuity
theorem [Yo95, Chapter IV] together with further -though elementary-
computations.

We distinguish two cases, according to whether the diffeomorphism is
quadratically tangent to the identity at all fixed points or not.

###  3.1 The case of quadratic tangencies to the identity at all fixed points

Fix $\varepsilon>0$, denote again

$C^{\prime}:=|L|\cdot\left\|\frac{D^{2}f}{Df}\right\|_{{}_{\infty}},$

and let

$\delta:=\frac{2\varepsilon}{e^{C^{\prime}}\left\|\frac{D^{2}f}{Df}\right\|_{\infty}}.$

There are only finitely many components of $f$ of length $>\delta$. Denote
them by $J_{1},\ldots,J_{k}$. By our assumption, for the restriction
$f|_{J_{i}}$ of $f$ to each $J_{i}$, the result from Section 2.1 gives

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f|_{J_{i}})}{n^{2}}=0.$

This implies that there exists $N=N_{\varepsilon}$ such that

$\frac{Df^{n}(x)}{n^{2}}\leq\varepsilon\quad\mbox{ for all }n\geq N\mbox{ and
all }x\in J_{1}\cup\ldots\cup J_{k}.$ (20)

Now, if $J$ is any other component of $f$, then $|J|\leq\delta$. By the choice
of $\delta$, Lemma 1.1 gives

$Df^{n}(x)\leq(n\sqrt{\varepsilon}+1)^{2}\quad\mbox{ for all }n\geq 1\mbox{
and all }x\in J.$ (21)

Putting (20) and (21) together, we obtain

$\limsup_{n\to\infty}\frac{\sup_{x}Df^{n}(x)}{n^{2}}\leq\varepsilon.$

Since this holds for every $\varepsilon>0$, we conclude that
$\Gamma^{n}_{+}(f)/n^{2}$ converges to 0.

###  3.2 The case of a non-quadratic tangency to the identity at a fixed point

Here we assume that $f$ has at least one component $I_{*}$ with a
topologically repelling fixed point at which $f$ is not quadratically tangent
to the identity. By the case of no interior fixed points, we have

$\lim_{n\to\infty}\frac{\Gamma_{+}^{n}(f|_{I_{*}})}{n^{2}}=\frac{1}{2}D^{2}f(r_{I_{*}})\cdot\sup_{y\in
I_{*}}X_{I_{*}}(y)>0.$

Fix $\varepsilon>0$ such that

$\varepsilon<\frac{1}{2}D^{2}f(r_{I_{*}})\cdot\sup_{y\in I_{*}}X_{I_{*}}(y),$
(22)

and define again

$C^{\prime}:=|L|\cdot\left\|\frac{D^{2}f}{Df}\right\|_{{}_{\infty}},\qquad\delta:=\frac{2\varepsilon}{e^{C^{\prime}}\left\|\frac{D^{2}f}{Df}\right\|_{\infty}}.$

As in the previous section, there are finitely many components of $f$ of
length $>\delta$, say $J_{1},\ldots,J_{k}$. On any other component $J$,
inequality (21) holds, which means that

$Df^{n}(x)\leq\left(n\sqrt{\varepsilon}+1\right)^{2}\quad\mbox{for all
}x\notin J_{1}\cup\ldots\cup J_{k}\mbox{ and all }n\geq 1.$ (23)

In particular, by the case of diffeomorphisms with no interior fixed points,
for each such $J$ we have

$\frac{D^{2}f(r_{J})}{2}\cdot\sup_{y\in
J}X_{f,J}(y)=\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f|_{J})}{n^{2}}\leq\varepsilon.$
(24)

Since

$\lim_{n\to\infty}\frac{\Gamma_{+}^{n}(f|_{I_{*}})}{n^{2}}>\varepsilon,$

(23) implies that

$\lim_{n\to\infty}\frac{\Gamma_{+}^{n}(f)}{n^{2}}=\max_{1\leq i\leq
k}\lim_{n\to\infty}\frac{\Gamma_{+}^{n}(f|_{J_{i}})}{n^{2}}.$

By the case with no interior fixed points, this gives

$\lim_{n\to\infty}\frac{\Gamma_{+}^{n}(f)}{n^{2}}=\max_{1\leq i\leq
k}\left[\frac{1}{2}D^{2}f(r_{J_{i}})\cdot\sup_{y\in
J_{i}}X_{J_{i}}(y)\right].$

In virtue of (22) and (24), this finally give

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{2}}=\sup_{I\in\mathcal{C}(f)}\left[\frac{1}{2}D^{2}f(r_{I})\cdot\sup_{y\in
I}X_{I}(y)\right].$

##  4 An example

To better explain the pertinence of the example that follows, let us first
state a result that deals with non infinitely flat fixed points, with the
constraint that the topologically repelling fixed point is (at least)
quadratically tangent to the identity. For simplicity, we state it for
$C^{\infty}$ diffeomorphisms, yet it easily extends to $C^{2}$ diffeomorphisms
with tame behavior near the fixed points, according to the hypothesis below.
Also, we will only consider diffeomorphisms with no interior fixed point,
leaving the treatment of the general case to the reader. The proposition below
is very close to a result from [Bo04], yet slightly more general. Note that it
still holds for $k=1$, where it becomes a particular case of the Main Theorem.

######  Proposition 4.1.

Let $f$ be a $C^{\infty}$ diffeomorphism of a closed interval $L$ with no
fixed point inside. Suppose that $f$ is tangent to the identity exactly up to
order $k\geq 2$ at the topologically repelling fixed point, and that it is not
infinitely flat at the topologically contracting fixed point. Then

$\lim_{n\to\infty}\frac{\Gamma^{n}_{+}(f)}{n^{\frac{k+1}{k}}}=\sqrt[k]{\frac{k^{k+1}\cdot
Df^{k+1}(r_{L})}{(k+1)!}}\cdot\sup_{y\in L}X_{L}(y).$

Sketch of Proof. Up to two key steps, the proof is very similar to that of the
Main Theorem. To fix ideas, we assume that $f$ sends all the interior points
of $L:=[a,b]$ to the right, hence $r_{L}=a$.

  * •

The first difference concerns the asymptotic of the orbits (of the inverse
map) near the topologically repelling fixed point. Namely, in this case, for
all $y\in(a,b)$ one has

$\lim_{n\to\infty}n^{\frac{1}{k}}\,(f^{-n}(y)-a)=\sqrt[k]{\frac{(k+1)\cdot(k-1)!}{Df^{k+1}(a)}},$

and the convergence is uniform on compact subsets of $(a,b)$. The proof is
similar to that of Lemma 2.2. One first reduces to study germs of the form
$g(x)=x-sx^{k+1}+o(x^{k+1})$, with $s>0$. Then, one conjugates by
$I_{k}(z):=1/z^{\frac{1}{k}}$ to obtain that $G:=I\circ g\circ I^{-1}$ is of
the form $G(z)=z+sk+o(1)$, from where one easily deduces the announced
asymptotic.

  * •

The second difference lies in the growth of the derivatives of the iterates
near the topologically contracting fixed point. Indeed, in a neighborhood of
this point, the derivative must be strictly increasing, hence smaller than 1
in its interior. This allows showing that Lemma 2.1 still holds, with a much
simpler proof for the second half of it. Namely, for a point $x_{n}$ that
realizes $\Gamma_{+}^{n}(f)$, the point $y_{n}:=f^{n}(x_{n})$ cannot lie on
$[f(B),b]$ if $Df$ is strictly smaller than 1 on $(B,b)$ and $n$ is large
enough. Indeed, if $y_{n}>f(B)$ and $Df$ is strictly larger than 1 on
$(a,x_{n})$, then $Df^{n}(f^{-1}(x_{n}))>Df^{n}(x_{n})$.

Using these two facts, the proof is concluded via the very same final
argument. We leave the details to the reader. $\hfill\square$

What we next show is that, quite surprisingly, the proposition above does no
longer hold without the hypothesis made for the topologically contracting
fixed point. The example we present is actually the inverse of the map
constructed by Watanabe in [Wa04]. The interesting fact is that this inverse
map still gives strong (yet subquadratic) infinitesimal expansion along the
iterates in certain regions close to the topologically contracting fixed
point, which is somewhat counter-intuitive. For the sake of clearness, we
state this as a proposition.

######  Proposition 4.2.

For each $k\geq 2$, there exists a $C^{\infty}$ diffeomorphism $f$ of the
interval with no fixed point in the interior and such that:

  * •

it is infinitely flat at the topologically contracting fixed point;

  * •

$k$ is its exact order of tangency to the identity at the topologically
repelling fixed point;

  * •

for every $0<\tau<2$, one has

$\limsup_{n\rightarrow\infty}\frac{\Gamma^{n}_{+}(f)}{n^{\tau}}=\infty.$

We give a proof for this for completeness. First, as it was discussed above,
what we need to build is a germ of a topologically contracting $C^{\infty}$
diffeomorphism that is infinitely flat at the fixed point yet there exists a
sequence of points $x_{n}$ converging to this point such that, for all
$\tau<2$,

$\limsup_{n\to\infty}\frac{Df^{n}(x_{n})}{n^{\tau}}=\infty.$ (25)

Indeed, one can easily extend this germ to a $C^{\infty}$ diffeomorphism of a
closed interval having no interior fixed point and with a tangency to the
identity of exact order $k$ at the other fixed point. By the previous
discussion, such a diffeomorphism satisfies the properties of the proposition
above.

For simplicity of the computations, our germ will be defined around the
origin. We consider the vector field $X(x)=Y(x)+Z(x)$, where for small $x>0$
we let

$Y(x):=\frac{1}{2}\left(\cos\frac{1}{x}-1\right)\exp\left(-\frac{1}{x}\right),\qquad
Z(x):=-\exp\left(-\exp\left(\frac{3}{x}\right)\right).$

We extend them to the origin by letting $X(0)=Y(0)=Z(0)=0$. We leave to the
reader to check that $X,Y,Z$ are smooth and infinitely flat at the origin, and
that $X(x)<0$ for small $x>0$. (Note that these are the negative of the vector
fields considered in [Wa04].) We denote by $f^{t}$ (resp. $h^{t}$) the flow
associated to $X$ (resp. $Y$). We claim that there exist a sequence $(t_{k})$
of times that goes to infinity and a sequence $(c_{k})$ of points converging
to the origin such that, for large-enough $k$, one has

$Df^{t_{k}}(c_{k})\geq\frac{1}{2}\cdot\frac{{t_{k}^{2}}}{\log(t_{k})}.$ (26)

Although the times $t_{k}$ are not necessarily integers, the $C^{1}$
continuity of the flow of $X$ implies (25) for the germ of diffeomorphism
$f:=f^{1}$ just replacing $t_{k}$ by its integer part.

We are thus reduced to show (26). We start by giving a lower bound for the
derivative of the flow associated to $Y$. As before, we denote $h:=h^{1}$.

Claim 1: There exists a sequence of times $(v_{k})$ that goes to infinity and
a sequence of points $(c_{k})$ converging to the origin such that

$Dh^{v_{k}}(c_{k})\geq\frac{v_{k}^{2}}{\log(v_{k})}.$ (27)

To show this, for each $k\geq 1$ we let $a_{k}:=\frac{1}{2\pi k}$. One readily
checks that

$Y(a_{k})=0,\qquad DY(a_{k})=0,\qquad D^{2}Y(a_{k})=-\frac{1}{2}(2\pi
k)^{4}\exp(-2\pi k)<0.$

We denote $\varepsilon_{k}:=D^{2}Y(a_{k}).$ A straightforward computation
shows that there exists $M\in(0,2\pi)$ such that, if we denote
$b_{k}:=\frac{1}{2\pi k+M}<a_{k}$, then for each $k\geq 1$ and all
$x\in[b_{k},a_{k}]$, it holds

$2\varepsilon_{k}\leq D^{2}Y(x)\leq\frac{1}{2}\varepsilon_{k}.$ (28)

For each $k\geq 1$, we consider the vector fields $\bar{Y}_{k},\tilde{Y}_{k}$
around $a_{k}$ given by $\bar{Y}_{k}(x):=\varepsilon_{k}(x-a_{k})^{2}$ and
$\tilde{Y}_{k}(x):=\frac{\varepsilon_{k}}{4}(x-a_{k})^{2}.$ By (28), for all
$x\in[b_{k},a_{k}]$,

$\bar{Y}_{k}(x)\leq Y(x)\leq\tilde{Y}_{k}(x).$ (29)

Note that these vector fields induce affine flows, that we denote
$\bar{h}^{t}$ and $\tilde{h}^{t}$ accordingly. In concrete terms,

$\bar{h}^{t}(x)=a_{k}+\frac{x-a_{k}}{1-\varepsilon_{k}t(x-a_{k})},\qquad
D\bar{h}^{t}(x)=\frac{1}{(1-\varepsilon_{k}t(x-a_{k}))^{2}}.$ (30)

In particular,

$D\bar{h}^{t}(h^{-t}(b_{k}))=\frac{1}{D\bar{h}^{-t}(b_{k})}=(1+\varepsilon_{k}t(b_{k}-a_{k}))^{2}\geq\varepsilon_{k}^{2}(a_{k}-b_{k})^{2}t^{2}.$
(31)

For each $k\geq 1$, we consider the times

$u_{k}:=\exp\left(\frac{1}{4\varepsilon_{k}^{2}(a_{k}-b_{k})^{2}}\right)$ (32)

and we let $c_{k}\in[b_{k},a_{k}]$ be defined as
$c_{k}:=\bar{h}^{-u_{k}}(b_{k})$. Since
$\varepsilon_{k}^{2}(a_{k}-b_{k})^{2}\rightarrow 0$ as $k$ goes to infinity,
we have that $u_{k}\rightarrow\infty$ as $k$ goes to infinity. For each $k\geq
1$, we let $v_{k},w_{k}$ be the (unique) times for which

$h^{v_{k}}(c_{k})=\tilde{h}^{w_{k}}(c_{k})=b_{k}.$

Observe that (29) implies that $w_{k}\leq v_{k}\leq u_{k}.$

We can now estimate the derivative of the flow $h^{t}$. Indeed, using the
definition of $c_{k}$ and inequality (29), we obtain

$Dh^{v_{k}}(c_{k})=\frac{Y(h^{v_{k}}(c_{k}))}{Y(c_{k})}=\frac{Y(b_{k})}{Y(c_{k})}\geq\frac{\bar{Y}_{k}(b_{k})}{\tilde{Y}_{k}(c_{k})}=4\cdot\frac{\bar{Y}_{k}(\bar{h}^{u_{k}}(c_{k}))}{\bar{Y}_{k}(c_{k})}=4\cdot
D\bar{h}^{u_{k}}(c_{k}).$ (33)

The right-side expression equals
$4\\!\cdot\\!D\bar{h}^{u_{k}}(\bar{h}^{-u_{k}}(b_{k}))$, which, by (31) and
(32), satisfies

$4\cdot D\bar{h}^{u_{k}}(\bar{h}^{-u_{k}}(b_{k}))\geq
4\cdot\varepsilon_{k}^{2}(b_{k}-a_{k})^{2}\cdot
u_{k}^{2}.=\frac{u_{k}^{2}}{\log u_{k}}.$ (34)

Finally, since $u_{k}\geq v_{k}$ and the function $x\mapsto\frac{x^{2}}{\log
x}$ is increasing, from (33) and (34) we deduce that

$Dh^{v_{k}}(c_{k})\geq\frac{v_{k}^{2}}{\log(v_{k})}.$ (35)

This shows (27).

Claim 2: For sufficiently large $k$, we have that $Y(c_{k})\leq Z(c_{k})$.

Observe that (29) implies that

$Y(c_{k})\geq\bar{Y}_{k}(c_{k})=\varepsilon_{k}(a_{k}-c_{k})^{2}.$ (36)

By the definitions of $\varepsilon_{k}$ and $u_{k}$, we have that
$\varepsilon_{k}\rightarrow 0$ and $u_{k}(a_{k}-b_{k})\rightarrow\infty$ as
$k$ goes to infinity. In particular, for large-enough $k$,

$u_{k}(a_{k}-b_{k})(1+\varepsilon_{k})\geq 1.$

By (30), for large-enough $k$, this yields

$a_{k}-c_{k}=a_{k}-\bar{h}^{-u_{k}}(b_{k})=-\frac{(b_{k}-a_{k})}{1+\varepsilon_{k}u_{k}(b_{k}-a_{k})}\geq\frac{1}{u_{k}}\geq\frac{1}{\exp(\exp(4\pi
k))}.$

Using (36), the definition of the vector field $Z$ and the fact that
$c_{k}<a_{k},$ one finally obtains

$Y(c_{k})\leq\frac{\varepsilon_{k}}{\exp(2\exp(4\pi
k))}\leq\frac{-1}{\exp(\exp(6\pi k))}=Z(a_{k})\leq Z(c_{k}).$

Claim 3: If we let $t_{k}$ be defined so that $f^{t_{k}}(c_{k})=b_{k}$, then
(26) holds for sufficiently large $k$.

As before, we have that $t_{k}\leq v_{k}$ and $t_{k}\rightarrow\infty$ as $k$
goes to infinity. Using (35) and the previous claim, we can estimate the
derivative of the flow of $X$ as follows:

$Df^{t_{k}}(c_{k})=\frac{X(b_{k})}{X(c_{k})}\geq\frac{1}{2}\cdot\frac{Y(b_{k})}{Y(c_{k})}=\frac{1}{2}\cdot
Dh^{v_{k}}(c_{k})\geq\frac{v_{k}^{2}}{2\cdot\log
v_{k}}\geq\frac{t_{k}^{2}}{2\cdot\log t_{k}}.$

##  5 Some questions

It is a consequence of the Main Theorem (actually, of Polterovich-Sodin’s
result) that the growth of the second derivative of a $C^{2}$ diffeomorphism
of a closed interval having only parabolic fixed points is at most polynomial
of degree 5. Indeed, using the affine derivative $D^{2}/D$ and its chain rule

$\frac{D^{2}(f\circ g)}{D(f\circ g)}=\frac{D^{2}f}{Df}\circ g\cdot
Dg+\frac{D^{2}g}{Dg},$

we obtain

$\frac{D^{2}f^{n}}{Df^{n}}(x)=\sum_{i=0}^{n-1}\frac{D^{2}f}{Df}(f^{i}(x))\cdot
Df^{i}(x).$

Therefore, if $\|Df^{n}\|_{\infty}\leq Cn^{2}$ holds for all $n\geq 1$, then

$\left\|\frac{D^{2}f^{n}}{Df^{n}}\right\|_{\infty}\leq\left\|\frac{D^{2}f}{Df}\right\|_{\infty}\cdot\left(1+\sum_{i=1}^{n-1}Ci^{2}\right)\leq
C^{\prime}n^{3}\left\|\frac{D^{2}f}{Df}\right\|_{\infty},$ (37)

hence

$\left\|D^{2}f^{n}\right\|_{\infty}\leq
C^{\prime\prime}n^{3}\|Df^{n}\|_{\infty}\leq C^{\prime\prime\prime}n^{5}.$

For a $C^{3}$ diffeomorphisms we can imitate the computations above using the
Schwarzian derivative $S:=D^{3}/D-3/2\,(D^{2}/D)^{2}$ and its chain rule

$S(fg)=S(f)\circ g\cdot(Dg)^{2}+S(g).$

Indeed, this gives

$S(f^{n})=\sum_{i=0}^{n-1}S(f)\circ f^{i}\cdot(Df^{i})^{2},$

hence

$\|S(f^{n})\|_{\infty}\leq\|S(f)\|_{\infty}\cdot\left(1+\sum_{i=1}^{n-1}(Ci^{2})^{2}\right)\leq\bar{C}^{\prime}n^{5}.$

Using (37) and the the definition of $S$, this easily implies that

$\|D^{3}f^{n}\|_{\infty}\leq\bar{C}^{\prime\prime}n^{7}.$

We do not know whether the estimates above are sharp (we suspect they are
not). We state all of this as part of a more general question.

Question 1. Given a $C^{\infty}$ diffeomorphisms of the closed interval having
only parabolic fixed points and an integer $k\geq 2$, what is the maximal
possible growth for $\|D^{k}f^{n}\|_{\infty}$ as $n$ goes to infinite ? In
which cases it is achieved ?

Note that, by combining Polterovich-Sodin’s estimate with Arbogast’s
formula111This is mostly known as Faà di Bruno formula; see [Cr05] on this.,
one easily concludes by induction that the growth above is at most polynomial.
However, it is unclear whether the degree thus detected is optimal.

It is worth mentioning that looking at norms different from
$\|\cdot\|_{\infty}$ is also pertinent. For example, a change of variable in
the chain rule for the affine derivative immediately yields222Note that, since
$\frac{D^{2}f}{Df}=D(\log Df)$, we have
$\|\frac{D^{2}f}{Df}\|_{{}_{1}}=\mathrm{var}(\log Df)$, hence the triangle
inequality (38) is nothing but Denjoy’s inequality $\mathrm{var}(\log D(f\circ
g))\leq\mathrm{var}(\log Df)+\mathrm{var}(\log Dg)$.

$\left\|\frac{D^{2}(f\circ g)}{D(f\circ
g)}\right\|_{1}\leq\left\|\frac{D^{2}f}{Df}\right\|_{1}+\left\|\frac{D^{2}g}{Dg}\right\|_{1},$
(38)

hence $\|\frac{D^{2}f^{n}}{Df^{n}}\|_{{}_{1}}$ grows at most linearly.
Actually, there is a nice description of those $f$ for which this growth is
sublinear: they must have only parabolic fixed points and must arise as the
time-1 map of a $C^{1}$ vector field [EN21].

If we go to the third derivative along these ideas, a good candidate to
replace the Schwarzian derivative333The explicit relation comes from the
equality $Sf(x)=6\,\lim_{y\to x}c(f)(x,y)$, which holds for every $C^{3}$
diffeomorphism $f$. of a diffeomorphism of a closed interval $L$ is the
Liouville cocycle, which is the function $c(f)$ defined on $L\times L$ as

$c(f)(x,y):=\frac{Df(x)\,Df(y)}{(f(x)-f(y))^{2}}-\frac{1}{(x-y)^{2}}.$

Note that this is defined even if $f$ is $C^{1}$. However, one can ensure that
$c(f)$ belongs to $L^{1}(I\times I)$ only if $f$ is of class
$C^{2+\varepsilon}$ for some $\varepsilon>0$; see [Na06]. This cocycle
satisfies the chain rule

$c(f\circ g)(x,y)=c(f)(g(x),g(y))\cdot Dg(x)\,Dg(y)+c(g)(x,y),$

which after taking $L^{1}$ norms gives, via a change of variable,

$\|c(f\circ g)\|_{1}\leq\|c(f)\|_{1}+\|c(g)\|_{1}.$

As a consequence, $\|c(f^{n})\|_{1}$ grows at most linearly. We do not know
any characterization of those $f$ for which this grow is sublinear.

Acknowledgments. The authors are indebted to Hélène Eynard-Bontemps for her
interest on this article and several valuable hints and remarks. This work was
supported by the ECOS project 23003 “Small spaces under action”. The authors
also acknowledge support of the Institut Henri Poincaré (UAR 839 CNRS-Sorbonne
Université) and LabEx CARMIN (ANR-10-LABX-59-01); in particular, the first-
named author was funded by the CIMPA-CARMIN program.

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Leonardo Dinamarca Opazo & Andrés Navas

Universidad de Santiago de Chile

Alameda 3363, Santiago, Chile

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