# Smooth Linearization of Nonautonomous Coupled Systems

Lucas Backes  Departamento de Matemática, Universidade Federal do Rio Grande
do Sul, Av. Bento Gonçalves 9500, CEP 91509-900, Porto Alegre, RS, Brazil.
<EMAIL_ADDRESS>and  Davor Dragičević  Faculty of Mathematics,
University of Rijeka, Croatia<EMAIL_ADDRESS>
###### Abstract.

In a joint work with Palmer [3] we have formulated sufficient conditions under
which there exist continuous and invertible transformations of the form
$H_{n}(x,y)$ taking solutions of a coupled system

$x_{n+1}=A_{n}x_{n}+f_{n}(x_{n},y_{n}),\quad y_{n+1}=g_{n}(y_{n}),$

onto the solutions of the associated partially linearized uncoupled system

$x_{n+1}=A_{n}x_{n},\quad y_{n+1}=g_{n}(y_{n}).$

In the present work we go one step further and provide conditions under which
$H_{n}$ and $H_{n}^{-1}$ are smooth in one of the variables $x$ and $y$. We
emphasise that our conditions are of a general form and do not involve any
kind of dichotomy, nonresonance or spectral gap assumptions for the linear
part which are present on most of the related works.

###### Key words and phrases:

linearization; differentiability; nonautonomous coupled systems

###### 2020 Mathematics Subject Classification:

37C15

##  1\. Introduction

###  1.1. Autonomous linearization

The classical Grobman-Hartman theorem [15, 16, 17, 18] is one of the most
important results in the qualitative theory of dynamical systems and
differential equations. It asserts that a $C^{1}$-diffeomorphism of
$\mathbb{R}^{n}$ is topologically equivalent (or conjugated) to the associated
linear part on a neighborhood of a hyperbolic fixed point. This result was
later generalized to Banach spaces independently by Palis [22] and Pugh [24].
It is well-known that in general, the conjugacy is only locally Hölder
continuous [29] and that it can fail to exhibit higher regularity even for
$C^{\infty}$ dynamics. On the other hand, topological conjugacy is often not
sufficient since for example it can fail to distinguish a node from a focus as
pointed out by van Strien [32].

Therefore, it was important to study the problem of formulating sufficient
conditions that would ensure that the conjugacy in the Grobman-Hartman theorem
exhibits higher regularity properties. The first results in this direction are
due to Sternberg [30, 31] who proved that $C^{k}$ ($k\geq 1$) diffeomorphisms
can be $C^{r}$ linearized near hyperbolic fixed points, where the integer
$r>k$ depends on $k$ and so-called nonresonant conditions. Thus, in order to
obtain $C^{r}$-linearization his results required that the dynamics exhibits
higher regularity. Later Belitskii [4, 5] gave conditions for $C^{k}$
linearization of $C^{k,1}$ ($k\geq 1$) diffeomorphisms under appropriate
nonresonant conditions. His results was partially generalized to infinite-
dimensional setting in [14, 27, 36]. For other relevant results in this
direction, we refer to [13, 28, 34, 35] and references therein.

###  1.2. Nonautonomous topological linearization

So far, we have only discussed the case of autonomous dynamics. The first
version of the Grobman-Hartman theorem for nonautonomous dynamics was obtained
by Palmer [23]. Let us formulate the version of his result in the case of
discrete time which was obtained in [1]. Assume that
$(A_{n})_{n\in\mathbb{Z}}$ is a sequence of bounded and invertible linear
operators acting on a Banach space $X=(X,|\cdot|)$. Furthermore, suppose that
$f_{n}\colon X\to X$, $n\in\mathbb{Z}$ is a sequence of (nonlinear) maps. We
can consider the associated nonlinear and nonautonomous difference equation

$x_{n+1}=A_{n}x_{n}+f_{n}(x_{n})\quad n\in\mathbb{Z},$ (1)

as well as its linear part given by

$x_{n+1}=A_{n}x_{n}\quad n\in\mathbb{Z}.$ (2)

Assume that the following conditions hold:

  * •

the sequence $(A_{n})_{n\in\mathbb{Z}}$ admits an exponential dichotomy (see
[9]);

  * •

there exist $M,c>0$ such that $|f_{n}(x)|\leq M$ and

$|f_{n}(x)-f_{n}(y)|\leq c|x-y|,$

for $n\in\mathbb{Z}$ and $x,y\in X$.

Then, provided that $c$ is sufficiently small, equations (1) and (2) are
topologically equivalent. Moreover, under additional mild conditions the
conjugacies are locally Hölder continuous [29].

More recently, several authors obtained extensions of Palmer’s theorem by
weakening the hyperbolicity requirement for the linear part (2), which also
required changing the assumptions dealing with nonlinear terms $f_{n}$. We
refer to [2, 19, 20, 21, 25, 26] and references therein.

###  1.3. Nonautonomous linearization: higher regularity of conjugacies

Motivated by previously described results dealing with autonomous dynamics, in
recent years several authors dealt with the problem of formulating sufficient
conditions under which the conjugacies in the nonautonomous linearization
exhibit higher regularity.

To the best of our knowledge, the first results in this direction are those in
[6] (see also [7]) dealing with the case when the linear part admits an
exponential contraction. In [10], the authors have established the
nonautonomous versions of Sternberg’s results by formulating sufficient
conditions under which the conjugacies are of order $C^{r}$, $r\geq 1$. These
conditions require that the linear part (2) admits an exponential dichotomy
and that appopriate nonresonant conditions expressed in terms of the so-called
Sacker-Sell spectrum hold. Similar in nature (although different in
techniques) are the result established in [11, 12] which give conditons for
$C^{1}$-linearization of nonautonomous systems. There, it is required that (2)
admits a nonuniform exponential dichotomy and that appropriate spectral-gap
conditions hold.

Finally, we mention the recent paper [8] in which the authors formulate
sufficient conditions for $C^{1}$-linearization of nonautonomous discrete
systems. The main novelties when compared with the previous works are the
following:

  * •

(2) is assumed to admit a nonuniform dichotomy which is not necessarily
exponential;

  * •

no nonresonance or spectral-gap conditions are required.

The main idea of [8] is that the lack of hyperbolicity and spectral data can
be compensated by appropriate assumptions related to the size of nonlinear
terms $f_{n}$ in (1). However, it should be emphasized that when restricted to
the setting of (for example) [11], the result in [8] can fail to be applicable
or could yield weaker results from that in [11]. We refer to Remark 3.10 for a
detailed discussion.

###  1.4. Contributions of the present paper

In the present paper (following [3, 29, 33]), we study the coupled
nonautonomous and nonlinear system

$x_{n+1}=A_{n}x_{n}+f_{n}(x_{n},y_{n}),\quad y_{n+1}=g_{n}(y_{n}),$ (3)

as well as the associate uncoupled system

$x_{n+1}=A_{n}x_{n},\quad y_{n+1}=g_{n}(y_{n}).$ (4)

Here, $(A_{n})_{n\in\mathbb{Z}}$ is a sequence of bounded and invertible
linear operators on a Banach space $X$, $g_{n}\colon Y\to Y$, $n\in\mathbb{Z}$
is a sequence of homeomorphisms of a Banach space $Y$, and $f_{n}\colon
X\times Y\to X$, $n\in\mathbb{Z}$ is a sequence of nonlinear maps. Observe
that when $Y$ is a trivial Banach space, (3) and (4) reduce to (1) and (2),
respectively.

In [3], the authors have formulated very general conditions under which (3)
and (4) are topologically conjugated. The goal of the present paper is to
formulate conditions under which these conjugacies are smooth. Our approach is
inspired by that in the previously described paper by Castañeda and Jara [8].
We stress that our arguments require a nontrivial modification of those in [8]
as those need to be combined with techniques developed in [3]. In addition, we
do not impose any dichotomy conditions for the linear part in (4).

##  2\. Preliminaries

Let $(X,|\cdot|_{X})$ and $(Y,|\cdot|_{Y})$ be two arbitrary Banach spaces.
For the sake of simplicity both norms $|\cdot|_{X}$ and $|\cdot|_{Y}$ will be
denoted simply by $|\cdot|$. By $\mathcal{B}(X)$ we denote the space of all
bounded operators on $X$ equipped with the operator norm, which we will also
denote by $|\cdot|$.

Let $(A_{n})_{n\in\mathbb{Z}}$ be a sequence of invertible operators in
$\mathcal{B}(X)$ and $f_{n}\colon X\times Y\to X$, $n\in\mathbb{Z}$ a sequence
of maps. Furthermore, let $(g_{n})_{n\in\mathbb{Z}}$ be a sequence of
homeomorphisms on $Y$. We consider the associated coupled system

$x_{n+1}=A_{n}x_{n}+f_{n}(x_{n},y_{n}),\quad y_{n+1}=g_{n}(y_{n})$ (5)

as well the uncoupled system

$x_{n+1}=A_{n}x_{n},\quad y_{n+1}=g_{n}(y_{n}).$ (6)

Let $k\mapsto(x_{1}(k,n,x),y(k,n,y))$ denote the solution of (6) which equals
$(x,y)$ when $k=n$. Similarly, let $k\mapsto(x_{2}(k,n,x,y),y(k,n,y))$ be the
solution of (5) with value $(x,y)$ at $k=n$.

For $m,n\in\mathbb{Z}$, set

$\mathcal{A}(m,n)=\begin{cases}A_{m-1}\cdots A_{n}&\text{for $m>n$;}\\\
\operatorname{Id}&\text{for $m=n$;}\\\ A_{m}^{-1}\cdots A_{n-1}^{-1}&\text{for
$m<n$,}\end{cases}$

where $\operatorname{Id}$ denotes the identity operator on $X$. Let
$(P_{n})_{n\in\mathbb{Z}}$ be a sequence in $\mathcal{B}(X)$. We emphasize
that the maps $P_{n}$ do not need to be projections. Then we define

$\mathcal{G}(m,n)=\begin{cases}\mathcal{A}(m,n)P_{n}&\text{for $m\geq n$;}\\\
-\mathcal{A}(m,n)(\operatorname{Id}-P_{n})&\text{for $m<n$.}\end{cases}$ (7)

###  2.1. Hypothesis

In order to facilitate future reference, we gather in this subsection all the
hypothesis we are going to use in our results. We group them into two
categories: basic and advanced conditions.

####  2.1.1. Basic conditions:

  1. (BC1)

there exist sequences $(\mu_{n})_{n\in\mathbb{Z}}$ and
$(\gamma_{n})_{n\in\mathbb{Z}}$ in $[0,\infty)$ such that

$|f_{n}(x,y)|\leq\mu_{n}\quad\text{and}\quad|f_{n}(x_{1},y)-f_{n}(x_{2},y)|\leq\gamma_{n}|x_{1}-x_{2}|,$

for $n\in\mathbb{Z}$, $x,x_{1},x_{2}\in X$ and $y\in Y$;

  2. (BC2)

$N:=\sup_{m\in\mathbb{Z}}\sum^{\infty}_{n=-\infty}|\mathcal{G}(m,n)|\mu_{n-1}<\infty;$

  3. (BC3)

$q:=\sup_{m\in\mathbb{Z}}\sum^{\infty}_{n=-\infty}|\mathcal{G}(m,n)|\gamma_{n-1}<1;$

  4. (BC4)

$|A_{n}^{-1}|\gamma_{n}<1$ for every $n\in\mathbb{Z}$.

The basic conditions (BC1), (BC2), (BC3) and (BC4) ensure that the systems (3)
and (4) are topologically equivalent (note that condition (BC4) guarantees
that the solutions of (3) are defined globally). More precisely, we have the
following result.

######  Theorem 2.1 (Theorem 3.1 of [3]).

Suppose that (BC1), (BC2), (BC3) and (BC4) are satisfied. Then, there exists a
sequence of continuous functions $H_{n}\colon X\times Y\to X\times Y$,
$n\in\mathbb{Z}$ of the form $H_{n}(x,y)=(x+h_{n}(x,y),y)$, where
$\sup_{n,x,y}|h_{n}(x,y)|<\infty$, such that if $n\mapsto(x_{n},y_{n})$ is a
solution of (6), then $n\mapsto H_{n}(x_{n},y_{n})$ is a solution of (5). In
addition, there exists a sequence of continuous functions $\bar{H}_{n}\colon
X\times Y\to X\times Y$, $n\in\mathbb{Z}$ of the form
$\bar{H}_{n}(x,y)=(x+\bar{h}_{n}(x,y),y)$, where
$\sup_{n,x,y}|\bar{h}_{n}(x,y)|<\infty$, such that if $n\mapsto(x_{n},y_{n})$
is a solution of (5), then $n\mapsto\bar{H}_{n}(x_{n},y_{n})$ is a solution of
(6). Moreover, $H_{n}$ and $\bar{H}_{n}$ are inverses of each other, that is,

$H_{n}(\bar{H}_{n}(x,y))=(x,y)=\bar{H}_{n}(H_{n}(x,y)),$

for $n\in\mathbb{Z}$ and $(x,y)\in X\times Y$.

Furthermore, under some additional assumptions it was proved in [3] that the
conjugacies $H_{n}(x,y)$ and $\bar{H}_{n}(x,y)$ are Hölder continuous with
respect to the variables $x$ and $y$. Our objective now is to formulate
sufficient conditions under which those maps exhibit higher regularity
properties. For this purpose we introduce the next set of conditions.

####  2.1.2. Advanced conditions:

  1. (AC1)

for every $y\in Y$, the map $x\to f_{n}(x,y)$ is $C^{r}$, $r\geq 1$, for every
$n\in\mathbb{Z}$;

  2. (AC2)

given $n\in\mathbb{Z}$,

$K_{n}:=\sum_{k<n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\right)<+\infty$

and

$J_{n}:=\sum_{k>n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})\right)<+\infty;$

  3. (AC3)

given $n\in\mathbb{Z}$,

$K_{n}+J_{n}+|\mathcal{G}(n,n+1)|\gamma_{n}<1;$

  4. (AC4)

there exists a sequence $(\rho_{n})_{n\in\mathbb{Z}}$ in $[0,\infty)$ such
that

$|f_{n}(x,y_{1})-f_{n}(x,y_{2})|\leq\rho_{n}|y_{1}-y_{2}|,$

for $n\in\mathbb{Z}$, $x\in X$ and $y_{1},y_{2}\in Y$;

  5. (AC5)

there exist sequences $(\tau_{n})_{n\in\mathbb{Z}}$ and
$(\sigma_{n})_{n\in\mathbb{Z}}$ in $[0,\infty)$ such that

$|g_{n}(y_{1})-g_{n}(y_{2})|\leq\tau_{n}|y_{1}-y_{2}|\text{ and
}|g^{-1}_{n}(y_{1})-g^{-1}_{n}(y_{2})|\leq\sigma_{n}|y_{1}-y_{2}|,$

for every $n\in\mathbb{Z}$ and $y_{1},y_{2}\in Y$.

  6. (AC6)

$\sigma_{n}\rho_{n}\leq 1$ for every $n\in\mathbb{Z}$;

  7. (AC7)

the map $y\to f_{n}(x,y)$ is $C^{r}$, $r\geq 1$, for every $n\in\mathbb{Z}$;

  8. (AC8)

the map $y\to g_{n}(y)$ is a $C^{r}$-diffeomorphism, $r\geq 1$, for every
$n\in\mathbb{Z}$;

  9. (AC9)

given $n\in\mathbb{Z}$,

$\sum_{k\in\mathbb{Z}}|\mathcal{G}(n,k+1)|\left(\gamma_{k}M_{k,n}+\rho_{k}D_{k,n}\right)<+\infty$

where $D_{k,n}$ and $M_{k,n}$ are given by (27) and (28) respectively.

######  Remark 2.2.

The advanced conditions (AC1)-(AC9) will play a fundamental role in improving
the regularity of the topological conjugacies given by Theorem 2.1. More
precisely,

  * •

provided that (AC1) holds and that (AC2) is satisfied for $n\in\mathbb{Z}$, we
have that the conjugacy $\bar{H}_{n}$ given by Theorem 2.1 is $C^{1}$ in the
first variable (see Theorem 3.3). In addition, we can ensure that (AC2) holds
simultaneously for every $n\in\mathbb{Z}$ (see Example 3.5);

  * •

provided that (AC1) holds and that (AC3) is satisfied for $n\in\mathbb{Z}$, we
have that the conjugacy $H_{n}$ given by Theorem 2.1 is $C^{1}$ in the first
variable (see Theorem 3.4). In addition, we can ensure that (AC3) holds
simultaneously for every $n\in\mathbb{Z}$ (see Examples 3.6 and 3.7);

  * •

if (AC1), (AC4), (AC5), (AC6), (AC7), (AC8) hold and (AC9) is satisfied for
$n\in\mathbb{Z}$, then $\bar{H}_{n}$ is $C^{1}$ in the second variable (see
Theorem 3.16);

  * •

suppose that (AC1), (AC4), (AC5), (AC6), (AC7) and (AC8) hold and that (AC3)
and (AC9) are satisfied for $n\in\mathbb{Z}$. Then, $H_{n}$ is $C^{1}$ in the
second variable (see Theorem 3.17). In addition, we can ensure that (AC9)
holds simultaneously for every $n\in\mathbb{Z}$ (see Example 3.18).

##  3\. Main results

###  3.1. Regularity of conjugacies with respect to the first variable

We are interested in formulating sufficient conditions under which conjugacies
$H_{n}$ and $\bar{H}_{n}$ are smooth with respect to the first variable.

We start by studying the regularity properties of the map $\xi\to
x_{2}(k,n,\xi,\eta)$. Consider

$C_{k,n}=\begin{cases}\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})&\text{for
$k>n$;}\\\ 1&\text{for $k=n$;}\\\
\prod_{j=k}^{n-1}\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}&\text{for
$k<n$.}\end{cases}$

######  Lemma 3.1.

Suppose (BC1) and (BC4) are satisfied. Then, for any $\xi,\zeta\in X$,
$\eta\in Y$ and $n,k\in\mathbb{Z}$, we have that

$|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\zeta,\eta)|\leq C_{k,n}|\xi-\zeta|.$

###### Proof.

Fix $n\in\mathbb{Z}$. By (5) we have that

$\begin{split}x_{2}(n+1,n,\xi,\eta)&=A_{n}x_{2}(n,n,\xi,\eta)+f_{n}(x_{2}(n,n,\xi,\eta),y(n,n,\eta))\\\
&=A_{n}\xi+f_{n}(\xi,\eta).\\\ \end{split}$ (8)

Consequently, using (BC1),

$\begin{split}|x_{2}(n+1,n,\xi,\eta)-x_{2}(n+1,n,\zeta,\eta)|&=|A_{n}\xi+f_{n}(\xi,\eta)-A_{n}\zeta-
f_{n}(\zeta,\eta)|\\\
&\leq\left(|A_{n}|+\gamma_{n}\right)|\xi-\zeta|.\end{split}$

Thus, using that

$x_{2}(k,n,\xi,\eta)=x_{2}(k,m,x_{2}(m,n,\xi,\eta),y(m,n,\eta))\quad k\geq
m\geq n,$

and proceeding inductively we conclude that

$\begin{split}|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\zeta,\eta)|&\leq\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})|\xi-\zeta|\\\
&=C_{k,n}|\xi-\zeta|,\end{split}$

for every $k>n$.

We now consider the case when $k<n$. Given $(\xi,\eta)\in X\times Y$ and
$j\in\mathbb{Z}$, let us consider

$F_{j}(\xi,\eta)=A_{j}\xi+f_{j}(\xi,\eta)$

and

$T_{j}(\xi,\eta)=A_{j}^{-1}\xi-A_{j}^{-1}f_{j}(T_{j}(\xi,\eta),\eta).$ (9)

It is easy to check that $F_{j}(T_{j}(\xi,\eta),\eta)=\xi$. In addition, we we
have that

$\begin{split}T_{j}(F_{j}(\xi,\eta),\eta)&=A_{j}^{-1}F_{j}(\xi,\eta)-A_{j}^{-1}f_{j}(T_{j}(F_{j}(\xi,\eta),\eta))\\\
&=A_{j}^{-1}(A_{j}\xi+f_{j}(\xi,\eta))-A_{j}^{-1}f_{j}(T_{j}(F_{j}(\xi,\eta),\eta))\\\
&=\xi+A_{j}^{-1}f_{j}(\xi,\eta)-A_{j}^{-1}f_{j}(T_{j}(F_{j}(\xi,\eta),\eta)),\end{split}$

which implies that

$|T_{j}(F_{j}(\xi,\eta),\eta)-\xi|\leq|A_{j}^{-1}|\gamma_{j}|T_{j}(F_{j}(\xi,\eta),\eta)-\xi|.$

By (BC4), we conclude that $T_{j}(F_{j}(\xi,\eta),\eta)=\xi$. Moreover, since

$\begin{split}\xi&=A_{n-1}x_{2}(n-1,n,\xi,\eta)+f_{n-1}(x_{2}(n-1,n,\xi,\eta),y(n-1,n,\eta))\\\
&=F_{n-1}(x_{2}(n-1,n,\xi,\eta),y(n-1,n,\eta)),\end{split}$

it follows from the previous observations that

$T_{n-1}(\xi,y(n-1,n,\eta))=x_{2}(n-1,n,\xi,\eta).$ (10)

Consequently,

$\begin{split}&|x_{2}(n-1,n,\xi,\eta)-x_{2}(n-1,n,\zeta,\eta)|\\\
&=|T_{n-1}(\xi,y(n-1,n,\eta))-T_{n-1}(\zeta,y(n-1,n,\eta))|.\end{split}$

Now, using (BC1), for any $y\in Y$ we have that

$\begin{split}&|T_{n-1}(\xi,y)-T_{n-1}(\zeta,y)|\\\
&=|A_{n-1}^{-1}(\xi-\zeta)+A_{n-1}^{-1}\left(f_{n-1}(T_{n-1}(\zeta,y),y)-f_{n-1}(T_{n-1}(\xi,y),y)\right)|\\\
&\leq|A_{n-1}^{-1}||\xi-\zeta|+|A_{n-1}^{-1}|\gamma_{n-1}|T_{n-1}(\zeta,y)-T_{n-1}(\xi,y)|.\end{split}$

Therefore, using (BC4) we conclude that

$\begin{split}|T_{n-1}(\xi,y)-T_{n-1}(\zeta,y)|&\leq\frac{|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}|\xi-\zeta|,\end{split}$

which combined with the previous observations implies that

$\begin{split}|x_{2}(n-1,n,\xi,\eta)-x_{2}(n-1,n,\zeta,\eta)|&\leq\frac{|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}|\xi-\zeta|.\end{split}$

Consequently, by proceeding inductively as in the case when $k>n$, we conclude
that

$\begin{split}|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\zeta,\eta)|&\leq
C_{k,n}|\xi-\zeta|\end{split}$

for every $k<n$. The case when $k=n$ is obvious and therefore the proof is
complete. ∎

######  Proposition 3.2.

Assume that (BC1), (BC4) and (AC1) hold. Then, the map $\xi\to
x_{2}(k,n,\xi,\eta)$ is of class $C^{r}$, $r\geq 1$, for every
$n,k\in\mathbb{Z}$ and $\eta\in Y$.

###### Proof.

Fix $n\in\mathbb{Z}$. By (5), we have that

$\begin{split}x_{2}(n+1,n,\xi,\eta)&=A_{n}\xi+f_{n}(\xi,\eta).\\\ \end{split}$

Consequently, it follows from (AC1) that $\xi\to x_{2}(n+1,n,\xi,\eta)$ is of
class $C^{r}$. Thus, proceeding inductively, we conclude that $\xi\to
x_{2}(k,n,\xi,\eta)$ is of class $C^{r}$ for every $k>n$.

Next, we claim that the map $\xi\to T_{j}(\xi,\eta)$ is of class $C^{r}$,
where $T_{j}$ is given by (9). Indeed, by (BC1) we have that

$\begin{split}\left|\frac{\partial f_{j}(u,v)}{\partial
u}\right|&=\lim_{\delta\to
0}\frac{|f_{j}(u+\delta,v)-f_{j}(u,v)|}{|\delta|}\\\ &\leq\lim_{\delta\to
0}\frac{\gamma_{j}|\delta|}{|\delta|}\\\ &=\gamma_{j},\end{split}$ (11)

where $\frac{\partial f_{j}(\cdot,\cdot)}{\partial u}$ denotes the derivative
of $f_{j}$ with respect to the first variable. Thus, by (BC4), it follows that
the operator $\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(\xi,\eta)}{\partial\xi}$ is invertible. In particular, we can consider

$L:=\left(\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta)}{\partial
u}\right)^{-1}A_{j}^{-1}=\left(A_{j}+\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta)}{\partial u}\right)^{-1}.$ (12)

Now, observing that

$\begin{split}&T_{j}(\xi+\delta,\eta)-T_{j}(\xi,\eta)-L\delta\\\
&=A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi+\delta,\eta),\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta\\\
&=A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi+\delta,\eta),\eta)-f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)\right)\\\
&\phantom{=}-A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta,\\\
\end{split}$

it follows from (BC1) that

$\begin{split}&\left|T_{j}(\xi+\delta,\eta)-T_{j}(\xi,\eta)-L\delta\right|\\\
&\leq\left|A_{j}^{-1}\left(f_{j}(T_{j}(\xi+\delta,\eta),\eta)-f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)\right)\right|\\\
&\phantom{=}+\left|A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta\right|\\\
&\leq\gamma_{j}|A_{j}^{-1}|\left|T_{j}(\xi+\delta,\eta)-T_{j}(\xi,\eta)-L\delta\right|\\\
&\phantom{=}+\left|A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta\right|.\\\
\end{split}$

Using (BC4), we obtain that

$\begin{split}&(1-\gamma_{j}|A_{j}^{-1}|)\left|T_{j}(\xi+\delta,\eta)-T_{j}(\xi,\eta)-L\delta\right|\\\
&\leq\left|A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta\right|.\\\
\end{split}$

Therefore, since by (AC1) $u\to f_{j}(T_{j}(\xi,\eta)+Lu,\eta)$ is
differentiable,

$\frac{\partial f_{j}(T_{j}(\xi,\eta)+Lu,\eta)}{\partial
u}\bigg{|}_{u=0}=\frac{\partial f_{j}(T_{j}(\xi,\eta),\eta)}{\partial u}L$

and since

$\begin{split}&A_{j}^{-1}\frac{\partial f_{j}(T_{j}(\xi,\eta),\eta)}{\partial
u}L+L\\\ &=\left(\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta)}{\partial u}\right)L\\\
&=\left(\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta)}{\partial
u}\right)\left(\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta)}{\partial u}\right)^{-1}A_{j}^{-1}\\\
&=A_{j}^{-1},\end{split}$

we get that

$\begin{split}\lim_{\delta\to 0}\frac{\left|A_{j}^{-1}\delta-
A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\eta)+L\delta,\eta)-f_{j}(T_{j}(\xi,\eta),\eta)\right)-L\delta\right|}{|\delta|}=0.\\\
\end{split}$

Thus, combining the previous observations we conclude that

$\begin{split}\lim_{\delta\to
0}\frac{\left|T_{j}(\xi+\delta,\eta)-T_{j}(\xi,\eta)-L\delta\right|}{|\delta|}=0.\\\
\end{split}$

In particular, $T_{j}(\xi,\eta)$ is differentiable with respect to $\xi$ and
$\frac{\partial T_{j}(\xi,\eta)}{\partial\xi}=L$. Moreover, from (12) and
(AC1), we have that $\xi\to T_{j}(\xi,\eta)$ is $C^{r}$.

Consequently, since by (10) we have that
$T_{n-1}(\xi,y(n-1,n,\eta))=x_{2}(n-1,n,\xi,\eta)$, it follows that $\xi\to
x_{2}(n-1,n,\xi,\eta)$ is $C^{r}$. Finally, proceeding inductively we conclude
that $\xi\to x_{2}(k,n,\xi,\eta)$ is $C^{r}$ for every $k<n$. Therefore, since
the desired conclusion clearly holds for $k=n$, the proof of the proposition
is completed. ∎

######  Theorem 3.3.

Suppose that (BC1), (BC2), (BC3), (BC4) and (AC1) hold. Moreover, given
$n\in\mathbb{Z}$ suppose that hypothesis (AC2) is satisfied for $n$. Then, the
map $\xi\to\bar{H}_{n}(\xi,\eta)$ is $C^{1}$.

###### Proof.

Fix $n\in\mathbb{Z}$. We start by recalling (see the proof of [3, Theorem
3.1]) that

$\bar{H}_{n}(\xi,\eta)=(\xi+\bar{h}_{n}(\xi,\eta),\eta),$

where

$\bar{h}_{n}(\xi,\eta)=-\sum_{k\in\mathbb{Z}}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta)),$
(13)

for $(\xi,\eta)\in X\times Y$. Then, it remains to prove that the map
$\xi\to\bar{h}_{n}(\xi,\eta)$ is $C^{1}$.

By hypothesis (AC1) and Proposition 3.2, we have that each of the terms
involved in the series (13) is differentiable with respect to $\xi$. Moreover,
by (11) we have that

$\begin{split}\left|\frac{\partial f_{k}(u,v)}{\partial
u}\right|&\leq\gamma_{k}.\end{split}$

Similarly, using Lemma 3.1 we conclude that

$\begin{split}\left|\frac{\partial
x_{2}(k,n,\xi,\eta)}{\partial\xi}\right|&=\lim_{\delta\to
0}\frac{|x_{2}(k,n,\xi+\delta,\eta)-x_{2}(k,n,\xi,\eta)|}{|\delta|}\\\
&\leq\lim_{\delta\to 0}\frac{C_{k,n}|\delta|}{|\delta|}\\\
&=C_{k,n}.\end{split}$

Combining these observations with (AC2), it follows that

$\begin{split}&\sum_{k\in\mathbb{Z}}\left|\frac{\partial}{\partial\xi}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))\right|\\\
&\leq\sum_{k\in\mathbb{Z}}|\mathcal{G}(n,k+1)|\left|\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial u}\right|\left|\frac{\partial
x_{2}(k,n,\xi,\eta)}{\partial\xi}\right|\\\
&\leq\sum_{k\in\mathbb{Z}}|\mathcal{G}(n,k+1)|\gamma_{k}C_{k,n}<+\infty.\\\
\end{split}$ (14)

In particular, the series

$-\sum_{k\in\mathbb{Z}}\frac{\partial}{\partial\xi}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))$

converges uniformly. Thus, it coincides with
$\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\xi}$ and
$\xi\to\bar{h}_{n}(\xi,\eta)$ is $C^{1}$. Consequently,
$\xi\to\bar{H}_{n}(\xi,\eta)$ is $C^{1}$ as claimed. ∎

######  Theorem 3.4.

Suppose hypothesis (BC1), (BC2), (BC3), (BC4) and (AC1) are satisfied.
Moreover, given $n\in\mathbb{Z}$ assume that the hypothesis (AC3) is satisfied
for $n$. Then, the map $\xi\to H_{n}(\xi,\eta)$ is $C^{1}$.

###### Proof.

Recall that by Theorem 2.1, $H_{n}(\xi,\eta)=(\xi+h_{n}(\xi,\eta),\eta)$.
Hence, it remains to prove that the map $\xi\to h_{n}(\xi,\eta)$ is $C^{1}$.
We start by observing that hypothesis (AC3) and the estimates from the proof
of Theorem 3.3 (see (14)) imply that
$\left|\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\xi}\right|<1$, for every
$(\xi,\eta)\in X\times Y$. In particular, the operator
$\operatorname{Id}+\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\xi}$ is
invertible and we may consider

$R:=-\left(\operatorname{Id}+\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
u}\right)^{-1}\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
u},$ (15)

where $\frac{\partial\bar{h}_{n}(\cdot,\cdot)}{\partial u}$ denotes the
derivative of $\bar{h}_{n}(\cdot,\cdot)$ with respect to the first variable.

We now claim that $\xi\to h_{n}(\xi,\eta)$ is differentiable and that
$\frac{\partial h_{n}(\xi,\eta)}{\partial\xi}=R$. We first note that Theorem
2.1 implies that $\bar{H}_{n}(H_{n}(\xi,\eta))=(\xi,\eta)$, and consequently

$h_{n}(\xi,\eta)=-\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta),\quad\text{for every
$(\xi,\eta)\in X\times Y$.}$ (16)

Moreover, since by Theorem 3.3 $\bar{h}_{n}$ is differentiable, we have that

$\begin{split}&h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)\\\
&=-\bar{h}_{n}(\xi+\delta+h_{n}(\xi+\delta,\eta),\eta)+\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)\\\
&=-\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
u}\left(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)\right)\\\
&\phantom{=}+\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)),\\\
\end{split}$

where $l(x)=\mathbf{o}(x)$ means that $\lim_{x\to 0}\frac{|l(x)|}{|x|}=0$.
Thus,

$\begin{split}&\left(\operatorname{Id}+\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
u}\right)\left(h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)\right)\\\
&=-\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial u}\delta\\\
&\phantom{=}+\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)),\\\
\end{split}$

which implies that

$\begin{split}h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)=R\delta+\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)).\\\
\end{split}$ (17)

On the other hand, since $\xi\to h_{n}(\xi,\eta)$ is continuous,

$\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))\leq\frac{1}{2}|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|,$

whenever $|\delta|$ is sufficiently small. Consequently, by taking the norm on
both sides of (17) we get that

$\begin{split}|h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|&\leq|R||\delta|+\frac{1}{2}|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|\\\
&\leq\left(|R|+\frac{1}{2}\right)|\delta|+\frac{1}{2}|h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|,\\\
\end{split}$

whenever $|\delta|$ is sufficiently small. In particular, there exist $C>0$
and $\rho>0$ such that

$\frac{|h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|}{|\delta|}\leq
C,\quad\text{for $|\delta|<\rho$.}$

Thus, whenever $|\delta|<\rho$, we have that

$\begin{split}&\frac{\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))}{|\delta|}\\\
&=\frac{\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))}{|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|}\cdot\frac{|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|}{|\delta|}\\\
&\leq(C+1)\frac{\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))}{|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|}.\end{split}$
(18)

Consequently, since $\xi\to h_{n}(\xi,\eta)$ is continuous, we have that

$\lim_{\delta\to 0}|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|=0,$

and thus

$\lim_{\delta\to
0}\frac{\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))}{|\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)|}=0.$

This fact combined with (18) implies that

$\mathbf{o}(\delta+h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta))=\mathbf{o}(\delta)$

which together with (17) yields that

$\begin{split}h_{n}(\xi+\delta,\eta)-h_{n}(\xi,\eta)=R\delta+\mathbf{o}(\delta).\\\
\end{split}$

Hence, $\xi\to h_{n}(\xi,\eta)$ is differentiable and $\frac{\partial
h_{n}(\xi,\eta)}{\partial\xi}=R$. Moreover, it follows from the expression of
$R$ given in (15) and the fact that $\xi\to\bar{h}_{n}(\xi,\eta)$ is $C^{1}$
that $\xi\to h_{n}(\xi,\eta)$ is also $C^{1}$. The proof of the theorem is
completed. ∎

######  Example 3.5.

It is easy to construct examples where property (AC2) is satisfied for every
$n\in\mathbb{Z}$ simultaneously. For instance, suppose that

$M:=\sup_{k\in\mathbb{Z}}\max\\{|A_{k}|,|A_{k}^{-1}|,|P_{k}|\\}<+\infty.$

Without any loss of generality, we may assume that $M\geq 1$. For
$k\in\mathbb{Z}$, take $\gamma_{k}$ such that
$0\leq\gamma_{k}\leq\frac{1}{(2M)^{2|k|+1}}$. Observe that
$\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\leq 2M$ for $k\in\mathbb{Z}$.
Consequently,

$\begin{split}K_{n}&=\sum_{k<n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\right)\\\
&\leq\sum_{k<n}M^{n-k}\frac{1}{(2M)^{2|k|}}(2M)^{n-k}\\\
&\leq(2M)^{2|n|}\sum_{k<n}\frac{1}{2^{|k|}}<+\infty,\end{split}$

for every $n\in\mathbb{Z}$. Similarly, since $|A_{k}|+\gamma_{k}\leq 2M$, it
follows that

$\begin{split}J_{n}&=\sum_{k>n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})\right)\\\
&\leq\sum_{k>n}M^{k+2-n}\frac{1}{(2M)^{2|k|}}(2M)^{k-n}\\\ &\leq
M^{2}(2M)^{2|n|}\sum_{k>n}\frac{1}{2^{|k|}}<+\infty,\end{split}$

for $n\in\mathbb{Z}$.

We conclude that Theorem 3.3 is applicable under mild assumptions on the
linear part in (6), which in particular does not have to exhibit any
asymptotic behaviour. For example, we can take $A_{k}=P_{k}=\operatorname{Id}$
for each $k\in\mathbb{Z}$.

As for (AC3), fix $n\in\mathbb{Z}$ and choose $n_{0}\in\mathbb{N}$ such that
$M^{2}(2M)^{2|n|}\sum_{k\in\mathbb{Z}}\frac{1}{2^{|k|+n_{0}}}<1$. Then, if
$\gamma_{k}<\frac{1}{(2M)^{2|k|+n_{0}}}$ for every $k\in\mathbb{Z}$, our
assumptions combined with the previous estimates implies that (AC3) is
satisfied for $n$.

We now present examples under which (AC3) holds for each $n\in\mathbb{Z}$.

######  Example 3.6.

We consider the case when $X$ is a product of a Banach space
$\mathcal{X}=(\mathcal{X},|\cdot|_{\mathcal{X}})$ with itself, i.e.
$X=\mathcal{X}\times\mathcal{X}$. Then, $X$ is a Banach space with respect to
the norm
$|(x_{1},x_{2})|:=\max\\{|x_{1}|_{\mathcal{X}},|x_{2}|_{\mathcal{X}}\\}$,
$x_{1},x_{2}\in\mathcal{X}$.

Take an arbitrary $\lambda>0$. Let us consider sequences of bounded linear
operators $(A_{n})_{n\in\mathbb{Z}}$ and $(P_{n})_{n\in\mathbb{Z}}$ acting on
$X$ given by

$A_{n}=\begin{pmatrix}e^{\lambda}\operatorname{Id}&0\\\
0&e^{-\lambda}\operatorname{Id}\\\ \end{pmatrix}\quad\text{and}\quad
P_{n}=\begin{pmatrix}0&0\\\ 0&\operatorname{Id}\\\ \end{pmatrix},$

for $n\in\mathbb{Z}$, where $\operatorname{Id}$ denotes the identity operator
on $\mathcal{X}$.

Then,

$\mathcal{G}(m,n)=\begin{cases}\begin{pmatrix}0&0\\\
0&e^{-\lambda(m-n)}\operatorname{Id}\end{pmatrix}&\text{for $m\geq n$;}\\\
-\begin{pmatrix}e^{-\lambda(n-m)}\operatorname{Id}&0\\\ 0&0\\\
\end{pmatrix}&\text{for $m<n$.}\end{cases}$

Consequently,

$|\mathcal{G}(m,n)|=e^{-\lambda|m-n|}.$

Set

$M:=\prod_{j\in\mathbb{Z}}\left(1+e^{-\lambda|j|}\right)<+\infty.$

Now, let us consider a sequence $(\gamma_{k})_{k\in\mathbb{Z}}$ of nonnegative
numbers such that:

$\gamma_{k}\leq\frac{1}{e^{\lambda(|k|+1)}+e^{\lambda}}\quad\text{for
$k\in\mathbb{Z}$},$ (19)

and

$\sum_{k\in\mathbb{Z}}\gamma_{k}<\frac{1}{e^{\lambda}M}.$ (20)

Observe that it follows from (19) that

$\frac{1}{1-\gamma_{k}}\leq\frac{1}{1-e^{\lambda}\gamma_{k}}\leq
1+e^{-\lambda|k|},\quad k\in\mathbb{Z}.$ (21)

Then, using (21) we have that

$\begin{split}K_{n}&=\sum_{k<n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\right)\\\
&\leq\sum_{k<n}e^{-\lambda(n-k-1)}\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{e^{\lambda}}{1-e^{\lambda}\gamma_{j}}\right)\\\
&\leq\sum_{k<n}e^{-\lambda(n-k-1)}\gamma_{k}e^{\lambda(n-k)}\left(\prod_{j=k}^{n-1}\frac{1}{1-e^{\lambda}\gamma_{j}}\right)\\\
&\leq
e^{\lambda}\sum_{k<n}\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{1}{1-e^{\lambda}\gamma_{j}}\right)\\\
&\leq
e^{\lambda}\sum_{k<n}\gamma_{k}\left(\prod_{j=k}^{n-1}\left(1+e^{-\lambda|j|}\right)\right)\\\
&\leq e^{\lambda}M\sum_{k<n}\gamma_{k}.\\\ \end{split}$

Similarly, since $1+e^{-\lambda}\gamma_{j}\leq
1+\gamma_{j}\leq\frac{1}{1-\gamma_{j}}$, it follows from (21) that

$\begin{split}J_{n}&=\sum_{k>n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})\right)\\\
&\leq\sum_{k>n}e^{-\lambda(k+1-n)}\gamma_{k}\left(\prod_{j=n}^{k-1}(e^{\lambda}+\gamma_{j})\right)\\\
&=\sum_{k>n}e^{-\lambda(k+1-n)}\gamma_{k}e^{\lambda(k-n)}\left(\prod_{j=n}^{k-1}(1+e^{-\lambda}\gamma_{j})\right)\\\
&\leq e^{-\lambda}M\sum_{k>n}\gamma_{k}\\\ &\leq
e^{\lambda}M\sum_{k>n}\gamma_{k}.\end{split}$

Finally, since $M\geq 1$ we have that
$|\mathcal{G}(n,n+1)|\gamma_{n}=e^{-\lambda}\gamma_{n}\leq
e^{\lambda}M\gamma_{n}$. Thus, from (20) we conclude that

$K_{n}+J_{n}+|\mathcal{G}(n,n+1)|\gamma_{n}\leq
e^{\lambda}M\sum_{n\in\mathbb{Z}}\gamma_{n}<1,$

for each $n\in\mathbb{Z}$.

######  Example 3.7.

Let $X$ be as in Example 3.6. Furthermore, take a sequence of isometries
$(B_{n})_{n\in\mathbb{Z}}$ acting $\mathcal{X}$ and let
$(\theta_{n})_{n\in\mathbb{Z}}$ be any sequence of numbers satisfying
$\theta_{n}\geq 1$ for every $n\in\mathbb{Z}$. Moreover, suppose that
$\theta_{n}\leq\theta_{n+1}$ and that there exists a constant $T\geq 1$ such
that $\frac{\theta_{n+1}}{\theta_{n}}\leq T$ for every $n\in\mathbb{Z}$. We
now consider sequences of bounded linear operators $(A_{n})_{n\in\mathbb{Z}}$
and $(P_{n})_{n\in\mathbb{Z}}$ acting on $X$ given by

$A_{n}=\begin{pmatrix}\frac{\theta_{n}}{\theta_{n+1}}\operatorname{Id}&0\\\
0&B_{n}\\\ \end{pmatrix}\quad\text{and}\quad
P_{n}=\begin{pmatrix}\operatorname{Id}&0\\\ 0&0\\\ \end{pmatrix},$

for $n\in\mathbb{Z}$. Then,

$\mathcal{G}(m,n)=\begin{cases}\begin{pmatrix}\frac{\theta_{n}}{\theta_{m}}\operatorname{Id}&0\\\
0&0\\\ \end{pmatrix}&\text{for $m\geq n$;}\\\ -\begin{pmatrix}0&0\\\
0&\mathcal{B}(m,n)\\\ \end{pmatrix}&\text{for $m<n$,}\end{cases}$

where

$\mathcal{B}(m,n)=\begin{cases}B_{m-1}\cdots B_{n}&\text{for $m>n$;}\\\
\operatorname{Id}&\text{for $m=n$;}\\\ B_{m}^{-1}\ldots B_{n-1}^{-1}&\text{for
$m<n$.}\\\ \end{cases}$

Consequently,

$|\mathcal{G}(m,n)|=\begin{cases}\frac{\theta_{n}}{\theta_{m}}&\text{for
$m\geq n$;}\\\ 1&\text{for $m<n$.}\end{cases}$

Set

$M:=\prod_{j\in\mathbb{Z}}\left(1+\frac{1}{2^{|j|}}\right)<+\infty.$

Now, let us consider a sequence $(\gamma_{k})_{k}$ of nonnegative numbers such
that:

$\gamma_{k}<\frac{1}{T(2^{|k|+1}+1)}\quad\text{for $k\in\mathbb{Z}$,}$ (22)

and

$\sum_{k\in\mathbb{Z}}\gamma_{k}<\frac{1}{TM}.$ (23)

Observe that it follows from (22) that

$\frac{1}{1-\gamma_{k}}\leq\frac{1}{1-T\gamma_{k}}\leq
1+\frac{1}{2^{|k|}},\quad k\in\mathbb{Z}.$ (24)

Then, observing that $|A^{-1}_{j}|=\frac{\theta_{j+1}}{\theta_{j}}$ and using
(24), we get that

$\begin{split}K_{n}&=\sum_{k<n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\right)\\\
&=\sum_{k<n}\frac{\theta_{k+1}}{\theta_{n}}\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{\left(\frac{\theta_{j+1}}{\theta_{j}}\right)}{1-\left(\frac{\theta_{j+1}}{\theta_{j}}\right)\gamma_{j}}\right)\\\
&\leq\sum_{k<n}\frac{\theta_{k+1}}{\theta_{n}}\gamma_{k}\left(\prod_{j=k}^{n-1}\frac{\theta_{j+1}}{\theta_{j}}\left(\frac{1}{1-T\gamma_{j}}\right)\right)\\\
&=\sum_{k<n}\frac{\theta_{k+1}}{\theta_{n}}\gamma_{k}\frac{\theta_{n}}{\theta_{k}}\left(\prod_{j=k}^{n-1}\frac{1}{1-T\gamma_{j}}\right)\\\
&\leq\sum_{k<n}TM\gamma_{k}.\\\ \end{split}$

Similarly, observing that $|A_{j}|=1$ and
$1+\gamma_{j}\leq\frac{1}{1-\gamma_{j}}\leq\frac{1}{1-T\gamma_{j}}$, we have
that

$\begin{split}J_{n}&=\sum_{k>n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})\right)\\\
&\leq\sum_{k>n}\gamma_{k}\left(\prod_{j=n}^{k-1}(1+\gamma_{j})\right)\\\ &\leq
M\sum_{k>n}\gamma_{k}\\\ &\leq TM\sum_{k>n}\gamma_{k}.\end{split}$

Finally, since $M,T\geq 1$, it follows that
$|\mathcal{G}(n,n+1)|\gamma_{n}=\gamma_{n}\leq TM\gamma_{n}$. Thus, from (23)
we conclude that

$K_{n}+J_{n}+|\mathcal{G}(n,n+1)|\gamma_{n}\leq
TM\sum_{k\in\mathbb{Z}}\gamma_{k}<1,$

for every $n\in\mathbb{Z}$.

###  3.2. An important particular case

We will now restrict to the case when $Y$ is a trivial Banach space, i.e.
$Y=\\{0\\}$. In this case, (5) reduces to

$x_{n+1}=A_{n}x_{n}+f_{n}(x_{n}),$ (25)

where $f_{n}\colon X\to X$, while (6) is equivalent to

$x_{n+1}=A_{n}x_{n}.$ (26)

The following result (originally established in [26]) is a consequence of
Theorem 2.1.

######  Corollary 3.8.

Suppose that (BC1), (BC2) and (BC3) are satisfied. Then, there exists a
sequence of continuous functions $H_{n}\colon X\to X$, $n\in\mathbb{Z}$ of the
form $H_{n}(x)=x+h_{n}(x)$, where $\sup_{n,x}|h_{n}(x)|<\infty$, such that if
$n\mapsto x_{n}$ is a solution of (26), then $n\mapsto H_{n}(x_{n})$ is a
solution of (25). In addition, there exists a sequence of continuous functions
$\bar{H}_{n}\colon X\to X$, $n\in\mathbb{Z}$ of the form
$\bar{H}_{n}(x)=x+\bar{h}_{n}(x)$, where $\sup_{n,x}|\bar{h}_{n}(x)|<\infty$,
such that if $n\mapsto x_{n}$ is a solution of (25), then
$n\mapsto\bar{H}_{n}(x_{n})$ is a solution of (26). Moreover, $H_{n}$ and
$\bar{H}_{n}$ are inverses of each other.

The following result is a direct consequence of Theorems 3.3 and 3.4.

######  Corollary 3.9.

Suppose hypothesis (BC1), (BC2), (BC3), (BC4) and (AC1) are satisfied.
Moreover, assume that the hypothesis (AC3) is satisfied for each
$n\in\mathbb{Z}$. Then, $H_{n}$ and $\bar{H}_{n}$ are of class $C^{1}$ for
every $n\in\mathbb{Z}$.

######  Remark 3.10.

We would like to compare Corollary 3.9 with the smooth linearization result
established in [11, Theorem 2]. We stress that [11, Theorem 2] requires that
(26) admits a (nonuniform) exponential dichotomy. In particular, it is
applicable when the sequence $(A_{n})_{n\in\mathbb{Z}}$ is as in Example 3.6
and when $(\gamma_{k})_{k\in\mathbb{Z}}$ is a constant sequence, i.e.
$\gamma_{k}=c$, $k\in\mathbb{Z}$, where $c$ is sufficiently small. We note
that this situation is not covered with Corollary 3.9. Indeed, for any
$n\in\mathbb{Z}$ we have that

$\begin{split}J_{n}&=\sum_{k>n}|\mathcal{G}(n,k+1)|\gamma_{k}\left(\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j})\right)\\\
&=c\sum_{k>n}e^{-\lambda(k+1-n)}\left(\prod_{j=n}^{k-1}(e^{\lambda}+c)\right)\\\
&\geq c\sum_{k>n}e^{-\lambda(k+1-n)}e^{\lambda(k-n)}\\\ &=+\infty,\end{split}$

and thus (AC3) cannot hold.

On the other hand, as illustrated by Example 3.7, Corollary 3.9 is applicable
in situations when (26) does not admit an exponential dichotomy. We conclude
that Corollary 3.9 and [11, Theorem 2] complement each other.

######  Remark 3.11.

We note that a result similar to Corollary 3.9 has been established in [8,
Theorem 3.2] under some additional assumptions on (26) and assuming that $X$
is finite-dimensional.

###  3.3. Regularity with respect to the second variable

We are now interested in formulating sufficient conditions under which $H_{n}$
and $\bar{H}_{n}$ are smooth in the second variable.

We start studying the regularity properties of the maps $\eta\to y(k,n,\eta)$
and $\eta\to x_{2}(k,n,\xi,\eta)$. Consider

$D_{k,n}=\begin{cases}\prod_{j=n}^{k-1}\tau_{j}&\text{for $k>n$;}\\\
1&\text{for $k=n$;}\\\ \prod_{j=k}^{n-1}\sigma_{j}&\text{for
$k<n$,}\end{cases}$ (27)

and

$M_{k,n}=\begin{cases}\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j}+\max\\{\rho_{j},\tau_{j}\\})&\text{for
$k>n$;}\\\ 1&\text{for $k=n$;}\\\
\prod_{j=k}^{n-1}\left(\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}+\sigma_{j}\right)&\text{for
$k<n$.}\end{cases}$ (28)

######  Lemma 3.12.

Suppose (AC5) is satisfied. Then, for any $\eta,\tilde{\eta}\in Y$ and
$n,k\in\mathbb{Z}$,

$|y(k,n,\eta)-y(k,n,\tilde{\eta})|\leq D_{k,n}|\eta-\tilde{\eta}|.$

###### Proof.

This follows easily from the definition of $y(k,n,\eta)$ and hypothesis (AC5).
∎

######  Lemma 3.13.

Suppose (BC1), (BC4), (AC4), (AC5) and (AC6) are satisfied. Then, for any
$\xi\in X$, $\eta,\tilde{\eta}\in Y$ and $n,k\in\mathbb{Z}$, we have that

$|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\xi,\tilde{\eta})|\leq
M_{k,n}|\eta-\tilde{\eta}|.$

###### Proof.

Fix $n\in\mathbb{Z}$. By (5) we have that

$\begin{split}x_{2}(n+1,n,\xi,\eta)&=A_{n}\xi+f_{n}(\xi,\eta).\\\ \end{split}$
(29)

Consequently, using (AC4) we have that

$\begin{split}|x_{2}(n+1,n,\xi,\eta)-x_{2}(n+1,n,\xi,\tilde{\eta})|&=|f_{n}(\xi,\eta)-f_{n}(\xi,\tilde{\eta})|\\\
&\leq\rho_{n}|\eta-\tilde{\eta}|.\end{split}$

Moreover, using Lemmas 3.1 and 3.12 and the previous observation we get that

$\begin{split}&|x_{2}(n+2,n,\xi,\eta)-x_{2}(n+2,n,\xi,\tilde{\eta})|\\\
&=|x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\eta),y(n+1,n,\eta))\\\
&\phantom{=}-x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\tilde{\eta}),y(n+1,n,\tilde{\eta}))|\\\
&\leq|x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\eta),y(n+1,n,\eta))\\\
&\phantom{\leq}-x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\tilde{\eta}),y(n+1,n,\eta))|\\\
&\phantom{=}+|x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\tilde{\eta}),y(n+1,n,\eta))\\\
&\phantom{=}-x_{2}(n+2,n+1,x_{2}(n+1,n,\xi,\tilde{\eta}),y(n+1,n,\tilde{\eta}))|\\\
&\leq C_{n+2,n+1}|x_{2}(n+1,n,\xi,\eta)-x_{2}(n+1,n,\xi,\tilde{\eta})|\\\
&\phantom{\leq}+\rho_{n+1}|y(n+1,n,\eta)-y(n+1,n,\tilde{\eta})|\\\ &\leq
C_{n+2,n+1}\rho_{n}|\eta-\tilde{\eta}|+\rho_{n+1}\tau_{n}|\eta-\tilde{\eta}|\\\
&\leq\left(|A_{n+1}|+\gamma_{n+1}+\max\\{\rho_{n+1},\tau_{n+1}\\}\right)\max\\{\rho_{n},\tau_{n}\\}|\eta-\tilde{\eta}|\\\
&\leq\left(|A_{n+1}|+\gamma_{n+1}+\max\\{\rho_{n+1},\tau_{n+1}\\}\right)\left(|A_{n}|+\gamma_{n}+\max\\{\rho_{n},\tau_{n}\\}\right)|\eta-\tilde{\eta}|.\end{split}$

Using that

$x_{2}(k,n,\xi,\eta)=x_{2}(k,m,x_{2}(m,n,\xi,\eta),y(m,n,\eta)),$

and proceeding inductively we conclude that

$\begin{split}|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\xi,\tilde{\eta})|&\leq\prod_{j=n}^{k-1}(|A_{j}|+\gamma_{j}+\max\\{\rho_{j},\tau_{j}\\})|\eta-\tilde{\eta}|\\\
&=M_{k,n}|\eta-\tilde{\eta}|,\end{split}$

for every $k>n$.

We now consider the case when $k<n$. Let $T_{j}:X\times Y\to X$ be given by
(9). Given $\xi\in X$ and $\eta,\tilde{\eta}\in Y$, using (BC1) we get that

$\begin{split}|T_{j}(\xi,\eta)-T_{j}(\xi,\tilde{\eta})|&=\left|A_{j}^{-1}\left(f_{j}(T_{j}(\xi,\tilde{\eta}),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\eta)\right)\right|\\\
&\leq|A_{j}^{-1}|\left|f_{j}(T_{j}(\xi,\tilde{\eta}),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\tilde{\eta})\right|\\\
&\phantom{=}+|A_{j}^{-1}|\left|f_{j}(T_{j}(\xi,\eta),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\eta)\right|\\\
&\leq|A_{j}^{-1}|\gamma_{j}\left|T_{j}(\xi,\tilde{\eta})-T_{j}(\xi,\eta)\right|\\\
&\phantom{=}+|A_{j}^{-1}|\left|f_{j}(T_{j}(\xi,\eta),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\eta)\right|,\\\
\end{split}$ (30)

which implies that

$\begin{split}\left(1-\gamma_{j}|A_{j}^{-1}|\right)|T_{j}(\xi,\eta)-T_{j}(\xi,\tilde{\eta})|&\leq|A_{j}^{-1}|\left|f_{j}(T_{j}(\xi,\eta),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\eta)\right|\\\
&\leq\rho_{j}|A_{j}^{-1}|\left|\eta-\tilde{\eta}\right|.\end{split}$

Thus, combining this observation with (BC4) we conclude that

$\begin{split}|T_{j}(\xi,\eta)-T_{j}(\xi,\tilde{\eta})|&\leq\frac{\rho_{j}|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}\left|\eta-\tilde{\eta}\right|.\end{split}$

Consequently, by (10), Lemma 3.12 and the the previous observation, we get
that

$\begin{split}&|x_{2}(n-1,n,\xi,\eta)-x_{2}(n-1,n,\xi,\tilde{\eta})|\\\
&=|T_{n-1}(\xi,y(n-1,n,\eta))-T_{n-1}(\xi,y(n-1,n,\tilde{\eta}))|\\\
&\leq\frac{\rho_{n-1}|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}\left|y(n-1,n,\eta)-y(n-1,n,\tilde{\eta})\right|\\\
&\leq\frac{\sigma_{n-1}\rho_{n-1}|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}\left|\eta-\tilde{\eta}\right|.\\\
\end{split}$

Moreover, using Lemmas 3.1 and 3.12, the previous observation and (AC6), we
obtain that

$\begin{split}&|x_{2}(n-2,n,\xi,\eta)-x_{2}(n-2,n,\xi,\tilde{\eta})|\\\
&=|x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\eta),y(n-1,n,\eta))\\\
&\phantom{=}-x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\tilde{\eta}),y(n-1,n,\tilde{\eta}))|\\\
&\leq|x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\eta),y(n-1,n,\eta))\\\
&\phantom{\leq}-x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\tilde{\eta}),y(n-1,n,\eta))|\\\
&\phantom{=}+|x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\tilde{\eta}),y(n-1,n,\eta))\\\
&\phantom{=}-x_{2}(n-2,n-1,x_{2}(n-1,n,\xi,\tilde{\eta}),y(n-1,n,\tilde{\eta}))|\\\
&\leq C_{n-2,n-1}|x_{2}(n-1,n,\xi,\eta)-x_{2}(n-1,n,\xi,\tilde{\eta})|\\\
&\phantom{\leq}+\frac{\sigma_{n-2}\rho_{n-2}|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}|y(n-1,n,\eta)-y(n-1,n,\tilde{\eta})|\\\
&\leq\frac{|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}\frac{\sigma_{n-1}\rho_{n-1}|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}|\eta-\tilde{\eta}|+\frac{\sigma_{n-2}\rho_{n-2}|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}\sigma_{n-1}|\eta-\tilde{\eta}|\\\
&\leq\frac{|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}\frac{|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}|\eta-\tilde{\eta}|+\frac{|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}\sigma_{n-1}|\eta-\tilde{\eta}|\\\
&\leq\left(\frac{|A_{n-2}^{-1}|}{1-\gamma_{n-2}|A_{n-2}^{-1}|}+\sigma_{n-2}\right)\left(\frac{|A_{n-1}^{-1}|}{1-\gamma_{n-1}|A_{n-1}^{-1}|}+\sigma_{n-1}\right)|\eta-\tilde{\eta}|.\end{split}$

Using that

$x_{2}(k,n,\xi,\eta)=x_{2}(k,m,x_{2}(m,n,\xi,\eta),y(m,n,\eta)),$

and proceeding recursively we conclude that

$\begin{split}|x_{2}(k,n,\xi,\eta)-x_{2}(k,n,\xi,\tilde{\eta})|&\leq\prod_{j=k}^{n-1}\left(\frac{|A_{j}^{-1}|}{1-\gamma_{j}|A_{j}^{-1}|}+\sigma_{j}\right)|\eta-\tilde{\eta}|\\\
&=M_{k,n}|\eta-\tilde{\eta}|,\end{split}$

for every $k<n$.

Finally, since $x_{2}(n,n,\xi,\eta)=\xi$ the proof is complete. ∎

######  Lemma 3.14.

Suppose (AC8) is satisfied. Then, the map $\eta\to y(k,n,\eta)$ is of class
$C^{r}$, $r\geq 1$, for every $n,k\in\mathbb{Z}$.

###### Proof.

This is a direct consequence of the definition and hypothesis (AC8). ∎

######  Proposition 3.15.

Suppose (BC1), (BC4), (AC1), (AC7) and (AC8) are satisfied. Then the map
$\eta\to x_{2}(k,n,\xi,\eta)$ is of class $C^{r}$, $r\geq 1$, for every
$\xi\in X$ and $n,k\in\mathbb{Z}$.

###### Proof.

Fix $n\in\mathbb{Z}$. By (5) we have that

$\begin{split}x_{2}(n+1,n,\xi,\eta)&=A_{n}\xi+f_{n}(\xi,\eta).\\\ \end{split}$

Consequently, by (AC7) it follows that $\eta\to x_{2}(n+1,n,\xi,\eta)$ is of
class $C^{r}$. Thus, using that

$x_{2}(k,n,\xi,\eta)=x_{2}(k,m,x_{2}(m,n,\xi,\eta),y(m,n,\eta)),$

Proposition 3.2, Lemma 3.14, hypothesis (AC1) and (AC7) and proceeding
inductively, it follows that $\eta\to x_{2}(k,n,\xi,\eta)$ is of class $C^{r}$
for every $k>n$.

We now consider the case when $k<n$. Let $T_{j}:X\times Y\to X$ be given by
(9). As in the proof of Proposition 3.2, the result will follow once we prove
that $\eta\to T_{j}(\xi,\eta)$ is of class $C^{r}$. We start observing that
$\eta\to T_{j}(\xi,\eta)$ is continuous. Indeed, by (30) we get that

$\begin{split}\left(1-\gamma_{j}|A_{j}^{-1}|\right)|T_{j}(\xi,\eta)-T_{j}(\xi,\tilde{\eta})|&\leq|A_{j}^{-1}|\left|f_{j}(T_{j}(\xi,\eta),\tilde{\eta})-f_{j}(T_{j}(\xi,\eta),\eta)\right|.\\\
\end{split}$

Consequently, by (BC4) and (AC7), it follows that
$|T_{j}(\xi,\eta)-T_{j}(\xi,\tilde{\eta})|\to 0$ whenever
$\tilde{\eta}\to\eta$. We conclude that $\eta\to T_{j}(\xi,\eta)$ is
continuous.

Let us now consider

$\tilde{L}:=-\left(\operatorname{Id}+A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta))}{\partial u}\right)^{-1}A_{j}^{-1}\frac{\partial
f_{j}(T_{j}(\xi,\eta),\eta))}{\partial v},$

where $\frac{\partial f_{j}(\cdot,\cdot)}{\partial u}$ and $\frac{\partial
f_{j}(\cdot,\cdot)}{\partial v}$ denote the derivatives of $f_{j}$ with
respect to the first and second variable, respectively. Observe that, since
$|A_{j}^{-1}\frac{\partial f_{j}(T_{j}(\xi,\eta),\eta))}{\partial u}|<1$, the
operator $\tilde{L}$ is well defined. Now, since $\eta\to T_{j}(\xi,\eta)$ is
continuous, one may proceed similarly to the proof of Theorem 3.4, and obtain
that $\eta\to T_{j}(\xi,\eta)$ is differentiable and $\frac{\partial
T_{j}(\xi,\eta)}{\partial\eta}=\tilde{L}$. Thus, from our hypothesis and the
definition of $\tilde{L}$ it follows that $\eta\to T_{j}(\xi,\eta)$ is in fact
of class $C^{r}$. Consequently, combining this observation with (10) and Lemma
3.14, it follows that $\eta\to x_{2}(k,n,\xi,\eta)$ is of class $C^{r}$ for
every $k<n$. Finally, since $x_{2}(n,n,\xi,\eta)=\xi$ for any $\eta$ the
desired conclusion holds in the case $k=n$ also. ∎

######  Theorem 3.16.

Suppose hypothesis (BC1), (BC2), (BC3), (BC4), (AC1), (AC4), (AC5), (AC6),
(AC7) and (AC8) are satisfied. Moreover, given $n\in\mathbb{Z}$ assume that
hypothesis (AC9) is satisfied for $n$. Then, the map
$\eta\to\bar{H}_{n}(\xi,\eta)$ is $C^{1}$.

###### Proof.

We proceed as in the proof of Theorem 3.3. Fix $n\in\mathbb{Z}$ and recall
that

$\bar{H}_{n}(\xi,\eta)=(\xi+\bar{h}_{n}(\xi,\eta),\eta)$

where

$\bar{h}_{n}(\xi,\eta)=-\sum_{k\in\mathbb{Z}}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta)),$
(31)

for $(\xi,\eta)\in X\times Y$ (see the proof of [3, Theorem 3.1]). Thus, it
remains to prove that the map $\eta\to\bar{h}_{n}(\xi,\eta)$ is $C^{1}$.

By hypothesis (AC1) and (AC7), Lemma 3.14 and Proposition 3.15 we have that
each of the terms involved in the series (31) are differentiable with respect
to $\eta$. Moreover,

$\begin{split}\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial\eta}&=\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial u}\frac{\partial
x_{2}(k,n,\xi,\eta)}{\partial\eta}\\\ &\phantom{=}+\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial v}\frac{\partial
y(k,n,\eta)}{\partial\eta},\end{split}$ (32)

where $\frac{\partial f_{k}(\cdot,\cdot)}{\partial u}$ and $\frac{\partial
f_{k}(\cdot,\cdot)}{\partial v}$ denote the derivative of $f_{k}$ with respect
to the first and second variables, respectively. Furthermore, by (11) we have
that

$\begin{split}\left|\frac{\partial f_{k}(u,v)}{\partial
u}\right|&\leq\gamma_{k}.\end{split}$

Similarly, using respectively (AC4), Lemma 3.12 and Lemma 3.13, we get that

$\begin{split}\left|\frac{\partial f_{k}(u,v)}{\partial
v}\right|\leq\rho_{k},\quad\left|\frac{\partial
y(k,n,\eta)}{\partial\eta}\right|\leq D_{k,n}\text{ and }\left|\frac{\partial
x_{2}(k,n,\xi,\eta)}{\partial\eta}\right|\leq M_{k,n}.\end{split}$

Combining these facts with (32), we conclude that

$\begin{split}\left|\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial\eta}\right|\leq\gamma_{k}M_{k,n}+\rho_{k}D_{k,n}.\end{split}$
(33)

By (AC9) and (33), we have that

$\begin{split}&\sum_{k\in\mathbb{Z}}\left|\frac{\partial}{\partial\eta}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))\right|\\\
&\leq\sum_{k\in\mathbb{Z}}|\mathcal{G}(n,k+1)|\left|\frac{\partial
f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))}{\partial\eta}\right|\\\
&\leq\sum_{k\in\mathbb{Z}}|\mathcal{G}(n,k+1)|\left(\gamma_{k}M_{k,n}+\rho_{k}D_{k,n}\right)<+\infty.\\\
\end{split}$

In particular, the series

$-\sum_{k\in\mathbb{Z}}\frac{\partial}{\partial\eta}\mathcal{G}(n,k+1)f_{k}(x_{2}(k,n,\xi,\eta),y(k,n,\eta))$

converges uniformly. Thus, it coincides with
$\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\eta}$ and
$\eta\to\bar{h}_{n}(\xi,\eta)$ is $C^{1}$. Consequently,
$\eta\to\bar{H}_{n}(\xi,\eta)$ is $C^{1}$ as claimed. The proof of the theorem
is completed. ∎

######  Theorem 3.17.

Suppose hypothesis (BC1), (BC2), (BC3), (BC4), (AC1), (AC4), (AC5), (AC6),
(AC7) and (AC8) are satisfied. Moreover, given $n\in\mathbb{Z}$ assume that
hypothesis (AC3) and (AC9) are satisfied for $n$. Then, the map $\eta\to
H_{n}(\xi,\eta)$ is $C^{1}$.

###### Proof.

The proof is similar to the proof of Theorem 3.4. By Theorem 2.1,
$H_{n}(\xi,\eta)=(\xi+h_{n}(\xi,\eta),\eta)$ we need to prove that the map
$\eta\to h_{n}(\xi,\eta)$ is $C^{1}$. As in the proof of Theorem 3.4 we get
that $\left|\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\xi}\right|<1$ for
every $(\xi,\eta)\in X\times Y$. Hence, the operator
$\operatorname{Id}+\frac{\partial\bar{h}_{n}(\xi,\eta)}{\partial\xi}$ is
invertible. Let

$\tilde{R}:=-\left(\operatorname{Id}+\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
u}\right)^{-1}\frac{\partial\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)}{\partial
v},$ (34)

where $\frac{\partial\bar{h}_{n}(\cdot,\cdot)}{\partial u}$ and
$\frac{\partial\bar{h}_{n}(\cdot,\cdot)}{\partial v}$ denote the derivative of
$\bar{h}_{n}(\cdot,\cdot)$ with respect to the first and second variables,
respectively.

Now, using that $h_{n}(\xi,\eta)=-\bar{h}_{n}(\xi+h_{n}(\xi,\eta),\eta)$ (see
(16)) together with the differentiability of $\bar{h}_{n}$ with respect to the
first and second variable (see Theorems 3.3 and 3.16) and the continuity of
$\eta\to h_{n}(\xi,\eta)$, we may proceed as in the proof of Theorem 3.4 and
conclude that $\eta\to h_{n}(\xi,\eta)$ is differentiable and that
$\frac{\partial h_{n}(\xi,\eta)}{\partial\xi}=\tilde{R}$. Moreover, it follows
from the expression of $\tilde{R}$ given in (34) and the fact that
$\xi\to\bar{h}_{n}(\xi,\eta)$ and $\eta\to\bar{h}_{n}(\xi,\eta)$ are $C^{1}$
that $\eta\to h_{n}(\xi,\eta)$ is also $C^{1}$ as claimed. ∎

######  Example 3.18.

Let us discuss an example under which (AC9) holds. Suppose that

$M:=\sup_{k\in\mathbb{Z}}\max\\{|A_{k}|,|A_{k}^{-1}|,|P_{k}|,\tau_{k},\sigma_{k}\\}<+\infty.$

Furthermore, assume that

$\rho_{k}\leq\tau_{k},\quad\gamma_{k}\leq\frac{1}{(3M)^{2|k|+1}}\quad\text{and}\quad\rho_{k}\leq\frac{1}{2^{|k|}M^{2|k|}},$

for $k\in\mathbb{Z}$. By arguing as in Example 3.5, one can easily show that
(AC9) holds for each $n\in\mathbb{Z}$.

##  4\. Acknowledgements

We would like to thank the referees for their useful comments that helped us
to improve our paper. L.B. was partially supported by a CNPq-Brazil PQ
fellowship under Grant No. 306484/2018-8. D. D. was supported in part by
Croatian Science Foundation under the project IP-2019-04-1239 and by the
University of Rijeka under the projects uniri-prirod-18-9 and uniri-
prprirod-19-16.

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