# SMOOTHNESS OF CLASS $C^{2}$ OF NONAUTONOMOUS LINEARIZATION without spectral
conditions

N. Jara  Universidad de Chile, Departamento de Matemáticas. Casilla 653,
Santiago, Chile<EMAIL_ADDRESS>
###### Abstract.

We prove that smoothness of nonautonomous linearization is of class $C^{2}.$
Our approach admits the existence of stable and unstable manifolds determined
by a family of nonautonomous hyperbolicities. Moreover, our goal is reached
without spectral conditions.

###### Key words and phrases:

Nonautonomus hyperbolicity, Nonautonomous differential equation, Smooth
linearization

###### 2010 Mathematics Subject Classification:

37C60, 37D25

This research has been partially supported by ANID, Beca Nacional de Magister
22200774.

##  1\. Introduction

The linearization problem was initially regarded locally by P. Hartman [9,
Theorem I] in the context of autonomous equations, later a specific type of
those equations, namely a linear and a quasi-linear system, were studied by C.
Pugh [13] in order to achieve a global homeomorphism between the flows of the
systems. Next this result was adapted to the nonautonomous case by K. J.
Palmer [11], who implemented the concepts of exponential dichotomy in order to
obtain its linearization problem.

The homoeomorphisms with its properties that K.J. Palmer obtained are known as
the concept of topological equivalence which is useful to describe the
asymptotic behaviour of solutions, even when they may be unknown, for example,
to find locally or global attractors, or more generally, stable and unstable
manifolds. Nevertheless, there are more dynamical properties that cannot be
described with purely topological tools and require a study of the the
differentiability of such homeomorphisms.

The study of the smoothness of homomorphisms has been a problem with a vast
analysis in both the autonomous and nonautonomous contexts: In the autonomous
framework, S. Sternberg [16, 17] proved that dynamical system given by
$C^{r}-$diffeomorphism can be $C^{k}-$linearized locally around hyperbolic
fixed points that satisfy a nonresonant condition on its spectrum. This
results were later improved by G. R. Belickiĭ [1, 2], still locally but giving
more explicit conditions for the derivatives of the conjugation. S. van Strein
[18] was the first to obtain a similar result without imposing resonance
conditions, but his proof turned out to be wrong, as stated by V. Rayskin
[14].

As far as the author has been able to ascertain of studying the
differentiability of the topological conjugacy between a linear and a
quasilinear nonautonomous system was first regarded in the work [6] by Á.
Castañeda and G. Robledo, who ensured the homeomorphisms are $C^{2}$
preserving orientation diffeomorphisms if the linear system are exponentially
asymptotically stable on $\mathbb{R}$ and verified some conditions. This
result was later improved in [3] to include more general decay ratios other
than exponential but in $\mathbb{R}^{+}$. In these results authors were able
to construct a global diffeomorphism in the same fashion as K.J. Palmer.

L. V. Cuong et. al. [7] obtained a smooth linearization in a similar way as S.
Sternberg when the linear part admits an exponential dichotomy on whole real
line. Recently, D. Dragičević et. al. [8] have proved Strein’s statement to be
true, and further extended it to the nonautonomous case while also improving
Cuong’s result, under the assumption that the linear system satisfy a
nonuniform hyperbolicity and other technical conditions, which do not include
nonresonance conditions but impose spectral bounds.

In this paper we develop ideas hold many similarities with Á. Castañeda, P.
Monzón and G. Robledo [5], since both study the topological equivalence of a
linear and a quasi-linear systems on the positive half real line. We improve
their result in two senses: first we present a wider family of dichotomies
accepted for the linear system, including both the exponential and the
nonuniform exponential under some technical conditions that will be expressed
later. Secondly, and we allow the existence of nonempty unstable manifolds for
our linear system. We do not rely on of spectral bounds or nonresonant
conditions, but only in the interlacing of the properties of nonautonomous
hyperbolicity of the linear part, and boundedness and lipschitzness of the
nonlinearities.

However, our work does not regard the continuity of the homeomorphism of
topological equivalence as a function of two variables, a property that Á.
Castañeda et al. called continuous topological equivalence [5, Definition
1.4]. We neither obtain a strong topological equivalence, a characterization
introduced by J. Shi [15, Definition 2.4], which corresponds to the uniform
continuity of the homeomorphism and was also achieved on [5, Theorem 2.1].
Nevertheless, as a byproduct of the existence of an uniform bound for the
derivative of the homeomorphism at any fixed time, we obtain a weaker version
of this.

In the third section of this work we present and prove the topological
equivalence between our systems, admitting a wide family of dichotomies,
similar to those described by P. Gonzalez et. al. [4] for the discrete
framework, work from which we borrow important tools, adapted for the
continuous framework.

In the fourth section we prove under technical conditions that the
homeomorphism that we previously constructed is a $C^{1}-$diffeomorphism and
give an explicit form for its derivative. Furthermore, we give examples of
concrete dichotomies that satisfy our conditions. Finally, in the fifth
section we give enough conditions to ensure the second class of
differentiability for the topological equivalence, as well as an explicit way
to achieve those conditions with an specific dichotomy.

##  2\. Preliminaries

We study the systems

(2.1) $\dot{x}=A(t)x,$

and

(2.2) $\dot{y}=A(t)y+f(t,y).$

Denote $t\mapsto x(t,\tau,\xi)$ y $t\to y(t,\tau,\eta)$ to the solutions of
(2.1) and (2.2) that pass through $\xi$ and $\eta$ respectively on $t=\tau$.
We also denote $X(t,s)$ the transition matrix of (2.1) such that for $t=s$ is
the identity. Moreover, $A\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathcal{M}_{d}(\mathbb{R})$ is continuous, non
singular (i.e. has invertible images) and uniformly bounded, that is to say,
there exists $M>1$ such that

$\max\left\\{\sup_{s\in\mathbb{R}^{+}}\mathinner{\\!\left\lVert
A(s)\right\rVert},\sup_{s\in\mathbb{R}^{+}}\mathinner{\\!\left\lVert
A^{-1}(s)\right\rVert}\right\\}=M.$

The function $f\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ is such that
there exist sequences $\mathfrak{u},\mathfrak{v}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathbb{R}^{+}$ that satisfy that for every
$s\in\mathbb{R}^{+}$ and every pair
$(y,\tilde{y})\in\mathbb{R}^{d}\times\mathbb{R}^{d}$

$|f(s,y)-f(s,\tilde{y})|\leq\mathfrak{v}(s)|y-\tilde{y}|\text{ ;
}|f(s,y)|\leq\mathfrak{u}(s).$

Furthermore, allow us to consider the following hypothesi:

  * (c1)

(2.1) admits a non uniform dichotomy, i.e. there are two invariant
complementary projectors $P(\cdot)$ and $Q(\cdot)$ such that $P(t)+Q(t)=I$ for
every $t\geq 0$, a continuous function $K\mathrel{\mathop{\mathchar
58\relax}}[0,+\infty[\to[0,\infty[$ and a decreasing $C^{1}$ function
$h\mathrel{\mathop{\mathchar 58\relax}}[0,\infty[\to]0,1]$, such that $h(0)=1$
and $\lim_{t\to\infty}h(t)=0$ and they satisfy

$\left\\{\begin{array}[]{lc}\mathinner{\\!\left\lVert
X(t,s)P(s)\right\rVert}\leq K(s)\left({\frac{h(t)}{h(s)}}\right),&\forall
t\geq s\geq 0\\\ \\\ \mathinner{\\!\left\lVert X(t,s)Q(s)\right\rVert}\leq
K(s)\left({\frac{h(s)}{h(t)}}\right),&\forall 0\leq t\leq
s.\end{array}\right.$

  * (c2)

$\int_{0}^{t}\mathinner{\\!\left\lVert
X(t,s)P(s)\right\rVert}\mathfrak{u}(s)ds+\int_{t}^{\infty}\mathinner{\\!\left\lVert
X(t,s)Q(s)\right\rVert}\mathfrak{u}(s)ds\leq\mathfrak{p}<\infty\text{ , for
every }t\in\mathbb{R}^{+}.$

  * (c3)

$\int_{0}^{t}\mathinner{\\!\left\lVert
X(t,s)P(s)\right\rVert}\mathfrak{v}(s)ds+\int_{t}^{\infty}\mathinner{\\!\left\lVert
X(t,s)Q(s)\right\rVert}\mathfrak{v}(s)ds\leq\mathfrak{q}<1\text{ , for every
}t\in\mathbb{R}^{+}.$

  * (c4)

The map $u\mapsto f(t,u)$ and its derivatives respect to $u$ up to the order
$r$ ($r\geq 1$) are continuous functions of
$(t,u)\in\mathbb{R}^{+}\times\mathbb{R}^{d}$ and
$\sup_{u\in\mathbb{R}^{d}}\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(t,u)\right\rVert}<+\infty$ is bounded.

  * (c5)

For every fixed $\tau\in\mathbb{R}^{+}$, the functions $K,h$ and
$\mathfrak{v}$ satisfy

$\int_{\tau}^{\infty}K(s)h(s)\mathfrak{v}(s)\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)<+\infty.$

######  Remark 2.1.

The projectors $P$ and $Q$ have been called invariant for (2.1), which means
that for every $t,s\in\mathbb{R}^{+}$ they verify:

$P(t)X(t,s)=X(t,s)P(s)\text{ and }Q(t)X(t,s)=X(t,s)Q(s).$

######  Definition 2.2.

Green’s operator associated to (2.1) and the dichotomy (c1) is the matrix
function $\mathcal{G}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\times\mathbb{R}^{+}\to\mathcal{M}_{d}(\mathbb{R})$
given by:

$\mathcal{G}(t,s)=\left\\{\begin{array}[]{cc}X(t,s)P(s)&\forall t\geq s\geq
0,\\\ \\\ -X(t,s)Q(s)&\forall 0\leq t<s,\end{array}\right.$

and it is easily deduced that

$\frac{\partial\mathcal{G}}{\partial t}(t,s)=A(t)\mathcal{G}(t,s).$

######  Remark 2.3.

With this notation, conditions (c2) and (c3) are rewritten as:

  * (c2)

$\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{u}(s)ds\leq\mathfrak{p}<\infty\text{
, for every }t\in\mathbb{R}^{+}.$

  * (c3)

$\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)ds\leq\mathfrak{q}<1\text{
, for every }t\in\mathbb{R}^{+}.$

##  3\. Topological equivalence

In this section we prove that the linear system (2.1) and (2.2) are
topologically equivalents on the positive half line. In order to state this
result, we first recall the following definition.

######  Definition 3.1.

Let $J\subset\mathbb{R}$ an interval. Systems (2.1) and (2.2) are
$J$-topologically equivalent if there is a function
$H\mathrel{\mathop{\mathchar 58\relax}}J\times\mathbb{R}^{d}\to\mathbb{R}^{d}$
that satisfies

  * i)

If $x(t)$ is solution of (2.1), then $H[t,x(t)]$ is solution of (2.2).

  * ii)

$H(t,u)-u$ is bounded on $J\times\mathbb{R}^{d}$.

  * iii)

For every fixed $\tau\in J$, the map $u\mapsto H(\tau,u)$ is an homeomophism
of $\mathbb{R}^{d}$.

Moreover, the function $u\mapsto G(\tau,u)=H^{-1}(\tau,u)$ verifies conditions
ii) and iii) and maps solutions of (2.2) on solutions of (2.1).

######  Theorem 3.2.

If conditions (c1), (c2) and (c3) hold, then (2.1) and (2.2) are topologically
equivalent on $\mathbb{R}^{+}$.

###### Proof.

We develop the proof in several steps.

Step 1: Auxiliary functions. We define $w^{*}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathbb{R}^{d}$ for
$(\tau,\eta)\in\mathbb{R}^{+}\times\mathbb{R}^{d}$ by

$\displaystyle w^{*}(t;(\tau,\eta))$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,\eta))ds$
$\displaystyle=$
$\displaystyle-\int_{0}^{t}X(t,s)P(s)f(s,y(s,\tau,\eta))ds+\int_{t}^{\infty}X(t,s)Q(s)f(s,y(s,\tau,\eta))ds.$

We also define $T\mathrel{\mathop{\mathchar
58\relax}}BC(\mathbb{R}^{+},\mathbb{R}^{d})\to
BC(\mathbb{R}^{+},\mathbb{R}^{d})$ for
$(\tau,\xi)\in\mathbb{R}^{+}\times\mathbb{R}^{d}$ by

$T(\phi)(t;(\tau,\xi))=\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,\tau,\xi)+\phi(s))ds.$

$T$ is well defined by (c2). Using condition (c3) we obtain

$\displaystyle\left|T(\phi)(t;(\tau,\xi))-T(\psi)(t;(\tau,\xi))\right|$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\left|\phi(s)-\psi(s)\right|\mathfrak{v}(s)ds$
$\displaystyle\leq$
$\displaystyle\mathfrak{q}\mathinner{\\!\left\lVert\phi-\psi\right\rVert}_{\infty},$

hence, using Banach’s fixed point Theorem, we obtain the existence of an
unique fixed point

$z^{*}(t;(\tau,\xi))=\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,\tau,\xi)+z^{*}(s;(\tau,\xi)))ds.$

It is easy to verify $t\mapsto w^{*}(t;(\tau,\eta))$ is solution to the
initial value problem

$\left\\{\begin{array}[]{ccl}\dot{w}(t)&=&A(t)w(t)-f(t,y(t,\tau,\eta))\\\ \\\
w(0)&=&-{\int}_{0}^{\infty}X(0,s)Q(s)f(s,y(s,\tau,\eta))ds,\end{array}\right.$

while $t\mapsto z^{*}(t;(\tau,\xi))$ is respectively solution to the initial
value problem:

$\left\\{\begin{array}[]{ccl}\dot{z}(t)&=&A(t)z(t)-f(t,x(t,\tau,\xi)+z(t))\\\
\\\
z(0)&=&-{\int}_{0}^{\infty}X(0,s)Q(s)f(s,x(s,\tau,\xi)+z^{*}(s;(\tau,\xi)))ds,\end{array}\right.$

furthermore, by using (c2), the maps $t\mapsto w^{*}(t;(t,\eta))$ and
$t\mapsto z^{*}(t;(t,\xi))$ are uniformly bounded.

Step 2: Construct maps $H$ and $G$. By uniqueness of solutions

(3.2) $x(t,\tau,\xi)=x(t,s,x(s,\tau,\xi))\text{ , for every
}t,s,\tau\in\mathbb{R}^{+},$

and

(3.3) $z^{*}(t;(\tau,\xi))=z^{*}(t;(s,x(s,\tau,\xi)))\text{ , for every
}t,s,\tau\in\mathbb{R}^{+}.$

For every fixed $t\in\mathbb{R}^{+}$ we define
$H(t,\cdot)\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{d}\to\mathbb{R}^{d}$ and
$G(t,\cdot)\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{d}\to\mathbb{R}^{d}$ by

(3.4)
$\left\\{\begin{array}[]{ccl}H(t,\xi)&=&\xi+{\int}_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,t,\xi)+z^{*}(s;(t,\xi)))ds\\\
\\\ &=&\xi+z^{*}(t;(t,\xi))\end{array}\right.$

and

(3.5)
$\left\\{\begin{array}[]{ccl}G(t,\eta)&=&\eta-{\int}_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,t,\eta))ds\\\
\\\ &=&\eta+w^{*}(t;(t,\eta)).\end{array}\right.$

Using (c2), it follows immediately that $H(t,\xi)-\xi$ and $G(t,\eta)-\eta$
are bounded on $\mathbb{R}^{+}\times\mathbb{R}^{d}$, hence $H$ and $G$ satisfy
condition ii) from Definition 3.1. In order to study additional properties of
the map $G$, note that for $\tau\geq t$

$y(t,\tau,\eta)=X(t,\tau)\eta-\int_{t}^{\tau}X(t,s)f(s,y(s,\tau,\eta))ds,$

or equivalently

$\displaystyle X(\tau,t)y(t,\tau,\eta)$ $\displaystyle=$
$\displaystyle\eta-\int_{t}^{\tau}X(\tau,s)f(s,y(s,\tau,\eta))ds$
$\displaystyle=$
$\displaystyle\eta-\int_{t}^{\tau}X(\tau,s)P(s)f(s,y(s,\tau,\eta))ds-\int_{t}^{\tau}X(\tau,s)Q(s)f(s,y(s,\tau,\eta))ds,$

in particular, for $t=0$ we obtain

$\displaystyle X(\tau,0)y(0,\tau,\eta)$ $\displaystyle=$
$\displaystyle\eta-\int_{0}^{\tau}X(\tau,s)P(s)f(s,y(s,\tau,\eta))ds-\int_{0}^{\tau}X(\tau,s)Q(s)f(s,y(s,\tau,\eta))ds$
$\displaystyle=$
$\displaystyle\eta-\int_{0}^{\tau}X(\tau,s)P(s)f(s,y(s,\tau,\eta))ds$
$\displaystyle+\int_{\tau}^{\infty}X(\tau,s)Q(s)f(s,y(s,\tau,\eta))ds-\int_{0}^{\infty}X(\tau,s)Q(s)f(s,y(s,\tau,\eta))ds$
$\displaystyle=$
$\displaystyle\eta-\int_{0}^{\infty}\mathcal{G}(\tau,s)f(s,y(s,\tau,\eta))ds$
$\displaystyle-X(\tau,0)\int_{0}^{\infty}X(0,s)Q(s)f(s,y(s,\tau,\eta))ds$
$\displaystyle=$ $\displaystyle G(\tau,\eta)-X(\tau,0)w^{*}(0;(\tau,\eta)),$

thus

(3.6)
$G(\tau,\eta)=X(\tau,0)\left\\{y(0,\tau,\eta)+w^{*}(0;(\tau,\eta))\right\\}.$

Step 3: $H$ maps solutions of (2.1) on solutions of (2.2) and $G$ maps
solutions of (2.2) on solutions of (2.1). By uniqueness of solutions, by
simple differentiation we obtain

(3.7) $H[t,x(t,\tau,\xi)]=y(t,\tau,H(\tau,\xi))$

and

(3.8) $G[t,y(t,\tau,\eta)]=x(t,\tau,G(\tau,\eta))=X(t,\tau)G(\tau,\eta),$

hence both maps satisfy condition i) of Definition 3.1 respectively.

Step 4: $u\mapsto G(t,u)$ and $u\mapsto H(t,u)$ are bijective for every fixed
$t\geq 0$.

First we show $H(t,G(t,\eta))=\eta$ for every $t\geq 0$. Using (3.4) and (3.5)

$\displaystyle H(t,G[t,y(t,\tau,\eta)])$ $\displaystyle=$ $\displaystyle
G[t,y(t,\tau,\eta)]$
$\displaystyle+\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,t,G[t,y(t,\tau,\eta)])+z^{*}(s;(t,G[t,y(t,\tau,\eta)])))$
$\displaystyle=$ $\displaystyle
y(t,\tau,\eta)-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,\eta))ds$
$\displaystyle+\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,t,G[t,y(t,\tau,\eta)])+z^{*}(s;(t,G[t,y(t,\tau,\eta)]))).$

Define $\omega(t)=\left|H[t,G[t,y(t,\tau,\eta)]]-y(t,\tau,\eta)\right|$. Note
that

$\displaystyle\omega(t)\leq\left|H[t,G[t,y(t,\tau,\eta)]]-G[t,y(t,\tau,\eta)]\right|+\left|G[t,y(t,\tau,\eta)]-y(t,\tau,\eta)\right|<\infty,$

since both $H$ and $G$ satisfy condition ii) of Definition 3.1. Thus, using
(c3), along with the previous expression and the identities (3.3), (3.4) and
(3.8), for an arbitrary $t\in\mathbb{R}^{+}$ we have

$\displaystyle\omega(t)$ $\displaystyle=$
$\displaystyle\left|\int_{0}^{\infty}\mathcal{G}(t,s)\left\\{f(s,x(s,t,G[t,y(t,\tau,\eta)])+z^{*}(s;(t,G[t,y(t,\tau,\eta)])))-f(s,y(s,\tau,\eta))\right\\}ds\right|$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)\left|x(s,t,G[t,y(t,\tau,\eta)])+z^{*}(s;(t,G[t,y(t,\tau,\eta)]))-y(s,\tau,\eta)\right|ds$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)\left|x(s,t,x(t,\tau,G(\tau,\eta)))+z^{*}(s;(t,x(t,\tau,G(\tau,\eta))))-y(s,\tau,\eta)\right|ds$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)\left|x(s,\tau,G(\tau,\eta))+z^{*}(s;(s,x(s,\tau,G(\tau,\eta))))-y(s,\tau,\eta)\right|ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)\left|H[s,G[s,y(s,\tau,\eta)])]-y(s,\tau,\eta)\right|ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}(s)\omega(s)ds\leq\mathfrak{q}\cdot\sup_{s\in\mathbb{R}^{+}}\\{\omega(s)\\}.$

Thus $\omega(t)=0$ for every $t\in\mathbb{R}^{+}$, otherwise we get a
contradiction. In particular, taking $t=\tau$ we obtain
$H(\tau,G(\tau,\eta))=\eta$. Now we show $G(t,H(t,\xi))=\xi$. Indeed, using
(3.2), (3.3), (3.4), (3.5) and (3.7) we get

$\displaystyle G[t,H[t,x(t,\tau,\xi)]]$ $\displaystyle=$ $\displaystyle
H[t,x(t,\tau,\xi)]-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,t,H[t,x(t,\tau,\xi)]))ds$
$\displaystyle=$ $\displaystyle
H[t,x(t,\tau,\xi)]-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,t,y(t,\tau,H(\tau,\xi))))ds$
$\displaystyle=$ $\displaystyle
H[t,x(t,\tau,\xi)]-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,H(\tau,\xi)))ds$
$\displaystyle=$ $\displaystyle
x(t,\tau,\xi)+\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,t,x(t,\tau,\xi))+z^{*}(s;(t,x(t,\tau,\xi))))ds$
$\displaystyle-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,H(\tau,\xi)))ds$
$\displaystyle=$ $\displaystyle
x(t,\tau,\xi)+\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,\tau,x(\tau,\tau,\xi))+z^{*}(s;(s,x(s,\tau,\xi))))ds$
$\displaystyle-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,H(\tau,\xi)))ds$
$\displaystyle=$ $\displaystyle x(t,\tau,\xi)$
$\displaystyle+\int_{0}^{\infty}\mathcal{G}(t,s)\left\\{f(s,x(s,\tau,\xi)+z^{*}(s;(s,x(s,\tau,\xi))))-f(s,y(s,\tau,H(\tau,\xi)))\right\\}ds$
$\displaystyle=$ $\displaystyle
x(t,\tau,\xi)+\int_{0}^{\infty}\mathcal{G}(t,s)\left\\{f(s,H(s,x(s,\tau,\xi)))-f(s,y(s,\tau,H(\tau,\xi)))\right\\}ds$
$\displaystyle=$ $\displaystyle
x(t,\tau,\xi)+\int_{0}^{\infty}\mathcal{G}(t,s)\left\\{f(s,y(s,\tau,H(\tau,\xi)))-f(s,y(s,\tau,H(\tau,\xi)))\right\\}ds$
$\displaystyle=$ $\displaystyle x(t,\tau,\xi),$

evaluating on $t=\tau$ we obtain $G(\tau,H(\tau,\xi))=\xi$.

Step 5: $u\mapsto G(t,u)$ is continuous.

It is enough to show $\eta\mapsto w^{*}(t;(t,\eta))$ is continuous for every
$t\geq 0$, since $G(t,\eta)=\eta+w^{*}(t;(t,\eta))$. Let
$\eta\in\mathbb{R}^{d}$ and
$\\{\eta_{n}\\}_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ be a sequence such that
$\lim_{n\to\infty}\eta_{n}=\eta$. Fix $t,\tau\in\mathbb{R}^{+}$ and define
$(a_{n})_{n\in\mathbb{N}}$ the sequence of functions over $\mathbb{R}^{+}$
given by

$a_{n}(s)=\mathcal{G}(t,s)f(s,y(s,\tau,\eta_{n})),$

notice that

$|a_{n}(s)|\leq\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{u}(s)\text{
, for every }s\in\mathbb{R}^{+}\text{ and }n\in\mathbb{N}.$

On the other hand, as $u\mapsto f(s,u)$ and $\xi\mapsto y(s,\tau,\xi)$ are
continuous, it is clear $(a_{n})_{n\in\mathbb{N}}$ converges pointwise to
$a\mathrel{\mathop{\mathchar 58\relax}}\mathbb{R}^{+}\to\mathbb{R}^{d}$ given
by

$a(s)=\mathcal{G}(t,s)f(s,y(s,\tau,\eta)).$

Thus, by Lebesgue’s dominated convergence Theorem we have

$\displaystyle\lim_{n\to\infty}w^{*}(t;(\tau,\eta_{n}))$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}-\int_{0}^{\infty}\mathcal{G}(t,s)f(s,y(s,\tau,\eta_{n}))ds$
$\displaystyle=$
$\displaystyle-\lim_{n\to\infty}\int_{0}^{\infty}a_{n}(s)ds=-\int_{0}^{\infty}a(s)ds=w^{*}(t;(\tau,\eta)).$

Hence $\eta\mapsto w^{*}(t;(\tau,\eta))$ is continuous and in particular
$\eta\mapsto w^{*}(t;(t,\eta))$ is also continuous.

Step 6: $u\mapsto H(t,u)$ is continuous.

It is enough to show $\xi\mapsto z^{*}(t;(t,\xi))$ is continuous for every
$t\geq 0$.

Let $\xi\in\mathbb{R}^{d}$ and a sequence
$\\{\xi_{n}\\}_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ such that
$\lim_{n\to\infty}\xi_{n}=\xi$. Let $u\mapsto\phi(t;(\tau,u))$ be a continuous
function for $t,\tau\in\mathbb{R}^{+}$ fixed. We define

$b_{n}(s)=\mathcal{G}(t,s)f(s,x(s,\tau,\xi_{n})+\phi(s;(\tau,\xi_{n}))),$

note it satisfies

$|b_{n}(s)|\leq\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{u}(s)\text{
, para todo }s\in\mathbb{R}^{+}\text{ y }n\in\mathbb{N},$

and

$\lim_{n\to\infty}b_{n}(s)=\mathcal{G}(t,s)f(s,x(s,\tau,\xi)+\phi(s;(\tau,\xi)))\mathrel{\mathop{\mathchar
58\relax}}=b(s).$

Using Lebesgue’s dominated convergence Theorem we have

$\displaystyle\lim_{n\to\infty}(T\phi)(t;(\tau,\xi_{n}))$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{0}^{\infty}\mathcal{G}(t,s)f(s,x(s,\tau,\xi_{n})+\phi(s;(\tau,\xi_{n})))ds$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{0}^{\infty}b_{n}(s)ds=\int_{0}^{\infty}b(s)ds=(T\phi)(t;(\tau,\xi)),$

thus $\xi\mapsto(T\phi)(t;(\tau,\xi))$ is continuous, and hence it’s fixed
point $\xi\mapsto z^{*}(t;(\tau,\xi))$ is continuous as well, which allows us
to conclude $H$ is continuous and so it is an homeomorphism. In conclusion
(2.1) and (2.2) are topologically equivalent on $\mathbb{R}^{+}$. ∎

##  4\. Differentiability of topological equivalence under a dichotomy

In this section we prove that the topological equivalence is of class $C^{1}.$
Our approach does not impose resonance conditions or spectral gaps.

We recall of the definition, introduced on [4], of $C^{r}-$ topologically
equivalent on the positive half line .

######  Definition 4.1.

The systems (2.1) and (2.2) are $C^{r}$-topologically equivalent on
$\mathbb{R}^{+}$ if they are topologically equivalent on $\mathbb{R}^{+}$ with
the map $u\mapsto H(t,u)$, which is a diffeomorphism of class $C^{r}$, with
$r\geq 1$, for every fixed $t\geq 0$.

######  Lemma 4.2.

If conditions (c1)-(c5) hold, with $r=1$ on (c4), then $\eta\mapsto
w^{*}(0;(\tau,\eta))$ defined on (3) is a $C^{1}$ map.

###### Proof.

Fix $\eta\in\mathbb{R}^{d}$ and let
$(\delta_{n})_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ be a properly convergent
to zero sequence. Fix $\tau\in\mathbb{R}^{+}$ and define

$\varphi_{n}(s)=\mathcal{G}(0,s)\frac{f(s,y(s,\tau,\eta+\delta_{n}))-f(s,y(s,\tau,\eta))-\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\delta_{n}}{|\delta_{n}|}.$

As $\eta\mapsto y(s,\tau,\eta)$ is continuous, then
$\lim_{n\to\infty}y(s,\tau,\eta+\delta_{n})=y(s,\tau,\eta)$. Using (c4) it
follows from classic results of differentiability respect to the initial
conditions (see for example Theorem 4.1 in [10]) that $\eta\mapsto
y(s,\tau,\eta)$ is differentiable. Again by (c4) we obtain

$\lim_{n\to\infty}\varphi_{n}(s)=0.$

Also note

$\displaystyle\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,u)\right\rVert}=\lim_{\delta\to
0}\frac{|f(s,u+\delta)-f(s,u)|}{|\delta|}\leq\lim_{\delta\to
0}\mathfrak{v}(s)=\mathfrak{v}(s),$

hence

$\displaystyle|\varphi_{n}(s)|$ $\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\frac{\left|f(s,y(s,\tau,\eta+\delta_{n}))-f(s,y(s,\tau,\eta))\right|+\left|\frac{\partial
f}{\partial u}(sy(s,\tau,\eta))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\delta_{n}\right|}{|\delta_{n}|}$
$\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\frac{\mathfrak{v}(s)\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|+\mathfrak{v}(s)\left|\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\delta_{n}\right|}{|\delta_{n}|}$
$\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\mathfrak{v}(s)\left(\frac{\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|}{|\delta_{n}|}+\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}\right).$

Let $\tilde{\eta}\in\mathbb{R}^{d}$ and $s\geq\tau$. We know

$y(s,\tau,\eta)=y(\tau,\tau,\eta)+\int_{\tau}^{s}\dot{y}(r,\tau,\eta)dr=\eta+\int_{\tau}^{s}\dot{y}(r,\tau,\eta)dr,$

hence

$y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})=\eta-\tilde{\eta}+\int_{\tau}^{s}\dot{y}(r,\tau,\eta)-\dot{y}(r,\tau,\tilde{\eta})dr.$

Defining $\mathfrak{z}(s)=\dot{y}(s,\tau,\eta)-\dot{y}(s,\tau,\tilde{\eta})$,
the previous expression implies

$|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|\leq|\eta-\tilde{\eta}|+\int_{\tau}^{s}|\mathfrak{z}(r)|dr.$

Note

$\dot{y}(s,\tau,\eta)-\dot{y}(s,\tau,\tilde{\eta})=A(s)\left[y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})\right]+f(s,y(s,\tau,\eta))-f(s,y(s,\tau,\tilde{\eta})),$

from where it follows that

$\displaystyle|\mathfrak{z}(s)|$ $\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert
A(s)\right\rVert}|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|+|f(s,y(s,\tau,\eta))-f(s,y(s,\tau,\tilde{\eta}))|$
$\displaystyle\leq$ $\displaystyle\mathinner{\\!\left\lVert
A(s)\right\rVert}|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|+\mathfrak{v}(s)|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|$
$\displaystyle\leq$ $\displaystyle\left(\mathinner{\\!\left\lVert
A(s)\right\rVert}+\mathfrak{v}(s)\right)|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|$
$\displaystyle\leq$ $\displaystyle\left(\mathinner{\\!\left\lVert
A(s)\right\rVert}+\mathfrak{v}(s)\right)\left[|\eta-\tilde{\eta}|+\int_{\tau}^{s}|\mathfrak{z}(r)|dr\right].$

Define
$\mathcal{Z}(s)=|\eta-\tilde{\eta}|+\int_{\tau}^{s}|\mathfrak{z}(r)|dr$. Note
$s>\tau\Rightarrow\mathcal{Z}\neq 0$, thus, the previous expression becomes

$\frac{\mathcal{Z}^{\prime}(s)}{\mathcal{Z}(s)}\leq\left(\mathinner{\\!\left\lVert
A(s)\right\rVert}+\mathfrak{v}(s)\right),$

from where

$\log\left(\mathcal{Z}(s)\right)-\log\left(\mathcal{Z}(\tau)\right)=\log\left(\frac{\mathcal{Z}(s)}{|\eta-\tilde{\eta}|}\right)\leq\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr,$

hence

$|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})|\leq|\eta-\tilde{\eta}|+\int_{\tau}^{s}|z(r)|dr=\mathcal{Z}(s)\leq|\eta-\tilde{\eta}|\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right),$

so, for $s\geq\tau$

$\frac{|y(s,\tau,\eta)-y(s,\tau,\eta+\delta_{n})|}{|\delta_{n}|}\leq\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right),$

which in particular implies

$\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}\leq\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right).$

In conclusion, for $s\geq\tau$ we have

$\displaystyle|\varphi_{n}(s)|$ $\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\mathfrak{v}(s)\left(\frac{\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|}{|\delta_{n}|}+\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}\right)$ $\displaystyle\leq$
$\displaystyle
2\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\mathfrak{v}(s)\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)$ $\displaystyle\leq$ $\displaystyle
2K(s)h(s)\mathfrak{v}(s)\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right).$

On the other hand, on the compact $[0,\tau]$ the sequence of continuous
functions
$s\mapsto\frac{\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|}{|\delta_{n}|}$
converge pointwise to $s\mapsto\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}$ when $n\to\infty$, which is also
continuous, hence the convergence is uniform, i.e.  there exist
$\hat{n}\in\mathbb{N}$ such that $n\geq\hat{n}$ and $s\in[0,\tau]$ we have

$\frac{\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|}{|\delta_{n}|}+\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}\leq
2\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}+1.$

Hence, for $n\geq\hat{n}$ and $s\in[0,\tau]$ we have

$\displaystyle|\varphi_{n}(s)|$ $\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\mathfrak{v}(s)\left(\frac{\left|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)\right|}{|\delta_{n}|}+\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}\right)$ $\displaystyle\leq$
$\displaystyle
K(s)h(s)\mathfrak{v}(s)\left(2\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}+1\right).$

Summarizing, if we define $\mathcal{F}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathbb{R}$ by

$\mathcal{F}(s)=\left\\{\begin{array}[]{cc}K(s)h(s)\mathfrak{v}(s)\left(2\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}+1\right)&\tau\geq s\geq 0,\\\ \\\
2K(s)h(s)\mathfrak{v}(s)\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)&\forall
s\geq\tau,\end{array}\right.$

we get $|\varphi_{n}(s)|\leq\mathcal{F}(s)$ for every $n\geq\hat{n}$. Note
that using (c5)

$\displaystyle\int_{0}^{\infty}\mathcal{F}(s)ds$ $\displaystyle\leq$
$\displaystyle\int_{0}^{\tau}K(s)h(s)\mathfrak{v}(s)\left(2\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}+1\right)ds$
$\displaystyle+2\int_{\tau}^{\infty}K(s)h(s)\mathfrak{v}(s)\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)ds<+\infty.$

Thus, using Lebesgue’s dominated convergence Theorem we obtain

$\lim_{n\to\infty}\frac{w^{*}(0;(\tau,\eta+\delta_{n}))-w^{*}(0;(\tau,\eta))+\left[\int_{0}^{\infty}\mathcal{G}(0,s)\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)ds\right]\delta_{n}}{|\delta_{n}|}$

$=\lim_{n\to\infty}\int_{0}^{\infty}-\varphi_{n}(s)ds=-\int_{0}^{\infty}\left(\lim_{n\to\infty}\varphi_{n}(s)\right)ds=0,$

hence $\eta\mapsto w^{*}(0;(\tau,\eta))$ is differentiable. ∎

######  Corollary 4.3.

If conditions (c1)-(c5) hold, with $r=1$ on (c4), then for every fixed
$\tau\in\mathbb{R}^{+}$ we have

$\frac{\partial
w^{*}(0;(\tau,\eta))}{\partial\eta}=-\int_{0}^{\infty}\mathcal{G}(0,s)\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)ds.$

######  Theorem 4.4.

If conditions (c1)-(c5) hold, with $r=1$ on (c4), then the systems (2.1) and
(2.2) are $C^{1}$-topologically equivalent on $\mathbb{R}^{+}$.

###### Proof.

By Theorem 3.2 we know they are topologically equivalent. By Lemma 4.2 we know
$\eta\mapsto w^{*}(0;(\tau,\eta))$ is a $C^{1}$ map and by classic results of
differentiability respect to the initial conditions (see for example Theorem
4.1 in [10]) we know $\eta\mapsto y(0,\tau,\eta)$ is as well a $C^{1}$ map,
which along with the expression (3.6) allows us to conclude $\eta\mapsto
G(\tau,\eta)$ is a $C^{1}$ map.

Furthermore, as $G$ is a topological equivalence, then $\xi\mapsto
G(\tau,\xi)-\xi$ is bounded, thus $G(\tau,\xi)\to\infty$ when
$|\xi|\to\infty$. This fact, along with the previous discussion implies by
[12, Corollary 2.1] that $\xi\mapsto G(\tau,\xi)$ is a $C^{1}$ diffeomorphism.
Moreover, as $G(\tau,H(\tau,\xi))=\xi$, we have

$\frac{\partial G}{\partial\xi}(\tau,H(\tau,\xi))\frac{\partial
H}{\partial\xi}(\tau,\xi)=I,$

and so $\frac{\partial H}{\partial\xi}(\tau,\xi)=\left[\frac{\partial
G}{\partial\xi}(\tau,H(\tau,\xi))\right]^{-1}$, which completes the proof. ∎

######  Corollary 4.5.

If (c4) is satisfied with $r=1$. Furthermore, suppose (2.1) admits an
exponential dichotomy, i.e. $P$ and $Q$ are constant complementary projectors,
$K(s)=K>0$ for every $s\in\mathbb{R}^{+}$ and $h(s)=e^{-\lambda s}$, with
$\lambda>0$. Suppose $\mathfrak{v}(s)=\mathfrak{v}>0$ and
$\mathfrak{u}(s)=\mathfrak{u}>0$ for every $s\in\mathbb{R}^{+}$. Then, if
$2K\mathfrak{v}<\lambda$ and $M+\mathfrak{v}<\lambda$, the systems (2.1) and
(2.2) are $C^{1}$-topologically equivalent on $\mathbb{R}^{+}$.

###### Proof.

For an arbitrary $t\in\mathbb{R}^{+}$ we have

$\displaystyle\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{u}ds$
$\displaystyle=$ $\displaystyle\int_{0}^{t}\mathinner{\\!\left\lVert
X(t,s)P\right\rVert}\mathfrak{u}ds+\int_{t}^{\infty}\mathinner{\\!\left\lVert
X(t,s)Q\right\rVert}\mathfrak{u}ds$ $\displaystyle\leq$
$\displaystyle\int_{0}^{t}e^{-\lambda(t-s)}K\mathfrak{u}ds+\int_{t}^{\infty}e^{-\lambda(s-t)}K\mathfrak{u}ds$
$\displaystyle\leq$ $\displaystyle K\mathfrak{u}\frac{1-e^{-\lambda
t}}{\lambda}+\frac{K\mathfrak{u}}{\lambda}\leq\frac{2K\mathfrak{u}}{\lambda},$

thus (c2) is satisfied. Analogously

$\int_{0}^{\infty}\mathinner{\\!\left\lVert\mathcal{G}(t,s)\right\rVert}\mathfrak{v}ds\leq\frac{2K\mathfrak{v}}{\lambda}<1,$

thus (c3) is verified. Finally, for an arbitrary $\tau\in\mathbb{R}^{+}$ we
have

$\displaystyle{\int_{\tau}^{\infty}}K\mathfrak{v}e^{-\lambda
s}\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}dr\right)ds$ $\displaystyle\leq$
$\displaystyle{\int_{\tau}^{\infty}}K\mathfrak{v}e^{-\lambda
s}\exp\left(\int_{\tau}^{s}M+\mathfrak{v}dr\right)ds$ $\displaystyle=$
$\displaystyle{\int_{\tau}^{\infty}}K\mathfrak{v}e^{-\lambda
s}e^{(M+\mathfrak{v})(s-\tau)}ds$ $\displaystyle=$ $\displaystyle
e^{-(M+\mathfrak{v})\tau}K\mathfrak{v}{\int_{\tau}^{\infty}}e^{(M+\mathfrak{v}-\lambda)s}ds<+\infty,$

so (c5) is satisfied. Applying Theorem 4.4 the systems are
$C^{1}$-topologically equivalent on $\mathbb{R}^{+}$. ∎

######  Corollary 4.6.

Suppose (2.1) admits a nonuniform exponential dichotomy, i.e. there are two
complementary invariant projectors $P(\cdot)$ and $Q(\cdot)$ and constants
constanes $C,\lambda,\varepsilon_{1}>0$ such that

$\left\\{\begin{array}[]{lc}\mathinner{\\!\left\lVert
X(t,s)P(s)\right\rVert}\leq Ce^{-\lambda(t-s)+\varepsilon_{1}s},&\forall t\geq
s\geq 0\\\ \\\ \mathinner{\\!\left\lVert X(t,s)Q(s)\right\rVert}\leq
Ce^{\lambda(t-s)+\varepsilon_{1}s},&\forall 0\leq t\leq s.\end{array}\right.$

Furthermore, suppose that for each $t\in\mathbb{R}^{+}$ $u\mapsto f(t,u)$ is a
$C^{1}$ map such that $u\mapsto\frac{\partial f}{\partial u}(t,u)$ is a
bounded map that satisfies

(4.1) $|f(s,u)|\leq\kappa e^{-\varepsilon_{0}s}$

and

(4.2) $\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,u)\right\rVert}\leq\nu e^{-\varepsilon_{1}s},$

for given $\nu,\kappa>0$ y $\varepsilon_{0}>\varepsilon_{1}-\lambda$. Then, if
$M<\lambda$, for $\nu>0$ being small enough the systems (2.1) and (2.2) are
$C^{1}$-topologically equivalent on $\mathbb{R}^{+}.$

###### Proof.

Condition (c1) is easily verified with $K(s)=Ce^{\varepsilon_{1}s}$ and
$h(s)=e^{-\lambda s}$. Note that condition (4.2) implies

$|f(s,y)-f(s,\tilde{y})|\leq\nu e^{-\varepsilon_{1}s}|y-\tilde{y}|,$

hence our general conditions are verified with $\mathfrak{v}(s)=\nu
e^{-\varepsilon_{1}s}$ and $\mathfrak{u}(s)=\kappa e^{-\varepsilon_{0}s}$.
Thus condition (c2) is satisfied inmedietly and, for small enough $\nu$, (c3)
is as well, meanwhile (c4) is granted by hypothesis. Denote

$\Psi_{\tau}(s)=\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right).$

It is easy to see $s\geq\tau\geq t$ implies $\Psi_{t}(s)\leq\Psi_{\tau}(s)$,
hence for a fixed $\tau\in\mathbb{R}^{+}$

$\Psi_{\tau}(s)\leq\Psi_{0}(s)\leq e^{(M+\nu)s}.$

Thus

$\displaystyle\int_{\tau}^{\infty}K(s)h(s)\mathfrak{v}(s)\Psi_{\tau}(s)ds$
$\displaystyle\leq$ $\displaystyle\int_{0}^{\infty}C\nu e^{-\lambda
s}\Psi_{0}(s)ds$ $\displaystyle\leq$ $\displaystyle
C\nu\int_{0}^{\infty}e^{(M+\nu-\lambda)s}ds,$

which, as $M<\lambda$, is finite as soon as $\nu$ is small enough,
particularly $0<\nu<\lambda-M$. Thus, condition (c5) is verified. Applying
Theorem 4.4 the corollary follows. ∎

##  5\. Second class of differentiability for the topological equivalence

We once again consider the expression (3.6) in order to study the second
derivative of the homeomorphism of topological equivalence. Taking in account
classic results of differentiability respect to the initial conditions (see
for example Theorem 4.1 in [10]), we know the map $\eta\mapsto y(0,\tau,\eta)$
has the same class of differentiability as $u\mapsto f(\tau,u)$ for every
$\tau\in\mathbb{R}^{+}$ when conditions (c1)-(c4) are satisfied; hence the
class of differentiability of the homeomorphism relies on the
differentiability of the map $\eta\mapsto w^{*}(0;(\tau,\eta))$.

######  Lemma 5.1.

Suppose conditions (c1)-(c5) hold, with $r=2$ on (c4). Furthermore, suppose
there are functions $\mathfrak{V}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathbb{R}^{+}$ and
$\pi_{\tau}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{R}^{+}\to\mathbb{R}^{+}$ such that

(5.1) $\mathinner{\\!\left\lVert\frac{\partial^{2}f}{\partial
u^{2}}(s,u)\right\rVert}\leq\mathfrak{V}(s)\text{ , for every
}s\in\mathbb{R}^{+}$

and

(5.2)
$\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}\leq\pi_{\tau}(s)\text{
, for every }s\geq\tau.$

If for every fixed $\tau\in\mathbb{R}^{+}$ the previous functions verify

(5.3)
${\int_{\tau}^{\infty}}\left(K(s)h(s)\left\\{\pi_{\tau}(s)\mathfrak{v}(s)+\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2}\right\\}\right)ds<+\infty,$

then $\eta\mapsto w^{*}(0;(\tau,\eta))$ is a $C^{2}$ map.

###### Proof.

We follow the same strategy as in Lemma 4.2. In Corollary 4.3 we established

$\frac{\partial
w^{*}(0;(\tau,\eta))}{\partial\eta}=-\int_{0}^{\infty}\mathcal{G}(0,s)\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta))\frac{\partial y}{\partial\eta}(s,\tau,\eta),$

for every fixed $\tau\in\mathbb{R}^{+}$, which is granted by conditions
(c1)-(c5). Denote

$\mathcal{O}_{\tau,\eta}(s)=\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)+\frac{\partial^{2}f}{\partial
u^{2}}(s,y(s,\tau,\eta))\left(\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right)^{2}.$

Fix $\eta\in\mathbb{R}^{d}$ and let
$(\delta_{n})_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ be a properly convergence
to zero sequence. Choose $\tau\in\mathbb{R}^{+}$ and define the sequence of
functions $(\mathfrak{F}_{n,\tau,\eta})_{n\in\mathbb{N}}$ over
$\mathbb{R}^{+}$ given by

$\mathfrak{F}_{n,\tau,\eta}(s)=\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta+\delta_{n}))\frac{\partial
y}{\partial\eta}(s,\tau,\eta+\delta_{n})-\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial y}{\partial\eta}(s,\tau,\eta).$

We now define $(\psi_{n})_{n\in\mathbb{N}}$ by

$\psi_{n}(s)=\mathcal{G}(0,s)\frac{\mathfrak{F}_{n,\tau,\eta}(s)-\mathcal{O}_{\tau,\eta}(s)\delta_{n}}{|\delta_{n}|},$

which by (c4) verifies

$\lim_{n\to\infty}\psi_{n}(j)=0.$

On the other hand, it is easy to deduce from (5.1) and (5.2) that

$\displaystyle\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,u)-\frac{\partial f}{\partial u}(s,\tilde{u})\right\rVert}$
$\displaystyle\leq$ $\displaystyle\mathfrak{V}(s)|u-\tilde{u}|\text{ , for
every }s\in\mathbb{R}^{+}$

and

$\displaystyle\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)-\frac{\partial
y}{\partial\eta}(s,\tau,\tilde{\eta})\right\rVert}$ $\displaystyle\leq$
$\displaystyle\pi_{\tau}(s)|\eta-\tilde{\eta}|\text{ , for }s\geq\tau.$

Thus, for $s\geq\tau$

$\displaystyle\mathinner{\\!\left\lVert\mathcal{O}_{\tau,\eta}(s)\right\rVert}$
$\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}\mathinner{\\!\left\lVert\frac{\partial
f}{\partial
u}(s,y(s,\tau,\eta))\right\rVert}+\mathinner{\\!\left\lVert\frac{\partial^{2}f}{\partial
u^{2}}(s,y(s,\tau,\eta))\right\rVert}\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}^{2}$ $\displaystyle\leq$
$\displaystyle\pi_{\tau}(s)\mathfrak{v}(s)+\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2},$

and

$\displaystyle\mathinner{\\!\left\lVert\mathfrak{F}_{n,\tau,\eta}(s)\right\rVert}$
$\displaystyle\leq$ $\displaystyle\mathinner{\\!\left\lVert\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta+\delta_{n}))\frac{\partial
y}{\partial\eta}(s,\tau,\eta+\delta_{n})-\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta+\delta_{n}))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}$
$\displaystyle+\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta+\delta_{n}))\frac{\partial
y}{\partial\eta}(s,\tau,\eta)-\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial y}{\partial\eta}(s,\tau,\eta)\right\rVert}$
$\displaystyle\leq$ $\displaystyle\mathinner{\\!\left\lVert\frac{\partial
f}{\partial
u}(s,y(s,\tau,\eta+\delta_{n}))\right\rVert}\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta+\delta_{n})-\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}$
$\displaystyle+\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta+\delta_{n}))-\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\right\rVert}\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}$ $\displaystyle\leq$
$\displaystyle\mathfrak{v}(s)\pi_{\tau}(s)|\delta_{n}|+\mathfrak{V}(s)|y(s,\tau,\eta+\delta_{n})-y(s,\tau,\eta)|\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)$ $\displaystyle\leq$
$\displaystyle\left(\mathfrak{v}(s)\pi_{\tau}(s)+\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2}\right)|\delta_{n}|.$

Now, consider the sequence of continuous functions
$s\mapsto\dfrac{\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)-\frac{\partial
y}{\partial\eta}(s,\tau,\eta+\delta_{n})\right\rVert}}{|\delta_{n}|}$ defined
on $[0,\tau]$. As they converge pointwise to the continuous function
$s\mapsto\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}$
on the compact domain $[0,\tau]$ the convergence is uniform. Thus, there is
$\hat{n}\in\mathbb{N}$ such that

$\dfrac{\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)-\frac{\partial
y}{\partial\eta}(s,\tau,\eta+\delta_{n})\right\rVert}}{|\delta_{n}|}\leq\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}+2\text{
for every }0\leq s\leq\tau;n\geq\hat{n}.$

Thus, for $0\leq s\leq\tau$ and $n\geq\hat{n}$

$\displaystyle\mathinner{\\!\left\lVert\mathfrak{F}_{n,\tau,\eta}(s)\right\rVert}$
$\displaystyle\leq$
$\displaystyle\left(\mathfrak{v}(s)\left(\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}+2\right)+\mathfrak{V}(s)\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}^{2}\right)|\delta_{n}|,$

while for $s\in[0,\tau]$

$\displaystyle\mathinner{\\!\left\lVert\mathcal{O}_{\tau,\eta}(s)\right\rVert}$
$\displaystyle\leq$
$\displaystyle\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}\mathfrak{v}(s)+\mathfrak{V}(s)\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}^{2}.$

Let us define

$\mathscr{F}_{\tau}(s)\mathrel{\mathop{\mathchar
58\relax}}=\left\\{\begin{array}[]{lc}2\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\left(\mathfrak{v}(s)\left(\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}+1\right)+\mathfrak{V}(s)\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}^{2}\right)&,0\leq s\leq\tau\\\ \\\
2\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\left(\mathfrak{v}(s)\pi_{\tau}(s)+\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2}\right)&,s\geq\tau.\end{array}\right.$

It is easy to see
$\mathinner{\\!\left\lVert\psi_{n}(s)\right\rVert}\leq\mathscr{F}_{\tau}(s)$
for every $s\in\mathbb{R}^{+}$ and $n\geq\hat{n}$. Now, using (5.3) we have

$\displaystyle\int_{0}^{\infty}\mathscr{F}_{\tau}(s)ds$ $\displaystyle\leq$
$\displaystyle
2\int_{0}^{\tau}\mathinner{\\!\left\lVert\mathcal{G}(0,s)\right\rVert}\left(\mathfrak{v}(s)\left(\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}+1\right)+\mathfrak{V}(s)\mathinner{\\!\left\lVert\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right\rVert}^{2}\right)ds$
$\displaystyle+2\int_{\tau}^{\infty}K(s)h(s)\left(\mathfrak{v}(s)\pi_{\tau}(s)+\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2}\right)ds<+\infty.$

Finally, using Lebesgue’s dominated convergence Theorem we obtain

$\displaystyle\lim_{n\to\infty}\frac{\frac{\partial
w^{*}}{\partial\eta}(0;(\tau,\eta+\delta_{n}))-\frac{\partial
w^{*}}{\partial\eta}(0;(\tau,\eta))+\left[\int_{0}^{\infty}\mathcal{G}(0,s)\mathcal{O}_{\tau,\eta}(s)ds\right]\delta_{n}}{|\delta_{n}|}$
$\displaystyle=$
$\displaystyle-\lim_{n\to\infty}\int_{0}^{\infty}\psi_{n}(s)ds$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}\left(\lim_{n\to\infty}\psi_{n}(s)\right)=0,$

which implies $\eta\mapsto\frac{\partial w^{*}}{\partial\eta}(0;(\tau,\eta))$
is differentiable, hence $\eta\mapsto w^{*}(0;(\tau,\eta))$ is a $C^{2}$ map.
∎

######  Corollary 5.2.

If conditions from Lemma hold 5.1, then for every fixed
$\tau\in\mathbb{R}^{+}$

$\frac{\partial^{2}w^{*}(0;(\tau,\eta))}{\partial\eta^{2}}=-\int_{0}^{\infty}\mathcal{G}(0,s)\left[\frac{\partial
f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)+\frac{\partial^{2}f}{\partial
u^{2}}(s,y(s,\tau,\eta))\left(\frac{\partial
y}{\partial\eta}(s,\tau,\eta)\right)^{2}\right]ds.$

######  Theorem 5.3.

If conditions (c1)-(c5) hold, with $r=2$ on (c4), and conditions from Lemma
5.1 are satisfied, then (2.1) and (2.2) are $C^{2}$-topologically equivalent
on $\mathbb{R}^{+}$.

Note that Theorem 5.3 follows easily in the same fashion as the proof of
Theorem 4.4. Now we proceed to study some technical results in order to
establish a concrete example of above Theorem.

######  Lemma 5.4.

If conditions (c1)-(c4) hold, with $r=2$ on (c4), then

$\frac{\partial}{\partial s}\frac{\partial
y}{\partial\eta}(s,\tau,\eta)=\frac{\partial}{\partial\eta}\frac{\partial
y}{\partial s}(s,\tau,\eta).$

###### Proof.

Note that

$\displaystyle\frac{\partial}{\partial s}\frac{\partial
y}{\partial\eta}(s,\tau,\eta)$ $\displaystyle=$
$\displaystyle\frac{\partial}{\partial
s}\left(X(s,\tau)+\frac{\partial}{\partial\eta}\int_{\tau}^{s}X(s,r)f(r,y(r,\tau,\eta))dr\right)$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial
s}\left(X(s,\tau)+\int_{\tau}^{s}X(s,r)\frac{\partial f}{\partial
u}(r,y(r,\tau,\eta))\frac{\partial y}{\partial\eta}(r,\tau,\eta)dr\right)$
$\displaystyle=$ $\displaystyle
A(s)\left(X(s,\tau)+\int_{\tau}^{s}X(s,r)\frac{\partial f}{\partial
u}(r,y(r,\tau,\eta))\frac{\partial
y}{\partial\eta}(r,\tau,\eta)dr\right)+\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial y}{\partial\eta}(s,\tau,\eta)$
$\displaystyle=$ $\displaystyle A(s)\frac{\partial
y}{\partial\eta}(s,\tau,\eta)+\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\frac{\partial y}{\partial\eta}(s,\tau,\eta)$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial\eta}\frac{\partial
y}{\partial s}(s,\tau,\eta),$

where the second equality is verified using Lebesgue’s dominated convergence
Theorem. ∎

######  Corollary 5.5.

If conditions (c1)-(c5) hold, with $r=2$ on (c4), then $t\mapsto
z(t,\tau,\eta)\mathrel{\mathop{\mathchar 58\relax}}=\frac{\partial
y}{\partial\eta}(t,\tau,\eta)$ is solution to the matrix initial value
problem:

$\left\\{\begin{array}[]{ccl}z^{\prime}(t)&=&\left[A(t)+\dfrac{\partial
f}{\partial u}(t,y(t,\tau,\eta)\right]z(t)\\\ \\\
z(\tau)&=&I.\end{array}\right.$

######  Lemma 5.6.

Suppose conditions (c1)-(c5) hold, with $r=2$ on (c4). If
$s\mapsto\mathfrak{V}(s)$ satisfies (5.1), then for every $s\geq\tau$

$\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}\leq\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(p)\right\rVert}+\mathfrak{v}(p)dp\right)\cdot\int_{\tau}^{s}\left\\{\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right\\}dp.$

###### Proof.

Let $\eta,\tilde{\eta}\in\mathbb{R}^{n}$ and fix $\tau\in\mathbb{R}^{+}$.
Denote $s\mapsto z(s,\tau,\eta)\mathrel{\mathop{\mathchar
58\relax}}=\frac{\partial y}{\partial\eta}(s,\tau,\eta)$ and $s\mapsto
z(s,\tau,\tilde{\eta})\mathrel{\mathop{\mathchar 58\relax}}=\frac{\partial
y}{\partial\eta}(s,\tau,\tilde{\eta})$. By Corollary 5.5 we know

$z^{\prime}(s,\tau,\eta)=\left[A(s)+\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\right]z(s,\tau,\eta),$

hence

$\displaystyle z^{\prime}(s,\tau,\eta)-z^{\prime}(s,\tau,\tilde{\eta})$
$\displaystyle=$ $\displaystyle
A(s)\left(z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right)+\frac{\partial
f}{\partial u}(s,y(s,\tau,\eta))z(s,\tau,\eta)$ $\displaystyle-\frac{\partial
f}{\partial u}(s,y(s,\tau,\tilde{\eta}))z(s,\tau,\tilde{\eta})$
$\displaystyle=$ $\displaystyle\left[A(s)+\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))\right]\left(z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right)$
$\displaystyle+\left[\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))-\frac{\partial f}{\partial
u}(s,y(s,\tau,\tilde{\eta}))\right]z(s,\tau,\tilde{\eta}).$

Thus

$\displaystyle\mathinner{\\!\left\lVert
z^{\prime}(s,\tau,\eta)-z^{\prime}(s,\tau,\tilde{\eta})\right\rVert}$
$\displaystyle\leq$ $\displaystyle\mathinner{\\!\left\lVert
A(s)+\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta)\right\rVert}\mathinner{\\!\left\lVert
z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right\rVert}$
$\displaystyle+\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,y(s,\tau,\eta))-\frac{\partial f}{\partial
u}(s,y(s,\tau,\tilde{\eta}))\right\rVert}\mathinner{\\!\left\lVert
z(s,\tau,\tilde{\eta})\right\rVert}$ $\displaystyle\leq$
$\displaystyle\left(\mathinner{\\!\left\lVert
A(s)\right\rVert}+\mathfrak{v}(s)\right)\mathinner{\\!\left\lVert
z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right\rVert}$
$\displaystyle+\mathfrak{V}(s)\left|y(s,\tau,\eta)-y(s,\tau,\tilde{\eta})\right|\mathinner{\\!\left\lVert
z(s,\tau,\tilde{\eta})\right\rVert}$ $\displaystyle\leq$
$\displaystyle\left(\mathinner{\\!\left\lVert
A(s)\right\rVert}+\mathfrak{v}(s)\right)\mathinner{\\!\left\lVert
z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right\rVert}$
$\displaystyle+\mathfrak{V}(s)\exp\left(2\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)|\eta-\tilde{\eta}|,$

or equivalently, for $p\in[\tau,s]$

$\displaystyle\frac{\mathinner{\\!\left\lVert
z^{\prime}(p,\tau,\eta)-z^{\prime}(p,\tau,\tilde{\eta})\right\rVert}}{|\eta-\tilde{\eta}|}$
$\displaystyle\leq$ $\displaystyle\left(\mathinner{\\!\left\lVert
A(p)\right\rVert}+\mathfrak{v}(p)\right)\frac{\mathinner{\\!\left\lVert
z(p,\tau,\eta)-z(p,\tau,\tilde{\eta})\right\rVert}}{|\eta-\tilde{\eta}|}+\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right).$

On the other hand

$\displaystyle z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})$ $\displaystyle=$
$\displaystyle
z(\tau,\tau,\eta)+\int_{\tau}^{s}z^{\prime}(p,\tau,\eta)dp-z(\tau,\tau,\tilde{\eta})-\int_{\tau}^{s}z^{\prime}(p,\tau,\tilde{\eta})dp$
$\displaystyle=$
$\displaystyle\int_{\tau}^{s}z^{\prime}(p,\tau,\eta)-z^{\prime}(p,\tau,\tilde{\eta})dp,$

hence

$\phi(s)\mathrel{\mathop{\mathchar 58\relax}}=\frac{\mathinner{\\!\left\lVert
z(s,\tau,\eta)-z(s,\tau,\tilde{\eta})\right\rVert}}{|\eta-\tilde{\eta}|}\leq\int_{\tau}^{s}\frac{\mathinner{\\!\left\lVert
z^{\prime}(p,\tau,\eta)-z^{\prime}(p,\tau,\tilde{\eta})\right\rVert}}{|\eta-\tilde{\eta}|}dp,$

thus, we can deduce

$\displaystyle\phi(s)$ $\displaystyle\leq$
$\displaystyle\int_{\tau}^{s}\left\\{\left(\mathinner{\\!\left\lVert
A(p)\right\rVert}+\mathfrak{v}(p)\right)\phi(p)+\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right\\}dp,$

or alternatively, writing for $s\geq\tau$

$\alpha(s)=\mathinner{\\!\left\lVert A(s)\right\rVert}+\mathfrak{v}(s)$

and

$\beta(s)=\int_{\tau}^{s}\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)dp,$

we have

$\phi(s)\leq\beta(s)+\int_{\tau}^{s}\alpha(p)\phi(p)dp$

by using Gronwall’s Lemma we have

$\displaystyle\phi(s)$ $\displaystyle\leq$
$\displaystyle\beta(s)\exp\left(\int_{\tau}^{s}\alpha(p)dp\right)$
$\displaystyle=$
$\displaystyle\int_{\tau}^{s}\left\\{\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right\\}dp\cdot\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(p)\right\rVert}+\mathfrak{v}(p)dp\right),$

which by definition of
$\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}$,
concludes the proof. ∎

######  Corollary 5.7.

Suppose conditions (c1)-(c5) hold, with $r=2$ on (c4). If $\mathfrak{v}(s)=\nu
e^{-\varepsilon_{1}s}$ and $\mathfrak{V}(s)=\zeta e^{-\varepsilon_{2}s}$
satisfies (5.1), then for $s\geq\tau$

$\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}\leq\frac{\zeta
e^{-3\tau(M+\nu)}}{2(M+\nu)-\varepsilon_{2}}\left[e^{(3(M+\nu)-\varepsilon_{2})s}-e^{(2(M+\nu)-\varepsilon_{2})\tau+(M+\nu)s}\right].$

###### Proof.

By Lemma 5.6 we have

$\displaystyle\mathinner{\\!\left\lVert\frac{\partial^{2}y}{\partial\eta^{2}}(s,\tau,\eta)\right\rVert}$
$\displaystyle\leq$
$\displaystyle\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(p)\right\rVert}+\mathfrak{v}(p)dp\right)\cdot\int_{\tau}^{s}\left\\{\mathfrak{V}(p)\exp\left(2\int_{\tau}^{p}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right\\}dp$ $\displaystyle\leq$
$\displaystyle e^{(s-\tau)(M+\nu)}\int_{\tau}^{s}\left\\{\zeta
e^{-\varepsilon_{2}p}e^{2(p-\tau)(M+\nu)}\right\\}dp$ $\displaystyle=$
$\displaystyle\zeta
e^{(s-3\tau)(M+\nu)}\frac{e^{(2(M+\nu)-\varepsilon_{2})s}-e^{(2(M+\nu)-\varepsilon_{2})\tau}}{2(M+\nu)-\varepsilon_{2}}$
$\displaystyle=$ $\displaystyle\frac{\zeta
e^{-3\tau(M+\nu)}}{2(M+\nu)-\varepsilon_{2}}\left[e^{(3(M+\nu)-\varepsilon_{2})s}-e^{(2(M+\nu)-\varepsilon_{2})\tau+(M+\nu)s}\right]$

∎

######  Theorem 5.8.

Suppose (2.1) admits a nonuniform exponential dichotomy, i.e. there are two
complementary invariant projectors $P(\cdot)$, $Q(\cdot)$ and constants
$C,\lambda,\varepsilon_{1}>0$ such that

$\left\\{\begin{array}[]{lc}\mathinner{\\!\left\lVert
X(t,s)P(s)\right\rVert}\leq Ce^{-\lambda(t-s)+\varepsilon_{1}s},&\forall t\geq
s\geq 0\\\ \\\ \mathinner{\\!\left\lVert X(t,s)Q(s)\right\rVert}\leq
Ce^{\lambda(t-s)+\varepsilon_{1}s},&\forall 0\leq t\leq s.\end{array}\right.$

Furthermore, suppose that for each $t\in\mathbb{R}^{+}$ $u\mapsto f(t,u)$ is a
$C^{2}$ map satisfying

$|f(s,u)|\leq\kappa e^{-\varepsilon_{0}s},$
$\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,u)\right\rVert}\leq\nu e^{-\varepsilon_{1}s}$

and

(5.4) $\mathinner{\\!\left\lVert\frac{\partial f}{\partial
u}(s,u)-\frac{\partial f}{\partial u}(s,\tilde{u})\right\rVert}\leq\zeta
e^{-\varepsilon_{2}s}|u-\tilde{u}|,$

for given $\kappa,\nu,\zeta>0$ and $\varepsilon_{0}>\varepsilon_{1}-\lambda$.
If $3M<\lambda+\varepsilon_{2}$, $2M<\lambda+\varepsilon_{2}-\varepsilon_{1}$
and $M<\lambda$, then for small enough $\nu>0$ the systems (2.1) and (2.2) are
$C^{2}$-topologically equivalent on $\mathbb{R}^{+}.$

###### Proof.

By Corollary 4.6 we know conditions (c1)-(c5) are verified, this the systems
are $C^{1}$-topologically equivalent on $\mathbb{R}^{+}$, with
$\mathfrak{v}(s)=\nu e^{-\varepsilon_{1}s}$. It is easy to see (5.4) implies
(5.1), with $\mathfrak{V}(s)=\zeta e^{-\varepsilon_{2}s}$. As in Corollary
4.6, denote

$\Psi_{\tau}(s)=\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right),$

and note that $s\geq\tau\geq t$ implies $\Psi_{t}(s)\leq\Psi_{\tau}(s)$, thus
for every fixed $\tau\in\mathbb{R}^{+}$

$\Psi_{\tau}(s)^{2}\leq\Psi_{0}(s)^{2}\leq e^{2(M+\nu)s}.$

Hence

$\displaystyle\int_{\tau}^{\infty}K(s)h(s)\mathfrak{V}(s)\Psi_{\tau}(s)^{2}ds$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}K(s)h(s)\mathfrak{V}(s)\Psi_{0}(s)^{2}ds$
$\displaystyle\leq$ $\displaystyle
C\zeta\int_{0}^{\infty}e^{[-\lambda-\varepsilon_{2}+\varepsilon_{1}+2(M+\nu)]s}ds.$

As $2M<\lambda+\varepsilon_{2}-\varepsilon_{1}$, then for a small enough $\nu$
we have

(5.5)
$\int_{\tau}^{\infty}K(s)h(s)\mathfrak{V}(s)\left[\exp\left(\int_{\tau}^{s}\mathinner{\\!\left\lVert
A(r)\right\rVert}+\mathfrak{v}(r)dr\right)\right]^{2}ds<+\infty.$

Now, by Corollary 5.7, we know

$\pi_{\tau}(s)\mathrel{\mathop{\mathchar 58\relax}}=\frac{\zeta
e^{-3\tau(M+\nu)}}{2(M+\nu)-\varepsilon_{2}}\left[e^{(3(M+\nu)-\varepsilon_{2})s}-e^{(2(M+\nu)-\varepsilon_{2})\tau+(M+\nu)s}\right],$

satisfies condition (5.2) from Lemma 5.1. Note that

$\displaystyle\int_{\tau}^{\infty}K(s)h(s)\pi_{\tau}(s)\mathfrak{v}(s)ds$
$\displaystyle=$
$\displaystyle\int_{\tau}^{\infty}\mathfrak{K}_{\tau}\left[e^{(3(M+\nu)-\varepsilon_{2}-\lambda)s}-\frac{e^{(M+\nu-\lambda)s}}{e^{(\varepsilon_{2}-2(M+\nu))\tau}}\right]ds<+\infty,$

for a small enough $\nu$, where $\mathfrak{K}_{\tau}=\frac{C\zeta\nu
e^{-3\tau(M+\nu)}}{2(M+\nu)-\varepsilon_{2}}$. The previous argument, along
with (5.5) imply that conditions (5.3) from Lemma 5.1 is satisfied. Finally,
applying Theorem 5.3 the result follows. ∎

######  Remark 5.9.

In the previous result, if
$\varepsilon_{0}=\varepsilon_{1}=\varepsilon_{2}\mathrel{\mathop{\mathchar
58\relax}}=\varepsilon$, then the conditions
$\varepsilon_{0}>\varepsilon_{1}-\lambda$, $3M<\lambda+\varepsilon_{2}$,
$2M<\lambda+\varepsilon_{2}-\varepsilon_{1}$ and $M<\lambda$, may be reduced
to $2M<\lambda$ and $3M<\lambda+\varepsilon$.

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# Semiannihilator small submodules with respect to an arbitrary submodule

S. Rajaee ∗ ,  F. Farzalipour  and  M. Poyan

###### Abstract.

In this paper we introduce a new concept namely $T$-semiannihilator small
($T$-sa-small for short) submodules of an $R$-module $M$ with respect to an
arbitrary submodule $T$ of $M$ which is a generalization of the concept of
semiannihilator small submodules. We say that a submodule $N$ of $M$ is
$T$-sa-small in $M$ provided for each submodule $X$ of $M$ such that
$T\subseteq N+X$ implies that ${\rm Ann}(X)\ll(T:M)$. We investigate some new
results concerning to this class of submodules. Among various results we prove
that for a prime module $M$ over a semisimple ring $R$, if $N$ is a sa-small
submodule of $M$, then for every submodule $T$ of $M$ such that $N\subsetneq
T$, $N$ is also a $T$-sa-small submodule of $M$. For a faithful finitely
generated multiplication module $M$, we prove that $N$ is a $T$-sa-small
submodule of $M$ if and only if $(N:M)$ is a $(T:M)$-sa-small ideal of $R$.

MSC(2010): Primary:13C13, 13C99, 16D80.
Keywords: Small submodule; annihilator small submodule; semiannihilator small
submodule.
$*$Corresponding author

##  1\. Introduction

Throughout this paper, $R$ will denote a commutative ring with identity $1\neq
0$ and $M$ a nonzero unital $R$-module. We use the notations $\subseteq$ and
$\leq$ to denote inclusion and submodule. A nonempty subset $S$ of $R$ is said
to be a multiplicatively closed set (briefly, m.c.s) of $R$ if $0\notin S$,
$1\in S$ and $st\in S$ for each $s,t\in S$. For an $R$-module $M$, the set of
all submodules of $M$, denoted by ${\rm L}(M)$ and also ${\rm L}^{*}(M)={\rm
L}(M)\setminus\\{M\\}$. As usual, the rings of integers and integers modulo
$n$ will be denoted by $\mathbb{Z}$ and $\mathbb{Z}_{n}$, respectively. A
module $M$ on a ring $R$ (not necessarily commutative) is called prime if for
every nonzero submodule $K$ of $M$, ${\rm Ann}(K)={\rm Ann}(M)$. An $R$-module
$M$ is called a multiplication module, if every submodule $N$ of $M$ has the
form $N=IM$ for some ideal $I$ of $R$, and in this case, $N=(N:_{R}M)M$, see
[5]. A submodule $N$ of $M$ is called small (superfluous) which is denoted by
$N\ll M$, if for every submodule $L$ of $M$, $N+L=M$, implies that $L=M$.
Clearly, the zero submodule of every nonzero module is superfluous. The
Jacobson radical of $M$, denoted by ${\rm J}(M)$ is the intersection of all
maximal submodules of $M$ and also it is the sum of all small submodules of
$M$, i.e., ${\rm J}(M)=\cap_{\mathfrak{M}\in{\rm
Max}(M)}\mathfrak{M}=\sum_{N\ll M}N$. If $M$ does not have maximal submodules,
we put ${\rm J}(M)=M$. Consequently, if ${\rm J}(M)$ is a small submodule of
$M$, then ${\rm J}(M)$ is the largest small submodule of $M$. If $M$ is a
finitely generated nonzero module, then $M\neq{\rm J}(M)$ and ${\rm J}(M)$ is
the largest small submodule of $M$. We refer the reader to [3, 11] for the
basic properties and more information on small submodules. We know that if $M$
is a semisimple module, then the zero submodule is the only small submodule of
$M$ and $M$ is the only essential submodule of $M$.

Gilmer [7, p.60] has defined the concept of cancellation ideal to be the ideal
$I$ of $R$ which satisfies the following: Whenever $AI=BI$ with $A$ and $B$
are ideals of $R$ implies $A=B$. Mijbass in [10] has generalized this concept
to modules. An $R$-module $M$ is called a cancellation module whenever $IM=JM$
with $I$ and $J$ are ideals of $R$ implies $I=J$.

We recall that $R$ is a von Neumann regular ring (associative, with 1, not
necessarily commutative) if for every element $a$ of $R$, there is an element
$b\in R$ with $a=aba$. These rings are characterized by the fact that every
left $R$-module is flat.

##  2\. Preliminaries and Notations

In [2], the authors introduced the concept of annihilator-small submodules of
any right $R$-module $M$. For an unitary right $R$-module $M$ on an
associative ring $R$ with identity they called a submodule $K$ of $M_{R}$
annihilator-small if $K+T=M$, such that $T$ is a submodule of $M_{R}$, implies
that $\ell_{S}(T)=0$, where $\ell_{S}(T)$ indicates the left annihilator of
$T$ over $S={\rm End}(M_{R})$.

A submodule $N$ of $M$ is called an $R$-annihilator-small (breifly R-a-small)
submodule of $M$, if $N+X=M$ for some submodule $X$ of $M$, implies that ${\rm
Ann}(X)=0$, see [1]. We use the notation $N\ll^{a}M$ to indicate this concept.
In [12], the author introduced the concept of a semiannihilator small
submodule $N$ of a module $M$ on a commutative ring $R$ with identity $1\neq
0$ such that $N$ is called semiannihilator small (sa-small for short), denoted
by $N\ll^{sa}M$, if for every submodule $L$ of $M$ with $N+L=M$ implies that
${\rm Ann}(L)\ll R$. It is clear that every R-a-small submodule is sa-small,
but the converse is not true. Also a sa-small submodule need not be small, for
example consider $M=\mathbb{Z}$ as a $\mathbb{Z}$-module, every proper
submodule of $\mathbb{Z}$ is a sa-small submodule whereas the zero submodule
of $\mathbb{Z}$ is the only small submodule of $\mathbb{Z}$. Let $M$ be a
faithful $R$-module, then clearly every small submodule is an R-a-small
submodule and hence it is a sa-small submodule. The converse is not true,
because consider the $\mathbb{Z}_{8}$-module $M=\mathbb{Z}_{8}$, then
$\mathbb{Z}_{8}=\mathbb{Z}_{8}+\langle\overline{2}\rangle$ and ${\rm
Ann}(\langle\overline{2}\rangle)=\langle\overline{4}\rangle\neq\langle\overline{0}\rangle$,
whereas ${\rm Ann}(\langle\overline{2}\rangle)\ll\mathbb{Z}_{8}$.

Similarly, an ideal $I$ is a sa-small ideal of a ring $R$, if it is a sa-small
submodule of $R$ as an $R$-module, see [12, Definition 2.1]. We know that an
ideal $I$ of $R$ is small in $R$ if and only if $I\subseteq{\rm J}(R)$.
Therefore a submodule $N$ of a module $M$ is sa-small in $M$ if $N+X=M$ for a
submodule $X$ of $M$, implies that ${\rm Ann}(X)\ll{\rm J}(R)$, see [12,
Proposition 2.3]. A non-trivial $R$-module $M$ is called a semiannihilator
hollow (sa-hollow for short) module if every proper submodule of $M$ is sa-
small in $M$, see [12, Definition 3.1]. An $R$-epimorphism $f:M\rightarrow N$
is called a sa-small epimorphism whenever ${\rm Ker}(f)\ll^{sa}M$.

An $R$-module $M$ is said to be a comultiplication module if for every
submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N={\rm
Ann}_{M}(I)$. An $R$-module $M$ satisfies the double annihilator condition
(DAC for short), if for each ideal $I$ of $R$, $I={\rm Ann}_{R}({\rm
Ann}_{M}(I))$. Also $M$ is said to be a strong comultiplication module, if $M$
is a comultiplication module which satisfies DAC, see [4].

##  3\. Sa-small submodules w.r.t. an arbitrary submodule

In this section we generalize the concept of sa-small submodules to the
$T$-sa-small submodules of $M$ with respect to an arbitrary submodule $T$ of
$M$. Moreover, we investigate some new other properties of the sa-small
submodules of an $R$-module $M$ and we will generalize these properties to
this new class of submodules of $M$.

######  Definition 3.1.

Let $M$ be an $R$-module and let $T$ be an arbitrary submodule of $M$.

  * (i)

We say that a submodule $N$ of $M$ is a $T$-semiannihilator small (berifly,
$T$-sa-small) submodule of $M$, denoted by $N\ll_{T}^{sa}M$, provided for
every nonzero submodule $X\leq M$ such that $T\subseteq N+X$ implies that
${\rm Ann}(X)\ll(T:_{R}M)={\rm Ann}_{R}(M/T)$. Equivalently, if for a
submodule $X$ of $M$, ${\rm Ann}(X)$ is not small in $(T:_{R}M)$, then
$T\nsubseteq N+X$.

  * (ii)

We say that $M$ is a $T$-sa-small hollow module if every submodule $N$ of $M$
is a $T$-sa-small submodule of $M$. In particular, for an arbitrary ideal $A$
of $R$, we say that an ideal $I$ of $R$ is an $A$-sa-small ideal of $R$ if $I$
is an $A$-sa-small submodule of $R$ as an $R$-module. We shall denote the sum
of all $T$-sa-small submodules of $M$ by ${\rm J}_{T}^{sa}(M)$.

  * (iii)

Let $f:M\rightarrow N$ be an $R$-epimorphism and let $T$ be an arbitrary
submodule of $M$, we say that $f$ is a $T$-sa-small epimorphism in case ${\rm
Ker}(f)\ll_{T}^{sa}M$.

We denote the set of all small (resp. sa-small, $T$-sa-small) submodules of
$M$ by ${\rm S}(M)$ (resp. ${\rm S}^{sa}(M)$, ${\rm S}_{T}^{sa}(M)$).

######  Note 3.2.

Let $M$ be an $R$-module and let $T$ be an arbitrary submodule of $M$.

  * (i)

If we take $T=M$, then the notions of $T$-sa-small submodules and sa-small
submodules are equal.

  * (ii)

Let $T=0$, then for submodules $N,X$ of $M$, since $0\subseteq N+X$ hence
$N\ll_{0}^{sa}M$ implies that ${\rm Ann}(X)\ll{\rm Ann}(M)$. Now since ${\rm
Ann}(M)\subseteq{\rm Ann}(X)$ therefore ${\rm Ann}(X)={\rm Ann}(M)$. This is
impossible because a nonzero module $M$ is never small in itself. It concludes
that a nonzero module $M$ has no any $0$-sa-small submodule.

  * (iii)

If $T\neq 0$, then $N\ll^{sa}_{T}M$ implies that $T\nsubseteq N$, otherwise,
$T\subseteq N+0$ conclude that $R={\rm Ann}(0)\ll(T:M)$ which is impossible.
Let $M$ be a finitely generated $R$-module and let $T$ be a nonzero proper
arbitrary submodule of $M$ then by Zorn’s lemma there exists a maximal
submodule $N$ of $M$ such that $T\subseteq N$. This implies that $N$ is not a
$T$-sa-small submodule of $M$.

######  Theorem 3.3.

Let $M$ be an $R$-module and let $T$ be a submodule of $M$. The following
assertions hold.

  * (i)

Every $T$-sa-small submodule of $M$ is a sa-small submodule of $M$.

  * (ii)

If $M$ is a faithful prime $R$-module, then $M$ is a $T$-sa-small hollow
module.

  * (iii)

If $M$ is a prime module on a semisimple ring $R$ and $N\ll^{sa}M$, then for
every submodule $T\varsupsetneq N$ of $M$, $N\ll_{T}^{sa}M$.

###### Proof.

(i) Assume that $N\ll_{T}^{sa}M$ and $N+K=M$ for some module $K$ of $M$. Since
$T\subseteq N+K$ and $N\ll^{sa}_{T}M$ hence ${\rm Ann}(K)\ll(T:M)\leq R$. By
virtue of [11, Remark 2.8, (2)], ${\rm Ann}(K)\ll R$ this implies that
$N\ll^{sa}M$.
(ii) By hypothesis, $T\subseteq N+X$ implies that $0={\rm Ann}(X)={\rm
Ann}(M)$ which is small in $(T:M)$.
(iii) By [11, Proposition 3.5], $M$ is a semisimple module, hence
$M=T+T^{\prime}$ for some submodule $T^{\prime}$ of $M$. Suppose that
$N\ll^{sa}M$ and $T\subseteq N+K$ for some submodule $K$ of $M$. This implies
that $M=T+T^{\prime}\subseteq N+K+T^{\prime}$ and so $N+K+T^{\prime}=M$. Since
$N\ll^{sa}M$ hence ${\rm Ann}(K+T^{\prime})\ll R$. Since $M$ is prime hence

${\rm Ann}(K+T^{\prime})={\rm Ann}(K)\subseteq(T:M)\subseteq(T:N)=R.$

By virtue of [11, Remark 2.8, (3)], ${\rm Ann}(K)\ll(T:M)$ since $R$ is
semisimple and the proof is complete. ∎

The following example shows that in general the concepts of small submodules
and sa-small submodules are indepentent of each other. In (iii) we show that
the converse of Theorem 3.3 is not true in general and also in (iv) we show
that for $R$-modules $M,M^{\prime}$, if $f:M\rightarrow M^{\prime}$ is an
$R$-epimorphism, then the image of a $T$-sa-small submodule of $M$ need not be
an $f(T)$-sa-small submodule in $M^{\prime}$.

######  Example 3.4.

Let $\mathbb{Z}$ and $\mathbb{Z}_{n}$ be the rings of integers and integers
modulo $n$, respectively.

  * (i)

Consider $M=\mathbb{Z}_{6}$ as a $\mathbb{Z}$-module. Since ${\rm
Ann}(\mathbb{Z}_{6})=6\mathbb{Z}$ is not a small ideal of $\mathbb{Z}$, hence
$\langle\overline{0}\rangle\notin{\rm S}^{sa}(\mathbb{Z}_{6})$ whereas
$\langle\overline{0}\rangle\in{\rm S}(\mathbb{Z}_{6})$. We note that the only
nonzero proper submodules of $\mathbb{Z}_{6}$ are
$N=\langle\overline{2}\rangle$ and $L=\langle\overline{3}\rangle$ where
$N+L=\mathbb{Z}_{6}$ and both of ${\rm Ann}(N)=3\mathbb{Z}$ and ${\rm
Ann}(L)=2\mathbb{Z}$ are not small ideals of $\mathbb{Z}$. It concludes that
${\rm S}^{sa}(\mathbb{Z}_{6})=\emptyset$ whereas ${\rm
S}(\mathbb{Z}_{6})=\\{\langle\overline{0}\rangle\\}$.

  * (ii)

Take $M=\mathbb{Z}$ as a $\mathbb{Z}$-module. We know that
$k\mathbb{Z}+s\mathbb{Z}=\mathbb{Z}$ if and only if $(k,s)=1$. Since for every
submodule $s\mathbb{Z}$ of $\mathbb{Z}$, ${\rm
Ann}_{\mathbb{Z}}(\mathbb{Z})={\rm
Ann}_{\mathbb{Z}}(s\mathbb{Z})=0\ll\mathbb{Z}$ hence
$k\mathbb{Z}\ll^{sa}\mathbb{Z}$ for every proper submodule $k\mathbb{Z}$ of
$\mathbb{Z}$. It concludes that ${\rm S}^{sa}(\mathbb{Z})={\rm
L}^{*}(\mathbb{Z})$ whereas ${\rm S}(\mathbb{Z})=\\{0\\}$.

  * (iii)

We take the $\mathbb{Z}$-module $M=2\mathbb{Z}\times\mathbb{Z}_{8}$. Then
$N=\langle(0,\bar{0})\rangle$ is a sa-small submodule of $M$ but $N$ is not a
$T$-sa-small submodule of $M$ for submodule $T=\langle(0,\bar{4})\rangle$ of
$M$ since $T\subseteq N+\langle(0,\bar{2})\rangle$ whereas ${\rm
Ann}(\langle(0,\bar{2})\rangle)=4\mathbb{Z}$ is not small in $(T:M)=0$.

  * (iv)

Consider the natural $\mathbb{Z}$-epimorphism
$\pi:\mathbb{Z}\longrightarrow\mathbb{Z}_{8}$ where $\pi(n)=\overline{n}$.
Take $N=0$ and $T=2\mathbb{Z}$. Clearly $0\ll_{2\mathbb{Z}}^{sa}\mathbb{Z}$,
because we have $2\mathbb{Z}\subseteq 0+2\mathbb{Z}$ and also
$2\mathbb{Z}\subseteq 0+\mathbb{Z}$ and then

$0={\rm
Ann}(2\mathbb{Z})\ll(2\mathbb{Z}:_{\mathbb{Z}}\mathbb{Z})=2\mathbb{Z},$
$0={\rm Ann}(\mathbb{Z})\ll(2\mathbb{Z}:_{\mathbb{Z}}\mathbb{Z})=2\mathbb{Z}.$

But $\pi(N)=\pi(0)=\langle\bar{0}\rangle$ is not $\pi(T)$-sa-small submodule
of $\mathbb{Z}_{8}$ since $\pi(T)=\langle\bar{2}\rangle$ and
$\langle\bar{2}\rangle\subseteq\langle\bar{0}\rangle+\mathbb{Z}_{8}$ whereas
${\rm Ann}(\mathbb{Z}_{8})=8\mathbb{Z}$ is not small in
$(\langle\bar{2}\rangle:\mathbb{Z}_{8})=2\mathbb{Z}$.

######  Note 3.5.

Let $M$ be an $R$-module and let $T$ be an arbitrary submodule of $M$. For
every submodule $K$ of $M$, $K\ll^{sa}_{T}M$ if and only if $R$-epimorphism
$p_{{}_{K}}:M\rightarrow M/K$ is a $T$-sa-small epimorphism.

A submodule $N$ of an $R$-module $M$ is said to be completely irreducible if
$N=\cap_{i\in I}N_{i}$ where $\\{N_{i}\\}_{i\in I}$ is a family of submodules
of $M$, then $N=N_{i}$ for some $i\in I$. It is easy to see that every
submodule of $M$ is an intersection of completely irreducible submodules of
$M$.

######  Theorem 3.6.

Let $M$ be an $R$-module and $N\leq M$.

  * (i)

Let $T\lneq T^{\prime}\lneq M$. If $N\ll^{sa}_{T}M$, then
$N\ll^{sa}_{T^{\prime}}M$.

  * (ii)

If $T=T_{1}\,\cap\cdots\,\cap T_{s}$ and $N\ll^{sa}_{T}M$, then
$N\ll^{sa}_{T_{i}}M$ for any $1\leq i\leq s$. Conversely, if $T$ is a
completely irreducible submodule and for any $1\leq i\leq s$,
$N\ll^{sa}_{T_{i}}M$, then $N\ll^{sa}_{T}M$.

  * (iii)

If $T\leq K$ and $N\ll^{sa}_{T}M$, then $N\ll^{sa}_{T}K$.

  * (iv)

If $T=T_{1}+\cdots+T_{n}$ for some submodules $T_{i}$ of $M$ and
$N\leq_{T_{i}}^{sa}M$ for some $1\leq i\leq n$, then $N\leq_{T}^{sa}M$.

###### Proof.

(i) Assume that $T^{\prime}\subseteq N+X$ for some submodule $X$ of $M$, then
$T\subseteq N+X$ and so ${\rm Ann}(X)\ll(T:M)\leq(T^{\prime}:M)$. By [11,
Remark 2.8, (2)], ${\rm Ann}(X)\ll(T^{\prime}:M)$ as we needed.
(ii) It is clear by (i). Conversely, since $T$ is a completely irreducible
submodule hence there exists $1\leq i\leq s$ such that $T_{i}=T$ and the proof
is complete again using by (i).
(iii) Assume that $T\subseteq N+L$ for some submodule $L$ of $K$. Then ${\rm
Ann}(L)\ll(T:M)\leq(T:K)$ and this implies that ${\rm Ann}(L)\ll(T:K)$ and
therefore $N\ll^{sa}_{T}K$.
(iv) The proof is straightforward by (i). ∎

######  Theorem 3.7.

The following assertions hold.

  * (i)

${\rm S}(R)\subseteq{\rm S}^{sa}(R)$.

  * (ii)

If $R$ is an Artinian ring and $I\ll_{{\rm J}(R)}^{sa}R$, then for every
maximal ideal $\mathfrak{m}$ of $R$, $I\ll_{\mathfrak{m}}^{sa}R$.

  * (iii)

Let $M=N\oplus K$ be a multiplication module such that $N,K$ are finitely
generated submodules of $M$. Then $N$ and $K$ are not sa-small in $M$.

###### Proof.

(i) Let $I\in{\rm S}(R)$ and $I+J=R$ for some ideal $J$ of $R$. Then $J=R$ and
hence ${\rm Ann}(J)={\rm Ann}(R)=0$ which is a small ideal of $R$. Therefore
$I\in{\rm S}^{sa}(R)$.
(ii) Assume that ${\rm
Max}(R)=\\{\mathfrak{m}_{1},\cdots,\mathfrak{m}_{s}\\}$, then ${\rm
J}(R)=\cap_{i=1}^{s}\mathfrak{m}_{i}$. The proof is clear by Theorem 3.6 (i).
(iii) By [6, Corrollary 2.3], ${\rm Ann}(N)+{\rm Ann}(K)=R$. If $N\ll^{sa}M$,
then since $M=N+K$ hence ${\rm Ann}(K)\ll R$ and so ${\rm Ann}(N)=R$. It
concludes that $N=0$ which is impossible. ∎

######  Theorem 3.8.

Let $M$ be an $R$-module. The following assertions hold.

  * (i)

$M\notin{\rm S}^{sa}(M)$.

  * (ii)

$0\in{\rm S}^{sa}(M)$ if and only if ${\rm Ann}(M)\ll R$.

  * (iii)

If $R$ is a simple ring, then ${\rm L}^{*}(M)={\rm S}^{sa}(M)$.

  * (iv)

Let $\mathfrak{m}\in{\rm Max}(R)\cap{\rm S}^{sa}(R)$. If
$x\notin\mathfrak{m}$, then ${\rm Ann}(Rx)\subseteq{\rm J}(R)$.

  * (v)

If $M$ is a prime module with ${\rm Ann}(M)\ll R$, then ${\rm L}^{*}(M)={\rm
S}^{sa}(M)$. Moreover, in this case, ${\rm S}({M})\subseteq{\rm S}^{sa}(M)$.

###### Proof.

(i) Assume that $M\in{\rm S}^{sa}(M)$, then the equality $M+0=M$, implies that
${\rm Ann}(0)=R$ is a small ideal of $R$ which is impossible.
(ii) The proof is straightforward.
(iii) Let $R$ be a simple ring and let $N$ be a proper submodule of $M$ such
that $N+X=M$ for some submodule $X$ of $M$. If ${\rm Ann}(X)=0$, then ${\rm
Ann}(X)\ll R$ and the proof is complete. If ${\rm Ann}(X)=R$, then $X=0$ and
hence $N=M$ which is contradiction.
(iv) Suppose that $x\in R-\mathfrak{m}$, then $\mathfrak{m}+Rx=R$ hence ${\rm
Ann}(Rx)\ll R$ since $\mathfrak{m}\in{\rm S}^{sa}(R)$. This implies that ${\rm
Ann}(Rx)\subseteq{\rm J}(R)$.
(v) It is clear. ∎

######  Corollary 3.9.

Every proper submodule of a faithful prime module $M$ is a sa-small submodule
of $M$.

######  Theorem 3.10.

Let $R$ be a commutative ring. The following statments are true.

  * (i)

If $R$ has a nonzero comaximal sa-small ideal, then $R$ is not semisimple.

  * (ii)

Every sa-small hollow semisimple ring is simple.

  * (iii)

Let $R$ be a sa-small hollow ring. If $x\in{\rm Z}(R)$, then $R\neq Rx+Ry$ for
some element $y\in R$ and also $1$ is the only nonzero idempotent element of
$R$.

  * (iv)

If $R$ is a von Neumann regular ring, then none of the finitely generated
ideals of $R$ is a sa-small ideal of $R$.

  * (v)

Let $R$ be an integral domain and let $M$ be a faithful multiplication module,
then ${\rm L}^{*}(M)={\rm S}^{sa}(M)$.

###### Proof.

(i) Suppose that $I$ is a nonzero comaximal sa-small ideal of $R$, then there
exists an ideal $J$ of $R$ with $I+J=R$. We claim that $I\cap J\neq 0$. Assume
$I\cap J=0$ and so $IJ=0$. Hence $R=I+J\subseteq{\rm Ann}(J)+{\rm Ann}(I)$,
then ${\rm Ann}(I)+{\rm Ann}(J)=R$. Now $I\ll^{sa}R$, implies that ${\rm
Ann}(J)\ll R$ and thus ${\rm Ann}(I)=R$. It concludes that $I=0$ which is a
contradiction. We infer that $I$ is not a direct summand of $R$ and so $R$ is
not semisimple.
(ii) Suppose that $I$ is a nonzero ideal of $R$, then $I$ is a direct summand
of $R$ since $R$ is semisimple. By (i), $I$ is not a sa-small ideal of $R$ and
this is contradiction because $R$ is a sa-small hollow ring. It concludes that
$R$ has no any nonzero ideal and so $R$ is a simple ring.
(iii) Assume that $x\in{\rm Z}(R)$, then there exists an element $0\neq y\in
R$ such that $xy=0$. Let $R=Rx+Ry$, then $Ry\subseteq{\rm Ann}(Rx)\ll R$. This
implies that $Ry\ll R$ hence $Rx=R$ and so $x\in{\rm U}(R)$ which is
contradiction.
Now let $e$ be an idempotent element of $R$. Since $R(1-e)+Re=R$ therefore
$R(1-e)\subseteq{\rm Ann}(Re)\ll R$. This implies that $R(1-e)\ll R$ and so
$Re=R$. It concludes that $e=1$.
(iv) The proof follows from the fact that since $R$ is a von Neumann regular
ring hence every finitely generated ideal $I$ of $R$ is a direct summand of
$R$ such that $I$ is generated by an idempotent. Suppose that $R=I\oplus J$,
then $IJ=0$ and by (i) $I$ can not be a sa-small ideal.
(v) Take $N=0$, then $0+M=M$ and so ${\rm Ann}(M)=0$ is small in $R$. Now
sssume that $N$ is a nonzero submodule of $M$, then $N=IM$ for some nonzero
ideal $I$ of $R$. Let $r\in{\rm Ann}(N)$, then $r(IM)=0$ and so $rI=0$. It
concludes that $r=0$ since $R$ is an integral domain and so ${\rm Ann}(N)=0$
which is a small ideal of $R$. ∎

######  Proposition 3.11.

Let $M$ be an $R$-module with $N\leq K\leq M$ and let $T$ be an arbitrary
submodule of $M$. The following statments are true.

  * (i)

Let $M$ be a strong comultiplication module and $N\ll^{sa}M$. Then for every
nonzero submodule $L$ of $M$ with $N+L=M$, $L\leq_{e}M$.

  * (ii)

If $K\ll^{sa}_{T}M$, then $N\ll^{sa}_{T}M$.

  * (iii)

Assume $\\{N_{\lambda}\\}_{\lambda\in\Lambda}$ be a family of submodules of
$M$. If $N_{t}\ll^{sa}_{T}M$ for some $t\in\Lambda$, then
$\cap_{\lambda\in\Lambda}N_{\lambda}\ll^{sa}_{T}M$.

  * (iv)

Let $T\subseteq K$ and $N\ll^{sa}_{T}K$, then $N\ll^{sa}_{T}M$.

  * (v)

Suppose that $M$ is a multiplication module. If $N\ll^{sa}M$ (resp.
$N\ll_{T}^{sa}M$), then $(N:M)\ll^{sa}R$ (resp. $(N:M)\ll_{(T:M)}^{sa}R)$. The
converse is true if $M$ is also a finitely generated faithful module.
Furthermore, in this case, ${\rm J}^{sa}_{T}(M)={\rm J}^{sa}_{(T:M)}(R)M$.

  * (vi)

Assume that $M$ and $M^{\prime}$ are $R$-modules and $f:M\rightarrow
M^{\prime}$ is an $R$-epimorphism. If
$N^{\prime}\ll_{T^{\prime}}^{sa}M^{\prime}$ for some submodule $T^{\prime}$ of
$M^{\prime}$, then $f^{-1}(N^{\prime})\ll^{sa}_{f^{-1}(T^{\prime})}M$.

  * (vii)

Assume that $T\supsetneq N$ is an arbitrary submodule of $M$. If
$K/N\ll_{T/N}^{sa}M/N$, then $K\ll_{T}^{sa}M$ and $N\ll_{T}^{sa}M$.

  * (viii)

Let $M$ be a Noetherian $R$-module. If $S$ is a m.c.s. of $R$ and $S^{-1}N$ is
an $S^{-1}T$-sa-small submodule of $S^{-1}R$-module $S^{-1}M$, then $N$ is a
$T$-sa-small submodule of $M$.

###### Proof.

(i) Since $N\ll^{sa}M$ and $N+L=M$ hence ${\rm Ann}(L)\ll R$. By virtue of
[13, Theorem 2.5], $L=(0:_{M}{\rm Ann}_{R}(L))$ is an essential submodule of
$M$.
(ii) Assume that $T\subseteq N+X$ for some submodule $X$ of $M$. Therefore
$T\subseteq K+X$ and since $K\ll^{sa}_{T}M$ hence ${\rm Ann}(X)\ll(T:M)$ and
the proof is complete.
(iii) It is clear by (ii).
(iv) Let $L\leq M$ and $T\subseteq N+L$, then by the modular law
$T\subseteq(N+L)\cap K=N+(L\cap K)$. Since $N\ll^{sa}_{T}K$, hence ${\rm
Ann}(L)\subseteq{\rm Ann}(L\cap K)\ll(T:_{R}M)$ and this implies that ${\rm
Ann}(L)\ll(T:_{R}M)$ and the proof is complete.
(v) ($\Rightarrow$) Let $(N:M)+J=R$ for some ideal $J$ of $R$, then
$(N:M)M+JM=M$. Since $M$ is multiplication hence $N+JM=M$ and so
${\rm{Ann}}(JM)\ll R$. From ${\rm{Ann}}(J)\subseteq{\rm{Ann}}(JM)$ we infer
that ${\rm{Ann}}(J)\ll R$ and so $(N:M)\ll^{sa}R$.
($\Leftarrow$) Suppose that $(N:M)\ll^{sa}R$ and $N+K=M$ for some submodule
$K$ of $M$. Thus $(N:M)M+(K:M)M=RM$ and since $M$ is a finitely generated
faithful multiplication module hence $M$ is a cancellation module and so
$(N:M)+(K:M)=R$. It concludes that ${\rm{Ann}}(K:M)\ll R$. By hypothesis,
since $M$ is a faithful module hence ${\rm{Ann}}(K:M)={\rm{Ann}}(K)\ll R$ and
the proof is complete.

Now assume that $N\ll_{T}^{sa}M$ and $(T:M)\subseteq(N:M)+J$ for some ideal
$J$ of $R$. We show that ${\rm Ann}(J)\ll((T:_{R}M):_{R}R)=(T:_{R}M)$. Since
$M$ is a multiplication module hence $T=(T:M)M\subseteq(N:M)M+JM=N+JM$. Now
the hypothesis $N\ll_{T}^{sa}M$ implies that ${\rm Ann}(J)\subseteq{\rm
Ann}(JM)\ll(T:_{R}M)$ and so ${\rm Ann}(J)\ll(T:_{R}M)$ as we needed.
Conversely, let $(N:M)\ll_{(T:M)}^{sa}R$ and $T\subseteq N+K$ for some
submodules $N,K$ of $M$. Since $M$ is multiplication there exist ideals $I,J$
of $M$ such that $N=IM=(N:M)M$ and $K=JM=(K:M)M$. Therefore
$T=(T:M)M\subseteq(N:M)M+JM=((N:M)+J)M$ and so $(T:M)\subseteq(N:M)+J$,
because $M$ is a cancellation module. From the hypothesis
$(N:M)\ll_{(T:M)}^{sa}R$, we infer that ${\rm Ann}(J)\ll(T:M)$. Since $M$ is
faithful hence ${\rm Ann}(J)={\rm Ann}(JM)\ll(T:M)$ and the proof is complete.
For the second part we note that

$\displaystyle{\rm J}^{sa}_{T}(M)$
$\displaystyle:=\sum_{N\ll^{sa}_{T}M}N=\sum_{N\ll^{sa}_{T}M}(N:M)M$
$\displaystyle:=\left(\sum_{(N:M)\ll^{sa}_{(T:M)}R}(N:M)\right)M={\rm
J}^{sa}_{(T:M)}(R)M.$

(vi) Suppose that $f^{-1}(T^{\prime})\subseteq f^{-1}(N^{\prime})+L$ for some
submodule $L$ of $M$. Then since $f$ is an $R$-epimorphism hence

$T^{\prime}=f(f^{-1}(T^{\prime}))\subseteq f(f^{-1}(N^{\prime})+L)\subseteq
N^{\prime}+f(L).$

Since $N^{\prime}\ll_{T^{\prime}}^{sa}M^{\prime}$ hence ${\rm
Ann}(f(L))\ll(T^{\prime}:M^{\prime})$ and therefore

${\rm Ann}(L)\subseteq{\rm
Ann}(f(L))\ll(T^{\prime}:_{R}M^{\prime})=(f^{-1}(T^{\prime}):_{R}M).$

It implies that ${\rm Ann}(L)\ll(f^{-1}(T^{\prime}):_{R}M)$ and the proof is
complete.
(vii) Assume that $K/N\ll_{T/N}^{sa}M/N$ and also $T\subseteq K+L$ for some
submodule $L$ of $M$. Then $T/N\subseteq(K+L)/N=K/N+L/N$ implies that ${\rm
Ann}(L/N)\ll(T/N:M/N)=(T:M)$. Since ${\rm Ann}(L)\subseteq{\rm
Ann}(L/N)\ll(T:M)$ hence ${\rm Ann}(L)\ll(T:M)$. It conclude that
$K\leq_{T}^{sa}M$ and by (ii), $N\leq_{T}^{sa}M$.
(viii) We recall that if $X$ is a finitely generated submodule of $M$, then
$S^{-1}(0:_{R}X)=(S^{-1}0:_{S^{-1}R}S^{-1}X)$. Suppose that $T\subseteq N+X$
we show that ${\rm Ann}_{R}(X)\ll(T:_{R}M)$. Then we have

$S^{-1}T\subseteq S^{-1}(N+X)=S^{-1}N+S^{-1}X.$

By hypothesis ${\rm Ann}_{S^{-1}R}(S^{-1}X)\ll(S^{-1}T:_{S^{-1}R}S^{-1}M)$.
Therefore

$\displaystyle{\rm Ann}_{S^{-1}R}(S^{-1}X):$
$\displaystyle=(S^{-1}0:_{S^{-1}R}S^{-1}X)=S^{-1}(0:_{R}X)$ $\displaystyle\ll$
$\displaystyle S^{-1}(T:_{R}M)=(S^{-1}T:_{S^{-1}R}S^{-1}M).$

This implies that ${\rm Ann}_{R}(X)\ll(T:_{R}M)$ and the proof is complete. ∎

######  Corollary 3.12.

Let $M$ be a faithful finitely generated multiplication $R$-module and let $T$
be an arbitrary submodule of $M$. Then $M$ is a $T$-sa-hollow module if and
only if $R$ is a $(T:M)$-sa-hollow ring.

###### Proof.

The proof is straightforward by Proposition 3.11 (v). ∎

######  Corollary 3.13.

Let $(R,\mathfrak{m})$ be a local ring and let $A$ be an arbitrary ideal of
$R$. If $\mathfrak{m}\in{\rm S}^{sa}_{A}(R)$, then $I\in{\rm S}^{sa}_{A}(R)$
for every ideal $I$ of $R$.

###### Proof.

The proof is straightforward by Proposition 3.11 (ii). ∎

######  Theorem 3.14.

Let $f:M\rightarrow M^{\prime}$ be an $R$-epimorphism and let $T^{\prime}$ be
a submodule of $M^{\prime}$. If $M^{\prime}$ is a $T^{\prime}$-sa-hollow
module, then $M$ is an $f^{-1}(T^{\prime})$-sa-hollow module.

###### Proof.

Assume that $K$ is a submodule of $M$. Then
$f(K)\leq_{T^{\prime}}^{sa}M^{\prime}$ since $M^{\prime}$ is a
$T^{\prime}$-sa-hollow module. By Proposition 3.11 (vi), $f^{-1}(f(K))$ is an
$f^{-1}(T^{\prime})$-sa-small submodule of $M$. Since $K\subseteq
f^{-1}(f(K))$, so by Proposition 3.11 (ii), $K$ is also an
$f^{-1}(T^{\prime})$-sa-small submodule of $M$. ∎

The following example shows that the converse of Proposition 3.11 (vii), is
not true.

######  Example 3.15.

Consider the $\mathbb{Z}$-module $M=\mathbb{Z}$. Take $T=2\mathbb{Z}$,
$K=4\mathbb{Z}$ and $N=8\mathbb{Z}$. Then $8\mathbb{Z}$ is an
$2\mathbb{Z}$-sa-small submodule of $\mathbb{Z}$, because if
$2\mathbb{Z}\subseteq 8\mathbb{Z}+k\mathbb{Z}$ for some submodule
$k\mathbb{Z}$ of $\mathbb{Z}$, then either $(k,8)=2$ or $(k,8)=1$. In any
case, $0={\rm
Ann}(k\mathbb{Z})\ll(2\mathbb{Z}:_{\mathbb{Z}}\mathbb{Z})=2\mathbb{Z}$, but
$K/N=4\mathbb{Z}/8\mathbb{Z}$ is not $2\mathbb{Z}/8\mathbb{Z}$-sa-small
submodule of $\mathbb{Z}/8\mathbb{Z}$, because if
$2\mathbb{Z}/8\mathbb{Z}\subseteq
4\mathbb{Z}/8\mathbb{Z}+k\mathbb{Z}/8\mathbb{Z}$ for some submodule
$k\mathbb{Z}/8\mathbb{Z}$ of $\mathbb{Z}/8\mathbb{Z}$, then
$2\mathbb{Z}\subseteq 4\mathbb{Z}+k\mathbb{Z}=(4,k)\mathbb{Z}$. If $k=2$, then
${\rm Ann}(2\mathbb{Z}/8\mathbb{Z})=4\mathbb{Z}$ is not small in
$(2\mathbb{Z}/8\mathbb{Z}:_{\mathbb{Z}}\mathbb{Z}/8\mathbb{Z})=2\mathbb{Z}$.
If $k=1$, then ${\rm Ann}(\mathbb{Z}/8\mathbb{Z})=8\mathbb{Z}$ is not small in
$(2\mathbb{Z}/8\mathbb{Z}:_{\mathbb{Z}}\mathbb{Z}/8\mathbb{Z})=2\mathbb{Z}$.

We recall that for an ideal $I$ of a ring $R$ the radical of $I$ is defined by
${\rm{rad}}(I)=\\{x\in R\,\mid\,x^{n}\in I\,\,{\rm for}\,\,{\rm
some}\,n\in\mathbb{N}\\}$. Let $N$ be a proper submodule of $M$. Then, the
prime radical of $N$, denoted by ${\rm{rad}}(N)$ is defined to be the
intersection of all prime submodules of $M$ containing $N$, and in case $N$ is
not contained in any prime submodule then ${\rm{rad}}(N)$ is defined to be
$M$.

######  Lemma 3.16.

If $I\ll^{sa}R$, then ${\rm{rad}}(I)\ll^{sa}R$.

###### Proof.

Let ${\rm{rad}}(I)+J=R$ for some ideal $J$ of $R$. Since
${\rm{rad}}(I)+J\subseteq{\rm{rad}}(I)+{\rm{rad}}(J)$, so
${\rm{rad}}(I)+{\rm{rad}}(J)=R$. This implies that ${\rm{rad}}(I+J)=R$ and so
$I+J=R$. Hence ${\rm{Ann}}(J)\ll R$ since $I\ll^{sa}R$ and so
${\rm{rad}}(I)\ll^{sa}R$. ∎

######  Proposition 3.17.

Let $M$ be a finitely generated faithful multiplication $R$-module. If
$N\ll^{sa}M$, then ${\rm{rad}}(N)\ll^{sa}M$.

###### Proof.

By virtue of [9, Theorem 4], ${\rm{rad}}(N)={\rm{rad}}(N:M)\,M$ and so
$({\rm{rad}}(N):M)={\rm{rad}}(N:M)$. Since $N\ll^{sa}M$, then by Proposition
3.11 (v), $(N:M)\ll^{sa}R$ and by Lemma 3.16, ${\rm{rad}}(N:M)\ll^{sa}R$. It
concludes that $({\rm{rad}}(N):M)\ll^{sa}R$. Again using Proposition 3.11 (v)
we find that ${\rm{rad}}(N)\ll^{sa}M$. ∎

######  Theorem 3.18.

Let $N$ be a nonzero sa-small submodule of $M$ and $K\leq M$ with
$(N:K)+(K:N)=R$, then $N\cap K\neq 0$.

###### Proof.

Suppose that $N\cap K=0$, then $(N:K)+(K:N)={\rm Ann}(K)+{\rm Ann}(N)=R$.
Since $N\in{\rm S}^{sa}(M)$ hence ${\rm Ann}(K)\ll R$ and so ${\rm Ann}(N)=R$
which is contradiction. ∎

######  Theorem 3.19.

Let $K,H$ be submodules of $M$.

  * (i)

If $K+H\ll^{sa}_{T}M$, then $K\ll^{sa}_{T}M$ and $H\ll^{sa}_{T}M$.

  * (ii)

If $M$ is a prime module and $K\ll^{sa}_{T}M$ and $K+H\neq M$, then
$K+H\ll^{sa}_{T}M$.

###### Proof.

(i) It is clear by Proposition 3.11 (ii).
(ii) Let $T\subseteq K+H+X$ for some submodule $X$ of $M$. Since
$K\ll^{sa}_{T}M$ hence ${\rm Ann}(X)={\rm Ann}(H+X)\ll(T:M)$. ∎

######  Theorem 3.20.

Let $R$ be a semisimple hollow ring and let $M_{1},M_{2}$ be $R$-modules.
Suppose that $N_{1}\ll^{sa}_{T_{1}}M_{1}$ and $N_{2}\ll^{sa}_{T_{2}}M_{2}$ for
submodules $T_{1}\leq M_{1}$ and $T_{2}\leq M_{2}$, then $N_{1}\oplus
N_{2}\ll^{sa}_{T_{1}\oplus T_{2}}M_{1}\oplus M_{2}$.

###### Proof.

Suppose that

$T_{1}\oplus T_{2}\subseteq(N_{1}\oplus N_{2})+(L_{1}\oplus
L_{2})=(N_{1}+L_{1})\oplus(N_{2}+L_{2}).$

Then $T_{1}\subseteq N_{1}+L_{1}$ and $T_{2}\subseteq N_{2}+L_{2}$ hence ${\rm
Ann}(L_{1})\ll(T_{1}:M_{1})$ and ${\rm Ann}(L_{2})\ll(T_{2}:M_{2})$. By [8,
Lemma 2], if $S\subseteq E\subseteq F$ and $S\ll F$ such that $E$ is a direct
summand of $F$, then $S\ll E$. Take $S={\rm Ann}(L_{1})\cap{\rm Ann}(L_{2})$
and $F=(T_{1}:M_{1})\cap(T_{2}:M_{2})$, then

$\displaystyle{\rm Ann}(L_{1}\oplus L_{2})$ $\displaystyle={\rm
Ann}(L_{1})\cap{\rm Ann}(L_{2})\ll(T_{1}:M_{1})\cap(T_{2}:M_{2})$
$\displaystyle=(T_{1}\oplus T_{2}:M_{1}\oplus M_{2}).$

∎

######  Theorem 3.21.

Let $f:M\rightarrow N$ be a monomorphism and let $T$ be an arbitrary submodule
of $M$. If $K\ll^{sa}_{T}M$, then $f(K)\ll_{f(T)}^{sa}f(M)$.

###### Proof.

Assume that $f(T)\subseteq f(K)+L$ for some submodule $L$ of $f(M)$. We show
that ${\rm Ann}(L)\ll(f(T):_{R}f(M))$. We have

$T=f^{-1}(f(T))\subseteq f^{-1}(f(K)+L)=K+f^{-1}(L)\leq M.$

Then ${\rm Ann}(L)\subseteq{\rm Ann}(f^{-1}(L))\ll(T:_{R}M)=(f(T):_{R}f(M))$.
∎

######  Theorem 3.22.

Let $f:N\rightarrow K$ be a monomorphism and let $T$ be an arbitrary submodule
of $N$. If $g:K\rightarrow M$ is an $f(T)$-sa-small epimorphism, then $g\circ
f:N\rightarrow M$ is also a $T$-sa-small epimorphism.

###### Proof.

Assume that $T\subseteq{\rm Ker}(g\circ f)+X$ for some submodule $X$ of $N$.
We show that ${\rm Ann}(X)\ll(T:N)$. Since ${\rm Ker}(g\circ f)=f^{-1}({\rm
Ker}g)$ hence

$f(T)\subseteq f({\rm Ker}(g\circ f))+f(X)\subseteq{\rm Ker}g+f(X).$

Since ${\rm Ker}g\ll^{sa}_{f(T)}K$, hence

${\rm Ann}(X)\subseteq{\rm Ann}(f(X))\ll(f(T):_{R}K).$

We have $(f(T):_{R}K)\subseteq(T:_{R}N)$, because if $r\in(f(T):_{R}K)$ and
$x\in N$, then $rf(x)=f(rx)\in f(T)$. Since $f$ is monomorphism hence $rx\in
f^{-1}(f(T))=T$. It concludes that $r\in(T:N)$. Therefore ${\rm
Ann}(X)\subseteq{\rm Ann}(f(X))\ll(T:_{R}N)$ and the proof is complete. ∎

We recall that an $R$-module $F$ is called flat if whenever $N\rightarrow
K\rightarrow L$ is an exact sequence of $R$-modules, then $F\otimes
N\rightarrow F\otimes K\rightarrow F\otimes L$ is an exact sequence as well.
An $R$-module $F$ is called faithfully flat, whenever $N\rightarrow
K\rightarrow L$ is an exact sequence of $R$-modules if and only if $F\otimes
N\rightarrow F\otimes K\rightarrow F\otimes L$ is an exact sequence.

######  Theorem 3.23.

Let $F$ be a faithfully flat $R$-module and let $M$ be an $R$-module. Assume
that $N\leq M$ and $T$ is an arbitrary submodule of $M$. Then the following
statements hold.

  * (i)

$N$ is a sa-small submodule of $M$ if and only if $F\otimes N$ is a sa-small
submodule of $F\otimes M$.

  * (ii)

$N$ is a $T$-sa-small submodule of $M$ if and only if $F\otimes N$ is a
$F\otimes T$-sa-small submodule of $F\otimes M$.

###### Proof.

(i) ($\Rightarrow$) Let $N\leq^{sa}M$ and $F\otimes N+F\otimes K=F\otimes M$
for some submodule $F\otimes K$ of $F\otimes M$. Then $F\otimes(N+K)=F\otimes
N+F\otimes K=F\otimes M$. Thus $0\rightarrow F\otimes(N+K)\rightarrow F\otimes
M\rightarrow 0$ is an exact sequence. Since $F$ is faithfully flat, so
$0\rightarrow N+K\rightarrow M\rightarrow 0$ is an exact sequence. Therefore
$N+K=M$ and this implies that ${\rm Ann}(K)\ll R$ since $N\leq^{sa}M$. We have
${\rm Ann}(K)={\rm Ann}(F\otimes K)$, because if $r\in{\rm Ann}(K)$, then
$rK=0$. Thus $r(F\otimes K)=F\otimes rK=0$ and so $r\in{\rm Ann}(F\otimes K)$.
If $r\in{\rm Ann}(F\otimes K)$, then $0\rightarrow F\otimes rK\rightarrow 0$
is an exact sequence and so $0\rightarrow rK\rightarrow 0$ is exact since $F$
is a faithfully flat $R$-module. Therefore $rK=0$ and so $r\in{\rm Ann}(K)$.
It concludes that ${\rm Ann}(F\otimes K)\ll R$, so $F\otimes
N\leq^{sa}F\otimes M$.
($\Leftarrow$) Let $N+K=M$ for some submodule $K$ of $M$. Then

$F\otimes(N+K)=F\otimes N+F\otimes K=F\otimes M.$

Hence ${\rm Ann}(K)={\rm Ann}(F\otimes K)\ll R$. Therefore $N\leq^{sa}M$, as
needed.

(ii) ($\Rightarrow$) Let $F\otimes T\subseteq F\otimes N+F\otimes K$ for some
submodule $F\otimes K$ of $F\otimes M$. Thus $F\otimes T\subseteq
F\otimes(N+K)$ and so $0\rightarrow F\otimes T\rightarrow F\otimes(N+K)$ is
exact. Therefore $0\rightarrow T\rightarrow N+K$ is also exact since $F$ is a
faithfully flat $R$-module. This implies that $T\subseteq N+K$ and since
$N\ll^{sa}_{T}M$ hence ${\rm Ann}(K)\ll(T:_{R}M)$. It is easy to see that
$(T:_{R}M)=(F\otimes T:_{R}F\otimes M)$ since $F$ is faithfully flat. Hence
${\rm Ann}(F\otimes K)\ll(F\otimes T:_{R}F\otimes M)$ and so $F\otimes
N\ll_{F\otimes T}^{sa}F\otimes M$.
($\Leftarrow$) Let $T\subseteq N+K$. Then $F\otimes T\subseteq
F\otimes(N+K)=F\otimes N+F\otimes K$. Now since $F\otimes N\ll_{F\otimes
T}^{sa}F\otimes M$ hence ${\rm Ann}(F\otimes K)\ll(F\otimes T:_{R}F\otimes
M)$. It concludes that ${\rm Ann}(K)\ll(T:_{R}M)$ and so $N\ll^{sa}_{T}M$. ∎

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Saeed Rajaee
Department of Mathematics, Payame Noor University, P.O.Box 19395-3697, Tehran,
Iran.
Email: saeed_<EMAIL_ADDRESS>
Farkhondeh Farzalipour
Department of Mathematics, Payame Noor University, P.O.Box 19395-3697, Tehran,
Iran.
Email: f_<EMAIL_ADDRESS>
Marzieh Poyan
Department of Mathematics, PhD student of Mathematical Sciences, Payame Noor
University, P.O.Box 19395-3697, Tehran, Iran.
Email: r_<EMAIL_ADDRESS>