text:
\begin{document}

\title{Stability in Distribution  of Neutral Stochastic Functional Differential Equations  with Infinite Delay
}

\author{Hussein K. Asker ${}^{a,b}$ \footnote{ Email:husseink.askar@uokufa.edu.iq }  \\[0.2cm]
{\small a: Department of Mathematics, Faculty of Computer Science and Mathematics, }\\
{\small Kufa University, Al-Najaf, Iraq }\\
{\small b:Department of Mathematics, Swansea University, Singleton Park, SA2 8PP, UK }\\
 }

\maketitle
\begin{abstract}
In this paper, we investigate stability in distribution of neutral stochastic functional differential equations with infinite delay (NSFDEwID) at the state space\\
 \begin{equation*}
 C_{r}=\{{\varphi\in C((-\infty,0];R^{d}):\|\varphi\|_{r}=\sup_{-\infty<\theta\leq0}e^{r\theta}\lvert\varphi(\theta)\rvert} < \infty, \quad r > 0 \}.
 \end{equation*}
 We drive a sufficient strong monotone condition for the existence and uniqueness of the global solutions of NSFDEwID in the state space $ C_{r} $. We also address  the stability of  the solution map $ x_{t} $  and illustrate the theory with an example.
\end{abstract}
\noindent \textbf{Keywords}: Neutral stochastic functional differential equations, Infinite delay, Solution map, Stability in distribution.
\section{Introduction}
 Recently, the asymptotic properties of neutral stochastic functional differential equations (NSFDEs) have been addressed by many authors. Kolmanovskii and Myshkis  \cite{KM} have  studied the theory of existence and uniqueness,  and  the stability of  the solutions of  these equations. Mao  \cite{M1, M2} has investigated  the moment exponential stability and the almost sure exponential stability.  Bao and Hou \cite{BH}, Under a non-Lipschitz condition and a weakened linear growth condition, the existence and uniqueness of mild solutions to neutral partial functional SDEs have been investigated. Tan et al. \cite{TJS} by the weak convergence approach, they have reviewed stability in distribution for NSFDEs. Huang and Deng  \cite{HD} have studied Razumikhin-type theorems on general $ p $-th moment asymptotic stability. Kolmanovskii et al. \cite{KMM} have dealt with NSDDE with Markovian switching. Liu and Xia \cite{LX} have established some results which are more effective and relatively easy to verify to obtain the required stability, Hu and Wang \cite{hw} have concerned with the NSFDEs with Markovian switching and derived some sufficient conditions for stability in distribution, to name a few. On the other hand,  infinite NSFDEs have also been  received a great attention, many remarkable results got on the existence and uniqueness and the asymptotic behaviour of NSFDEwID.  Ren and Xia \cite{RX}, for example, have studied existence and uniqueness of solutions with infinite delay at phase space $ BC((- \infty; 0]; \mathbb{R}^{d}) $ which denotes the family of bounded continuous  $ \mathbb{R}^{d} $-value functions  with norm $ \lVert \varphi\lVert = \sup_{ - \infty < \theta \le 0} \rvert \varphi (\theta)\rvert $ under non-Lipschitz condition. Prato and Zabczyk \cite{dz} devoted their work to asymptotic properties of solution. Boufoussi and Hajji \cite{BH2, BH1}, by using successive approximation, they have proved existence and uniqueness result for a class of NFSDEs in Hilbert spaces. Bao \cite{Bao} has considered the existence and uniqueness of solutions in $ L^{p}(\Omega, C_{h}) $ space. Caraballo et al. \cite{CRT} have analysed the almost sure exponential stability and ultimate boundedness of the solutions. For Wei and Cai \cite{WC} by choosing $ C_{g} $ as  phase space, the existence and uniqueness of the solutions to neutral stochastic functional differential equations with infinite delay have obtained under non-Lipschitz condition, weakened liner growth condition and contractive condition. Chen and Banas\cite{CB},  have devoted their work to obtain some sufficient conditions for the exponential stability as well as almost surely exponential stability for mild solution of neutral stochastic partial differential equations with delays by establishing an integral-inequality. Zhou, Y. \cite{Z} has studied the stability property of stochastic differential equations in Hilbert spaces. Bao and Cao \cite{BC} have examined the existence and uniqueness of mild solutions. Bao et al. \cite{BYY, BYYW} have investigated the ergodicity for several kinds of FSDEs with variable delays. and Wu et al. \cite{FYM} have been studied the ergodicity of underlying processes and establishes existence of the invariant measure for SFDEs with infinite delay.\\
The state space for functional differential equations (FDEs) with infinite delay plays a key role in solving a specific problem. In general, the state space will be Banach space of functions or equivalence class of functions see \cite{KS}. Mohammed \cite{M} has examined solution maps of SFDEs with finite delay on appropriate phase spaces and proved that the solution maps have Markov property. Based on the Markov property of solution maps of SFDEs with finite delay, Bao et al.\cite{BYY, BYYW} have examined the ergodicity, while Wu et al. \cite{FYM} investigates existence and uniqueness of solutions, Markov properties, and ergodicity of SFDEs with infinite delay by using phase state $ C_{r} $.\\
Motivated by the discussion above, in this paper, to guarantee the wellposedness of solutions of NSFDEwID, boundedness of solutions  from different initial data and further asymptotic properties including the mean-square boundedness and convergence of the solutions from different initial data we have imposed appropriate strong monotone condition. We have also studied, the stability in distribution of the solution data. Finally, we have introduced an example to describe the theory that we addressed.
\section{Perliminary}
Throughout this paper, unless otherwise specified, we use the following notation. Let  $ R^{d} $ denote the usual $d$-dimensional Euclidean space and $ \lvert \cdotp \lvert $   the Euclidean norm. If $A$ is a vector or a matrix, its transpose is denoted by $ A^{T} $; and $  \lvert A \rvert = \sqrt{\mbox{trace} (A^{T}A)} $ denotes its trace norm. Denote by $ x^{T}y $ the inner product of  $x$ and $y $ in $\R^{d} $. Let  $ C((-\infty,0];R^{d}) $ denote the family of all  continuous functions from $ (-\infty,0] $  to $ R^{d}.$ We choose the state space with the fading memory  defined as follows: for given positive number $ r $,
\begin{equation}\label{2.1}
C _{r} = \Big \{ \varphi \in C((-\infty,0];R^{d}): \|\varphi \| _{r} = \sup_{ - \infty < \theta \leq 0} e^{r\theta} \arrowvert \varphi (\theta) \arrowvert < \infty  \Big \}.
\end{equation}
 $ C_{r} $ is a Banach space with norm $ \|\cdot \| _{r}$, which is introduced in  \cite{Kuang and Smith}, contains the Banach space of bounded and continuous functions and for any $0<r_{1}\leq r_{2} < \infty$, $C_{r_{1}} \subset C_{r_{2}} $. Let $(\Omega,\mathcal{F}, \mathbb{P}) $  be a complete probability space with a filtration $ {\{{\mathcal{F}_{t}\}_{t\in [0,+\infty)}}} $ satisfying the usual conditions (i.e. it is right continuous and $ {{\mathcal{F}_{0}}} $  contains all P-null sets). Let $  1_{B} $ denote the indicator function of a set B. $  M^{2}([ a, b];R^{d})$ is a family of process $ \{ x(t) \}_{a \leq t \leq b}  $ in $ \mathcal{L}^{2}([a, b]; \mathbb{R}^{d}) $ such that $ \mathbb{E} \int_{a}^{b} \rvert x(t)\rvert^{2} dt < \infty $. Consider a $ d $-dimensional neutral stochastic functional differential equations with infinite delay
\begin{equation}\label{2.2}
d\{x(t) - D(x_{t})\} = b(x_{t})dt + \sigma (x_{t})dw(t), \quad \text{on} \quad t \geq 0,
\end{equation}
with the initial data
\begin{equation}\label{2.3}
x_{0} = \xi = \{\xi(\theta):-\infty<\theta\leq0 \} \in C_{r},
\end{equation}
where
\begin{equation*}
x_{t} = \{ x(t+\theta): - \infty < \theta \leq 0\}
\end{equation*}
 and $ b, D:  C_{r} \to\mathbb R^{d} $;      $ \sigma : C_{r} \to\mathbb R^{d\times m} $ are Borel measurable, $ w(t) $  is an $ m $-dimensional Brownian motion. It should be pointed out that $ x(t) \in  R^{d}  $ is a point, while $ x_{t}  \in C_{r}$
 is a continuous function on the interval $ (-\infty, t] $ taking values in $ R^{d}. $ Now, we  give the definition of the solutions for the equation \eqref{2.2}.
\begin{defn}\label{d1}
    \cite{FYM,Mo} Set $ {\mathcal{F}_{t}} = {\mathcal{F}_{0}} $ for $ -\infty< t \leq 0 $ and let x(t),   $ -\infty< t \leq \tau _{e}$ be continuous $ {R}^{d} $-value and $ {\mathcal{F}_{t}} $-adapted process.  $ x(t) $ is called a $   \textbf {local  strong  solution}   $ of \eqref{2.2} with initial data  $ \xi \in C_{r} $ if $ x(t)= \xi(t) \quad on - \infty < t \leq 0 $ and for all $ t \geq 0 $,
    \begin{equation*}
    x (t \wedge \tau _{k}) =   \xi (0) + D ( x _{t} \wedge \tau _{k} ) - D ( \xi) + \int_{0}^{t \wedge \tau _{k}} b (x_{s}) ds + \int_{0}^{t \wedge \tau _{k}} \sigma ( x_{s}) dw(s) \quad a.s.,
    \end{equation*}
    for each $ \ell \geq 1 $, where $ \{ \tau _{\ell} \} _{\ell \geq 1} $  is a non-decreasing sequence of stopping times such that  $ \tau _{\ell} \longrightarrow \tau _{e} $ almost surely as $ \ell \longrightarrow \infty $. If moreover, $ \lim \sup_{t \longrightarrow \tau _{e} } \arrowvert x(t) \arrowvert  = \infty   $  is satisfied a.s. when $ \tau _{e} < \infty  $ a.s., x(t) $ ( - \infty < t < \tau _{e} ) $ is called a \textbf{maximal local strong solution}  and $  \tau _{e} $ is called the explosion time. It is called a global solution when $ \tau _{e} = \infty $. A maximal local strong solution x(t), $ - \infty < t < \tau _{e} $ is said to be unique if for any other maximal local strong solution $ \overline{x} (t) $, $ - \infty < t < \overline {\tau } _{e} $, we have $ \tau _{e} = \overline {\tau } _{e} $ and $ x(t) = \overline{x}(t) $ for $ - \infty < t < \tau _{e} $ almost surely.
\end{defn}
\section{Existence and uniqueness of solutions}
In this section we study the existence and uniqueness of the global solutions of the NSFDEwID \eqref{2.2}, mean-square boundedness and convergence of the solutions from different initial data. Let $ \mathcal{P}(C_{r} )  $ denotes the family of all probability measures on $ (C_{r},\mathcal{B}(C_{r} )) $. Denote $ \mathcal{C}_{b}(C_{r}) $ the set of all bounded continuous functional. For any $ F \in \mathcal{C}_{b}(C_{r}) $, $ F:C_{r} \rightarrow \mathbb{R} $  and $ \pi(\cdot) \in  \mathcal{P}(C_{r} ) $, let $ \pi(F ):=\int_{C_{r}} F(\phi) \pi(d\phi)$. $ M_{0} $ stands for the set of probability measures  on $ (- \infty, 0] $, namely, for any $ \mu \in M_{0} $, $ \int_{- \infty}^{0} \mu (d\theta) = 1  $. For any $ r > 0 $, let us further define $ {M}_{r}  $  as follows:
\begin{equation}
M _{r} : = \Big\{ \mu \in M_{0}; \mu ^{(r)}:=\int_{- \infty}^{0} e^{-r\theta} \mu (d\theta) < \infty \Big\}.
\end{equation}
Obviously, there exist many such probability measures, see \cite{FYM, huwuhuang} for details.
\begin{rem}\label{remark 1}
\cite{FYM} Fix $ r_{_0 } > 0 $ and $  \mu \in M _{r_{0}} $. For any $ r \in (0, r_{0}) $, $ \mu^{(r)} $ is continuously non-decreasing on $ r $ and satisfies $ \mu^{(r_{0})} > \mu^{(r)} > \mu^{(0)} = 1$, and $ M_{r_{0}} \subset M_{r} \subset M_{0} $.
\end{rem}
In order to examine the stability in distribution of the equation \eqref{2.2} we assume the following conditions.
\begin{description}
\item[(H1)]  For $ \mu_{1} \in M _{2r}$ and $ \varphi \in C_{r} $ there exist $ k \in (0, 1)  $ with $  \mu^{(2r)}_{1} < 1 $  such that
\begin{equation}\label{4.6}
\rvert D (\varphi)-D(\phi) \rvert ^{2} \leq k \int_{- \infty}^{0} \rvert \varphi(\theta)-\phi(\theta) \rvert ^{2} \mu_{1} (d\theta) \quad \text{and} \quad D(0)=0.
\end{equation}
\item[(H2)]  Let $ b $ be a continuous functions. Assume there exist constants $ \lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}>0 $, and probability measures $ \mu_{2}, \mu_{3} \in M_{2r} $ such that for any $ \varphi , \phi \in C_{r} $
\begin{equation}\label{4.7}
\begin{split}
\Big[ \varphi (0) - \phi (0) - \big( D(\varphi) - D (\phi) \big) \Big]^{T} \Big[ b ( \varphi) - b ( \phi ) \Big] &\leq  - \lambda_{1} \rvert \varphi (0) - \phi (0) \rvert ^{2} \\
& + \lambda_{2} \int_{-\infty}^{0} \rvert \varphi (\theta) - \phi (\theta) \rvert ^{2} \mu _{2} (d \theta),
\end{split}
\end{equation}
and for any function $ \sigma $
\begin{equation}\label{4.8}
\big \rvert \sigma (\varphi) - \sigma (\phi) \big\rvert ^{2} \leq \lambda_{3} \big\rvert \varphi (0) - \phi (0) \big\rvert ^{2} + \lambda_{4} \int_{- \infty}^{0} \big \rvert \varphi (\theta) - \phi (\theta)   \big \rvert ^{2} \mu _{3} (d\theta).
\end{equation}
\end{description}
\begin{rem}
	For simplicity, we suppose that $ \mu_{1} = \mu_{2} = \mu_{3} = \mu $.
\end{rem}
\begin{rem}
	It is easy to get from \eqref{4.6}, \eqref{4.7} and \eqref{4.8}, that
	\begin{description}
		\item[i)]  \begin{equation}\label{i}
		\rvert D (\varphi) - D (\phi)\rvert ^{2} \leq k \mu^{(2r)}\norm{\varphi-\phi}.
		\end{equation}
		\item[ii)] \begin{equation}\label{ii}
		\begin{split}
		\Big[ \varphi (0) - \phi (0) - \big( D(\varphi) - D (\phi) \big) \Big]^{T} \Big[ b ( \varphi) - b ( \phi ) \Big] &\leq  (- \lambda_{1}+ \lambda_{2} \mu^{(2r)})\norm{\varphi-\phi}.
		\end{split}
		\end{equation}
		\item[iii)]		\begin{equation}\label{iii}
		\big \rvert \sigma (\varphi) - \sigma (\phi) \big\rvert ^{2} \leq (\lambda_{3}+ \lambda_{4} \mu^{(2r)})\norm{\varphi-\phi}.
		\end{equation}
	\end{description}
\end{rem}
So, the assumptions \noindent{\bf (H1)} and \noindent{\bf (H2)} guarantees that the equation \eqref{2.2} admits a unique local strong solution $ \{x(t;\xi)\}_{t>- \infty} $ with initial data $ \xi \in C_{r} $. So, for global solution and further asymptotic properties including the mean-square boundedness and convergence of the solutions from different initial data, we impose that conditions.
\begin{lem}\label{lemma 2}
	\cite{Mo}   Let $ p > 1 $, $ \varepsilon > 0 $ and $ a, b \in R $. Then $ \rvert a + b \rvert ^{p} \leq \big [ 1 + \varepsilon ^{\frac{1}{p-1}} \big]^{p-1}  \Big( \rvert a \rvert ^{p} + \dfrac{\rvert b \rvert ^{p}}{\varepsilon} \Big) $.
\end{lem}
\begin{rem}
When $ \phi \equiv 0, D(\phi) \equiv 0 $, $ \varepsilon_{1},\varepsilon_{2}>0 $, by the Lemma \ref{lemma 2} and the assumption \noindent{\bf (H2)}, we get the strong monotone condition:
\begin{equation} \label{monotone}
\begin{split}
2\Big[ \varphi (0)  -  D(\varphi)  \Big]^{T} \Big[ b ( \varphi)  \Big]  +  \big \rvert \sigma (\varphi)  \big\rvert ^{2}\leq  - \alpha_{1} \rvert \varphi (0)  \rvert ^{2}  + \alpha_{2} \int_{-\infty}^{0} \rvert \varphi (\theta)  \rvert ^{2} \mu  (d \theta) + N ,
\end{split}
\end{equation}
where, $ \alpha_{1} = 2 \lambda_{1} - 2 \varepsilon_{1} -\dfrac{\lambda_{3}}{1-\varepsilon_{2}}  $,
$ \alpha_{2} =  2 \lambda_{2} +  2k\varepsilon_{1} +\dfrac{\lambda_{4}}{1-\varepsilon_{2}} $, $ N =\dfrac{1}{\varepsilon}_{1} \rvert b(0) \rvert ^{2} + \dfrac{1}{\varepsilon}_{2}  \rvert \sigma (0)\rvert^{2}  $.
\end{rem}
\begin{lem}\label{l1} For any $ \xi \in C_{r} $, under the assumption $ \bf(H1) $ we have,
\begin{equation*}
\rvert \xi(0) -  D(\xi) ) \rvert^{2}\leq M  \rVert \xi \rVert^{2}_{r},
\end{equation*}
where $ M=(1+k) ( 1  + \mu^{(2r)})  $
\end{lem}
\noindent{\bf Proof:} By the Lemma \ref{lemma 2} and the assumption $ \bf(H1) $ we have,
    \begin{equation*}
    \begin{split}
    \rvert \xi(0) -  D(\xi) ) \rvert^{2}  &\leq (1+\varepsilon) \rvert \xi(0) \rvert^{2} + ( 1 +\dfrac{1}{\varepsilon} )  \rvert D(\xi) \rvert^{2}\\
    &  \leq (1+\varepsilon) \sup_{ - \infty < \theta \le 0} e^{2r\theta }\rvert \xi(\theta) \rvert^{2} + k ( 1 +\dfrac{1}{\varepsilon} )  \int_{- \infty}^{0}  e^{2r\theta} \rvert \xi(\theta) \rvert ^{2} e^{-2r\theta} \mu (d\theta)\\
    &  \leq (1+\varepsilon) \sup_{ - \infty < \theta \le 0} e^{2r\theta }\rvert \xi(\theta) \rvert^{2} + k ( 1 +\dfrac{1}{\varepsilon} )  \int_{- \infty}^{0} \sup_{ - \infty < \theta \le 0} e^{2r\theta} \rvert \xi(\theta) \rvert ^{2} e^{-2r\theta} \mu (d\theta)\\
    & \leq (1+\varepsilon) \rVert \xi \rVert^{2}_{r} + k ( 1 +\dfrac{1}{\varepsilon} ) \int_{- \infty}^{0} \rVert \xi \rVert^{2} e^{-2r\theta} \mu (d\theta) \\
    & = (1+\varepsilon) \rVert \xi \rVert^{2}_{r} + k ( 1 +\dfrac{1}{\varepsilon} )  \rVert \xi \rVert^{2}  \mu^{(2r)}.\\
    \end{split}
    \end{equation*}
Set $ \varepsilon=k $, so that
 \begin{equation}\label{ln1}
\begin{split}
\rvert \xi(0) -  D(\xi) ) \rvert^{2}  &\leq (1+k) ( 1  + \mu^{(2r)})    \rVert \xi \rVert^{2}_{r}=M\rVert \xi \rVert^{2}_{r}.\\
\end{split}
\end{equation}
 \begin{lem}\label{l2} Under $ \bf(H1) $ let $ \xi,\eta \in C_{r} $, then
\begin{equation}\label{l6}
\rvert \xi(0) - \eta(0) - ( D(\xi) - D ( \eta )) \rvert^{2}  \leq M \rVert \xi - \eta \rVert_{r}^{2},
\end{equation}
where, $ M=(1+k) ( 1  + \mu^{(2r)}) $.\\
The proof is similar to that of Lemma \ref{l1}, we omit it here.
 \end{lem}
\begin{lem}\label{Lf1} Let $\xi\in
	C_r((-\infty,0];\R^d)$ and $ \bf(H1) $ holds, then for $ t>0 $ there exist positive constants $k_1$
	and $k_2$ such that
	\begin{equation}\label{lc1}
	\begin{split}
	\sup_{0<s\leq t}|\xi(s)|^2&\leq
	k_1e^{-2rs}\|\xi\|^2_r+k_2\sup_{0<s\leq t}|\xi(s)-D(\xi_s)|^2,
	\end{split}
	\end{equation}
	where $k_1=\frac{k
		\mu^{(2r)}}{1-k}$ and $k_2=\frac{1}{(1-k)^{2}}$.
\end{lem}
\noindent{\bf Proof:}
By the lemma \ref{lemma 2} and condition $ \bf(H1) $, for any
$\varepsilon>0$,
\begin{equation*}
\begin{split}
&|\xi(t)|^2=|D(\xi_t)+\xi(t)-D(\xi_t)|^2\\
&\leq(1+\varepsilon)\Big(\frac{|D(\xi_t)|^2}{\varepsilon}+|\xi(t)-D(\xi_t)|^2\Big)\\
&\leq(1+\varepsilon)\Big(\frac{k^2}{\varepsilon}\int_{-\infty}^0|\xi(t+\theta)|^2\mu
d(\theta) +|\xi(t)-D(\xi_t)|^2\Big).
\end{split}\end{equation*}
Taking $\varepsilon=\frac{k}{1-k}$ and using the definition of norm in
the phase space $C_r((-\infty,0];\R^d)$, we obtain
\begin{equation*}
\begin{split}
\sup_{0<s\leq t}|\xi(s)|^2 &\leq k\int_{-\infty}^0\sup_{0<s\leq
	t}|\xi(s+\theta)|^2\mu (d\theta)+\frac{1}{(1-k)}\sup_{0<s\leq
	t}|\xi(s)-D(\xi_s)|^2\\
& =k\int_{-\infty}^{-s}\sup_{0<s\leq	t}|\xi(s+\theta)|^2\mu (d\theta) +k\int_{-s}^0\sup_{0<s\leq	t}|\xi(s+\theta)|^2\mu (d\theta) \\
&+\frac{1}{(1-k)}\sup_{0<s\leq	t}|\xi(s)-D(\xi_s)|^2\\

& \leq k\mu^{(2r)} e^{-2rs}\|\xi\|_r^2 +k\sup_{0<s\leq t}|\xi(s)|^2+\frac{1}{(1-k)}\sup_{0<s\leq t}|\xi(s)-D(\xi_s)|^2.
\end{split}\end{equation*}
Subsequently,

\begin{equation*}
\begin{split}
\sup_{0<s\leq t}|\xi(s)|^2 &\leq \frac{k\mu^{(2r)}}{1-k}
e^{-2rs}\|\xi\|_r^2 +\frac{1}{(1-k)^{2}}\sup_{0<s\leq
	t}|\xi(s)-D(\xi_s)|^2\\
& = k_1 e^{-2rs}\|\xi\|_r^2 +k_2\sup_{0<s\leq
	t}|\xi(s)-D(\xi_s)|^2,\\

\end{split}
\end{equation*}
where $k_1=\frac{k \mu^{(2r)}}{1-k}$ and
$k_2=\frac{1}{(1-k)^{2}}$. The proof is complete.\hfill $\Box$\\
\begin{lem}\label{Lf2} Let  $r>0$ and $\xi, \eta\in
	C_r((-\infty,0];\R^d)$. Let condition $ \bf(H1) $ hold. Then there
	exist positive constants $k_3, k_{4}$, such that
  \begin{equation*}
\sup_{0\leq s\leq t}|\xi(s)-\eta(s)|^2 \leq  k_3e^{-2rs}\|\xi - \eta\|^2_r+
k_4\sup_{0\leq s\leq t}|\xi(s)-\eta(s)-D(z_s)+D(\eta_s)|^2,
  \end{equation*}
  where $k_3=\frac{k\mu^{(2r)}}{(1-k)}$ and
  $k_4=\frac{1}{(1-k)^{2}}$.
\end{lem}

Since the proof is similar to that of Lemma \ref{Lf2}, we will not give details here.
\begin{rem}
In the next theorem, by the initial data $ \xi $, we mean the initial function or initial segment process.  So that $ \xi $ is a function not a fixed constant. To highlight the initial segment process, we denote by $ x(t; \xi) $ and $ x_{t}(\xi) $  the solution and the solution map of \eqref{2.2}, respectively.
\end{rem}
\begin{thm}\label{4.}
    Under assumptions  $ \bf(H1) $ and $ \bf(H2) $,
    \begin{description}
        \item[(i)] For any initial data $  \xi \in C_{r}  $ the NSFDEwID \eqref{2.2} has a global solution $ x(t) $ almost surely, which is continuous and $ \mathcal{F}_{t} $-adapted.
        \item[(ii)]  If $ \lambda_{1}, \lambda_{2}, \lambda_{3}$ and $ \lambda_{4} $ satisfy $ 2 \lambda_{1} > 73 \lambda_{3}  + 2 \lambda_{2} \mu ^{(2r)} +73 \lambda_{4} \mu ^{(2r)} $, then there exist constants $ C_{1}, C_{2} > 0 $ and $ \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1} - 73 \lambda_{3} - 2 \lambda_{2} \mu ^{(2r)} -73 \lambda_{4}\mu ^{(2r)}\big]\wedge 2r \big) $ such that for any initial data $ \xi \in C_{r} $,
        \begin{equation}
        \mathbb{E} \big(  \rvert x(t;\xi)  \rvert ^{2} \big) \leq C_{1} + C_{2}\mathbb{E}  \rVert \xi \rVert^{2}_{r} e^{-\lambda t},
        \end{equation}
where $ C_{1}= \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big) $,\\$ C_{2}=\Bigg\{k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)   \Bigg] \Bigg\} $ and $ \varepsilon_{1}, \varepsilon_{2} $ are both sufficiently small constants such that
\begin{equation*}
 \Bigg[2 \lambda_{1}- M\lambda - 2\varepsilon_{1}- \dfrac{73 \lambda_{3}}{1-\varepsilon_{2}} -\Big(  2 \lambda_{2}+2k\varepsilon_{1}+ \dfrac{73 \lambda_{4}}{1-\varepsilon_{2}}  \Big) \mu^{(2r)}     \Bigg] > 0,
\end{equation*}
namely, solution $ x(t;\xi)  $   is mean-square bounded.
        \item[(iii)] Under conditions in $ \bf(ii) $, for different initial dat $ \xi $ and $ \eta $, the corresponding solution $ x(t; \xi) $ and $ x(t;\eta) $ satisfy
         \begin{equation}
         \mathbb{E} \Big( \sup_{ 0 < s \le t}  \rvert  x(t;\xi) - x(t; \eta) \rvert^{2} \Big)  \leq C_{3} \mathbb{E}\norm{\xi -\eta}e^{-\lambda t},
         \end{equation}
        where  $ C_{3}=\Bigg\{k_{3}+2k_{4} \Bigg[ M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2} + 73 \lambda_{4} )       \Big)                \Bigg] \Bigg\}. $
    \end{description}
\end{thm}
\noindent{\bf Proof:} We divide the proof into three steps:\\
\noindent{\bf Step 1 ( Proof of (i).):}
The proof is similar to that of \cite[Theorem 3.2]{FYM}, we here
only highlight the difficulty from the neutral term. By the conditions  $ \bf(H1) $ and $ \bf(H2) $, the coefficients are local Lipschitz continuous, then there exists a uinque maximal local strong solution $ x(t) $ to \eqref{2.2} on $ t \in (- \infty, \tau_{\ell}) $. In order to prove
that $x(t)$ is a global solution,  we need only prove that $
\tau_{e} = \infty $ almost surely. Define $  \tau_{\ell} = \inf \big\{ t \geq 0 : \rvert x(t) \rvert \geq \ell
\big\}  $, then $ \tau_{\ell} $ is increasing as $ \ell  \rightarrow  \infty $, let $ \lim\limits_{\ell \rightarrow \infty} \tau_{\ell}=
 \tau_{\infty} \leq \tau_{e} $ almost surly. If we can
show $ \tau_{\infty} = \infty $ a.s., then  $ \tau_{e} = \infty $
a.s., which implies that $ x(t) $ is actually global. This is
equivalent to prove that for any $ T > 0 $, $ \mathbb{P} (
\tau_{\ell} \leq T) \rightarrow 0 $ as $ \ell \rightarrow \infty $.\\
Using $ \bf(H1) $, one can derive that
\begin{equation}\label{4.16}
\begin{split}
&\sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (\mathbb{E} \rvert  x(s) \rvert^{2} \Big)  \leq ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)\\
& \quad + k ( 1+ \dfrac{1}{\varepsilon}) \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (\mathbb{E}\int_{- \infty}^{0}  \rvert x(s+\theta) \rvert ^{2}  \mu (d\theta) \Big)\\
& \leq ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)+ k ( 1+ \dfrac{1}{\varepsilon}) \mathbb{E} \Big[\int_{- \infty}^{-s}  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big ( \rvert x(s+\theta) \rvert ^{2} \Big)  \mu (d\theta) \\
&\quad + \int_{- s}^{0}  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big ( \rvert x(s+\theta) \rvert ^{2} \Big)  \mu (d\theta)\Big]\\
&= ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)\\
& \quad+ k ( 1+ \dfrac{1}{\varepsilon}) \mathbb{E} \Big[\int_{- \infty}^{-s}  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (e^{2r(s+\theta)} \rvert x(s+\theta) \rvert ^{2} \Big) e^{-2r(s+\theta)} \mu (d\theta) \\
&\quad + \int_{- s}^{0}  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big ( \rvert x(s+\theta) \rvert ^{2} \Big)  \mu (d\theta)\Big]\\
&\leq ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)\\
&+ k ( 1+ \dfrac{1}{\varepsilon}) \mathbb{E} \Big[\sup_{ -\infty <  s+\theta \le t \wedge \tau_{\ell}-s}  \Big (e^{2r(s+\theta)} \rvert x(s+\theta) \rvert ^{2} \Big)\int_{- \infty}^{-s}   e^{-2r(s+\theta)} \mu (d\theta) \\

&\quad + \sup_{ -\theta <  s+\theta \le t \wedge \tau_{\ell}}  \Big ( \rvert x(s+\theta) \rvert ^{2} \Big)\int_{- s}^{0}    \mu (d\theta)\Big]\\
&\leq ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)\\
&\quad+ k ( 1+ \dfrac{1}{\varepsilon}) \mathbb{E} \Big[\sup_{ -\infty <  s+\theta \le 0}  \Big (e^{2r(s+\theta)} \rvert x(s+\theta) \rvert ^{2} \Big)\int_{- \infty}^{-s}   e^{-2r(s+\theta)} \mu (d\theta)\\
& \quad + \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big ( \rvert x(s) \rvert ^{2} \Big)\int_{- \infty}^{0}    \mu (d\theta)\Big]\\
&= ( 1 + \varepsilon)  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big)+ k e^{-2rs} ( 1+ \dfrac{1}{\varepsilon}) \mu^{(2r)}\mathbb{E} \lVert \xi \lVert _{r}^{2}  \\
& \quad+ k ( 1+ \dfrac{1}{\varepsilon}) \sup_{ 0  <  s \le t \wedge \tau_{\ell}} \Big( \mathbb{E}  \rvert x(s) \rvert ^{2}   \Big) .
\end{split}
\end{equation}
Choosing $ \varepsilon > \dfrac{k}{1-k} $  implies $ \gamma=k ( 1+ \dfrac{1}{\varepsilon})<1 $, we arrive at
\begin{equation}\label{4.17}
\begin{split}
\sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (\mathbb{E} \rvert & x(s) \rvert^{2} \Big)  \leq  \dfrac{1 + \varepsilon}{(1- \gamma )}  \sup_{ 0 <  s \le t \wedge \tau_{\ell}}  \Big (  \mathbb{E} \rvert  x(s)  - D   ( x_{s})\rvert ^{2} \Big) + \dfrac{\gamma e^{-2rs}  \mu^{(2r)}}{1- \gamma}\mathbb{E} \rVert \xi \rVert^{2}_{r}.
\end{split}
\end{equation}
Applying the It\^o  formula to $ |x(t) - D(x_{t})|^{2} $, by
\eqref{ln1} and the monotone condition \eqref{monotone} yields for
any $ t \in [0, T] $,
\begin{equation}\label{s1}
\begin{split}
\mathbb{E} \rvert & x(t \wedge \tau_{\ell}) - D   ( x_{t \wedge \tau_{\ell}})\rvert^{2} = \mathbb{E} \rvert x(0) - D ( \xi) \rvert^{2} + 2 \mathbb{E} \int_{0}^{t \wedge \tau_{\ell}} [ x(s) - D (x_{s})]^{T} b( x_{s }) ds\\
& \qquad + \mathbb{E} \int_{0}^{t \wedge \tau_{\ell}} \rvert \sigma ( x_{s}) \rvert ^{2} ds\\
 &  \leq M \mathbb{E} \rVert \xi \rVert^{2}_{r} -\alpha_{1}   \mathbb{E} \int_{0}^{t \wedge \tau_{\ell}} \rvert x (s)  \rvert ^{2} ds + \alpha_{2} \mathbb{E} \int_{0}^{t \wedge \tau_{\ell}} \int_{-\infty}^{0} \rvert x (s+\theta) \rvert ^{2} \mu (d \theta) ds  +  N .
\end{split}
\end{equation}
By the fact [(3.12), from \cite{FYM}], we have
\begin{equation}\label{u}
\begin{split}
&\int_{0}^{t \wedge \tau_{\ell}} \int_{- \infty}^{0}   \rvert x(s+\theta) \rvert ^{2} \mu (d\theta) ds \leq \dfrac{1}{2r} \lVert \xi \lVert _{r} ^{2} \mu ^{(2r)}  +  \int_{0}^{t }  \rvert x(s \wedge \tau_{\ell}) \rvert ^{2}  ds.
\end{split}
\end{equation}
 Substituting \eqref{u} into \eqref{s1}, one has
\begin{equation}\label{4.14}
\begin{split}
\mathbb{E} \rvert x(t \wedge \tau_{\ell}) - D ( x_{t \wedge \tau_{\ell}})\rvert^{2}  &  \leq   L_{1} + J_{1}  \int_{0}^{t } \mathbb{E} \rvert x(s \wedge \tau_{\ell}) \rvert ^{2}  ds,
\end{split}
\end{equation}
where, $ L_{1} =\Big[ \Big(M + \dfrac{\alpha_{2}\mu^{(2r)}}{2r}  \Big) \mathbb{E} \rVert \xi \rVert^{2}_{r} + N \Big] $ and $  J_{1} =  - \alpha_{1}+\alpha_{2} $.\\
Therefore, by substituting   \eqref{4.14} into \eqref{4.17} we obtain
\begin{equation}\label{4.18}
\begin{split}
\mathbb{E} \rvert x(t \wedge \tau_{\ell}) \rvert^{2}  &  \leq  \dfrac{(1 + \varepsilon)L_{1}}{(1- \gamma )}  +   \dfrac{J_{1}(1 + \varepsilon)}{(1- \gamma)}   \mathbb{E} \int_{0}^{t}    \rvert  x(s \wedge \tau_{\ell} ) \rvert ^{2} ds \\
& +\dfrac{\gamma e^{-2rs}  \mu^{(2r)}}{1- \gamma} \mathbb{E} \rVert \xi \rVert^{2}_{r}.
\end{split}
\end{equation}
Hence, by applying  Gronwall's inequality yields
\begin{equation}\label{4.19}
\begin{split}
\mathbb{E} \rvert x(t \wedge \tau_{\ell}) \rvert^{2}  &  \leq   L_{2}      e^{J_{2}t} , \\
\end{split}
\end{equation}
where, $ L_{2} = \dfrac{(1 + \varepsilon)L_{1}}{(1- \gamma )}  +\dfrac{\gamma e^{-2rs}  \mu^{(2r)}}{1- \gamma} \mathbb{E} \rVert \xi \rVert^{2}_{r} $, $ J_{2} = \dfrac{J_{1}(1 + \varepsilon)}{(1- \gamma )} . $

According to the definition of $ \tau_{\ell}, $ we have
\begin{equation}
\begin{split}
\mathbb{E} \Big (   \rvert x( T \wedge \tau_{\ell}) \rvert^{2} \Big)&  =    \mathbb{E} \Big ( \rvert x( T \wedge \tau_{\ell}) \rvert^{2}  I_{\{ T \leq \tau_{\ell} \}}\Big) + \mathbb{E} \Big ( \rvert x( T \wedge \tau_{\ell}) \rvert^{2}  I_{\{ T \geq \tau_{\ell} \}}\Big)\\
& \ge \ell^{2} \mathbb{P} \big( \tau_{\ell} \leq T \big).
\end{split}
\end{equation}
Consequently, by \eqref{4.19}
\begin{equation*}
\begin{split}
\mathbb{P} ( \tau_{\ell} \leq T) & \leq  \dfrac{L_{2}      e^{J_{2}t}}{\ell^{2}}.
\end{split}
\end{equation*}
This implies
\begin{equation*}
\limsup_{\ell \rightarrow \infty} \mathbb{P} ( \tau_{\ell} \leq T ) = 0,
\end{equation*}
which means that \eqref{2.2} has a unique global solution $ x(t) $ on $ [0, \infty) $ almost surely.\\

\noindent{\bf Step 2 ( Proof of (ii).):} For any $ \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1} - 73 \lambda_{3} - 2 \lambda_{2} \mu ^{(2r)} -73 \lambda_{4}\mu ^{(2r)}\big]\wedge 2r \big) $ applying the It\^o  formula to $ e ^{\lambda t}|x(t) - D(x_{t})|^{2}, $   together with the  \eqref{ln1}, the Lemma \ref{lemma 2} with $ \varepsilon = k $, the monotone condition \eqref{monotone}  and the assumption  \noindent{\bf (H1)},     yields for any $ t \in [0, T] $ that,
\begin{equation}\label{4.42}
\begin{split}
&\mathbb{E}  \Big ( \sup_{ 0 < s\le t}  e ^{\lambda s} \rvert x(s)  - D ( x_{s})\rvert^{2} \Big)  = \mathbb{E} \rvert x(0) - D ( \xi) \rvert^{2} +  \mathbb{E} \Big ( \sup_{ 0 <  s \le t}\int_{0}^{s} e ^{\lambda u} \Big [ \lambda \rvert x(u) - D (x_{u}) \rvert^{2} \\
&  + 2[x(u) - D (x_{u})] ^{T} b( x_{u }) + \rvert \sigma (x_{u}) \rvert^{2} \Big] du \Big) + 2\mathbb{E} \Big ( \sup_{ 0 <  s \le t}\int_{0}^{s} e ^{\lambda u}  [x(u) - D (x_{u})] ^{T} \sigma (x_{u}) dw(u) \Big) \\
& \leq M \mathbb{E}  \rVert \xi \rVert^{2}_{r} + \lambda  (1+k) \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e^{\lambda u}  \rvert x(u)\rvert ^{2} du \Big) +  \lambda(1 + \dfrac{1}{k})  \mathbb{E} \Big( \sup_{ 0 <  s \le t} \int_{0}^{s} e^{\lambda u}  \rvert D (x_{u}) \rvert^{2} du \Big)\\
&   +  \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e ^{\lambda u} \Big[ - \alpha_{1} \big\rvert x (u)  \big\rvert ^{2} + \alpha _{2} \int_{- \infty}^{0} \big \rvert x (u+\theta) \big \rvert ^{2} \mu  (d\theta) +  N \Big] du \Big)\\
& + 2\mathbb{E} \Big ( \sup_{ 0 <  s \le t}\int_{0}^{s} e ^{\lambda u}  [x(u) - D (x_{u})] ^{T} \sigma (x_{u}) dw(u) \Big) \\
& \leq M \mathbb{E}  \rVert \xi \rVert^{2}_{r} +   ((1+k)\lambda-\alpha_{1}) \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e^{\lambda u}  \rvert x(u)\rvert ^{2} du \Big)\\
&   + ((1+k)\lambda+\alpha_{2}) \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e ^{\lambda u}   \int_{- \infty}^{0} \big \rvert x (u+\theta) \big \rvert ^{2} \mu  (d\theta)du\Big) + \dfrac{ N}{\lambda}(e^{\lambda t}-1)  \\
& + 2\mathbb{E} \Big ( \sup_{ 0 <  s \le t}\int_{0}^{s} e ^{\lambda u}  [x(u) - D (x_{u})] ^{T} \sigma (x_{u}) dw(u) \Big). \\
\end{split}
\end{equation}

By [(3.13), \cite{FYM}], we have
\begin{equation}\label{4.24}
\begin{split}
\int_{0}^{t} \int_{- \infty}^{0} e^{\lambda s} & \rvert x(s +\theta) \rvert ^{2} \mu (d\theta) ds

\leq \dfrac{1}{2r - \lambda} \rVert \xi \rVert^{2} _{r}  \mu ^{(2r)} + \mu  ^{(2r)} \int_{0}^{t } e^{\lambda s} \rvert x(s) \rvert ^{2}  ds .
\end{split}
\end{equation}
Now, using the Burkholder-Davis-Gundy inequality, the Lemma \ref{lemma 2}, \noindent{\bf (H1)}  and the condition \eqref{4.8},  we obtain
\begin{equation}\label{w}
\begin{split}
2\mathbb{E} \Big ( & \sup_{ 0 <  s \le t}  \int_{0}^{s} e ^{\lambda u}  [x(u) - D (x_{u})] ^{T} \sigma (x_{u}) dw(u) \Big)  \\
& \leq 8\sqrt{2} \mathbb{E} \Big ( \int_{0}^{t} e ^{2 \lambda s} \rvert [x(s) - D (x_{s})] ^{T} \sigma (x_{s})\rvert^{2} ds \Big) ^{\frac{1}{2}}\\
& \leq 12 \mathbb{E} \Big ( \int_{0}^{t} e ^{2 \lambda s} \rvert x(s) - D (x_{s}) \rvert ^{2}  \rvert \sigma (x_{s})\rvert^{2} ds \Big)^{\frac{1}{2}}\\
& \leq 12 \mathbb{E} \Big[ \Big ( \sup_{ 0 <  s \le t}  e ^{ \lambda s} \rvert x(s) - D (x_{s}) \rvert ^{2} \Big)^{\frac{1}{2}} \Big( \int_{0}^{t} e ^{ \lambda s} \rvert \sigma (x_{s})\rvert^{2} ds \Big)^{\frac{1}{2}} \Big]\\
&\leq \dfrac{1}{2}  \mathbb{E} \Big ( \sup_{ 0 <  s \le t}  e ^{ \lambda s } \rvert x(s) - D (x_{s}) \rvert ^{2} \Big) + 72 \mathbb{E}  \int_{0}^{t} e ^{ \lambda s} \rvert \sigma (x_{s})\rvert^{2} ds\\
& \leq \dfrac{1}{2} \mathbb{E} \Big ( \sup_{ 0 <  s \le t}  e ^{ \lambda s } \rvert x(s) - D (x_{s}) \rvert ^{2} \Big)+ \dfrac{72 \lambda_{3}}{1-\varepsilon_{2}} \mathbb{E}  \int_{0}^{t} e ^{ \lambda s}  \big\rvert x (s)  \big\rvert ^{2} ds\\
&  +  \dfrac{72  \lambda_{4}}{1-\varepsilon_{2}} \mathbb{E}  \int_{0}^{t} \int_{-\infty}^{0}  e ^{ \lambda s} \big\rvert x ( s + \theta)  \big\rvert ^{2} \mu  (d \theta) ds  +  \dfrac{72 }{ \lambda \varepsilon_{2} } \rvert \sigma (0)\rvert^{2} ( e^{\lambda t} - 1)   .
\end{split}
\end{equation}

By taking in account, the value of $ N, \alpha_{1}$ and $\alpha_{2} $,  substitute \eqref{4.24} and \eqref{w}  into \eqref{4.42} with $ \lambda \leq 2r$, we arrive at
\begin{equation}\label{4.25}
	\begin{split}
		\mathbb{E}& \Big ( \sup_{ 0 < s\le t}  e ^{\lambda s}  \rvert x(s) - D ( x_{s})\rvert^{2} \Big)  \leq 2 \Bigg[ M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}}        \Big)           \Bigg]  \mathbb{E}  \rVert \xi \rVert^{2}_{r}\\
		& +  \dfrac{2}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big) e^{\lambda t}  + 2 \Bigg[(1+k)\lambda - 2 \lambda_{1}+2\varepsilon_{1}+ \dfrac{73 \lambda_{3}}{1-\varepsilon_{2}} \\
		 & +\Big( (1+k)\lambda + 2 \lambda_{2}+2k\varepsilon_{1}+ \dfrac{73 \lambda_{4}}{1-\varepsilon_{2}}  \Big) \mu^{(2r)}   \Bigg] \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e ^{\lambda u} \big\rvert x (u)  \big\rvert ^{2} du \Big)  .\\
	\end{split}
\end{equation}
Consequently, by the Lemma \ref{Lf1}, we have
\begin{equation}\label{4.26}
\begin{split}
\mathbb{E}\Big ( \sup_{ 0 < s\le t}&  e ^{\lambda s}  \rvert x(s) \rvert^{2} \Big)  \leq\Bigg\{k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)                    \Bigg] \Bigg\} \mathbb{E}  \rVert \xi \rVert^{2}_{r}\\
& +  \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big) e^{\lambda t}  - 2k_{2} \Bigg[ 2 \lambda_{1}-M\lambda  -2\varepsilon_{1}- \dfrac{73 \lambda_{3}}{1-\varepsilon_{2}} \\
& -\Big(  2 \lambda_{2}+2k\varepsilon_{1}+ \dfrac{73 \lambda_{4}}{1-\varepsilon_{2}}  \Big) \mu^{(2r)}     \Bigg] \mathbb{E} \Big ( \sup_{ 0 <  s \le t} \int_{0}^{s} e ^{\lambda u} \big\rvert x (u)  \big\rvert ^{2} du \Big)  .
\end{split}
\end{equation}
Note that $ 2 \lambda_{1} > 73 \lambda_{3}  + 2 \lambda_{2} \mu ^{(2r)} +73 \lambda_{4} \mu ^{(2r)} $,  $ k\in(0, 1) $ and
\begin{equation*}
  \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1} - 73 \lambda_{3} - 2 \lambda_{2} \mu ^{(2r)} -73 \lambda_{4}\mu ^{(2r)}\big]\wedge 2r \big),
\end{equation*}
we can choose $ \varepsilon_{1}, \varepsilon_{2} $ sufficiently small such that:\\
$ \Bigg[2 \lambda_{1}- M\lambda - 2\varepsilon_{1}- \dfrac{73 \lambda_{3}}{1-\varepsilon_{2}} -\Big(  2 \lambda_{2}+2k\varepsilon_{1}+ \dfrac{73 \lambda_{4}}{1-\varepsilon_{2}}  \Big) \mu^{(2r)}     \Bigg] > 0 $, thus
\begin{equation}\label{4.27}
\begin{split}
\mathbb{E}& \Big ( \sup_{ 0 < s\le t}    \rvert x(s) \rvert^{2} \Big) \leq C_{1}+C_{2}\mathbb{E}  \rVert \xi \rVert^{2}_{r}e^{-\lambda t},\\
\end{split}
\end{equation}
where,
$ C_{1}= \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big) $ and \\$ C_{2}=\Bigg\{k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)                    \Bigg] \Bigg\} $.\\
This means that the solution $   x(t;\xi)  $ is mean-square bounded.\hfill $\Box$\\

\noindent{\bf Step 3 ( Proof of (iii).):} Let $  x(t; \xi) $ and $ x(t; \eta) $ be two different solutions with two different initial date  $ \xi, \eta $ to the equation \eqref{2.2}, then:
\begin{equation}
\begin{split}
d(x(t; \xi) - x(t; \eta) - D(x_{t }(\xi)) + D(x_{t }(\eta)) )& = \{b(x_{t }(\xi)) - b(x_{t }(\eta))\}dt \\
& + \{\sigma (x_{t }(\xi)) - \sigma (x_{t }(\eta))\}dw(t).
\end{split}
\end{equation}
For simplicity, let $ z(t) = x(t; \xi) - x(t; \eta) $ , $ \bar D(t) = D(x_{t }(\xi)) - D(x_{t }(\eta)) $ , $ \bar{b}(t) = b(x_{t }(\xi)) - b(x_{t }(\eta)) $ and $ \bar{\sigma}(t) = \sigma (x_{t }(\xi)) - \sigma (x_{t }(\eta)) $
with the initial data $ \xi - \eta $. For $ \lambda > 0  $ defined in (ii), applying the  It\^{o}   formula to $ e ^{\lambda t}\mathbb{E}\rvert z(t) -  \bar D(u)\rvert^{2} $ with \eqref{l6} gives:
\begin{equation}
\begin{split}
\mathbb{E} \Big( &\sup_{ 0 < s \le t} e^{ \lambda s} \rvert z ( s ) -  \bar D(s)\rvert^{2} \Big)  \leq M      \mathbb{E}  \rVert \xi - \eta \rVert_{r}^{2}\\
& \quad + \mathbb{E} \Big( \sup_{ 0 < s \le t} \int_{0}^{s} e^{\lambda u} \big[\lambda   \rvert z(u) - \bar D(u) \rvert ^{2} +2[z(u) -\bar D(u)  ] ^{T}  \bar{b}(z_{u}) + \rvert \bar{\sigma}(z_{u}) \rvert^{2}\big] du\\
& \quad + 2 \mathbb{E} \Big( \sup_{ 0 < s\le t} \int_{0}^{s} e^{\lambda u}[z(u) - \bar D(t) ] ^{T}  \bar{\sigma}(z_{u}) dw(u\Big).\\
\end{split}
\end{equation}
By the Lemma \ref{lemma 2} and the assumption $ \bf(H1) $, with $ \varepsilon=k $, we have
\begin{equation}
\rvert z(u) - \bar D(u) \rvert ^{2}\leq(1+k)   \rvert z(u)\lvert^{2}+(1+k)\int_{- \infty}^{0} \rvert z(u+\theta)\rvert^{2} \mu (d\theta),
\end{equation}
and by assumption $ \bf(H2) $,
\begin{equation}
2[z(u) -\bar D(u)  ] ^{T}  \bar{b}(z_{u})\leq - 2\lambda_{1}\lvert z(u)  \rvert^{2}+ 2 \lambda_{2}\int_{- \infty}^{0} \rvert z(u+\theta)\rvert^{2} \mu (d\theta),
\end{equation}
\begin{equation}
\begin{split}
&\lvert\bar{\sigma}(z_{u}) \rvert^{2} \leq \lambda_{3} \rvert z(u) \rvert^{2} +\lambda_{4} \int_{- \infty}^{0} \rvert z(u+\theta)\rvert^{2} \mu (d\theta),\\
\end{split}
\end{equation}
similar to \eqref{w}, with the condition \eqref{4.8}  we have
\begin{equation}
\begin{split}
&2 \mathbb{E} \Big( \sup_{ 0 < s\le t} \int_{0}^{s} e^{\lambda u}[z(u) - \bar D(t) ] ^{T}  \bar{\sigma}(z_{u}) dw(u\Big)  \leq\dfrac{1}{2}\mathbb{E} \Big( \sup_{ 0 < s \le t} e^{ \lambda s} \rvert z ( s ) -  \bar D(s)\rvert^{2} \Big)\\
& + 72\lambda_{3} \mathbb{E}\Big( \sup_{ 0 < s\le t}  \int_{0}^{s} e ^{ \lambda u}  \rvert z(u) \rvert^{2}du\Big) + 72\lambda_{4} \mathbb{E}\Big( \sup_{ 0 < s\le t}  \int_{0}^{s} \int_{- \infty}^{0}  e ^{ \lambda u}\rvert z(u+\theta)\rvert^{2} \mu (d\theta)du\Big).
\end{split}
\end{equation}
Consequently, by considering the fact \eqref{4.24} and the Lemma \ref{Lf2}, we get that
\begin{equation*}\label{4.30}
\begin{split}
\mathbb{E} &\Big( \sup_{ 0 < s \le t} e^{ \lambda s}  \rvert z ( s ) \rvert^{2} \Big)  \leq\Bigg\{k_{3}+2k_{4} \Bigg[ M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2} + 73 \lambda_{4} )       \Big)                \Bigg] \Bigg\}\mathbb{E}  \rVert \xi - \eta \rVert_{r}^{2} \\
& - 2k_{4} \Bigg[ 2 \lambda_{1}-73 \lambda_{3} -  M\lambda-2 \lambda_{2} \mu^{(2r)}  - 73 \lambda_{4}  \mu^{(2r)}   \Bigg]\mathbb{E}  \int_{0}^{t} e^{\lambda s}   \rvert z(s)\lvert^{2}ds .
\end{split}
\end{equation*}
Note that $ 2 \lambda_{1} > 73 \lambda_{3} +2 \lambda_{2} \mu^{(2r)}  + 73 \lambda_{4}  \mu^{(2r)}  $ and  \\ $ \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1}-73 \lambda_{3} -2 \lambda_{2} \mu^{(2r)}  - 73 \lambda_{4}  \mu^{(2r)}\big] \wedge 2r \big) ,$  $ k\in(0, 1) $,  yields  that:\\
$ \Big[ 2 \lambda_{1}-73 \lambda_{3} -  M\lambda-2 \lambda_{2} \mu^{(2r)}  - 73 \lambda_{4}  \mu^{(2r)} \Big] > 0 $, implies to
\begin{equation}
\begin{split}
\mathbb{E}& \Big( \sup_{ 0 < s \le t}  \rvert  x(t;\xi) - x(t; \eta) \rvert^{2} \Big)  \leq C_{3} \mathbb{E}\norm{\xi -\eta}e^{-\lambda t},
\end{split}
\end{equation}
where, $ C_{3}=\Bigg\{k_{3}+2k_{4} \Bigg[ M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2} + 73 \lambda_{4} )       \Big)                \Bigg] \Bigg\} $.\\
The proof is therefore complete. \hfill $\Box$\\
\section{Stability in Distribution }
In this section, we shall study the stability in distribution for the  segment process  $ \{x_{ t}\}_{t \geq 0} $. We need the following Lemma.

\begin{lem}
		If the process $ \{x (t)\}_{t \geq 0} $ is the unique solution of the equation \eqref{2.2}, then the segment process $ \{ x_{t}\}_{t \geq 0 } $ is a strong homogeneous Markov process on $ C_{r} $:

	$ \mathbb{P}(x_{t} \in A) \mid \mathcal{F}_{s} ) = \mathbb{P}(x_{t} \in A) \mid x_{s} ) $ $ P $-a.s., for all $ 0 \leq s \leq t $ and Borel sets $  A \in \mathcal{B}(C_{r}) $.
\end{lem}

\noindent{\bf Proof:} We divide the proof into two steps ( the proof is similer to ( Theorem 4.2, \cite{FYM} )).

\textbf{Step 1: Strong Markov Property}: For all $ 0 \leq s \leq t < \infty $ and  a finite stopping time $ \tau > 0 $ be,   we consider the equation
\begin{equation*}
\begin{split}
x (t + \tau) &= x(\tau) + \big( D ( x_{t +\tau}) - D (x_{\tau}) \big) + \int_{\tau}^{t } b(x_{s+\tau})ds + \int_{\tau}^{t} \sigma(x_{s+\tau})dw(s)\\
& =x(\tau) + \big( D ( x_{t +\tau}) - D (x_{\tau}) \big) + \int_{\tau}^{t+\tau } b(x_{s})ds + \int_{\tau}^{t+\tau} \sigma(x_{s})dw(s),
\end{split}
\end{equation*}
By the definition of $ x_{t} $, we have $ x_{t}(t_{0}, x_{t_{0}})  = x ( t+\theta, t_{0} ; x(t_{0}+\theta)) $ with $ x_{t_{0}} (t_{0}, x_{t_{0}} ) = x_{t_{0}} $, for any $ t_{0}\leq t $. Note that $ w(t) $ is a strong Markov process with independent increment. It follows that $ \mathcal{F}_{\tau} $ is independent of $ \mathcal{G}_{\tau}=\sigma\{w(\tau+s) - w(\tau)\} $ for any $ s>0 $. Also, note that $ x_{t}(\tau, \xi) $ depends completely on the increments $ w(\tau+s) - w(\tau) $  and so is $ \mathcal{G}_{\tau}$-measurable when $ x_{\tau}=\xi $ is given. Hence $ x_{t}(\tau, \xi) $ is independent of $ \mathcal{F}_{\tau} $ for any $ t>0 $. For any $ A \subset C_{r} $, we therefore have
\begin{equation*}
\begin{split}
\mathbb{E} ( 1_{\{x_{t+\tau} (0, \xi)\}\in A} \mid \mathcal{F}_{\tau}) &= \mathbb{E} ( 1_{\{x_{t} (\tau, x_{\tau})\}\in A} \mid \mathcal{F}_{\tau}) \\
& = \mathbb{E} ( 1_{\{x_{t} (\tau, \xi)\}\in A}\rvert\xi=x_{\tau} )\\
&= \mathbb{P}(x_{\tau}=\xi, x_{t}(\tau, \xi)\in A\rvert\xi=x_{t})\\
&  = \mathbb{E} ( 1_{\{x_{t} (\tau, \xi)\}\in A} \mid x_{\tau})
\end{split}
\end{equation*}
Toward that standard technique, it takes after that for at whatever bounded Borel measurable function $ \varphi: C_{r} \rightarrow \mathbb{R} $,
\begin{equation*}
\mathbb{E} ( \varphi ( x_{t+\tau} ) \mid \mathcal{F}_{\tau} ] = \mathbb{E} ( \varphi ( x_{t+\tau} ) \mid x_{\tau} ),
\end{equation*}
which ends the proof and yields that $ x_{t} $ is a strong Markov process.\hfill $\Box$\\

\textbf{Step 2: Homogeneity:} According to the definition of transition probability:
\begin{equation*}
P( \xi , u; A, t+ u) = \mathbb{P} ( x _{t+u} ( u, \xi) \in A),
\end{equation*}
where $  x _{t+u} ( u, \xi)  $ is determined by the solution $ x(t) $:
\begin{equation*}
x( t+u) = \xi(0) + D ( x_{t +u}) - D(\xi) + \int_{u}^{t+u} b(x_{s}) ds + \int_{u}^{t+u} \sigma(x_{s}) dw(s).
\end{equation*}
This equation is equivalent to
\begin{equation}\label{M-1}
x( t+u) = \xi(0) + D ( x_{t +u}) - D(\xi) + \int_{u}^{t} b(x_{s+u}) ds + \int_{u}^{t} \sigma(x_{s+u}) d\widetilde{w}(s),
\end{equation}
where $ \widetilde{w}(s) = w(s+u) - w(u) $ is clearly an $ d $-dimensional Brownian motion. So , we have
\begin{equation}\label{M-2}
x( t) = \xi(0) + D ( x_{t }) - D(\xi) + \int_{0}^{t} b(x_{s}) ds + \int_{0}^{t} \sigma(x_{s}) dw(s),
\end{equation}
with $  x_{0 } = \xi $. Comparing equations \eqref{M-1} with \eqref{M-2} and noting that $ x_{t+u} $ and $ x_{t} $ completely depends on $ x(t+u) $ and $ x(t) $ and their history, we see by the week uniqueness that $ \{ x_{t+u}\}_{t \geq 0}  $ are identical in probability law. Consequently,
\begin{equation*}
\mathbb{P}( x _{t+u} ( u, \xi) \in A ) = \mathbb{P} ( x_{t} ( 0, \xi)  \in A),
\end{equation*}
namely,
\begin{equation*}
P( \xi , u; A, t+ u) = P ( \xi, 0; A, t).
\end{equation*}
This complete the proof.\hfill $\Box$\\
\subsection{Stability in Distribution of  $ x_{t} $}
In order to address the stability of $ x_t $ in this subsection, let we highlight some notations, see \cite{TJS} for details. For  $ t\geq 0 $, denoted by $ P(\xi;t, \cdot) $ the transition probability of $ \{x_t\}_{t \geq 0}$ and for any $ P_{1}, P_{2}\in \mathcal{P}(C_{r}) $, define the metric $ d_{\mathbb{L}}  $ by
\begin{equation*}
	d_{\mathbb{L}} (P_{1}, P_{2})=\sup_{ g \in \mathbb{L}}\Bigg| \int_{ C_{r} }g(\xi)(P_{1}(d\xi)- \int_{ C_{r} }g(\eta)(P_{2}(d\eta)\Bigg|,
\end{equation*}
where $ \mathbb{L}= \{g: C_{r}\rightarrow \mathbb{R}: \rvert g(\xi)-g(\eta) \rvert \leq \lVert \xi-\eta \lVert_{\infty}\quad \text{and}\quad \rvert g(\cdot)\rvert\leq1\quad \text{for}\quad \xi,\eta \in C_{r} \} $.
\begin{defn}
 \cite{TJS}. The process $ x_{t}(\xi) $ is said to be stable in distribution if there exists a probability measure $ \pi \in \mathcal{P}(C_{r}) $ such that $ P(\xi;t, \cdot) $ converges weakly to $ \pi $ as $ t \rightarrow \infty $ for any $ \xi \in C_{r} $, that is,
\begin{equation*}
\lim\limits_{t\rightarrow \infty}d_{\mathbb{L}}(P(\xi;t, \cdot),\pi(\cdot))=0, \quad \text{for all}\quad \xi \in  C_{r}.
\end{equation*}
In this case, \eqref{2.2} is said to be stable in distribution.
\end{defn}
We start by investigating some sufficient conditions on the stability in distribution for the segment process $ x_{t} $ on $ t\geq 0 $.

\begin{lem}\label{p1}
Under assumptions $ \bf(H1) $  and $ \bf(H2) $, if $ \lambda_{1}, \lambda_{2}, \lambda_{3}$ and $ \lambda_{4} $ satisfy $ 2 \lambda_{1} > 73 \lambda_{3}  + 2 \lambda_{2} \mu ^{(2r)} +73 \lambda_{4} \mu ^{(2r)} $, then there exist constants $ C_{4}, C_{5} > 0 $ and $ \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1} - 73 \lambda_{3} - 2 \lambda_{2} \mu ^{(2r)} -73 \lambda_{4}\mu ^{(2r)}\big]\wedge 2r \big) $ such that for any initial data $ \xi \in C_{r} $,
\begin{equation}\label{4.36}
\mathbb{E} \rVert x_{t} \rVert _{r}^{2} \leq C_{4} + C_{5} e ^{- \lambda t},
\end{equation}
where
\begin{equation*}
\begin{split}
& C_{4} = \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big),\\
& C_{5} =\Bigg\{1+k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)  \Bigg] \Bigg\} \mathbb{E}  \rVert \xi \rVert^{2}_{r}.\\
\end{split}
\end{equation*}
\end{lem}
\noindent{\bf Proof:} Noting that $ \lambda < 2r $, therefore, correspond to the definition of the norm $ \rVert \cdot \rVert _{r} $, it is easy to see that:
\begin{equation}\label{5.8}
\begin{split}
\mathbb{E} \rVert x_{t} \rVert _{r}^{2}

& =  e ^{- \lambda t} \mathbb{E} \rVert  \xi \rVert^{2}_{r} +  \mathbb{E} \Big( \sup_{ 0 < s \le t} \rvert x ( s ) \rvert^{2} \Big).
\end{split}
\end{equation}
From \eqref{4.27}, we have
\begin{equation}\label{5.4}
\begin{split}
\mathbb{E}& \Big ( \sup_{ 0 < s\le t} \rvert x(s) \rvert^{2} \Big)  \leq \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big) \\
& + \Bigg\{k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)  \Bigg] \Bigg\}e^{-\lambda t} \mathbb{E}  \rVert \xi\rVert^{2}_{r}.\\
\end{split}
\end{equation}
By substituting \eqref{5.4} into \eqref{5.8}, we get
\begin{equation}\label{key}
\begin{split}
\mathbb{E} \rVert x_{t} \rVert _{r}^{2} & \leq C_{4} + C_{5} e ^{- \lambda t} ,\\
\end{split}
\end{equation}
 where
\begin{equation*}
\begin{split}
& C_{4} = \dfrac{2k_{2}}{\lambda} \Big( \dfrac{73 \rvert \sigma (0) \rvert^{2}}{\varepsilon_{2}}   +  \dfrac{\rvert b(0) \rvert^{2}}{\varepsilon_{1}} \Big),\\
& C_{5} = \Bigg\{1+k_{1}+2k_{2} \Bigg[M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2}+ 2k\varepsilon_{1} +  \dfrac{73\lambda_{4}}{1-\varepsilon_{2}} \Big)  \Bigg] \Bigg\} \mathbb{E}  \rVert \xi \rVert^{2}_{r}               .\\
\end{split}
\end{equation*}
This means that, the solution map $ x_{t} $ is mean-square bounded and giving the desired  assertion \eqref{4.36}.\hfill $\Box$
\begin{lem}\label{p2}
Under assumptions $ \bf(H1) $ and $ \bf(H2) $, if $ \lambda_{1}, \lambda_{2}, \lambda_{3}$ and $ \lambda_{4} $ satisfy $ 2 \lambda_{1} > 73 \lambda_{3}  + 2 \lambda_{2} \mu ^{(2r)} +73 \lambda_{4} \mu ^{(2r)} $, then there exist constant $ C_{6} > 0 $ and $ \lambda \in \big( 0,  \dfrac{1}{M}\big[2 \lambda_{1} - 73 \lambda_{3} - 2 \lambda_{2} \mu ^{(2r)} -73 \lambda_{4}\mu ^{(2r)}\big]\wedge 2r \big) $ such that
for the different initial data $ \xi $ and $ \eta  \in C_{r}$, the corresponding solution maps $  x_{t}(\xi) $ and $ x_{t}(\eta) $ satisfy,
\begin{equation}\label{5.2}
\mathbb{E} \rVert x_{t}(\xi) -  x_{t}(\eta) \rVert^{2} _{r} \leq C_{6} \mathbb{E} \rVert \xi -  \eta \rVert^{2} _{r} e^{-\lambda t},\\
\end{equation}
where
\begin{equation*}
C_{6} = \Bigg\{1+k_{3}+2k_{4} \Bigg[ M + \dfrac{\mu^{(2r)}}{2r - \lambda} \Big(  (1 +k) \lambda+ 2 \lambda_{2} + 73 \lambda_{4} )       \Big)                \Bigg] \Bigg\}.
\end{equation*}
Namely, the solution map $ x_{t} $ is mean-square bounded and solution maps from different initial data are convergent.
\end{lem}
Since the proof is comparable to that of Lemma \ref{p1}, we'll not explicate here.

\begin{thm}
	Under assumptions $ \bf(H1) $ and $ \bf(H2) $, \eqref{2.2} is stable in distribution.
\end{thm}
\noindent{\bf Prove:} Since \eqref{key} and \eqref{5.2} hold, the proof is standard, we omit here.
\section{Example }
 In this section we give an example in order to verify the theory that we studied. Let we consider the one-dimensional neutral stochastic differential equation with infinity delay:
 \begin{equation}\label{ex}
 d[x(t) - D(x_{t})] = - c x(t) dt + \Big( cos(x(t))  + \int_{- \infty}^{0} x_{t} (\theta) \mu(d\theta) \Big) dw(t), \qquad t \geq 0,
 \end{equation}
 with initial value $ x(t) = \xi $ when $ t \in ( - \infty, 0] $. Where $ c > 0 $, $ w(t) $ is a scalar Brownian motion, while $ D : C_{r} \rightarrow R $ is defined by:
  \begin{equation*}
  D(\phi) = \dfrac{1}{2} \int_{ - \infty }^{0} \phi(\theta) \mu(d \theta).
  \end{equation*}
So that by  H\"{o}lder inequality we have
\begin{equation*}
\lvert D(\varphi) - D(\phi)\lvert ^{2} \leq \dfrac{1}{4} \int_{ - \infty }^{0}\lvert \varphi(\theta) - \phi(\theta) \lvert ^{2} \mu(d \theta).
\end{equation*}
Define $ b(\phi) = - c \phi, \sigma(\phi) = cos(\phi)  + \int_{- \infty}^{0} \phi (\theta) \mu(d\theta) $. It is easy to show that:
\begin{equation}\label{ex 1}
\begin{split}
\Big[ \varphi (0) - \phi (0) - \big( D(\varphi) - & D (\phi) \big) \Big]^{T} \Big[ b ( \varphi) - b ( \phi ) \Big] =   \Big[ \varphi (0) - \phi (0) \\
& \qquad + \dfrac{1}{2}\int_{ - \infty }^{0} \big(\varphi (\theta) - \phi (\theta )\big) \mu(d \theta) \Big]^{T} \Big[ - c\varphi  + c \phi\Big]\\
& \leq -c \rvert \varphi (0) - \phi (0) \rvert ^{2}  + \dfrac{c}{4} \int_{-\infty}^{0} \rvert \varphi (\theta) - \phi (\theta) \rvert ^{2} \mu  (d \theta),
\end{split}
\end{equation}
and
\begin{equation} \label{ex 2}
\begin{split}
\big \rvert \sigma (\varphi) - \sigma (\phi) \big\rvert ^{2} & =  \big\rvert cos(\varphi)  - cos(\phi)   +  \int_{- \infty}^{0} ( \varphi (\theta) - \phi (\theta)   ) \mu  (d\theta) \big\lvert^{2} \\
& \leq( 1+\varepsilon) \big\rvert cos(\varphi)  - cos(\phi) \big\rvert^{2} + \dfrac{1+\varepsilon}{\varepsilon} \big\rvert \int_{- \infty}^{0} ( \varphi (\theta) - \phi (\theta)   ) \mu  (d\theta) \big\lvert^{2}\\
& \leq (1+\varepsilon) \big\rvert \varphi(0)  - \phi(0) \big\rvert^{2} + \dfrac{1+\varepsilon}{\varepsilon}  \int_{- \infty}^{0} \big\rvert \varphi (\theta) - \phi (\theta)  \big\lvert^{2}  \mu  (d\theta). \\
\end{split}
\end{equation}
Thus, from \eqref{ex 1} and \eqref{ex 2} we get that: $ \lambda_{1}=c, \lambda_{2}=\dfrac{c}{4},\lambda_{3}=1+\varepsilon$ and $\lambda_{4}=\dfrac{1+\varepsilon}{\varepsilon} $, hence for any $ \varepsilon > 0 $ and $ \mu^{(2r)} \in M_{2r} $, the solution map $ x_{t} $ of the equation \eqref{ex} satisfies all the properties that we studied whenever $ c > \dfrac{4+ 146 (1+\varepsilon) + 146 (1+\varepsilon^{-1}) \mu^{(2r)}}{4 - \mu^{(2r)} } $.

\section*{Acknowledgement}
I would like to thank Professor Chenggui Yuan for his expert advice, as well as Dr. Jianhai Bao for his encouragement and advice throughout this project. This research was supported by Kufa University and the Iraqi Ministry of Higher Education and Scientific Research.


TITLE: How to properly apply the Lie Series
QUESTION [7 upvotes]: I am trying to solve this problem from Symmetry Methods for Differential Equations A Beginner's Guide (Peter E. Hydon).
Use the Lie Series
$$F(\hat{x},\hat{y})=\sum_{j=0}^{\infty}\frac{\varepsilon^j}{j!}X^j F(x,y)$$
where
$$X = \xi(x,y)\partial_x +\eta(x,y)\partial_y$$
to verify, that
$$\hat{x}=\exp{(\varepsilon X)}x,$$
$$\hat{y}=\exp{(\varepsilon X)}y$$
holds for
$$a) X = x\partial_x-y\partial_y$$
$$b) X = x^2\partial_x+xy\partial_y$$
$$c) X = -y\partial_x+x\partial_y$$
I don't even know where to start ... . A step by step solution for one of the $X$ would be nice :).

REPLY [0 votes]: An alternative and much more effective approach is proposed in this second answer.
References:

Exponential of a function times derivative

How to derive these Lie Series formulas

In the first reference, the following formula is proved:
$$
x_1(t) = e^{t\,g(x)\frac{d}{dx}} x \quad \Longleftrightarrow \quad
\dot{x}_1(t) = g(x_1(t)) \quad \mbox{with} \quad x = x_1(0)
$$
But I've found in my old notes that there is a far more general result.
For any one $(t)$ parameter Lie group the following theorem holds:
$$
{\bf x}_1(t) = e^{t X} {\bf x} = e^{t\,{\bf g(x)}\cdot\nabla} {\bf x}
\quad \Longleftrightarrow \quad
\dot{{\bf x}}_1(t) = {\bf g}({\bf x}_1(t)) \quad \mbox{with} \quad {\bf x} = {\bf x}_1(0)
$$
Meaning that the problem of finding the Lie series for the independent variables ${\bf x}$
can be reduced to solving a system of ordinary differential equations.
In the two dimensional case:
$$
X = {\bf g(x)}\cdot \nabla = \xi(x,y)\,\partial_x +\eta(x,y)\,\partial_y
$$
So we only have to solve the ODE system:
$$
\left\{\begin{matrix}\dot{x}_1 = \xi(x_1,y_1) \\ \dot{y}_1 = \eta(x_1,y_1)\end{matrix}\right.
$$
With boundary conditions, always the same:
$$
\left\{\begin{matrix} x_1(0) = x \\ y_1(0) = y\end{matrix}\right.
$$
Now let's do it for the operators at hand.
$a)\; X = x\, \partial_x-y\, \partial_y$
Accompanying ODE system:
$$
\left\{\begin{matrix}\dot{x}_1 = x_1 \\ \dot{y}_1 = -y_1\end{matrix}\right.
$$
Together with the boundary conditions giving a solution as has been found in the first answer:
$$
\left\{\begin{matrix} x_1 = x\,e^t \\ y_1 = y\,e^{-t} \end{matrix}\right.
$$
$b)\; X = x^2\,\partial_x+xy\,\partial_y$
Accompanying ODE system:
$$
\left\{\begin{matrix}\dot{x}_1 = x_1^2 \\ \dot{y}_1 = x_1 y_1\end{matrix}\right.
$$
Solve the first equation and substitute the solution into the second one.
Solve again and apply the boundary conditions:
$$
\left\{\begin{matrix} x_1 = {\Large \frac{x}{1-t\,x}} \\ y_1 = {\Large \frac{y}{1-t\,x}} \end{matrix}\right.
$$
This solution deserves some attention because it is not in the first answer.
$c)\; X = -y\, \partial_x+x\, \partial_y$
Accompanying ODE system:
$$
\left\{\begin{matrix}\dot{x}_1 = -y_1 \\ \dot{y}_1 = x_1\end{matrix}\right.
$$
Two separate equations for $x_1$ and $y_1$ can be found from this:
$$
\left\{\begin{matrix}\ddot{x}_1 + x_1 = 0 \\ \ddot{y}_1 + y_1 = 0 \end{matrix}\right. \quad \Longrightarrow \quad
\left\{\begin{matrix}x_1 = A\cos(t) + B\sin(t) \\ y_1 = C\cos(t) + D\sin(t) \end{matrix}\right.
$$
Employing the original ODE : $\;\dot{y}_1=x_1$ gives $\;-C = B\;$ and $\;D = A\;$ .
At last, apply the boundary conditions:
$$
\left\{\begin{matrix}x_1 = x\cos(t) - y \sin(t) \\ y_1 = x\sin(t) + y \cos(t)\end{matrix}\right.
$$
As has been found in the first answer too.


TITLE: how come square root of 2 times itself equals 2
QUESTION [0 upvotes]: since square root of $2$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $2$.  It is like saying $0.928041\dots$ times itself is some finite number (like $1.64$ not going $3$ dots in the end).

REPLY [0 votes]: In the real world, you only want to be accurate up to a point.  Within an inch, or a millimetre, or much less if you are building a computer chip.
Two people might agree that two numbers are close enough to being equal, but a third might disagree.
If two numbers are so close that everyone agrees they are close enough - so they agree to every decimal point - then we say they are equal.  So $1/3=0.333333333...$


TITLE: Show that $\int_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^n}$ is finite
QUESTION [1 upvotes]: Show that $\int_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^n}$ is finite.

Here $|x|^2 := x_1^2+\dots+x_n^2$ is the norm of $x$. The statement is clearly true for $n=1$. For $n=2$ we have $$ \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(1+x_1^2+x_2^2)^2}\le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(x_1^2+x_2^2)^2} \le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{x_1^2+x_2^2} \le\,\, ?$$ How do I proceed with this. I noticed that the integrandum is even, and I feel like I need to find an upper bound for this integral that is the $n$-th power of an integral that is finite and that can be easily computed (for example the $n=1$-integral); something like $$ \int_{\mathbb{R}} \frac{dx_1}{1+x_1^2}\int_{\mathbb{R}} \frac{dx_2}{1+x_2^2}\cdots\int_{\mathbb{R}} \frac{dx_n}{1+x_n^2}.$$ Thanks in advance.

REPLY [1 votes]: $\frac 1 {1+|x|^{2}} = \frac 1 {1+x_1^{2}+x_2^{2}+...+x_n^{2}} \leq \frac  1 {1+x_i^{2}}$ for each $i$. Multiply these $n$ inequalities and integrate.


TITLE: Is this inference correct using the existential quantifier?
QUESTION [1 upvotes]: Suppose that one has the statement $\forall x\exists y :P(x,y)$. Given an $a$, then it follows that there is a $b$ such that $P(a,b)$.
My question is: Am I allowed to call this object $b_a$ or $b(a)$ or any other version that behaves like a function symbol? If so, why or why not?
Why am I asking this? Suppose that $X$ is a set and we have $\forall x \in X \exists y \in X: P(x,y)$ and suppose that we are allowed to introduce $b_a$ for the appropriate object as explained above. Then we know that there exists the set $A$ such that $$\forall z : z \in A \iff \exists a,b \in X : z=(a,b) \wedge =b_a$$ or in other words $$A=\{(a,b_a):a \in X\}.$$
If this were allowed, I wouldn't need the axiom of choice to construct the set $A$. However, if this wasn't allowed, I would need the axiom of choice, since I can't (or at least I don't know how to) construct such a set from the statement $\forall x \in X \exists y \in X:P(x,y)$. More precisely, I don't see how one could prove the existence of a set $A$ that consists of pairs of the elements $x$ of $X$ with exactly one element $y$ of $X$ that satisfy $P(x,y)$. If the object $y$ is unique, then this is easy: $$\forall z : z \in A \iff \exists x,y \in X: z=(x,y) \wedge P(x,y)$$ or $$A=\{(x,y) \in X \times X: P(x,y)\}$$
One possible explanation? I think in formal logic this is not allowed, but I am not sure how it is in more informal maths other than formal logic, or even why its not allowed in formal logic. One potential reason I came up with is the following. Suppose one has an object $x$ and introduces a function symbol $f(x)$ for an object $y$ that satisfies $P(x,y)$. Then, if $y_1,y_2$ are two objects that satisfy $P(x,y_1),P(x,y_2)$, one might be tempted to incorrectly infer: $P(x,y_1)$ implies $y_1=f(x)$ and $P(x,y_2)$ implies $y_2=f(x)$, thus $y_2=y_1$, where one used the intuition that function symbols should be unique objects, which they of course are not in this case. However, with enough caution, such mistakes shouldn't happen. Therefore I am not entirely sure why its not allowed to introduce those.

REPLY [1 votes]: There is a precise sense in which it is acceptable to do this even without any form of the axiom of choice.
Suppose we are working in first-order logic over vocabulary $V$. And suppose we have some set of axioms $T$ and some proposition $P(x, y)$ in the variables $x, y$, and we have $T \vdash \forall x \exists y P(x, y)$. Then we can define $V' = V \cup \{f\}$, where $f$ is a unary function symbol, and define $T' = T \cup \{\forall x P(x, f(x))\}$. Then this new deductive system is a conservative extension. More formally, for any sentence $\phi$ in the language of $V$, we have $T \vdash \phi$ if, and only if, $T' \vdash \phi$.
This can be proved even without using the axiom of choice, as it is purely a syntactic fact. However, it is more convenient to prove it using the axiom of choice and the completeness theorem. For if we have any $V$-structure $M$ such that $M \models T$, then we have $M \models \forall x \exists y P(x, y)$. We then apply the axiom of choice to produce some $f^M$ such that $\forall x \in M (P^M(x, y))$. Adding $f^M$ to $M$ gives us $M'$, which is a $V'$ structure which is a model of $T'$. And of course, for any sentence $\phi$ in the language of $V$, we have $M' \models \phi$ if and only if $M \models \phi$. Thus, we see that semantically, $T'$ is a conservative extension of $T$ - for all $\phi$ in the language of $V$, we see that $T \models \phi$ iff $T' \models \phi$. Apply the completeness theorem to get the syntactic version.
In fact, we only require the ultrafilter lemma to get that $T'$ is semantically a conservative extension of $T$. So this fact does not imply the axiom of choice. However, without the axiom of choice, we cannot, in general, take a model $M$ of $T$ and extend it to a model $M'$ of $T'$, where both models are exactly the same $V$-structure. Instead, what we see is that if $M$ is a $V$-structure modelling $T$ such that $M \models \neg \phi$, then there exists some $M'$ which is a model of $T'$ and of $\neg \phi$, but there is no reason to believe that $M$ and $M'$ should be nicely related to each other by being the same $V$-structure.
In particular, if we have $\forall x \in X \exists y \in X (P(x, y))$, then we have $X \models \forall x \exists y P(x, y)$. So if we wish to make deductions in first-order logic about $X$, we could, within the first-order system, add a function symbol $a \mapsto b_a$ and the axiom $\forall x (P(x, b_x))$. But (in the absence of the axiom of choice) this would not actually mean there exists such a function $X \to X$.


TITLE: Biggest n that can be solved in one second
QUESTION [0 upvotes]: I have been given the following problem:
What is the largest n for which one can solve in one second a problem that requires $(\log_2(n))^2$
elementary operations, where each elementary operation is carried out in $10^{−12}$ second?
I have the following solution:
$(\log_2(n))^2 = 10^{12}$
$\log_2(n) = 10^{6}$
$n = 2^{10^6}$
Is my reasoning (and ultimately, the solution) correct?

REPLY [2 votes]: Yes your computations and your answer looks correct.


TITLE: An integral for $2\pi+e-9$
QUESTION [9 upvotes]: Motivation
Lucian asked about the almost-integer $2\pi+e\approx9$ in a comment to a partially answered why question about $e\approx H_8$. This is more involved than approximations to $\pi$ and logarithms because two transcendental constants are included, as in $e^\pi-\pi\approx20$.
Tried so far
An answer can be crafted from integrals related to $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx =\frac{22}{7}-\pi$$
$$\frac{1}{14}\int_0^1 x^2(1-x)^2e^xdx=e-\frac{19}{7}$$
to obtain

$$\int_0^1 x^2 (1-x)^2 \left(\frac{e^x}{14}-\frac{2 x^2 (1-x)^2}{1+x^2}\right) dx = 2\pi+e-9 $$

The visual representation of this integral provided by WolframAlpha shows that $2\pi+e-9$ is positive and small (the integrand is between $0$ and $0.004$ for $0<x<1$), although this is not immediate from the analytic expression.
Moreover, two maxima appear, instead of the single one that is usual in this type of integrals.
Question

Is there a simpler integral with positive integrand in (0,1) that proves $2\pi+e\approx 9$?

REPLY [6 votes]: Approximations $e\approx \frac{163}{60} $ and $ \pi \approx \frac{377}{120}$ are related to the integrals
$$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$
and
$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi.$$
Combining both, we can build
$$\frac{1}{2} \int_0^1 (1 - x)^2 \left(e^x - 1 - x - \frac{x^2}{2} - \frac{2 x^5 (1 - x)^4}{1 + x^2}\right) dx = 2\pi+e-9,$$
which explains the result with nonnegative small integrand and a single maximum in $(0,1)$.
WolframAlpha link
This sets the lower bound
$$9<2\pi+e$$
For an upper bound, we may take the integral from a failed attempt to match $9$
$$\int_0^1 x^4(1-x)^4\left(-\frac{e^x}{24024}+\frac{2}{1+x^2}\right) dx = 9+\frac{4}{1001}-2\pi-e$$
and write
$$2\pi+e<9+\frac{4}{1001}$$
Finally,
$$9<2\pi+e<9+\frac{4}{1001}$$
In short, $2\pi+e$ is close to $9$ because
$\pi\approx\dfrac{377}{120}$
from above,
$e \approx \dfrac{163}{60}$
from below and
$$2·\frac{377}{120}+\frac{163}{60}=9$$


TITLE: Dilemma evaluating integral after sinusoidal substitution
QUESTION [1 upvotes]: To solve the definite integral
$$ I = \int_{-a}^{a} \frac{dx}{\pi \sqrt{a^2-x^2}}$$
I used the substitution $x = a \sin \theta$ and tried to solve the integral without its interval definition, which yields
$$ I = \frac{\theta}{\pi} $$
Now according to what I have been taught, from this point on I can do either of two things:

write the result in terms of $x$ and evaluate with the initial limits; or
transform $x$'s limits into limits for $\theta$ and evaluate.

Selecting the first option listed, the result of the evaluation is 1, but selecting the second option I get as result
$$ x = a \Rightarrow \theta = \frac{\pi}{2} + 2\pi n$$
$$x=-a\Rightarrow \theta = \frac{3 \pi}{2} + 2 \pi k$$
$$ I = \frac{\theta}{\pi} = \left(\frac{1}{2} + 2n\right) - \left(\frac{3}{2} + 2k \right) = 2(n-k)-1$$
(where $n$ and $k$ are integers.) Why would $n-k=1$ so this result would also be correct?

REPLY [0 votes]: The Problem is, you have to exactly specify the bijection (one-to-one and onto) used for the substitution.
$$u(\theta) = a \sin(\theta)$$
maps bijective to $[-a,a]$ for several choices of ${\rm Dom}_u$. The most straight-forward of these is
$$u: \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ] \to [-a,a]$$
But the only limitations on the domain are

$\left.\sin\right|_D$ must be a bijection
If $D = [d_0, d_1]$ then $\sin(d_0) \in \{-1,1\} \ni \sin(d_1) \neq \sin(d_0)$ so that the endpoints $u(d_0), u(d_1) \in \{-a,a\}$ and by intermediate value theorem $u(D) = [-a,a]$ since $\sin(D) = [-1,1]$

So to find the range of theta start with a maximum at $\sin(d_0)$, i.e.
$$\cos(d_0) = 0 \Leftrightarrow d_0 = \frac{\pi}{2} + k\pi, \qquad k\in \mathbb{Z}$$
and take an interval of length $\pi$, so that $\sin(d_1) = \sin(d_0 + \pi) = -\sin(d_0)$. You get
$$D = [\frac{\pi}{2} + k\pi, \frac{\pi}{2} + (k+1) \pi]$$
In your above notation $"n-k" = (k+1)-k = 1$ so your two solutions are the same (correct).


TITLE: Power Series With Bernoulli Numbers
QUESTION [6 upvotes]: The exercise reads: Express the power series for $\large \frac{z}{\sin (z)} = \frac{2 i z}{e^{iz} - e^{-iz}} $ in terms of Bernoulli numbers.
I am given in a previous exercise that the Bernoulli numbers are defined by
$$ \frac{z}{e^z - 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n, $$
where $ B_n $ is the $ n^{th} $ Bernoulli number. I've been looking for a clever way to write $ \;\large\frac{1}{e^{iz} - e^{-iz}} \;$ in the form of a linear combination involving terms of the form $\; \large\frac{1}{e^z - 1}\; $, but haven't had any luck.  Any pointers would be greatly appreciated!

REPLY [8 votes]: If you start with $f(z):=\frac{z}{e^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}$, consider what $f(z)+f(-z)$ is. You'll get a sum of only even bernoulli numbers. Now notice that
$f(z)+f(-z)=\frac{z(e^{t/2}+e^{-t/2})}{(e^{t/2}-e^{-t/2})}$
which you can reverse engineer to verify (multiply by the right ratio which equuals 1). This essentially gives you an expansion for $\cot(z)$ in terms of the even Bernoulli numbers. However you want $\csc(z)$. This is amenable by recalling that
$\cot(z)-\cot(2z)=\csc(2z)$
So subtract the power series to get your result.
The correct answer is:
$$\csc(z)=\sum_{n=0}^\infty \frac{(-1)^{n+1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}$$


TITLE: The union of a set with a set that has the finite intersection property has F.I.P
QUESTION [1 upvotes]: Let $I$ be a set and let $C$ be a collection of subsets of $I$ with the finite intersection
property. Prove that for every $A \subset I$, either $C \cup \{A\}$ or $C \cup \{A^{c}\}$ has the finite intersection property.

I am trying to prove the above statement and I am just not sure where to even start. I know that since C has the finite intersection property that the intersection of any finite subcollection of C is nonempty.
Do i need to look at different cases? If $A \in C$ and $A \notin C$.
If $A \in C$ does that automatically mean $C \cup \{A\}$ has F.I.P. or no?
And if $A \notin C$ does that imply $A^c \in C$ since $A \subset I$?

REPLY [2 votes]: I’ll use $\mathscr{C}$ instead of $C$ for the family with the finite intersection property.
Yes, if $A\in\mathscr{C}$, then $\mathscr{C}\cup\{A\}=\mathscr{C}$, so it has the finite intersection property. However, $A\notin\mathscr{C}$ does not imply that $I\setminus A\in\mathscr{C}$. Suppose, for instance, that $A_n=\{k\in\Bbb N:k\ge n\}$ for each $n\in\Bbb N$, and $\mathscr{C}=\{A_n:n\in\Bbb N\}$; it’s easy to check that $\mathscr{C}$ has the finite intersection property. Let $A=\{2n:n\in\Bbb N\}$, the set of even natural numbers, so that $\Bbb N\setminus A$ is the set of odd natural numbers; then neither $A$ nor $\Bbb N\setminus A$ belongs to $\mathscr{C}$.
HINT: I suggest trying to prove the contrapositive: show that if neither $\{A\}\cup\mathscr{C}$ nor $\{I\setminus A\}\cup\mathscr{C}$ has the finite intersection property, then $\mathscr{C}$ does not have the finite intersection property. Start by noticing that if $\{A\}\cup\mathscr{C}$ does not have the finite intersection property, then there is a finite $\mathscr{C}_0\subseteq\mathscr{C}$ such that $A\cap\bigcap\mathscr{C}_0=\varnothing$. There’s a further hint in the spoiler box if you need it.

 If $A\cap X=\varnothing=B\cap X$, then $$(A\cup B)\cap X=(A\cap X)\cup(B\cap X)=\varnothing\,.$$


TITLE: Counterexample for direct generalization of the one-dimensional Mean Value Theorem
QUESTION [0 upvotes]: It is written in our textbook that the Mean Value Theorem is only correct in one-dimensional spaces and that its direct generalization is not true in higher dimensions. Instead it has used an inequality in the generalization of the Mean Value Theorem.
I want to find a counterexample to a direct generalization of the theorem, say a fuction $f: U \rightarrow \mathbb{R}^n$ where $U$ is an open set and $[p,q] \subset U \subset \mathbb{R}^m$ such that there is no point $\theta \in [p,q]$ such that $$f(q)-f(p)=(Df)_{\theta}(q-p).$$ By $(Df)_\theta$ I mean derivative of $f$ in $\theta$.
Any idea for finding this counterexample would be appreciated.

REPLY [5 votes]: Take $f(x) = (\cos x, \sin x)$ on $[0,2\pi]$. Then, take $q=x>0$, $p=0$ and suppose
$$(\cos x-1, \sin x) =f(x)-f(0) = f'(\xi)x  = (-\sin \xi , cos \xi)x $$
for some $\xi \in (0,x)$. This is impossible, since the magnitude of the right hand side is $x$, while the magnitude of the left hand side is
$$\sqrt{(\cos x-1)^2+\sin^2 x} = \sqrt{2- 2 \cos x} = 2 \sin \frac{x}{2} < x$$