S. V. Lototsky]Department of Mathematics, USC
Los Angeles, CA 90089

# A Sharp Rate of Convergence in the Functional Central Limit Theorem with
Gaussian Input

S. V. Lototsky [<EMAIL_ADDRESS>https://dornsife.usc.edu/sergey-
lototsky/

###### Abstract.

When the underlying random variables are Gaussian, the classical Central Limit
Theorem (CLT) is trivial, but the functional CLT is not. The objective of the
paper is to investigate the functional CLT for stationary Gaussian processes
in the Wasserstein-1 metric on the space of continuous functions. Matching
upper and lower bounds are established, indicating that the convergence rate
is slightly faster than in the Lévy-Prokhorov metric.

###### Key words and phrases:

Brownian Bridge, Gauss-Markov Process, Kantorovich-Rubinstein metric

###### 2020 Mathematics Subject Classification:

Primary 60F17; Secondary 60G10, 60G15.

##  1\. Introduction

By the Central Limit Theorem, given a collection $\\{\xi_{k},\ k\geq 1\\}$ of
independent and identically distributed random variables, each with mean zero
and variance one, the sequence $S_{n}=n^{-1/2}\sum_{k=1}^{n}\xi_{k}$ converges
in distribution, as $n\to\infty$, to the standard Gaussian random variable.
The Berry-Esseen bound [16, Theorem 15.51] gives the rate of convergence in
the Kolmogorov metric:

$\sup_{x\in{\mathbb{R}}}|F_{n}(x)-\Phi(x)|\leq\frac{{\mathbb{E}}|\xi_{1}|^{3}}{\sqrt{n}},$
(1.1)

where $F_{n}$ is the cumulative distribution function of $S_{n}$ and $\Phi$ is
the cumulative distribution function of the standard Gaussian random variable.
While the rate $1/\sqrt{n}$ is sharp in general, it can be improved by
imposing additional conditions on the random variables $\xi_{k}$. For example,
if ${\mathbb{E}}\xi_{k}^{3}=0$ and ${\mathbb{E}}\xi_{k}^{4}<\infty$, then the
left-hand side of (1.1) is of order $1/n$; cf [21, Theorem 5.2.1]. Of course,
if each $\xi_{k}$ is standard normal, then the left-hand side of (1.1) is
zero.

The functional version of the Central Limit Theorem, also known as the Donsker
invariance principle [16, Theorem 21.43], establishes weak convergence of the
sequence of processes

$S_{n}(t)=n^{-1/2}\sum_{k=1}^{\lfloor nt\rfloor}\xi_{k}+\frac{nt-\lfloor
nt\rfloor}{\sqrt{n}}\xi_{\lfloor nt\rfloor+1},\ t\geq 0,\ n\geq 1,$ (1.2)

to the standard Brownian motion $W$. An analog of (1.1) becomes a bound on the
distance between the distributions of $S_{n}$ and $W$ on the space of
continuous functions ${\mathcal{C}}(0,T)$ in the Lévy-Prokhorov metric.
Compared to (1.1), the corresponding rate of convergence depends on
integrability properties of $\xi_{k}$ in a more complicated way: if
${\mathbb{E}}|\xi_{1}|^{p}<\infty$, $p>2$, then the rate $n^{-(p-2)/(2(p+1))}$
is sharp; if ${\mathbb{E}}e^{t\xi_{1}}<\infty,\ |t|<\delta,\ \delta>0,$ then
the rate $\ln n/\sqrt{n}$ is sharp. For details, see [7, Chapter 1]; earlier
works on the subject include [17, 18, 25].

A more general approach to investigating the rate of convergence is to find a
bound on

$\sup_{\varphi\in\mathcal{G}}|{\mathbb{E}}\varphi(S_{n})-{\mathbb{E}}\varphi(W)|$
(1.3)

for a suitable class $\mathcal{G}$ of functions
$\varphi:{\mathcal{C}}(0,T)\to{\mathbb{R}}$. Barbour [3, Theorem 1] used an
infinite-dimensional version of Stein’s method to establish the benchmark
result

$|{\mathbb{E}}\varphi(S_{n})-{\mathbb{E}}\varphi(W)|\leq\frac{C}{\sqrt{n}}\Big{(}\sqrt{\ln
n}+{\mathbb{E}}|\xi_{1}|^{3}\Big{)}$ (1.4)

for a certain (rather restrictive) class $\mathcal{G}$; the restrictive nature
of this class ensures that there is no contradiction with [7, Chapter 1] or
[25]. For various other $\mathcal{G}$, there are bounds of the form

$|{\mathbb{E}}\varphi(S_{n})-{\mathbb{E}}\varphi(W)|\leq\frac{C}{{n}^{r}},\
0<r<\frac{1}{2};$ (1.5)

cf. [6] and references therein.

Unlike (1.1), the left-hand side of (1.4) will not be zero even if the random
variables $\xi_{k}$ are Gaussian, as long as $\mathcal{G}$ is rich enough to
capture the infinite-dimensional nature of the problem. In fact, for certain
$\mathcal{G}$, one can use (1.3) to define a metric on the space of
distributions. For example, if $\mathcal{G}$ is the collection of bounded
Lipschitz continuous functions, then convergence in the corresponding bounded
Lipschitz metric is equivalent to weak convergence, that is, convergence in
the Lévy-Prokhorov metric; cf. [16, Theorem 13.16] or [9, Theorem 11.3.3].
Removing the boundedness condition (for example, to include linear
functionals) leads to the Wasserstein-1 metric, which is the subject of this
paper.

Recall that, for two probability measures $\mu$, $\nu$ on a complete separable
metric space $E$ with distance function $\rho$ and the corresponding Borel
sigma-algebra $\mathcal{B}(E)$,

  * •

the bounded Lipschitz metric is

$\mathrm{d}_{{}_{BL}}(\mu,\nu)=\sup_{\varphi}\left|\int_{E}\varphi
d\mu-\int_{E}\varphi d\nu\right|,$ (1.6)

with supremum over functions $\varphi:E\to\mathbb{R}$ such that, for all
$x,y\in E$, $|\varphi(x)-\varphi(y)|\leq\rho(x,y)$ and $|\varphi(x)|\leq 1$;

  * •

the Wasserstein-1 metric $\mathrm{d}_{{w}}(\mu,\nu)$, also known as the
Kantorovich-Rubinstein metric, is

$\mathrm{d}_{{w}}(\mu,\nu)=\sup_{\varphi}\left|\int_{E}\varphi
d\mu-\int_{E}\varphi d\nu\right|,$ (1.7)

with supremum over functions $\varphi:E\to\mathbb{R}$ such that, for all
$x,y\in E$, $|\varphi(x)-\varphi(y)|\leq\rho(x,y)$;

  * •

the Lévy-Prokhorov metric is

$\mathrm{d}_{{}_{LP}}(\mu,\nu)=\inf\\{\varepsilon>0:\mu(A)\leq\nu(A^{\varepsilon})+\varepsilon,\
A\in\mathcal{B}(E)\\},$ (1.8)

where $A^{\varepsilon}=\\{x\in E:\inf\limits_{y\in
A}\rho(x,y)\leq\varepsilon\\}.$

We have $\mathrm{d}_{{}_{BL}}(\mu,\nu)\leq\mathrm{d}_{{w}}(\mu,\nu)$ (by
definition),
$\mathrm{d}_{{}_{LP}}(\mu,\nu)\leq\sqrt{\mathrm{d}_{{}_{BL}}(\mu,\nu)}$ ([9,
Proof of Theorem 11.3.3]), and $\mathrm{d}_{{}_{BL}}(\mu,\nu)\leq
4\mathrm{d}_{{}_{LP}}(\mu,\nu)$ ([9, Corollary 11.6.5]). In particular,
convergence in the Wasserstein-1 metric implies weak convergence, that is,
convergence in either Lévy-Prokhorov or bounded Lipschitz metric; the converse
is not always true [9, p. 421]; in fact, the diagram in [11] suggests that the
Wasserstein-1 metric $\mathrm{d}_{{w}}$ is the strongest possible for the CLT-
type problems in function spaces. Still, a sharp rate of convergence in one
metric does not directly lead to a sharp rate in any other metric.

The invariance principle can hold if independence requirement for the random
variables $\xi_{k}$ is relaxed, for example, to a strictly stationary and
ergodic martingale difference [20, Theorem 9.1.1], or a stationary Markov
process satisfying Doebling’s condition [13] [where a bound of the type (1.5)
is also established]. In continuous time, if $X=X(t),\ t\in{\mathbb{R}},$ is a
strictly stationary process with mean zero and covariance function
$R(t)={\mathbb{E}}\big{(}X(t)X(0)\big{)}$ satisfying
$\int_{-\infty}^{+\infty}R(t)\,dt=1$, then, under some additional conditions
of weak dependence, the sequence of processes

$S_{n}(t)=\frac{1}{\sqrt{n}}\int_{0}^{nt}X(s)\,ds,\ n=1,2,\ldots,t\in[0,1],$
(1.9)

converges weakly to the standard Brownian motion; cf. [20, Theorem 9.2.1] or
[14, Theorem VIII.3.79].

The objective of this paper is to show that if $X$ is a stationary Gauss-
Markov process, in either discrete or continuous time, then the Wasserstein-1
distance between $S_{n}$ and $W$ in the space of continuous functions is of
order $\big{(}n^{-1}\ln n\big{)}^{1/2}$. In other words, if $S_{n}$ is
Gaussian, then the convergence rate in Wasserstein-1 metric is slightly faster
than the Lévy-Prokhorov rate $\ln n/\sqrt{n}$. This difference does not
contradict the results from [7, Chapter 1] and [25], and the discrepancy by a
$\sqrt{\ln n}$ factor can be explained as follows: in the limit $\sigma\to
0+$, the distance between a Gaussian distribution with mean zero and variance
$\sigma^{2}$ and a point mass at zero is of order $\sigma$ in the
Wasserstein-1 metric, but it is of order $\sigma\sqrt{|\ln\sigma|}$ in the
Lévy-Prokhorov metric.

Section 2 discusses (the easier) continuous-time case (1.9). Discrete-time
case, a generalization of (1.2) for a stationary Gaussian sequence
$\\{\xi_{k},\ k\geq 0\\}$, is in Section 3. In Section 4, the results are
applied to weak approximation for some ordinary differential equations with
additive noise. Section 5 is a summary. Traditionally, models (1.2) and (1.9)
are studied on a bounded time interval $[0,T]$, and the index parameter
$n=1,2,\ldots$ is discrete. In this paper, the time interval is $(0,+\infty)$
and the index parameter $\kappa\geq 1$ is not necessarily an integer.

The following two properties of Gaussian processes will be used on several
occasions:

1\. The Borell-TIS inequality [1, Theorem 2.1.1]: If $X=X(t),\
t\in\mathcal{T},$ is a zero-mean Gaussian process indexed by the set
$\mathcal{T}$, and
$\mathbb{P}\left(\sup\limits_{t\in\mathcal{T}}X(t)<\infty\right)=1$, then
${\mathbb{E}}\sup\limits_{t\in\mathcal{T}}|X(t)|<\infty$ and, with
$X^{*}={\mathbb{E}}\sup\limits_{t\in\mathcal{T}}X(t),\ \
\sigma_{X}^{2}=\sup\limits_{t\in\mathcal{T}}{\mathbb{E}}X^{2}(t)$,

$\mathbb{P}\Big{(}\sup\limits_{t\in\mathcal{T}}X(t)-X^{*}>x\Big{)}\leq
e^{-x^{2}/(2\sigma_{X}^{2})},\ x>0.$ (1.10)

2\. The Fernique-Sudakov inequality [1, Theorem 2.2.3]: If $X=X(t),\ Y=Y(t)\
t\in\mathcal{T},$ are zero-mean Gaussian processes indexed by the set
$\mathcal{T}$, and, for all $t,s\in\mathcal{T}$,
${\mathbb{E}}|X(t)-X(s)|^{2}\leq{\mathbb{E}}|Y(t)-Y(s)|^{2}$, then

${\mathbb{E}}\sup\limits_{t\in\mathcal{T}}X(t)\leq{\mathbb{E}}\sup\limits_{t\in\mathcal{T}}Y(t).$
(1.11)

##  2\. Continuous Time

Let $X=X(t),\ t\in{\mathbb{R}},$ be a (continuous version of a) stationary
Gaussian process with mean zero and covariance

${\mathbb{E}}X(t)X(s)=e^{-2|t-s|}.$

In particular, $X(t)$ is a standard Gaussian random variable for every $t$.
Equivalent characterizations of $X$ are as follows:

$\displaystyle X(t)$ $\displaystyle=e^{-t}W(e^{2t});$ (2.1) $\displaystyle
X(t)$ $\displaystyle=2\int_{-\infty}^{t}e^{-2(t-s)}dW(s);$ (2.2)
$\displaystyle dX(t)$ $\displaystyle=-2X(t)dt+2\,dW(t),\ t\geq 0.$ (2.3)

In (2.1), (2.2), and (2.3), $W=W(t),\ t\geq 0,$ is a standard Brownian motion;
in (2.2), when $t<0$, $W(t)=V(-t)$ for an independent copy $V$ of $W$. The
initial condition $X(0)$ in (2.3) is a standard Gaussian random variable
independent of $W$.

Given a real number $\kappa\geq 1$, we define

$W^{\kappa}(t)=\frac{1}{\sqrt{\kappa}}\int_{0}^{\kappa t}X(s)\,ds,\ \ t\geq
0.$ (2.4)

Denote by ${\mathcal{C}}_{(0)}$ the collection of continuous functions
$f=f(t)$ on $[0,+\infty)$ such that

$f(0)=0,\ \lim_{t\to+\infty}\frac{|f(t)|}{t}=0.$

Endowed with the norm

$\|f\|_{(0)}=\sup_{t>0}\frac{|f(t)|}{1+t},$

${\mathcal{C}}_{(0)}$ becomes a separable Banach space; cf. [8, Section 1.3].
We have $\mathbb{P}(W\in{\mathcal{C}}_{(0)})=1$ (either by the law of large
numbers for square integrable martingales or using that $t\mapsto tW(1/t)$ is
a standard Brownian motion), and also
$\mathbb{P}(W^{\kappa}\in{\mathcal{C}}_{(0)})=1$, $\kappa\geq 1$, because the
ergodic theorem implies

$\lim_{t\to\infty}\frac{1}{t}\int_{0}^{t}X(s)\,ds={\mathbb{E}}X(0)=0$

with probability one.

######  Proposition 2.1.

There exists a constant $C_{X}$ such that, for every $\kappa\geq 1$ and every
function $\varphi:{\mathcal{C}}_{(0)}\to{\mathbb{R}}$ satisfying

$|\varphi(f)-\varphi(g)|\leq\|f-g\|_{(0)},$ (2.5)

we have

$\Big{|}{\mathbb{E}}\varphi\big{(}W^{\kappa}\big{)}-{\mathbb{E}}\varphi(W)\Big{|}\leq
C_{X}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$ (2.6)

###### Proof.

Using (2.3),

$X(t)=X(0)e^{-2t}+2\int_{0}^{t}e^{-2(t-s)}\,dW(s).$

Changing the order of integration (stochastic Fubini theorem [23, Theorem
IV.46]),

$W^{\kappa}(t)=\kappa^{-1/2}W(\kappa t)+\frac{X(0)-X(\kappa
t)}{2\sqrt{\kappa}}.$ (2.7)

Define

$V^{\kappa}(t)=\kappa^{-1/2}W(\kappa t),\ \ \ X^{\kappa}(t)=\frac{X(\kappa
t)-X(0)}{2}.$ (2.8)

Because $t\mapsto V^{\kappa}(t)$, $t\geq 0$, is standard Brownian motion for
every $\kappa>0$, we have
${\mathbb{E}}\varphi\big{(}W\big{)}={\mathbb{E}}\varphi\big{(}V^{\kappa}\big{)}$
and then, using (2.5),

$|{\mathbb{E}}\varphi\big{(}W^{\kappa}\big{)}-{\mathbb{E}}\big{(}\varphi(W)\big{)}\Big{|}\leq\frac{{\mathbb{E}}\|X^{\kappa}\|_{(0)}}{\sqrt{\kappa}}.$

Next, by [5, Proposition 2.1],

$\lim_{T\to+\infty}\frac{\max_{0\leq t\leq T}X(t)}{\sqrt{2\ln T}}=1,$ (2.9)

with probability one. Using the same arguments as in [19, Proof of Proposition
10.2], we conclude from (2.9) that

$\limsup_{T\to+\infty}\frac{\max_{0\leq t\leq T}|X(t)|}{\sqrt{2\ln T}}\leq 2,$
(2.10)

and then continuity of $X$ implies that the random variable

$\zeta=\sup_{t>0}\frac{|X(t)-X(0)|}{2\sqrt{\ln(2+t)}}$

is finite with probability one. Indeed, if

$T^{*}=\sup\left\\{t\geq 0:\frac{\max_{0\leq t\leq
T}|X(t)-X(0)|}{2\sqrt{2\ln(2+T)}}>5\right\\}$

then, by (2.10), $\mathbb{P}(T^{*}<\infty)=1$, so that

$\zeta\leq\max_{0\leq t\leq
T^{*}}\frac{|X(t)-X(0)|}{2\sqrt{\ln(2+t)}}+5<\infty.$

Moreover, because $t\mapsto X(t)-X(0)$ is a Gaussian process with mean zero,
the Borell-TIS inequality (1.10) implies

${\mathbb{E}}\zeta^{p}<\infty$ (2.11)

for all $p>0$.

If $t>0$ and $\kappa\geq 1$, then, by direct computation,

$(2+\kappa t)\leq(1+\kappa)(1+t),\ \frac{1}{2}\leq\sqrt{\ln(1+\kappa)},\
\frac{\sqrt{\ln(1+t)}}{1+t}\leq\frac{1}{2}.$

As a result,

$\displaystyle\frac{\sqrt{\ln(2+\kappa t)}}{1+t}$
$\displaystyle\leq\frac{\sqrt{\ln(1+t)}}{1+t}+\sqrt{\ln(1+\kappa)}\leq
2\sqrt{\ln(1+\kappa)},$ $\displaystyle\|X^{\kappa}\|_{(0)}$
$\displaystyle=\sup_{t>0}\frac{|X(\kappa t)-X(0)|}{2(1+t)}$
$\displaystyle\leq\left(\sup_{t,\kappa>0}\frac{|X(\kappa
t)-X(0)|}{2\sqrt{\ln(2+\kappa
t)}}\right)\left(\sup_{t>0}\frac{\sqrt{\ln(2+\kappa t)}}{1+t}\right)\leq
2\zeta\sqrt{\ln(1+\kappa)},$

and (2.6) follows with

$C_{X}={\mathbb{E}}\left[\sup_{t>0}\frac{|X(t)-X(0)|}{\sqrt{\ln(2+t)}}\right].$
(2.12)

$\Box$

Denote by $\boldsymbol{\mu}_{0}$ and $\boldsymbol{\mu}_{\kappa}$ the measures
on ${\mathcal{C}}_{(0)}$ generated by the processes $W$ and $W^{\kappa}$. The
following is the main result of this section, showing that the convergence
rate $\kappa^{-1/2}\sqrt{\ln\kappa}$ is sharp for the Wasserstein-1 metric.

######  Theorem 2.2.

There exist positive constants $C_{X}$ and $c_{X}$ such that, for every
$\kappa\geq 1$,

$c_{X}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}\leq\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa},\boldsymbol{\mu}_{0})\leq
C_{X}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$ (2.13)

###### Proof.

The upper bound in (2.13) follows from (1.7) and Proposition 2.1. To establish
the lower bound, we use (1.7) with a particular $\varphi$.

If $\varphi:{\mathcal{C}}_{(0)}\to{\mathbb{R}}$ is a bounded linear
functional, then (2.7) and (2.8) imply

${\mathbb{E}}\varphi(W^{\kappa})-{\mathbb{E}}\varphi(W)=\kappa^{-1/2}{\mathbb{E}}\varphi(X^{\kappa}).$
(2.14)

For $f\in{\mathcal{C}}_{(0)},$ define

$\varphi_{\kappa}:f\mapsto\frac{f(t^{*}_{\kappa})}{2},$

where

$t^{*}_{\kappa}=\arg\max_{0\leq t\leq 1}X^{\kappa}(t).$

In particular, $t^{*}_{\kappa}\in[0,1]$. Then $\varphi_{\kappa}$ is a bounded
linear functional on ${\mathcal{C}}_{(0)}$:

$|\varphi_{\kappa}(f)|\leq\frac{\max_{0\leq t\leq 1}|f(t)|}{2}\leq\max_{0\leq
t\leq 1}\frac{|f(t)|}{1+t}\leq\|f\|_{(0)}.$

Therefore, by (1.7) and (2.14),

$\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa},\boldsymbol{\mu}_{0})\geq\kappa^{-1/2}|{\mathbb{E}}\varphi_{\kappa}(X^{\kappa})|=\frac{{\mathbb{E}}\max_{0\leq
t\leq 1}X^{\kappa}(t)}{2\sqrt{\kappa}}.$ (2.15)

Next, define

$\zeta_{\kappa}=\frac{\max\limits_{0\leq t\leq
1}X^{\kappa}(t)}{\sqrt{\ln(1+\kappa)}}=\frac{\max\limits_{0\leq
t\leq\kappa}X(t)-X(0)}{2\sqrt{\ln(1+\kappa)}},\ \ \kappa\geq 1;$

the second equality follows from (2.8). By (2.11), the family
$\\{\zeta_{\kappa},\ \kappa\geq 1\\}$ is uniformly integrable, so that (2.9)
implies

$\lim_{\kappa\to\infty}{\mathbb{E}}\zeta_{\kappa}=\frac{1}{\sqrt{2}},$ (2.16)

which, in turn, means

$\inf_{\kappa\geq 1}{\mathbb{E}}\zeta_{\kappa}>0.$ (2.17)

The lower bound in (2.13), with

$c_{X}=\frac{1}{2}\inf_{\kappa\geq
1}\frac{{\mathbb{E}}\left[\max\limits_{0\leq
t\leq\kappa}X(t)\right]}{\sqrt{\ln(1+\kappa)}},$ (2.18)

now follows from (2.15) and (2.17), because ${\mathbb{E}}X(0)=0$. $\Box$

To get a better idea about numerical values of $C_{X}$ and $c_{X}$, we need

######  Proposition 2.3.

Let $X=X(t),\ t\in{\mathbb{R}},$ be a stationary Gaussian process with mean
zero and covariance $e^{-2|t-s|}$. Define the random variable

$\bar{\eta}=\max_{0\leq t\leq 1}X(t).$ (2.19)

Then

$\displaystyle 1.2$ $\displaystyle<{\mathbb{E}}\bar{\eta}<3.2;$ (2.20)
$\displaystyle\mathbb{P}(\bar{\eta}>x)$ $\displaystyle\leq
e^{-(x-3.2)^{2}/2},\ x\geq 3.2.$ (2.21)

###### Proof.

Let $B=B(t)$, $0\leq t\leq 1$, be the standard Brownian bridge and $W=W(t)$,
$0\leq t\leq 1$, a standard Brownian motion. Then

$\displaystyle{\mathbb{E}}|B(t)-B(s)|^{2}$ $\displaystyle=|t-s|-|t-s|^{2},\
{\mathbb{E}}|X(t)-X(s)|^{2}=2(1-e^{-2|t-s|}),$
$\displaystyle{\mathbb{E}}|W(t)-W(s)|^{2}=|t-s|,$

and, because

$x-x^{2}\leq 1-e^{-2x}\leq 2x,\ x\geq 0,$

inequality (1.11) implies

$2\,{\mathbb{E}}\max_{0\leq t\leq 1}B(t)\leq{\mathbb{E}}\bar{\eta}\leq
4\,{\mathbb{E}}\max_{0\leq t\leq 1}W(t).$

It is well known (e.g. [9, Section 12.3]) that

$\mathbb{P}\left(\max_{0\leq t\leq 1}B(t)>x\right)=e^{-2x^{2}},\ \
\mathbb{P}\left(\max_{0\leq t\leq
1}W(t)>x\right)=\frac{2}{\sqrt{2\pi}}\int_{x}^{+\infty}e^{-t^{2}/2}dt.$ (2.22)

Then

$\displaystyle{\mathbb{E}}\max_{0\leq t\leq 1}B(t)$
$\displaystyle=\int_{0}^{+\infty}e^{-2x^{2}}\,dx=\frac{\sqrt{2\pi}}{4}>0.6,$
$\displaystyle{\mathbb{E}}\max_{0\leq t\leq 1}W(t)$
$\displaystyle=\frac{2}{\sqrt{2\pi}}\int_{0}^{+\infty}xe^{-x^{2}/2}\,dx=\sqrt{\frac{2}{\pi}}<0.8,$

and (2.20) follows. After that, (2.21) is a re-statement of the Borell-TIS
inequality (1.10). $\Box$

We can now show that the number $C_{X}$ defined in (2.12) satisfies

$1.4<C_{X}<14.$ (2.23)

For the lower bound, note that

$C_{X}\geq{\mathbb{E}}\left[\sup_{t>0}\frac{X(t)-X(0)}{\sqrt{\ln(2+t)}}\right],$

whereas

$\sup_{t>0}\frac{X(t)-X(0)}{\sqrt{\ln(2+t)}}\geq\limsup_{T\to\infty}\frac{\max\limits_{0\leq
t\leq T}\big{(}X(t)-X(0)\big{)}}{\sqrt{\ln(2+T)}},$

and it remains to apply (2.16).

For the upper bound in (2.23), start by writing

$C_{X}\leq{\mathbb{E}}\left[\sup_{t>0}\frac{|X(t)|}{\sqrt{\ln(2+t)}}\right]+\frac{{\mathbb{E}}|X(0)|}{\sqrt{\ln
2}};\ \ \ \frac{{\mathbb{E}}|X(0)|}{\sqrt{\ln 2}}=\sqrt{\frac{2}{\pi\ln
2}}\approx 0.96.$ (2.24)

Next, let $\bar{X}$ be the process consisting of iid copies of $X(t),\
t\in[0,1),$ on each of the intervals $[k-1,k)$, $k=1,2,\ldots$. Then
${\mathbb{E}}X^{2}(t)={\mathbb{E}}\bar{X}^{2}(t)$,
${\mathbb{E}}X(t)X(s)\geq{\mathbb{E}}\bar{X}(t)\bar{X}(s)$, and so

${\mathbb{E}}\left[\sup_{t>0}\frac{|X(t)|}{\sqrt{\ln(2+t)}}\right]\leq
2{\mathbb{E}}\left[\sup_{t>0}\frac{X(t)}{\sqrt{\ln(2+t)}}\right]\leq
2{\mathbb{E}}\left[\sup_{t>0}\frac{\bar{X}(t)}{\sqrt{\ln(2+t)}}\right],$
(2.25)

where the first inequality follows from [19, Proposition 10.2], and the
second, from (1.11). On the other hand, if $\bar{\eta}_{k}$, $k\geq 1$, are
iid copies of the random variable $\bar{\eta}$ from (2.19), then

${\mathbb{E}}\left[\sup_{t>0}\frac{\bar{X}(t)}{\sqrt{\ln(2+t)}}\right]\leq{\mathbb{E}}\left[\sup_{k\geq
1}\frac{\bar{\eta}_{k}}{\sqrt{\ln(1+k)}}\right].$ (2.26)

Using (2.21) with $x>5$,

$\displaystyle\mathbb{P}\Big{(}\sup_{k\geq
1}\frac{\bar{\eta}_{k}}{\sqrt{\ln(1+k)}}>x\Big{)}$
$\displaystyle\leq\sum_{k\geq
1}\mathbb{P}\Big{(}\bar{\eta}_{k}>x\sqrt{\ln(1+k)}\Big{)}$
$\displaystyle\leq\sum_{k\geq
1}\frac{1}{(1+k)^{-(x-3.2)^{2}/2}}\leq\frac{2}{(x-3.2)^{2}-2},$

and therefore

${\mathbb{E}}\left[\sup_{k\geq
1}\frac{\bar{\eta}_{k}}{\sqrt{\ln(1+k)}}\right]\leq
5+2\int_{5}^{+\infty}\frac{dx}{(x-3.2)^{2}-2}<6.5.$

Then (2.23) follows from (2.24) – (2.26).

Next, we will show that the number $c_{X}$ defined in (2.18) satisfies

$0.2<c_{X}<0.8.$ (2.27)

Indeed, the upper bound follows immediately from (2.16). For the lower bound,
start by noting that, for $N\leq\kappa<N+1$,

${\mathbb{E}}\left[\max\limits_{0\leq
t\leq\kappa}X(t)\right]\geq{\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}X(k)\right]$

and ${\mathbb{E}}|X(k)-X(m)|^{2}=2(1-e^{-2})>1$. Now take iid Gaussian
$Y_{k}$, $k=1,\ldots,N$ with mean zero and variance $1/2$. Then
${\mathbb{E}}|Y(k)-Y(m)|^{2}=1$ and so

${\mathbb{E}}|X(k)-X(m)|^{2}\geq{\mathbb{E}}|Y(k)-Y(m)|^{2}.$

By (1.11),

${\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}X(k)\right]\geq{\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}Y(k)\right],$

and, by [19, Lemma 10.2],

${\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}Y(k)\right]\geq 0.4\sqrt{\ln N},$

leading to the lower bound in (2.27).

To summarize, we can write (2.13) in a more explicit form

$0.2\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}\leq\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa},\boldsymbol{\mu}_{0})\leq
14\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2},\ \kappa\geq 1.$

Theorem 2.2 can be used to study convergence on a bounded interval. For $T>0$,
let ${\mathcal{C}}(0,T)$ be the space of continuous functions on $[0,T]$ with
the sup norm, and denote by $\boldsymbol{\mu}_{0}^{T}$ and
$\boldsymbol{\mu}_{\kappa}^{T}$ the measures on ${\mathcal{C}}(0,T)$ generated
by the processes $W$ and $W^{\kappa}$.

######  Theorem 2.4.

There exist positive constants $C_{X,T}$ and $c_{{}_{X,T}}$ such that, for
every $\kappa\geq 1$,

$c_{{}_{X,T}}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}\leq\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa}^{T},\boldsymbol{\mu}_{0}^{T})\leq
C_{X,T}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$ (2.28)

###### Proof.

If $f\in{\mathcal{C}}_{(0)}$, then, for $0<t<T$,
$|f(t)|\leq(1+T)|f(t)|/(1+t)$, so that

$\|f\|_{{\mathcal{C}}(0,T)}=\max_{0\leq t\leq T}|f(t)|\leq(1+T)\|f\|_{(0)}$

and the upper bound in (2.28) follows from the upper bound in (2.13), with
$C_{X,T}=(1+T)C_{X}$.

The lower bound, with

$c_{{}_{X,T}}=\frac{1}{2}\inf_{\kappa\geq
1}\frac{{\mathbb{E}}\left[\max_{0\leq t\leq
T}X^{\kappa}(t)\right]}{\sqrt{\ln(1+\kappa)}},$

follows after repeating the corresponding steps in the proof of Theorem 2.2.
$\Box$

##  3\. Discrete Time

Consider a stationary Gaussian sequence $X=\\{X_{n},\ n\geq 0\\},$ with
${\mathbb{E}}X_{n}=0$ and ${\mathbb{E}}X_{k+n}X_{k}=(1-a)a^{n}/(1+a),\ n,k\geq
0$, where $a\in(-1,1)$ and $a=0$ corresponds to the sequence of iid standard
Gaussian random variables. The variance of $X_{n}$ is chosen so that, for all
$a\in(-1,1)$, the covariance function $R(n)={\mathbb{E}}X_{k+n}X_{n}$ of $X$
satisfies

$R(0)+2\sum_{n=1}^{\infty}R(n)=1.$

Using a collection $\xi_{k},\ k=0,\pm 1,\pm 2,\ldots$ of iid standard normal
random variables, we get the discrete-time analogs of (2.2) and (2.3):

$\displaystyle X_{n}$ $\displaystyle=(1-a)\sum_{k=-\infty}^{n}a^{n-k}\xi_{k},$
(3.1) $\displaystyle X_{n+1}$ $\displaystyle=aX_{n}+\xi_{n+1};$ (3.2)

in (3.2), the initial condition $X_{0}$ is independent of $\xi_{k},\ k\geq 1,$
and is a normal random variable with mean $0$ and variance $(1-a)/(1+a)$.

For $x>0$, let $\lfloor x\rfloor$ denote the largest integer that is less than
or equal to $x$. Define the processes $W^{\kappa}$ by

$W^{\kappa}(t)=\frac{1}{\sqrt{\kappa}}\sum_{n=1}^{\lfloor\kappa
t\rfloor}X_{n}+\frac{\kappa t-\lfloor\kappa
t\rfloor}{\sqrt{\kappa}}X_{\lfloor\kappa t\rfloor+1},\ t\geq 0,\ \kappa\geq
1.$ (3.3)

The second term on the right-hand side of (3.3) ensures that $W^{\kappa}$ is a
continuous function of $t$.

The case $a=0$, that is, the Gaussian version of the original Donsker theorem,
is of special interest; the corresponding process $W^{\kappa}$ will be denoted
by $S_{\kappa}$:

$S_{\kappa}(t)=\frac{1}{\sqrt{\kappa}}\sum_{n=1}^{\lfloor\kappa
t\rfloor}\xi_{n}+\frac{\kappa t-\lfloor\kappa
t\rfloor}{\sqrt{\kappa}}\xi_{\lfloor\kappa t\rfloor+1},\ t\geq 0,\ \kappa\geq
1.$ (3.4)

We have $\mathbb{P}(W^{\kappa}\in{\mathcal{C}}_{(0)})=1$ for every $\kappa\geq
1$, and $a\in(-1,1)$, because the ergodic theorem implies

$\lim_{t\to\infty}\frac{1}{\lfloor\kappa t\rfloor}\sum_{n=1}^{\lfloor\kappa
t\rfloor}X_{n}={\mathbb{E}}X_{0}=0$

with probability one.

Let $W=W(t),\ t\geq 0$, be a standard Brownian motion, and denote by
$\boldsymbol{\mu}_{0}$ and $\boldsymbol{\mu}_{\kappa}$ the measures on
${\mathcal{C}}_{(0)}$ generated by the processes $W$ and $W^{\kappa}$. The
following is the discrete-time analog of Theorem 2.2.

######  Theorem 3.1.

There exist positive constants $C_{a}$ and $c_{a}$ such that, for every
$\kappa\geq 1$,

$c_{a}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}\leq\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa},\boldsymbol{\mu}_{0})\leq
C_{a}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$ (3.5)

###### Proof.

The steps are the same as in the proof of Theorem 2.2.

Substituting (3.1) in (3.3) and changing the order of summation,

$W^{\kappa}(t)=S_{\kappa}(t)+\frac{X^{\kappa}(t)}{\sqrt{\kappa}},$ (3.6)

where $S_{\kappa}$ is from (3.4) and

$X^{\kappa}(t)=\left((a-a^{\lfloor\kappa
t\rfloor})\sum_{n=-\infty}^{0}a^{-n}\xi_{n}-\sum_{n=1}^{\lfloor\kappa
t\rfloor}a^{\lfloor\kappa t\rfloor-n}\xi_{n}\right).$

As a result, it is enough to establish (3.5) when $a=0$:

$c_{0}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}\leq\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa},\boldsymbol{\mu}_{0})\leq
C_{0}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$ (3.7)

Then, similar to the continuous time case, we see that

${\mathbb{E}}\|X^{\kappa}\|_{(0)}\leq\bar{C}_{a}\sqrt{\ln(1+\kappa)},$

with a suitable constant $\bar{C}_{a}$, and then (3.5) follows from (3.7) with
$C_{a}=C_{0}+\bar{C}_{a}$ and $c_{a}=c_{0}$.

To prove (3.7), we choose the random variables $\xi_{k}$ in (3.4) as the
increments of the Brownian motion $W$:

$\frac{\xi_{n}}{\sqrt{\kappa}}=W(n/\kappa)-W((n-1)/\kappa).$ (3.8)

Then, for $(n-1)/\kappa\leq t\leq n/\kappa$, the process $t\mapsto
S_{\kappa}-W$ is a Brownian bridge, and, for every function
$\varphi:{\mathcal{C}}_{(0)}\to{\mathbb{R}}$ satisfying (2.5),

$|{\mathbb{E}}\varphi(S_{\kappa})-{\mathbb{E}}\varphi(W)|\leq{\mathbb{E}}\|B^{\kappa}\|_{(0)},$
(3.9)

where $B^{\kappa}$ is a collection of independent Brownian bridges on
$[(n-1)/\kappa,n/\kappa]$, $n=1,2,\ldots$.

Direct computations show that, for $N=1,2,\ldots$,

$\displaystyle\sqrt{\kappa}\;{\mathbb{E}}\max_{0\leq t\leq
N/\kappa}|B^{\kappa}(t)|\leq 8\sqrt{{\ln(N+1)}},$ (3.10)
$\displaystyle\sqrt{\kappa}\;{\mathbb{E}}\max_{0\leq t\leq
N/\kappa}B^{\kappa}(t)\geq 0.3\sqrt{{\ln(N+1)}};$ (3.11)

the numbers $0.3$ and $8$ do not necessarily provide optimal bounds. Indeed,
let $B=B(t),\ t\in[0,1],$ be the standard Brownian bridge, and let

$\eta=\max_{0\leq t\leq 1}B(t).$

Then

$\sqrt{\kappa}\;{\mathbb{E}}\max_{0\leq t\leq
N/\kappa}|B^{\kappa}(t)|={\mathbb{E}}\max_{k=1,\ldots,N}|\eta_{k}|,$

where $\eta_{k},\ k=1,\ldots,N$, are iid copies of $\eta$. Also,

${\mathbb{E}}\max_{k=1,\ldots,N}\eta_{k}\leq{\mathbb{E}}\max_{k=1,\ldots,N}|\eta_{k}|\leq
2{\mathbb{E}}\max_{k=1,\ldots,N}\eta_{k}.$

To derive (3.10), we repeat the arguments from the proof of Lemma 10.1 in [19]
using (2.22) and conclude that
${\mathbb{E}}\max\limits_{k=1,\ldots,N}\eta_{k}\leq 4\sqrt{\ln(N+1)}$.
Similarly, for (3.11), we repeat the proof of Lemma 10.2 in [19].

Next, denote by $\bar{B}$ the process $B^{\kappa}$ corresponding to
$\kappa=1$, that is, the collection of independent standard Brownian bridges
on $[n-1,n]$, $n\geq 1$. Then we get the upper bound in (3.7), with

$C_{0}={\mathbb{E}}\left[\sup_{t>0}\frac{|\bar{B}(t)|}{\sqrt{\ln(2+t)}}\right]<14,$

by combining (3.9) and (3.10); the upper bound on $C_{0}$ is from (2.23),
because, by (1.11), $C_{0}\leq C_{X}$. The lower bound in (3.7), with

$c_{0}=\frac{1}{2}\inf_{\kappa\geq 1}\frac{{\mathbb{E}}\left[\max_{0\leq
t\leq\kappa}\bar{B}(t)\right]}{\sqrt{\ln(1+\kappa)}}>0.2,$

follows from (3.11) after the same arguments as in the proof of Theorem 2.2;
the lower bound on $c_{0}$ follows from (3.11). $\Box$

For $T>0$, let ${\mathcal{C}}(0,T)$ be the space of continuous functions on
$[0,T]$ with the sup norm, and denote by $\boldsymbol{\mu}_{0}^{T}$ and
$\boldsymbol{\mu}_{\kappa}^{T}$ the measures on ${\mathcal{C}}(0,T)$ generated
by the processes $W$ and $W^{\kappa}$. The discrete-time version of Theorem
2.4 is obvious. When $a=0$, and there is no continuous-time analog, we also
have the following result (cf. [2, Proposition 2.1]).

######  Proposition 3.2.

If $\boldsymbol{\mu}_{\kappa}^{T}$ is the measure on ${\mathcal{C}}(0,T)$
generated by the process $S_{\kappa}$ from (3.4), then

$\lim_{\kappa\to\infty}\sqrt{\frac{\kappa}{\ln\kappa}}\,\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa}^{T},\boldsymbol{\mu}_{0}^{T})={\sqrt{2}}.$
(3.12)

###### Proof.

Using the random variables $\eta_{k}$ from the proof of Theorem 3.1,

$\limsup_{\kappa\to\infty}\sqrt{\frac{\kappa}{\ln\kappa}}\;\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa}^{T},\boldsymbol{\mu}_{0}^{T})\leq\lim_{N\to\infty}\frac{{\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}|\eta_{k}|\right]}{\sqrt{\ln
N}},$

and

$\liminf_{\kappa\to\infty}\sqrt{\frac{\kappa}{\ln\kappa}}\;\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa}^{T},\boldsymbol{\mu}_{0}^{T})\geq\lim_{N\to\infty}\frac{{\mathbb{E}}\left[\max\limits_{k=1,\ldots,N}\eta_{k}\right]}{\sqrt{\ln
N}}.$

By (2.22) and [24, Theorem 1],

$\lim_{N\to\infty}\frac{\max\limits_{k=1,\ldots,N}\eta_{k}}{\sqrt{\ln
N}}=\lim_{N\to\infty}\frac{\max\limits_{k=1,\ldots,N}|\eta_{k}|}{\sqrt{\ln
N}}={\sqrt{2}}$

with probability 1; then uniform integrability [22, Theorem 2.1] implies
(3.12). $\Box$

##  4\. Applications

Let $W$ be a standard Brownian motion and let $W^{\kappa}$ be the process from
(2.4) or (3.3). Consider a continuous mapping
$\Psi:{\mathcal{C}}_{(0)}\to{\mathcal{C}}_{(0)}$. Denote by
$\boldsymbol{\mu}_{0,\psi}$ and $\boldsymbol{\mu}_{\kappa,\psi}$ the measures
on ${\mathcal{C}}_{(0)}$ generated by the processes $\Psi(W)$ and
$\Psi(W^{\kappa})$.

######  Proposition 4.1.

We have weak convergence
$\lim_{\kappa\to\infty}\boldsymbol{\mu}_{\kappa,\psi}=\boldsymbol{\mu}_{0,\psi}$.
Moreover, if there exists a number $C_{\psi}$ such that, for all
$f,g\in{\mathcal{C}}_{(0)}$,

$\|\Psi(f)-\Psi(g)\|_{(0)}\leq C_{\psi}\|f-g\|_{(0)},$ (4.1)

then

$\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa,\psi},\boldsymbol{\mu}_{0,\psi})\leq\bar{C}_{X}C_{\psi}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2},$
(4.2)

with $\bar{C}_{X}=C_{X}$ from (2.12) in continuous time and
$\bar{C}_{X}=C_{a}$ from (3.5) in discrete time.

###### Proof.

Weak convergence follows by the continuous mapping theorem (e.g. [4, Theorem
2.7]). To establish (4.2), we use either (2.6) or the upper bound in (3.5) and
note that if $\varphi:{\mathcal{C}}_{(0)}\to{\mathbb{R}}$ satisfies (2.5),
then

$|\varphi(\Psi(f))-\varphi(\Psi(g))|\leq C_{\psi}\|f-g\|_{(0)}.$

$\Box$

Example 1. Given ${\alpha}>0$, let $Y^{\kappa},Y$ be the solutions of

$Y^{\kappa}(t)=-{\alpha}\int_{0}^{t}Y^{\kappa}(s)\,ds+W^{\kappa}(t),\ \
Y(t)=-{\alpha}\int_{0}^{t}Y(s)\,ds+W(t),\ \ t\geq 0.$

Then $Y^{\kappa},Y\in{\mathcal{C}}_{(0)}$, and the corresponding measures
$\boldsymbol{\nu}_{\kappa},\boldsymbol{\nu}$ on ${\mathcal{C}}_{(0)}$ satisfy

$\mathrm{d}_{{w}}(\boldsymbol{\nu}_{\kappa},\boldsymbol{\nu})\leq
2\bar{C}_{X}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2},\ \kappa\geq 1;$

the constant $\bar{C}_{X}$ is from Proposition 4.1.

Indeed, by direct computation,

$Y^{\kappa}(t)=\Psi_{\alpha}(W^{\kappa})(t),\ \ Y(t)=\Psi_{\alpha}(W)(t),$

where

$\Psi_{\alpha}:f(t)\mapsto
f(t)-{\alpha}\int_{0}^{t}e^{-{\alpha}(t-s)}f(s)\,ds,\ \
f\in\mathcal{C}_{(0)},$ (4.3)

is a linear operator. To see that $\Psi_{\alpha}$ maps ${\mathcal{C}}_{(0)}$
to itself, note that, for every $t>T>0$,

$\displaystyle\frac{|\Psi_{\alpha}(f)(t)|}{1+t}$
$\displaystyle\leq\frac{|f(t)|}{1+t}+\frac{{\alpha}}{1+t}\int_{0}^{t}e^{-{\alpha}(t-s)}|f(s)|\,ds$
$\displaystyle\leq\frac{|f(t)|}{1+t}+\frac{{\alpha}}{1+t}\int_{0}^{T}e^{-{\alpha}(t-s)}|f(s)|\,ds$
$\displaystyle+\frac{{\alpha}e^{-{\alpha}t}}{1+t}\int_{T}^{t}(1+s)e^{{\alpha}s}\,\frac{|f(s)|}{1+s}\,ds.$

If $\lim_{t\to\infty}|f(t)|/(1+t)=0$, then, for every $\varepsilon>0$, we can
find $T$ so that $|f(s)|/(1+s)<\varepsilon$, $s>T$. As a result, keeping in
mind that

$\frac{{\alpha}}{1+t}\int_{T}^{t}(1+s)e^{{\alpha}s}\,ds\leq{\alpha}\int_{0}^{t}e^{{\alpha}s}\,ds\leq
e^{{\alpha}t},$

we compute

$\limsup_{t\to\infty}\frac{|\Psi_{\alpha}(f)(t)|}{1+t}\leq\varepsilon$

and conclude that $\lim_{t\to\infty}|\Psi_{\alpha}(f)(t)|/(1+t)=0$. Similarly,

$\|\Psi_{\alpha}(f)\|_{(0)}\leq\|f\|_{(0)}\left(1+\alpha\int_{0}^{+\infty}e^{-\alpha
t}\,dt\right)=2\|f\|_{(0)},$ (4.4)

so that (4.1) holds with $C_{\psi}=2$. $\Box$

The analog of Proposition 4.1 on a bounded interval is as follows. Let
$\Psi^{T}$ be a continuous mapping of ${\mathcal{C}}(0,T)$ to itself. Denote
by $\boldsymbol{\mu}_{0,\psi}^{T}$ and $\boldsymbol{\mu}_{\kappa,\psi}^{T}$
the measures on ${\mathcal{C}}(0,T)$ generated by the processes $\Psi^{T}(W)$
and $\Psi^{T}(W^{\kappa})$.

######  Proposition 4.2.

We have weak convergence
$\lim_{\kappa\to\infty}\boldsymbol{\mu}_{\kappa,\psi}^{T}=\boldsymbol{\mu}_{0,\psi}^{T}$.
Moreover, if there exists a number $C_{\psi}^{T}$ such that, for all
$f,g\in{\mathcal{C}}(0,T)$,

$\|\Psi^{T}(f)-\Psi^{T}(g)\|_{{\mathcal{C}}(0,T)}\leq
C_{\psi}^{T}\|f-g\|_{{\mathcal{C}}(0,T)},$ (4.5)

then

$\mathrm{d}_{{w}}(\boldsymbol{\mu}_{\kappa,\psi}^{T},\boldsymbol{\mu}_{0,\psi}^{T})\leq(1+T)\bar{C}_{X}C_{\psi}^{T}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2},$

with $\bar{C}_{X}=C_{X}$ from (2.12) in continuous time and
$\bar{C}_{X}=C_{a}$ from (3.5) in discrete time.

Example 2. Let the function $b=b(x),\ x\in{\mathbb{R}},$ satisfy

$|b(x)-b(y)|\leq K|x-y|,\ x,y\in{\mathbb{R}},$ (4.6)

and let $Y^{\kappa},Y$ be the solutions of

$Y^{\kappa}(t)=\int_{0}^{t}b\big{(}Y^{\kappa}(s)\big{)}\,ds+W^{\kappa}(t),\ \
Y(t)=\int_{0}^{t}b\big{(}Y(s)\big{)}\,ds+W(t),\ \ 0\leq t\leq T.$

If $\boldsymbol{\nu}_{\kappa}^{T},\boldsymbol{\nu}^{T}$ are the corresponding
measures on ${\mathcal{C}}(0,T)$, then

$\mathrm{d}_{{w}}(\boldsymbol{\nu}_{\kappa}^{T},\boldsymbol{\nu}^{T})\leq(1+T)\bar{C}_{X}e^{KT}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$

Indeed, for $f\in{\mathcal{C}}(0,T)$, define $\Psi^{T}(f)(t)=y(t)$ as the
solution of

$y(t)=\int_{0}^{t}b\big{(}y(s)\big{)}\,ds+f(t),\ 0\leq t\leq T.$

By direct computation (e.g. [10, Chapter 4, Lemma 1.1]), we have (4.5) with
$C_{\psi}^{T}=e^{KT}$.

$\Box$

Example 3. Let us combine Examples 1 and 2. Take a positive number ${\alpha}$
and a function $b=b(x)$ satisfying (4.6), and let $Y^{\kappa},Y$ be the
solutions of

$\displaystyle Y^{\kappa}(t)$
$\displaystyle=-{\alpha}\int_{0}^{t}Y^{\kappa}(s)\,ds+\int_{0}^{t}b\big{(}Y^{\kappa}(s)\big{)}\,ds+W^{\kappa}(t),$
$\displaystyle Y(t)$
$\displaystyle=-{\alpha}\int_{0}^{t}Y(s)\,ds+\int_{0}^{t}b\big{(}Y(s)\big{)}\,ds+W(t),\
\ t\geq 0.$

If ${\alpha}>K$, then $Y^{\kappa},Y\in{\mathcal{C}}_{(0)}$ and, for the
corresponding measures $\boldsymbol{\nu}_{\kappa},\boldsymbol{\nu}$,

$\mathrm{d}_{{w}}(\boldsymbol{\nu}_{\kappa},\boldsymbol{\nu})\leq\frac{2{\alpha}\bar{C}_{X}}{{\alpha}-K}\left(\frac{\ln(1+\kappa)}{\kappa}\right)^{1/2}.$
(4.7)

Indeed, for $f\in{\mathcal{C}}_{(0)}$, define $\Psi(f)(t)=y(t)$ as the
solution of

$y(t)=-{\alpha}\int_{0}^{t}y(s)\,ds+\int_{0}^{t}b\big{(}y(s)\big{)}\,ds+f(t),\
t\geq 0.$

Using variation of parameters formula and (4.3),

$\Psi(f)(t)=\int_{0}^{t}e^{-{\alpha}(t-s)}b\big{(}\Psi(f)(s)\big{)}\,ds+\Psi_{a}(f)(t).$

Then, similar to Example 1, we conclude that $\Psi$ maps ${\mathcal{C}}_{(0)}$
to itself. In particular, using (4.6) and (4.4),

$\Psi(f)(t)-\Psi(g)(t)=\int_{0}^{t}e^{-{\alpha}(t-s)}\Big{(}b\big{(}\Psi(f)(s)\big{)}-b\big{(}\Psi(g)(s)\big{)}\Big{)}\,ds+\Psi_{\alpha}(f-g)(t)$

so that

$\|\Psi(f)-\Psi(g)\|_{(0)}\leq
K\|\Psi(f)-\Psi(g)\|_{(0)}\int_{0}^{\infty}e^{-as}\,ds+2\|f-g\|_{(0)}.$

As a result, if $\alpha>K$, then

$\|\Psi(f)-\Psi(g)\|_{(0)}\leq\frac{2{\alpha}}{{\alpha}-K}\|f-g\|_{(0)},$

and (4.7) follows from (4.2). $\Box$

##  5\. Concluding Remarks

A proof of the functional Central Limit Theorem for processes of the type
(2.4) or (3.3) usually includes the following steps:

  1. (1)

A Gordin-type decomposition [12], when $W^{\kappa}$ is written as a sum of a
martingale and an a “small” correction;

  2. (2)

A coupling argument, when $W^{\kappa}$ is constructed on the same probability
space as $W$;

  3. (3)

A Skorokhod embedding for the martingale component of $W^{\kappa}$.

Each step leads to an approximation error; in particular, [17, 18] developed a
systematic procedure, now known as the KMT approximation, to minimize the
error due to the Skorokhod embedding. When the underlying processes are
Gaussian, some of the approximation errors are not present.

In continuous time, the first two steps are the equality (2.7). There is no
need for Skorokhod embedding because the martingale component is the Brownian
motion. In discrete time, the first step is the equality (3.6), whereas (3.8)
represents coupling and the Skorokhod embedding. For convergence in the space
of continuous functions, the $\sqrt{\ln\kappa}$ correction to the classical
rate $1/\sqrt{\kappa}$ comes from the growth of the maximum of iid standard
Gaussian random variables.

Keeping in mind that rate of convergence in the functional CLT can depend both
on the underlying functional space and on the distance between the measures on
that space, the rate $1/\sqrt{\kappa}$ is possible to achieve. For example, by
considering $W$ and $S_{\kappa}$ [from (3.4)] as processes in $L_{1}(0,T)$, as
opposed to ${\mathcal{C}}(0,T)$, direct computations [2, Proposition 2.1]
yield

${\mathbb{E}}\int_{0}^{1}|S_{\kappa}(t)-W(t)|\,dt=\frac{1}{\sqrt{\kappa}}\int_{0}^{1}{\mathbb{E}}|B(t)|\,dt=\sqrt{\frac{2}{\pi\,\kappa}}\int_{0}^{1}\sqrt{t(1-t)}\,dt=\sqrt{\frac{\pi}{32\,\kappa}},$

that is, the Wasserstein-1 distance between $S_{\kappa}$ and $W$ in
$L_{1}(0,T)$ is of order $1/\sqrt{\kappa}$; see also [3, Remark 1].

Given the variety of function spaces that can support $W$ and $W^{\kappa}$, as
well as the variety of ways to measure the distance between the corresponding
probability distributions [11], identifying all situations with a sharp
$1/\sqrt{\kappa}$ bound becomes an interesting challenge. For
$\mathrm{d}_{{w}}(W,W^{\kappa})$ in the space of continuous functions with the
sup norm, there is strong evidence that convergence cannot be faster than
$\sqrt{\ln\kappa/\kappa}$: the results of this paper demonstrate it in the
Gaussian case, and, by [15, Corollary 4.4], the simple symmetric random walk
cannot beat this rate either.

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# A NOTE ON TWO CLASSES OF HYPERIDEALS

M. Anbarloei  Department of Mathematics, Faculty of Sciences, Imam Khomeini
International University, Qazvin, Iran<EMAIL_ADDRESS>
###### Abstract.

Let $R$ be a commutative multiplicative hyperring with 1. In this paper, we
define the concepts of r-hyperideal and n-hyperideal of the hyperring $R$
which are two new classes of hyperideals. Several properties of them are
provided. A hyperideal $I$ of a multiplicative hyperring $R$ is called an
r-hyperideal of $R$, if for all $x,y\in R$, $xoy\subseteq I$ and
$ann(x)=\\{0\\}$, then $y\in I$. Also, a hyperideal $I$ of a multiplicative
hyperring $R$ is called an n-hyperideal of $R$, if for all $x,y\in R$,
$xoy\subseteq I$ and $x\notin r(0)$, then $y\in I$.

###### Key words and phrases:

Prime hyperideal, r-hyperideal, n-hyperideal, Hyperring.

###### 2000 Mathematics Subject Classification:

20N20

##  1\. Introduction

Throughout this paper our ring $R$ is a commutative multiplicative hyperring
with 1. Hyperstructures, as a natural generalization of ordinary algebraic
structures, were first initiated by Marty in 1934 [10], when he defined the
hypergroups and began to investigate their properties with applications to
groups, algebraic functions and rational fractions. Later on, many researchers
have observed that the theory of hyperstructures also have many applications
in both applied and pure sciences which a comprehensive review of this theory
can be found in [6] and [13]. In [13], Corsini and Leoreanu-Fotea have
collected numerous applications of algebraic hyperstructures, especially those
from the last fifteen years to the following subjects: geometry, lattices,
binary relations, rough sets and fuzzy sets, cryptography, automata, median
algebras, codes, artificial intelligence, relation algebras, and
probabilities.

Motivated from the concept of the prime hyperideals, in this paper, we
introduce and study two new classes of hyperideals called r-hyperideals and
n-hyperideals. Let $R$ be a commutative multiplicative hyperring. Several
properties of them are provided. A hyperideal $I$ of a multiplicative
hyperring $R$ is called an r-hyperideal of $R$, if for all $x,y\in R$,
$xoy\subseteq I$ and $ann(x)=\\{0\\}$, then $y\in I$. Also, We define a proper
hyperideal $I$ of $R$ as an n-hyperideal if for all $x,y\in R$, $xoy\subseteq
I$ and $x\notin r(0)$, then $y\in I$.
The paper is orgnized as follows. In Section 2, we have given some basic
definitions and results of multiplicative hyperrings which we need to develop
our paper. In Section 3, we introduce the concept of r-hyperideals and give
some basic properties of them. For example we show (Theorem 3.6) that if $R$
is a reduced hyperring and $P$ is a minimal hyperideal of $R$ such that $s\in
R$ is an idempotent element, then $P+ann(s)$ is an r-hyperideal. Section 4
continues the study of the properties of r-hyperideals. In particular, we
discuss the relations between prime hyperideals and r-hyperideals. In Section
5, we introduce the notion of n-hyperideals and give some basic properties of
them. For example, we show (Theorem 5.5) that if $I$ is a proper hyperideal of
$R$, then the following statements are equivalent:
(1) $I$ is an n-hyperideal of $R$.
(2) $I=(I:a)$ for every $a\notin r(0)$.
(3) $I_{1}oI_{2}\subseteq I$ such that $I_{1}\cap(R-r(0))\neq\varnothing$, for
hyperideals $I_{1},I_{2}$ of $R$, implies $I_{2}\subseteq I$.
It is shown (Theorem 5.11) that if $S$ is an n-multiplicatively closed subset
of $R$ and $K$ is an hyperideal of $R$ disjoint from $S$, then there exists a
hyperideal $I$ which is maximal in the set of all hyperideals of $R$ disjoint
from $S$, containing $K$. Any such hyperideal $I$ is an n-hyperideal of $R$.
In the final section, we investigate the stability of n-hyperideals in various
hyperring-theoric constructions.

##  2\. Preliminaries

Recall first the basic terms and definitions from the hyperring theory. A
hyperoperation on a non-empty set $G$ is a map $"o":G\times G\longrightarrow
P^{*}(G)$, where $P^{*}(G)$ is the set of all the nonempty subsets of $G$. An
algebraic system $(G,o)$ is called a hypergroupoid. A hypergroupoid $(G,o)$ is
called a hypergroup if it satisfies the following:

(1) $ao(boc)=(aob)oc$, for all $a,b,c\in G$.

(2) $aoG=Goa=G$, for all $a\in G$.

A hypergroupoid with the associative hyperoperation is called a
semihypergroup.
The notion of multiplicative hyperring as one of the important classes of
hyperrings was introduced by Rota [12].

A non-empty set $R$ with a operation + and a hyperoperation $o$ is called a
multiplicative hyperring if it satisfies the following:

(1)$(R,+)$ is an abelian group;

(2)$(R,o)$ is a semihypergroup;

(3) for all $a,b,c\in R$, we have $ao(b+c)\subseteq aob+aoc$ and
$(b+c)oa\subseteq boa+coa$;

(4) for all $a,b\in R$, we have $ao(-b)=(-a)ob=-(aob)$.

For any two nonempty subsets $A$ and $B$ of $R$ and $x\in R$, we define

$AoB=\bigcup_{a\in A,\ b\in B}aob,\ \ \ \ Aox=Ao\\{x\\}.$

Let R be a commutative multiplicative hyperring. Then

(i) An element $e\in R$ is said to be identity if $a\in aoe$ for all $a\in R$.

(ii) An element $e\in R$ is said to be scalar identity if $a=aoe$ for all
$a\in R$.

######  Definition 2.1.

A non-empty subset $I$ of a multiplicative hyperring $R$ is a hyperideal

(1) If $a,b\in I$, then $a-b\in I$;

(2) If $x\in I$ and $r\in R$, then $rox\subseteq I$.

######  Definition 2.2.

[9] Let $(R_{1},+_{1},o_{1})$ and $(R_{2},+_{2},o_{2})$ be two multiplicative
hyperrings. A mapping from $R_{1}$ into $R_{2}$ is said to be a good
homomorphism if for all $x,y\in R_{1}$, $\phi(x+_{1}y)=\phi(x)+_{2}\phi(y)$
and $\phi(xo_{1}y)=\phi(x)o_{2}\phi(y)$.

######  Definition 2.3.

[8] A nonzero proper hyperideal $P$ of $R$ is called a prime hyperideal if
$xoy\subseteq P$ for $x,y\in R$ implies that $x\in P$ or $y\in P$.

The intersection of all prime hyperideals of $R$ containing $I$ is called the
prime radical of $I$, being denoted by $r(I)$. Therefore, $r(0)$ is the prime
radical of the zero hyperideal in the hyperring $R$. If the multiplicative
hyperring $R$ does not have any prime hyperideal containing $I$, we define
$r(I)=R$.

Let C be the class of all finite products of elements of $R$, i.e., ${\bf
C}=\\{r_{1}or_{2}o...or_{n}\ :\ r_{i}\in R,n\in\mathbb{N}\\}\subseteq
P^{\ast}(R)$. A hyperideal $I$ of $R$ is said to be a C-hyperideal of $R$ if,
for any $A\in{\bf C},A\cap I\neq\varnothing$ implies $A\subseteq I$. Let I be
a hyperideal of $R$. Let $D=\\{r\in R:r^{n}\subseteq I\ for\ some\
n\in\mathbb{N}\\}$. Then $D\subseteq r(I)$, the prime radical of $I$. The
equality holds when $I$ is a C-hyperideal of $R$([8], proposition 3.2). In
this paper, we assume that all hyperideals are C-hyperideals.

######  Definition 2.4.

[8] A nonzero proper hyperideal $Q$ of $R$ is called a primary hyperideal if
$xoy\subseteq Q$ for $x,y\in R$ implies that $x\in Q$ or $y\in r(Q)$.

Since $r(Q)=P$ is a prime hyperideal of $R$ by Propodition 3.6 in [8], $Q$ is
referred to as a P-primary hyperideal of $R$.

######  Definition 2.5.

[3] Let $R$ be a multiplicative hyperring. The $n\times n$ matrix with entries
in $R$ is called a hypermatix and the set of all hypermatices of $R$ is
denoted by $M_{n}(R)$.
Now, for all $A=(A_{ij})_{n\times n},B=(B_{ij})_{n\times n}\in
P^{*}(M_{n}(R));A\subseteq B$ means $A_{ij}\subseteq B_{ij}$.

######  Definition 2.6.

[3] Let $R$ be a commutative multiplicative hyperring and $e$ be an identity
of $R$. An element $x\in R$ is called invertible, if there exists $y\in R$,
such that $e\in xoy$. So, $R$ is called an invertible if every element of $R$
is invertible.

######  Definition 2.7.

[1] An element $x\in R$ is said to be a zero divizor, if there exists $0\neq
y\in R$ such that $\\{0\\}=xoy$. The set of all zero divizor elements in $R$
is denoted by $Z(R)$.

######  Definition 2.8.

A hyperring $R$ is called an integral hyperdomain, if for all $x,y\in R$,
$0\in xoy$ implies that $x=0$ or $y=0$.

######  Definition 2.9.

A hyperring $R$ is said to be a reduced hyperring if it has no nilpotent
elements. That is, if $x^{n}=0$ for $x\in R$ and a natural number $n$, then
$x=0$.

######  Definition 2.10.

[2] Let $R$ be a multiplicative hyperring. An element $r\in R$ is called
regular if there exists $x\in R$ such that $r\in r^{2}ox$. $R$ is called
regular multiplicative hyperring, if all the elements in $R$ are regular
elements. The set of all regular elements in $R$ is denoted by $V(R)$.

######  Definition 2.11.

Let $I,J$ be two hyperideals of $R$ and $x\in R$. Then define:

$(I:J)=\\{r\in R:roJ\subseteq I\\}$ $ann(x)=\\{y\in R:xoy=\\{0\\}\\}$

##  3\. $r$-hyperideals

In this section, we introduce the notion of r-hyperideals in a commutative
multiplicative hyperring and we investigate many properties of them.

######  Definition 3.1.

A hyperideal $I$ of a multiplicative hyperring $R$ is called an r-hyperideal
of $R$, if for all $x,y\in R$, $xoy\subseteq I$ and $ann(x)=\\{0\\}$, then
$y\in I$.

######  Lemma 3.1.

Let $I$ be a hyperideal of $R$. Then
1) $I$ is an r -hyperideal if and only if $I_{1}oI_{2}\subseteq I$ for some
hyperideals $I_{1},I_{2}$ of $R$ such that $I_{1}\cap V(R)\neq\varnothing$
implies that $I_{2}\subseteq I$ .
2) If either $IoI_{1}=IoI_{2}$ or $I\cap I_{1}=I\cap I_{2}$ for some
r-hyperideals $I_{1},I_{2}$ of $R$ such that $I\cap V(R)\neq\varnothing$, then
$I_{1}=I_{2}$ .
3) If $IoJ$ is an r -hyperideal of $R$ for some hyperideal $J$ of $R$ such
that $J\cap V(R)\neq\varnothing$, then $I=IoJ$ and so $I$ is an r -hyperideal.

###### Proof.

It is clear. ∎

If $S=V(R)$, then we denote the hyperring $S^{-1}R$ by $T(R)$. Let
$\phi:R\longrightarrow T(R)$ be the natural homomorphism. Put
$\phi^{-1}(J)=J^{c}$, for each hyperideal $J$ in $T(R)$. For more details
about notion of contraction, see [3].

######  Theorem 3.1.

Let R be a commutative multiplicative hyperring and let $I$ be a hyperideal of
$R$. Then the following are equivalent:
1) $I$ is an r-hyperideal.
2) $<a>\cap I=aoI$ , for all $a\in V(R)$.
3) $I=(I:a)$, for all $a\in V(R)\backslash I$.
4) $I=J^{c}$, for some hyperideal $J$ in $T(R)$.

###### Proof.

$(1)\Longrightarrow(2)$ Let $I$ is an r-hyperideal. It is clear that
$aoI\subseteq<a>\cap I$. Let $x\in<a>\cap I$. There exists some $s\in R$ such
that $x\in aos$. Since $a\in V(R)$, we have for some $y\in R$, $a\in a^{2}oy$.
Hence $x\in aos\subseteq a^{2}oyos=ao(aosoy)\subseteq aoI$, since $I\cap
V(R)=\varnothing$.
$(2)\Longrightarrow(3)$ It is clear that $I\subseteq(I:a)$. Now let
$x\in(I:a)$. Then $aox\subseteq I$ and so $aox\subseteq<a>\cap I$. By part
(2), we have $aox\subseteq aoI$. Thus $x\in I$.
$(3)\Longrightarrow(4)$ and $(4)\Longrightarrow(1)$ follow from Theorem 4.12
in [3]. ∎

######  Theorem 3.2.

The intersection of a nonempty set of r-hyperideals in $R$, is an
r-hyperideal.

###### Proof.

Let $\\{I_{i}\\}_{i\in\Omega}$ be a nonempty set of r-hyperideals in $R$ and
$I=\bigcap_{i\in\Omega}I_{i}$. Clearly $I$ is a hyperideal. Now, assume that
for $x,y\in R$, $xoy\subseteq I$ such that $ann(x)=\\{0\\}$. Then we have
$xoy\subseteq I_{i}$ for every $i\in\Omega$. Since $I_{i}$ is an r-hyperideal
for every $i\in\Omega$ and $ann(x)=\\{0\\}$, we obtain $y\in I_{i}$. It means
$y\in I$. ∎

######  Theorem 3.3.

If $I$ is an r-hyperideal, then $I\subseteq Z(R)$.

###### Proof.

Let $I\nsubseteq Z(R)$. Then there exists $x\in I$ such that $x\notin Z(R)$.
It means that $ann(x)=\\{0\\}$. Since $I$ is an r-hyperideal and $xo1\subseteq
I$ with $ann(x)=\\{0\\}$, we conclude that $1\in I$ which is a contradiction.
Thus $I\subseteq Z(R)$. ∎

######  Lemma 3.2.

For $0\neq x\in R$, $ann(x)$ is an r-hyperideal of $R$.

###### Proof.

Let $aob\subseteq ann(x)$ and $ann(a)=\\{0\\}$ for some $0\neq x\in R$. Thus
$aobox=\\{0\\}$. Since $ann(a)=\\{0\\}$, we have $box=\\{0\\}$. It means $b\in
ann(x)$. Hence $ann(x)$ is an r-hyperideal of $R$. ∎

######  Theorem 3.4.

Let $R$ is a commutative multiplicative hyperring with 1. Then the following
are equivalent:
1) $R$ is an integral hyperdomain.
2) $\\{0\\}$ is the only r-hyperideal of $R$.
3) For every $x,y\in R$, $ann(xoy)=ann(x)\cup ann(y)$.

###### Proof.

$(1)\Longrightarrow(2)$ Assume that $I$ is a nonzero proper r-hyperideal of
$R$. Then there exists $0\neq x\in I$. Since $R$ is an integral hyperdomain,
then $ann(x)=\\{0\\}$. Since $xo1\subseteq I$ and $ann(x)=\\{0\\}$, we
conclude that $1\in I$ which is a contradiction. Hence $\\{0\\}$ is the only
r-hyperideal of $R$.
$(2)\Longrightarrow(3)$ follows from Lemma 3.2.
$(3)\Longrightarrow(1)$ Let $xoy=\\{0\\}$ , for some $x,y\in R$. By the
hypothesis, we have $1\in ann(x)\cup ann(y)$. Hence $x=0$ or $y=0$. Thus $R$
is an integral hyperdomain. ∎

######  Theorem 3.5.

Let $x+y=1$ for some $x,y\in R$. Then $ann(x)+ann(y)$ is an r -hyperideal.

###### Proof.

Let $aob\subseteq ann(x)+ann(y)$ and $ann(a)=\\{0\\}$. It is easy to see that
$aoboxoy=\\{0\\}$. Therefore we have $xoyob=\\{0\\}$, Since $ann(a)=\\{0\\}$.
Hence $box\subseteq ann(y)$ and $boy\subseteq Ann(x)$. Therefore,
$b=bo1=bo(x+y)\subseteq box+boy\subseteq ann(y)+ann(x)$. Consequently,
$ann(x)+ann(y)$ is an r-hyperideal of $R$. ∎

######  Theorem 3.6.

Let $R$ be a reduced hyperring and $P$ be a minimal hyperideal of $R$. Suppose
that $s\in R$ is an idempotent element. Then $P+ann(s)$ is an r-hyperideal.

###### Proof.

Let $aob\subseteq P+ann(s)$ and $ann(a)=\\{0\\}$. Therefore for each element
$t\in aob$, there exists $\alpha\in P$ and $\beta\in ann(s)$ such that
$t=\alpha+\beta$. It is clear that there exists $x\notin P$ with
$xos=\\{0\\}$. Therefore we have $xosoaob=\\{0\\}$. Since $ann(a)=\\{0\\}$, we
get $xosob=\\{0\\}\subseteq P$. Hence $sob\subseteq P$. Since $sob+1ob-
sob\subseteq P+ann(s)$ and $b=1ob$, we have $b\in P+ann(s)$. Consequently,
$P+ann(s)$ is an r-hyperideal of $R$. ∎

##  4\. the relations between $r$-hyperideals and prime hyperideals

In this section, we continue the study of basic properties of r-hyperideals
begun in the previous section.

######  Theorem 4.1.

Every maximal r-hyperideal of $R$ is a prime hyperideal.

###### Proof.

Let $M$ be a maximal r-hyperideal of $R$. Let $aob\subseteq M$ and $a\notin
M$. It is easy to see that $(M:a)$ is an r-hyperideal of $R$. Moreover, we
have $M\subseteq(M:a)$ and $b\in(M:a)$. Since $M$ is a maximal r-hyperideal of
$R$, we get $(M:a)=M$. Thus $b\in M$. ∎

######  Theorem 4.2.

Let $P$ be a prime hyperideal of $R$. $P$ is an r-hyperideal of $R$ if and
only if it consists entirely of zerodivisors .

###### Proof.

$\Longrightarrow$ is clear by Theorem 3.3
$\Longleftarrow$ Let $P$ consist entirely of zerodivisors. Let $xoy\subseteq
P$ for some $x,y\in R$ and $ann(x)=\\{0\\}$. Since $P$ is a prime hyperideal
of $R$ and $P$ consists entirely of zerodivisors, we get $y\in P$.
Consequently, $P$ is an r-hyperideal of $R$. ∎

######  Theorem 4.3.

Let $R$ be a commutative multiplicative hyperring with 1 and let
$P_{1},\cdot\cdot\cdot,P_{n}$ be incomparable prime hyperideals of $R$. If
$P=\bigcap_{i=1}^{n}P_{i}$ is an r-hyperideal of $R$, then $P_{i}$ is an
r-hyperideal for all $1\leq i\leq n$.

###### Proof.

Let $xoy\subseteq P_{t}$ for some $1\leq t\leq n$ and $ann(x)=\\{0\\}$. We
must show $y\in P_{t}$. Suppose that $z\in(\Pi_{k\neq t}P_{k})\backslash
P_{t}$. Then we have $xoyoz\subseteq P$. By the assumption, we have
$yoz\subseteq P$. It means $yoz\subseteq P_{t}$. Since $P_{t}$ is a prime
hyperideal of $R$ and $z\notin P_{t}$, we conclude that $y\in P_{t}$. ∎

Recall that a nonzero hyperideal $I$ of $R$ is called essential if $I\cap
J\neq\\{0\\}$ for every nonzero hyperideal $J$ of $R$.

######  Theorem 4.4.

Let $R$ be a reduced hyperring. If an r-hyperideal $I$ of $R$ is not
essential, then there exists a minimal prime hyperideal $P$ with $I\subseteq
P$ such that $P$ is a maximal r-hyperideal.

###### Proof.

Let $I$ be an r-hyperideal of $R$ that is not essential. Then there exists a
nonzero hyperideal $J$ of $R$ such that $I\cap J\neq\\{0\\}$. Thus there
exists a minimal prime hyperideal $P$ of $R$ such that $J\nsubseteq P$, since
$R$ is a reduced hyperring. By Zorn,s Lemma, we infer that there exists a
maximal r-hyperideal $M$ with $I\subseteq M$ such that $J\cap M=\\{0\\}$. It
means $MoJ=\\{0\\}$. Let $a\in J\backslash P$. Thus $Moa=\\{0\\}\subseteq P$.
Since $P$ is a prime hyperideal, then $M\subseteq P$. By Theorem 4.1, $M$ is a
prime hyperideal. Thus $M=P$ and $I\subseteq P$. ∎

######  Theorem 4.5.

Let $I,I_{1},...,I_{n}$ be hyperideals of $R$. Let $I\subseteq
I_{1}\cup...\cup I_{n}$ such that no $I_{i}$ can be removed from the union. If
one of the $I_{i}^{,}$s, say $I_{1}$, is an r-hyperideal and the others have
regular elements, then $I\subseteq I_{1}$.

###### Proof.

Since no $I_{i}$ can be removed from the union $I_{1}\cup...\cup I_{n}$, then
there exist $x\in I$ but $x\notin I_{2}\cup...\cup I_{n}$. Hence $x\in I_{1}$.
Let $y\in I\cap I_{2}\cap...\cap I_{n}$. Therefore $x+y\notin I_{2}\cup...\cup
I_{n}$. It is clear that $x+y\in I\subseteq I_{1}\cup...\cup I_{n}$. This
implies that $y\in I_{1}$. It means $I\cap I_{2}\cap...\cap I_{n}\subseteq
I_{1}$. Since $I_{2}o...oI_{n}\subseteq I_{2}\cap...\cap I_{n}$, then we have
$IoI_{2}o...oI_{n}\subseteq I_{1}$. By Lemma 5.3 (1), we have $I\subseteq
I_{1}$, since $I_{2}\cap...\cap I_{n}\cap V(R)\neq\varnothing$. ∎

######  Corollary 4.1 (Prime Avoidance Theorem).

Let $I,P_{1},...,P_{n}$ be hyperideals of $R$. Let $I\subseteq
P_{1}\cup...\cup P_{n}$ such that no $P_{i}$ can be removed from the union. If
one of the $P_{i}^{,}$s, say $P_{1}$, is a minimal prime hyperideal of $R$ and
the others have regular elements, then $I\subseteq P_{1}$.

A nonempty subset $S$ of $R$ is called an r-multiplicatively closed subset of
$R$ if $S$ contains at least a regular element $u\neq 1$, $0\notin S$, $1\in
S$ and for all regular elements $r\in S$ and all $a\in S$, $roa\subseteq S$ .

######  Remark 1.

Let $S$ be an r-multiplicatively closed subset of $R$ and $T$ be
multiplicatively closed subset of $R$ containing a regular element. Put
$D=S\cup T\cup\\{u\ |u\in sot,\ s\in S,t\in T\\}$. Then $D$ is an
r-multiplicatively closed subset of $R$.

######  Corollary 4.2.

$I$ is an r-hyperideal of $R$ if and only if $R-I$ is an r-multiplicatively
closed subset of $R$.

######  Theorem 4.6.

Let $S$ be an r-multiplicatively closed subset of $R$ and $K$ be an hyperideal
of $R$ disjoint from $S$. Then there exists a hyperideal $I$ which is maximal
in the set of all hyperideals of $R$ disjoint from $S$, containing $K$. Any
such hyperideal $I$ is an r-hyperideal of $R$.

###### Proof.

Let $\Delta$ be the set of all hyperideals of $R$ disjoint from $S$,
containing $K$. Then $\Delta\neq\varnothing$, since $K\in\Delta$ .So $\Delta$
is a partially ordered set with respect to set inclusion relation. By Zorn,s
lemma, there exists a hyperideal $I$ which is maximal in $\Delta$. Let
$xoy\subseteq I$ for some $x,y\in R$ and $ann(x)=\\{0\\}$. Let $y\notin I$.
Since $y\in(I:x)$ then $(I:x)$ properly contains $I$. Now let $u\in(I:x)\cap
S$. Hence $uox\subseteq I$. By Remark 1, we have $x\in V(R)\subseteq S$.
Therefore $uox\subseteq S$ which means $uox\subseteq I\cap S$. This is
conradictory to the fact that $I\cap S=\varnothing$. Hence $(I:x)\cap
S=\varnothing$. Consequently, $y\in I$. Thus $I$ is an r-hyperideal of $R$. ∎

##  5\. $n$-hyperideals

In this section, we introduce the notion of n-hyperideals in a commutative
multiplicative hyperring and we investigate many properties of them.

######  Definition 5.1.

A hyperideal $I$ of a multiplicative hyperring $R$ is called an n-hyperideal
of $R$, if for all $x,y\in R$, $xoy\subseteq I$ and $x\notin r(0)$, then $y\in
I$.

######  Theorem 5.1.

Every n-hyperideal of $R$ is an r-hyperideal.

###### Proof.

Let $I$ be an n-hyperideal of $R$. Assume that $xoy\subseteq I$ with
$ann(x)=\\{0\\}$ for some $x,y\in R$. Thus $x\notin r(0)$. Since $I$ is an
n-hyperideal of $R$, we get $y\in I$ and therefore $I$ is an r-hyperideal. ∎

######  Theorem 5.2.

Let $<0>$ be a primary hyperideal of $R$. Then the n-hyperideals and
r-hyperideals are equivalent.

###### Proof.

It is clear. ∎

######  Theorem 5.3.

Let $I$ be an n-hyperideal of $R$. Then $I\subseteq r(0)$.

###### Proof.

Let $I\nsubseteq r(0)$. It means that there exists an $x\in I$ but $x\notin
r(0)$. Since $I$ is a C-hyperideal and $x\in xo1$, then we have $xo1\subseteq
I$. Since $I$ is an n-hyperideal of $R$, we have $1\in I$ which means $I=R$.
This is a contradiction. Thus $I\subseteq r(0)$. ∎

######  Theorem 5.4.

The intersection of a nonempty set of n-hyperideals in $R$, is an
n-hyperideal.

###### Proof.

Let $\\{I_{i}\\}_{i\in\Omega}$ be a nonempty set of n-hyperideals in $R$ and
$I=\bigcap_{i\in\Omega}I_{i}$. Assume that for $x,y\in R$, $xoy\subseteq I$
such that $x\notin r(0)$. Then we have $xoy\subseteq I_{i}$ for every
$i\in\Omega$. Since $I_{i}$ is an n-hyperideal for every $i\in\Omega$ and
$x\notin r(0)$, we obtain $y\in I_{i}$. It means $y\in I$. ∎

######  Theorem 5.5.

Let $I$ be a proper hyperideal of $R$. Then the following statements are
equivalent:
(1) $I$ is an n-hyperideal of $R$.
(2) $I=(I:a)$ for every $a\notin r(0)$.
(3) $I_{1}oI_{2}\subseteq I$ such that $I_{1}\cap(R-r(0))\neq\varnothing$, for
hyperideals $I_{1},I_{2}$ of $R$, implies $I_{2}\subseteq I$.

###### Proof.

$(1)\Longrightarrow(2)$ Since $I\subseteq(I:x)$, we must show $(I:x)\subseteq
I$. Let $y\in(I:x)$ and $x\notin r(0)$. Therefore $xoy\subseteq I$. Since $I$
is an n-hyperideal of $R$, we get $y\in I$. Consequently, $I=(I:x)$.
$(2)\Longrightarrow(3)$ Let $I_{1}oI_{2}\subseteq I$ such that
$I_{1}\cap(R-r(0))\neq\varnothing$. Thus there exists an $x\in I_{1}$ but
$x\notin r(0)$. Hence $xoI_{2}\subseteq I$. It means that
$I_{2}\subseteq(I:x)=I$.
$(3)\Longrightarrow(1)$ Assume that $xoy\subseteq I$ but $x\notin r(0)$ for
some $x,y\in R$. If we consider $I_{1}=<x>$ and $I_{2}=<y>$, then we are done.
∎

######  Corollary 5.1.

Let $L$ be a hyperideal of R such that $L\cap(R-r(0))\neq\varnothing$. If
$I,J$ are n-hyperideals of $R$ such that $IoL=JoL$, then $I=J$.

######  Theorem 5.6.

Let $I$ be a prime hyperideal of $R$. $I$ is an n-hyperideal of $R$ if and
only if $I=r(0)$.

###### Proof.

$\Longrightarrow$ Let $I$ be a prime hyperideal of $R$ such that $I$ is
n-hyperideal. The inclusion $r(0)\subseteq I$ always holds. By Theorem 5.3, we
obtain $I\subseteq r(0)$. Consequently, $I=r(0)$.
$\Longleftarrow$ Let $xoy\subseteq I$ such that $x\notin r(0)$ for some
$x,y\in R$. Since $I$ is a prime ideal, we conclude that $y\in I$. It means
$I$ is an n-hyperideal of $R$. ∎

######  Corollary 5.2.

$r(0)$ is a prime hyperideal of $R$ if and only if it is an n-hyperideal of
$R$.

###### Proof.

$\Longrightarrow$ follows from Theorem 5.6.
$\Longleftarrow$ Assume that $xoy\subseteq r(0)$ such that $x\notin r(0)$. We
get $y\in r(0)$, because $r(0)$ is an n-hyperideal of $R$. It means $I$ is a
prime hyperideal. ∎

######  Theorem 5.7.

Let $I$ be an n-hyperideal of $R$ and $T$ a nonempty subset of $R$ such that
$T\nsubseteq I$, then $(I:T)$ is an n-hyperideal of $R$.

###### Proof.

Let $xoy\subseteq(I:T)$ but $x\notin r(0)$. It means that $xoyot\subseteq I$
for every $t\in T$. Therefore for $a\in yot$, $xoa\subseteq xoyot\subseteq I$.
Since $I$ is an n-hyperideal of $R$, we get $a\in I$. Since $I$ is a
C-hyperideal and $yot\cap I\neq\varnothing$, then $yot\subseteq I$. So
$y\in(I:T)$. ∎

######  Theorem 5.8.

Let $I$ be a maximal n-hyperideal of $R$. Then $I=r(0)$.

###### Proof.

It is sufficient to show $I$ is a prime hyperideal of $R$. Then by Theorem
5.6, we are done. Let $xoy\subseteq I$ for some $x,y\in R$ such that $x\notin
I$. By Theorem 5.5, $(I:x)$ is an n-hyperideal of $R$. Thus by maximality of
$I$ we have $y\in(I:x)=I$. Consequently, $I$ is a prime hyperideal. ∎

######  Theorem 5.9.

$r(0)$ is a prime hyperideal of $R$ if and only if there exists an
n-hyperideal of $R$.

###### Proof.

$\Longrightarrow$ follows from Corollary 5.2.
$\Longleftarrow$ Let $\Delta$ be the set of all n-hyperideals of $R$ and $I$
an n-hyperideal of $R$. Then $\Delta\neq\varnothing$, since $I\in\Delta$. So
$\Delta$ is a partially ordered set with respect to set inclusion relation.
Now we take the chain $J_{1}\subseteq J_{2}\subseteq...\subseteq
J_{n}\subseteq...$ in $\Delta$. Let $J=\bigcup_{i=1}^{\infty}J_{i}$ and let
$xoy\subseteq J$ for some $x,y\in R$ such that $x\notin r(0)$. It means that
there exists $t\in\mathbb{N}$ such that $xoy\subseteq J_{t}$. Since $J_{t}$ is
an n-hyperideal, we get $y\in J_{t}\subseteq J$. Hence $J$ is an upper bound
of the chain, then $\Delta$ has a maximal element $L$, by Zorn,s Lemma. Thus
$L=r(0)$, by Theorem 5.8 and so it is prime hyperideal of $R$. ∎

A nonempty subset $S$ of $R$ with $R-r(0)\subseteq S$ is called an
n-multiplicatively closed subset of $R$ if for all $a\in R-r(0)$ and all $b\in
S$, $aob\subseteq S$ .

######  Theorem 5.10.

Let $I$ be a hyperideal of the hyperring $R$. Then $I$ is an n-hyperideal of
$R$ if and only if $R-I$ is an n-multiplicatively closed subset of $R$.

###### Proof.

$\Longrightarrow$ By Theorem 5.3, $I\subseteq r(0)$. Thus $R-r(0)\subseteq
R-I$. Let $a\in R-r(0)$ and $b\in R-I$. If $aob\subseteq I$, then we get $b\in
I$ which is a contradiction. Thus $aob\subseteq R-I$. It means that $R-I$ is
an n-multiplicatively closed subset of $R$.
$\Longleftarrow$ Let $xoy\subseteq I$ for some $x,y\in R$ such that $x\notin
r(0)$. If $y\notin I$, then $xoy\in R-I$. This is a contradiction. Hence $y\in
I$ and then $I$ is an n-hyperideal of $R$. ∎

######  Theorem 5.11.

Let $S$ be an n-multiplicatively closed subset of $R$ and $K$ be an hyperideal
of $R$ disjoint from $S$. Then there exists a hyperideal $I$ which is maximal
in the set of all hyperideals of $R$ disjoint from $S$, containing $K$. Any
such hyperideal $I$ is an n-hyperideal of $R$.

###### Proof.

Let $\Delta$ be the set of all hyperideals of $R$ disjoint from $S$,
containing $K$. Then $\Delta\neq\varnothing$, since $K\in\Delta$. So $\Delta$
is a partially ordered set with respect to set inclusion relation. As in the
proof of Theorem 5.9 we can show that $\Delta$ is bounded above. Hence by
Zorn,s Lemma, there exists a hyperideal $I$ which is maximal in $\Delta$. Let
$xoy\subseteq I$ for some $x,y\in R$. If $x\notin r(0)$ and $y\notin I$, then
we have $y\in(I:x)$ and $(I:x)$ properly contain $I$ . So, by maximality of
$I$ in $\Delta$, we get $(I:x)\cap S\neq\varnothing$. It means that there
exists an $s$ in $S$ with $s\in(I:x)$. Thus $sox\subseteq I$. Since $S$ is an
n-multiplicatively closed subset of $R$ and $x\notin r(0)$, then we have
$sox\subseteq S$. It implies that $S\cap I\neq\varnothing$ which is
contradictory to the fact that $I\in\Delta$. Therefore $I$ is an n-hyperideal
of $R$. ∎

######  Theorem 5.12.

Let $I\subseteq\bigcup_{i=1}^{n}I_{i}$ for some hyperideals $I_{1},...,I_{n}$
of $R$. Suppose that there exists $1\leq t\leq n$ such that $I_{t}$ is an
n-hyperideal and others are without nilpotent elements. If
$I\nsubseteq\bigcup_{i\neq t}I_{i}$, then $I\subseteq I_{t}$.

###### Proof.

Without destroying our assumption, we may assume that $t=1$. So
$I\nsubseteq\bigcup_{i=2}^{n}I_{i}$. It means that there exists $a\in I$ such
that $a\notin\bigcup_{i=2}^{n}I_{i}$. It implies that $a\in I$. Take $b\in
I\cap(\bigcap_{i=2}^{n})$. For every $2\leq i\leq n$, we have $a\notin I_{i}$
and $b\in I_{i}$. It means $a+b\notin I_{i}$, for every $2\leq i\leq n$. So
$a+b\in I$ but $a+b\notin\bigcup_{i=2}^{n}I_{i}$. Thus we have $a+b\in I_{1}$
and then $b\in I_{1}$ since $a\in I_{1}$. It follows that
$I\cap(\bigcap_{i=2}^{n}I_{i})\subseteq I_{1}$. Therefore
$(R-r(0))\cap(\Pi_{i=2}^{n})\neq\varnothing$, Since the product of non-
nilpotent elements is non-nilpotent. Also, we have $Io(\Pi_{i=2}^{n})\subseteq
I_{1}$. Hence, by Theorem 5.5, we conclude that $I\subseteq I_{1}$. ∎

######  Theorem 5.13.

(1) If $R$ is a reduced hyperring that is not integral hyperdomain, then $R$
has no n-hyperideal.
(2) Let $R$ be a reduced hyperring. Then 0 is an n-hyperideal of $R$ if and
only if $R$ is an integral hyperdomain.

###### Proof.

(1) By the assumption, $r(0)=0$ is not a prime hyperideal of $R$. Thus $0$ is
not an n-hyperideal, by Corollary 5.2. Also, let $I$ be a nonzero n-hyperideal
of $R$. Then $I\subseteq r(0)=0$ by Theorem 5.3. It means $I=0$. This is a
contradiction.
(2) $\Longrightarrow$ follows by (1).
$\Longleftarrow$ Let $R$ be an integral hyperdomain. Thus $r(0)=0$ is a prime
hyperideal. Hence $0$ is an n-hyperideal of $R$ by Corollary 5.2. ∎

######  Theorem 5.14.

$0$ is the only n-hyperideal of $R$ if and only if $R$ is an integral
hyperdomain.

###### Proof.

$\Longrightarrow$ Let $0$ be the only n-hyperideal of $R$. Then $r(0)$ is
prime hyperideal and n-hyperideal by Corollary 5.2 and Theorem 5.9. Therefore
$r(0)=0$ is a prime hyperideal. Thus $R$ is an integral hyperdomain.
$\Longleftarrow$ Assume that $I$ is an n-hyperideal of $R$. Then $I\subseteq
r(0)=0$ by Theorem 5.3. It means $I=0$. ∎

######  Example 5.1.

Let $(\mathbb{Z},+,\textperiodcentered)$ be the ring of integers. For all
$x,y\in\mathbb{Z}$, we define $xoy=\\{x\cdot a\cdot y:a\in A\\}$ which
$"\cdot"$ is ordinary multiplication and $A=\\{5,7\\}$. Then
$(\mathbb{Z},+,o)$ is a multiplicative hyperring (see [8]). We can show the
hyperring by $\mathbb{Z}_{A}$. This hyperring is an integral hyperdomain. Thus
$0$ is the only n-hyperideal of $\mathbb{Z}_{A}$.

######  Theorem 5.15.

$R$ is an invertible if and only if $R$ is a regular hyperring and $0$ is an
n-hyperideal.

###### Proof.

$\Longrightarrow$ Let $R$ be an invertible. Clearly, $R$ is a regular
hyperring. Also, $0$ is an n-hyperideal by Theorem 5.14.
$\Longleftarrow$ Let $R$ be a regular hyperring and $0$ be an n-hyperideal of
$R$. It is clear that $r(0)=0$. Let $0\neq r\in R$. Since $R$ is a regular
hyperring, then there exists $x\in R$ such that $r\in r^{2}ox$. Therefore
$0\in r^{2}ox-r$. Since $0$ is an n-hyperideal then $R$ is an integral
hyperdomain, by Theorem 5.13. Then $0\in r(rox-e)$ implies that $r=0$ which a
contradiction or $0\in rox-e$. This means $e\in rox$. Thus $R$ is an
invertible. ∎

##  6\. Stability of $n$-hyperideals

In this section, we investigate the stability of n-hyperideals in various
hyperring-theoric constructions.

######  Theorem 6.1.

Let $R_{1}$ and $R_{2}$ be multiplicative hyperrings and
$\phi:R_{1}\longrightarrow R_{2}$ be a good homomorphism. Then the following
statements hold :
(1) If $\phi$ is a monomorphism and $I_{2}$ is an n-hyperideal of $R_{2}$ ,
then $\phi^{-1}(I_{2})$ is an n- hyperideal of $R_{1}$.
(2) If $\phi$ is an epimorphism and $I_{1}$ is an n-hyperideal of $R_{1}$
containing $Ker(\phi)$, then $\phi(I_{1})$ is an n-hyperideal of $R_{2}$ .

###### Proof.

(1) Let $xoy\subseteq\phi^{-1}(I_{2})$ with $x,y\in R_{1}$ such that $x\notin
r(0_{R_{1}})$. Thus we have $\phi(xoy)=\phi(x)o\phi(y)\subseteq I_{2}$. Since
$\phi$ is a monomorphism , we conclude that $\phi(x)$ is not in
$r(0_{R_{2}})$. It means that $\phi(y)\in I_{2}$, since $I_{2}$ is an n-
hyperideal of $R_{2}$. Hence $y\in\phi^{-1}(I_{2})$. Therefore
$\phi^{-1}(I_{2})$ is an n-hyperideal of $R_{1}$.
(2) Let $x_{2}oy_{2}\subseteq\phi(I_{1})$ with $x_{2},y_{2}\in R_{2}$ such
that $x_{2}\notin r(0_{R_{2}})$. Since $\phi$ is an epimorphism, there exist
$x_{1},y_{1}\in R_{1}$ such that $\phi(x_{1})=x_{2},\phi(y_{1})=y_{2}$ and so
$\phi(x_{1}oy_{1})=x_{2}oy_{2}\subseteq\phi(I_{1})$. Now take any $u\in
x_{1}oy_{1}$. Then we get $\phi(u)\in\phi(x_{1}oy_{1})\subseteq\phi(I_{1})$
and so $\phi(u)=\phi(w)$ for some $w\in I_{1}$. This implies that
$\phi(u-w)=0\in(0)$, that is, $u-w\in Ker(\phi)\subseteq I_{1}$ and so $u\in
I_{1}$. Since $I_{1}$ is a ${\bf C}$-hyperideal of $R_{1}$, we conclude that
$x_{1}oy_{1}\subseteq I_{1}$. Since $I_{1}$ is an n-hyperideal of $R_{1}$ and
$x_{1}\notin r(0_{R_{1}})$, then we have $y_{1}\in I_{1}$. It means that
$\phi(y_{1})=y_{2}\in\phi(I_{1})$.Consequently, $\phi(I_{1})$ is an
n-hyperideal of $R_{2}$. ∎

######  Corollary 6.1.

Let $I,J$ be proper hyperideals of $R$ such that $J\subseteq I$ and $I$ be an
n-hyperideal of $R$. Then $I/J$ is an n-hyperideal of $R/J$.

###### Proof.

Define $f:R\longrightarrow R/J$ by $f(r)=r+J$. Clearly, $f$ is a good
epimorphism. Since $Ker(f)=J\subseteq I$ and $I$ is an n-hyperideal of $R$ ,
then the claim follows from Theorem 6.1 (1). ∎

######  Corollary 6.2.

Let $I,J$ be proper hyperideals of $R$ such that $J\subseteq I$. If $I/J$ is
an n-hyperideal of $R/J$ with $J\subseteq r(0)$, then $I$ is an n-hyperideal
of $R$.

###### Proof.

Let $xoy\subseteq I$ for some $x,y\in R$ such that $x\notin r(0)$. It is clear
that $(x+J)o(y+J)=\\{z+J:z\in xoy\\}\subseteq I/J$ and $x+J\notin r(0_{R/I})$.
Thus $y+J\in I/J$, Since $I/J$ is an n-hyperideal of $R/J$. Therefore $y\in
I$. It means that $I$ is an n-hyperideal of $R$. ∎

######  Corollary 6.3.

Let $I,J$ be proper hyperideals of $R$ such that $J\subseteq I$. If $J$ is an
n-hyperideal of $R$ and $I/J$ is an n-hyperideal of $R/J$, then $I$ is an
n-hyperideal of $R$.

###### Proof.

It is obvious by Theorem 5.3 and Corollary 6.2. ∎

######  Theorem 6.2.

Let $R$ be a multiplicative hyperring with scalar identity 1 and $I$ be a
hyperideal of $R$. If $M_{n}(I)$ is an n-hyperideal of $M_{n}(R)$, then $I$ is
an n-hyperideal of $R$.

###### Proof.

Suppose that for $x,y\in R$ , $xoy\subseteq I$ such that $x\notin r(0_{R})$.
Then

$\begin{pmatrix}xoy&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\ 0&0&\cdots&0\end{pmatrix}\subseteq M_{n}(I)$

It is clear that

$\begin{pmatrix}xoy&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}=\begin{pmatrix}x&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\begin{pmatrix}y&0&\cdots&0\\\ 0&0&\cdots&0\\\
\vdots&\vdots&\ddots\vdots\\\ 0&0&\cdots&0\end{pmatrix}.$

Since $M_{n}(I)$ is an n-hyperideal of $M_{n}(R)$ and

$\begin{pmatrix}x&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\notin r(0_{M_{n}(R)})$

then we have

$\begin{pmatrix}y&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\ddots\vdots\\\
0&0&\cdots&0\end{pmatrix}\in M_{n}(I).$

It follows that $y\in I$. Therefore $I$ is an n-hyperideal of $R$.

∎

######  Theorem 6.3.

Let $T$ be a subhyperring of $R$. If I is an n-hyperideal of $R$ such that
$T\nsubseteq I$, then $I\cap T$ is an n-hyperideal of $T$.

###### Proof.

Define $j:T\longrightarrow R$ by $j(t)=t$. It is clear that $j^{-1}(I)=I\cap
T$. Thus $I\cap T$ is an n-hyperideal of $T$, by 6.1 (1). ∎

######  Theorem 6.4.

Let $R_{1}$ and $R_{2}$ be two multiplicative hyperrings. If $I=I_{1}\times
I_{2}$ is an n-hyperideal of $R=R_{1}\times R_{2}$, then $I=R$.

###### Proof.

It is clear that $(1_{R_{1}},0_{R_{2}})o(0_{R_{1}},1_{R_{2}})\subseteq I$ but
none of the elements are in $r(0_{R_{1}\times R_{2}})$. Since $I$ is an
n-hyperideal of $R$, then $(1_{R_{1}},0_{R_{2}})$ and $(0_{R_{1}},1_{R_{2}})$
are in $I$. Thus $I=R$. ∎

Let $(R,+,o)$ be a hyperring. We define the relation $\gamma$ on $R$ as
follows:
$a\gamma b$ if and only if $\\{a,b\\}\subseteq U$ where $U$ is a finite sum of
finite products of elements of R, i.e.,$a\gamma b$ if and only if there exist
$z_{1},...,z_{n}\in R$ such that $\\{a,b\\}\subseteq\sum_{j\in J}\prod_{i\in
I_{j}}z_{i};\ \ I_{j},J\subseteq\\{1,...,n\\}$

We denote the transitive closure of $\gamma$ by $\gamma^{\ast}$. The relation
$\gamma^{\ast}$ is the smallest equivalence relation on a multiplicative
hyperring $(R,+,o)$ such that the quotient $R/\gamma^{\ast}$, the set of all
equivalence classes, is a fundamental ring. Let $\mathfrak{U}$ be the set of
all finite sums of products of elements of R we can rewrite the definition of
$\gamma^{\ast}$ on $R$ as follows:
$a\gamma^{\ast}b$ if and only if there exist $z_{1},...,z_{n}\in R$ with
$z_{1}=a,z_{n+1}=b$ and $u_{1},...,u_{n}\in\mathfrak{U}$ such that
$\\{z_{i},z_{i+1}\\}\subseteq u_{i}$ for $i\in\\{1,...,n\\}$. Suppose that
$\gamma^{\ast}(a)$ is the equivalence class containing $a\in R$. Then, both
the sum $\oplus$ and the product $\odot$ in $R/\gamma^{\ast}$ are defined as
follows:$\gamma^{\ast}(a)\oplus\gamma^{\ast}(b)=\gamma^{\ast}(c)$ for all
$c\in\gamma^{\ast}(a)+\gamma^{\ast}(b)$ and
$\gamma^{\ast}(a)\odot\gamma^{\ast}(b)=\gamma^{\ast}(d)$ for all
$d\in\gamma^{\ast}(a)o\gamma^{\ast}(b)$ Then $R/\gamma^{\ast}$ is a ring,
which is called a fundamental ring of $R$ (see also [13]).

######  Theorem 6.5.

Let $R$ be a multiplicative hyperring with scalar identity $1$. Then the
hyperideal $I$ of $R$ is n-hyperideal if and only if $I/\gamma^{\ast}$ be an
n-ideal of $R/\gamma^{\ast}$.

###### Proof.

($\Longrightarrow$) Let for $x,y\in R/\gamma^{\ast},\ x\odot y\in
I/\gamma^{\ast}$ such that $x\notin r(0_{R/\gamma^{\ast}})$. Thus, there exist
$a,b\in R$ such that $x=\gamma^{\ast}(a),y=\gamma^{\ast}(b)$ and $x\odot
y=\gamma^{\ast}(a)\odot\gamma^{\ast}(b)=\gamma^{\ast}(aob)$. So,
$\gamma^{\ast}(a)\odot\gamma^{\ast}(b)=\gamma^{\ast}(aob)\in I/\gamma^{\ast}$,
then $aob\subseteq I$. Since $I$ is an n-hyperideal and $a\notin r(0_{R})$,
then $b\in I$. Hencey $y=\gamma^{\ast}(b)\in I/\gamma^{\ast}$. Thus
$I/\gamma^{\ast}$ is an n-ideal of $R/\gamma^{\ast}$.
($\Longleftarrow$) Suppose that $aob\subseteq I$ for $a,b\in R$ such that
$a\notin r(0_{R})$, then $\gamma^{\ast}(a),\gamma^{\ast}(b)\in
R/\gamma^{\ast}$ and
$\gamma^{\ast}(a)\odot\gamma^{\ast}(b)=\gamma^{\ast}(aob)\in I/\gamma^{\ast}$.
Since $I/\gamma^{\ast}$ is an n-ideal of $R/\gamma^{\ast}$ and
$\gamma^{\ast}(a)\notin r(0_{R/\gamma^{\ast}})$, then we have
$\gamma^{\ast}(b)\in I/\gamma^{\ast}$. It means that $b\in I$. Hence $I$ is an
n-hyperideal of $R$. ∎

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