Common fixed points of monotone ρ-nonexpansive semigroup in modular spaces

Abstract

In this paper, we consider the class of monotone ρ-nonexpansive semigroups and give existence and convergence results for common fixed points. First, we prove that the set of common fixed points is nonempty in uniformly convex modular spaces and modular spaces. Then we introduce an iteration algorithm to approximate a common fixed point for the same class of semigroups.

1 Introduction

We prove the existence and convergence to a common fixed point of monotone ρ-nonexpansive semigroups in modular spaces. Recall that a family ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ is called a semigroup on a subset C of a modular space $X_{\rho }$ if

• (i) $T_{0} (x)=x$ for all $x \in C$.

• (ii) $T_{s+t} = T_{s} \circ T_{t}$ for all positive s, t.

The theory of semigroups is very interesting in mathematics and applications. As a situation, in the theory of dynamical systems the space $X_{\rho }$ on which the semigroup ${\mathcal{S}}$ is defined represents the states space, and the mapping

$$\begin{aligned} \begin{aligned} &\mathbb{R}_{+} \times C \longrightarrow C, \\ & (t,s ) \longmapsto T_{t} (x) \end{aligned} \end{aligned}$$

represents the evolution function of the dynamical system (see [9, 11]).

The problem of the existence of common fixed points for semigroup is still in its infancy. Kozlowski [9] has demonstrated the existence of common fixed points for semigroups of monotone contractions and monotone ρ-nonexpansive mappings in Banach spaces. Afterward, Bashar et al. [3] generalized Kozlowski’s work in Banach spaces. In the case of monotone nonexpansive semigroups, they proved the following theorem.

Theorem 1.1

([3])

Let $(X, \Vert \cdot \Vert )$ be a Banach space uniformly convex in every direction. Let C be a weakly compact convex nonempty subset of X, and let ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ be a monotone nonexpansive semigroup defined on C. Assume that there exists $x_{0} \in C$ such that $x_{0} \leq T_{t}(x_{0})$ (resp., $T_{t}(x_{0}) \leq x_{0}$) for all $t \geq 0$. Then there exists a common fixed point $z \in \mathit{Fix} ({\mathcal{S}})$ such that $x_{0} \leq z$ (resp., $z \geq x_{0}$).

Under the frame of modular function spaces, Kozlowski [7] has shown that the set of common fixed points of any ρ-nonexpansive semigroups, acting on a ρ-closed convex and ρ-bounded subset of a uniformly convex modular function space $L_{\rho }$, is nonempty ρ-closed and convex (see Theorem 6.5 in [8]).

For finding a common fixed point of a nonexpansive mapping, Halpern [5] has introduced in Hilbert spaces H the following explicit iteration scheme for elements $u \in H$ and $x_{0} \in H$:

$$ x_{n+1}= \alpha _{n} u +(1-\alpha _{n}) T(x_{n}) \quad \text{for all }n \geq 0, $$

(1)

where $(\alpha _{n})_{n}$ is a sequence in $(0, 1)$. Subsequently, many mathematicians paid their attention to studying the convergence of Halpern iteration for semigroups of various nonlinear mappings in different spaces and under different conditions.

2 Methods

In 2002, Xu [15] showed, under certain assumptions on semigroups, the strong convergence of the modified Halpern iteration given by

$$ x_{n+1}= \alpha _{n} u + (1-\alpha _{n}) T_{t_{n}}(x_{n}) $$

(2)

for all $t_{n} > 0$ (see also Wangkeeree et al. [14] for the asymptotically nonexpansive semigroups, Yao et al. [16] for the nonexpansive semigroups, and Song et al. [12] for ρ-nonexpansive semigroups in Hilbert spaces).

Motivated by the results cited, we begin by generalizing Theorem 1.1 for the monotone ρ-nonexpansive semigroups in uniformly convex and uniformly convex in every direction modular spaces. Next, we define a new iteration algorithm for monotone ρ-nonexpansive semigroups as follows:

$$ \textstyle\begin{cases} x_{n+1}= (1-\alpha _{n} ) T_{t_{n}}(x_{n}) + \alpha _{n} T_{t_{n}} (y_{n}), \\ y_{n} = (1-\beta _{n} )x_{n} + \beta _{n} T_{t_{n}} (x_{n}), \end{cases} $$

where the sequences $(t_{n})_{n} \subset \mathbb{R}_{+}$ and $(\alpha _{n})_{n}$, $(\beta _{n})_{n} \subset (0,1)$ satisfy some conditions. This process generalizes the work of [2]. Later, we show under some assumptions that the sequence $(x_{n})_{n}$ ρ-converges to a common fixed point of a monotone ρ-nonexpansive semigroup.

3 Results and discussion

Throughout this work, X stands for a real vector space.

Definition 3.1

([1])

A function $\rho : X \longrightarrow [0,+\infty ]$ is called a modular if the following holds:

• (1) $\rho (x ) =0$ if and only if $x=0$;

• (2) $\rho (- x )= \rho (x )$;

• (3) $\rho (\alpha x + (1-\alpha ) y ) \leq \rho (x ) +\rho (y )$ for all $\alpha \in [ 0,1 ]$ and $x, y\in X$.

If (3) is replaced by

$$ \rho \bigl(\alpha x + (1-\alpha ) y \bigr)\leq \alpha \rho (x )+ (1-\alpha )\rho (y ) $$

for all $\alpha \in [0,1 ]$ and $x, y\in X$, then ρ is called a convex modular.

A modular ρ defines the corresponding modular space, that is, the vector space

$$ X_{\rho }= \Bigl\{ x\in X : \lim_{\lambda \to 0} \rho ( \lambda x )=0 \Bigr\} . $$

Let ρ be a convex modular. Then the modular space $X_{\rho }$ is equipped with a norm called the Luxemburg norm, defined by

$$ \Vert x \Vert _{\rho }=\inf \biggl\{ \lambda >0 : \rho \biggl( \frac{x}{\lambda } \biggr) \leq 1 \biggr\} . $$

We now give the basic definitions.

Definition 3.2

([1])

Let ρ be a modular defined on a vector space X.

• (1) We say that a sequence $(x_{n} )_{n\in \mathbb{N}} \subset X_{\rho }$ is ρ-convergent to $x\in X_{\rho }$ if and only if $\rho (x_{n} -x )$ converges to 0 as n goes to infinity. Note that the limit is unique.

• (2) A sequence $(x_{n} )_{n} \subset X_{\rho }$ is called ρ-Cauchy if $\rho (x_{n} -x_{m} ) \longrightarrow 0$ as n, $m \longrightarrow + \infty$.

• (3) We say that $X_{\rho }$ is ρ-complete if any ρ-Cauchy sequence is ρ-convergent.

• (4) A subset C of $X_{\rho }$ is said ρ-closed if the ρ-limit of a ρ-convergent sequence of C always belongs to C.

• (5) A subset C of $X_{\rho }$ is said to be ρ-bounded if

  $$ \delta _{\rho } (C ) =\sup \bigl\{ \rho (x-y ) :x,y \in C \bigr\} < \infty . $$

• (6) A subset K of $X_{\rho }$ is said to be ρ-compact if any sequence $(x_{n} )_{n}$ of C has a subsequence that ρ-converges to a point $x\in C$.

• (7) We say that ρ satisfies the Fatou property if

  $$ \rho (x-y )\leq {\lim_{n \to +\infty }}\rho (x-y_{n} ) $$

  for any x whenever $(y_{n} )_{n}$ ρ-converges to y in $X_{\rho }$.

Note that the ρ-convergence does not imply the ρ-Cauchy condition. Also, $x_{n}\overset{\rho }{\longrightarrow }x$ does not imply in general that $\lambda x_{n}\overset{\rho }{\longrightarrow } \lambda x$ for every $\lambda >1$.

An important property associated with a modular, which plays a powerful role in modular spaces, is the $\Delta _{2}$-condition and the $\Delta _{2}$-type condition.

Definition 3.3

([1, 10])

Let ρ be a modular defined on a vector space X. We say that ρ satisfies

• (i) the $\Delta _{2}$-condition if $\rho (2x_{n})\longrightarrow 0$ whenever $\rho (x_{n})\longrightarrow 0$ as $n\to +\infty$;

• (ii) the $\Delta _{2}$-type condition, if there exists $K>0$ such that $\rho (2x)\leq K \rho (x)$.

Definition 3.4

Let ρ be a modular, and let C be a nonempty subset of the modular space $X_{\rho }$. A mapping $T : C\longrightarrow C$ is said to be

• (a) monotone if $T (x ) \leq T (y )$ for all x, $y\in C$ such that $x\leq y$;

• (b) monotone ρ-nonexpansive if T is monotone such that

  $$ \rho \bigl(T (x ) -T (y ) \bigr)\leq \rho (x-y ) $$

  for all x, $y\in X_{\rho }$ such that $x\leq y$.

Recall that $T : C \longrightarrow C$ is said to be ρ-continuous if $(T (x_{n} ) )_{n}$ ρ-converges to $T (x )$ whenever $(x_{n} )_{n}$ ρ-converges to x. It is not true that a monotone ρ-nonexpansive mapping is ρ-continuous since this result is not true in general when ρ is a norm.

We further assume that ρ is a convex modular.

Definition 3.5

([1])

Let ρ be a modular, and let $r>0$ and $\varepsilon >0$. Define, for $i\in \{1, 2\}$,

$$ D_{i} (r,\varepsilon )= \biggl\{ (x,y ) \in X_{\rho } \times X_{\rho }: \rho (x )\leq r , \rho (y )\leq r , \rho \biggl( \frac{x-y}{i} \biggr)\geq r \varepsilon \biggr\} . $$

If $D_{i} (r,\varepsilon ) \neq \emptyset$, then let

$$ \delta _{i} (r,\varepsilon )= \inf \biggl\{ 1-\frac{1}{r} \rho \biggl(\frac{x+y}{2} \biggr) : (x,y ) \in D_{i} \biggr\} . $$

If $D_{i} (r,\varepsilon ) = \emptyset$, then we set $\delta _{i} (r,\varepsilon )=1$.

• (i) We say that ρ satisfies uniform convexity (UCi) if for all $r>0$ and $\varepsilon >0$, we have $\delta _{i} (r,\varepsilon )>0$.

• (ii) We say that ρ satisfies unique uniform convexity (UUCi) if for all $s\geq 0$ and $\varepsilon >0$, there exists $\eta (s,\varepsilon )>0$ such that

  $$ \delta _{i} (r,\varepsilon )>\eta (s,\varepsilon )\quad \text{for } r>s. $$

• (iii) We say that ρ is strictly convex (SC) if for all x, $y\in X_{\rho }$ such that $\rho (x )=\rho (y )$ and

  $$ \rho \biggl(\frac{x+y}{2} \biggr)= \frac{\rho (x )+\rho (y )}{2}, $$

  we have $x=y$.

The following proposition characterizes the relationship between the above notions.

Proposition 3.6

([1])

• (a) $(UUCi )$ implies $(UCi )$ for $i=1,2$;

• (b) $\delta _{1} (r,\varepsilon ) \leq \delta _{2} (r, \varepsilon )$ for $r>0$ and $\varepsilon >0$;

• (c) $(UC1 )$ implies $(UC2 )$;

• (d) $(UC2 )$ implies $(SC )$;

• (e) $(UUC1 )$ implies $(UUC2 )$.

In the following definition, we introduce the uniform convexity in every direction (UCED) of a modular.

Definition 3.7

Let ρ be a modular. We say that ρ is uniformly convex in every direction (UCED) if for any $r>0$ and nonzero $z\in X_{\rho }$, we have

$$ \delta (r,z )=\inf \biggl\{ 1-\frac{1}{r} \rho \biggl(x+ \frac{z}{2} \biggr) : \rho (x )\leq r , \rho (x+z )\leq r \biggr\} > 0. $$

We say that ρ satisfies unique uniform convexity in every direction (UUCED) if there exists $\eta (s,z )>0$ for $s\geq 0$ and nonzero $z\in X_{\rho }$ such that

$$ \delta (r,z )>\eta (s,z ) \quad \text{for } r>s. $$

Proposition 3.8

• (a) $(UCi )$ (resp., $(UUCi )$) implies $(UCED )$ (resp., $(UUCED )$) for $i=1,2$;

• (b) $(UUCED )$ implies $(UCED )$;

• (c) $(UCED )$ implies $(SC )$.

Proof

It is quite easy to show (a) and (b). To prove (c), let x, $y \in X_{\rho }$ be such that $x \neq y$.

First, if $\rho (x )\neq \rho (y )$, then there is nothing to prove. Otherwise, we assume that $\rho (x ) =\rho (y )=r >0$ and consider $z=y-x (\neq 0)$. Hence $\rho (x+z )=\rho (y )=r$. Since ρ is $(UCED )$, $\delta (r,z ) >0$, which implies

$$ 1-\frac{1}{r} \rho \biggl(x+\frac{z}{2} \biggr) \geq \delta (r,z ) >0. $$

Thus

$$ \rho \biggl(\frac{x+y}{2} \biggr)= \rho \biggl(x+\frac{y-x}{2} \biggr) \leq \bigl(1-\delta (r,z ) \bigr)r < r, $$

that is, $\rho (\frac{x+y}{2} ) < r = \frac{\rho (x )+\rho (y )}{2}$ □

The following property plays a similar role as the reflexivity in Banach spaces for modular spaces.

Definition 3.9

([10])

Let ρ be a modular. We say that the modular space $X_{\rho }$ satisfies property (R) if for every decreasing sequence $(C_{n} )_{n\in \mathbb{N}}$ of nonempty ρ-closed convex and ρ-bounded subsets of $X_{\rho }$, we have

$$ \bigcap_{n\in \mathbb{N}} C_{n} \neq \emptyset . $$

Lemma 3.10

([1])

Let ρ be a convex modular satisfying the Fatou property. Assume that $X_{\rho }$ is ρ-complete and ρ is (UUC2). Then $X_{\rho }$ satisfies property (R).

Proposition 3.11

([1])

Let ρ be a convex modular. Assume that $X_{\rho }$ is ρ-complete and ρ is (UUC2). Let C be a ρ-closed convex and ρ-bounded nonempty subset of $X_{\rho }$. Let $(C_{i} )_{i\in I}$ be a family of ρ-closed convex nonempty subsets of C such that $\underset{i\in F}{\cap } C_{i}$ is nonempty for any finite subset F of I. Then

$$ \bigcap_{i\in I}C_{i} \neq \emptyset . $$

The ρ-type function is a powerful technical tool to prove the existence of a fixed point.

Definition 3.12

([6])

Let $(x_{n} )_{n}$ be a sequence in $X_{\rho }$, and let K be a nonempty subset of $X_{\rho }$.

The function $\tau : K \longrightarrow [0, \infty ]$ defined by

$$ \tau (x )=\limsup_{n\to \infty }\rho (x_{n}-x ) $$

is called a ρ-type function.

The next definition is an adaptation of the definition of ρ-type functions to a one-parameter family of mappings.

Definition 3.13

([6])

Let $C\subset X_{\rho }$ be convex ρ-bounded. A function $\tau : C\longrightarrow \overline{\mathbb{R}_{+}}$ is called a ρ-type function (or shortly a type) if there exists a one-parameter family $\lbrace T_{t} : t\geq 0 \rbrace$ of elements of a nonempty subset K of $X_{\rho }$ such that for all $x\in K$,

$$ \tau (y )=\limsup_{t\to \infty }\rho \bigl(T_{t}(x)-y \bigr) $$

for all $y \in K$.

A sequence $(c_{n} )_{n} \subset K$ is called a minimizing sequence of τ if

$$ \lim_{n\to +\infty } \tau (c_{n} )=\inf _{x\in K}\tau (x ). $$

Note that the ρ-type function τ is convex since ρ is convex.

Recall the definition of the uniform continuity of a modular.

Definition 3.14

A modular ρ is said to be uniformly continuous if for any $\varepsilon >0$ and $R>0$, there exists $\eta >0$ such that

$$ \bigl\vert \rho (y ) -\rho (x+y ) \bigr\vert \leq \varepsilon $$

whenever $\rho (x ) \leq \eta$ and $\rho (y ) \leq R$.

The following lemma plays an important role in the proof of the next fixed point theorem. To prove it, we use the ideas of the proof of Lemma 3.5 in [3].

Lemma 3.15

Let ρ be a convex modular uniformly continuous and (UUCED). Assume that the modular space $X_{\rho }$ satisfies property (R). Let C be a ρ-closed ρ-bounded convex nonempty subset of $X_{\rho }$. Let K be a nonempty ρ-closed convex subset of C. Let $(x_{k} )_{k\in \mathbb{N}}$ be a sequence in C and consider the ρ-type function $\tau : K \longrightarrow [0,+\infty ]$ defined by

$$ \tau (y ) =\limsup_{k\to +\infty }\rho (x_{k}-y ). $$

(3)

Then τ has a unique minimum point in K.

A subset $P \subset X_{\rho }$ is called a pointed ρ-closed convex cone if P is a nonempty ρ-closed subset of $X_{\rho }$ satisfying the following properties:

• (i) $P+P \subset P$,

• (ii) $\lambda P \subset P$ for all $\lambda \in {\mathbb{R}}_{+}$,

• (iii) $P\cap (-P ) =\{ 0 \}$.

Using P, we define an ordering on $X_{\rho }$ by

$$ x\leq y \quad \text{if and only if}\quad y-x\in P. $$

We further suppose that the modular space $X_{\rho }$ is equipped with the partial order defined by P.

3.1 Common fixed point results for a monotone ρ-nonexpansive semigroup

Before we state our main results, let us recall the definition of a monotone ρ-Lipschitz semigroup.

Definition 3.16

Let C be a nonempty subset of a modular space $X_{\rho }$. A one-parameter family ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ of mappings from C into C is said to be a monotone semigroup on C if it satisfies the following conditions:

• (i) $T_{0} (x) = x$ for all $x\in C$,

• (ii) $T_{s+t} = T_{s} \circ T_{t}$ for all s, $t\geq 0$,

• (iii) $T_{t}$ is monotone for all $t\geq 0$, that is, $T_{t} x \leq T_{t} y$ for all x, $y\in C$ such that $x\leq y$.

Definition 3.17

A semigroup ${\mathcal{S}}$ is said to be a monotone ρ-Lipschitz semigroup if ${\mathcal{S}}$ is monotone and there exists $k\geq 0$ such that

$$ \rho \bigl(T_{t} (x) -T_{t}(y) \bigr) \leq k \rho (x-y ) $$

for all x, $y\in C$ such that $x\leq y$ and all $t\geq 0$.

If $k<1$, then ${\mathcal{S}}$ is said to be a monotone ρ-contraction semigroup. If $k=1$, then ${\mathcal{S}}$ is said to be a monotone ρ-nonexpansive semigroup. The set of all common fixed points of ${\mathcal{S}}$ is defined by

$$ \mathit{Fix} ({\mathcal{S}} )= \bigl\lbrace x\in C : T_{t} (x )=x \text{ for all } t\geq 0 \bigr\rbrace = \bigcap_{t\geq 0} \mathit{Fix} (T_{t} ). $$

The following lemma generalizes the minimizing sequence property for type functions generated by a sequence to the case of type functions defined by a one-parameter family $\lbrace h_{t} : t\geq 0 \rbrace$. To prove it, we use the ideas of the proof of Lemma 7.11 of [6].

Lemma 3.18

Let ρ be a convex modular satisfying the Fatou property and (UUC1), and let $X_{\rho }$ be a ρ-complete modular space. Let C be a nonempty ρ-closed convex subset of $X_{\rho }$. Let ${\mathcal{S}}$ be a monotone ρ-nonexpansive semigroup on C. Fix $x_{0} \in C$ and consider the function $\varphi : C \longrightarrow \mathbb{R}_{+}$ given by

$$ \varphi (y )= \limsup_{t\to +\infty } \rho \bigl(T_{t} ( x_{0} )-y \bigr) =\inf_{s\geq 0}\sup _{t\geq s} \rho \bigl(T_{t} ( x_{0} )-y \bigr). $$

Then every minimizing sequence of φ ρ-converges to the same limit.

Theorem 3.19

Let ρ be a convex modular satisfying the Fatou property and (UUC1). Let C be a nonempty ρ-closed convex ρ-bounded subset of a ρ-complete modular space $X_{\rho }$. Let ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ be a monotone ρ-nonexpansive semigroup such that $T_{t}$ is ρ-continuous for any $t\geq 0$. Assume that there exists $x_{0} \in C$ such that $x_{0} \leq T_{t}(x_{0})$ (resp., $T_{t} (x_{0}) \leq x_{0}$) for all $t\geq 0$. Then there exists a common fixed point $z\in Fix ({\mathcal{S}} )$ such that $x_{0}\leq z$ (resp., $z\leq x_{0}$).

Proof

Without loss of generality, we assume that $x_{0} \leq T_{t}x_{0}$ for all $t>0$. By the definition of the partial order and Proposition 3.11 we have that

$$ K=\bigcap_{t\geq 0} \bigl[T_{t}(x_{0} ), \rightarrow \bigr) \cap C $$

is nonempty. In fact, using Proposition 3.11, it suffices to prove that

$$ \bigcap_{t\in F} \bigl[T_{t}(x_{0}), \rightarrow \bigr) \cap C $$

is nonempty for any finite subset $F= \lbrace t_{0}, \ldots , t_{n} \rbrace$ of $\mathbb{R}_{+}$, where $t_{i}$ are arbitrarily chosen in $\mathbb{R}_{+}$.

Let $x= T_{t_{0} + \cdots + t_{n}}(x_{0}) \in C$. Since ${\mathcal{S}}$ is a monotone semigroup and $x_{0} \leq T_{t}(x_{0})$ for all $t\geq 0$, we have $T_{s} (x_{0}) \leq T_{s+t}(x_{0})$ for all s, $t\geq 0$. Hence

$$ T_{t_{i}}(x_{0})\leq x $$

for all $i \in \lbrace 1,\ldots , n \rbrace$, that is, $x\in [ T_{t_{i}}(x_{0}) , \rightarrow ) \cap C$. Thus $\bigcap_{t_{i}\in F} [T_{t_{i}}(x_{0}), \rightarrow ) \cap C$ is nonempty for all $n\geq 0$. Moreover, K is ρ-closed convex.

Furthermore, K is invariant by ${\mathcal{S}}$. Indeed, let $x\in K$ and t, $s\geq 0$. If $t\geq s$, then $t-s\geq 0$. Hence $T_{t-s}(x_{0})\leq x$ implies $T_{t}(x_{0})\leq T_{s}(x)$. If $t< s$, then $\varepsilon = s-t >0$. Since $x_{0} \leq x$, we have

$$ x_{0} \leq T_{\varepsilon }(x_{0}) \leq T_{\varepsilon }(x) \quad \Longrightarrow \quad T_{t} (x_{0}) \leq T_{t+\varepsilon } (x) = T_{s} (x). $$

Thus $T_{t}(x_{0}) \leq T_{s}(x)$ for all t, $s\geq 0$. Then $T_{s} (x) \in K$ for all $s\geq 0$. Therefore ${\mathcal{S}} (K ) \subset K$.

Consider the function $\varphi : K \longrightarrow [0,+\infty [$ defined by

$$ \varphi (y) = \limsup_{t\to +\infty }\rho \bigl(T_{t} (x_{0}) - y \bigr)= \inf_{s\geq 0}\sup _{t\geq s} \rho \bigl(T_{t} (x_{0}) - y \bigr). $$

Since K is ρ-bounded, $\varphi _{0}= \inf_{y\in C} \varphi (y) <\infty$. Thus for any $n\geq 1$, there exists $z_{n} \in K$ such that

$$ \varphi _{0} \leq \varphi (z_{n}) \leq \varphi _{0} + \frac{1}{n}. $$

Then $(z_{n} )_{n}$ is a minimizing sequence of φ and ρ-converges to $z \in K$ by Lemma 3.18. To prove that $z\in \mathit{Fix} ({\mathcal{S}} )$, it suffices to show that $(T_{t} (z_{n} ) )_{n}$ is also a minimizing sequence of φ for any $t\geq 0$.

Fix s, $\eta \geq 0$, and let $t\geq s+\eta$ and $y\in K$. Then $T_{t-s}(x_{0})\leq y$. As ${\mathcal{S}}$ is a monotone ρ-nonexpansive semigroup, we have

$$\begin{aligned} \rho \bigl(T_{s} \bigl(T_{t-s}(x_{0}) \bigr)-T_{s}(y) \bigr) & = \rho \bigl(T_{t}(x_{0}) - T_{s} (y) \bigr) \\ & \leq \rho \bigl( T_{t-s}(x_{0}) -y \bigr) \\ & \leq \sup_{\bar{t}\geq \eta } \rho \bigl(T_{\bar{t}}(x_{0})-y \bigr). \end{aligned}$$

Hence

$$ \sup_{t\geq \eta } \rho \bigl(T_{t}(x_{0}) - T_{s} (z) \bigr) \leq \sup_{t\geq s+\eta } \rho \bigl(T_{t}(x_{0}) - T_{s} (z) \bigr) \leq \sup_{\bar{t}\geq \eta } \rho \bigl(T_{\bar{t}}(x_{0}) - z \bigr). $$

Taking the $\inf_{\eta \geq 0}$ in the previous inequality, we get

$$ \inf_{\eta \geq 0} \sup_{t\geq \eta } \rho \bigl(T_{t}(x_{0}) - T_{s} (z) \bigr) \leq \inf_{\eta \geq 0}\sup_{\bar{t}\geq \eta } \rho \bigl(T_{\bar{t}}(x_{0}) - z \bigr), $$

which implies

$$ \varphi \bigl(T_{s}(y) \bigr) \leq \inf_{\eta \geq 0} \sup_{ \bar{t}\geq \eta } \rho \bigl( T_{\bar{t}}(x_{0}) - y \bigr). $$

Since η is arbitrary positive, we have

$$ \varphi \bigl( T_{s}(y) \bigr) \leq \varphi (y ) $$

for any $s\geq 0$. Therefore $( T_{s} (z_{n}) )_{n}$ is also a minimizing sequence of φ for all $s\geq 0$.

By Lemma 3.18 we get that $( T_{s} (z_{n}) )_{n}$ ρ-converges to z for all $s\geq 0$. Since $T_{s}$ is ρ-continuous for all $s\geq 0$, $( T_{s} (z_{n}) )_{n}$ ρ-converges to $T_{s} (z)$ for all $s\geq 0$. By the uniqueness of the limit we conclude that $z= T_{s}(z)$ for all $s\geq 0$. Then z is a common fixed point of the semigroup ${\mathcal{S}}$. □

Example 3.20

Let $(p_{n} )_{n\geq 1}$ be a sequence of real numbers such that $1\leq p_{n} <\infty$ for all $n \geq 1$. Consider the vector space

$$ \ell _{ (p_{n} )} = \Biggl\lbrace (x_{n} )_{n} \in \mathbb{R}^{\mathbb{N}^{*}} : \sum_{n=1}^{+\infty } \frac{1}{p_{n}} \vert \lambda x_{n} \vert ^{p_{n}} < \infty \text{ for some } \lambda >0 \Biggr\rbrace , $$

where the modular ρ is given by $\rho ( (x_{n} )_{n} ) = \sum_{n=0}^{+\infty } \frac{1}{p_{n}} \vert x_{n}\vert ^{p_{n}}$ for all $(x_{n} )_{n} \in \ell _{(p_{n})}$. Suppose that $1< p^{-} =\inf_{n\geq 1}p_{n} \leq p_{n} \leq \sup_{n\geq 1}p_{n} = p^{+} <\infty$ for all $n\geq 1$.

According to [13], the modular ρ is convex and satisfies (UUC1), and the space $\ell _{ (p_{n} )}$ under the Luxemburg norm $\Vert \cdot \Vert _{\rho }$ endowed by the modular ρ is a Banach space. Moreover, ρ satisfies the $\Delta _{2}$-type condition. In fact, let $(x_{n} )_{n} \in \ell _{ (p_{n} )}$. Since $p_{n} \leq p^{+}$, we have ${\sum_{n=1}^{q} \frac{2^{p_{n}}}{p_{n}}\vert x_{n}\vert ^{p_{n}}} \leq {\sum_{n=1}^{q} \frac{2^{p^{+}}}{p_{n}}\vert x_{n} \vert ^{p_{n}}}$ for all $q\in \mathbb{N}^{*}$. Taking ${\lim_{n\to \infty }}$, we have

$$ \rho (2x )= \sum_{n=1}^{+\infty } \frac{2^{p_{n}}}{p_{n}} \vert x_{n} \vert ^{p_{n}} \leq 2^{p^{+}} \sum_{n=1}^{+\infty } \frac{1}{p_{n}} \vert x_{n} \vert ^{p_{n}} = 2^{p^{+}}\rho (x ). $$

Recall that if ρ satisfies the $\Delta _{2}$-type condition, then $\Vert \cdot \Vert _{\rho }$ convergence is equivalent to ρ-convergence (see [6]). Thus $\ell _{ (p_{n} )}$ is a ρ-complete modular space. Moreover, ρ satisfies the Fatou property.

Consider the partial ordering ⪯ defined by

$$ (x_{n} )_{n} \preceq (y_{n} )_{n} \quad \Longleftrightarrow\quad x_{n}\leq y_{n} , \quad \forall n\geq 1, $$

for all $(x_{n} )_{n}$ and $(y_{n} )_{n}$ in $\ell _{ (p_{n} )}$.

Let $C= B_{\rho } (0,r )$ be the ρ-closed ball of $\ell _{ (p_{n} )}$ centered at 0 with radius $r>1$; it is ρ-bounded. Let the family $\mathcal{S}= \lbrace T_{t} : t\geq 0 \rbrace$ of mappings be given by

$$ \begin{aligned} &T_{t} : C \longrightarrow C, \\ & (x_{n} )_{n} \longmapsto T_{t} \bigl( (x_{n} )_{n} \bigr)= \bigl(e^{-t}x_{1}, e^{-2t}x_{2}, \ldots \bigr) \end{aligned} $$

for all $t\geq 0$. It easy to verify that $\mathcal{S}$ is a monotone ρ-nonexpansive semigroup and $T_{t}$ is ρ-continuous for all $t\geq 0$. As an example, we consider $p_{n}= \frac{4n^{2}}{n^{2}+1}$ for $n\geq 1$. We have $p^{-}= 2$ and $p^{+}=4$. Let $x^{0}= (x_{n}^{0} )_{n\geq 1} = (\frac{1}{2^{n}} )_{n\geq 1}$. We have $x^{0} \in C$ and $T_{t} (x^{0} ) \preceq x^{0}$ for all $t\geq 0$. Then by Theorem 3.19 there exists a common fixed point $z = 0$ such that $z \preceq x_{0}$.

The next lemma is a generalization of Lemma 3.15 for ρ-type functions defined by a given one-parameter family of mappings.

Lemma 3.21

Let ρ be a convex modular uniformly continuous and (UUCED), and let $X_{\rho }$ be a modular space satisfying property (R). Let C be a nonempty ρ-closed convex ρ-bounded subset of $X_{\rho }$, let ${\mathcal{S}}$ be a monotone ρ-nonexpansive semigroup on C, and let K be a ρ-closed convex subset of C. Fix $x_{0} \in C$ and consider the function $\varphi : C \longrightarrow \mathbb{R}_{+}$ given by

$$ \varphi (y )= \limsup_{t\to +\infty } \rho \bigl(T_{t} ( x_{0} )-y \bigr) =\inf_{s\geq 0}\sup _{t\geq s} \rho \bigl(T_{t} ( x_{0} )-y \bigr). $$

Then there exists a unique $z\in K$ such that $\varphi (z )= \inf_{y\in K} \varphi (y )$.

Proof

Fix $x_{0}\in C$. Since C is ρ-bounded, $\varphi _{0} = \inf_{y\in K}\varphi (y) <\infty$. First, assume that $\varphi _{0}>0$. Let $\varepsilon >0$. There exists $y\in K$ such that $\varphi (y)\leq \varphi _{0}+\varepsilon$. Then, for $\varepsilon =\frac{1}{n}$ with $n\geq 1$, there exists $y_{n}\in K$ such that $\varphi (y_{n})\leq \varphi _{0}+\frac{1}{n}$.

For any $n\geq 1$, set

$$ K_{n} = \biggl\lbrace y\in K : \varphi (y)\leq \varphi _{0}+ \frac{1}{n} \biggr\rbrace . $$

$( K_{n} )_{n}$ is a sequence of nonempty ρ-closed convex and ρ-bounded subsets. Indeed, for all $n\geq 1$, $K_{n}$ is ρ-closed since φ is a ρ-lower semicontinuous function. In fact, let $(y_{n} )_{n}$ in K ρ-converge to $y\in K$. Then

$$ \varphi (y )\leq \liminf_{n\to +\infty }\varphi (y_{n} ). $$

Indeed, fix $\varepsilon >0$ and $R=\operatorname{diam}_{\rho } (C )>0$. Using the uniform continuity of ρ, there exists $\eta >0$ such that

$$ \bigl\vert \rho (y)-\rho (x+y) \bigr\vert \leq \varepsilon $$

(4)

whenever $\rho (x)\leq \eta$ and $\rho (y)\leq R$. Since $(y_{n} )_{n}$ ρ-converges to y, there exists $n_{0}>0$ such that

$$ \rho (y_{n} -y) \leq \eta $$

for any $n\geq n_{0}$. Moreover, for $s\geq 0$, let $t\geq s$. As $x_{0}\in C$, then $T_{t}(x_{0}) \in C$. Thus $\rho ( T_{t} (x_{0}) -y )\leq R$. Therefore by (4)

$$ \bigl\vert \rho \bigl( T_{t} (x_{0}) -y \bigr) - \rho \bigl( T_{t} (x_{0}) -y +y - y_{n} \bigr) \bigr\vert \leq \varepsilon $$

for any $n\geq n_{0}$ and $t\geq s$. Hence

$$ \bigl\vert \rho \bigl( T_{t} (x_{0}) -y \bigr) - \rho \bigl( T_{t} (x_{0}) - y_{n} \bigr) \bigr\vert \leq \varepsilon $$

for any $n\geq n_{0}$ and $t\geq s$. In particular,

$$ \rho \bigl( T_{t} (x_{0}) -y \bigr) \leq \rho \bigl( T_{t} (x_{0}) - y_{n} \bigr) + \varepsilon $$

for any $n\geq n_{0}$ and $t\geq s$. This implies

$$ \sup_{t\geq s} \rho \bigl( T_{t} (x_{0}) -y \bigr) \leq \sup_{t \geq s}\rho \bigl( T_{t} (x_{0}) - y_{n} \bigr) + \varepsilon $$

for any $n\geq n_{0}$. Since $s\geq 0$ is arbitrary, we have

$$ \varphi (y) = \inf_{s\geq 0}\sup_{t\geq s} \rho \bigl( T_{t} (x_{0}) -y \bigr) \leq \inf _{s\geq 0}\sup_{t\geq s}\rho \bigl( T_{t} (x_{0}) - y_{n} \bigr) + \varepsilon = \varphi (y_{n})+\varepsilon $$

for any $n\geq n_{0}$. Hence

$$ \varphi (y)\leq \liminf_{n\to +\infty }\varphi (y_{n} )+ \varepsilon $$

for any $\varepsilon >0$. Consequently, ${\varphi (y)\leq \liminf_{n\to +\infty }\varphi (y_{n} )}$, that is, φ is ρ-lower semicontinuous. Then $K_{n}$ is ρ-closed for all $n\geq 1$.

For all $n\geq 1$, $K_{n}$ is convex since φ is convex. Moreover, $K_{n}$ is ρ-bounded, and the sequence $(K_{n} )_{n}$ is decreasing.

By property (R) the set ${K_{\infty }= \underset{n\geq 1}{\bigcap } K_{n}}$ is nonempty ρ-closed convex. Furthermore,

$$ K_{\infty }= \bigl\lbrace y\in K : \varphi (y)=\varphi _{0} \bigr\rbrace . $$

Indeed, if $y\in K_{\infty }$ then $y\in K_{n}$ for all $n\geq 1$. Thus $\varphi (y)\leq \varphi _{0} + \frac{1}{n}$ for all $n\geq 1$. Hence $\varphi (y) \leq \varphi _{0}$. Since $\varphi _{0} \leq \varphi (y)$, we have $\varphi (y)=\varphi _{0}$.

Next, we prove that $K_{\infty }$ is reduced to one point. Let $z_{1}$, $z_{2} \in K_{\infty }$ be such that $z_{1} \neq z_{2}$. Set $z=z_{1}+z_{2}$ and let $\varepsilon >0$. By the definition of φ there exists $s_{0}\geq 0$ such that

$$ \sup_{t\geq s_{0}}\rho \bigl(T_{t}(x_{0})-z_{i} \bigr)) \leq \varphi _{0} + \varepsilon ,\quad i=1,2. $$

Thus

$$ \rho \bigl(T_{t}(x_{0})-z_{i} \bigr) \leq \varphi _{0} + \varepsilon ,\quad i=1,2, $$

for all $t\geq s_{0}$.

Fixing $t\geq s_{0}$, we have

$$ \rho \bigl(T_{t}(x_{0})-z_{1} \bigr) \leq \varphi _{0} + \varepsilon\quad \text{and}\quad \rho \bigl(T_{t}(x_{0})-z_{1} + z_{1}-z_{2} \bigr) =\rho \bigl(T_{t}(x_{0})-z_{2} \bigr) \leq \varphi _{0} + \varepsilon . $$

Since ρ is (UUCED), there exists $\eta ( \varphi _{0} ,z )>0$ such that

$$ 1-\frac{1}{\varphi _{0}+\varepsilon }\rho \biggl(T_{t}(x_{0})-z_{1}+ \frac{z_{1}-z_{2}}{2} \biggr)\geq \delta (\varphi _{0}+ \varepsilon , z ) \geq \eta (\varphi _{0} ,z ). $$

Hence

$$ \rho \biggl(T_{t}(x_{0})-\frac{z_{1}+z_{2}}{2} \biggr) \leq \bigl(1- \eta (\varphi _{0} ,z ) \bigr) (\varphi _{0}+ \varepsilon ). $$

Since t is arbitrarily fixed such that $t\geq s_{0}$, we have

$$ \sup_{t\geq s_{0}}\rho \biggl(T_{t}(x_{0})+ \frac{z_{1}+z_{2}}{2} \biggr) \leq \bigl(1-\eta (\varphi _{0} ,z ) \bigr) (\varphi _{0}+\varepsilon ). $$

Therefore

$$ \varphi \biggl(\frac{z_{1}+z_{2}}{2} \biggr)\leq \bigl(1-\eta (\varphi _{0} ,z ) \bigr) (\varphi _{0}+ \varepsilon ). $$

As ε goes to 0⁺, we get

$$ \varphi \biggl(\frac{z_{1}+z_{2}}{2} \biggr)\leq \bigl(1-\eta (\varphi _{0} ,z ) \bigr)\varphi _{0}. $$

Since $K_{\infty }$ is convex, $\frac{z_{1}+z_{2}}{2}\in K_{\infty }$. Therefore

$$ \varphi _{0}= \varphi \biggl( \frac{z_{1}+z_{2}}{2} \biggr) \leq \bigl(1-\eta (\varphi _{0} ,z ) \bigr)\varphi _{0} < \varphi _{0}, $$

a contradiction. Then $K_{\infty }$ is reduced to one point. To finish the proof, we show that $K_{\infty }$ is reduced to one point if $\varphi =0$. For x, $y\in K$, we have

$$ \rho \biggl(\frac{x-y}{2} \biggr) \leq \frac{\varphi (x)+\varphi (y)}{2}. $$

In fact, let $s\geq 0$. Then for every $t\geq s$,

$$ \rho \biggl(\frac{x-y}{2} \biggr) \leq \rho \biggl( \frac{x-T_{t}(x_{0})}{2} \biggr)+\rho \biggl( \frac{T_{t}(x_{0})-y}{2} \biggr), $$

and then

$$ \rho \biggl(\frac{x-y}{2} \biggr) \leq \frac{\varphi (x)+\varphi (y)}{2} $$

for all x, $y\in K$. Especially, for x, $y\in K_{\infty }$,

$$ \rho \biggl(\frac{x-y}{2} \biggr) \leq \frac{\varphi (x)+\varphi (y)}{2}= \varphi _{0}=0 . $$

Thus $x=y$. In both cases, we have shown that $K_{\infty }$ is reduced to one point. As a result, φ has a unique minimum point in K. □

The next result is a generalization of Theorem 1.1 in uniformly convex in every direction modular spaces.

Theorem 3.22

Let ρ be a convex modular uniformly continuous and (UUCED), and let $X_{\rho }$ be a modular space satisfyiong property (R). Let C be a nonempty ρ-closed convex ρ-bounded subset of $X_{\rho }$. Let ${\mathcal{S}}$ be a monotone ρ-nonexpansive semigroup on C. Assume that there exists $x_{0} \in C$ such that $x_{0} \leq T_{t}(x_{0})$ (resp., $T_{t} (x_{0}) \leq x_{0}$) for all $t\geq 0$. Then there exists a common fixed point $z\in \mathit{Fix} ({\mathcal{S}} )$ such that $x_{0}\leq z$ (resp., $z\leq x_{0}$).

Proof

Without loss of generality, we assume that $x_{0} \leq T_{t}(x_{0})$ for all $t\geq 0$. Let $(s_{n} )_{n}$ be a nondecreasing sequence in $\mathbb{R}_{+}$ such that $s_{0}=0$ and ${\lim_{n}s_{n}=+\infty }$.

For all $n\geq 0$, set

$$ K_{n} = \bigcap_{t\geq s_{n}} \bigl[T_{t}(x_{0}), \rightarrow \bigr) \cap C. $$

$(K_{n} )_{n}$ is a decreasing sequence of ρ-closed convex and ρ-bounded subsets of C. In fact, for all $h\geq 0$, $x_{0} \leq T_{h}(x_{0})$. In particular, for $h=s_{n}$, we have $x_{0} \leq T_{s_{n}}(x_{0})$. Let $t\geq s_{n}$. Then $T_{t}(x_{0}) \leq T_{t+s_{n}}(x_{0})\leq T_{2s_{n}}(x_{0})=x$. Then $x=T_{2s_{n}}(x_{0})\in K_{n}$. Hence $K_{n}$ is nonempty for all $n\geq 0$.

For all $n\geq 0$, $K_{n}$ is ρ-closed. Indeed, let $(y_{p} )_{p}$ be a sequence in $K_{n}$ that ρ-converges to $y\in C$. For all $p\geq 0$, $y_{p} \in K_{n}$, that is, $T_{t}(x_{0})\leq y_{p}$ for all $t\geq s_{n}$ and $p\geq 0$. Then $y_{p} - T_{t}(x_{0}) \in P$ for all $t\geq s_{n}$ and $p\geq 0$. Since

$$ \lim_{p\to +\infty } \rho \bigl(y_{p} - T_{t}(x_{0}) -y +T_{t}(x_{0}) \bigr)=\lim_{n\to +\infty }\rho (y_{p} -y )=0 $$

and P is ρ-closed, we have $y-T_{t}(x_{0}) \in P$ for all $t\geq s_{n}$, that is, $y\in K_{n}$.

$K_{n}$ is convex and ρ-bounded, since P is convex and $K_{n} \subset C$. Moreover, $(K_{n})_{n}$ is decreasing since $(s_{n})_{n}$ is increasing.

By property (R) the set ${ K=\bigcap_{n\geq 0} K_{n}}$ is nonempty ρ-closed and convex.

K is invariant by ${\mathcal{S}}$. Indeed, let $x\in K$; then $T_{t}(x_{0})\leq x$ for all $n\geq 0$ and $t\geq s_{n}$. Letting $\eta \geq 0$, let us prove that $T_{t}(x_{0})\leq T_{\eta }(x)$ for all $t\geq s_{n}$.

Let $t\in \mathbb{N}$. If $\eta >t$, then $\varepsilon =\eta -t >0$, where $t\geq s_{n}$. Since $x_{0} \leq x$, $x_{0} \leq T_{\varepsilon }(x_{0})\leq T_{\varepsilon }(x)$. Hence $T_{t}(x_{0})\leq T_{\eta }(x)$ for all $t\geq s_{n}$. Thus $T_{\eta }(x) \in K_{n}$.

If $\eta \leq t$, then $t-\eta \geq 0$, where $t\geq s_{n}$, which implies $T_{t-\eta }(x_{0})\leq x$, because $x\in K_{0}$. Then $T_{t}(x_{0})\leq T_{\eta }(x)$. Hence $T_{\eta }(x)\in K_{n}$. In both cases, we have ${\mathcal{S}} (K )\subset K$.

Consider the function $\varphi : K \longrightarrow \mathbb{R}_{+}$ defined by

$$ \varphi (y)= \limsup_{t\to +\infty }\rho \bigl(T_{t}(x_{0})-y \bigr). $$

By Lemma 3.21φ has a unique minimum point $z\in K$.

Fix s, $\eta \geq 0$ and let $t\geq s+\eta$. As ${\mathcal{S}}$ is a monotone ρ-nonexpansive semigroup, we have

$$\begin{aligned} \rho \bigl( T_{t}(x_{0})-T_{s}(z) \bigr) & = \rho \bigl( T_{s} \bigl( T_{t-s}(x_{0}) \bigr)-T_{s}(z) \bigr) \\ & \leq \rho \bigl( T_{t-s}(x_{0})-z \bigr) \\ & \leq \sup_{\bar{t}\geq \eta } \rho \bigl(T_{\bar{t}}(x_{0})-z \bigr), \end{aligned}$$

which implies

$$ \varphi \bigl(T_{s}(y) \bigr) \leq \inf_{\eta \geq 0} \sup_{ \bar{t}\geq \eta } \rho \bigl( T_{\bar{t}}(x_{0}) - y \bigr). $$

Then $\varphi (T_{s}(z) )\leq \varphi (z)$ for all $s\geq 0$. Thus $T_{s}(z)$ is also a minimum point of φ for all $s\geq 0$. By the uniqueness of z, $T_{s}(z)=z$ for all $s\geq 0$. Therefore z is a common fixed point of ${\mathcal{S}}$. □

3.2 Convergence theorems for common fixed point of a monotone semigroup

First, we introduce the notion of uniformly asymptotic regular semigroups.

Definition 3.23

Let C be a subset of $X_{\rho }$. A semigroup ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ on C is said to be uniformly asymptotic regular (u.a.r.) if for any $s\geq 0$ and any ρ-bounded subset K of C, we have

$$ \lim_{t\to +\infty } \sup_{x\in K} \rho \bigl(T_{s} \bigl(T_{t}(x) \bigr)-T_{t}(x) \bigr)=0. $$

Example 3.24

Let $X_{\rho }= \mathbb{R}^{2}$, and let the modular ρ be defined by $\rho (x)=x_{1}^{2} +x_{2}^{2}$ for $x=(x_{1},x_{2})$ in $X_{\rho }$. Let $C= [0,A ]\times [0,A ]$, where $A>0$. Consider the one-parameter family ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ defined by

$$\begin{aligned} &T : C \longrightarrow C, \\ & x \longmapsto T_{t}(x)=e^{-t}x \end{aligned}$$

for all $t\geq 0$. It is quite easy to show that ${\mathcal{S}}$ is a semigroup. Moreover, ${\mathcal{S}}$ is u.a.r. In fact, let $s\geq 0$, and let K be a ρ-bounded subset of C. Then

$$\begin{aligned} \begin{aligned} \lim_{t\to +\infty } \sup_{x\in K} \rho \bigl(T_{s} \bigl(T_{t}(x) \bigr)-T_{t}(x) \bigr) & = \lim_{t\to +\infty } \sup_{x\in K} \rho \bigl(e^{-s} \bigl(e^{-t}x \bigr)-e^{-t}x \bigr) \\ & = \lim_{t\to +\infty } \sup_{x\in K} \bigl(e^{-t} \bigr)^{2} \bigl(e^{-s} - 1 \bigr)^{2} \bigl(x_{1}^{2}+x_{2}^{2} \bigr) =0. \end{aligned} \end{aligned}$$

Next, we give some properties of the partial order defined on the modular space $X_{\rho }$ by a ρ-closed convex cone P.

Definition 3.25

We say that a partial order ≤ is ρ-closed if for any two sequences $(x_{n} )_{n}$ and $(y_{n} )_{n}$ in $X_{\rho }$ such that $x_{n}\leq y_{n}$ for all $n\geq 0$ that ρ-converge to x and y, respectively, then $x\leq y$.

Proposition 3.26

Let ρ be a convex modular satisfying the $\Delta _{2}$-type condition. The partial order defined by a ρ-closed convex cone P ($x\leq y \Longleftrightarrow y-x\in P$ for x and y in $X_{\rho }$) is ρ-closed.

Proof

Let $(x_{n} )_{n}$ and $(y_{n} )_{n}$ be two sequences in $X_{\rho }$ that ρ-converge to x and y, respectively, such that $x_{n} \leq y_{n}$ for all $n\geq 0$.

We have $y_{n} -x_{n} \in P$ for all $n\geq 0$. Moreover, for all $n\geq 0$,

$$\begin{aligned} \rho \bigl( (y_{n} -x_{n} )- (y-x ) \bigr) & = \rho \bigl( (y_{n} -y )+ (x-x_{n} ) \bigr) \\ & \leq \frac{k}{2} \rho (y_{n} -y )+ \frac{k}{2} \rho (x_{n}-x ) \underset{n\to +\infty }{ \longrightarrow } 0. \end{aligned}$$

Hence $(y_{n} -x_{n} )_{n}$ ρ-converges to $y-x$. Since the cone P is ρ-closed, we have $y-x \in P$, which equivalent to $x\leq y$. □

Remark 3.27

The partial order “≤” defined by a ρ-closed convex cone P satisfies the following property:

$$ \text{If } (x_{n})_{n} \text{ is a nondecreasing sequence such that } x_{n} \overset{\rho }{\longrightarrow } x, \text{then } x_{n} \leq x \text{ for all }n. $$

Indeed, fix arbitrary $n_{0} \in \mathbb{N}$. Since $(x_{n})_{n}$ is a nondecreasing sequence, $x_{n_{0}} \leq x_{n}$ for all $n\geq n_{0}$, which is equivalent to $x_{n} -x_{n_{0}} \in P$. Hence

$$ x_{n} \in x_{n_{0}}+P. $$

Since P is ρ-closed, so is $x_{n_{0}}+P$. Therefore $x\in x_{n_{0}}+P$ implies $x - x_{n_{0}} \in P$. Thus $x_{n_{0}} \leq x$ for all $n_{0}\geq 0$. Hence

$$ x_{n} \leq x \quad \text{for all } n \geq 0. $$

Lemma 3.28

([4])

Let ρ be a convex modular (UUC1), and let $X_{\rho }$ be a modular space. Let $R>0$ and $(\alpha _{n} )_{n} \subset [a,b ]$ with $0< a\leq b<1$. Let $(u_{n})_{n}$ and $(v_{n})_{n}$ be two sequences in $X_{\rho }$. Assume that

$$ \textstyle\begin{cases} {\limsup_{n\to +\infty }\rho (u_{n} )} \leq R, \\ {\limsup_{n \to +\infty }\rho (v_{n} )} \leq R, \\ {\lim_{n\to +\infty }\rho (\alpha _{n} u_{n} + (1-\alpha _{n} ) v_{n} )}=R. \end{cases} $$

Then

$$ \lim_{n\to +\infty } \rho (u_{n}-v_{n} )=0. $$

We further define a new iteration algorithm for monotone ρ-nonexpansive semigroups in modular spaces. Our iteration process is defined as follows: for $x_{0} \in C$ such that $x_{0} \leq T_{s} x_{0}$ for all $s\geq 0$,

$$ (Si)\quad \textstyle\begin{cases} x_{n+1}= (1-\alpha _{n} ) T_{t_{n}}(x_{n}) + \alpha _{n} T_{t_{n}} (y_{n}), \\ y_{n} = (1-\beta _{n} )x_{n} + \beta _{n} T_{t_{n}} (x_{n}), \end{cases} $$

where $(\alpha _{n} )_{n}$ and $(\beta _{n} )_{n}$ are two sequences in $(0,1)$ such that $0< a\leq \alpha _{n}\leq b<1$ and $0< c\leq \beta _{n}\leq d<1$, and $(t_{n} )_{n} \subset \mathbb{R}_{+}$ is a nondecreasing sequence such that ${\lim_{n}} t_{n} = +\infty$ and $T_{t_{n}} (x) \leq T_{t_{n+1}} (x)$ for all $x\in C$.

The sequence $(x_{n} )_{n}$ is nondecreasing, and for all $n\geq 0$,

$$ x_{n} \leq T_{t_{n}} (x_{n}) \leq x_{n+1}\leq T_{t_{n+1}} (x_{n+1}). $$

(5)

Indeed, for $n=0$, we have $x_{0} \leq T_{t_{0}}(x_{0})$. By the convexity of the order interval $[ x_{0} , T_{t_{0}}(x_{0}) ]$ we have

$$ x_{0} \leq y_{0}\leq T_{t_{0}}(x_{0}). $$

(6)

Using the monotonicity of $T_{t_{0}}$, we have

$$ x_{0} \leq y_{0}\leq T_{t_{0}}(x_{0}) \leq T_{t_{0}}(y_{0}). $$

(7)

By the convexity of the order interval $[ T_{t_{0}}x_{0} , T_{t_{0}}y_{0} ]$ we have

$$ x_{0} \leq y_{0}\leq T_{t_{0}}(x_{0})\leq x_{1} \leq T_{t_{0}}(y_{0}). $$

(8)

Hence by the condition on $(t_{n})_{n}$ and the monotonicity of $T_{t_{1}}$ we have

$$ T_{t_{0}}(y_{0})\leq T_{t_{1}}(y_{0}) \leq T_{t_{1}}(x_{1}). $$

(9)

By (8) and (9) we have

$$ x_{0} \leq T_{t_{0}}(x_{0})\leq x_{1} \leq T_{t_{1}}(x_{1}). $$

Assume that inequality (5) is true for $n\geq 0$. Let us prove that

$$ x_{n+2} \leq T_{t_{n+2}} (x_{n+2}) \leq x_{n+3}\leq T_{t_{n+3}} (x_{n+3}). $$

(10)

We have $x_{n+1} \leq T_{t_{n+1}} (x_{n+1})$, and by the convexity of the order interval $[ x_{n+1} , T_{t_{n+1}} (x_{n+1}) ]$ we get

$$ x_{n+1} \leq y_{n+1} \leq T_{t_{n+1}} (x_{n+1}). $$

(11)

Since $T_{t_{n+1}}$ is monotone, we have

$$ x_{n+1} \leq y_{n+1} \leq T_{t_{n+1}} (x_{n+1}) \leq T_{t_{n+1}}(y_{n+1}). $$

Using the convexity of the order interval $[ T_{t_{n+1}} (x_{n+1}) , T_{t_{n+1}}(y_{n+1}) ]$ and the condition on $(t_{n} )_{n}$, we get

$$ x_{n+1} \leq T_{t_{n+1}} (x_{n+1}) \leq x_{n+2} \leq T_{t_{n+2}}(x_{n+2}). $$

(12)

By the same way we prove that

$$ x_{n+2} \leq T_{t_{n+2}}(x_{n+2}) \leq x_{n+3} \leq T_{t_{n+3}}(x_{n+3}). $$

(13)

Hence, for all $n\geq 0$,

$$ x_{n} \leq T_{t_{n}}(x_{n}) \leq x_{n+1} \leq T_{t_{n+1}}(x_{n+1}). $$

Remark 3.29

As an example of a sequence $(t_{n})_{n}$, we can consider the sequence $(2^{n}s )_{n}$ where $s>0$. In fact, for $n=0$, we show as before that $x_{0} \leq T_{s}(x_{0})\leq x_{1}\leq T_{2s}(x_{1})$. Next, we assume that $x_{n} \leq T_{2^{n} s}(x_{n})\leq x_{n+1}\leq T_{2^{n+1} s}(x_{n+1})$. As before, we get

$$ x_{n} \leq y_{n} \leq T_{2^{n} s}(x_{n}) \leq x_{n+1} \leq T_{2^{n}s}(y_{n}). $$

Moreover, $T_{2^{n}s}(y_{n}) \leq T_{2^{n}s} ( T_{2^{n}s}(y_{n}) ) = T_{2^{n}s + 2^{n} s}(y_{n}) =T_{2^{n+1}s}(y_{n})\leq T_{2^{n+1}s}(x_{n+1})$. Hence for all $n\geq 0$,

$$ x_{n} \leq y_{n} \leq T_{2^{n} s}(x_{n}) \leq x_{n+1} \leq T_{2^{n}s}(y_{n}) \leq T_{2^{n+1}s}(x_{n+1}). $$

Then for all $n\geq 0$,

$$ x_{n} \leq T_{2^{n} s}(x_{n}) \leq x_{n+1} \leq T_{2^{n+1}s}(x_{n+1}). $$

Lemma 3.30

Let ρ be a convex modular (UUC1) satisfying the $\Delta _{2}$-type condition, and let C be a ρ-closed convex ρ-bounded subset of a modular space $X_{\rho }$. Let ${\mathcal{S}}$ be a monotone ρ-nonexpansive semigroup on C, and let $x_{0} \in C$ be such that $x_{0} \leq T_{s} (x_{0})$ for all $s\geq 0$. Let $p\in {\mathcal{F}}ix ({\mathcal{S}} )$ be such that $x_{0} \leq p$. Then

$$ \lim_{n}\rho \bigl(x_{n} - T_{t_{n}}(x_{n}) \bigr)=0. $$

Proof

It obvious that $x_{n} \leq p$ and $y_{n} \leq p$ for all $n\geq 0$. Moreover,

$$\begin{aligned} \begin{aligned} \rho (x_{n+1} -p )& = \rho \bigl((1-\alpha _{n})T_{t_{n}}(x_{n}) + \alpha _{n} T_{t_{n}}(y_{n}) - p \bigr) \\ &\leq (1-\alpha _{n})\rho \bigl(T_{t_{n}}(x_{n}) -p \bigr) +\alpha _{n} \rho \bigl( T_{t_{n}}(y_{n}) - p \bigr) \\ & \leq (1-\alpha _{n})\rho (x_{n} -p ) + \alpha _{n} \rho ( y_{n} - p ), \end{aligned} \end{aligned}$$

(14)

or

$$\begin{aligned} \begin{aligned} \rho (y_{n} -p )& = \rho \bigl((1-\beta _{n})x_{n} + \beta _{n} T_{t_{n}}(x_{n}) - p \bigr) \\ &\leq (1-\beta _{n})\rho (x_{n} -p ) +\beta _{n} \rho \bigl( T_{t_{n}}(x_{n}) - p \bigr) \\ & \leq (1-\beta _{n})\rho (x_{n} -p ) + \beta _{n} \rho ( x_{n} - p ) = \rho ( x_{n} - p ). \end{aligned} \end{aligned}$$

(15)

From (14) and (15) we have $\rho ( x_{n+1} - p ) \leq \rho ( x_{n} - p )$. Hence the sequence $(\rho ( x_{n} - p ) )_{n}$ is decreasing in $\mathbb{R}_{+}$. Then ${\lim_{n}} \rho ( x_{n} - p ) =R \geq 0$ exists.

If $R=0$, then there is nothing to prove. Indeed, for all $n\geq 0$,

$$\begin{aligned} \rho (x_{n} -T_{t_{n}}x_{n} ) & \leq \frac{k}{2}\rho (x_{n} -p ) + \frac{k}{2}\rho \bigl(T_{t_{n}}(x_{n}) -p \bigr) \\ & \leq k \rho (x_{n} -p ) \underset{n\to +\infty }{\longrightarrow } 0. \end{aligned}$$

If $R >0$, then we put $u_{n}=x_{n} -p$ and $v_{n}= T_{t_{n}} (x_{n}) -p$ in Lemma 3.28. Then

$$ \limsup_{n\to +\infty }\rho (x_{n} -p ) = R $$

and

$$ \limsup_{n\to +\infty }\rho \bigl( T_{t_{n}}(x_{n}) -p \bigr) \leq \limsup_{n\to +\infty }\rho (x_{n} -p ) = R. $$

Moreover,

$$ \rho (x_{n+1} -p ) \leq (1-\alpha _{n}) \rho (x_{n} -p ) + \alpha _{n} \rho (y_{n} -p ), $$

which implies

$$ \frac{\rho (x_{n+1} -p )- \rho (x_{n} -p )}{\alpha _{n}} \leq \rho (y_{n} -p )-\rho (x_{n} -p ). $$

Since $0< a\leq \alpha _{n} \leq b<1$, we have $\frac{1}{b} \leq \frac{1}{\alpha _{n}}\leq \frac{1}{a}$. Thus by the previous inequality we have

$$ \frac{\rho (x_{n+1}-p ) - \rho ( x_{n} -p )}{a} \leq \frac{\rho (x_{n+1}-p ) - \rho ( x_{n} -p )}{\alpha _{n}} \leq \rho (y_{n} -p )-\rho (x_{n} -p ) $$

because $\rho (x_{n+1}-p )\leq \rho (x_{n}-p )$. Consequently, as n goes to infinity, we have

$$ R \leq \liminf_{n\to +\infty } \rho (y_{n} -p ). $$

(16)

Otherwise, $\rho (y_{n} -p ) \leq \rho (x_{n} -p )$, and then

$$ \limsup_{n\to +\infty } \rho (y_{n} -p ) \leq R. $$

(17)

By (16) and (17),

$$ R \leq \liminf_{n\to +\infty } \rho (y_{n} -p ) \leq \limsup_{n\to +\infty } \rho (y_{n} -p )\leq R, $$

and thus

$$ \lim_{n\to +\infty } \rho (y_{n} -p ) = R, $$

that is,

$$ \lim_{n\to +\infty } \rho \bigl((1-\alpha _{n}) (x_{n} -p) + \alpha _{n} \bigl(T_{t_{n}}(x_{n}) -p \bigr) \bigr) = R. $$

By Lemma 3.28,

$$ \lim_{n} \rho \bigl(x_{n} - T_{t_{n}} (x_{n}) \bigr)=0. $$

 □

Lemma 3.31

Let ρ be a convex modular satisfying the $\Delta _{2}$-type condition. Let C be a nonempty ρ-closed convex subset of a modular space $X_{\rho }$, and let $T : C \longrightarrow C$ be a monotone ρ-nonexpansive mapping. Suppose $(x_{n} )_{n}$ is a sequence in C such that there exists a subsequence $(x_{\varphi } (n ) )_{n}$ that ρ-converges to $x\in C$, $x_{\varphi (n)} \leq T(x_{\varphi (n)}) \leq x$ (or $x \leq T(x_{\varphi (n)}) \leq x_{\varphi (n)}$) for all integer $n\geq 0$, and

$$ \lim_{n\to +\infty } \rho \bigl(x_{\varphi (n)} - T(x_{\varphi (n)}) \bigr)=0. $$

(18)

Then x is a fixed point of T.

Proof

Without loss of generality, we assume that $x_{\varphi (n)} \leq T(x_{\varphi (n)}) \leq x$ for all $n\geq 0$. Since T is monotone ρ-nonexpansive, we have

$$ \rho \bigl(T (x_{\varphi (n)}) - T(x) \bigr) \leq \rho (x_{ \varphi (n)}-x ). $$

(19)

Hence

$$ \begin{aligned} \rho \biggl(\frac{x-T(x)}{2} \biggr) & \leq \frac{1}{2} \rho \bigl(x-T(x_{ \varphi (n)}) \bigr)+ \frac{1}{2}\rho \bigl(T(x_{\varphi (n)}) - T(x) \bigr) \\ & \leq \frac{1}{2} \rho \bigl(x-T(x_{\varphi (n)}) \bigr)+ \frac{1}{2}\rho (x_{\varphi (n)} - x ). \end{aligned} $$

(20)

Moreover,

$$ \rho \bigl(x-T(x_{\varphi (n)}) \bigr)\leq \frac{k}{2} \rho (x-x_{ \varphi (n)} ) + \frac{k}{2}\rho \bigl(x_{\varphi (n)}-T(x_{ \varphi (n)}) \bigr). $$

(21)

Therefore from (18), (20), and (21) we have

$$ \rho \biggl(\frac{x-T(x)}{2} \biggr)\leq \frac{k}{4}\rho (x-x_{ \varphi (n)} )+\frac{k}{4}\rho \bigl(x_{\varphi (n)} - T(x_{ \varphi (n)}) \bigr) + \frac{1}{2}\rho (x_{\varphi (n)} -x ) \underset{n\to +\infty }{\longrightarrow } 0. $$

Hence x is a fixed point of T. □

We further use the fixed point sets with the partial order $\mathcal{F}^{\leq }_{x} ({\mathcal{S}} )$ and $\mathcal{F}^{\geq }_{x} ({\mathcal{S}} )$ given by

$$ \mathcal{F}^{\leq }_{x} ({\mathcal{S}} )= \bigl\lbrace p \in \mathcal{F}ix ({\mathcal{S}} ) : p\leq x \bigr\rbrace \quad \text{for some } x $$

and

$$ \mathcal{F}^{\geq }_{x} ({\mathcal{S}} )= \bigl\lbrace p \in \mathcal{F}ix ({\mathcal{S}} ) : p\geq x \bigr\rbrace \quad \text{for some } x, $$

respectively. Next, we study the convergence of the iteration (Si) for monotone ρ-nonexpansive semigroups u.a.r. ${\mathcal{S}}$ in uniformly convex modular spaces.

Theorem 3.32

Let ρ be a convex modular (UUC1) satisfying the $\Delta _{2}$-type condition. Let C be a nonempty convex ρ-sequentially compact and ρ-bounded subset of a modular space $X_{\rho }$. Let ${\mathcal{S}}= \lbrace T_{t} : t\geq 0 \rbrace$ be a monotone ρ-nonexpansive u.a.r. semigroup on C. Assume that there exists $x_{0}\in C$ such that $x_{0} \leq T_{t} (x_{0})$ for all $t\geq 0$ and that $\mathcal{F}^{\geq }_{x_{0}} ({\mathcal{S}} )$ is nonempty. Then the sequence $(x_{n} )_{n}$ defined by the iterations (Si) ρ-converge to a common fixed point of the semigroup ${\mathcal{S}}$.

Proof

Fix $p\in \mathcal{F}_{x_{0}}^{\geq } ({\mathcal{S}} )$. Without loss of generality, assume that $x_{0} \leq T_{t} (x_{0})$ for all $t\geq 0$. We have $x_{n} \leq p$ for all $n\geq 0$, and by Lemma 3.30

$$ \lim_{n\to +\infty } \rho \bigl(x_{n} - T_{t_{n}}(x_{n}) \bigr)=0. $$

(22)

Let us prove that

$$ \lim_{n\to +\infty } \rho \bigl(x_{n} - T_{s} (x_{n}) \bigr) =0 \quad \text{for all } s\geq 0. $$

(23)

For all $n\geq 0$,

$$\begin{aligned}& \begin{aligned} \rho \bigl(x_{n+1} - T_{s}(x_{n+1}) \bigr) & = \rho \bigl((1- \alpha _{n})T_{t_{n}}(x_{n}) + \alpha _{n} T_{t_{n}}(y_{n}) - T_{s} (x_{n+1}) \bigr) \\ & \leq (1-\alpha _{n}) \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} (x_{n+1}) \bigr) + \alpha _{n} \rho \bigl(T_{t_{n}}(y_{n}) - T_{s} (x_{n+1}) \bigr), \end{aligned} \end{aligned}$$

(24)

$$\begin{aligned}& \begin{aligned} \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} (x_{n+1}) \bigr) & = \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} T_{t_{n}}(x_{n}) + T_{s} T_{t_{n}}(x_{n}) - T_{s} (x_{n+1}) \bigr) \\ & \leq \frac{k}{2} \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} T_{t_{n}}(x_{n}) \bigr) + \frac{k}{2} \rho \bigl( T_{s} T_{t_{n}}(x_{n}) - T_{s} (x_{n+1}) \bigr) \\ & \leq \frac{k}{2} \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} T_{t_{n}}(x_{n}) \bigr) + \frac{k}{2} \rho \bigl( T_{t_{n}}(x_{n}) - x_{n+1} \bigr). \end{aligned} \end{aligned}$$

(25)

Since ${\mathcal{S}}$ is u.a.r., for any ρ-bounded subset K of C, we have

$$ \lim_{n\to +\infty } \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} T_{t_{n}}(x_{n}) \bigr) \leq \lim _{n\to +\infty } \sup_{x\in K} \rho \bigl(T_{t_{n}}(x) - T_{s} T_{t_{n}}(x) \bigr) = 0. $$

Hence

$$ {\lim_{n\to +\infty } \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} T_{t_{n}}(x_{n}) \bigr)} = 0. $$

(26)

Moreover, for all $n\geq 0$,

$$\begin{aligned} \rho \bigl( T_{t_{n}}(x_{n}) - x_{n+1} \bigr) & = \rho \bigl( (1- \alpha _{n}) T_{t_{n}}(x_{n}) + \alpha _{n} T_{t_{n}}(y_{n}) - T_{t_{n}}(x_{n}) \bigr) \\ & \leq \alpha _{n} \rho \bigl(T_{t_{n}}(y_{n}) - T_{t_{n}}(x_{n}) \bigr) \\ & \leq \alpha _{n} \rho (y_{n} - x_{n} ) \\ & \leq \alpha _{n} \rho \bigl((1-\beta _{n})x_{n} + \beta _{n} T_{t_{n}}(x_{n}) - x_{n} \bigr) \\ & \leq \alpha _{n} \beta _{n} \rho \bigl( T_{t_{n}}(x_{n}) - x_{n} \bigr). \end{aligned}$$

From the hypothesis on $(\alpha _{n})_{n}$ and $(\beta _{n})_{n}$ and (22) we get

$$ \lim_{n} \rho \bigl(x_{n+1} - T_{t_{n}}(x_{n}) \bigr) = 0. $$

(27)

Using (25), (26), and (27), we have

$$ \lim_{n} \rho \bigl(T_{t_{n}}(x_{n}) - T_{s} (x_{n+1}) \bigr)=0. $$

(28)

Otherwise,

$$ \begin{aligned} \rho \bigl(T_{t_{n}}(y_{n}) -T_{s} (x_{n+1}) \bigr) & = \rho \bigl(T_{t_{n}}(y_{n}) - T_{s} T_{t_{n}} (y_{n}) + T_{s} T_{t_{n}} (y_{n}) - T_{s} (x_{n+1}) \bigr) \\ & \leq \frac{k}{2} \rho \bigl(T_{t_{n}}(y_{n}) - T_{s} T_{t_{n}} (y_{n}) \bigr) + \frac{k}{2} \rho \bigl( T_{s} T_{t_{n}}( y_{n}) - T_{s} (x_{n+1}) \bigr) \\ & \leq \frac{k}{2} \rho \bigl(T_{t_{n}}(y_{n}) - T_{s} T_{t_{n}} (y_{n}) \bigr) + \frac{k}{2} \rho \bigl( T_{t_{n}} (y_{n}) - x_{n+1} \bigr). \end{aligned} $$

(29)

Since ${\mathcal{S}}$ is u.a.r., we have

$$ \lim_{n} \rho \bigl(T_{t_{n}}(y_{n}) - T_{s} T_{t_{n}} (y_{n} ) \bigr) = 0. $$

(30)

Moreover,

$$\begin{aligned} \rho \bigl( T_{t_{n}} (y_{n}) - x_{n+1} \bigr) & = \rho \bigl( T_{t_{n}} (y_{n}) - (1-\alpha _{n}) T_{t_{n}}(x_{n}) - \alpha _{n} T_{t_{n}}(y_{n}) \bigr) \\ & \leq (1-\alpha _{n})\rho (x_{n} - y_{n} ) \\ & \leq (1-\alpha _{n}) (1-\beta _{n}) \rho \bigl(x_{n} -T_{t_{n}}(x_{n}) \bigr). \end{aligned}$$

Therefore

$$ \lim_{n} \rho \bigl( T_{t_{n}} (y_{n}) - x_{n+1} \bigr) = 0. $$

(31)

By (29), (30), and (30)

$$ \lim_{n} \rho \bigl(T_{t_{n}}(y_{n}) -T_{s} (x_{n+1}) \bigr) =0. $$

(32)

From (24), (28), and (32) we have

$$ \lim_{n} \rho \bigl( x_{n+1} - T_{s} (x_{n+1}) \bigr)= 0. $$

(33)

Since C is ρ-sequentially compact, $(x_{n})_{n}$ has a subsequence $(x_{\varphi (n)})_{n}$ ρ-converging to a point $x\in C$ such that $x_{\varphi (n)} \leq x$. Moreover, by (33)

$$ \lim_{n} \rho \bigl( x_{\varphi ( n)} - T_{s} (x_{\varphi ( n)}) \bigr)= 0. $$

Hence by Lemma 3.31x is a fixed point of $T_{s}$ for all $s\geq 0$. Then x is a common fixed point of the semigroup ${\mathcal{S}}$.

To complete the proof, we prove that $(x_{n})_{n}$ ρ-converges to x.

Let $(x_{\psi (n)})_{n}$ be another subsequence of $(x_{n})_{n}$ that ρ-converges to y. For each $\varphi (n)$, there exists a large enough $\psi (n)$ such that $x_{\varphi (n)} \leq x_{\psi (n)}$. Then by Proposition 3.26 we have $x\leq y$. In the same way, we get $y\leq x$. Therefore $x=y$.

Hence the sequence $(x_{n})_{n}$ has a unique cluster point x, and since C is ρ-sequentially compact, $(x_{n}) _{n}$ ρ-converges to x. □

Example 3.33

Let the space $\mathbb{R}$ be equipped with the convex modular $\rho (x )=\vert x\vert ^{2}$ for $x\in \mathbb{R}$. It is quite easy to see that ρ is (UUC1) and satisfies the Fatou property and $\Delta _{2}$-type condition. Consider the usual partial ordering defined on $\mathbb{R}$, that is, $x\preceq y$ if and only if $y-x\in [0,\infty [$. Let $C= [0,1 ]$ be a ρ-sequentially compact, ρ-bounded, and convex subset of $\mathbb{R}$.

Let the family $\mathcal{S}= \lbrace T_{t} : t\geq 0 \rbrace$ be such that

$$\begin{aligned} &T_{t} : C \longrightarrow C, \\ & x \longmapsto T_{t}(x)= f \bigl(5^{-t}f^{-1} (x ) \bigr), \end{aligned}$$

where $f (w )=1-w$ for all $w\in C$. It easy to verify that $\mathcal{S}$ is a monotone ρ-nonexpansive semigroup and uniformly asymptotic regular (u.a.r.).

Let $x_{0}= 0 \in C$. We have $x_{0} \preceq T_{t}(x_{0})=1-5^{-t}$ for all $t\geq 0$. Moreover, $\mathcal{F}^{\geq }_{x_{0}} ({\mathcal{S}} )= \lbrace 1 \rbrace$ is nonempty. Let $\alpha _{n} = \alpha \in (0,1)$, $\beta _{n}=\beta \in (0,1)$, and $t_{n}=2^{n}$ for all $n\geq 0$. By induction on n we construct the sequence $(x_{n} )_{n\geq 0}$ given as follows:

$$ x_{n+1}= \biggl(1-\frac{1}{5^{2^{n}}} \biggr) \biggl(1+ \frac{\alpha \beta }{5^{2^{n}}} \biggr) + \frac{x_{n}}{5^{2^{n}}} \biggl(1-\alpha \beta \biggl(1-\frac{1}{5^{2^{n}}} \biggr) \biggr) $$

(34)

for $n\geq 0$. In fact, for $n=0$, we have $x_{0}=0$ and $T_{t_{0}}(x_{0})=1-\frac{1}{5}$. Then, using the iteration $(Si)$, we get $y_{0} = (1-\frac{1}{5} )\beta$. Thus $x_{1}= (1-\frac{1}{5} ) (1+\frac{\alpha \beta }{5} )$. Assume that (34) is true until the order n. Let us prove that

$$ x_{n+2}= \biggl(1-\frac{1}{5^{2^{n+1}}} \biggr) \biggl(1+ \frac{\alpha \beta }{5^{2^{n+1}}} \biggr) + \frac{x_{n+1}}{5^{2^{n+1}}} \biggl(1-\alpha \beta \biggl(1- \frac{1}{5^{2^{n+1}}} \biggr) \biggr). $$

Using the iteration $(Si)$, we obtain $y_{n+1}= \beta (1-\frac{1}{5^{2^{n+1}}} )+ (1- \beta (1-\frac{1}{5^{2^{n+1}}} ) )x_{n+1}$. Thus

$$\begin{aligned} x_{n+2} & = \alpha T_{t_{n+1}}(y_{n+1}) + (1-\alpha ) T_{t_{n+1}}(x_{n+1}) \\ & = \alpha \biggl( 1-\frac{1}{5^{2^{n+1}}} ( 1-y_{n+1} ) \biggr) + (1-\alpha ) \biggl( 1-\frac{1}{5^{2^{n+1}}} ( 1-x_{n+1} ) \biggr) \\ & = \biggl(1-\frac{1}{5^{2^{n+1}}} \biggr)+ \frac{x_{n+1}}{5^{2^{n+1}}}+ \frac{\alpha }{5^{2^{n+1}}} (y_{n+1} - x_{n+1} ) \\ & = \biggl(1-\frac{1}{5^{2^{n+1}}} \biggr)+ \frac{x_{n+1}}{5^{2^{n+1}}}+ \frac{\alpha \beta }{5^{2^{n+1}}} (T_{t_{n+1}}x_{n+1} - x_{n+1} ) \\ & = \biggl(1-\frac{1}{5^{2^{n+1}}} \biggr) \biggl(1+ \frac{\alpha \beta }{5^{2^{n+1}}} \biggr) + \frac{x_{n+1}}{5^{2^{n+1}}} \biggl(1-\alpha \beta \biggl(1- \frac{1}{5^{2^{n+1}}} \biggr) \biggr). \end{aligned}$$

Therefore by Theorem 3.32 the sequence $(x_{n} )_{n}$ ρ-converges to 1.

4 Conclusion

We have established some existence results for monotone ρ-nonexpansive semigroups in modular spaces. Then we proposed an iteration scheme with some convergence results for the class of uniformly asymptotic regular monotone ρ-nonexpansive semigroups. Our results of existence are generalizations of several results mentioned in the introduction and the reference sections of this paper.