Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 66, pp. 1–14.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

FIXED POINT THEOREM FOR MIXED MONOTONE NEARLY

ASYMPTOTICALLY NONEXPANSIVE MAPPINGS AND

APPLICATIONS TO INTEGRAL EQUATIONS

HAMZA EL BAZI, ABDELLATIF SADRATI

Abstract. This work concerns the existence of a fixed point for mixed mono-
tone nearly asymptotically nonexpansive mappings. We extend and generalize

some well-known results concerning nearly asymptotically nonexpansive map-

pings in a uniformly convex hyperbolic metric space. As application of our
results, we study the existence of solutions for an integral equation.

1. Introduction

The mixed monotone operators were introduced by Guo and Lakshmikantham
in 1987, and have been applied in many areas since then; see, e.g., [8, 17, 22, 23]. In
recent years, new generalizations of nonexpansive mappings have been discovered
and their fixed point theory has been studied by many authors. Some interesting
generalizations of nonexpansive mappings can be found in [3, 7, 12, 13, 14, 15, 19].
Recall that the theory of fixed point of nonexpansive mapping extends the classical
theory of successive approximations for strict contractions.

Recently, a new direction has been developed when the nonexpansive mapping
is monotone and defined in partially ordered hyperbolic metric spaces. One can see
for example some various work containing those new results in [1, 6, 18] and the
references therein.

Our goal in the present work is twofold. First, we introduce a new class of
mixed monotone defined as mixed monotone nearly asymptotically nonexpansive
mappings and we extend to it the fixed point results for monotone nearly asymptot-
ically nonexpansive mappings, obtained by Agarwal et al. in [3]. Second, we apply
our result to prove the existence of solutions for a nonlinear integral equation.

This article is organized as follows. In section 2, we give definitions and basic
results which will be used later. In Section 3, we study the existence of fixed
point for mixed monotone mapping that is nearly asymptotically nonexpansive. In
section 4, we illustrate our results with an application.

2020 Mathematics Subject Classification. 47H05, 47H09, 47H10, 45G10.
Key words and phrases. Fixed point theorem; hyperbolic metric space; integral equation

mixed monotone operator; monotone nearly asymptotically nonexpansive operator.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted April 6, 2022. Published September 26, 2022.

1



2 H. EL BAZI, A. SADRATI EJDE-2022/66

2. Definitions and preliminaries

The following notation will be used. R is the set of real numbers, R+ is the set
of nonnegative real numbers and N is the set of nonnegative integers. Let (Ω, d)
be a metric space endowed with a partial order �. We will say that (Ω, d,�) is a
partial ordered metric space and we will say that x, y ∈ Ω are comparable whenever
x � y or y � x. When referring to the partial order on the product space Ω × Ω,
we understand the following partial order

(x, y), (u, v) ∈ Ω× Ω, (x, y) - (u, v) ⇔ x � u, y � v.
A metric space

(
Ω, d

)
is a convex metric space if any two points x, y ∈ Ω are

endpoints of a unique metric segment [x, y], we shall denote by w = αx⊕ (1− α)y
the unique point of [x, y] which satisfies

d(x,w) = (1− α)d(x, y) and d(y, w) = αd(x, y), for α ∈ [0, 1].

From the definition of convex metric space, we have (i) 0x⊕1y = y, (ii) 1x⊕0y = x,
(iii) αx⊕ (1− α)x = x.

A convex metric space
(
Ω, d

)
is hyperbolic if

d(αx⊕ (1− α)y, αu⊕ (1− α)v) ≤ αd(x, u) + (1− α)d(y, v),

for all x, y, u, v ∈ Ω and α ∈ [0, 1].
Let (Ω, d) be a convex metric space. Consider the following metric in Ω× Ω

∂
(
(x, y), (u, v)

)
=
(
d2(x, u) + d2(y, v)

)1/2
, for every (x, y), (u, v) ∈ Ω× Ω.

And for all (x, y), (u, v) ∈ Ω× Ω, we have

∂
((
αx⊕ (1− α)u, αy ⊕ (1− α)v

)
, (x, y)

)
=
(
d2
(
αx⊕ (1− α)u, x

)
+ d2

(
αy ⊕ (1− α)v, y

))1/2
=
(

(1− α)2d2
(
u, x

)
+ (1− α)2d2

(
v, y
))1/2

= (1− α)∂
(
(x, y), (u, v)

)
.

Similarly, we have ∂
((
αx⊕ (1− α)u, αy ⊕ (1− α)v

)
, (u, v)

)
= α∂

(
(x, y), (u, v)

)
.

Thus, Ω× Ω is a convex metric space, and for every (x, y), (u, v) ∈ Ω× Ω

α(x, y)⊕ (1− α)(u, v)
)

=
(
αx⊕ (1− α)u, αy ⊕ (1− α)v

)
.

Definition 2.1 ([10]). Let
(
Ω, d

)
be a hyperbolic metric space. If for any z ∈ Ω,

r > 0 and ε > 0,

δ(r, ε) = inf
{

1− 1

r
d
(
z,

1

2
x⊕ 1

2
y
)

: d(x, z) ≤ r, d(y, z) ≤ r, d(x, y) ≥ rε
}
> 0,

then
(
Ω, d

)
is said to be uniformly convex hyperbolic metric space.

Definition 2.2. Let
(
Ω, d,�

)
be a partial order metric space. A mapping A :

Ω×Ω→ Ω is said to be mixed monotone if A is nondecreasing in the first argument
and nonincreasing in the second argument, i.e.,

x1, x2, y ∈ Ω; x1 � x2 =⇒ A(x1, y) � A(x2, y),

x, y1, y2 ∈ Ω; y1 � y2 =⇒ A(x, y2) � A(x, y1).

The following theorem is a metric version of the parallelogram identity.



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 3

Theorem 2.3 ([10]). Let (Ω, d) be a hyperbolic uniformly convex metric space. For
any a ∈ Ω, for each r > 0 and for each ε > 0, set

Ψ(r, ε) = inf
{1

2
d2(a, x) +

1

2
d2(a, y)− d2(a,

1

2
x⊕ 1

2
y)
}
, (2.1)

where the infimum is taken over all x, y ∈ Ω such that d(a, x) ≤ r, d(a, y) ≤ r and
d(x, y) ≥ rε. Then Ψ(r, ε) > 0 for any r > 0 and each ε > 0. Moreover, for a fixed
r > 0, we have

(i) Ψ(r, 0) = 0;
(ii) Ψ(r, ε) is a nondecreasing function of ε;

(iii) If limn→+∞Ψ(r, tn) = 0, then limn→+∞ tn = 0.

Recall that the result below is one of the important properties of complete uni-
formly convex hyperbolic metric space and known as property (R).

Proposition 2.4 ([10]). Let
(
Ω, d

)
be a uniformly convex hyperbolic metric space

and {Cn}n∈N be a decreasing sequence of nonempty, closed, bounded and convex
subsets of Ω. Then, ∩n∈NCn 6= ∅.

Through this article, we will assume that all order intervals are closed and convex.
It is worth noting that an order interval is any of the subsets

[a,→) = {x ∈ Ω : a � x}, (←, b] = {x ∈ Ω : x � b}, [a, b] = [a,→) ∩ (←, b],
for every a, b ∈ Ω.

3. Fixed point theorem

In this section, we give sufficient conditions so that a mixed monotone nearly
asymptotically nonexpansive mapping has fixed point. Similarly to the definition
of monotone nearly asymptotically nonexpansive mapping [2], we introduce the
following definition for mixed monotone mapping.

Definition 3.1. Let (Ω, d,�) be a partially ordered metric space, A : Ω× Ω→ Ω
be a map and {an} be a fixed sequence in [0,+∞) with an → 0. Then the map A
is said to be mixed monotone nearly Lipschitzian mapping with respect to an if A
is mixed monotone and for each n ∈ N∗, there exists a constant Kn ≥ 0, such that

d (An(x, y), An(u, v)) ≤ Kn

(1

2
d(x, u) +

1

2
d(y, v) + an

)
, (3.1)

for every comparable elements (x, y), (u, v) ∈ Ω × Ω, for any n ∈ N A0(x, y) = x
and An+1(x, y) = A

(
An(x, y), An(y, x)

)
for all x, y ∈ Ω. The infimum of constants

Kn for which the last inequality (3.1) holds is denoted by η(An) and called the
nearly Lipschitz constant. The mixed monotone nearly Lipschitzian mapping A
with sequence {an, η(An)} is said to be mixed monotone nearly asymptotically
nonexpansive if

(i) η(An) ≥ 1 for all n ∈ N∗ and
(ii) limn→∞ η(An) = 1.

Recall that a map T : Ω → Ω is said to be monotone nearly asymptotically
nonexpansive if it is nondecreasing and for any n ≥ 0 there exists Kn ≥ 1 and
an ≥ 0, such that

(i) limn→∞Kn = 1 and
(ii) limn→∞ an = 0.



4 H. EL BAZI, A. SADRATI EJDE-2022/66

And for every comparable element x, y ∈ Ω, for any n ≥ 0, we have

d (Tn(x), Tn(y)) ≤ Kn (d(x, y) + an) . (3.2)

Proposition 3.2. Let (Ω, d) be a uniformly convex hyperbolic metric space. If
{Cn} is a decreasing sequence of bounded, nonempty, closed and convex subsets of(
Ω× Ω, ∂

)
, then ∩n∈NCn 6= ∅.

Proof. Let P1 : Ω × Ω → Ω defined by P1(x, y) = x for every (x, y) ∈ Ω × Ω.
Since for each n ≥ 0, Cn is a convex subset in Ω × Ω, it follows that for every
(x, y), (u, v) ∈ Cn, for each α ∈ [0, 1] and n ≥ 0,

αP1(x, y)⊕ (1− α)P1(u, v) = αx⊕ (1− α)u

= P1

(
αx⊕ (1− α)u, αy ⊕ (1− α)v

)
= P1

(
α(x, y)⊕ (1− α)(u, v)

)
,

which implies that αP1(x, y) ⊕ (1 − α)P1(u, v) ∈ P1(Cn). Thus P1(Cn) is convex.
Moreover, it is clear that if {Cn} is a decreasing sequence of bounded, nonempty
and closed subsets of Ω × Ω then {P1(Cn)} is a decreasing sequence of bounded,
nonempty and closed subsets of Ω. Hence, ∩n≥0P1(Cn) 6= ∅, therefore for any
n ∈ N there exist yn ∈ Ω such that (x, yn) ∈ Cn, since {Cn} is decreasing,
closed and convex subsets, then for any n ∈ N, {x} × conv{yk : k ≥ n} ⊂ Cn,
with conv{yk : k ≥ n} is the convex envelope of {yk : k ≥ n}. Thus for any
n ∈ N, {x} × ∩n≥0conv{yk : k ≥ n} ⊂ ∩n∈NCn. Since {conv{yk : k ≥ n}} is a
decreasing sequence of bounded, nonempty, closed and convex subsets of Ω, then
∩n≥0conv{yk : k ≥ n} 6= ∅, therefore ∩n∈NCn 6= ∅. �

Lemma 3.3. Let C and D be a nonempty, closed and convex subsets of a uniformly
convex hyperbolic metric space (Ω, d). Let τ : C ×D → [0,+∞) be a type function,
i.e., there exist bounded sequences {xn}, {yn} ∈ Ω such that

τ(x, y) = lim sup
n→∞

[d2(xn, x) + d2(yn, y)]1/2, (x, y) ∈ C ×D. (3.3)

Then τ is continuous, and since Ω is hyperbolic, τ is convex, i.e., the subset {(x, y) ∈
C × D : τ(x, y) ≤ r} is convex for any r ≥ 0. Moreover, there exists a unique
minimum point (a, b) ∈ C ×D such that

τ(a, b) = inf{τ(x, y); (x, y) ∈ C ×D} = τ0.

In addition, if {(an, bn)} is a minimizing sequence in C×D, i.e., limn→∞ τ(an, bn) =
τ(a, b), then {an} converges to a and {bn} converges to b.

Proof. The continuity of τ is obvious. Let us prove the convexity of τ . Since (Ω, d)
is hyperbolic, and using the discrete Minkowski inequality, we have for any α ∈ [0, 1]
and all (x, y), (u, v) ∈ C ×D,

τ
(
α(x, y)⊕ (1− α)(u, v)

)
= τ(αx⊕ (1− α)u, αy ⊕ (1− α)v)

= lim sup
n→∞

[d2(xn, αx⊕ (1− α)u) + d2(yn, αy ⊕ (1− α)v)]1/2

≤ lim sup
n→∞

[(αd(xn, x) + (1− α)d(xn, u))2 + (αd(yn, y) + (1− α)d(yn, v))2]1/2

≤ lim sup
n→∞

(
[(α2d2(xn, x) + (α2d2(yn, y)]1/2



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 5

+ [(1− α)2d2(xn, u) + (1− α)2d2(yn, v)]1/2
)

≤ ατ(x, y) + (1− α)τ(u, v),

which implies that τ is convex.
Now, denote τ0 = inf{τ(x, y) : (x, y) ∈ C ×D}. Then for any n ≥ 1, the subset

Cn = {(x, y) ∈ C ×D : τ(x, y) ≤ τ0 + 1
n} is nonempty, closed, bounded and convex

subset of C ×D. From proposition 3.2, we have C∞ = ∩n≥1Cn 6= ∅. Obviously we
have

C∞ = {(u, v) ∈ C ×D : τ(u, v) = τ0}.
To prove that C∞ is reduced to one point (a, b), let (a1, b1) and (a2, b2) be in C∞. If
we assume that (a1, b1) 6= (a2, b2), then we must have τ0 > 0. From the definition of
τ , we obtain that for any α ∈ (0, τ0), there exists n0 ≥ 1 such that for any n ≥ n0,

d(xn, a1) ≤ τ0 + α and d(xn, a2) ≤ τ0 + α.

Since d(a1, a2) ≥ (τ0 + α)
d(a1, a2)

2τ0
, by Theorem 2.3 we have

d2(xn,
1

2
a1 ⊕

1

2
a2) ≤ 1

2
d2(xn, a1) +

1

2
d2(xn, a2)−Ψ(τ0 + α,

1

2τ0
d(a1, a2)).

Analogously, we obtain

d2(yn,
1

2
b1 ⊕

1

2
b2) ≤ 1

2
d2(yn, b1) +

1

2
d2(yn, b2)−Ψ(τ0 + α,

1

2τ0
d(b1, b2)).

It follows that,

d2(xn,
1

2
a1 ⊕

1

2
a2) + d2(yn,

1

2
b1 ⊕

1

2
b2)

≤ 1

2
[d2(xn, a1) + d2(yn, b1)] +

1

2
[d2(xn, a2) + d2(yn, b2)]

− [Ψ(τ0 + α,
1

2τ0
d(a1, a2)) + Ψ(τ0 + α,

1

2τ0
d(b1, b2))].

Hence,

τ20 ≤ τ20 − [Ψ(τ0 + α,
1

2τ0
d(a1, a2)) + Ψ(τ0 + α,

1

2τ0
d(b1, b2))].

Therefore, [Ψ(τ0 + α, 1
2τ0
d(a1, a2)) + Ψ(τ0 + α, 1

2τ0
d(b1, b2))] = 0. Thus, by the

property (iii) in Theorem 2.3 we obtain a1 = a2 and b1 = b2.
Next, let {(an, bn)} ⊂ C × D be a minimizing sequence of τ . Since {xn} and

{yn} are bounded sequences, then {an} and {bn} are also bounded. Therefore,
there exists R > 0 such that

max{d(xn, ak), d(yn, bk), d(xn, a), d(yn, b)} ≤ R,
for every n, k ∈ N.

Once again, by Theorem 2.3 we have

d2(xn,
1

2
ak ⊕

1

2
a) ≤ 1

2
d2(xn, ak) +

1

2
d2(xn, a)−Ψ(R,

1

R
d(ak, a)),

d2(yn,
1

2
bk ⊕

1

2
b) ≤ 1

2
d2(yn, bk) +

1

2
d2(yn, b)−Ψ(R,

1

R
d(bk, b)).

When n goes to infinity, taking the limit-sup, we obtain

τ2
(1

2
ak ⊕

1

2
a,

1

2
bk ⊕

1

2
b
)



6 H. EL BAZI, A. SADRATI EJDE-2022/66

≤ 1

2

(
τ2(ak, bk) + τ2(a, b)

)
−Ψ(R,

1

R
d(ak, a))−Ψ(R,

1

R
d(bk, b)),

for every k ∈ N, which implies

Ψ(R,
1

R
d(ak, a)) + Ψ(R,

1

R
d(bk, b)) ≤

1

2
(τ2(ak, bk) + τ2(a, b))− τ20 .

Consequently, limk→∞Ψ(R, 1
Rd(ak, a)) = 0 and limk→∞Ψ(R, 1

Rd(bk, b)) = 0. The
property (iii) in Theorem 2.3 gives limk→∞ d(ak, a) = limk→∞ d(bk, b) = 0. �

The proof of the following lemma is similar to the previous proof.

Lemma 3.4. Let C be a nonempty, closed and convex subset of a uniformly convex
hyperbolic metric space (Ω, d). Let τ : C → [0,+∞) be a pseudo-type function, i.e.,
there exist bounded sequences {xn}, {yn} ∈ Ω such that

τ(x) = lim sup
n→∞

[d2(xn, x) + d2(yn, x)]1/2, x ∈ C. (3.4)

Then τ is continuous, and since Ω is hyperbolic, τ is convex, i.e., the subset {x ∈
C : τ(x) ≤ r} is convex for any r ≥ 0. Moreover, there exists a unique minimum
point a ∈ C such that

τ(a) = inf{τ(x);x ∈ C} = τ0.

In addition, if {an} is a minimizing sequence in C, i.e., limn→∞ τ(an) = τ(a), then
{an} converges to a.

Theorem 3.5. Let (Ω, d,�) be a complete uniformly convex partially ordered hy-
perbolic metric space, and C be a nonempty, convex, bounded and closed subset
of Ω. Let T : C × C → C × C be a continuous monotone nearly asymptotically
nonexpansive mapping. If there exist (x0, x

0) ∈ C × C such that

(x0, x
0) - T (x0, x

0), (3.5)

then T has a fixed point (x∗, y∗) ∈ C2, that is, T (x∗, y∗) = (x∗, y∗).

Proof. From (3.5) and the monotonicity of T we have Tn(x0, x
0) - Tn+1(x0, x

0),
for all n ∈ N. By the convexity of C and proposition 3.2, we conclude that

C∞ = ∩
(

[Tn(x0, x
0),→) ∩ C × C

)
6= ∅.

Let (x, y) ∈ C∞, then Tn+1(x0, x
0) - T (x, y), for all n ≥ 1. It follows that

Tn(C∞) ⊂ C∞. Consider the type function τ : C∞ → [0,+∞) generated by
the sequences {Tn1 (x0, x

0)} and {Tn2 (x0, x
0)}, with Tni (x0, x

0) = Pi
(
Tn(x0, x

0)
)
.

Using Lemma 3.3, there exists a unique minimum point (x∗, y∗) ∈ C∞, that is,
τ(x∗, y∗) = inf{τ(x, y); (x, y) ∈ C∞}. In addition, we have T p(x∗, y∗) ∈ C∞ for all
p ∈ N.

The same reasoning as in the proof of [2, Theorem 3.1] gives T (x∗, y∗) = (x∗, y∗).
�

Note that the previous theorem is a generalization to Ω×Ω of [2, theorem 3-1].
In addition the next consequence prove the existence of a coupled fixed point of a
continuous mixed monotone nearly asymptotically nonexpansive mapping.

Corollary 3.6. Let (Ω, d,�) be a complete uniformly convex partially ordered hy-
perbolic metric space, and C be a nonempty, convex, bounded and closed subset of



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 7

Ω. Let A : C × C → C be a continuous mixed monotone nearly asymptotically
nonexpansive mapping. If there exist x0, x

0 ∈ C such that

x0 � A(x0, x
0) and A(x0, x0) � x0, (3.6)

then A has a coupled fixed point (x∗, y∗) ∈ C2, that is, A(x∗, y∗) = x∗ and
A(y∗, x∗) = y∗.

Proof. We shall prove that the hypotheses of the previous theorem are verified with
the operator T defined by T (x, y) = (A(x, y), A(y, x)), for all (x, y) ∈ C × C.

Let (x, y) - (u, v) in C × C i.e., x � u and v � y. Using the mixed monotone
property of A, we show easily that T (x, y) - T (u, v).

Now, for any two comparable elements (x, y) , (u, v) in C × C, we have

∂
(
Tn(x, y), Tn(u, v)

)
=
(
d2
(
An(x, y), An(u, v)

)
+ d2

(
An(y, x), An(v, u)

))1/2
≤
(
K2
n

(1

2
d(x, u) +

1

2
d(y, v) + an

)2
+K2

n

(1

2
d(x, u) +

1

2
d(y, v) + an

)2)1/2
≤ Kn

(
2
(1

2
d(x, u) +

1

2
d(y, v) + an

)2)1/2
≤ Kn

((
d2(x, u) + d2(y, v)

)1/2
+
√

2an

)
.

Since A is continuous,
√

2an → 0 and Kn → 1, when n → +∞. Thus, T is a
continuous monotone nearly asymptotically nonexpansive mapping with respect to√

2an. By Theorem 3.5, T has a fixed point (x∗, y∗) ∈ C × C, i.e., T (x∗, y∗) =(
A(x∗, y∗), A(y∗, x∗)

)
= (x∗, y∗), that is, A(x∗, y∗) = x∗ and A(y∗, x∗) = y∗. �

Theorem 3.7. Assume that all hypotheses of corollary 3.6 hold. If

x0 � A(x0, x
0) � A(x0, x0) � x0, (3.7)

then A has a fixed point x∗ in C with x0 � x∗ � x0.

Proof. We construct successively the iterative sequences

xn = An(x0, x
0) and xn = An(x0, x0).

Since x0 � x0, according to the inequality (3.7) and the mixed monotonicity of A,
it is easy to show by induction that for any n ∈ N

x0 � x1 � · · · � xn � · · · � xn � · · · � x1 � x0.
By the convexity of C and Proposition 2.4, we conclude that

C0 = ∩n≥0[xn, x
n] 6= ∅.

Let x ∈ C0, i.e., for all n ∈ N, xn � x � xn, then A(xn, x
n) � A(x, x) � A(xn, xn),

and hence xn+1 � A(x, x) � xn+1, for all n ≥ 0. It follows that Ap(x, x) ∈ C0

for all x ∈ C0 and for any p ∈ N. Now we consider the pseudo-type function
τ : C0 → [0,+∞) generated by the sequences {An(x0, x

0)} and {An(x0, x0)}.
Using Lemma 3.4, there exists a unique minimum point x∗ ∈ C0, that is, τ(x∗) =

inf{τ(x);x ∈ C0}. By Definition 3.1 and the discrete Minkowski inequality, we have

τ (Ap(x∗, x∗))

= lim sup
n→∞

[
d2
(
An(x0, x

0), Ap(x∗, x∗)
)

+ d2
(
An(x0, x0), Ap(x∗, x∗)

)]1/2



8 H. EL BAZI, A. SADRATI EJDE-2022/66

≤ η(Ap) lim sup
n→∞

[(1

2
d
(
An(x0, x

0), x∗
)

+
1

2
d
(
An(x0, x0), x∗

)
+ ap

)2
+
(1

2
d
(
An(x0, x0), x∗

)
+

1

2
d
(
An(x0, x

0), x∗
)

+ ap

)2]1/2
≤ η(Ap) lim sup

n→∞

[
d2
(
An(x0, x0), x∗

)
+ d2

(
An(x0, x

0), x∗
)]1/2

+
√

2η(Ap)ap.

Since τ(x∗) is minimum,

τ(x∗) ≤ τ (Ap(x∗, x∗)) ≤ η(Ap)τ(x∗) +
√

2η(Ap)ap.

Also, from Definition 3.1, we obtain

lim
p→∞

τ (Ap(x∗, x∗)) = τ(x∗),

which gives by Lemma 3.4 that limp→∞Ap(x∗, x∗) = x∗. Using the continuity of
A,

τ(x∗) = τ
(
A
(

lim
p→∞

Ap−1(x∗, x∗), lim
p→∞

Ap−1(x∗, x∗)
))

= τ
(
A(x∗, x∗)

)
.

By the uniqueness of minimum point, we obtain A(x∗, x∗) = x∗. Thus, x∗ is fixed
point of A. Obviously x∗ ∈ C0 ⊂ [x0, x

0], hence x0 � x∗ � x0. �

As a consequence of the previous theorem, we obtain in the following the exis-
tence of a fixed point for a continuous nonincreasing nearly asymptotically nonex-
pansive mapping.

Corollary 3.8. Let (Ω, d,�) be a complete uniformly convex partially ordered hy-
perbolic metric space and C be a nonempty convex, bounded and closed subset of Ω.
Let T : C → C be a continuous nonincreasing nearly asymptotically nonexpansive
mapping. If there exists x0, x

0 in C, such that x0 � T (x0) � T (x0) � x0, then T
has a fixed point x∗ ∈ C.

Proof. Let A : C × C → C be the mapping defined by A(x, y) = T (y), for every
x, y ∈ C. Then we have

x0 � T (x0) = A(x0, x
0) � A(x0, x0) = T (x0) � x0.

Moreover A is a mixed monotone in C, since T is continuous and nearly asymptoti-
cally nonexpansive, then according to theorem 3.7, T has a fixed point x∗ ∈ C. �

Example 3.9. Let C = {x = (x1, x2, x3, . . . ) ∈ `∞ : xi ∈ [−2, 2] for all i ≥ 1}.
The order relation x = (x1, x2, x3, . . . ) � y = (y1, y2, y3, . . . ) is defined by xi ≤ yi
for all i ≥ 1. Let {kn} be a nonincreasing sequences, such that kn ≥ 1 and
limn→∞ kn = 1. Consider the mapping T : C × C → C defined by

T (x, y) =
(

0,
k1
2

(x1 − y1),
k2
2k1

(x2 − y2),
k3
2k2

(x3 − y3), . . .
)
, x, y ∈ C.

Obviously, T is a continuous mixed monotone mapping, and

Tn(x, y) =
(

0, . . . , 0,
kn
2

(x1 − y1),
kn+1

2k1
(x2 − y2),

kn+2

2k2
(x3 − y3), . . .

)
,

for n ≥ 2, where 0 appears n times on the right side. Then, we have for all
x, y, u, v ∈ C

‖Tn(x, y)− Tn(u, v)‖ ≤ kn
(1

2
‖x− u‖+

1

2
‖y − v‖+ an

)
.



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 9

For any sequences {an} ⊂ [0,+∞), satisfying limn→∞ an = 0. Thus, T is mixed
monotone nearly asymptotically nonexpansive with (0, 0, . . . , 0) fixed point.

4. Applications to integral equations

In this section, we will apply Theorem 3.7 to study the existence of solutions to
the integral equation

x(t) =

∫ 1

0

a(t, s)f(s, x(s))g(x(s))ds. (4.1)

Firstly, we make the following assumptions, with I =]0, 1[.

(H1) (i) a(·, ·) ∈ L2
(
I × I

)
, ∂1a(·, ·) ∈ L2

(
I × I

)
(where ∂1a(t, s) = ∂

∂ta(t, s)),
and there exist δ > 0 such that a(t, s) > δ almost everywhere t, s ∈ I
(or, a.e. t, s ∈ I for short).

(ii) f : [0, 1]× R+ → R+ and g : R+ → R+, with for every s ∈ I, f(s, ·) is
nondecreasing and g is nonincreasing.

(H2) There exist f1, f2 : R+ → R+, such that f1 ∈ L2(I) and f2 with f2a and
f2∂1a in L2(I2). Furthermore, there exist η > 0, M > 0, such that

(i) For all x ∈ [0, η], f1(s)x ≤ f(s, x), a.e. s ∈ I,
(ii) For all x ∈ R+, f(s, x) ≤ f2(s), a.e. s ∈ I,

(iii) There exist µ > 0 and ν > 0, such that µ ≤ g(x) ≤ ν, for all x ∈ R+.
(H3) There exists λ > 0 and f3 : R+ → R+ with f3a and f3∂1a in L2(I2), such

that

|f(s, x)− f(s, y)| ≤ f3(s)|x− y| and |g(x)− g(y)| ≤ λ|x− y|,

for all x, y ∈ R+ and a.e. s ∈ I.

Next, we recall some notion and results about positive cone. Let E be a real
Banach space. A closed convex set P in E is called a convex cone if the following
conditions are satisfied.

(1) If x ∈ P , then λx ∈ P for all λ ∈ R+,
(2) If x ∈ P , and −x ∈ P , then x = 0.

A cone P induces a partial ordering � in E by x � y if and only if y − x ∈ P . We
denote by P̊ the interior set of P . A cone P is called a solid cone if P̊ 6= ∅.

Recall that H1(I) is the space of all equivalence classes of functions x ∈ L2(I)
having derivatives in the sense of distributions x′ ∈ L2(I). Set

P = {x ∈ H1(I) : x(s) ≥ 0 a.e. s ∈ I}. (4.2)

Thus, P is a positive cone in H1(I).

Theorem 4.1 ([5]). The embedding i : H1(I)→ C([0, 1]) is compact.

Using the above theorem, one can prove easily the following lemma.

Lemma 4.2. The cone P defined in (4.2) is solid and P = i−1(K), where K is
the positive cone of C([0, 1]) given by K = {x ∈ C([0, 1]) : x(s) ≥ 0, s ∈ [0, 1]}.

Lemma 4.3. Let P be the positive cone of H1(I) given by (4.2). Then, the topo-
logical interior of P is defined by

P̊ = {x ∈ P : ∃α > 0 with x(s) ≥ α, a.e. s ∈ I}.



10 H. EL BAZI, A. SADRATI EJDE-2022/66

Proof. Suppose that there exists x ∈ P̊ while for any ε > 0 there exists Jε ⊂ I
such that measJε > 0 and x(s) < ε for each s ∈ Jε (here measJε is the Lebesgue
measure of Jε). It follows that there exists r > 0 such that any y ∈ H1(I) satisfying
‖x− y‖H1 < r, we have y ∈ P .

Choose r/2 < δ < r, then z = x− δ ∈ H1(I) and ‖x− z‖H1 = δ < r. However,
z(s) = x(s)− δ < x(s)− r

2 < 0 for each s ∈ J r
2
, which means that z /∈ P , which is

a contradiction.
Conversely, let x ∈ P such that x(s) ≥ α almost everywhere s ∈ I for some

α > 0. Then, by theorem 4.1 we have i(x) ≥ α > 0. Hence i(x) ∈
◦
K, which implies

that x ∈ i−1
( ◦
K
)
⊂ ˚̂
i−1(K). Thus, x ∈ P̊ . �

Lemma 4.4. Suppose a(·, ·) ∈ L2
(
I × I

)
, ∂1a(·, ·) ∈ L2

(
I × I

)
, and there exists

δ > 0 such that a(t, s) > δ a.e. t, s ∈ I. Then, the operator A : H1(I) → H1(I)
defined by

A(x)(t) =

∫ 1

0

a(t, s)x(s)ds,

is linear, compact and strongly positive.

Proof. It is clear that A is linear. Let x ∈ H1(I), from the inequality of Holder and

from Fubini’s theorem we prove that
∫ 1

0
a(t, s)x(s)ds and

∫ 1

0
∂1a(t, s)x(s)ds are in

L2(I).
And also we apply Fubini’s theorem and integration by parts on the integral∫ 1

0

(∫ 1

0

a(t, s)x(s)ds
)
ϕ′(t)dt, ϕ ∈ C∞0 (I).

With C∞0 (I) is the class of test functions. Moreover from the definition of the
derivative in the sense of distributions, we obtain∫ 1

0

∂1a(t, s)x(s)ds =
(∫ 1

0

a(t, s)x(s)ds
)′
. (4.3)

Therefore, A
(
H1(I)

)
⊂ H1(I).

Let A1 and A2 defined from L2(I) to L2(I) by

A1(x) =

∫ 1

0

a(·, s)x(s)ds, A2(x) =

∫ 1

0

∂1a(·, s)x(s)ds .

from Holder’s inequality and Fubini’s theorem it follows that A1, A2 are contin-
uous; thus there exist M1,M2 ≥ 0 such that ‖A1(x)‖L2(I) ≤ M1‖x‖L2(I) and
‖A2(x)‖L2(I) ≤M2‖x‖L2(I). Hence

‖A(x)‖H1(I) =
(
‖A1(x)‖2L2(I) + ‖A2(x)‖2L2(I)

)1/2
≤
(
M2

1 +M2
2

)1/2‖x‖L2(I)

≤
(
M2

1 +M2
2

)1/2‖x‖H1(I).

Therefore A is continuous.
For proving that A is compact, let {xn} be a bounded sequence in H1(I). It

follows that {xn} is bounded in L2(I). Hence, by [9, p. 188], we obtain that



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 11

A(x) =
∫ 1

0
a(·, s)x(s)ds and B(x) =

∫ 1

0
∂1a(·, s)x(s)ds are compact operators in

L2(I). Therefore, there exist a subsequence {xk} and x, y ∈ L2(I) such that∫ 1

0

a(·, s)xk(s)ds→ x,

∫ 1

0

∂1a(·, s)xk(s)ds→ y in L2(I)( as k →∞).

We shall prove that x ∈ H1(I) and x′ = y. Let ϕ ∈ C∞0 (I). Then∫ 1

0

(∫ 1

0

a(t, s)xk(s)ds
)
ϕ′(t)dt→

∫ 1

0

x(t)ϕ′(t)dt,∫ 1

0

(∫ 1

0

∂1a(t, s)xk(s)ds
)
ϕ(t)dt→

∫ 1

0

y(t)ϕ(t)dt,

as k →∞. By equality (4.3), we obtain∫ 1

0

(∫ 1

0

∂1a(t, s)xk(s)ds
)
ϕ(t)dt = −

∫ 1

0

(∫ 1

0

a(t, s)xk(s)ds
)
ϕ′(t)dt.

Thus ∫ 1

0

x(t)ϕ′(t)dt = −
∫ 1

0

y(t)ϕ(t)dt,

which implies that x′ = y. Also, we have∫ 1

0

a(·, s)xk(s)ds→ x and
(∫ 1

0

a(·, s)xk(s)ds
)′
→ x′

as k → ∞. This means that
∫ 1

0
a(·, s)xk(s)ds → x in H1(I) (as k → ∞). Thus A

is a compact operator from H1(I) to H1(I).
Finally, we show that A is strongly positive. Let x ∈ P \ {0}, i.e., there exists

J ⊂ I with
(
measJ > 0

)
such that x(s) > 0 for each s ∈ J . By theorem 4.1

i(x)(s) > 0 a.e. s ∈ J . Then there exists s0 ∈ J such that i(x)(s0) = x(s0) > 0,
since i(x) is continuous. Thus there exists ε > 0 and α > 0 such that i(x)(s) ≥ α,
for all ]s0 − ε, s0 + ε[. Since a(t, s) > δ a.e. t, s ∈ I, a(t, s)i(x)(s) > αδ a.e. (t, s) ∈
I×]s0 − ε, s0 + ε[. Therefore, A(x)(t) =

∫ 1

0
a(t, s)x(s)ds =

∫ 1

0
a(t, s)i(x)(s)ds ≥

2εαδ > 0 a.e. t ∈ I. Thus, Lemma 4.3 gives A(x) ∈ P̊ , and consequently A is
strongly positive. �

Let us now define the operators

L1(x)(t) = µ

∫ 1

0

a(t, s)f1(s)x(s)ds and L2(x)(t) = ν

∫ 1

0

a(t, s)f2(s)x(s)ds.

Theorem 4.5. Assume (H1)–(H3) and

(H4)
r(L1) ≥ 1 and r

(
L2

)
≤ 1,

where r(Li) = limn→∞ ‖Lni ‖
1
n (i = 1, 2) is the spectral radius of the linear operator

Li. Then equation (4.1) has a positive solution x∗ ∈ P .

Proof. First the cone P defined in (4.2) induces a partial ordering � in H1(I) by,
x � y if and only if y − x ∈ P , and since H1(I) is a Hilbert space, we have that
H1(I) is a complete uniformly convex partially ordered hyperbolic metric space.

From (H1) and Lemma 4.4, we obtain that Li is a linear compact and strongly
positive operator from H1(I) to H1(I), for i = 1, 2. Moreover, by (H4) there exist
m1 ≤ 1 and m2 ≥ 1 such that r(miLi) = 1. Thus, by Krein-Rutman’s theorem,



12 H. EL BAZI, A. SADRATI EJDE-2022/66

there exist xi ∈ P̊ verifying miLi(xi) = xi. It follows that x1 � L1(x1) and
L2(x2) � x2.

Using Theorem 4.1, there exists α > 0 such that αx1 � η, and by Lemma 4.3,
there exists β > 0 such that 1 � βx2. Let x0 = αx1, x0 = βx2. Then, by (H2) it is
easy to show that

x0 � L1(x0) � A(x0, x
0), A(x0, x0) � L2(x0) � x0.

With

A(x, y)(t) =

∫ 1

0

a(t, s)f(s, x(s))g(y(s)) ds. (4.4)

We impose on ν and λ the conditions

2ν2
(
‖f3.a‖2L2(I2) + ‖f3.∂1a‖2L2(I2)

)
≤ 1

4
,

2λ2
(
‖f2.a‖2L2(I2) + ‖f2.∂1a‖2L2(I2)

)
≤ 1

4
.

(4.5)

Let C = {x ∈ P : ‖x‖H1 ≤ ρ}, such that

ρ = sup
{
‖x0‖H1 : ‖x0‖H1 , ν

(
‖f2.a‖2L2(I2) + ‖f2.∂1a‖2L2(I2)

)1/2}
.

Now, we shall prove that A defined in 4.4 is a well-defined operator from C ×C to
C.

Recall that if x ∈ L2(I) then
∫ 1

0
a(·, s)x(s) ds ∈ L2(I) and

( ∫ 1

0
a(·, s)x(s) ds

)′ ∈
L2(I). Since for all x, y ∈ C, we have f(s, x(s))g(y(s)) ≤ νf2(s) a.e. s ∈ I, therefore
f(., x(.))g(y(.)) in L2(I), then A(x, y) ∈ H1(I). Moreover, f and g are non-negative
functions, and a(t, s) > δ a.e. t, s ∈ I, hence A(x, y) ∈ P . On the other hand, letting
x, y ∈ C,

‖A(x, y)‖2H1 =

∫ 1

0

(∫ 1

0

a(t, s)f(s, x(s))g(y(s)) ds
)2
dt

+

∫ 1

0

(∫ 1

0

∂1a(t, s)f(s, x(s))g(y(s)) ds
)2
dt

≤ ν2
∫ 1

0

(∫ 1

0

a(t, s)f2(s) ds
)2
dt+ ν2

∫ 1

0

(∫ 1

0

∂1a(t, s)f2(s) ds
)2
dt

≤ ν2
(
‖f2a‖2L2(I2) + ‖f2∂1a‖2L2(I2)

)
≤ ρ2.

Thus A(C × C) ⊂ C. And from the monotonicity of f and g, it is easy to show
that A is a mixed monotone operator.

Next, for all x, y, u, v ∈ C, we have

‖A(x, y)−A(u, v)‖2H1

=

∫ 1

0

(∫ 1

0

a(t, s)
(
f(s, x(s))g(y(s))− f(s, u(s))g(v(s))

)
ds
)2
dt

+

∫ 1

0

(∫ 1

0

∂1a(t, s)
(
f(s, x(s))g(y(s))− f(s, u(s))g(v(s))

)
ds
)2
dt

≤
∫ 1

0

(∫ 1

0

a(t, s)f(s, x(s))
(
g(y(s))− g(v(s))

)
ds

+

∫ 1

0

a(t, s)g(v(s))
(
f(s, x(s))− f(s, u(s))

)
ds
)2
dt



EJDE-2022/66 NEARLY ASYMPTOTICALLY NONEXPANSIVE MAPPINGS 13

+

∫ 1

0

(∫ 1

0

∂1a(t, s)f(s, x(s))
(
g(y(s))− g(v(s))

)
ds

+

∫ 1

0

∂1a(t, s)g(v(s)
(
f(s, x(s))− f(s, u(s))

)
ds
)2
dt

≤ 2

∫ 1

0

(∫ 1

0

a(t, s)f2(s)λ
∣∣y(s)− v(s)

∣∣ ds)2 dt
+ 2

∫ 1

0

(∫ 1

0

νa(t, s)f3(s)
∣∣x(s)− u(s)

∣∣ ds)2 dt
+ 2

∫ 1

0

(∫ 1

0

∂1a(t, s)f2(s)λ
∣∣y(s)− v(s)

∣∣ ds)2 dt
+ 2

∫ 1

0

(∫ 1

0

ν∂1a(t, s)f3(s)
∣∣x(s)− u(s)

∣∣ ds)2 dt
≤ 2λ2‖y − v‖2L2‖f2.a‖2L2(I2) + 2ν2‖x− u‖2L2‖f3a‖2L2(I2)

+ 2λ2‖y − v‖2L2‖f2∂1a‖2L2(I2) + 2ν2‖x− u‖2L2‖f3.∂1a‖2L2(I2)

≤ 2λ2‖y − v‖2H1

(
‖f2a‖2L2(I2) + ‖f2∂1a‖2L2(I2)

)
+ 2ν2‖x− u‖2H1

(
‖f3a‖2L2(I2) + ‖f3∂1a‖2L2(I2)

)
.

≤
(1

2
‖x− u‖H1

)2
+
(1

2
‖y − v‖H1

)2
.

Th last inequality above follows from (4.5). Hence, for all x, y, u, v ∈ C

‖A(x, y)−A(u, v)‖H1 ≤ 1

2
‖x− u‖H1 +

1

2
‖y − v‖H1 .

It follows that A is continuous. By induction we show that

‖An(x, y)−An(u, v)‖H1 ≤ 1

2
‖x− u‖H1 +

1

2
‖y − v‖H1

for all comparable elements (x, y), (u, v) ∈ C × C, and each n ∈ N∗. Consequently,
all the hypotheses of theorem 3.7 are satisfied for the operator A : C × C → C.
Therefore, there exist x∗ ∈ P , such that

x∗(t) =

∫ 1

0

a(t, s)f(s, x∗(s))g(x∗(s))ds.

�

References

[1] M. R. Alfuraidan, M. A. Khamsi; A fixed point theorem for monotone asmptotically nonex-

pansive mappings, Proc. Am. Math. Soc. 146 (2018), 2451–2456.
[2] S. Aggarwal, Izhar Uddin, J. Nieto; A fixed-point theorem for monotone nearly asymptotically

nonexpansive mappings, J. Fixed Point Theory Appl. 21, 91 (2019), 1–11.

[3] K. Aoyama, F. Kohsaka; Fixed point theorem for α-nonexpansive mappings in Banach spaces,
Nonlinear Anal. 74 (2011), 4387–4391.

[4] M. Bachar, M. A. Khamsi; Fixed points of monotone mappings and application to integral
equations, Fixed Point Theory Appl. 2015, 110 (2015), 1–7.

[5] H. Brezis; Functional Analysis, Sobolev Spaces and Partial Differential Equations, Sprenger,
2011.

[6] B. Dehaish, M. A. Khamsi; Browder and Gohde fixed point theorem for monotone nonex-
pansive mapping, Fixed Point Theory Appl. 20 (2016), 1–9.



14 H. EL BAZI, A. SADRATI EJDE-2022/66

[7] K. Goebel, W. A. Kirk; A fixed point theorem for asymptotically nonexpansive mappings,

Proceedings of the American Mathematical Society. 35, 1 (1972), 171–174.

[8] D. Guo, V. Lakshmikantham; Coupled fixed points of nonlinear operators with applications,
Nonlinear Anal. 11 (1987), 623–632.

[9] F. Hirsch, G. Lacombe: Éléments d’analyses fonctionnelle: cours et exercices avec réponses.
ISBN : 2-10-004571-7.

[10] M. A. Khamsi, A. R. Khan; Inequalities in metric spaces with applications, Nonlinear Anal-

ysis. 74, 12 (2011), 4036–4045.
[11] S. Khatoon, W. Cholamjiak, I. Uddin; A modified proximal point algorithm involving nearly

asymptotically quasi-nonexpansive mappings, Journal of Inequalities and Applications. 2021,

83, (2021), 1–20.
[12] W. A. Kirk, H. K. Xu; Asymptotic pointwise contractions, Nonlinear Anal. 89 (2008), 4706–

4712.
[13] E. Llorens-Fuster, E. Moreno Galvez; The fixed point theory for some generalized nonexpan-

sive mappings, Abstr. Appl. Anal. 2011 (2011), 1–15.

[14] R. Pandey, R. Pant, V. Rakocevic, R. Shukla; Approximating fixed points of a general class of
nonexpansive mappings in Banach spaces with applications, Results Math. 74 (2019), 1–24.

[15] R. Pant, R. Shukla; Approximating fixed points of generalized α-nonexpansive mappings in

Banach spaces, Numer. Funct. Anal. Optim. 38 (2017), 248–266.
[16] K. Sabiya, I. Uddin, B. Metin; A modified proximal point algorithm for a nearly asymptoti-

cally quasi-nonexpansive mapping with an application, Computational and Applied Mathe-

matics., 40(7), (2021), 1–19.
[17] Y. Sang; Existence And Uniqueness Of Fixed Point For Mixed Monotone Operators With

Perturbations, Electronic Journal of Differential Equations, 2013, 233 (2013), 1–16.

[18] R. Shukla, R. Pant, P. Kumam; On the α-nonexpansive mappings in partially ordered hyper-
bolic metric spaces, J. Math. Anal. 1 (2017), 1–15.

[19] T. Suzuki; Fixed point theorems and convergence theorems for some generalized nonexpan-
sive mappings, J. Math. Anal. Appl. 340, 2 (2008), 1088–1095.

[20] I. Uddin, C.Garodial, J. J. Nieto; Mann iteration for monotone nonexpansive mappings in

ordered CAT (0) space with an application to integral equations, Journal of Inequalities and
Applications. 2018, 339 (2018), 1–13.

[21] I. Uddin, S. Khatoon1, N. Mlaiki, T. Abdeljawad; A modified iteration for total asymptotically

nonexpansive mappings in Hadamard spaces, AIMS Mathematics. 6(5) (2021) , 4758–4770.
[22] Y. Wu, Z. Liang; Existence and uniqueness of fixed points for mixed monotone operators

with applications, Nonlinear Anal. 65 (2006), 1913–1924.

[23] Z. Zhitao; New Fixed Point Theorems of Mixed Monotone Operators and Applications,
Journal Of Mathematical Analysis And Applications. 204, 0439 (1996), 307–319.

Hamza El Bazi

MSISI Laboratory, AM2CSI Group, Department of Mathematics, FST, Erracidia, Uni-
versity Moulay Ismäıl of Meknes, BP 509, Boutalamine, 52000, Errachidia, Morocco

Email address: hamza.elbazi.uae@gmail.com

Abdellatif Sadrati
MSISI Laboratory, AM2CSI Group, Department of Mathematics, FST, Erracidia, Uni-

versity Moulay Ismäıl of Meknes, BP 509, Boutalamine, 52000, Errachidia, Morocco
Email address: abdo2sadrati@gmail.com


	1. Introduction
	2. Definitions and preliminaries
	3. Fixed point theorem
	4. Applications to integral equations
	References