EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5951
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Hierarchical Fixed Point Results for a Countable
Family of Strict Pseudo-Contractive Mappings in

Hadamard Manifolds

Prashant Patel1, Rahul Shukla2,∗

1 Department of Mathematics, School of Advanced Sciences, VIT-AP University, Inavolu,
Beside AP Secretariat, Amaravati, 522237, Andhra Pradesh, India
2 Department of Mathematical Sciences & Computing, Walter Sisulu University,
South Africa

Abstract. The aim of this paper is to present convergence results for a countable family of strict
pseudocontractive mappings in Hadamard manifolds. More precisely, we employ the shrinking
projection method to approximate common hierarchical fixed points of a countable family of strict
pseudocontractive mappings in the setting of Hadamard manifolds. We also present some nontrivial
examples to illustrate our result.

2020 Mathematics Subject Classifications: 47H10, 47H09

Key Words and Phrases: Hierarchical Fixed point, Hadamard Manifolds, Pseudocontractive
Mappings

1. Introduction

In 1966, Hartman and Stampacchia [1] introduced variational inequality theory as a
method for studying partial differential equations with applications, primarily in mechan-
ics. The variational inequality problem has a wide range of applications in some practical
problems arising in economics, transportation, network and structural analysis, elasticity,
engineering and mechanics, supply chain management, finance and game theory. In recent
years, many authors discussed variational inequality problems in the context of Banach
and Hilbert spaces [2–7]. To solve environmental projects concerning the transmission of
pollution in different kind of media we need to transfer pollution along certain bounded
surface areas. These restrictions lead to many boundary value problems on manifolds.
To overcome, this situation in 2003, Nemeth [8] introduced the variational inequalities in
Hadamard manifolds.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5951

Email addresses: prashant.patel9999@gmail.com, prashant.p@vitap.ac.in (P. Patel),
rshukla@wsu.ac.za (R. Shukla)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 2 of 15

A constrained optimization problem, where the constrained set is the solution set of
another optimization problem, is known as a bilevel programming problem. Over the past
thirty years, there has been extensive research on these challenges due to their relevance
in domains such as mechanics and network design. If the first level problem is one of
variational inequality and the second level problem is a collection of fixed points of a map-
ping, then the bilevel problem is known as a hierarchical variational inequality problem.
Stated otherwise, a variational inequality problem defined over the set of fixed points is
a hierarchical variational inequality problem, sometimes referred to as a hierarchical fixed
point problem.

In 2006, Moudafi and Mainge [9] introduced the hierarchical fixed point problem in
the setting of Hilbert space

Find ζ ∈ F (H) such that ⟨ζ −G(ζ), ζ − ν⟩ ≤ 0, for all ν ∈ F (G). (1)

Here G,H are nonexpansive mappings defined on a Hilbert space M. Later in 2010, Xu
[10] extended his work in the context of uniformly smooth Banach spaces. After that the
viscosity method was developed by a number of researchers to solve variational inequali-
ties defined on the set of fixed points of nonexpansive mapping in the context of Hilbert
or Banach spaces. These researchers replaced contraction mapping with weaker forms of
contraction mappings, such as pseudo-contraction mapping and weakly contraction map-
ping [3, 4, 11–14].
In 2020, Al-Homidan, presented a viscosity approach to solve the hierarchical fixed point
problem in the context of Hadamard manifolds defined on the set of fixed points of nonex-
pansive mapping and involving a nonexpansive mapping and another ϕ-contraction map-
ping. There are many authors who presented a viscosity approach for hierarchical varia-
tional inequality problems in the context of Hadamard manifolds [9, 15], and references
therein.

Motivated by the above works, the purpose of this paper is to introduce and analyse
a new algorithm for solving hierarchical variational inequality problems in the framework
of Hadamard manifolds. In this paper, we present a new algorithm to solve hierarchical
fixed point problem and present convergence result for a finite family of β-strict pseudo-
contractive mapping in Hadamard manifolds.

The manuscript is presented as follows: Section 2 contains basic definitions and facts.
Section 3 has the hierarchical variational inequality problem and the proposed algorithm
and its convergence analysis. Section 4 has some numerical examples which illustrates the
result presented in the manuscript.

2. Preliminaries

Now, we present some basic facts and definitions which are related to Hadamard man-
ifolds.



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 3 of 15

Let us consider, Γ is the differentiable and finite dimensional manifold. We write
the tangent space of Γ, for all ζ ∈ Γ as GζΓ. This GζΓ is also a vector space and its
dimension is same as the dimension of Γ. Also, the tangent bundle of Γ is denoted by
GΓ =

⋃
ζ∈Γ

GζΓ. If we define an inner product Rζ(·, ·) on the tangent space GζΓ then it is

said to be a Riemannian metric defined on GζΓ. If Γ can be endowed with a Riemannian
metric Rζ(·, ·) then we say Γ is a Riemannian manifold. We denote the corresponding
norm for the inner product on the tangent space GζΓ by ∥ · ∥ζ . A Riemannian manifold is
a manifold, which is differentiable endowed with a Riemannian metric R(·, ·). The length
of piecewise smooth curve Υ : [0, 1] → Γ joining ζ to η (i.e. Υ(0) = ζ and Υ(1) = η)

is given by L(Υ) =
1∫
0

∥Υ′
(µ)∥dµ. The Riemannian distance ρ(ζ, η) is the minimal length

over the set of all these curves joining ζ to η, which includes the original topology on Γ.

A Riemannian manifold Γ is said to be complete if for all ζ ∈ Γ, all geodesics starting
from ζ are defined for all µ ∈ R. A geodesic joining ζ to η is said to be minimal in Γ
if the length of the geodesic is equal to ρ(ζ, η). The Riemannian manifold Γ having the
Riemannian distance ρ is also a metric space (Γ, ρ).

Definition 1. Let us consider that Γ is a complete Riemannian manifold. We define the
exponential map expζ : GζΓ → Γ at point ζ ∈ Γ by expζ v = Υv(1, ζ) for all v ∈ GζΓ, where
Υv(·, ζ) is the geodesic with the velocity v and starting from the point ζ i.e. Υ′

v(0, ζ) = v
and Υv(0, ζ) = ζ [16].

We also know that for all µ ∈ R the exponential map expζ µv = Υv(µ, ζ). Here we
can also see for all zero tangent vector, exponential map expζ 0 = Υv(0, ζ) = ζ. The

exponential map expζ is differentiable on TζΓ for all ζ ∈ Γ and ρ(ζ, η) = ∥ exp−1
ζ η∥ for all

ζ, η ∈ Γ.

Definition 2. A Riemannian manifold of non positive sectional curvature is said to be a
Hadamard Manifold if it is complete and simply connected.

Lemma 1. [17].

(1) For all w, u, ς, η, , z ∈ Γ, 0 ≤ µ ≤ 1, following hold:

ρ(expη(1− µ) exp−1
η ς, z) ≤ µρ(η, z) + (1− µ)ρ(ς, z);

ρ2(expη(1− µ) exp−1
η ς, z) ≤ µρ2(η, z) + (1− µ)ρ2(ς, z)− µ(1− µ)ρ2(η, ς);

ρ(expη(1− µ) exp−1
η ς, expu(1− µ) exp−1

u ζ) ≤ µρ(η, u) + (1− µ)ρ(ς, ζ).

(2) Let Υ : [0, 1] → Γ be a geodesic joining points η to ς. Then

ρ(Υ(µ1),Υ(µ2)) = |µ1 − µ2|ρ(η, ς) for all µ1, µ2 ∈ [0, 1].



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 4 of 15

Proposition 1. [18]. expζ : GζΓ → Γ is said to be a diffeomorphism for all ζ ∈ Γ, any
pair of points ζ, η ∈ Γ, there exists a unique normalized geodesic Υ : [0, 1] → Γ joining the
points ζ = Υ(0) to η = Υ(1), in fact it is a minimal geodesic defined by

Υ(µ) = expζ µ exp−1
ζ η for all 0 ≤ µ ≤ 1.

Definition 3. The map G : B → B is called as

(i) nonexpansive
ρ(G(ς), G(η)) ≤ ρ(ς, η) for all ς, η ∈ B,

(ii) firmly nonexpansive, if for all ς, η ∈ B, function ϕ : [0, 1] → [0,+∞] defined by

ϕ(t) = ρ
(
expη t exp

−1
η G(η), expς t exp

−1
ς G(ς)

)
for all 0 ≤ t ≤ 1

is nonincreasing [18].

(iii) β-strict pseudocontractive if there exists β ∈ [0, 1) such that

ρ2(G(η), G(ς)) ≤ ρ2(η, ς) + β∥Pη,ς exp
−1
ς G(ς)− exp−1

η G(η)∥2, for all η, ς ∈ B.

Note that the mapping G is nonexpansive if and only if it is 0-strict pseudocontrac-
tive.

Suppose B is a geodesic convex and closed subset of the given Hadamard manifold
Γ. The projection map onto the geodesic convex and closed sets can also defined in the
setting of linear metric spaces. A projection mapping PB(·) : Γ → B is given by for any
ζ ∈ Γ

PB(ζ) = {w ∈ B : ρ(ζ, w) ≤ ρ(ζ, µ), for all µ ∈ B}.

Proposition 2. [19] Suppose B ̸= ∅ be closed and geodesic convex subset of Γ. Then we
have:

(1) PB is a single valued and firmly nonexpansive mapping;

(2) for all ζ ∈ Γ, ω = PB(ζ) if and only if R(exp−1
ω ζ, exp−1

ω ϑ) ≤ 0, for all ϑ ∈ B;

(3) if PB is firmly nonexpansive.

ρ2(ω, ν) + ρ2(ω, ζ) ≤ ρ2(ζ, ν), for all ζ ∈ Γ, ν ∈ B,

here ω = PB(ζ).

Lemma 2. [20] Suppose ∆(ζ1, ζ2, ζ3) is a geodesic triangle in Γ then there exists a triangle
∆(ζ1, ζ2, ζ3) in R2 for ∆(ζ1, ζ2, ζ3) such that ρ(ζi, ζi+1) = ∥ζi − ζi+1∥, indices are taken
modulo 3; and it is unique upto an isometry of R2.



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 5 of 15

Proposition 3. [18] Suppose ∆(ζ1, ζ2, ζ3) be a geodesic triangle in Γ. Then

ρ2(ζ1, ζ2) + ρ2(ζ2, ζ3)− 2R
(
exp−1

ζ2
ζ1, exp

−1
ζ2

ζ3

)
≤ ρ2(ζ3, ζ1), (2)

and
ρ2(ζ1, ζ2) ≤ R

(
exp−1

ζ1
ζ3, exp

−1
ζ1

ζ2

)
+R

(
exp−1

ζ2
ζ3, exp

−1
ζ2

ζ1

)
. (3)

Moreover, if θ is the angle at ζ1, then we have

R
(
exp−1

ζ1
ζ2, exp

−1
ζ1

ζ3

)
= ρ(ζ2, ζ1)ρ(ζ1, ζ3) cos(θ).

Lemma 3. [20] Suppose ∆(ζ1, ζ2, ζ3) be geodesic triangle in Γ, ∆(ζ1, ζ2, ζ3) its comparison
triangle.

(1) Suppose α1, α2, α3 and α1, α2, α3 be the angles of ∆(ζ1, ζ2, ζ3) and ∆(ζ1, ζ2, ζ3) at the
vertices ζ1, ζ2, ζ3 and ζ1, ζ2, ζ3, respectively. Then

α1 ≤ α1, α2 ≤ α2 and α3 ≤ α3.

(2) Suppose µ be any point on the geodesic connecting ζ1, ζ2 and µ its comparison point in
interval [ζ1, ζ2]. If ρ(ζ1, µ) = ∥ζ1−µ∥ and ρ(ζ2, µ) = ∥ζ2−µ∥ then ρ(ζ3, µ) ≤ ∥ζ3−µ∥.

Proposition 4. [21] Suppose B ̸= ∅ be a geodesic convex subset of a Hadamard manifold
Γ and G : B → B be a β-strictly pseudocontractive mapping with β ∈ (0, 1]. Then Gλ(ζ) =
expζ λ exp−1

ζ G(ζ) is a nonexpansive mapping for every λ ∈ (0, 1− β).

Remark 1. If G and Gλ are same as above proposition then F (G) = F (Gλ). It follows
from the following equivalence

Let ζ ∈ F (Gλ)

implies ζ = Gλ(ζ) ⇔ ζ = expζ λ exp
−1
ζ G(ζ)

⇔ 0 = exp−1
ζ G(ζ)

⇔ ζ = G(ζ).

Throughout the article we write the set of all single valued vector fields 𭟋 : Γ → GΓ
as Ξ(Γ) such that 𭟋(ζ) ∈ GζΓ for all ζ ∈ Γ. Let Ψ(Γ) denote the set of all multivalued
vector fields V : Γ → 2GΓ such that V (ζ) ⊆ GζΓ for all ζ ∈ Γ, and we denote the domain
of V by D(V ) = {ζ ∈ Γ : V (ζ) ̸= ∅}.

Definition 4. [22] Any vector field 𭟋 ∈ Ξ(Γ) is monotone if it satisfies

R
(
𭟋(ζ), exp−1

ζ ν
)
+R(𭟋(ν), exp−1

ν ζ) ≤ 0, for all ζ, ν ∈ Γ.



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 6 of 15

Definition 5. [23] A multivalued vector field V ∈ Ψ(Γ) is said to be monotone if for all
ζ, ν ∈ D(V )

R
(
u, exp−1

ζ ν
)
≤ R(w,− exp−1

ν ζ), for all , u ∈ V (ζ), w ∈ V (ν);

and maximal monotone if it is already monotone, for all ζ ∈ Γ, u ∈ GζΓ,

R
(
u, exp−1

ζ ν
)
≤ R(w,− exp−1

ν ζ), for all , ν ∈ D(V ), w ∈ V (ν) implies u ∈ V (ζ).

Definition 6. [22] Suppose G : Γ → Γ is a mapping and vector field 𭟋 ∈ Ξ(Γ) is defined
as

𭟋(ζ) = − exp−1
ζ G(ζ), for all ζ ∈ Γ,

is the complementary vector field.

Theorem 1. [22] The complimentary vector field 𭟋 = − exp−1G defined for any nonex-
pansive mapping G : Γ → Γ is always monotone.

3. Main Results

Suppose Γ is a Hadamard Manifold, Gi, Hi : Γ → Γ be two countable family of β-
strict pseudocontractive mappings with β ∈ (0, 1] and the set of common fixed points of
mappings Hi and Gi are given by

⋂
i F (Hi) and

⋂
i F (Gi). We define the hierarchical fixed

point problem as follows

Find ζ ∈
⋂
i

F (Hi) : R
(
exp−1

ζ Gλi
(ζ), exp−1

ζ ν
)
≤ 0, for all ν ∈

⋂
i

F (Hi). (4)

If
⋂
i
F (Hi) ̸= ∅, then using Proposition 2 (2) the above problem defined as

Find ζ ∈ Γ such that ζ = P⋂
i
F (Hi)Gλi

(ζ), (5)

here P⋂
i
F (Hi) is the metric projection of Γ onto

⋂
i
F (Hi). We denote the set of solution

of (4) as Φ =

{
ζ† ∈ Γ : ζ† = P⋂

i
F (Hi)Gλi

(ζ†)

}
.

Hierarchical Variational Inequality Problem: Suppose B ≠ ∅ is a geodesic convex,
closed subset of Γ and 𭟋 : B → HΓ be the monotone vector field. The problem is to find
ζ ∈ B:

R
(
𭟋(ζ), exp−1

ζ ν
)
≥ 0, for all ν ∈ B. (6)

Suppose Gi : Γ → Γ be a countable family of β-strict pseudocontractive mappings. Then
the complimentary vector field of the corresponding family of mappings Gλi

is monotone
by Theorem 1. Now, we can reduce the problem (4) as:

Find ζ ∈
⋂
i

F (Hi) : R
(
𭟋(ζ), exp−1

ζ ν
)
≥ 0, for all ν ∈

⋂
i

F (Hi). (7)



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 7 of 15

Where 𭟋 = − exp−1Gλi
is the complimentary vector field of Gλi

. In the context of normal
cone the set

⋂
i
F (Hi), we can easily see that problem (7) is equivalent to the following

Find ζ ∈
⋂
i

F (Hi) : 0 ∈ − exp−1
ζ Gλi

(ζ) +N⋂
i
F (Hi)(ζ). (8)

Here N⋂
i
F (Hi) is normal cone onto set

⋂
i
F (Hi) at ζ ∈

⋂
i
F (Hi), and defined as

N⋂
i
F (Hi)(ζ) =

{
µ ∈ Hλiζ

Γ : R
(
µ, exp−1

ζ ν
)
≤ 0, for all ν ∈

⋂
i

F (Hi)

}
.

Now, we present a new algorithm to approximate the solution of the problem (4). Suppose
Γ be a Hadamard Manifold and Gi, Hi : Γ → Γ be two countable family of β-strict
pseudocontractive mappings with β ∈ (0, 1]. Assume that ζ1 ∈ Γ and B1 = Γ, we can
generate sequences {ζn} and {ϑn} as follows:

ϑn = expGλi
(ζn)(1− δn) exp

−1
Gλi

(ζn)
Hλi

(ζn),

Bn+1 = {ν ∈ Bn : ρ(ϑn, ν) ≤ ρ(ζn, ν)},
ζn+1 = PBn+1(ζ1), for all n ≥ 1.

(9)

Here, Gλi
(ζ) = expζ λ exp−1

ζ Gi(ζ), Hλi
(ζ) = expζ λ exp−1

ζ Hi(ζ), {δn} ∈ (0, 1) satisfying
0 < δ1 ≤ δn ≤ δ2 < 1 and lim

n→+∞
δn = 0.

Theorem 2. Suppose Γ, Gi, Hi are same as defined above and Π = Φ
⋂
i
F (Gi) ̸= ∅. For

B1 = Γ the sequence generated by (9) converges to PΠ(ζ1).

Proof. Since Φ = F

(
P⋂

i
F (Hi)Gi

)
̸= ∅, it can be easily seen that Φ is geodesic convex

and closed and F (Gi) is also geodesic convex and closed. Therefore Π is also geodesic
convex and closed, indicating that PΠ(ζ1) is well defined. Now we prove that the set Bn

is closed and geodesic convex subset of Γ for all n ≥ 1. We prove this by mathematical
induction. If B1 = Γ, then it is geodesic convex and closed. Now, let us assume that Bn

is geodesic convex and closed subset in Γ for some n ≥ 2. Then we need to prove that
Bn+1 is a geodesic convex and closed subset of Γ. Since ν 7→ ρ(ζ, ν) is a convex geodesic
function, we can easily say that Bn+1 is a closed and geodesic convex subset of Γ.

Next we prove that Π ⊂ Bn for all n ≥ 1. We can easily see that Π ⊂ B1 = Γ. Now we
prove that Π ⊂ Bn for all n ≥ 2. If ζ† ∈ Π, then ζ† ∈ Φ and ζ† ∈

⋂
i
F (Gλi

). Now let

for a fix n ∈ N, ∆(Gλi
(ζn), Hλi

(ζn), ζ
†) ⊆ Γ be a geodesic triangle with vertices Gλi

(ζn),

Hλi
(ζn) and ζ†, and ∆(Gλi

(ζn), Hλi
(ζn), ζ†) ⊆ R2 a corresponding comparison triangle.

We have ρ(Gλi
(ζn) − Hλi

(ζn)) =
∥∥∥Gλi

(ζn)−Hλi
(ζn)

∥∥∥, ρ(Gλi
(ζn), ζ

†) =
∥∥∥Gλi

(ζn)− ζ†
∥∥∥,



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 8 of 15

ρ(Hλi
(ζn), ζ

†) =
∥∥∥Hλi

(ζn)− ζ†
∥∥∥. Suppose ϑn = δnGλi

(ζn) + (1 − δn)Hλi
(ζn) is the com-

parison point of ϑn. Using the nonexpansiveness of Gλi
, Hλi

and Lemma 3 (2), we get

ρ2(ϑn, ζ
†) = ∥ϑn − ζ†∥2

≤ δn

∥∥∥Gλi
(ζn)− ζ†

∥∥∥2 + (1− δn)
∥∥∥Hλi

(ζn)− ζ†
∥∥∥

− δn(1− δn)
∥∥∥Gλi

(ζn)−Hλi
(ζn)

∥∥∥
= δnρ

2
(
Gλi

(ζn), ζ
†
)
+ (1− δn)ρ

2
(
Hλi

(ζn), ζ
†
)

− δn(1− δn)ρ
2 (Gλi

(ζn), Hλi
(ζn))

≤ δnρ
2(ζn, ζ

†) + (1− δn)ρ
2(ζn, ζ

†)− δn(1− δn)ρ
2 (Gλi

(ζn), Hλi
(ζn))

≤ ρ2(ζn, ζ
†)− δn(1− δn)ρ

2 (Gλi
(ζn), Hλi

(ζn)) (10)

≤ ρ2(ζn, ζ
†).

We can say ζ† ∈ Bn+1 for all ζ† ∈ Π and hence Π ⊂ Bn for all n ≥ 1. Thus Bn is a
nonempty, geodesic convex and closed subset of Γ and Π ⊂ Bn+1 ⊂ Bn for all n ≥ 1.
Hence sequence {ζn} is well defined.

Now we prove that lim
n→+∞

ρ(ζn, ζ1) exists and the sequence {ζn} is bounded. Since we have

ζn = PBn(ζ1) and using the fact that Π ⊂ Bn+1 ⊂ Bn we get

ρ(ζn, ζ1) ≤ ρ(ζ†, ζ1), for all ζ† ∈ Π, n ≥ 1. (11)

Hence, we get the sequence {ζn} is bounded. Since ζn+1 ∈ Bn+1 ⊂ Bn and we get
ρ(ζn, ζ1) ≤ ρ(ζn+1, ζ1) and hence lim

n→+∞
ρ(ζn, ζ1) exists.

Next using Proposition 2 (3) we get

ρ2(ζn, ζn+1) ≤ ρ2(ζn+1, ζ1)− ρ2(ζn, ζ1). (12)

Applying summation on (12) and using (11), we have

N∑
n=1

ρ2(ζn+1, ζn) ≤
N∑

n=1

(
ρ2(ζn+1, ζ1)− ρ2(ζn, ζ1)

)
≤ ρ2(ζN+1, ζ1)− ρ2(ζ1, ζ1)

≤ ρ2(ζ†, ζ1),

and it gives us
N∑

n=1
ρ2(ζn+1, ζn) is convergent and hence

lim
n→+∞

ρ(ζn+1, ζn) = 0. (13)



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 9 of 15

Since from (9) we have ζn+1 = PBn+1(ζ1) ∈ Bn+1, so using the definition of Bn+1 we have
ρ(ϑn, ζn+1) ≤ ρ(ζn, ζn+1). Using the above equation, we have

lim
n→+∞

ρ(ϑn, ζn+1) = 0. (14)

Now
ρ(ϑn, ζn) ≤ ρ(ϑn, ζn+1) + ρ(ζn+1, ζn),

applying limit n → +∞ and using (13) and (14), we get

lim
n→+∞

ρ(ϑn, ζn) = 0. (15)

Using (10), we will have

δ1(1− δ2)ρ
2(Gλi

(ζn), Hλi
(ζn)) ≤ δn(1− δn)ρ

2(Gλi
(ζn), Hλi

(ζn))

≤ ρ2(ζn, ζ
†)− ρ2(ϑn, ζ

†)

≤ C1ρ(ζn, ϑn). (16)

Where C1 = sup
n≥1

{ρ(ζn, ζ†) + ρ(ϑn, ζ
†)}. Using (15), we get

lim
n→+∞

ρ(Gλi
(ζn), Hλi

(ζn)) = 0. (17)

Next we prove that the {ζn} is a Cauchy sequence in Γ and {ζn} converges to ζ ∈ Π. Since
PBn is firmly nonexpansive, ζm = PBm(ζ1) ∈ Bm ⊂ Bn for any n,m ∈ N and m > n. If we
take B = Bn, ζ = ζ1 and ϑ = ζm, we get

ρ2 (PBnζ1) = ρ2(ζn, ζm) ≤ ρ2(ζ1, ζm)− ρ2(ζn, ζ1),

Applying limit n,m → +∞ and using proposition 2 (3), we get

lim
n,m→+∞

ρ(ζn, ζm) → 0.

It gives us that the sequence {ζn} is a Cauchy sequence and since Γ is complete we can
assume that lim

n→+∞
ζn = ζ ∈ Γ.

Now we prove that ζ ∈ Π. From (9) we have

ρ(ϑn, Gλi
(ζn)) = δnρ(Gλi

(ζn), Hλi
(ζn)) ≤ δ2ρ(Gλi

(ζn), Hλi
(ζn)).

Applying limit n → +∞ and using (17), we get

lim
n→+∞

ρ(ϑn, Gλi
(ζn)) = 0. (18)

And
ρ(Gλi

(ζn), ζn) ≤ ρ(Gλi
(ζn), ϑn) + ρ(ϑn, ζn).



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 10 of 15

Applying limit n → +∞ to the above equation and using (18) and (15), we get

lim
n→+∞

ρ(Gλi
(ζn), ζn) = 0. (19)

Similarly,

ρ(Hλi
(ζn), ζn) ≤ ρ(Hλi

(ζn), Gλi
(ζn)) + ρ(Gλi

(ζn), ζn).

Applying limit n → +∞ to the above equation and using (18) and (19), we get

lim
n→+∞

ρ(Hλi
(ζn), ζn) = 0. (20)

Since Hλi
nonexpansive so it is demiclosed at 0 and hence ζ ∈

⋂
i
Hi, further from (19)

we can also say ζ ∈
⋂
i
Gi. Next we prove that ζ ∈ Φ. Suppose ζ∗ ∈

⋂
i
Hi is arbitrary

such that ζ ̸= ζ∗. Let the triangles ∆(Gλi
(ζn), Hλi

(ζn), ζ
∗), ∆(Gλi

(ζ∗), Hλi
(ζn), ζ

∗) and
∆(Gλi

(ζn), Gλi
(ζ∗), Hλi

(ζn)) then there exist comparison triangles ∆(Gλi
(ζn), Hλi

(ζn), ζ∗),
∆(Gλi

(ζ∗), Hλi
(ζn), ζ∗) and ∆(Gλi

(ζn), Gλi
(ζ∗), Hλi

(ζn)) such that

ρ(Gλi
(ζn), Hλi

(ζn)) =
∥∥∥Gλi

(ζn)−Hλi
(ζn)

∥∥∥, ρ(Gλi
(ζn), ζ

∗) =
∥∥∥Gλi

(ζn)− ζ∗
∥∥∥, ρ(Hλi

(ζn), ζ
∗) =∥∥∥Hλi

(ζn)− ζ∗
∥∥∥, ρ(Gλi

(ζ∗), ζ∗) =
∥∥∥Gλi

(ζ∗)− ζ∗
∥∥∥, ρ(Hλi

(ζn), ζ
∗) =

∥∥∥Hλi
(ζn)− ζ∗

∥∥∥, and
ρ(Gλi

(ζn), Gλi
(ζ∗)) =

∥∥∥Gλi
(ζn)−Gλi

(ζ∗)
∥∥∥. Suppose θ be the angle at ζ∗ in the triangle

∆(Gλi
(ζ∗), Hλi

(ζn), ζ
∗) and θ be the angle at ζ∗ in the comparison triangle

∆(Gλi
(ζ∗), Hλi

(ζn), ζ∗). Then θ ≤ θ, and thus, cos(θ) ≤ cos(θ). Since ϑn = δnGλi
(ζn) +

(1− δn)Hλi
(ζn) is the comparison point of ϑn. Using Proposition 3 and Lemma 3 (2), we

have

ρ2(ϑn, ζ
∗) ≤

∥∥ϑn − ζ∗
∥∥2

=
∥∥∥δnGλi

(ζn) + (1− δn)Hλi
(ζn)− ζ∗

∥∥∥
= δ2n

∥∥∥Gλi
(ζn)− ζ∗

∥∥∥2 + (1− δn)
2
∥∥∥Hλi

(ζn)− ζ∗
∥∥∥2

+ 2δn(1− δn)
〈
Gλi

(ζn)− ζ∗, Hλi
(ζn)− ζ∗

〉
= δ2n

∥∥∥Gλi
(ζn)− ζ∗

∥∥∥2 + (1− δn)
2
∥∥∥Hλi

(ζn)− ζ∗
∥∥∥2

+ 2δn(1− δn)
(〈

Gλi
(ζn)−Gλi

(ζ∗), Hλi
(ζn)− ζ∗

〉
+
〈
Gλi

(ζ∗)− ζ∗, Hλi
(ζn)− ζ∗

〉)
≤ δ2n

∥∥∥Gλi
(ζn)− ζ∗

∥∥∥2 + (1− δn)
2
∥∥∥Hλi

(ζn)− ζ∗
∥∥∥2

+ 2δn(1− δn)
(∥∥∥Gλi

(ζn)−Gλi
(ζ∗)

∥∥∥∥∥∥Hλi
(ζn)− ζ∗

∥∥∥
+
∥∥∥Gλi

(ζ∗)− ζ∗
∥∥∥∥∥∥Hλi

(ζn)− ζ∗
∥∥∥ cos(θ))

≤ δ2nρ
2 (Gλi

(ζn), ζ
∗) + (1− δn)

2ρ2 (Hλi
(ζn), ζ

∗)



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 11 of 15

+ 2δn(1− δn)ρ (Gλi
(ζn), Gλi

(ζ∗)) ρ(Hλi
(ζn), ζ

∗)

+ 2δn(1− δn)ρ (Gλi
(ζ∗), ζ∗) ρ(Hλi

(ζn), ζ
∗) cos(θ)

ρ2(ϑn, ζ
∗) ≤ δ2nρ

2(Gλi
(ζn), ζ

∗) + (1− δn)
2ρ2(ζn, ζ

∗) + 2δn(1− δn)ρ
2(ζn, ζ

∗)

+ 2δn(1− δn)R
(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
= δ2nρ

2(Gλi
(ζn), ζ

∗) + (1− δ2n)ρ
2(ζn, ζ

∗)

+ 2δn(1− δn)R
(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
= δ2nρ

2(Gλi
(ζn), ζ

∗) + ρ2(ζn, ζ
∗)

+ 2δn(1− δn)R
(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
,

and we get

0 ≤ δ2nρ
2(Gλi

, ζ∗) + ρ2(ζn, ζ
∗)− ρ2(ϑn, ζ

∗)

+ 2δn(1− δn)R
(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
≤ δ2nρ

2(Gλi
(ζn), ζ

∗) + C2ρ(ζn, ϑn) + 2δn(1− δn)R
(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
.

Here C2 = sup
n≥1

{ρ(ζn, ζ∗)+ρ(ϑn, ζ
∗)}. Since 0 < δ1 ≤ δn ≤ δ2 < 1 we can have 1

δn
≤ 1

δ1
.

Therefore above equation becomes

0 ≤ δn
2
ρ2(Gλi

(ζn), ζ
∗) +

C2

2δn
ρ(ζn, ϑn) + (1− δn)R

(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
≤ δn

2
ρ2(Gλi

(ζn), ζ
∗) +

C2

2δ1
ρ(ζn, ϑn) + (1− δn)R

(
exp−1

ζ∗ Gλi
(ζ∗), exp−1

ζ∗ Hλi
(ζn)

)
Since ζ ∈

⋂
i
F (Hλi

), lim
n→+∞

ζn = ζ and lim
n→+∞

δn = 0, and using (14), we get

R
(
exp−1

ζ∗ Gλi
(ζ∗),− exp−1

ζ∗ ζ
)
≤ 0,

i.e.
R

(
exp−1

ζ∗ Gλi
(ζ∗), Pζ∗,ζ exp

−1
ζ ζ∗

)
≤ 0, for all ζ∗ ∈

⋂
i

F (Hλi
). (21)

Since ζ ∈
⋂
i
F (Hλi

),
⋂
i
F (Hλi

) is geodesic convex, we get expζ t exp
−1
ζ µ ∈

⋂
i
F (Hλi

) for

any µ ∈
⋂
i
F (Hλi

) and t ∈ (0, 1). Replacing ζ∗ by ut = expζ t exp
−1
ζ µ ∈

⋂
i
F (Hλi

) where

µ ∈
⋂
i
F (Hλi

) and t ∈ (0, 1), in the equation (21), we get

R
(
exp−1

expζ t exp
−1
ζ µ

Gλi
(expζ t exp

−1
ζ µ), Put,ζ exp

−1
ζ expζ t exp

−1
ζ µ

)
≤ 0,



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 12 of 15

which gives us

R
(
exp−1

expζ t exp
−1
ζ µ

Gλi
(expζ t exp

−1
ζ µ), Put,ζt exp

−1
ζ µ

)
≤ 0.

Since the map P is linear, we get

R
(
exp−1

expζ t exp
−1
ζ µ

Gλi
(expζ t exp

−1
ζ µ), Put,ζ exp

−1
ζ µ

)
≤ 0.

Applying t → 0 then ut → ζ. Since expζ 0 = ζ for all ζ ∈ Γ, we get

R
(
exp−1

ζ Gλi
(ζ), Pζ,ζ exp

−1
ζ µ

)
≤ 0.

Thus
R

(
exp−1

ζ Gλi
(ζ), exp−1

ζ µ
)
≤ 0, for all µ ∈

⋂
i

F (Hλi
). (22)

Hence, ζ ∈ Φ, and we have ζ ∈ Π. Now finally we prove that the sequence {ζn} converges
to ζ = PΠ(ζ1). From ζn = PBn(ζ1), using Proposition 2 (1), we get

R
(
exp−1

ζn
ζ1, exp

−1
ζn

ν
)
≤ 0, for all ν ∈ Bn.

Using the fact that Π ⊂ Bn, we get

R
(
exp−1

ζn
ζ1, exp

−1
ζn

ν
)
≤ 0, for all ν ∈ Π. (23)

Applying limit n → +∞ in the above equation, we get

R
(
exp−1

ζ ζ1, exp
−1
ζ ν

)
≤ 0, for all ν ∈ Π,

i.e. lim
n→+∞

ζn = ζ = PΠ(ζ1). Thus the proof is complete.

4. Examples

In this section, we present a couple of examples which are not nonexpansive but does
satisfy strict pseudo-contraction. Later, we also present a nontrivial example which illus-
trates the Theorem 2.

Example 1. [24] Let Γ = R and G : R → R be a mapping defined as

G(ζ) = −
(
tan−1(ζ) +

sin(ζ) + cos(ζ)

4

)
, for all ζ ∈ R.

The, mapping G is a β-strict pseudocontractive mapping with β = 5
9 , but it is not a

nonexpansive mapping.



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 13 of 15

Example 2. [24] Let Γ = R2 and G : Γ → Γ be a mapping defined as

G(ζ1, ζ2) =
(
−2 tan−1(ζ1), 2 cot

−1(ζ2)
)
, for all (ζ1, ζ2) ∈ R2.

The mapping G is a β-strict pseudocontractive mapping with β = 3
4 but it is not a nonex-

pansive mapping.

Example 3. Define the inner product on R as

⟨ϑ, ζ⟩ = −ϑζ for all ϑ, ζ ∈ R.

Define
H = {ζ ∈ R : ⟨ζ, ζ⟩ = −1}.

Then this inner product induces Riemannian metric ρ, on tangent space TpH ⊂ TpR for
p ∈ H defined by

ρ(ϑ, ζ) = cosh−1(−⟨ϑ, ζ⟩) for all ϑ, ζ ∈ H.

Then (H, ρ) is the Hadamard manifold with sectional curvature −1 at any point.
Now, let Γ = H, G1(ζ) = −2ζ,G2(ζ) = −5ζ and H1(ζ) = 1

3ζ + 1
2 sin ζ,H2(ζ) =

−6ζ. Here mappings G1, G2 are 1
3 ,

2
3 strict pseudocontractive mappings, respectively.

And the mappings H1, H2 are 1
4 ,

5
7 strict pseudocontractive mappings, respectively. Also

2⋂
i=1

F (Gi)
2⋂

i=1
F (Hi)

⋂
Φ = {0}. It satisfies all the conditions of the Theorem 2, hence for

i = 1, 2 the sequence generated by (9) converges to PΠ(ζ1).

Acknowledgements

We are very much thankful to the reviewers for their constructive comments and sug-
gestions which have been useful for the improvement of this paper.

Conflicts of Interest

The authors declare that they have no conflicts of interests.

Funding

This work was supported by Directorate of Research and Innovation, Walter Sisulu
University, South Africa.

Data Availability

There is no any availability of data.



P. Patel, R. Shukla / Eur. J. Pure Appl. Math, 18 (2) (2025), 5951 14 of 15

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