




































©2024 Ada Academica https://adac.eeEur. J. Math. Anal. 4 (2024) 8doi: 10.28924/ada/ma.4.8
Hybrid Inertial Iterative Method for Fixed point, Variational Inequality and Generalized Mixed

Equilibrium Problems in Banach Space

Lawal Umar1,∗, Yusuf Ibrahim2, M.S. Lawan3

1Department of Mathematics, Federal College of Education, Zaria Kaduna, Nigeria
lawalu4@gmail.com

2Department of Mathematics, Saadatu Rimi University of Education, Kumbotso Kano, Nigeria
danustazz@gmail.com

3Department of Mathematics and Statistics, Kaduna Polytechnic Kaduna, Nigeria
mslawankh@yahoo.com

∗Correspondence: lawalu4@gmail.com

Abstract. In this paper, we introduced a hybrid inertial iterative method which converges stronglyto a common element of solution of generalized mixed equilibrium, variational inequality and fixedpoint problems in a two uniformly smooth and uniformly convex Banach space. Our hybrid inertialiterative method, techniques of proof and corollaries improves, extends and generalizes many resultsin the literature.

1. introduction
Let B denotes a real Banach space with B∗ as the dual space of B. We consider 〈τ1, j〉 as thevalue of the functional j ∈ B∗ at τ1 ∈ B and ‖ . ‖ as the norm of B or B∗. Let c 6= ∅ be subset of
B. A mapping J : B −→ 2B

∗ is called normalized duality provided that
Jτ1 = {τ2 ∈ B∗ : 〈τ2, τ1〉 = ‖τ1 ‖2= ‖τ2 ‖2},∀τ1 ∈ B.

We denotes the short form GMEP as generalized mixed equilibrium problem: Find v1 ∈ C suchthat
D(v1, v2) + 〈Gv1, v2 − v1〉+ ϑ(v1, v2)− ϑ(v1, v1) ≥ 0, ∀v2 ∈ C, (1.1)

where D, ϑ : C × C −→ R and G : C −→ B∗ denotes the bifunctions and a nonlinear mappingrespectively, also R is consider as the set of all real numbers. Then, Sol(GMEP (1.1)) is consideras the solution set of GMEP.(1.1).

Received: 16 Jan 2024.
Key words and phrases. Hybrid Inertial Iterative Method; Fixed point problem; Variational Inequality problem;Generalized Mixed Equilibrium Problem. 1

https://adac.ee
https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 2If G ≡ 0, GMEP (1.1) reduces to generalized equilibrium problem ( with GEP as the short form):Find v1 ∈ C such that
D(v1, v2) + ϑ(v1, v2)− ϑ(v1, v1) ≥ 0,∀v2 ∈ C. (1.2)

Then, Sol(GEP (1.2)) is represent the solution set of GEP (1.2).If G ≡ 0 and ϑ ≡ 0, GMEP (1.1) becomes equilibrium problem ( with EP as the short form) [3]:Find v1 ∈ C such that
D(v1, v2) ≥ 0,∀v2 ∈ C. (1.3)

Then, Sol(EP (1.3)) is consider as the solution set of EP.(1.3).If D ≡ 0 and ϑ ≡ 0, GMEP (1.1) reduces to variational inequality problem ( with V IP as theshort form): Find v1 ∈ C such that
〈Gv1, v2 − v1〉 ≥ 0,∀v2 ∈ C. (1.4)

Then, Sol(V IP (1.4)) is consider as the solution set of V IP (1.4).

Definition 1.1. Let T : C −→ C be a mapping [6], then(i) a point v1 ∈ C is called fixed point of T provided that F (T ) = {v1 ∈ C : Tv1 = v1} 6= ∅;(ii) a point v0 ∈ C is called an asymptotic fixed point of T provided that {vn} ⊂ C, vn ⇀ v0 suchthat
lim
n→∞

‖ vn − Tvn ‖= 0.

The set of asymptotic fixed point of T is denoted by F̂ (T );(iii) T is called quasi−φ−nonexpansive provided that φ(v0, T v) ≤ φ(v0, v) and F (T ) 6= ∅, ∀v ∈
C, v0 ∈ F (T );(iv) T is called quasi−φ−asymptotically nonexpansive provided that F (T ) 6= ∅ and there exists asequence {kn} ⊂ [1,∞) with kn −→ 1 as n →∞ such that

φ(v0, T
nv) ≤ knφ(v0, v), ∀v ∈ C, v0 ∈ F (T ), n ≥ 1.

Definition 1.2. A function T : C −→ B∗ is said to be [6] :(i) Monotone if 〈τ1 − τ2, T τ1 − Tτ2〉 ≥ 0, ∀τ1, τ2 ∈ B;(ii) γ−inverse strongly monotone (with i sm as short form) if ∃γ > 0 such that
〈τ1 − τ2, T τ1 − Tτ2〉 ≥ γ ‖ Tτ1 − Tτ2 ‖2, ∀τ1, τ2 ∈ B;

(iii) Lipschitz continuous if ∃L > 0 such that ‖ Tτ1 − Tτ2 ‖≤ L ‖ τ1 − τ2 ‖, ∀τ1, τ2 ∈ B. If T is
γ − i sm, then it is Lipschitz continuous with 1

γ
as a constant.

Definition 1.3. A mapping ΠC : B −→ C is called generalized projection [6], provided that ΠCτ1 =

v0, for any τ1 ∈ B and v0 be the solution of φ(v0, τ1) = inf
v∈C

φ(v , τ1).

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 3An inertial-type algorithm is a method for speeding the convergence of the sequence of an algorithmintroduced by Polyak [16]. Numerous problems have been approximated by using inertial algorithms( for more details see, [4, 5, 12] and the references therein). Mainge [13] proposed and studied thedevelopment of an inertial- type algorithm method as follows:{
un = ωn + θn(ωn − ωn−1),

ωn+1 = (1− δn)un + δnTun.Takahashi and Zembayashi [17] Proposed an iterative process which converges strongly to a commonelement of solution of equilibrium problem and fixed point problem of relatively nonexpansivemapping. Furthermore, the generalization of the proposed iterative process [17] have been carriedout by many researchers ( for more details see, [7, 8, 11, 18, 20] and the references therein). Kazmiand Ali [10] introduced an iterative algorithm for solving a common solution of EP.(1.3). and fixedpoint problemof quasi−φ− asymptotically nonexpansive mapping.
Alansari et al. [1] studied an inertial iterative method for finding a common solution of generalizedequilibrium, variational inequality and fixed point problems using the sequences {xn} and {zn}generated by the iterative algorithm:

x0 = x1, z0 ∈ C, C0 := C;

µn = xn + αn(xn − xn−1);

yn = ΠCJ
−1(Jµn − wnGµn);

un = J−1(δnJzn + (1− δn)JTyn);

zn+1 = Trnun;

Cn = {u ∈ C : φ(u, zn+1) ≤ δnφ(u, zn) + (1− δn)φ(u, µn);

Qn = 〈u ∈ C : xn − u, Jxn − Jx0〉 ≤ 0};
xn+1 = ΠCn∩Qnx0,∀n ≥ 0,where {αn} ⊂ (0, 1), {wn} ⊂ (0,∞), {δn} ⊂ [0, 1] and {rn} ⊂ [a,∞), for some a > 0. Then, {xn}converges strongly to $ = ΠΓx0.Farid et al. [6] proposed the following inertial algorithm for approximating a common solution ofgeneralized mixed equilibrium problem, variational inequality problem and fixed point problem forfamily of quasi−φ−nonexpansive mappings:

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 4



x0, x1 ∈ q, q1 := q;

ωn = xn + θn(xn − xn−1);

yn = ΠqJ
−1(Jωn − wnQωn);

vn = J−1(δn,0Jωn +

N∑
i=1

δn,iJTiωn);

zn = J−1(αnJyn + (1− αn)Jvn);

un = Trnzn;

qn = {u ∈ q : φ(u, un) ≤ φ(u, ωn);

Qn = 〈u ∈ q : xn − u, Jxn − Jx0〉 ≤ 0};
xn+1 = Πqn∩Qnx0,∀n ≥ 1.

Consider {δn,i} and {αn} ⊂ [0, 1], {wn} ⊂ (0,∞), {θn} ⊂ (0, 1) and {rn} ⊂ [a,∞), for some
a > 0. It has been proved that {xn} is a strong convergent to x̂ = ΠΩx0.Motivated and inspired by the work of Kazmi and Ali [10], Alansari et al. [1] and Farid et al. [6]. Weproposed a hybrid inertial iterative algorithm for approximating a common solution of GMEP.(1.1),

V IP (1.4) and fixed point problem for a family of two quasi−φ−asymptotically nonexpansive map-pings in two- uniformly convex and uniformly smooth Banach spaces. Our result extends andimproves the results of Kazmi and Ali [10], Alansari et al. [1] and Farid et al. [6], many results inthe literature.
2. Preliminaries

Let W = {τ1 ∈ B :‖ τ1 ‖= 1} be the unit sphere of B. If for any ε ∈ (0, 2] there exists δ > 0 suchthat ‖ τ1 − τ2 ‖≥ ε =⇒
‖ τ1 + τ2 ‖

2
≤ 1− δ, ∀τ1, τ2 ∈ W, then B is called uniformly convex. B is

called strictly convex if ‖ τ1 + τ2 ‖
2

< 1, ∀τ1, τ2 ∈ W and τ1 6= τ2. The space B is called smooth
if lim
t→0

‖ τ1 + tτ2 ‖ − ‖ τ1 ‖
t

exists, ∀τ1, τ2 ∈ W and also is said to be uniformly smooth if the limitis attained uniformly, ∀τ1, τ2 ∈ W.A function φ : B × B −→ R defined by
φ(τ1, τ2) =‖ τ1 ‖2 −2〈τ1, Jτ2〉+ ‖τ2 ‖2, ∀τ1, τ2 ∈ B.

is consider as Lyapunov functional. From the definition of φ, the following properties can be veri-fied [6]:
(L1) (‖ τ1 ‖ − ‖ τ2 ‖)2 ≤ φ(τ1, τ2) ≤ (‖ τ1 ‖ + ‖ τ2 ‖)2, ∀τ1, τ2 ∈ B;

(L2) φ(τ1, J
−1(λJτ2 + (1− λ)Jτ3)) ≤ λφ(τ1, τ2) + (1− λ)φ(τ1, τ3), ∀τ1, τ2, τ3 ∈ B,

(L3) φ(τ1, τ2) =‖ τ1 ‖ ‖ Jτ1 − Jτ2 ‖ + ‖ τ2 ‖ ‖ τ1 − τ2 ‖, ∀τ1, τ2 ∈ B.

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Eur. J. Math. Anal. 10.28924/ada/ma.4.8 5

Remark 2.1. Consider B as smooth, strictly convex and reflexive Banach space, then
φ(τ1, τ2) = 0⇐⇒ τ1 = τ2, ∀τ1, τ2 ∈ B.

Lemma 2.2. [9] Let C 6= ∅ be closed convex subset of a stricly convex, reflexive and smooth Banach
space B. Then, ∃ a unique element τ0 ∈ C such that φ(τ0, τ1) = inf

v∈C
φ(v , τ1), for τ1 ∈ B.

Lemma 2.3. [15] Let B be a uniformly convex and smooth Banach space, C ⊂ B be closed convex
and T : C −→ C be closed and quasi−φ−asymptotically nonexpansive mapping. Then, F (T ) is
closed and convex.

Lemma 2.4. [14] Let C 6= ∅ be closed convex subset of B and Q : C −→ B∗ be monotone and
hemicontinuous function. Then V IP (1.4). is closed and convex

Lemma 2.5. [19] Let B be a 2−uniformly convex and smooth Banach space. Then, τ1, τ2 ∈
B, φ(τ1, τ2) ≥ δ ‖ τ1 − τ2 ‖2, where 0 < δ ≤ 1 and called two-uniformly convex constant.

Lemma 2.6. [19] Let B be a two-uniformly convex Banach space, then

‖ τ1 − τ2 ‖≤
2

δ
‖ Jτ1 − Jτ2 ‖, ∀τ1, τ2 ∈ B,

where 0 < δ ≤ 1.

Lemma 2.7. [9] Let E be a smooth and uniformly convex Banach space and let {un} and {vn}
be sequences in E such that either {un} or {vn} is bounded. If lim

n→∞
φ(un, vn) = 0, then lim

n→∞
‖

un − vn ‖= 0.

Remark 2.8. By considering (L3), it is observe that the converse of Lemma 2.7 is true, providedthat {un} and {vn} are bounded
Lemma 2.9. [2] Let C 6= ∅ be closed convex subset of a stricly convex, reflexive and smooth Banach
space B. Then,

φ(v ,ΠCτ1) + φ(ΠCτ1, τ1) ≤ (v , τ1), ∀v ∈ C, τ1 ∈ B.

And, so for any τ1 ∈ B and v ∈ C,

u = ΠCτ1 ⇐⇒ 〈v − u, Jτ1 − Jv〉, ∀u ∈ C.

Assumption 1: Consider D : C × C −→ R as a bifunction satisfies the following assumptions [3]:
(D1) D(v , v) = 0,∀v ∈ C;

(D2) D is monotone, 1.e, D(v , u) +D(u, v) ≤ 0, ∀v , u ∈ C;

(D3) the mapping v 7→ D(v , u) is upper hemicontinuity, ∀ u ∈ C.
(D4) the mapping u 7→ D(v , u), u ∈ C is convex and lower semicontinuous.
Assumption 2: Also consider ϑ : C×C −→ R as a bifunction satisfying the following assumptions:

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 6

(ϑ1) ϑ is skew-symmetric, i.e., ϑ(v , v)− ϑ(v , u)− ϑ(u, v) + ϑ(u, u) ≥ 0,∀v , u ∈ C;

(ϑ2) ϑ is convex in the second argument;
(ϑ3) ϑ is continuous.
Lemma 2.10. [1, 6, 21] Let B a uniformly smooth, strictly convex and reflexive Banach space and
C ⊂ B be closed. Let G : C −→ B∗ be a continuous and monotone mapping, D : C × C −→ R be
a bifunction satisfying Assumptions 1 and ϑ : C ×C −→ R be a bifunction satisfying Assumptions
2. For any given number r > 0 and τ1 ∈ B, define a mapping Tr : B −→ C by

Tr (τ1) = {u ∈ C : D(u, v) + 〈v − u, Gu〉+
1

r
〈v − u, Ju − Jτ1〉+ ψ(u, v)− ψ(u, u) ≥ 0,∀y ∈ C},

∀v ∈ B.
The mapping Tr has the following properties:
(p1) Tr is single-valued;
(p2) Tr is a firmly nonexpansive - type mapping, for all τ1, τ2 ∈ B,

〈Trτ1 − Trτ2, JTrτ1 − JTrτ2〉 ≤ 〈Trτ1 − Trτ2, Jτ1 − Jτ2〉,

(p3) F (Tr ) = Sol(GMEP (1.1)) is closed convex set of C;
(p4) Tr is quasi−φ− nonexpansive;
(p5) φ(v0, Trτ1) + φ(Trτ1, τ1) ≤ φ(v0, τ1), ∀v0 ∈ F (Tr ), τ1 ∈ B.

Furthermore, consider the map Φ : B × B∗ −→ R, defined by
Φ(τ1, τ

∗
1 ) =‖ τ1 ‖2 −〈τ1, τ

∗
1 〉+ ‖ τ∗1 ‖2

Observe that Φ(τ1, τ
∗
1 ) = Φ(τ1, J

−1τ∗1 )

Lemma 2.11. [2] Let B be a strictly convex, smooth and reflexive Banach space. Then

Φ(τ1, τ
∗
1 ) + 2〈J−1τ∗1 − τ1, τ

∗
2 〉 ≤ Φ(τ1, τ

∗
1 + τ∗2 ), ∀τ1 ∈ B, τ∗1 , τ∗2 ∈ B∗.

3. Main Results
Theorem 3.1. Let C be a nonempty closed and convex subset of a 2−uniformly smooth and uniformly
convex Banach space B with B∗ as the dual space of B. Let Q :−→ B∗ be a γ−ism mapping
with γ ∈ (0, 1) as a constant. Let D : C × C −→ R be a bifunction satisfying Assumption 1,
ϑ : C × C −→ R be a bifunction satisfying Assumption 2 and G : C −→ B∗ be a monotone and
continuous mapping. Let Ti : C −→ C and Si : C −→ C, for each i = 1, 2, ..., N be two finite family
of closed li−Lipschitz continuous and uniformly quasi−φ−asymptotically nonexpansive mappings
such that Ω :=

(
∩Ni=1 F (Ti)

)
∩
(
∩Ni=1 F (Si)

)
∩ Sol

(
V IP (1.4)

)
∩ Sol

(
GMEP (1.1)

)
6= ∅. Let {xn}

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 7

generated by algorithm :

x0, x1 ∈ C, C1 := C,

ωn = xn + αn(xn − xn−1),

vn = ΠCJ
−1(Jωn − βnQωn),

yn = J−1(µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn);

zn = J−1(ηn,0Jvn +

N∑
i=1

ηn,iJS
n
i yn),

un = Trnzn,

Cn+1 = {u ∈ Cn : φ(u, un) ≤ k2
nφ(u, ωn)},

xn+1 = ΠCn+1
x0, ∀n ≥ 1,

(3.1)

where {αn} ⊂ (0, 1), {µn,i} ⊂ [0, 1] and {ηn,i} ⊂ (0, 1] satisfying the following conditions:

(S1)

N∑
i=0

µn,i = 1;

(S2)

N∑
i=0

ηn,i = 1;

(S3) lim sup
n→∞

ηn,0 < 1;

(S4) for same a > 0, rn ∈ [a,∞);

(S5) {βn} ⊂ (0,∞) satisfying the condition 0 < lim inf
n→∞

βn <
δ2γ

2
, where 0 < δ ≤ 1.

Then, {xn} converges strongly to $, where $ = ΠΩx0 is consider as the generalized projection
of $ onto Ω.

Proof. We consider the proof in the following steps:
Step 1 : We show that Cn+1 is closed and convex for each n ≥ 1 and {xn} is well defined.Observe clearly that C1 = C is closed and convex. Suppose that Cn is closed and convex for each
n ∈ N. Now, we know from 3.1 that for any u ∈ Cn,

φ(u, un) ≤ k2
nφ(u, ωn) ⇐⇒ (1− k2

n )
[
‖ u ‖2 −2(1− k2

n )〈u, Jun〉+ 2k2
n 〈u, Jωn − Jun〉

]
≤ k2

n ‖ ωn ‖2 − ‖ un ‖2 .

Then, Cn+1 is closed and convex. Implies that ΠCn+1
x0 is well defined ∀n ≥ 1, also {xn} is welldefined. Furthermore since Ω 6= ∅, by considering Lemma 2.3, 2.4 and 2.10 we conclude that Ω isclosed and convex, and so ΠΩx0 is well defined.

Step 2 : we show that Ω ⊂ Cn, ∀n ≥ 1. It is Obvious that Ω ⊂ C1 = C. Suppose that Ω ⊂ Cn forsome n ≥ 1. Let x̂ ∈ Ω, from the definition of φ, quasi−φ−asymptotically nonexpansive mapping

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Eur. J. Math. Anal. 10.28924/ada/ma.4.8 8of Si and convexity of ‖ . ‖2 we have the following estimate:
φ(x̂ , un) = φ(x̂ , Trnzn)

≤ φ(x̂ , zn) (3.2)
= φ

(
x̂ , J−1(ηn,0Jvn +

N∑
i=1

ηn,iJS
n
i yn)

)
= ‖ x̂ ‖2 −2(〈x̂ , ηn,0Jvn +

N∑
i=1

ηn,iJS
n
i yn〉) + ‖ηn,0Jvn +

N∑
i=1

ηn,iJS
n
i yn‖2

≤ ‖ x̂ ‖2 −2ηn,0〈x̂ , Jvn〉 − 2

N∑
i=1

ηn,i 〈x̂ , JSni yn〉+ ηn,0‖Jvn‖2

+

N∑
i=1

ηn,i‖JSni yn‖2

= ηn,0
(
‖x̂‖2 − 2〈x̂ , Jvn〉+ ‖vn‖2

)
+

N∑
i=1

ηn,i
(
‖x̂‖2 − 2〈x̂ , JSni yn〉+ ‖Sni yn‖2

)
= ηn,0φ(x̂ , vn) +

N∑
i=1

ηn,iφ(x̂ , Sni yn)

≤ ηn,0φ(x̂ , vn) + kn

N∑
i=1

ηn,iφ(x̂ , yn) (3.3)
Similarly, by quasi−φ−asymptotically nonexpansive of Ti , definition of φ and convexity of ‖ . ‖2,we estimate as follows:

φ(x̂ , yn) = φ
(
x̂ , J−1(µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn)

)
= ‖ x̂ ‖2 −2(〈x̂ , µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn〉)+ ‖ µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn ‖2

≤ ‖ x̂ ‖2 −2µn,0〈x̂ , Jωn〉 − 2

N∑
i=1

µn,i 〈x̂ , JT ni ωn〉+ µn,0‖Jωn‖2

+

N∑
i=1

µn,i‖JT ni ωn‖2

= µn,0
(
‖x̂‖2 − 2〈x̂ , Jωn〉+ ‖ωn‖2

)
+

N∑
i=1

µn,i
(
‖ x̂ ‖2 −2〈x̂ , JT ni ωn〉+ ‖T ni ωn‖2

)
= µn,0φ(x̂ , ωn) +

N∑
i=1

µn,iφ(x̂ , T ni ωn)

≤ µn,0φ(x̂ , ωn) + kn

N∑
i=1

µn,iφ(x̂ , ωn)

≤ knµn,0φ(x̂ , ωn) + kn

N∑
i=1

µn,iφ(x̂ , ωn)

= knφ(x̂ , ωn) (3.4)

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Eur. J. Math. Anal. 10.28924/ada/ma.4.8 9It has been observe from (3.3) and (3.4) that
φ(x̂ , un) ≤ ηn,0φ(x̂ , vn) + kn

N∑
i=1

ηn,i
[
knφ(x̂ , ωn)

]
= ηn,0φ(x̂ , vn) + k2

n

N∑
i=1

ηn,iφ(x̂ , ωn)

≤ k2
nηn,0φ(x̂ , vn) + k2

n

N∑
i=1

ηn,iφ(x̂ , ωn) (3.5)
Also, by Lemma 2.6 and 2.11, we estimate as:

φ(x̂ , vn) = φ
(
x̂ ,ΠCJ

−1(Jωn − βnQωn)
)

≤ φ
(
x̂ , J−1(Jωn − βnQωn)

)
= Φ

(
x̂ , Jωn − βnQωn)

≤ Φ
(
x̂ , (Jωn − βnQωn) + βnQωn

)
− 2〈J−1(Jωn − βnQωn)− x̂ , βnQωn〉

= Φ(x̂ , Jωn)− 2βn〈J−1(Jωn − βnQωn)− x̂ , Qωn〉

= φ(x̂ , ωn)− 2〈ωn − x̂ , Qωn〉 − 2βn〈J−1(Jωn − βnQωn)− ωn, Qωn〉

= φ(x̂ , ωn)− 2〈ωn − x̂ , Qωn −Qx̂〉 − 2βn〈J−1(Jωn − βnQωn)− ωn, Qωn〉

≤ φ(x̂ , ωn)− 2βnγ ‖ Qωn‖2 + 2βn ‖ J−1(Jωn −Qωn)− J−1Jωn‖‖Qωn‖2

≤ φ(x̂ , ωn)− 2βnγ ‖ Qωn ‖2 +
4β2
n

δ2
‖ Qωn ‖2

= φ(x̂ , ωn)− 2βn
(
γ −

2βn
δ2

)
‖ Qωn ‖2, (3.6)

if follows by combined with βn < δ2

2
that
φ(x̂ , vn) ≤ φ(x̂ , ωn) (3.7)

Now, putting (3.7) in (3.5) leads to
φ(x̂ , un) ≤ k2

nηn,0φ(x̂ , ωn) + k2
n

N∑
i=1

ηn,iφ(x̂ , ωn)

= (ηn,0 +

N∑
i=1

ηn,i)k
2
nφ(x̂ , ωn)

= k2
nφ(x̂ , ωn),

which gives
φ(x̂ , un) ≤ k2

nφ(x̂ , ωn), (3.8)
Therefore x̂ ∈ Cn+1, implies that Ω ⊂ Cn+1. Hence Ω ⊂ Cn, ∀n ≥ 1.

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 10

Step 3 : we show that {xn}, {ωn}, {vn}, {yn}, {zn} and {un} are bounded and {xn} is Cauchy.We consider xn = ΠCnx0 and Cn+1 ⊂ Cn, ∀n ≥ 1. Then from Lemma 2.9, we observe that
φ(xn, x0) ≤ φ(xn+1, x0)

Hence {φ(xn, x0)} is non decreasing. Also it has been observe that
φ(xn, x0) = φ(ΠCnx0, x0) ≤ φ(x̂ , x0)− φ(x̂ , xn) ≤ φ(x̂ , x0),

which gives that {φ(xn, x0)} is bounded and {xn} is also bounded. Therefore, since {φ(xn, x0)} nondecreasing. {φ(xn, x0)} convergent. Taking the advantage of {xn} as a bounded sequence impliesthat {ωn}, {vn}, {yn}, {zn} and {un} are all bounded. Also by Lemma 2.9, we have
φ(xm, xn) = φ(xm,ΠCnx0)

≤ φ(xm, x0)− φ(xn, x0) −→ 0 as n,m →∞. (3.9)
By Lemma 2.7, we have lim

n→∞
‖ xm − xn ‖= 0. Hence {xn} is a Cauchy sequence.

Step 4 : we show that xn −→ $, ωn −→ $, un −→ $, zn −→ $, yn −→ $and vn −→ $ (as n → ∞). Since {xn} is a Cauchy sequence, then by the closedness of C andthe completeness of B, we can assume that there exists $ ∈ C such that
lim
n→∞

xn = $. (3.10)
Now, setting m = n + 1 in (3.9), we obtain

lim
n→∞

φ(xn+1, xn) = 0. (3.11)
Using Lemma 2.7, we get

lim
n→∞
‖xn+1 − xn‖ = 0. (3.12)

We observe from (3.1) that
‖ ωn − xn ‖=‖ αn(xn − xn−1) ‖≤‖ xn − xn−1 ‖

Using (3.12), we arrive at
lim
n→∞
‖ωn − xn‖ = 0. (3.13)

By (3.10) and (3.13), we conclude that
lim
n→∞

ωn = $. (3.14)
Taking the advantage of Remark 2.8, (3.13) and boundedness of {ωn}, we get

lim
n→∞

φ(ωn, xn) = 0. (3.15)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 11Also, by (3.12) and (3.13), we obtain
lim
n→∞
‖xn+1 − ωn ‖= 0. (3.16)

Using Remark 2.8, we present (3.16) as
lim
n→∞

φ(xn+1, ωn) = 0. (3.17)
We observe from xn+1 = ΠCn+1

x0 ∈ Cn+1 ⊂ Cn and definition of Cn that
φ(xn+1, un) ≤ k2

nφ(xn+1, ωn)

Using (3.17,) we obtain
lim
n→∞

φ(xn+1, un) = 0.

Applying Lemma 2.7, we get
lim
n→∞

‖ xn+1 − un ‖= 0. (3.18)
Taking the advantage of triangular inequality, we present

‖xn − un‖ ≤ ‖xn − xn+1‖+ ‖xn+1 − un‖

By (3.12) and (3.18), we obtain
lim
n→∞

‖ xn − un ‖= 0. (3.19)
It follows from (3.10) and (3.19) that

lim
n→∞

un = $. (3.20)
Similarly, by definition of Cn and xn+1 = ΠCn+1

x0 ∈ Cn+1 ⊂ Cn, we also present that
φ(xn+1, zn) ≤ k2

nφ(xn+1, ωn)

By applying (3.17,) we arrive at
lim
n→∞

φ(xn+1, zn) = 0.

Using Lemma 2.7, we have
lim
n→∞

‖ xn+1 − zn ‖= 0. (3.21)
Taking into account that

‖xn − zn‖ ≤ ‖xn − xn+1‖+ ‖xn+1 − zn‖

Using (3.12) and (3.21), we get
lim
n→∞

‖ xn − zn ‖= 0. (3.22)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 12By considering(3.10) and (3.22), we obtain
lim
n→∞

zn = $. (3.23)
Also from the definition of Cn and xn+1 = ΠCn+1

x0 ∈ Cn+1 ⊂ Cn, we estimate as
φ(xn+1, yn) ≤ k2

nφ(xn+1, ωn)

By (3.17,) we get
lim
n→∞

φ(xn+1, yn) = 0.

It follows from Lemma 2.7 that
lim
n→∞

‖ xn+1 − yn ‖= 0. (3.24)
By triangular inequality, we obtain

‖xn − yn‖ ≤ ‖xn − xn+1‖+ ‖xn+1 − yn‖

Also by (3.12) and (3.24), we get
lim
n→∞

‖ xn − yn ‖= 0. (3.25)
Using (3.10) and (3.25), we obtain

lim
n→∞

yn = $. (3.26)
Finally, by considering xn+1 = ΠCn+1

x0 ∈ Cn+1 ⊂ Cn and definition of Cn, we present that
φ(xn+1, vn) ≤ k2

nφ(xn+1, ωn)

Applying (3.17,) we obtain
lim
n→∞

φ(xn+1, vn) = 0.

By Lemma 2.7, we get
lim
n→∞

‖ xn+1 − vn ‖= 0. (3.27)
We consider the following estimate using triangular inequality

‖xn − vn‖ ≤ ‖xn − xn+1‖+ ‖xn+1 − vn‖

Using (3.12) and (3.27), we obtain
lim
n→∞

‖ xn − vn ‖= 0. (3.28)
Using (3.10) and (3.28), we obtain

lim
n→∞

vn = $. (3.29)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 13

Step 4 : we show that ‖ ωn − T ni ωn ‖=‖ yn − Sni yn ‖= 0. Now, taking the advantage of J asuniformly continuity on bounded sets, then it follows from (3.16) and (3.24) that
‖ Jωn − Jxn+1 ‖=‖ Jxn+1 − Jyn ‖= 0. (3.30)

From (3.1), we observe that
‖Jxn+1 − Jyn ‖ = ‖ Jxn+1 −

(
µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn

)
‖

= ‖
N∑
i=1

µn,iJxn+1 −
N∑
i=1

µn,iJT
n
i ωn + µn,0Jxn+1 − µn,0Jωn ‖

= ‖
N∑
i=1

µn,i
(
Jxn+1 − JT ni ωn

)
+ µn,0

(
Jxn+1 − Jωn

)
‖

≥
N∑
i=1

µn,i ‖ Jxn+1 − JT ni ωn ‖ −µn,0 ‖ Jωn − Jxn+1 ‖,

this gives
‖ Jxn+1 − JT ni ωn ‖≤

1
N∑
i=1

µn,i

[
‖ Jxn+1 − Jyn ‖ +µn,0 ‖ Jωn − Jxn+1 ‖

]
.

By (3.30), we arrive at
lim
n→∞

‖ Jxn+1 − JT ni ωn ‖= 0.

As J−1 is uniform norm-to-norm continuous on bounded sets, we present that
lim
n→∞

‖ xn+1 − T ni ωn ‖= 0. (3.31)
Taking into account that

‖ ωn − T ni ωn ‖≤‖ ωn − xn+1 ‖ + ‖ xn+1 − T ni ωn ‖

By (3.16) and (3.31), we obtain
lim
n→∞

‖ ωn − T ni ωn ‖= 0. (3.32)
Similarly, we observe from (3.21), (3.27) and by continuity of J that

‖ Jxn+1 − Jzn ‖=‖ Jxn+1 − Jvn ‖= 0. (3.33)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 14Also by (3.1), we observe that
‖ Jxn+1 − Jzn ‖ = ‖ Jxn+1 −

(
ηn,0Jvn +

N∑
i=1

ηn,iJS
n
i yn
)
‖

= ‖
N∑
i=1

ηn,iJxn+1 −
N∑
i=1

ηn,iJS
n
i yn + ηn,0Jxn+1 − ηn,0Jvn ‖

= ‖
N∑
i=1

ηn,i
(
Jxn+1 − JSni yn

)
+ ηn,0

(
Jxn+1 − Jvn

)
‖

≥
N∑
i=1

ηn,i ‖ Jxn+1 − JSni yn ‖ −ηn,0 ‖ Jvn − Jxn+1 ‖,

this implies
‖Jxn+1 − JSni yn‖ ≤

1
N∑
i=1

ηn,i

[
‖Jxn+1 − Jzn‖+ ηn,0‖Jvn − Jxn+1 ‖

]
.

Also by (3.33), we get
lim
n→∞

‖ Jxn+1 − JSni yn ‖= 0.

Applying J−1 as uniform norm-to-norm continuous on bounded sets, we have
lim
n→∞

‖ xn+1 − Sni yn ‖= 0. (3.34)
By triangular inequality, we obtain

‖ yn − Sni yn ‖≤‖ yn − xn+1 ‖ + ‖ xn+1 − Sni yn ‖

By (3.24) and (3.34), we get
lim
n→∞

‖ yn − Sni yn ‖= 0. (3.35)
Therefore by (3.32) and (3.35), we conclude that

lim
n→∞

‖ ωn − T ni ωn ‖= lim
n→∞

‖ yn − Sni yn ‖= 0.

Step 5 : we show that $ ∈ Ω. To show this we claim as follows:
We claim that $ ∈ ( ∩Ni=1 F (Ti)

)
∩
(
∩Ni=1 F (Si)

)
. By triangular inequality for i ≥ 1, we have

‖ T ni ωn −$ ‖≤‖ T ni ωn − ωn ‖ + ‖ ωn −$ ‖ .

Using (3.14) and (3.32), we arrive at
lim
n→∞

‖ T ni ωn −$ ‖= 0. (3.36)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 15By the assumption that for each Ti is uniformly Li−Lipschitz continuous, we obtain
‖T n+1
i ωn − T ni ωn‖ ≤ ‖T n+1

i ωn − T n+1
i ωn+1‖+ ‖T n+1

i ωn+1 − ωn+1‖

+ ‖ωn+1 − ωn‖+ ‖ωn − T ni ωn‖

≤ (Li + 1)‖ωn+1 − ωn‖+ ‖T n+1
i ωn+1 − ωn+1‖+ ‖ωn − T ni ωn‖.

By (3.12) and (3.32,) we get
lim
n→∞
‖T n+1
i ωn − T ni ωn‖ = 0.

Which yields from (3.36) that
lim
n→∞
‖T n+1
i ωn −$‖ = 0, ∀i ≥ 1.

Consequently, we get Ti(T ni )ωn −→ $ ( as n →∞). In view of the closedness of Ti , we arrive at
Ti$ = $, ∀i ≥ 1. Thus $ ∈ ∩Ni=1F (Ti). Furthermore, following similar argument as above, onecan also claim that $ ∈ ∩Ni=1F (Si). Hence

$ ∈
(
∩ni=1 F (Ti)

)
∩
(
∩ni=1 F (Si)

)
.

Next, we claim that $ ∈ Sol(V IP (1.4)). Consider the triangular inequality
‖ ωn − zn ‖≤‖ ωn − xn ‖ + ‖ xn − zn ‖ .

Using (3.13) and (3.22,) leads to
lim
n→∞

‖ ωn − zn ‖= 0. (3.37)
From the uniform continuity of J on bounded set, we get

lim
n→∞

‖ Jωn − Jzn ‖= 0. (3.38)
Since x̂ ∈ Ω, then it follows from (3.2), (3.3), (3.4) and (3.6) that

φ(x̂ , zn) ≤ ηn,0
[
φ(x̂ , ωn)− 2βn

(
γ −

2βn
δ2

)
‖Qωn ‖2

]
+ kn

N∑
i=1

ηn,i
[
knφ(x̂ , ωn)

]
≤ k2

nηn,0φ(x̂ , ωn) + k2
n

N∑
i=1

ηn,iφ(x̂ , ωn)− 2βnηn,0
(
γ −

2βn
δ2

)
‖ Qωn ‖2

= k2
nφ(x̂ , ωn)− 2βnηn,0

(
γ −

2βn
δ2

)
‖ Qωn ‖2,

implies that
2βnηn,0

(
γ −

2βn
δ2

)
‖ Qωn ‖2≤ k2

nφ(x̂ , ωn)− φ(x̂ , zn) (3.39)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 16But
k2
nφ(x̂ , ωn)− φ(x̂ , zn) = k2

n

[
‖ x̂ ‖2 −2〈x̂ , Jωn〉+ ‖ ωn ‖2

]
−
[
‖ x̂ ‖2 −2〈x̂ , Jzn〉+ ‖ zn ‖2

]
= (k2

n − 1)‖x̂‖2 − 2(k2
n − 1)〈x̂ , Jzn〉 − 2k2

n 〈x̂ , Jωn − Jωn〉

+ k2
n‖ωn ‖2 − ‖ zn ‖2

= (k2
n − 1) ‖ x̂ ‖2 −2(k2

n − 1)〈x̂ , Jzn〉 − 2k2
n 〈x̂ , Jωn − Jzn〉

+ (k2
n − 1)‖ωn ‖2 + ‖ ωn ‖2 − ‖ zn ‖2

≤ | (k2
n − 1) ‖ x̂ ‖2| + | 2(k2

n − 1)〈x̂ , Jzn〉 | + | 2k2
n 〈x̂ , Jωn − Jzn〉 |

+ | (k2
n − 1)‖ωn ‖2| + |‖ ωn ‖2 + ‖ zn ‖2|

≤ (k2
n − 1) ‖ x̂ ‖2 +2(k2

n − 1) ‖ x̂ ‖ ‖ Jzn ‖ +2k2
n ‖ x̂ ‖ ‖ Jωn − Jzn ‖

+ (‖ ωn − zn ‖)(‖ ωn ‖ + ‖ zn ‖).

Since kn −→ 1 as n −→∞, then by (3.37) and (3.38,) we obtain
lim
n→∞

(
k2
nφ(x̂ , ωn)− φ(x̂ , zn)

)
= 0. (3.40)

Also since βnηn,0(γ − 2βn
δ2

)
> 0, by (3.39) and (3.40), we have

lim
n→∞

‖ Qωn ‖= 0. (3.41)
Taking the advantage of Q as γ − i sm and so 1

γ
−Lipschitz continuous. Therefore, it follows from(3.38) and (3.40) that $ ∈ Q−1(0). Hence, $ ∈ Sol(V IP (1.4)).

We also claim that $ ∈ Sol(GMEP (1.1)). Consider the triangular inequality
‖ un − zn ‖≤‖ un − xn ‖ + ‖ xn − zn ‖ .

By (3.19) and (3.22), we get
lim
n→∞

‖ un − zn ‖= 0.

From uniform continuity of J on bounded sets, we obtain
lim
n→∞

‖ Jun − Jzn ‖= 0. (3.42)
Since rn ≥ a and by (3.42), we have

lim
n→∞

‖ Jun − Jzn ‖
rn

= 0. (3.43)
Equation un = Trnzn implies that

H(un, v) +
1

rn
〈v − un, Jun − Jzn〉+ ϑ(v , un)− ϑ(un, un) ≥ 0, ∀v ∈ C.

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 17where
H(un, v) = D(un, v) + 〈Gun, v − un〉.

By applying Assumption (D2), we obtain
1

rn
〈v − un, Jun − Jzn〉 ≥ −H(un, v)− ϑ(v , un) + ϑ(un, un)

≥ H(v , un)− ϑ(v , un) + ϑ(un, un).

Letting n −→∞, by Assumption (D4) and (3.43), we get
H(v ,$)− ϑ(v ,$) + ϑ($,$) ≤ 0, ∀v ∈ C.

For all s ∈ (0, 1] and v ∈ C, setting vs := sv + (1− s)$. Therefore vs ∈ C and then,
H(vs ,$)− ϑ(vs ,$) + ϑ($,$) ≤ 0.

By Assumption (D1)− (D4), we estimate as
0 = H(vs , vs)

≤ sH(vs , v) + (1− s)H(vs ,$)

≤ sH(vs , v) + (1− s)
[
ϑ(vs ,$)− ϑ($,$)

]
≤ sH(vs , v) + (1− s)

[
ϑ(v ,$)− ϑ($,$)

]
As s > 0, from Assumption (D3), we conclude that

H($, v) + ϑ(v ,$)− ϑ($,$) ≥ 0, ∀v ∈ C.

Hence, $ ∈ Sol(GMEP (1.1)).

Step 6 : Finally we show that $ = ΠΩx0 and so xn −→ ΠΩx0 as n −→ ∞. Putting x∗ = ΠΩx0,since x∗ ∈ Ω ⊂ Cn and xn = ΠΩx0, we have
φ(xn, x0) ≤ φ(x∗, x0), ∀n ≥ 0.

Then
φ($, x0) = lim

n→∞
φ(xn, x0) ≤ φ(x∗, x0),

implies that $ = x∗ and since x∗ = ΠΩx0, then we conclude that xn −→ $ = ΠΩx0, as n →∞.This completes the proof. �

Corollary 3.2. Let C be a nonempty closed and convex subset of a 2−uniformly smooth and
uniformly convex Banach space B with B∗ as the dual space of B. Let D : C × C −→ R be
a bifunction satisfying Assumption 1, ϑ : C × C −→ R be a bifunction satisfying Assumption 2
and G : C −→ B∗ be a monotone and continuous mapping. Let Ti : C −→ C and Si : C −→
C, for each i = 1, 2, ..., N be two finite family of closed li−Lipschitz continuous and uniformly

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 18

quasi−φ−asymptotically nonexpansive mappings such that Ω :=
(
∩Ni=1 F (Ti)

)
∩
(
∩Ni=1 F (Si)

)
∩

∩Sol
(
GMEP (1.1)

)
6= ∅. Let {xn} generated by algorithm :

x0, x1 ∈ C, C1 := C,

ωn = xn + αn(xn − xn−1),

yn = J−1(µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn);

zn = J−1(ηn,0Jωn +

N∑
i=1

ηn,iJS
n
i yn),

un = Trnzn,

Cn+1 = {u ∈ Cn : φ(u, un) ≤ k2
nφ(u, ωn)},

xn+1 = ΠCn+1
x0, ∀n ≥ 1,

where {αn} ⊂ (0, 1), {µn,i} ⊂ [0, 1] and {ηn,i} ⊂ (0, 1] satisfying the following conditions:

(S1)

N∑
i=0

µn,i = 1;

(S2)

N∑
i=0

ηn,i = 1;

(S3) lim sup
n→∞

ηn,0 < 1;

(S4) for same a > 0, rn ∈ [a,∞).

Then, {xn} converges strongly to $, where $ = ΠΩx0 is consider as the generalized projection
of $ onto Ω.

Corollary 3.3. Let C be a nonempty closed and convex subset of a 2−uniformly smooth and
uniformly convex Banach space B with B∗ as the dual space of B. Let D : C × C −→ R be a
bifunction satisfying Assumption 1 and G : C −→ B∗ be a monotone and continuous mapping.
Let Ti : C −→ C and Si : C −→ C, for each i = 1, 2, ..., N be two finite family of closed
li−Lipschitz continuous and uniformly quasi−φ−asymptotically nonexpansive mappings such that
Ω :=

(
∩Ni=1 F (Ti)

)
∩
(
∩Ni=1 F (Si)

)
∩ ∩Sol

(
GEP (1.2)

)
6= ∅. Let {xn} generated by algorithm :

x0, x1 ∈ C, C1 := C,

ωn = xn + αn(xn − xn−1),

yn = J−1(µn,0Jωn +

N∑
i=1

µn,iJT
n
i ωn);

zn = J−1(ηn,0Jωn +

N∑
i=1

ηn,iJS
n
i yn),

un = Trnzn,

Cn+1 = {u ∈ Cn : φ(u, un) ≤ k2
nφ(u, ωn)},

xn+1 = ΠCn+1
x0, ∀n ≥ 1,

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 19

where {αn} ⊂ (0, 1), {µn,i} ⊂ [0, 1] and {ηn,i} ⊂ (0, 1] satisfying the following conditions:

(S1)

N∑
i=0

µn,i = 1;

(S2)

N∑
i=0

ηn,i = 1;

(S3) lim sup
n→∞

ηn,0 < 1;

(S4) for same a > 0, rn ∈ [a,∞).

Then, {xn} converges strongly to $, where $ = ΠΩx0 is consider as the generalized projection
of $ onto Ω.

4. Numerical Example
Let B = R and C = [0, 1]. Let Q : C → C be defined by Qu = 2u∀u ∈ C. Define ϑ : C×C → R,

D : C × C → R, G : C → R, Q : C → R, Ti : C → C and Si : C → C by ϑ(u, v) = 0,
D(u, v) = (u + v)(v − u), G(u) = u, Q(u) = 2u and Ti(u) = Si(u) = 1

i+1u, respectively.Setting {βn} = {0.9
2n }, rn = 1

2 , {αn} = 0.9, µ0,n = 1
2 , ∑Ni=1 µn,i = 1

2 such that ∑Ni=0 µi ,n = 1 and
η0,n = 1

3 , ∑Ni=1 ηn,i = 2
3 so that ∑Ni=0 ηi ,n = 1.Let {xn} be generated by the hybrid inertial iterative algorithm (3.1) converges to x∗ = {0} ∈ Ω.

Proof. Clearly ϑ and D satisfy assumptions 1 and 2, respectively, and G is continuous andmonotone so that Sol(GMEP (eq1.1)) = {0} 6= ∅, Sol(V IP (eq1.4)) = {0} 6= ∅. Obviously
Q is 1

2 − i sm, and Ti and Si are two finite families of closed 1-Lipschitz continuous and uni-formly quisi-φ-asymptotically nonexpansive mappings with F ix(Ti) = F ix(Si) = {0}. Thus
Ω = Sol(GMEP (eq1.1)) ∩ Sol(V IP (eq1.4)) ∩ F ix(Ti) ∩ F ix(Si) = {0} 6= ∅. Hence, the it-erative scheme (3.1) becomes the following scheme (4.1) after simplification:

x0, x1 ∈ C, C1 := C,

ωn = xn + 0.9(xn − xn−1),

yn = 1
2ωn + 1

2(n+1)ωn,

zn = 1
3yn + 2

3(n+1)vn,

un = 2zn
7 ,

Cn+1 =
[
0, un+ωn

2

]
,

xn+1 = ΠCn+1x0, ∀n ≥ 1,

where, f or ΠC a metr ic projection onto C,

vn = ΠC(ωn − βnQωn) =


0, ωn − 0.9

2n ωn < 0

1, ωn − 0.9
2n ωn > 1

ωn − 0.9
2n ωn, otherwise.

(4.1)

https://doi.org/10.28924/ada/ma.4.8


Eur. J. Math. Anal. 10.28924/ada/ma.4.8 20Finally, using the software Matlab 7.8.0, we have the following figure which shows that {xn}converges to {0} as n →∞.

Figure 1. Convergence of {xn} when x0 = 1.0 and x1 = 0.5

References
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https://doi.org/10.28924/ada/ma.4.8


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https://doi.org/10.28924/ada/ma.4.8

	1. introduction
	2. Preliminaries
	3. Main Results
	4. Numerical Example
	References

