©2021 Ada Academica https://adac.eeEur. J. Math. Anal. 1 (2021) 19-33doi: 10.28924/ada/ma.1.19
The Generalized Viscosity Implicit Rules of Asymptotically Nonexpansive Mappings in Hilbert

Spaces

Sang B Mendy, John T Mendy∗ , Alieu Jobe
University of the Gambia, Brikama Campus, Gambia

sangbm1@gmail.com, jt.mendy@utg.edu.gm, alieueejobe@gmail.com
∗Correspondence: jt.mendy@utg.edu.gm

Abstract. The generalized viscosity implicit rules of nonexpansive asymptotically mappings in Hilbertspaces are considered. The strong convergence theorems of the rules are proved under certain as-sumptions imposed on the sequences of parameters. An application of it in the convex minimizationproblem is considered. The results presented in this paper improve and extend some recent corre-sponding results in the literature.

1. Background

Let H be a real Hilbert space and M be a nonempty closed convex subset of H, T :M→Mbe a nonexpansive mapping with a nonempty fixed point set F(T )The following iteration method is known as the viscosity approximation method: for arbitrarilychosen u0 ∈M
un+1 = αnψ(un) + (1− αn)T un, n ≥ 0, (1.1)where ψ : M → M is a contraction and {αn} is a sequence in (0, 1). Under some certainconditions, the sequence {un} converges strongly to a point z ∈ F (T ) which solves the variationalinequality (V I)

〈(I − ψ)z, u − z〉 ≥ 0, u ∈ F (T ), (1.2)where I is the identity of H. Many authors studied iterative sequence for the implicit midpointrule because of it’s significant for solving ordinary differential equations; see [?]- [12], John T [9], [7]and the references therein. Recently, Xu et al [3] proposed the following viscosity implicit midpointrule (VIMR) for nonexpansive mappings:
un+1 = αnψ(un) + (1− αn)T

(un + un+1
2

)
, n ≥ 0, (1.3)

Received: 23 Aug 2021.
Key words and phrases. viscosity; Hilbert space; convex minimization; asymptotically nonexpansive mapping; varia-tional inequality; fixed point. 19

https://adac.ee
https://doi.org/10.28924/ada/ma.1.19
https://orcid.org/0000-0002-3774-0761


Eur. J. Math. Anal. 1 (2021) 20In 2015, Ke and Ma [4] proposed the generalized viscosity implicit rules of nonexpansive mappingsin Hilbert spaces as follows:
un+1 = αnψ(un) + (1− αn)T (snun + (1− sn)un+1), n ≥ 0, (1.4)

and
un+1 = αnun + βnψ(un) + γnT (snun + (1− sn)un+1), n ≥ 0, (1.5)

They proved that the generalized viscosity implicit rules 1.4 and 1.5 converge strongly to a fixedpoint of T under certain assumptions, which also solved the V I(1.1).
In 2016, motivated by the work of Xu [3], Zhao et al [5] proposed the following implicit midpointrule for asymptotically nonexpansive mappings:

un+1 = αnψ(un) + (1− αn)T n
(un + un+1

2

)
, n ≥ 0, (1.6)

where T is an asymptotically nonexpansive mapping. They proved that the sequence {un} con-verges strongly to a fixed point of T , which, in addition, also solves the V I(1.1).
In 2017, He et la [14] studied the following iterative

un+1 = αnψ(un) + (1− αn)T n(βnun + (1− βn)un+1), n ≥ 0 (1.7)
in the setting of a Hilbert space and proved that the sequence {un} converges strongly to u∗ =

PF (T )ψ(u∗) which is also the unique solution of the following V I
〈(I − ψ)u, v − u〉 ≥ 0,∀v ∈ F (T ) (1.8)

In this paper, we introduce and study the generalized viscosity implicit rules of asymptoticallynonexpansive mappings in Hilbert spaces. More precisely, we consider the following implicititerative algorithm: u1 ∈M
un+1 = αnun + βnψ(un) + γnT n

(
snun + (1− sn)un+1

)
∀n ∈ N

(1.9)
Under suitable conditions, we proved that the sequence {un} converge strongly to a fixed point ofthe asymptotically nonexpansive mapping T , which also solves the variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).

As applications, we apply our results to solve convexly constrained minimization problem. This wayresults in 1.5 are complemented, extended and generalized.



Eur. J. Math. Anal. 1 (2021) 212. Preliminaries

In the sequel, we always assume that H is a real Hilbert space and M is a nonempty, closed,and convex subset of H. The nearest point projection from H onto M, PM, is defined by
PM(u) := arg min

z∈M

∥∥∥u − z∥∥∥2, u ∈ H. (2.1)
Namely, PM(u) is the only point in M that minimizes the objective ∥∥∥u − z∥∥∥ over z ∈ M. and
PM(u) is characterized as follows:

PM(u) ∈M and
〈
u − PM(u), z − PM(u)

〉
≤ 0 f or al l z ∈M. (2.2)

Definition 2.1. . A mapping T :M→M is said to be:
a): α-inverse strongly monotone if there exists α > 0 satisfying

〈u − v , T u − T v〉 ≥ α‖Au − Av‖2 ∀u, v ∈M; (2.3)
b): L-Lipschitz continuous if there exists L ≥ 0 satisfying

‖T u − T v‖ ≤ L‖u − v‖ ∀u, v ∈M; (2.4)
c): nonexpansive if

‖T u − T v‖ ≤ ‖u − v‖ ∀u, v ∈M; (2.5)
d): asymptotically nonexpansive if there exists a sequence {kn} ⊂ [1,∞) with lim

n→∞
kn = 1 suchthat

‖T nu − T nv‖ ≤ kn‖u − v‖ ∀u, v ∈M and ∀n ∈ N; (2.6)
e): contraction if there exists the contractive constant α ∈ [0, 1) such that

‖T u − T v‖ ≤ α‖u − v‖ ∀u, v ∈M; (2.7)
Lemma 2.2. (The demiclosedness principle [10]) . Let H be a Hilbert space, M be a nonempty
closed convex subset of H, and T : M → M be a asymptotically nonexpansive mapping with
F ix(T ) 6= ∅. If {un} is a sequence in M such that {un} weakly converges to u and {(I − T )un}
converges strongly to 0, then u = T (u)

Lemma 2.3. Let H be a Hilbert space. Then for all θ, u, v ∈ H, the following inequality holds

‖u − θ‖2 ≤ ‖v − θ‖2 + 2〈u − v , u − θ〉

Lemma 2.4. [11]). Assume that {αn} is a sequence of nonnegative real numbers such that

αn+1 ≤ (1− λn)αn + δn

for all n ∈ N, where {λn} ⊆ (0, 1) and {δn} ⊆ R are two sequences satisfying the following
conditions:



Eur. J. Math. Anal. 1 (2021) 22

(i):
∞∑
n=1

λn =∞

(ii): lim sup
n→∞

δn
λn
≤ 0 or

∞∑
n=1

|δn| <∞

Then lim
n→∞

αn = 0

Then the sequence {αn} converges to 0.

3. Main Result

We now prove the following new result.
Theorem 3.1. Let M be a nonempty closed convex subset a real Hilbert space H, T : M →
M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that
limn→∞ kn = 1, F ix(T ) 6= ∅ and ψ : M → M be a contraction mapping with the contractive
constant α ∈ [0, 1). Define a sequence {un} in M as follows: u1 ∈M

un+1 = αnun + βnψ(un) + γnT n
(
snun + (1− sn)un+1

)
∀n ∈ N

(3.1)
where αn, βn, γn, sn ∈ (0, 1) satisfying the following conditions,

A1: αn + βn + γn = 1

A2:
∞∑
n=0

αn =∞

A3: 0 < ε ≤ sn ≤ sn+1 < 1 for all n ≥ 0

A4: lim
n→∞

γn = 1 and lim
n→∞

αn = lim
n→∞

βn = lim
n→∞

sn = 0

lim
n→∞

‖un − T nun‖ = 0

Then the sequence {un} strongly converges to a common fixed point q of T , which is also the
unique solution of the following variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).

We now show that algorithm 3.1 is well posed. Letting
Bn(u) = αnun + βnψ(un) + γnT n

(
snun + (1− sn)un

)
‖Bn(u)− Bn(v)‖ = ‖γnT n

(
snun + (1− sn)u

)
− γnT n

(
snun + (1− sn)v

)
‖

= ‖γnT n(1− sn)u − γnT n(1− sn)v‖

≤ γnkn(1− sn)‖u − v‖

Since lim
n→∞

sn = 0, lim
n→∞

kn = 1, lim
n→∞

γn = 1 and 0 < ε ≤ sn ≤ sn+1 < 1 for all n > 0, we mayassume that γnkn(1 − sn) ≤ 1 − ε for all n > 0. This implies that Bn is a contraction for each



Eur. J. Math. Anal. 1 (2021) 23

n. Therefore there exists a unique fixed point for Bn by Banach contraction principle, which alsoimplies that (3.1) is well-defined.
We now show that the sequence {un} is bounded.

Rewriting 3.1, we have
un+1 = βnψ(un) + αnun + (1− βn)vn (3.2)

where vn =
γnT n(snun + (1− sn)un+1)

1− βn

Remark 3.2. The real sequences that satisfies the above conditions are αn =
1

n
, βn =

1

n
and

γn = 1−
2

n

Proof. Our prove are in six steps. First we prove that the sequence {un} defined by 3.1 is bounded.
Step 1: Letting p ∈ F ix(T ), we have the following estimates
‖un+1 − p‖ = ‖βnψ(un) + αnun + (1− βn)vn − p‖

≤ βn‖ψ(un)− ψ(p)‖+ βn‖ψ(p)− p‖+ αn‖un − p‖+ (1− βn)‖vn − p‖

≤ (αβn + αn)‖un − p‖+ βn‖ψ(p)− p‖+ (1− βn)‖vn − p‖ (3.3)

‖vn − p‖ = ‖
γnT n(snun + (1− sn)un+1)

1− βn
− p‖

=
γnT nsn(un − p)

1− βn
+
γnT n(1− sn)(un+1 − p)

1− βn
‖

≤
γnknsn
1− βn

‖un − p‖+
γnkn(1− sn)

1− βn
‖un+1 − p‖ (3.4)

Putting 3.4 in 3.3, gives the following
‖un+1 − p‖ ≤ (αβn + αn)‖un − p‖+ βn‖ψ(p)− p‖

+ γnknsn‖un − P‖+ γnkn(1− sn)‖un+1 − p‖

(1− γnkn(1− sn))‖un+1 − p‖ ≤ (αβn + αn + γnknsn)‖un − p‖+ βn‖ψ(p)− p‖

‖un+1 − p‖ ≤
(αβn + αn + γnknsn)

1− γnkn(1− sn)
‖un − p‖

+
βn

1− γnkn(1− sn)
‖ψ(p)− p‖



Eur. J. Math. Anal. 1 (2021) 24Since γn, sn ∈ (0, 1), 1 − γnkn(1 − sn) > 0 and lim
n→∞

kn = 1. From the condition (A1), wehave
‖un+1 − p‖ ≤ 1−

1− αβn − αn − γnkn
1− γnkn(1− sn)

‖un − p‖

+
βn

1− γnkn(1− sn)
‖ψ(p)− p‖

]
‖un+1 − p‖ ≤ 1−

βn(1− α)

1− γnkn(1− sn)
‖un − p‖

+
βn(1− α)

1− γnkn(1− sn)

1

(1− α)
‖ψ(p)− p‖

]
‖un+1 − p‖ ≤ max

{
‖un − p‖,

1

(1− α)
‖ψ(p)− p‖

}
Therefore by mathematical induction, we have

‖un+1 − p‖ ≤ max
{
‖u0 − p‖,

1

(1− α)
‖ψ(p)− p‖

}
for all n ≥ N . Therefore {un} is bounded. Consequently,

{ψ(un)} and {vn}

are also bounded.
Step 2: We now prove that the sequence {un+1} converges to {un} as n →∞. That is lim

n→∞
‖un+1−

un‖ = 0

‖un+1 − un‖ = ‖un+1 − T nun + T nun − un‖

= ‖βnψ(un) + αnun + (1− βn)vn − (βn + αn + γn)T n + T nun − un‖

≤ ‖βnψ(un)− βnT nun‖+ ‖αnun − αnT nun‖

+‖(1− βn)vn − γnT n + T nun − un‖

≤ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖

+(1− βn)‖vn − γnT n‖+ ‖T nun − un‖ (3.5)
‖vn − γnT nun‖ = ‖

γnsn
1− βn

T nun +
γn(1− sn)

1− βn
T nun+1 − γnT nun‖

≤ ‖
γnsn

1− βn
‖T nun − T nun‖+

γn(1− sn)

1− βn
‖T nun+1 − T nun‖

≤
γn(1− sn)kn

1− βn
‖un+1 − un‖ (3.6)



Eur. J. Math. Anal. 1 (2021) 25Now putting 3.6 in 3.5, we have the following
‖un+1 − un‖ ≤ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖

+(1− βn)
[γn(1− sn)kn

1− βn
‖un+1 − un‖

]
+ ‖T nun − un‖

≤ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖

+γn(1− sn)kn‖un+1 − un‖
]

+ ‖T nun − un‖

≤ βn‖ψ(un)− T nun‖+ (αn + 1)‖un − T nun‖

+ γn(1− sn)kn‖un+1 − un‖[
1− γn(1− sn)kn

]
‖un+1 − un‖ ≤ βn‖ψ(un)− T nun‖+ (αn + 1)‖un − T nun‖

‖un+1 − un‖ ≤
βn

1− γn(1− sn)kn
‖ψ(un)− T nun‖

+
(αn + 1)

1− γn(1− sn)kn
‖un − T nun‖

Let M :> max
{
‖ψ(un)− T nun‖

}, then we have
‖un+1 − un‖ ≤

βnM

1− γn(1− sn)kn
+

(αn + 1)

1− γn(1− sn)kn
‖un − T nxn‖

‖un+1 − un‖ ≤
βnM

1− γn(1− sn)(1 + εαn)
+

(αn + 1)

1− γn(1− sn)(1 + εαn)
‖un − T nun‖

Since lim
n→∞

αn = lim
n→∞

βn = lim
n→∞

‖un − T nun‖ = 0, we then conclude that lim
n→∞

‖un+1 −
un‖ = 0

Step 3: Again we then show that lim
n→∞

∥∥∥un − T (un)
∥∥∥ = 0. Estimating as follows we have

‖un − T nun‖ = ‖un − un+1 + un+1 − T nun‖

≤ ‖un − un+1‖+ ‖un+1 − T nun‖

≤ ‖un − un+1‖+ ‖βnψ(un) + αnun + (1− βn)vn − T nun
∥∥∥

≤ ‖un − un+1‖+ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖+ (1− βn)‖vn − γnT nun‖(3.7)
‖vn − γnT nun‖ = ‖

γnT n(snun + (1− sn)un+1)

1− βn
− γnT nun‖

≤ ‖
γnsn

1− βn
‖T nun − T nun‖+

(1− sn)γn
1− βn

‖T nun+1 − T nun‖

≤
(1− sn)γnkn

1− βn
‖un+1 − un‖ (3.8)



Eur. J. Math. Anal. 1 (2021) 26Now substituting 3.8 into 3.7, gives the following estimation
‖un − T nun‖ ≤ ‖un − un+1‖+ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖

+ (1− βn)
((1− sn)γnkn

1− βn
‖un+1 − un‖

)
≤

(
1 + (1− sn)γnkn

)
‖un − un+1‖+ βn‖ψ(un)− T nun‖+ αn‖un − T nun‖

≤

(
1 + (1− sn)γnkn

)
1− αn

‖un − un+1‖+
βn

1− αn
‖ψ(un)− T nun‖

‖un − T nun‖ ≤

(
1 + (1− sn)γnkn

)
1− αn

‖un+1 − xn‖+
βnM

1− αn

Therefore from 3.1 condition A4, with lim
n→∞

‖un+1 − un‖ = 0, we can conclude that
lim
n→∞

‖un − T nun‖ = 0 (3.9)
But we know that from the following fact

lim
n→∞

‖un − T (un)‖ ≤ lim
n→∞

‖un − T nun‖+ lim
n→∞

‖T nun − T xn‖

≤ lim
n→∞

‖un − T nun‖+ lim
n→∞

k1‖T n−1un − un‖ (3.10)
Proving that lim

n→∞
‖T n−1un − un‖ = 0, we have the following estimation

‖T n−1(un)− un‖ = ‖un − T n−1(un)‖

= ‖βn−1ψ(un−1) + αn−1un−1 + (1− βn−1)vn−1

−(βn−1 + αn−1 + γn−1)T n−1un‖

= ‖βn−1ψ(un−1)− βn−1T n−1un + αn−1un−1 − αn−1T n−1un

+(1− βn−1)vn−1 − γn−1T n−1un‖

≤ βn−1‖ψ(un−1)− T n−1un‖+ αn−1‖un−1 − T n−1un‖

+(1− βn−1)‖vn−1 − γn−1T n−1xn‖ (3.11)

‖vn−1 − γn−1T n−1un‖ = ‖
γn−1T n−1(sn−1un−1 + (1− sn−1)un)

1− βn−1
− γn−1T n−1un‖

≤
γn−1kn−1sn−1

1− βn−1
‖|un − un−1‖ (3.12)



Eur. J. Math. Anal. 1 (2021) 27combining 3.12 and 3.11 we have the following
‖T n−1(un)− un‖ ≤ βn−1‖ψ(un−1)− T n−1un‖+ αn−1‖un−1 − T n−1un‖

+(1− βn−1)
[γn−1kn−1sn−1

1− βn−1
‖|un − un−1‖

]
≤ βn−1‖ψ(un−1)− T n−1un‖+ αn−1‖un−1 − T n−1un‖

+γn−1kn−1sn−1‖un − un−1‖

≤ βn−1‖ψ(un−1)− T n−1un‖+ αn−1‖un−1 − T n−1un‖

+γn−1kn−1sn−1‖|un − un−1‖

With the assumption of {αn}, {βn} and lim
n→∞

‖un+1 − un‖ = 0, we can conclude that
lim
n→∞

‖T n−1un − un‖ = 0 (3.13)
Therefore from 3.9 and 3.13, we can see from inequality 3.10, that

lim
n→∞

‖un − T (un)‖ = 0 (3.14)
Step 4: In this step, we will show that wω(xn) ⊆ F ix(T ), where

wω(un) := {u ∈ H : there exist a subsequence of {un} converges weakly to u}.Suppose that u ∈ wω(un). Then there exists a subsequence {uni} of {un} such that uni ⇀ xas i →∞ . From 3.14, we have
lim
i→∞

∥∥∥(I − T )xni

∥∥∥ = lim
n→∞

∥∥∥uni − T uni∥∥∥ = 0

. This implies that {(I − T )uni} converges strongly to 0. By using Lemma 2.2, we have
T u = u, and so u ∈ F ix(T ).

Step 5: In this step, we will show that
lim sup
n→∞
〈q − ψ(q), q − un〉 ≤ 0, (3.15)

where q ∈ F (T ) is the unique fixed point of PF (T ) ◦ ψ, that is, q = PF (T )(ψ(z)). Since
{un} is bounded, there exists a subsequence {uni} of {un} such that uni ⇀ u as i →∞ forsome u ∈ H and

lim sup
n→∞
〈q − ψ(q), q − un〉 = lim

i→∞
〈q − ψ(q), q − uni 〉 (3.16)

From Step 4, we get x ∈ F (T ). By using inequality 2.2, we obtain
lim sup
n→∞
〈q − ψ(q), q − un〉 = lim

i→∞
〈q − ψ(q), q − uni 〉

= 〈q − ψ(q), q − u〉 ≤ 0



Eur. J. Math. Anal. 1 (2021) 28

Step 6: Finally, setting ϕn = βnq + αnq + (1 − βn)vn we show that un → q as n → ∞. Again,take q ∈ F (T ) to be the unique fixed point of the contraction PF (T ) ◦ ψ. For each n ∈ N,consider
‖un+1 − q‖2 ≤ ‖ϕn − q‖2 + 2〈un+1 − ϕn, un+1 − q〉

= (1− βn)2‖vn − q‖2 + 2〈βn(ψ(un)− q) + αn(un − q), un+1 − q〉

≤ (1− βn)‖vn − q‖2 + 2〈βn(ψ(un)− ψ(q)) + βn(ψ(q)− q) + αn(un − q), un+1 − q〉

≤ (1− βn)2‖vn − q‖2 + 2βn‖ψ(xn)− ψ(q)‖‖un+1 − q‖+ 2αn‖un − q‖‖un+1 − q‖

+2βn〈ψ(q)− q, un+1 − q〉

≤ (1− βn)2‖vn − q‖2 + 2βnα‖un − q‖‖un+1 − q‖+ 2αn‖un − q‖‖un+1 − q‖

+2βn〈ψ(q)− q, un+1 − q〉

≤ (1− βn)2‖vn − q‖2 + (2βnα+ 2αn)‖un − q‖‖un+1 − q‖

+2βn〈ψ(q)− q, un+1 − q〉 (3.17)
For the fact that

‖vn − q‖2 =
∥∥∥γnT n(snun + (1− sn)un+1)

(1− βn)
− q
∥∥∥2

≤
γ2ns

2
nk
2
n

(1− βn)2
‖un − q‖2 +

γ2n(1− sn)2k2n
(1− βn)2

‖un+1 − q‖2

+
γ2nsn(1− sn)k2n

(1− βn)2
〈un − q, un+1 − q〉

≤
γ2ns

2
nk
2
n

(1− βn)2
‖un − q‖2 +

γ2n(1− sn)2k2n
(1− βn)2

‖un+1 − q‖2

+
γ2nsn(1− sn)k2n

(1− βn)2
‖un − q‖‖un+1 − q‖

+2αnβn

〈
ψ(un)− ψ(q), T n

(un + un+1
2

)
− q
〉 (3.18)

Now substituting 3.18 into 3.17, we have the following estimation
‖un+1 − q‖2 ≤ γ2ns

2
nk
2
n‖un − q‖2 + γ2n(1− sn)2k2n‖un+1 − q‖2

+γ2nsn(1− sn)k2n‖un − q‖‖un+1 − q‖

+2(βnα+ αn)‖un − q‖‖un+1 − q‖+ 2βn〈ψ(q)− q, un+1 − q〉

≤ γ2ns
2
nk
2
n‖un − q‖2 + γ2n(1− sn)2k2n‖un+1 − q‖2

+
[
γ2nsn(1− sn)k2n + 2(βnα+ αn)

]
‖un − q‖‖un+1 − q‖

+2βn〈ψ(q)− q, un+1 − q〉 (3.19)



Eur. J. Math. Anal. 1 (2021) 29Again using the fact that(
‖un − q‖ − ‖un+1 − q‖

)2
≤ ‖un − q‖2 − 2‖un − q‖‖un+1 − q‖

+‖un+1 − q‖2

Setting the left hand to zero, we have the following estimate
2‖un − q‖‖un+1 − q‖ ≤ ‖un − q‖2 + ‖un+1 − q‖2

‖un − q‖‖un+1 − q‖ ≤
1

2
‖un − q‖2 +

1

2
‖un+1 − q‖2 (3.20)

Putting inequality 3.20 in inequality 3.19, gives the following
‖un+1 − q‖2 ≤ γ2ns

2
nk
2
n‖un − q‖2 + γ2n(1− sn)2k2n‖un+1 − q‖2

+
γ2nsn(1− sn)k2n

2
‖un − q‖2 + (βnα+ αn)‖un − q‖2

+
γ2nsn(1− sn)k2n

2
‖un+1 − q‖2 + (βnα+ αn)‖un+1 − q‖2

+2βn〈ψ(q)− q, un+1 − q〉

‖un+1 − q‖2 ≤
[γ2nsnk2n (sn + 1) + 2(βnα+ αn)

2

]
‖un − q‖2

+
[γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

2

]
‖un+1 − q‖2

+2βn〈ψ(q)− q, un+1 − q〉

Thus we have(
1−

[γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

2

])
‖un+1 − q‖2

≤
[γ2nsnk2n (sn + 1) + 2(βnα+ αn)

2

]
‖un − q‖2

+ 2βn〈ψ(q)− q, un+1 − q〉

‖un+1 − q‖2 ≤
γ2nsnk

2
n (sn + 1) + 2(βnα+ αn)

2−
[
γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

]‖un − q‖2
+

4βn

2−
[
γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

]〈ψ(q)− q, un+1 − q〉

‖un+1 − q‖2 ≤
(

1−
2− γ2n(1− sn)2k2n (2− sn)− γ2nsnk2n (sn + 1)

2−
[
γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

])‖un − q‖2
+

4βn

2−
[
γ2n(1− sn)2k2n (2− sn) + 2(βnα+ αn)

]〈ψ(q)− q, un+1 − q〉



Eur. J. Math. Anal. 1 (2021) 30Therefore from condition lim
n→∞

αn = lim
n→∞

βn = lim
n→∞

sn = 0 in 3.1, we concludes that
‖un+1 − q‖2 ≤

(
1−

2− 2γ2nk
2
n

2− 2γ2nk
2
n

)
‖un − q‖2

lim
n→∞

‖un+1 − q‖2 = 0

This complete the proof. �

Theorem 3.3. Let M be a nonempty closed convex subset a real Hilbert space H, T : M →
M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that
limn→∞ kn = 1, F ix(T ) 6= ∅ and ω be a constant. Define a sequence {un} in M as follows: u1 ∈M

un+1 = αnun + βnω + γnT n
(
snun + (1− sn)un+1

)
∀n ∈ N

(3.21)
where αn, βn, γn, sn ∈ (0, 1) satisfying conditions A1− A4 and ψ(un) = ω

lim
n→∞

‖T nun − un‖ = 0

Then the sequence {un} strongly converges to a common fixed point q of T , which is also the
unique solution of the following variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).

Taking sn = 0The following corollaries holds:
Corollary 3.4. Let M be a nonempty closed convex subset a real Hilbert space H, T : M →
M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that
limn→∞ kn = 1, F ix(T ) 6= ∅ and ψ : M → M be a contraction mapping with the contractive
constant α ∈ [0, 1). Define a sequence {un} in M as follows:{

u1 ∈M
un+1 = αnun + βnψ(un) + γnT n(un+1) ∀n ∈ N

(3.22)
where αn, βn, γn ∈ (0, 1) satisfying conditions A1− A4 without lim

n→∞
sn = 0

lim
n→∞

‖T nun − un‖ = 0

Then the sequence {un} strongly converges to a common fixed point q of T , which is also the
unique solution of the following variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).



Eur. J. Math. Anal. 1 (2021) 31

Corollary 3.5. Let M be a nonempty closed convex subset a real Hilbert space H, T : M →
M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that
limn→∞ kn = 1, F ix(T ) 6= ∅ and u ∈M be a constant. Define a sequence {un} in M as follows:{

u1 ∈M
un+1 = αnun + βnω + γnT n(un+1) ∀n ∈ N

(3.23)
where αn, βn, γn ∈ (0, 1) satisfying conditions A1− A4 without lim

n→∞
sn = 0

lim
n→∞

‖T nun − un‖ = 0

Then the sequence {un} strongly converges to a common fixed point q of T , which is also the
unique solution of the following variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).

4. Application to convex minimization problems

In this section, we study the problem of finding a minimizer of a convex function Φ defined from areal Hilbert space M to R.Consider the optimization problem
min
x∈C

Φ(x) (4.1)where Φ : M → R is a convex and differentiable function. Assume 4.1 is consistent, and let
Ω 6= ∅ be its set of solutions. The gradient projection algorithm generates a sequence {un} via theiterative procedure:

un+1 = PM(un − δ∇Φ(u)) (4.2)if ∇Φ is θ−inverse strongly monotone mapping and δ(0, 2θ). The following basic results are wellknown.
Remark 4.1. It is well known that if Φ : M → R be a real-valued differentiable convex functionand u∗ ∈M, then the point u∗ is a minimizer of Φ on M if and only if dΦ(u∗) = 0.
Definition 4.2. A function Φ :M→ R is said to be strongly convex if there exists α > 0 such thatfor every u, v ∈M and λ ∈ (0, 1), the following inequality holds:

Φ(λu + (1− λ)v) ≤ λΦ(u) + (1− λ)Φ(v)− α‖u − v‖2. (4.3)
Lemma 4.3. Let E be normed linear space and Φ : M → R a real-valued differentiable convex
function. Assume that Φ is strongly convex. Then the differential map dΨ : M→M is strongly
monotone, i.e., there exists a positive constant k such that

〈dΦ(u)− dΦ(v), u − v〉 ≥ k‖u − v‖2 ∀ u, v ∈M. (4.4)
The prove of the following theorem follows from 3.1



Eur. J. Math. Anal. 1 (2021) 32

Theorem 4.4. Let M be a nonempty closed convex subset a real Hilbert space H. For the
minimization problem 4.1, assume that Φ is (Gateaux) differentiable and the gradient ∇Φ is a
θ−inverse-strongly monotone mapping for some positive real number θ. Let ψ : M → M be a
contraction with coefficient α ∈ [0, 1). For a given u1 ∈M, let {un} be a sequence generated by:{

u1 ∈M
un+1 = αnun + βnψ(un) + γnPM(1− δ∇Φ)(snun + (1− sn)(un+1)) ∀n ∈ N

(4.5)
where αn, βn, γn, sn ∈ (0, 1) satisfying the following conditions

A1: αn + βn + γn = 1

A2: lim
n→∞

k2n − 1

αn
= 0

A3:
∞∑
n=0

αn =∞

A4: lim
n→∞

γn = 1 and lim
n→∞

αn = lim
n→∞

βn = lim
n→∞

sn = 0

Then {un} converges strongly to a solution (u∗) of the minimization problem 4.1, which is also the
unique solution of the variational inequality

〈(I − ψ)u, p − u〉 ≥ 0 p ∈ F (T ).

Conflict of Interest:The authors declare that they have no competing interests.
Availability of data and materials:No data were used to support this study.
Funding:No funding was given towards this manuscript.
Authors Contributions:All authors have contributed equally and significantly in writing this paper and also readand approved the final manuscript.
Acknowledgement:The authors are very grateful to the editor and anonymous referees for their helpfulcomments.

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https://doi.org/10.1186/s13663-015-0439-6
http://doi.org/10.22436/jnsa.009.06.86
http://doi.org/10.22436/jnsa.009.06.86
https://doi.org/10.12988/ams.2017.718
http://doi.org/10.30538/oms2017.0011
https://doi.org/10.1112/S0024610702003332
https://doi.org/10.1112/S0024610702003332
https://dx.doi.org/10.1073/pnas.53.5.1100
https://doi.org/10.12988/ams.2017.718

	1. Background
	2. Preliminaries
	3. Main Result
	4. Application to convex minimization problems
	References