

























































©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)Tun,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) 20
In 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 ∈Mun+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) 21
2. 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 all 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,Tu −T v〉≥ α‖Au −Av‖2 ∀u,v ∈M; (2.3)
b): L-Lipschitz continuous if there exists L ≥ 0 satisfying

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

‖Tu −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

‖Tu −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
Fix(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,Fix(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 ∈Mun+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 < � ≤ 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 < � ≤ sn ≤ sn+1 < 1 for all n > 0, we mayassume that γnkn(1 − sn) ≤ 1 − � 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 ∈ Fix(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) 24
Since 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) 25
Now 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 + �αn)
+

(αn + 1)

1 −γn(1 − sn)(1 + �αn)
‖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) 26
Now 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→∞
‖Tn−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−1Tn−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) 27
combining 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) ⊆ Fix(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 −Tuni∥∥∥ = 0
. This implies that {(I −T )uni} converges strongly to 0. By using Lemma 2.2, we have
Tu = u, and so u ∈ Fix(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 =
∥∥∥γnTn(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),Tn

(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)k
2
n‖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)k

2
n + 2(βnα + αn)

]
‖un −q‖‖un+1 −q‖

+2βn〈ψ(q) −q,un+1 −q〉 (3.19)



Eur. J. Math. Anal. 1 (2021) 29
Again 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) 30
Therefore from condition lim

n→∞
αn = lim

n→∞
βn = lim

n→∞
sn = 0 in 3.1, we concludes that

‖un+1 −q‖2 ≤
(

1 −
2 − 2γ2nk2n
2 − 2γ2nk2n

)
‖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,Fix(T ) 6= ∅ and ω be a constant. Define a sequence {un} in M as follows: u1 ∈Mun+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,Fix(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,Fix(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.
References

[1] H. Attouch, Viscosity approximation methods for minimization problems, SIAM J. Optim. 6 (3) (1996) 769-806.
https://doi.org/10.1137/S1052623493259616.[2] A. Moudafi, Viscosity approximation methods for fixed-points problems, J. Math. Anal. Appl. 241 (1) (2000) 46-55.
https://doi.org/10.1006/jmaa.1999.6615.[3] H.K. Xu, M.A. Alghamdi, N. Shahzad, The viscosity technique for the implicit midpoint rule of nonexpansive mappingsin Hilbert spaces, Fixed Point Theory Appl. 2015 (2015) 41. https://doi.org/10.1186/s13663-015-0282-9.

https://doi.org/10.1137/S1052623493259616
https://doi.org/10.1006/jmaa.1999.6615
https://doi.org/10.1186/s13663-015-0282-9


Eur. J. Math. Anal. 1 (2021) 33
[4] Y. Ke, C. Ma, The generalized viscosity implicit rules of nonexpansive mappings in Hilbert spaces, Fixed PointTheory and Appl. 2015 (2015), 190. https://doi.org/10.1186/s13663-015-0439-6.[5] L.C. Zhao, S.S. Chang, C.F. Wen, Viscosity approximation methods for the implicit midpoint rule of asymptoticallynonexpansive mappings in Hilbert spaces, J. Nonlinear Sci. Appl. 9 (2016) 4478-4488. http://doi.org/10.22436/

jnsa.009.06.86.[6] S. He, Y. Mao, Z. Zhou, J.Q. Zhang, The generalized viscosity implicit rules of asymptotically nonexpansive mappingsin Hilbert spaces, Appl. Math. Sci. 11 (12) (2017) 549-560. https://doi.org/10.12988/ams.2017.718.[7] J.T. Mendy, S. Rahule, Viscosity like implicit methods for zeros of monotone operators in Banach spaces, KhayyamJ. Math. 2021.[8] S.F.A. Naqvi, M.S. Khan, On the viscosity rule for common fixed points of two nonexpansive mappings in Hilbertspaces, Open J. Math. Sci. 1 (1) (2017) 111-125. http://doi.org/10.30538/oms2017.0011.[9] J.T. Mendy, The viscosity iterative algorithms for the implicit double midpoint rule of nonexpansive mappings inHilbert spaces, Amer. J. Math. Anal. 8 (2020), 1-8.[10] K. Go eb el, W.A. Kirk, Topics in Me tric Fixed Point Theory, Cambridge Studies in Advanced Mathematics, vol. 28.Cambridge University Press, Cambridge (1990).[11] H.K. Xu, Iterative algorithms for nonlinear operators, J. Lond. Math. Soc. 66 (2) (2002) 240-256. https://doi.
org/10.1112/S0024610702003332.[12] F.E. Browder, Existence of periodic solutions for nonlinear equations of evolution, Proc. Natl. Acad. Sci. USA 53(5) (1965), 1100-1103. https://dx.doi.org/10.1073/pnas.53.5.1100.[13] S. Dhakal, W. Sintunavarat, The viscosity implicit midpoint rule for finding common fixed points of two asymptoticallynonexpansive mappings with applications, Thai J. Math. 17 (2019) 495-514.[14] S. He , Y. Mao , Z. Zhou, J.Q. Zhang, The generalized viscosity implicit rules of asymptotically nonexpansivemappings in Hilbert Spaces, Appl. Math. Sci. 11 (2017), 549-560. https://doi.org/10.12988/ams.2017.718.

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

