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Convergence of a Three-step Iteration Scheme to the Common Fixed Points of Mixed-Type

Total Asymtotically Nonexpansive Mappings in Uniformly Convex Banach Spaces

Imo Kalu Agwu1,∗, Donatus Ikechi Igbokwe1, Nathenial C. Ukeje2
1Department of Mathematics, Micheal Okpara University of Agriculture, Umudike, Umuahia Abia State

Nigeria
agwuimo@gmail.com, igbokwedi@yahoo.com

2Department of Mathematics and Statistics, University of Port-Harcourt, Port-Harcourt Rivers State
Nigeria

ukejechukwuebuanathan@gmail.com
∗Correspondence: agwuimo@gmail.com

Abstract. We propose a three-step iteration scheme of hybrid mixed-type for three total asymptoti-cally nonexpansive self mappings and three total asymptotically nonexpansive nonself mappings. Inaddition, we establish some weak convergence theorems of the scheme to the common fixed point ofthe mappings in uniformly convex Banach spaces. Our results extend and generalize numerous resultscurrently in literature.

1. Introduction
Let K be a nonempty subset of a real Banach space E. Let T : K −→ K be a nonlinear mapping,we denote the set of all fixed points of T by F (T ). The set of common fixed points of six mappings

S1,S2,S3,T1,T2 and T3 will be denoted by F = ∩3i=1(F (Ti ) ∩F (Si )).
Definition 1.1. A mapping T : K −→ K is said to asymptotically nonexpansive [6] if there exists
a sequence {kn} in [1,∞) with limn→∞kn = 1 such that

‖Tn(x) −Tn(y)‖≤ kn‖x −y‖,∀x,y ∈N (1.1). ln 1972, the class of asymptotically nonexpansive mapping was introduced by Goebel and Kirk [6].They proved that if K is a nonempty closed convex subset of a uniformly convex Banach space and
T is an asymptotically nonexpansive mapping of K, then T has a fixed point.

Received: 29 Aug 2021.
Key words and phrases. asymtotically nonexpansive mapping; total asymptotically nonexpansive nonself mapping;hybrid mixed type iteration scheme; common fixed point; uniformly convex Banach space; weak convergence.45

https://adac.ee
https://doi.org/10.28924/ada/ma.1.45


Eur. J. Math. Anal. 1 (2021) 46
Definition 1.2. A mapping T is said to be total asymptotically nonexpansive [1] if

‖Tn(x) −Tn(y)‖≤‖x −y‖ + µnφ(‖x −y‖) + νn,∀x,y ∈ K,∀n ∈N, (1.2)
where {µn}and {νn} are nonnegative real sequences such that µn → 0 and νn → 0 as n →∞
and φ is a strictly increasing continuous function φ : [0,∞) → [0,∞) with φ(0) = 0.

From the above definitions, we see that the class of total asymptotically nonexpansive mappingsincludes the class of asymptotically nonexpansive mapping as a special case; see [4] for moredetails. Each asymptotically nonexpansive mapping is total asymptotically nonexpansive mappingwith νn = 0,µn = kn − 1 f or all n ≥ 1,φ(t) = t,t ≥ 0.
Definition 1.3. A subset K of a Banach space E is said to be a retract of E if there exists a
continuous mapping P : E −→ K (called retraction) such that P (x) = x for all x ∈ K. If, in
addition P is nonexpansive, then P is said to be nonexpansive retraction of E. If P : E −→ K is
a retraction, then P2 = P. A retract of a Hausdorff space must be a closed subset. Every closed
convex subset of a uniformly convex Banach space is a retract.

In 2012, Yolacan and Kiziltune [18] defined the following:
Definition 1.4. Let K be a nonempty and closed convex subset of a Banach space E. A nonself
mapping T : K → E is said to be total asymptotically nonexpansive mapping if there exist
sequences k(1)n and k

(2)
n in [0,∞) with k

(1)
n → 0 and k

(2)
n → 0 as n →∞ and a strictly increasing

function φ : [0,∞) → [0,∞) with φ(0) = 0 such that

‖T (PT )n−1(x) −T (PT )n−1(y)‖≤‖x −y‖ + k1(n)φ(‖x −y‖) + k
(2)
n ,∀x,y ∈ K,n ∈N. (1.3)

Chidume et al. [3] studied the following iterative scheme in 2004:
x1 = x ∈ K

xn+1 = P (αnT (PT )
n−1xn + (1 −αn)xn),n ≥ 1, (1.4)

where {αn} is a sequence in (0, 1), K is a nonempty closed convex subset of of a real uniformlyconvex Banach space E, P is a nonexpansive retraction of E onto K, and proved some strong andweak convergence theorems for asymptotically nonexpansive nonself mappings in the intermediatesense in the framework of uniformly convex Banach spaces.ln 2006, Wang [17] generalised the iteration process (1.4) as follows:
x1 = x ∈ K,

xn+1 = P ((1 −αn)xn + αnT1(PT1)n−1yn),

yn = P ((1 −βn)xn + βnT2(PT2)n−1xn),n ≥ 1, (1.5)



Eur. J. Math. Anal. 1 (2021) 47
where T1,T2 : K −→ E are two asymptotically nonexpansive nonself mappings, {αn} and {βn} arereal sequences in [0, 1), and proved some weak and strong convergence theorems for asymptoticallynonexpansive nonself mappings.ln 2012, Guo et al [8] generalised the iteration process (1.5) as follows:

x1 = x ∈ K,

xn+1 = P ((1 −αn)Sn1xn + αnT1(PT1)
n−1yn),

yn = P ((1 −βn)Sn2xn + βnT2(PT2)
n−1xn),n ≥ 1, (1.6)

where T1,T2 : K −→ E are two asymptotically nonexpansive nonself mappings, S1,S2 : K −→ Eare two asymptotically nonexpansive self mappings and {αn},{βn} are real sequences in [0, 1),and proved some strong and weak convergence theorems for mixed-type asymptotically nonexpan-sive mappings.
Hybrid Mixed-Type Iteration SchemeLet E be a real uniformly convex Banach space, K a nonempty closed convex subset of E and
P : E −→ K a nonexpansive retraction of E onto K. Let S1,S2,S3 : K −→ K be three totalasymptotically nonexpansive self mappings and T1,T2,T3 : K −→ E be three total asymptoti-cally nonexpansive nonself mappings. Then, the hybrid iteration scheme for the above mentionedmappings is as follows:

x1 = x ∈ K;

xn+1 = P ((1 −αn)Sn1xn + αnT1(PT1)
n−1yn);

yn = P ((1 −βn)Sn2xn + βnT2(PT2)
n−1zn);

zn = P ((1 −γn)Sn3xn + γnT3(PT3)
n−1xn),

(1.7)

where {αn},{βn}, and {γn} are real sequences in [0, 1).The aim of this paper is to study this new hybrid mixed-type iteration scheme (1.7), prove demi-closedness principle for total asymptotically nonexpansive nonself map and establish some conver-gence theorems for mixed-type mappings in the setting of uniformly convex Banach spaces.
2. Preliminary

For the sake of convenience, we restate the following concepts and results:Let E be a Banach space with its dimension greater than or equal to 2. The modulus of convexityof E is a function δE(ε) : (0, 2] −→ (0, 2] defined by
δE(ε) = inf{1 −‖

1

2
(x + y)‖ : ‖x‖ = 1,‖y‖ = 1,ε = ‖x −y‖}.



Eur. J. Math. Anal. 1 (2021) 48
A Banach space E is uniformly convex if and if δE(ε) > 0, for all ε ∈ (0, 2].We recall the following:
Definition 2.1. (see [19]: Let % = {x ∈ E : ‖x‖ = 1} and let E? be the dual of E. The space E
has Gateaux differentiable norm if limn→∞

‖x+ty‖−‖x‖
t

exists ∀x,y ∈ %.

Definition 2.2. (see [19]: The space E has Frechet differentiable norm [15] if for each x ∈ %, the
limit of the norm above exists and is attained uniformly for all y ∈ %, and in this case, it is also
well known that

〈h,J(x)〉 +
1

2
‖x‖2 ≤

1

2
‖x + h‖2 ≤〈h,J(x)〉 +

1

2
‖x‖2 + b(‖x‖), (2.1)

∀x,y ∈ E, where J is the Frechet derivative of the functional 1
2
‖ · |2 at x ∈ E,〈·〉 is the pairing

between E and E? and b is an increasing function defined on [0,∞) such that limt→∞
b(t)
t

= 0.

Definition 2.3. : The space E has Opial condition [10] if for any sequence {xn} in E, xn converges
to x weakly, then it follows that lim supn→∞‖xn − x‖ < lim supn→∞‖xn − y‖ for all y ∈ E with
x 6= y .

Examples of Banach spaces satisfying Opial conditions are Hilbert spaces and all spaces lp(1 <
p < ∞). On the other hand, Lp[0,π] with 1 < p 6= 2 fails to satify Opial condition.
Definition 2.4. : A mapping T : K −→ K is said to be demiclosed at 0, if for any sequence {xn}
in K, the condition that xn converges weakly to x ∈ K and Txn converges strongly to 0 implies
Tx = 0.

Definition 2.5. : A Banach space has the Kadec-Klec property [14] if for every sequence xn in
E,xn → x weakly and ‖xn‖→‖x‖, then it follows that ‖xn −x‖→ 0.

Next, we state the following useful lemmas which will be needed in order to prove our mainresults.
Lemma 2.1. (see [16]): Let {αn}∞n=1,{βn}

∞
n=1 and {γn}

∞
n=1 be sequences of nonnegative numbers

satisfying the inequality:

αn+1 ≤ (1 + βn)αn + γn,∀n ≥ 1. (2.2)
If

∑∞
n=1βn < ∞ and

∑∞
n=1γn < ∞, then(1) limn→∞αn exists(2) ln particular, if {αn}∞n=1 has a subsequence which converges strongly to 0, then

limn→∞αn = 0.



Eur. J. Math. Anal. 1 (2021) 49
Lemma 2.2. (see [14]): Let E be a uniformly convex Banach space and 0 < p ≤ tn ≤ q < 1 for
each n ≥ 1. Suppose that {xn} and {yn} are sequences in E such that

lim sup
n→∞

‖xn‖≤ r, lim sup
n→∞

‖yn‖≤ r and lim
n→∞

‖tnxn + (1 − tn)yn‖ = r, (2.3)
hold for some r ≥ 0. Then limn→∞‖xn −yn‖ = 0.

Lemma 2.3. (see [14]): Let E be a real reflexive Banach space such that its dual E? has the
Kadec-Klec property. Let {xn} be a bounded sequence in E and p,q ∈ ωω(xn) ( where ωω(xn)
denotes the set of all weak subsequential limits of {xn}). Suppose limn→∞‖txn + (1 − t)p −q‖
exists for all t ∈ [0, 1]. Then, p = q.

Lemma 2.4. (see [14]): Let K be a nonempty convex subset of a uniformly convex Banach space E.
Then, there exists a strictly incraesing continous convex function φ : [0,∞) → [0,∞) with φ(0) = 0
such that for each Lipshitizian mapping T : C −→ C with the Lipschiz constant L,

‖tTx − (1 − t)Ty −T (tx − (1 − t)y)‖≤ Lφ−1(‖x −y‖−
1

L
‖Tx −Ty‖) (2.4)

for all x,y ∈ K and for all t ∈ [0.1].

Lemma 2.5. (see [2]) Let E be a uniformly convex Banach space, K a nonempty bounded close convex
subset of E. Then, there exists a strictly increasing continous convex function φ : [0,∞) −→ [0,∞)
with φ(0) = 0 such that for any Lipschitizian mapping T : K −→ E with Lipschitz constant L ≥ 1
and elements {xn}nj=i in K and any nonnegative numbers {tj}

n
j=1 with

∑n
j=1 tj = 1, the following

inequality holds:

‖T (
n∑
j=1

tjxj) −
n∑
j=1

tjTxj‖≤ Lφ−1{max1≤j,k≤n(‖xj −xk‖−L−1‖Txj −Txk‖)}

Lemma 2.6. (see [21]) If the sequence {xn}∞n=1 converges weakly to x, then there exists a sequence
of convex combination yj =

∑n(j)
k=1

λ
(j)
k
xk+j , λ

(j)
k
≥ 0 and

∑n(j)
k=1

λ(j) = 1, such that ‖yj − x‖ → 0.
as n →∞.

3. Main Results
Lemma 3.1. ( Demiclosedness Principle f or Nonself Total Asymptotically Nonexpansive
Maps ) Let K be a nonempty closed convex and bounded subset of a uniformly convex Banach space
E and T : K −→ E be L-Lipschitz continuous and total asymptotically nonexpansive mapping with
the function φ : [0,∞) −→ [0,∞) (such that φ(0) = 0) and nonnegative sequences {k(1)n },{k

(2)
n }

such that k(1)n ,k
(2)
n → 0 as n →∞. Then, I −T is demiclosed at 0.

Proof. Let {xn} converge weakly to ω ∈ K and {xn −Txn} converge strongly to 0. We prove that
(I −T )ω = 0. Clearly, {xn} is bounded. So, there exists ρ > 0 such that {xn}⊂ C = K ∩Bρ(0),where Bρ(0) is a closed ball in E with centre 0 and radius ρ. Thus, C is nonempty, closed ,



Eur. J. Math. Anal. 1 (2021) 50
bounded and convex subset in K.Claim: T (PT )n−1ω → ω as n → ∞. In fact, since {xn} converges weakly to ω, by Lemma6(see [21]), we have for all n > 1, there exists a convex combination

yn =

m(n)∑
i=1

t
(n)
i
xi+n, t

(n)
i
≥ 0 and m(n)∑

i=1

t
(n)
i

= 1 such that ‖yn −ω‖→ 0 as n →∞. (3.1)
Also, since {xn−Txn} converges to 0, then for any � > 0 and a positive integer m ≥ 1, there exists
N1 = N(�) > 0 such that

‖(I −T )xn‖ <
�

1 + m
,∀n ≥ N1. (3.2)

Hence, ∀n ≥ N1, using Definition 1.4 and the fact that P is nonexpansive , we have the followingestimates:For arbitrary but fixed j ≥ 1, we have
‖xn −T (PT )(j−1)xn‖ ≤ ‖(I −T )xn‖ + ‖(T −T (PT ))xn‖

+‖(T (PT ) −T (PT )2)xn‖

+‖(T (PT )2 −T (PT )3)xn‖

+ · · · + ‖(T (PT )j−2 −T (PT )j−1))xn‖

≤ ‖(I −T )xn‖ + (‖(I −T )xn‖ + µ
(1)
n φ(‖(I −T )xn‖)

+ξ
(1)
n ) + (‖(I −T )xn‖ + µ

(2)
n φ(‖(I −T )xn‖) + ξ

(2)
n )

+(‖(I −T )xn‖ + µ
(3)
n φ(‖(I −T )xn‖) + ξ

(3)
n )

+ · · · + (‖(I −T )xn‖ + µ
(j−1)
n φ(‖(I −T )xn‖) + ξ

(j−1)
n )

= ‖(I −T )xn‖ +
m−1∑
j=1

‖(I −T )xn‖ +
m−1∑
j=1

µ
(j)
n φ(‖(I −T )xn‖)

+

m−1∑
j=1

ξ
(j)
n

≤ m‖xn −Txn‖ + mµnφ(‖(I −T )xn‖) + mξn, (3.3)
where µn = max1≤j≤m−1{µ(j)n } and ξn = max1≤j≤m−1{ξ(j)n }.From (3.2) and (3.3), we get

‖xn −T (PT )j−1xn‖ < �. (3.4)
Now, since T : K −→ E is L-Lipschitizian and total asymptotically nonexpansive , so is T : C −→
E. Therefore, ∀j ≥ 1,T (PT )j−1 : C −→ E is Lipschitizian mapping with the Lipschitz constant
µj ≥ 1.



Eur. J. Math. Anal. 1 (2021) 51
In addition,
‖T (PT )j−1yn −yn‖ = ‖T (PT )j−1yn −

m(n)∑
i=1

t
(n)
i
T (PT )j−1xi+n +

m(n)∑
i=1

t
(n)
i
T (PT )j−1xi+n

−
m(n)∑
i=1

t
(n)
i
xi+n‖

≤ ‖T (PT )j−1yn −
m(n)∑
i=1

t
(n)
i
T (PT )j−1xi+n‖

+

m(n)∑
i=1

t
(n)
i
‖T (PT )j−1xi+n −xi+n‖. (3.5)

Using (3.4), we get
m(n)∑
i=1

t
(n)
i
‖T (PT )j−1xi+n −xi+n‖ < �,∀n ≥N. (3.6)

Furthermore, by Lemma 2.5, there exists a strictly increasing continous function φ : [0,∞) −→
[0,∞) with φ(0) = 0 such that for all n ≥N, we have

‖T (PT )j−1yn −
m(n)∑
i=1

t
(n)
i T (PT )

j−1xi+n‖ = ‖T (PT )j−1(
m(n)∑
i=1

t
(n)
i xi+n) −

m(n)∑
i=1

t
(n)
i T (PT )

j−1xi+n‖

≤ µjφ−1{max1≤j,k≤n(‖xi+n −xi+k‖

−µ−1j ‖T (PT )
J−1xi+n −T (PT )J−1xk+n‖)}

= µjφ
−1{max1≤j,k≤n(‖xi+n −T (PT )J−1xi+n

+T (PT )J−1xi+n −T (PT )J−1xk+n

+T (PT )J−1xk+n −xi+k‖

−µ−1j ‖T (PT )
J−1xi+n −T (PT )J−1xk+n‖)}

≤ µjφ−1{max1≤j,k≤n(‖xi+n −T (PT )J−1xi+n‖

+‖T (PT )J−1xi+n −T (PT )J−1xk+n‖

+‖T (PT )J−1xk+n −xi+k‖

−µ−1j ‖T (PT )
J−1xi+n −T (PT )J−1xk+n‖)}

≤ µjφ−1{max1≤j,k≤n(� + � + (1 −µ−1j )

×‖T (PT )J−1xi+n −T (PT )J−1xk+n‖)}

≤ µjφ−1{max1≤j,k≤n(� + � + (1 −µ−1j )µj

×‖xi+n −xk+n‖}

≤ µjφ−1{max1≤j,k≤n(� + � + (1 −µ−1j )µj

×(‖xi+n‖ + ‖xk+n‖}.



Eur. J. Math. Anal. 1 (2021) 52
Thus,

‖T (PT )j−1yn −
m(n)∑
i=1

t
(n)
i
T (PT )j−1xi+n‖≤ µjφ−1(� + � + 2r(1 −µ−1j )µj), (3.7)

since xi+n and xk+n are both in C.Also, (3.5), (3.6) and (3.7) imply that
‖T (PT )j−1yn −yn‖≤ µjφ−1(� + � + 2r(1 −µ−1j )µj). (3.8)

Taking lim supn→∞ on both sides of (3.8) and noting that � > 0 is arbitrary, we have that
lim sup
n→∞

‖T (PT )j−1yn −yn‖≤ µjφ−1(2r(1 −µ−1j )µj). (3.9)
On the other hand, for any j ≥ 1, it follows from (3.1) that
‖T (PT )j−1ω −ω‖ ≤ ‖T (PT )j−1ω −T (PT )j−1yn‖ + ‖T (PT )j−1yn −yn‖ + ‖yn −ω‖

≤ µj‖yn −ω‖ + ‖T (PT )j−1yn −yn‖ + ‖yn −ω‖. (3.10)
Taking lim supn→∞ on both sides of the above inequality and using (3.1) and (3.9), we have

‖T (PT )j−1ω −ω‖≤ µjφ−1(2r(1 −µ−1j )µj).

Again, taking lim supj→∞ on both sides of the above inequality, we have
lim sup
j→∞

‖T (PT )j−1ω −ω‖≤ φ−1(0) = 0,

which implies that ‖T (PT )j−1ω−ω‖→ 0 as j →∞, and hence proving our claim. By continuityof TP, we have that
lim
j→∞

TP (T (PT )j−1ω) = TPω = Tω = ω.

This completes the proof. �
Lemma 3.2. Let E be a uniformly convex Banach space and K a nonempty closed convex subset
of E. Let S1,S2,S3 : K −→ K be three total asymptotically nonexpansive self mapping with
sequences {k(1)n },{k

(2)
n },{k

(3)
n }∈ [1,∞),

{w(1)n },{w(2)}n,{w
(3)
n } ∈ [1,∞) and T1,T2,T3 : K −→ E are three total asymptotically nonex-

pansive nonself mappings with sequences {µ(1)n },{µ
(2)
n },{µ

(3)
n }∈ [1,∞),{ν

(1)
n },{ν

(2)
n },

{ν(3)n } ∈ [1,∞). Let {xn} be the sequence defined by (1.7), where {αn} and {βn} are real
sequences ∈ [0, 1). Suppose F = (F (Ti ) ∩F (Si )) 6= ∅. If the following conditions hold:i. ∑∞n=1k(1)n < ∞, ∑∞n=1k(2)n < ∞, ∑∞n=1k(3)n < ∞, ∑∞n=1µ(1)n < ∞, ∑∞n=1µ(2)n < ∞,∑∞

n=1µ
(3)
n < ∞,

∑∞
n=1ν

(1)
n < ∞,

∑∞
n=1ν

(2)
n < ∞,

∑∞
n=1ν

(3)
n < ∞,ii. There exists a constant M > 0 such thatΨ(t) = φ(t) ≤ Mt,t ≤ 0.

Then, limn∞‖xn −q‖ and limn∞d(xn −F )both exist for all q ∈ F .



Eur. J. Math. Anal. 1 (2021) 53
Proof. Set hn = max(k(1)n ,k(2)n ,k(3)n ,µ(1)n ,µ(2)n ,µ(3)n ),M = max(M1,M2,M3,M4,M5,M6) and θn =max(ν(1)n ,ν(2)n ,ν(3)n ,ω(1)n ,ω(2)n ,ω(3)n ). Then, ∑∞n=1hn < ∞ and ∑∞n=1θn < ∞. For any q ∈ F , itfollows from (3.1) that
‖zn −q‖ = |P ((1 −βn)Sn3xn + βnT3(PT3)

n−1xn) −P (q)‖

≤ ‖(1 −βn)Sn3xn + βnT3(PT
n−1
3 xn −q‖

= ‖(1 −βn)Sn3xn + βnq −q −βnq + βnT3(PT3)
n−1xn‖

= ‖(1 −βn)Sn3xn − (1 −βn)q + βn(T3(PT3)
n−1xn −q)‖

= ‖(1 −βn)(Sn3xn −q) + βn(T3(PT3)
n−1xn −q)‖ (3.11)

≤ (1 −βn)‖Sn3xn −q‖ + βn‖T3(PT3)
n−1xn −q‖

≤ (1 −βn)[‖xn −q‖ + k
(3)
n Ψ(‖xn −q‖) + ω

(3)
n ] + βn[‖xn −q‖ + µ

(3)
n φ(‖xn −q‖)

+ν
(3)
n ]

= (1 −βn)‖xn −q‖ + (1 −βn)hnΨ(‖xn −q‖) + (1 −βn)θn + βn‖xn −q‖

+βnhnφ(‖xn −q‖) + βnθn

≤ (1 −βn)(1 + hnM5)‖xn −q‖ + βn(1 + hnM6)‖xn −q‖ + θn

≤ (1 −βn)(1 + hnM)‖xn −q‖ + βn(1 + hnM)‖xn −q‖ + θn

≤ (1 + hnM)‖xn −q‖ + θn. (3.12)
Also, form (1.7), we get
‖yn −q‖ = |P ((1 −βn)Sn2xn + βnT2(PT2)

n−1zn) −P (q)‖

≤ ‖(1 −βn)Sn2xn + βnT2(PT2)
n−1xn −q‖

= ‖(1 −βn)Sn2xn + βnq −q −βnq + βnT2(PT2)
n−1zn‖

= ‖(1 −βn)Sn2xn − (1 −βn)q + βn(T2(PT2)
n−1zn −q)‖ (3.13)

= ‖(1 −βn)(Sn2xn −q) + βn(T2(PT2)
n−1zn −q)‖

≤ (1 −βn)‖Sn2xn −q‖ + βn‖T2(PT2)
n−1zn −q‖

≤ (1 −βn)[‖xn −q‖ + k
(2)
n Ψ(‖xn −q‖) + ω

(2)
n ] + βn[‖zn −q‖ + µ

(2)
n φ(‖xn −q‖)

+ν
(2)
n ]

= (1 −βn)‖xn −q‖ + (1 −βn)hnΨ(‖xn −q‖) + (1 −βn)θn + βn‖xn −q‖

+βnhnφ(‖zn −q‖) + βnθn

≤ (1 −βn)(1 + hnM3)‖xn −q‖ + βn(1 + hnM4)‖zn −q‖ + θn

≤ (1 −βn)(1 + hnM)‖xn −q‖ + βn(1 + hnM)‖zn −q‖ + θn. (3.14)



Eur. J. Math. Anal. 1 (2021) 54
Putting (3.12) into (3.14), we have
‖yn −q‖ ≤ (1 −βn)(1 + hnM)‖xn −q‖ + βn(1 + hnM)[(1 + hnM)‖xn −q‖ + θn] + θn

= (1 + hnM)[(1 −βn)‖xn −q‖ + βn((1 + hnM)‖xn −q‖ + θn)] + θn

= (1 + hnM)[(1 −βn + βn + βnhnM))‖xn −q‖ + θn)] + θn

≤ (1 + hnM)[1 + hnM))‖xn −q‖ + θn)] + θn

= (1 + hnM)
2‖xn −q‖ + (2 + hnM)θn. (3.15)

Again, using (1.7), we have
‖xn+1 −q‖ = |P ((1 −αn)Sn1xn + αnT1(PT1)

n−1yn) −P (q)‖

≤ ‖(1 −αn)Sn1xn + αnT1(PT1)
n−1yn −q‖

= ‖(1 −αn)Sn1xn + αnq −q −αnq + αnT1(PT1)
n−1yn‖

= ‖(1 −αn)Sn1xn − (1 −αn)q + αn(T1(PT1)
n−1yn −q)‖

= ‖(1 −αn)(Sn1xn −q) + αn(T1(PT1)
n−1yn −q)‖ (3.16)

≤ (1 −αn)‖Sn1xn −q‖ + αn‖T1(PT1)
n−1yn −q‖

≤ (1 −αn)[‖xn −q‖ + k
(1)
n Ψ(‖xn −q‖) + ω

(1)
n ] + αn[‖yn −q‖

+µ
(1)
n φ(‖yn −q‖) + ν

(1)
n ]

≤ (1 −αn)‖xn −q‖ + (1 −αn)hnΨ(‖xn −q‖) + (1 −αn)θn + αn‖yn −q‖

+αnhnφ(‖yn −q‖) + αnθn

≤ (1 −αn)(1 + hnM1)‖xn −q‖ + αn(1 + hnM2)‖yn −q‖ + θn

≤ (1 −αn)(1 + hnM)‖xn −q‖ + αn(1 + hnM)‖yn −q‖ + θn. (3.17)
Putting (3.15) into (3.17), we obtain

‖xn+1 −q‖ ≤ (1 −αn)(1 + hnM)‖xn −q‖ + αn(1 + hnM)[(1 + hnM)2‖xn −q‖

+(2 + hnM)θn] + θn]

= (1 + hnM)‖xn −q‖−αn(1 + hnM)‖xn −q‖ + αn(1 + hnM)3‖xn −q‖

+αn(1 + hnM)(2 + hnM)θn + θn

≤ [1 + (3 + 3hnM + h2nM
2)hnM]‖xn −q‖ + [1 + (1 + hnM)(2 + hnM]θn

= (1 + δn)‖xn −q‖ + ρn. (3.18)
where δn = 1 + (3 + 3hnM +h2nM2)hnM and ρn = [1 + (1 +hnM)(2 +hnM]θn. Since ∑∞n=1δn < ∞and ∑∞n=1ρn < ∞, it follows from lemma 2.1 that limn→∞‖xn −q‖ exists.



Eur. J. Math. Anal. 1 (2021) 55
Now taking the infimum over all q ∈ F in (3.18), we get

d(xn+1,F ) ≤ (1 + δn)d(xn,F ) + ρn,∀n ∈N. (3.19)
Again, since ∑∞n=1δn < ∞ and ∑∞n=1ρn < ∞, it follows from lemma 2.1 and (3.19) that
limn→∞d(xn,F ) exists. This completes the proof. �
Lemma 3.3. Let E be a uniformly convex Banach space and K a nonempty closed convex subset
of E. Let S1,S2,S3 : K −→ K be three total asymptotically nonexpansive self mapping with
sequences {k(1)n },{k

(2)
n },{k

(3)
n }∈ [1,∞),{w

(1)
n },{w(2)}n,{w

(3)
n }∈ [1,∞)

and T1,T2,T3 : K −→ E are three total asymptotically nonexpansive nonself mappings with se-
quences {µ(1)n },{µ

(2)
n },{µ

(3)
n } ∈ [1,∞),{ν

(1)
n },{ν

(2)
n },{ν

(3)
n } ∈ [1,∞). Let {xn} be the sequence

defined by (1.7), where {αn} and {βn} are real sequences ∈ [0, 1). Suppose F = (F (Ti )∩F (Si )) 6=
∅. If the following conditions hold:

i. ∑∞n=1k(1)n < ∞, ∑∞n=1k(2)n < ∞, ∑∞n=1k(3)n < ∞, ∑∞n=1µ(1)n < ∞, ∑∞n=1µ(2)n < ∞,∑∞
n=1µ

(3)
n < ∞,

∑∞
n=1ν

(1)
n < ∞,

∑∞
n=1ν

(2)
n < ∞,

∑∞
n=1ν

(3)
n < ∞,ii. ‖x−T1(PT1)n−1y‖≤‖Sn1x−T1(PT1)n−1y‖,‖x−T2(PT2)n−1y‖≤‖Sn2x−T2(PT2)n−1y‖,

‖x −T3(PT3)n−1y‖≤‖Sn3x −T3(PT3)
n−1y‖iii. There exists a constant M1,M2 > 0 such that Ψ(t) ≤ M1t, φ(t) ≤ M2t, t ≥ 0.

Then, limn∞‖xn −Sixn‖ = 0 and limn∞‖xn −Tixn‖ = 0, for i = 1, 2, , 3.

Proof. Set hn = max(k(1)n ,k(2)n ,k(3)n ,µ(1)n ,µ(2)n ,µ(3)n ),M = max(M1,M2,M3,M4,M5,M6) and θn =max(ν(1)n ,ν(2)n ,ν(3)n ,ω(1)n ,ω(2)n ,ω(3)n ). Then, ∑∞n=1hn < ∞ and ∑∞n=1θn < ∞. for any given q ∈ F ,
limn∞‖xn −q‖ exists by lemma 3.2. Now, assume that limn∞‖xn −q‖ = c. it follows from (3.15),(3.16) and the fact that ∑∞n=1hn < ∞ and ∑∞n=1θn < ∞ that

lim‖(1 −αn)(Sn1xn −q) + αnT1(PT1)
n−1yn −q)‖ = c. (3.20)

Also, we have
‖Sn1xn −q‖ ≤ ‖xn −q‖ + k

(1)
n Ψ(‖xn −q‖) + ω

(1)
n

≤ ‖xn −q‖ + k
(1)
n M‖xn −q‖) + ω

(1)
n

≤ (1 + k(1)n M)‖xn −q‖ + ω
(1)
n

≤ (1 + hnM)‖xn −q‖ + θn

⇒ lim sup‖Sn1xn −q‖ ≤ lim sup[(1 + hnM)‖xn −q‖ + θn] = c. (3.21)



Eur. J. Math. Anal. 1 (2021) 56
Furthermore,

‖T1(PT1)yn −q‖ ≤ ‖yn −q‖ + µ
(1)
n φ(‖yn −q‖) + ν

(1)
n

≤ ‖yn −q‖ + µ
(1)
n M‖yn −q‖) + ν

(1)
n

≤ (1 + µ(1)n M)‖yn −q‖ + ν
(1)
n

≤ (1 + hnM)‖yn −q‖ + θn

Taking limsup on both sides of (3.15), we obtain
lim sup‖yn−q‖≤ c and so lim sup‖T1(PT1)yn−q‖≤ lim sup[(1 + hnM)‖yn−q‖+ θn] ≤ c. Thus,

lim sup‖T1(PT1)yn −q‖≤ lim sup[(1 + hnM)‖yn −q‖ + θn] = c. (3.22)
Using lemma 2.2, we get

lim
n→∞

‖Sn1xn −T1(PT1)
n−1yn‖ = 0. (3.23)By condition (ii), it follows that

‖xn −T1(PT1)n−1yn‖≤‖Sn1xn −T1(PT1)
n−1yn‖,

and so from (3.23), we have
lim
n→∞

‖xn −T1(PT1)n−1yn‖ = 0. (3.24)Also, we have
‖Sn2xn −q‖ ≤ ‖xn −q‖ + k

(2)
n Ψ(‖xn −q‖) + ω

(2)
n

≤ ‖xn −q‖ + k
(2)
n M‖xn −q‖) + ω

(2)
n

≤ (1 + k(2)n M)‖xn −q‖ + ω
(2)
n

≤ (1 + hnM)‖xn −q‖ + θn

⇒ lim sup‖Sn2xn −q‖ ≤ lim sup[(1 + hnM)‖xn −q‖ + θn] = c. (3.25)Furthermore,
‖T2(PT2)zn −q‖ ≤ ‖zn −q‖ + µ(2)n φ(‖zn −q‖) + ν

(2)
n

≤ ‖zn −q‖ + µ(2)n M‖zn −q‖) + ν
(2)
n

≤ (1 + µ(2)n M)‖zn −q‖ + ν
(2)
n

≤ (1 + hnM)‖zn −q‖ + θn



Eur. J. Math. Anal. 1 (2021) 57
Taking lim sup on both sides of (3.12), we obtain lim supn→∞‖zn −q‖≤ c and so

lim sup‖T2(PT1)zn −q‖≤ lim sup[(1 + hnM)‖zn −q‖ + θn] ≤ c. (3.26)
(3.13), (3.25), (3.26) and lemma 2.2 imply

lim
n→∞

‖Sn2xn −T2(PT2)
n−1zn‖ = 0. (3.27)

(3.27) and condition (ii) yields
lim
n→∞

‖xn −T2(PT2)n−1zn‖ = 0. (3.28)
From (3.11), using the same argument as was used in obtaining (3.27) above, we get

lim
n→∞

‖Sn3xn −T3(PT3)
n−1xn‖ = 0. (3.29)

Now, we prove that
lim
n→∞

‖xn −T1(PT1)n−1xn‖ = lim
n→∞

‖xn −T2(PT2)n−1xn‖ lim
n→∞

‖xn −T3(PT3)n−1xn‖ = 0.

Indeed, since ‖xn −T3(PT3)n−1xn‖ ≤ ‖Sn3xn −T3(PT3)n−1xn‖, (by condition (ii)), it follows from(3.29) that
lim
n→∞

‖xn −T3(PT3)n−1xn‖ = 0. (3.30)
Since, P (Snxn) = Snxn and P : E −→ K is a nonexpansive retraction of E onto K, we get

‖zn −Sn3xn‖ = ‖P ((1 −γn)S
n
3xn + γnT3(PT3)

n−1xn) −Sn3xn‖

≤ ‖(1 −γn)Sn3xn + γnT3(PT3)
n−1xn −Sn3xn‖

= ‖−γn(Sn3xn −γnT3(PT3)
n−1xn)‖

= γn‖(Sn3xn −γnT3(PT3)
n−1xn)‖,

which by (3.29) gives
lim
n→∞

‖zn −Sn3xn‖ = 0. (3.31)Observe that
‖zn −xn‖ = ‖zn −Sn3xn + S

n
3xn −T3(PT3)

n−1xn + T3(PT3)
n−1xn −xn‖

≤ ‖zn −Sn3xn‖ + ‖S
n
3xn −T3(PT3)

n−1xn‖

+‖T3(PT3)n−1xn −xn‖. (3.32)
Thus, it follows from (3.29), (3.30),(3.31) and (3.32) that

lim
n→∞

‖zn −xn‖ = 0. (3.33)



Eur. J. Math. Anal. 1 (2021) 58
Again, observe that
‖Sn2xn −T2(PT2)

n−1xn‖ ≤ ‖Sn2xn −T2(PT2)
n−1zn‖ + ‖T2(PT2)n−1zn −T2(PT2)n−1xn‖

≤ ‖Sn2xn −T2(PT2)
n−1zn‖ + (‖zn −xn‖ + k

(2)
n φ(‖zn −xn‖) + ν

(2)
n

≤ ‖Sn2xn −T2(PT2)
n−1zn‖ + ‖zn −xn‖ + Mhn(‖zn −xn‖) + θn

= ‖Sn2xn −T2(PT2)
n−1zn‖ + (1 + Mhn)‖zn −xn‖ + θn. (3.34)

From (3.27),(3.33), (3.34) and the fact that ∑∞n=1θn < ∞, we get
lim
n→∞

‖Sn2xn −T2(PT2)
n−1xn‖ = 0. (3.35)

Since ‖xn−T2(PT2)n−1xn‖≤‖Sn2xn−T2(PT2)n−1xn‖ (by condition (ii), it follows from (3.35) that
lim
n→∞

‖xn −T2(PT2)n−1xn‖ = 0. (3.36)
Also, since P (Snxn) = Snxn and P : E −→ K is a nonexpansive retraction of E onto K, we get

‖yn −Sn2xn‖ = ‖P ((1 −βn)S
n
2xn + βnT2(PT2)

n−1zn) −Sn2xn‖

≤ ‖(1 −βn)Sn2xn + βnT2(PT2)
n−1zn −Sn2xn‖

= ‖−βn(Sn2xn −βnT2(PT2)
n−1zn)‖

= βn‖(Sn2xn −βnT2(PT2)
n−1zn)‖,

which by (3.27) gives
lim
n→∞

‖yn −Sn2xn‖ = 0. (3.37)Moreover, since
‖yn −xn‖ = ‖yn −Sn2xn + S

n
2xn −T2(PT2)

n−1zn + T2(PT2)
n−1xn −zn‖

≤ ‖yn −Sn2xn‖ + ‖S
n
2xn −T2(PT2)

n−1zn‖ + ‖T2(PT2)n−1zn −xn‖,

it follows from (3.27), (3.28) and (3.37) that
lim
n→∞

‖yn −xn‖ = 0. (3.38)
Observe that
‖Sn1xn −T1(PT1)

n−1xn‖ ≤ ‖Sn1xn −T1(PT1)
n−1yn‖ + ‖T1(PT1)n−1yn −T1(PT1)n−1xn‖

≤ ‖Sn1xn −T1(PT1)
n−1yn‖ + (‖yn −xn‖ + k

(1)
n Ψ(‖yn −xn‖) + ν

(1)
n

≤ ‖Sn2xn −T2(PT2)
n−1zn‖ + ‖yn −xn‖ + Mhn(‖zn −xn‖) + θn

= ‖Sn1xn −T1(PT1)
n−1yn‖ + (1 + Mhn)‖yn −xn‖ + θn. (3.39)

From (3.23), (3.38), (3.39) and the fact that ∑∞n=1θn < ∞
lim
n→∞

‖Sn1xn −T1(PT1)
n−1xn‖ = 0. (3.40)



Eur. J. Math. Anal. 1 (2021) 59
Now, since ‖xn−T1(PT1)n−1xn‖≤‖Sn1xn−T1(PT1)n−1xn‖ (by condition (ii), it follows from (3.40)that

lim
n→∞

‖xn −T1(PT1)n−1xn‖ = 0. (3.41)
From

‖xn+1 −Sn1xn‖ = ‖P [(1 −αn)S
n
1xn + αnT1(PT1)

n−1yn] −Sn1xn‖

≤ ‖(1 −αn)Sn1xn + αnT1(PT1)
n−1yn −Sn1xn‖

= ‖−αn(Sn1xn −T1(PT1)
n−1yn])‖

= αn‖Sn1xn −T1(PT1)
n−1yn]‖

and (3.23), we obtain
lim
n→∞

‖xn+1 −Sn1xn‖ = 0. (3.42)
From

‖xn+1 −T1(PT1)n−1yn‖≤‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T1(PT1)

n−1yn‖,

(3.23) and (3.42), we get
lim
n→∞

‖xn+1 −T1(PT1)n−1yn‖ = 0. (3.43)
Also, from (3.23), (3.24) and the inequality

‖Sn1xn −xn‖≤‖S
n
1xn −T1(PT1)

n−1yn‖ + ‖T1(PT1)n−1yn −xn‖,

we have
lim
n→∞

‖Sn1xn −xn‖ = 0. (3.44)
Again, from (3.41), (3.44) and the inequality

‖Sn1xn −T2(PT2)
n−1xn‖≤‖Sn1xn −xn‖ + ‖xn −T2(PT2)

n−1xn‖,

we have
lim
n→∞

‖Sn1xn −T2(PT2)
n−1xn‖ = 0. (3.45)



Eur. J. Math. Anal. 1 (2021) 60
Since

‖xn+1 −T2(PT2)n−1yn‖ ≤ ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T2(PT2)

n−1xn‖

+‖T2(PT2)n−1xn −T2(PT2)n−1yn‖

≤ ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T2(PT2)

n−1xn‖ + (‖xn −yn‖

+k
(2)
n φ(‖xn −yn‖) + ν

(2)
n )

≤ ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T2(PT2)

n−1xn‖ + ‖xn −yn‖

+Mhn‖xn −yn‖) + θn

= ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T2(PT2)

n−1xn‖

+(1 + Mhn)‖xn −yn‖) + θn,

it follows from (3.38), (3.42), (3.45) and the fact that ∑∞n=1θn < ∞ that
lim
n→∞

‖xn+1 −T2(PT2)n−1yn‖ = 0. (3.46)
Now, from (3.30), (3.41) and the inequality

‖Sn1xn −T3(PT3)
n−1xn‖≤‖Sn1xn −xn‖ + ‖xn −T3(PT3)

n−1xn‖,

we obtain
lim
n→∞

‖Sn1xn −T3(PT3)
n−1xn‖ = 0. (3.47)

Since
‖xn+1 −T3(PT3)n−1yn‖ ≤ ‖xn+1 −Sn1xn‖ + ‖S

n
1xn −T3(PT3)

n−1xn‖

+‖T3(PT3)n−1xn −T3(PT3)n−1yn‖

≤ ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T3(PT3)

n−1xn‖ + (‖xn −yn‖

+k
(3)
n φ(‖xn −yn‖) + ν

(3)
n )

≤ ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T3(PT3)

n−1xn‖ + ‖xn −yn‖

+Mhn‖xn −yn‖) + θn

= ‖xn+1 −Sn1xn‖ + ‖S
n
1xn −T3(PT3)

n−1xn‖

+(1 + Mhn)‖xn −yn‖) + θn

it follows from (3.38), (3.42), (3.47) and the fact that ∑∞n=1θn < ∞ that
lim
n→∞

‖xn+1 −T3(PT3)n−1yn‖ = 0. (3.48)



Eur. J. Math. Anal. 1 (2021) 61
Again, since (PTi)(PTI)n−2yn−1,xn ∈ K for i = 1, 2, 3 and T1,T2,T3 are three total asymptoti-cally nonexpansive nonself mappings, we have

‖Ti (PTi )n−1yn−1 −Tixn‖ = ‖Ti (PTi )(PTi )n−2yn−1 −Ti (Pxn)‖

≤ ‖(PTi )(PTi )n−2yn−1 −P (xn)‖

+k
(i)
n φ(‖(PTi )(PTi )n−2yn−1 −P (xn)‖) + ν

(i)
n

≤ ‖(PTi )(PTi )n−2yn−1 −P (xn)‖

+Mhn‖(PTi)(PTi)n−2yn−1 −P (xn)‖ + θn

= (1 + Mhn)‖(PTi )(PTi )n−2yn−1 −P (xn)‖ + θn

= (1 + Mhn)‖Ti (PTi )n−2yn−1 −xn‖ + θn. (3.49)
For i = 1.2.3,, it follows from (3.43), (3.46) and (3.48) that

lim
n→∞

‖Ti(PTi)n−1yn−1 −Tixn‖ = 0. (3.50)
Observe that

‖xn+1 −yn‖≤‖xn+1 −T1(PT1)n−1yn‖ + ‖T1(PT1)n−1yn −xn‖ + ‖xn −yn‖,

so that, by (3.24), (3.38) and (3.43), we get
lim
n→∞

‖xn+1 −yn‖ = 0. (3.51)
Next,observe, for i = 1, 2, 3, that

‖xn −Tixn‖ ≤ ‖xn −Ti (PTi )n−1xn‖ + ‖Ti (PTi )n−1xn −Ti (PTi )n−1yn−1‖

+‖Ti (PTi )n−1yn−1 −Tixn‖

≤ ‖xn −Ti (PTi )n−1xn‖ + [‖xn −yn−1‖ + k
(i)
n φ(‖xn −yn−1‖)

+ν
(i)
n ] + ‖Ti (PTi )n−1yn−1 −Tixn‖

≤ ‖xn −Ti (PTi )n−1xn‖ + ‖xn −yn−1‖ + k
(i)
n M‖xn −yn−1‖

+ν
(i)
n + ‖Ti (PT1)n−1yn−1 −Tixn‖

= ‖xn −Ti (PTi )n−1xn‖ + (1 + k
(i)
n M)‖xn −yn−1‖ + ν

(i)
n ]

+‖Ti (PTi )n−1yn−1 −Tixn‖

≤ ‖xn −Ti (PTi )n−1xn‖ + max[supn≥1(1 + k
(i)
n M)]‖xn −yn−1‖

+max[supn≥1]ν
(i)
n ] + ‖Ti (PTi )n−1yn−1 −Tixn‖

Thus, it follows from (3.30), (3.36), (3.41), (3.50) and (3.51) that limn→∞‖xn − Tixn‖ = 0, for
i = 1, , 2, 3.



Eur. J. Math. Anal. 1 (2021) 62
Finally, we prove that limn→∞‖xn −Sni xn‖ = 0, for i = 1, , 2, 3.Infact, by condition (ii), we have for i = 1, 2, 3, that

‖xn −Sni xn ≤‖xn −Ti (PTi )
n−1xn‖ + ‖Sni xn −Ti (PTi )

n−1xn‖

Thus, it follows from (3.29), (3.30), (3.36), (3.40), (3.41) and (3.45) that
lim
n→∞

‖xn −Sni xn‖ = 0, f or i = 1, 2, 3. (3.52)
This completes the proof of Lemma 3.3. �
Lemma 3.4. Under the assumption of Lemma 3.2, for all p1,p2 ∈ ∩3i1(F (Si ) ∩ F (Ti )), the limit
limn→∞‖xn + (1 − t)p1−p2‖ exists for all t ∈ [0, 1], where {xn} is the sequence defined by (1.7).
Proof. By lemma 3.2, limn→∞‖xn − q‖ exists for all q ∈ F and therefor {xn} is bounded. Let
an(t) = ‖xn + (1 − t)p1 − p2‖ exists for all t ∈ [0, 1]. Then, limn→∞a(0) = ‖p1 − p2‖ and
limn→∞a(1) = ‖xn −p2‖ exist by Lemma 3.2. It remains therefor to prove Lemma 3.4 for t ∈ (0, 1).For all x ∈ K, we define the mapping


Rn(x) = P [(1 −γn)Sn3 + γT3(PT3)

n−1xn];

Wn(x) = P [(1 −βn)Sn2 + βT2(PT2)
n−1xn];

Vn(x) = P [(1 −αn)Sn1 + αT1(PT1)
n−1xn],n ≥ 1.

(3.53)
Then, it follows that xn+1 = Vnxn,Vnp = p,∀p ∈ F . Now, from (3.12), (3.15) and (3.18) of Lemma3.2, we see that


‖Rn(x) −Rn(y)‖≤ (1 + hn)M‖x −y‖ + θn;

‖Wn(x) −Wn(y)‖≤ (1 + rn)M‖x −y‖ + δnθn;

‖Vn(x) −Vn(y)‖≤ (1 + en)M‖x −y‖ + θn = fn‖x −y‖ + gn,

(3.54)

where rn = 2hn+h2nM2,δn = 2+hnM,en = 3hnM+3h2nM2+h3nM3 and gn = (1+hnM)(2+hnM)θnwith ∑∞n=1en < ∞, ∑∞n=1gn < ∞ and fn = 1 + en. Since ∑∞n=1en < ∞, it follows that fn → 1 as
n →∞. Set

 Sn,m = Vn+m−1Vn+m−2 · · ·Vn, m ∈N;
bn,m = ‖Sn,m(txn + (1 − t)p1) −Sn.m(txm + (1 − t)p2‖.

(3.55)



Eur. J. Math. Anal. 1 (2021) 63
From (3.54) and (3.55), we have

‖Sn,m(x) −Sn,m(y)‖ = ‖Vn+m−1Vn+m−2 · · ·Vn(x) −Vn+m−1Vn+m−2 · · ·Vn(y)‖

≤ fn+m−1‖Vn+m−2Vn+m−3 · · ·Vn(x) −Vn+m−2Vn+m−3 · · ·Vn(y)‖

+gn+m−1

≤ (fn+m−1)(fn+m−2)‖Vn+m−3Vn+m−4 · · ·Vn(x)

−Vn+m−3Vn+m−4 · · ·Vn(y)‖ + gn+m−1 + gn+m−2...
≤ (

n+m−1∏
i=n

fi )‖x −y‖ +
n+m−1∑
i=n

gi

= Bn‖x −y‖ +
n+m−1∑
i=n

gi, (3.56)
for all x,y ∈ K, where Bn = ∏n+m−1i=n fi,Sn,mxn = xn and Sn,mp = p for all p ∈ F . Thus,

an+m(t) = ‖txn + (1 − t)p1 −p2‖

= ‖Sn,m(txn + (1 − t)p1 −p2‖

≤ bn,m + ‖Sn,m(txn + (1 − t)p1 −p2‖. (3.57)
By using Theorem 2.3 in [5], we have

bn,m ≤ ψ−1(‖(xn −u‖−‖xn+1 −Sn,mu‖)

= ψ−1(‖(xn −u‖−‖xn+1 −u + u −Sn,mu‖)

≤ ψ−1(‖(xn −u‖− (‖xn+1 −u‖ + ‖Sn,mu −u‖)), (3.58)
so that the sequence {bn.m} converges uniformly to 0, i.e, bn,m → 0 as n →∞. Since limn→Bn = 1and limn→∞bn,m = 0, it follows from (3.57) that lim supn→∞an(t) ≤ lim infb→∞bn.m ≤ lim infn→∞an(t).This shows that limn→∞an(t) exists, i.e, limn→∞‖txn + (1 − t)p1 − p2‖ exists for all t ∈ [0, 1].This completes the proof Lemma 3.4. �
Lemma 3.5. Under the assumption of Lemma 3.2, if E has Frechet differentiable norm,then for
all p1,Jp2 ∈ F = ∩3i=1(F (Ti ) ∩ F (Si )), the limn→∞(〈xn,J(p1 − p2)〉 exists, where {xn} is the
sequence defined by (1.7). If ωω(xn) denotes the set of all weak subsequential limits of {xn}, then
〈q1 −q2,J(p1 −p2〉 = 0 for all p1,p2 ∈ F and for all q1,q2 ∈ ωω(xn).

Proof. Suppose that x = p1 −p2 with p1 6= p2 and h = t(xn −p1) in(2.1). Then, we have
t(〈xn,J(p1 −p2)〉 +

1

2
‖p1 −p2‖2 ≤

1

2
‖txn + (1 − t)p1 −p2‖2

≤ t(〈xn,J(p1 −p2)〉 +
1

2
‖p1 −p2‖2 + b(t‖xn −p1‖)



Eur. J. Math. Anal. 1 (2021) 64
Since supn≥1‖xn −p‖≤ Q for some Q > 0, we have

t lim
n→∞

sup(〈xn,J(p1 −p2)〉 +
1

2
‖p1 −p2‖2 ≤

1

2
lim
n→∞

sup‖txn + (1 − t)p1 −p2‖2

≤ t lim
n→∞

inf (〈xn,J(p1 −p2)〉 +
1

2
‖p1 −p2‖2

+b(tQ)

That is, t limn→∞ sup(〈xn,J(p1 − p2)〉 ≤ t lim infn→∞(〈xn,J(p1 − p2)〉 + b(tQ). If t → 0, then
limn→∞〈xn − p1,J(p1 − p2)〉 exists for all p1,p2 ∈ F and for all q2,q2 ∈ ωω(xn); in particular,
(〈q1 −q2,J(p1 −p2)〉 = 0 for all q2,q2 ∈ ωω(xn). This completes the proof Lemma 3.5. �
Theorem 3.6. Under the assumption of Lemma 3.2, if E has Frechet differentiable norm, then the
sequence {xn} defined by (1.7) converges weakly to a common fixed point in F = ∩3i=1F (Ti )∩F (Si ).
Proof. By Lemma 3.5, (〈q1−q2,J(p1−p2)〉 = 0 for all q2,q2 ∈ ωω(xn). Therefore, ‖q?−p ?‖2 =
〈q? −p?,J(q? −p?)〉 = 0. This implies that p? = q?. Consequently, {xn} converges to a commonfixed point of F = ∩3i=1F (Ti ) ∩F (Si ). This completes the proof Theorem 3.6. �
Theorem 3.7. Under the assumption of Lemma 3.2, if the dual space E? of E has the Kadec Klec
(KK) property and the mappings I − Si and I − Ti for i = 1, 2, 3, where I denotes the identity
mapping, are demiclosed at zero, then the sequence {xn} defined by (1.7) converges weakly to a
common fixed point in F = ∩3i=1(F (Ti ) ∩F (Si )).

Proof. By Lemma 3.2 {xn} is bounded and since E is reflexive, there exists a subsequence {xnk} of
{xn} which converges weakly to some q? ∈ K. By Lemma 3.3, we have limn→∞‖xnk −Sixnk‖ = 0and limn→∞‖xnk −Tixnk‖ = 0 for i = 1, 2, 3. Since by hypothesis, the mappings I −Si and I −Tifor i = 1, 2, 3, where I denotes the identity mapping, are demiclosed at zero, Siq? = q? and
Tiq

? = q? for i = 1, 2, 3.; which means q? ∈ F = ∩3i=1(F (Ti ) ∩F (Si )). Now, we show that {xn}converges weakly to q?. Suppose {xnj} is another subsequence of {xn} which converges weakly to
p? ∈ K. By the same method as above, we have p? ∈ F and q? ∈ ωω(xn). By Lemma 3.4, the limit
limn→∞‖txn + (1 − t)q? −p?‖ exists for all t ∈ [0, 1] and so q? = p?. Thus, the sequence {xn}converges weakly to q? ∈ F . This completes the proof. �
Theorem 3.8. Under the assumption of Lemma 3.2, if E satisfies Opial’s condition and the mappings
I−Si and I−Ti for i = 1, 2, 3, where I denotes the identity mapping, are demiclosed at zero, then
the sequence {xn} defined by (1.7) converges weakly to a common fixed point in F = ∩3i=1(F (Ti )∩
F (Si )).

Proof. Let q? ∈ F . From Lemma 3.2, the squence {‖xn −p ?‖} is convergent and hence bounded.Since, E is uniformly convex , every bounded subset of E is weakly compact. Thus, the existsa subsequence {xnk} of {xn} which converges weakly to some q? ∈ K. By Lemma 3.3, we have



Eur. J. Math. Anal. 1 (2021) 65
limn→∞‖xnk −Sixnk‖ = 0 and limn→∞‖xnk −Tixnk‖ = 0 for i = 1, 2, 3. Since by hypothesis, themappings I −Si and I −Ti for i = 1, 2, 3, where I denotes the identity mapping, are demiclosedat zero, Siq? = q? and Tiq? = q? for i = 1, 2, 3.; which means q? ∈ F = ∩3i=1(F (Ti ) ∩ F (Si )).Finally, we show that {xn} converges weakly to q?. Suppose on the contrary that {xnj} is anothersubsequence of {xn} which converges weakly to p? ∈ K and q? 6= p? By Lemma 3.2, limn→∞‖xn−
q?‖ and limn→∞‖xn −p?‖ exist. By virtue of Opial’s condition on E, we obtain

lim
n→∞

‖xn −q?‖ = lim
n→∞

‖xnk −q
?‖

< lim
n→∞

‖xnk −p
?‖

= lim
n→∞

‖xn −p?‖

= lim
n→∞

‖xnj −p
?‖

< lim
n→∞

‖xnj −q
?‖

= lim
n→∞

‖xn −q?‖,

(3.59)
which is a contradiction, so q? = p? Therefore, the sequence {xn} defined by (1.7) converges weaklyto q? ∈ F . This completes the proof. �
Corollary 3.9. Let E be a uniformly convex Banach space and K a nonempty closed convex subset
of E. Let S1,S2,S3 : K −→ K be three generalize asymptotically nonexpansive self mapping with
sequences {k(1)n },{k

(2)
n },{k

(3)
n }∈ [1,∞),

{w(1)n },{w(2)}n,{w
(3)
n } ∈ [1,∞) and T1,T2,T3 : K −→ E are three generalize asymptotically

nonexpansive nonself mappings with sequences {µ(1)n },{µ
(2)
n },{µ

(3)
n }∈ [1,∞),{ν

(1)
n },{ν

(2)
n },

{ν(3)n }∈ [1,∞). Let {xn} be the sequence defined by (1.7), where {αn} and {βn} are real sequences
∈ [0, 1).. Suppose F = ∩3i=1(F (Ti ) ∩F (Si )) 6= 0. If the following conditions hold:i. ∑∞n=1k(1)n < ∞, ∑∞n=1k(2)n < ∞, ∑∞n=1k(3)n < ∞, ∑∞n=1µ(1)n < ∞, ∑∞n=1µ(2)n < ∞, ∑∞n=1µ(3)n <

∞,
∑∞
n=1ν

(1)
n < ∞,

∑∞
n=1ν

(2)
n < ∞,

∑∞
n=1ν

(3)
n < ∞,ii. There exists a constant M > 0 such that Ψ(t) = φ(t) ≤ Mt,t ≤ 0.

Then, limn∞‖xn −q‖ and limn∞d(xn −F ) both exist for all q ∈ F .

Corollary 3.10. Let E be a uniformly convex Banach space and K a nonempty closed convex subset
of E. Let S1,S2,S3 : K −→ K be three generalize asymptotically nonexpansive self mapping with
sequences {k(1)n },{k

(2)
n },{k

(3)
n }∈ [1,∞),

{w(1)n },{w(2)}n,{w
(3)
n } ∈ [1,∞) and T1,T2,T3 : K −→ E are three generalize asymptotically

nonexpansive nonself mappings with sequences {µ(1)n },{µ
(2)
n },{µ

(3)
n }∈ [1,∞),{ν

(1)
n },{ν

(2)
n },

{ν(3)n } ∈ [1,∞). Let {xn} be the sequence defined by (1.7), where {αn} and {βn} are real
sequences ∈ [0, 1). Suppose F = ∩3i=1(F (Ti ) ∩F (Si )) 6= 0. If the following conditions hold:



Eur. J. Math. Anal. 1 (2021) 66
i. ∑∞n=1k(1)n < ∞, ∑∞n=1k(2)n < ∞, ∑∞n=1k(3)n < ∞, ∑∞n=1µ(1)n < ∞, ∑∞n=1µ(2)n < ∞, ∑∞n=1µ(3)n <
∞,

∑∞
n=1ν

(1)
n < ∞,

∑∞
n=1ν

(2)
n < ∞,

∑∞
n=1ν

(3)
n < ∞,ii. ‖x−T1(PT1)n−1y‖≤‖Sn1x−T1(PT1)n−1y‖,‖x−T2(PT2)n−1y‖≤‖Sn2x−T2(PT2)n−1y‖,

‖x −T3(PT3)n−1y‖≤‖Sn3x −T3(PT3)
n−1y‖iii. There exists a constant M1,M2 > 0 such that Ψ(t) ≤ M1t, φ(t) ≤ M2t, t ≥ 0.

Then, limn∞‖xn −Sixn‖ = 0 and limn∞‖xn −Tixn‖ = 0, for i = 1, 2, , 3.

Abbreviations UsedNot applicable
Declaration:
Availability of Data and MaterialNot applicable
Competing lnterestThe authors declare that there is no conflict of interest.
FundingNo specific funding received for this work
Authors ContributionIKA and NCU wrote the paper while DII suggested the idea and did the analysis. The three authorsread and approved the final manuscript.
AcknowledgementThe authors thank the anonymous reviewers for their careful reading of this paper and approvedthe final manuscript.



Eur. J. Math. Anal. 1 (2021) 67
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https://doi.org/10.1155/FPTA/2006/10673
https://doi.org/10.1016/S0022-247X(03)00061-1
https://doi.org/10.1081/NFA-120039611
https://doi.org/10.1081/NFA-120039611
https://doi.org/10.1016/j.jmaa.2006.09.023
https://doi.org/10.1016/S0362-546X(99)00200-X
https://doi.org/10.1016/S0362-546X(99)00200-X
https://doi.org/10.1090/S0002-9939-1972-0298500-3
https://doi.org/10.1016/j.aml.2011.06.022
https://doi.org/10.1186/1687-1812-2012-224
https://doi.org/10.1016/S0895-7177(00)00199-0
https://doi.org/10.1016/S0895-7177(00)00199-0
https://doi.org/10.1016/j.jmaa.2005.10.062

	1. Introduction
	2. Preliminary
	3. Main Results
	References

