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@ Appl. Gen. Topol. 18, no. 2 (2017), 345-360doi:10.4995/agt.2017.7257
c© AGT, UPV, 2017

Convergence theorems for finding the split

common null point in Banach spaces

Suthep Suantai a, Kittipong Srisap b, Natthapong Naprang b,

Manatsawin Mamat b, Vithoon Yundon b and Prasit Cholamjiak b

a
Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200,

Thailand (suthep.s.cmu.ac.th)
b
School of Science, University of Phayao, Phayao 56000, Thailand. (prasitch2008@yahoo.com)

Communicated by E. A. Sánchez-Pérez

Abstract

In this paper, we introduce a new iterative scheme for solving the split

common null point problem. We then prove the strong convergence

theorem under suitable conditions. Finally, we give some numerical

examples for supporting our main results.

2010 MSC: 47H04; 47H10; 54H25.

Keywords: convergence theorem; split common null point problem; Banach
space; bounded linear operator.

1. Introduction

Let H1 and H2 be real Hilbert spaces and T : H1 → H2 a bounded linear
operator (we denote A∗ by its adjoint) . Let C and Q be nonempty, closed and
convex subsets of H1 and H2, respectively. The split feasibility problem is to
find x ∈ C such that T x ∈ Q. In order to solve the split feasibility problem
(SFP), Byrne [5] proposed the following iterative algorithm in the framework
of Hilbert spaces: x1 ∈ C and

(1.1) xn+1 = PC(xn − λT
∗(I − PQ)T xn), n ≥ 1,

which is often called the CQ algorithm, where λ > 0, PC and PQ are the
metric projections on C and Q, respectively. It was shown that the sequence

Received 09 February 2017 – Accepted 04 May 2017

http://dx.doi.org/10.4995/agt.2017.7257


S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

{xn} converges weakly to a solution of SFP. Since then several iterations have
been invented for solving the SFP (see, for example, [2, 11, 13, 17]).

Let A : H1 → 2
H1 and B : H2 → 2

H2 be set-valued mappings. Byrne et al.
[6] considered the problem of finding a point z in H1 such that

(1.2) z ∈ A−10 ∩ T −1(B−10),

where the set of null points of A is defined by A−10 = {z ∈ H1 : 0 ∈ Az}. We
know that A−10 is closed and convex. This problem is called the split common
null point problem and includes the spit feasibility problem as special cases;
see also [8].

In 1953, Mann [10] introduced the following iteration process. Let C be
a nonempty , closed and convex subset of a Banach space E. A mapping
T : C → C is called nonexpansive if

(1.3) ‖T x − T y‖ ≤ ‖x − y‖

for all x, y ∈ C. We denote by F(T ) the fixed point set of T . For an initial
point x1 ∈ C, an iteration process {xn} is defined recursively by

(1.4) xn+1 = αnxn + (1 − αn)T xn, n ∈ N,

where {αn} is a sequence in [0,1] and T is a nonexpansive mapping on C.
In 1967, Halpern [7] defined an iteration process as follows: Take x0, x1 ∈ C

arbitrarily and define {xn} recursively by

(1.5) xn+1 = αnx0 + (1 − αn)T xn, n ∈ N,

where {αn} is a sequence in [0, 1] and T is a nonexpansive mapping on C.
A mapping f : C → C is said to be a contraction if there exists α ∈ (0, 1)

such that

(1.6) ‖f(x) − f(y)‖ ≤ α‖x − y‖, ∀x, y ∈ C.

In 2000, Moudafi [12] introduced the following algorithm: For x1 ∈ C, define
the sequence {xn} by

(1.7) xn+1 = αnf(xn) + (1 − αn)T xn, n ∈ N,

where {αn} ⊂ (0, 1) and T is a nonexpansive mapping. This method is called
the viscosity approximation method.

Let H be a Hilbert space and let F be a strictly convex, reflexive and smooth
Banach space. Let JF be the duality mapping on F . Let C and D be nonempty,
closed and convex subsets of H and F , respectively. Let PC and PD be the
metric projections of H onto C and F onto D, respectively. Let T : H → F be
a bounded linear operator such that T 6= 0 and let T ∗ be the adjoint operator of
T . Suppose that C ∩ A−1D 6= ∅. In 2015, Alsulami and Takahashi [2] defined
the following algorithm: For any x1 ∈ H,

(1.8) xn+1 = βnxn + (1 − βn)PC(I − rT
∗
JF (T − PDT ))xn, n ∈ N,

where {βn} ⊂ [0, 1] and r ∈ (0, ∞). It was proved that if

(1.9) 0 < a ≤ βn ≤ b < 1 and 0 < r‖T ‖
2 < 2

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 346



Convergence theorems for finding the split common null point

for some a, b ∈ R, then {xn} converges weakly to z0 ∈ C ∩ T
−1D, where

z0 = limn→∞ PC∩T −1Dxn.
They introduced the following Halpern’s type iteration: For any x1 ∈ H,

(1.10)
xn+1 = βnxn + (1 − βn)(αnun + (1 − αn)PC(I − rT

∗
JF (I − PD)T )xn), n ∈ N,

where r ∈ (0, ∞), {αn} ⊂ (0, 1) and {βn} ⊂ (0, 1). It was proved that if

(1.11) 0 < r‖T ‖2 < 2, lim
n→∞

αn = 0,

(1.12)

∞
∑

n=1

αn = ∞ and 0 < a ≤ βn ≤ b < 1

where a, b ∈ R. Then {xn} converges strongly to a point z0 = C ∩ A
−1D, for

some z0 = PC∩A−1Du.
Recently, using the idea of Halpern’s iteration, Alofi et al. [1] proved the

following strong convergence theorem for finding a solution of the split common
null point problem in Banach spaces.

Theorem 1.1. Let H be a Hilbert space and let F be a uniformly convex and

smooth Banach space. Let JF be the duality mapping on F. Let A and B be

maximal monotone operators of H into 2H and F into 2F
∗

such that A−10 6=
∅ and B−10 6= ∅, respectively. Let Jλ be the resolvent of A for λ > 0 and let
Qµ be the metric resolvent of B for µ > 0. Let T : H → F be a bounded linear
operator such that T 6= 0 and let T ∗ be the adjoint operator of T . Suppose that
A−10 ∩ T −1(B−10) 6= ∅. Let {un} be a sequence in H such that un → u. Let
x1 = x ∈ H and let {xn} ⊂ H be a sequence generated by

xn+1 = βnxn + (1 − βn)(αnun + (1 − αn)Jλn (I − λnT
∗
JF (I − Qµn)T )xn)

(1.13)

for all n ∈ N, where {λn}, {µn} ⊂ (0, ∞), {αn} ⊂ (0, 1) and {βn} ⊂ (0, 1)
satisfy the following conditions

(1.14) 0 < a ≤ λn‖T ‖
2 ≤ b < 2, 0 < k ≤ µn, 0 < c ≤ βn ≤ d < 1,

(1.15) lim
n→∞

αn = 0 and
∞
∑

n=1

αn = ∞

for some a, b, c, d, k ∈ R. Then {xn} converges strongly to z0 ∈ A
−10 ∩

T −1(B−10), where z0 = PA−10∩T −1(B−10)u.

Motivated by the previous works, we introduce a new iterative scheme for
solving the split common null point problem. We then prove the strong con-
vergence theorem under suitable conditions. Finally, we give some numerical
examples for supporting our main results.

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 347



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

2. Preliminaries and lemmas

Let H be a real Hilbert space with inner product 〈·, ·〉 and norm ‖ · ‖,
respectively. For x, y ∈ H and λ ∈ R, we know from [15] that

(2.1) ‖x + y‖2 ≤ ‖x‖2 + 2〈y, x + y〉;

(2.2) ‖λx + (1 − λ)y‖2 = λ‖x‖2 + (1 − λ)‖y‖2 − λ(1 − λ)‖x − y‖2.

Furthermore, for x, y, u, v ∈ H,

(2.3) 2〈x − y, u − v〉 = ‖x − v‖2 + ‖y − u‖2 − ‖x − u‖2 − ‖y − v‖2.

The nearest point projection of a nonempty, closed and convex set C is denoted
by PC, that is, ‖x − PCx‖ ≤ ‖x − y‖ for all x ∈ H and y ∈ C. Such PC is
called the metric projection of H onto C. We know the metric projection PC
is firmly nonexpansive, i.e.,

(2.4) ‖PCx − PCy‖
2 ≤ 〈PCx − PCy, x − y〉

for all x, y ∈ H. Moreover 〈x − PCx, y − PCx〉 ≤ 0 holds for all x ∈ H and
y ∈ C; see [15].

Let E be a real Banach space with norm ‖ · ‖ and let E∗ be the dual space
of E. We denote the value of y∗ ∈ E∗ at x ∈ E by 〈x, y∗〉. When {xn} is a
sequence in E, we denote the strong convergence of {xn} to x ∈ E by xn → x
and the weak convergence by xn ⇀ x. The modulus δ of convexity of E is
defined by

δ(ǫ) = inf

{

1 −
‖x + y‖

2
: ‖x‖ ≤ 1, ‖y‖ ≤ 1, ‖x − y‖ ≥ ǫ

}

(2.5)

for every ǫ with 0 ≤ ǫ ≤ 2. A Banach space E is said to be uniformly convex
if δ(ǫ) > 0. It is known that a Banach space E is uniformly convex if and only
if for any two sequences {xn} and {yn} in E such that

lim
n→∞

‖xn‖ = lim
n→∞

‖yn‖ = 1 and lim
n→∞

‖xn − yn‖ = 2,(2.6)

limn→∞ ‖xn − yn‖ = 0 holds. A uniformly convex Banach space is strictly
convex and reflexive.

The duality mapping J from E into 2E
∗

is defined by

J(x) = {x∗ ∈ E∗ : 〈x, x∗〉 = ‖x‖2 = ‖x∗‖2}(2.7)

for every x ∈ E. Let U = {x ∈ E : ‖x‖ = 1}. The norm of E is said to be
Gâteaux differentiable if for each x, y ∈ U, the limit

(2.8) lim
t→0

‖x + ty‖ − ‖x‖

t

exists. In this case, E is called smooth. We know that E is smooth if and only
if J is a single-valued mapping of E into E∗. We also know that E is reflexive if
and only if J is surjective, and E is strictly convex if and only if J is one-to-one
Therefore, if E is a smooth, strictly convex and reflexive Banach space, then J

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 348



Convergence theorems for finding the split common null point

is a single-valued bijection and in this case, the inverse mapping J−1 coincides
with the duality mapping J∗ on E∗. For more details, see [14, 16].

Let C be a nonempty, closed and convex subset of a strictly convex and
reflexive Banach space E. Then we know that for any x ∈ E, there exists a
unique element z ∈ C such that ‖x − z‖ ≤ ‖x − y‖ for all y ∈ C. Putting
z = PCx, we call PC the metric projection of E onto C.

Lemma 2.1 ([16]). Let E be a smooth, strictly convex and reflexive Banach
space. Let C be a nonempty, closed and convex subset of E, and let x1 ∈ E and
z ∈ C Then, the following conditions are equivalent:

(1) z = PCx1;
(2) 〈z − y, J(x1 − z)〉 ≥ 0, ∀ y ∈ C.

Let E be a Banach space and let A be a mapping of E into 2E
∗

. The effective
domain of A is denoted by dom(A), that is, dom(A) = {x ∈ E : Ax 6= ∅}. A
multi-valued mapping A on E is said to be monotone if 〈x − y, u∗ − v∗〉 ≥ 0
for all x, y ∈ dom(A), u∗ ∈ Ax, and v∗ ∈ Ay. A monotone operator A on E is
said to be maximal if its graph is not properly contained in the graph of any
other monotone operator on E. The following theorem is due to Browder [4];
see also [14].

Lemma 2.2 ([4]). Let E be a uniformly convex and smooth Banach space and
let J be the duality mapping on E into E∗. Let A be a monotone operator of

E into 2E
∗

. Then A is maximal if and only for any r > 0,

(2.9) R(J + rA) = E∗,

where R(J + rA) is the range of J + rA.

Let E be a uniformly convex and smooth Banach space with a Gâteaux
differentiable norm and let A be a monotone operator of E into 2E

∗

. For all
x ∈ E and r > 0, we consider the following equation

(2.10) 0 ∈ J(xr − x) + rAxr.

This equation has a unique solution xr. We define Jr by xr = Jrx. Such Jr
where r > 0 are called the metric resolvent of A. In a Hilbert space H, the
metric resolvent Jr of A is simply called the resolvent of A. We also know the
following lemmas:

Lemma 2.3 ([3, 18]). Let {sn} be a sequence of nonnegative real numbers,
let {αn} be a sequence in [0, 1] with

∑

∞

n=1 αn = ∞, let {βn} be a sequence of
nonnegative real numbers with

∑

∞

n=1 βn < ∞ and {γn} be a sequence of real
numbers with lim supn→∞ γn ≤ 0. Suppose that

(2.11) sn+1 = (1 − αn)sn + αnγn + βn

for all n = 1, 2, .... Then limn→∞ sn = 0.

Lemma 2.4 ([9]). Let {Γn} be a sequence of real numbers that does not decrease
at infinity in the sense the there exists a subsequence {Γni} of {Γn} which

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 349



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

satisfies Γni < Γni+1 for all i ∈ N. Define the sequence {τ(n)}n>n0 of integers
as follows:

(2.12) τ(n) = max {k ≤ n : Γk < Γk+1} ,

where n0 ∈ N such that {k ≤ n0 : Γk < Γk+1} 6= ∅. Then, the following hold:
(i) τ(n0) ≤ τ(n0 + 1) ≤ ... and τ(n) → ∞;
(ii) Γτ(n) ≤ Γτ(n)+1 and Γn ≤ Γτ(n)+1, ∀n ≥ n0.

3. Main results

In this section, we prove strong convergence theorems for finding a solution
of the split common null point problem in Banach spaces.

Theorem 3.1. Let H be a Hilbert space and let F be a uniformly convex and

smooth Banach space. Let JF be the duality mapping on F. Let f : H → H
be a contraction. Let A and B be maximal monotone operators of H into 2H

and F into 2F
∗

, respectively. Let Jλ be the resolvent of A for λ > 0 and let
Qµ be the metric resolvent of B for µ > 0. Let T : H → F be a bounded linear
operator such that T 6= 0 and let T ∗ be the adjoint operator of T . Suppose that
A−10∩T −1(B−10) 6= ∅. Let x1 ∈ H and let {xn} ⊂ H be a sequence generated
by

xn+1 = αnf(xn) + βnxn + γnJλn(I − λnT
∗JF (I − Qµn)T )xn(3.1)

for all n ∈ N, where {µn}, {λn} ⊂ (0, ∞), {αn} ⊂ (0, 1), {βn} ⊂ (0, 1) and
{γn} ⊂ (0, 1) satisfy the following conditions:

(3.2) 0 < a ≤ λn‖T
2‖ ≤ b < 2, 0 < k ≤ µn, 0 < c ≤ γn ≤ d < 1,

(3.3) lim
n→∞

αn = 0 and

∞
∑

n=1

αn = ∞

for some a, b, c, d, k ∈ R. Then {xn} converges strongly to z0 ∈ A
−10 ∩

T −1(B−10), where z0 = PA−10∩T −1(B−10)f(z0).

Proof. Put zn = Jλn(I − λnT
∗JF (I − Qµn)T )xn for all n ∈ N and let z ∈

A−10 ∩ T −1(B−10). We have that z = Jλnz and T z = QµnT z for all n ∈ N.

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 350



Convergence theorems for finding the split common null point

Since Jλn is nonexpansive, we have

‖zn − z‖
2 = ‖Jλn(I − λnT

∗JF (I − Qµn)T )xn − Jλnz‖
2

≤ ‖xn − λnT
∗JF (I − Qµn)T )xn − z‖

2

= ‖xn − z‖
2 − 2λn〈xn − z, T

∗JF (I − Qµn)T xn〉

+λ2n‖T
∗JF (I − Qµn)T xn‖

2

≤ ‖xn − z‖
2 − 2λn〈T xn − T z, JF (I − Qµn)T xn〉

+ λ2n‖T ‖
2‖(I − Qµn)T xn‖

2

= ‖xn − z‖
2 − 2λn〈T xn − QµnT xn, JF (I − Qµn)T xn〉

−2λn〈QµnT xn − T z, JF (I − Qµn)T xn〉

+ λ2n‖T ‖
2‖(I − Qµn)T xn‖

2

= ‖xn − z‖
2 − 2λn‖T xn − QµnT xn‖

2

−2λn〈QµnT xn − T z, JF (I − Qµn)T xn〉

+ λ2n‖T ‖
2‖(I − Qµn)T xn‖

2

≤ ‖xn − z‖
2 − 2λn‖T xn − QµnT xn‖

2 + λ2n‖T ‖
2‖(I − Qµn)T xn‖

2

= ‖xn − z‖
2 + λn(λn‖T ‖

2 − 2)‖(I − Qµn)T xn‖
2.(3.4)

Since 0 < λn‖T ‖
2 < 2, it follows that ‖zn − z‖ ≤ ‖xn − z‖ for all n ∈ N. So we

obtain

‖xn+1 − z‖ = ‖αnf(xn) + βnxn + γnzn − z‖

≤ αn‖f(xn) − z‖ + βn‖xn − z‖ + γn‖xn − z‖

≤ αnα‖xn − z‖ + αn‖f(z) − z‖ + (1 − αn)‖xn − z‖

= (1 − αn(1 − α))‖xn − z‖ + αn‖f(z) − z‖.(3.5)

By induction, we conclude that {xn} is bounded. So are {T xn}, {zn} and {yn}.
Put z0 = PA−10∩T −1(B−10)f(z0). We see that

xn+1 − xn = αn(f(xn) − xn) + γn(zn − xn),(3.6)

which implies that

xn+1 − xn − αn(f(xn) − xn) = γn(zn − xn).(3.7)

It follows that

〈xn+1 − xn − αn(f(xn) − xn), xn − z0〉 = γn〈zn − xn, xn − z0〉

= −γn〈xn − zn, xn − z0〉.(3.8)

From (2.3), we obtain

2〈xn − zn, xn − z0〉 = ‖xn − z0‖
2 + ‖zn − xn‖

2 − ‖zn − z0‖
2

≥ ‖xn − z0‖
2 + ‖zn − xn‖

2 − ‖xn − z0‖
2

= ‖zn − xn‖
2.(3.9)

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 351



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

From (3.8) and (3.9), we obtain

2〈xn+1 − xn, xn − z0〉 = 2αn〈f(xn) − xn, xn − z0〉 − 2γn〈xn − zn, xn − z0〉

≤ 2αn〈f(xn) − xn, xn − z0〉 − γn‖zn − xn‖
2.(3.10)

Using (2.3) and (3.10), we have
(3.11)
‖xn+1−z0‖

2−‖xn−xn+1‖
2−‖xn−z0‖

2 ≤ 2αn〈f(xn)−xn, xn−z0〉−γn‖zn−xn‖
2.

Putting Γn = ‖xn − z0‖
2 for all n ∈ N, we see that

(3.12) Γn+1 − Γn − ‖xn − xn+1‖
2 ≤ 2αn〈f(xn) − xn, xn − z0〉 − γn‖zn − xn‖

2
.

We note that

‖xn+1 − xn‖ = ‖αnf(xn) + βnxn + γnzn − xn‖

≤ αn‖f(xn) − xn‖ + γn‖zn − xn‖.(3.13)

This shows that

‖xn+1 − xn‖
2 ≤ (αn‖f(xn) − xn‖ + γn‖zn − xn‖)

2

= α2n‖f(xn) − xn‖
2 + 2αnγn‖f(xn) − xn‖‖zn − xn‖

+ γ2n‖zn − xn‖
2
.(3.14)

Hence by (3.12) and (3.14), we have

Γn+1 − Γn ≤ αn(αn‖f(xn) − xn‖
2 + 2γn‖f(xn) − xn‖‖zn − xn‖)

+ γ2n‖zn − xn‖
2

+ 2αn〈f(xn) − xn, xn − z0〉 − γn‖zn − xn‖
2

= αn(αn‖f(xn) − xn‖
2 + 2γn‖f(xn) − xn‖‖zn − xn‖)

+ γn(γn − 1)‖zn − xn‖
2

+ 2αn〈f(xn) − z0, xn − z0〉 − 2αn‖xn − z0‖
2.(3.15)

So we obtain

Γn+1 − Γn + γn(1 − γn)‖zn − xn‖
2 ≤ αn(αn‖f(xn) − xn‖

2

+ 2γn‖f(xn) − xn‖‖zn − xn‖)

+ 2αn〈f(xn) − z0, xn − z0〉

− 2αn‖xn − z0‖
2.(3.16)

We next split the proof into two cases.
Case 1: Suppose that there exists a natural number N such that Γn+1 ≤ Γn

for all n ≥ N. In this case, limn→∞ Γn exists and then limn→∞(Γn+1−Γn) = 0.
Since limn→∞ αn = 0 and 0 < c ≤ γn ≤ d < 1, by (3.16), we have

lim
n→∞

‖zn − xn‖ = 0.(3.17)

From (3.13) we have

lim
n→∞

‖xn+1 − xn‖ = 0.(3.18)

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 352



Convergence theorems for finding the split common null point

We next show that lim supn→∞〈f(z0) − z0, zn − z0〉 ≤ 0. Put

(3.19) l = lim sup
n→∞

〈f(z0) − z0, zn − z0〉.

Then without loss of generality, there exists a subsequence {zni} of {zn} such
that l = limi→∞〈f(z0)−z0, zni −z0〉 and {zni} converges weakly to some point
w ∈ H. Since ‖xn − zn‖ → 0, we also have that {xni} converges weakly to
w ∈ H. On the other hand, from (3.4) we have

λn(2 − λn‖T ‖
2)‖(I − Qµn)T xn‖

2 ≤ ‖xn − zn‖
2 − ‖zn − z‖

2

≤ ‖xn − zn‖(‖xn − z‖ + ‖zn − z‖).(3.20)

Then since ‖xn − zn‖ → 0 and 0 < a ≤ λn‖T ‖
2 ≤ b < 2,

(3.21) lim
n→∞

‖T xn − QµnT xn‖ = 0.

Since {xni} converges weakly to w ∈ H and T is bounded and linear, we
also have {T xni} converges weakly to T w. Using this and limn→∞ ‖T xn −
QµnT xn‖ = 0, we have that Qµni T xni ⇀ T w. Since Qµn is the metric resolvent

of B for µn > 0, we have that
JF (T xn−Qµn T xn)

µn
∈ BQµnT xn for all n ∈ N. By

the monotonicity of B we obtain

0 ≤

〈

u − Qµni T xni, v
∗ −

JF (T xni − Qµni T xni)

µni

〉

(3.22)

for all (u, v∗) ∈ B. We observe that ‖JF (T xni − Qµni T xni)‖ = ‖T xni −

Qµni T xni‖ → 0 as i → ∞. Since 0 < k ≤ µni, it follows that 0 ≤ 〈u−T w, v
∗ −

0〉 for all (u, v∗) ∈ B. Because B is maximal monotone, we have T w ∈ B−10.
This implies that w ∈ T −1(B−10). Using zn = Jλn(xn − λnT

∗JF (T xn −
QλnT xn)), we obtain

zn = Jλn (xn − λnT
∗
JF (T xn − QµnT xn))(3.23)

⇔ xn − λnT
∗JF (T xn − QµnT xn) ∈ zn + λnAz

⇔ xn − zn − λnT
∗
JF (T xn − QµnT xn) ∈ λnAzn

⇔
1

λn
(xn − zn − λnT

∗JF (T xn − QµnT xn)) ∈ Azn.

Since A is monotone, we have that for (u, v) ∈ A,
〈

zn − u,
1

λn
(xn − zn − λnT

∗JF (T xn − QµnT xn)) − v

〉

≥ 0(3.24)

which implies that
〈

zn − u,
xn − zn

λn
− T ∗JF (T xn − QµnT xn)) − v

〉

≥ 0.(3.25)

Replacing n by ni, we have
〈

zni − u,
xni − zni

λni
− T ∗JF (T xni − Qµni T xni) − v

〉

≥ 0.(3.26)

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 353



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

Since xni −zni → 0, 0 < a ≤ λni‖T ‖
2, zni ⇀ w and T

∗JF (T xn −Qµni T xni) →

0, we get that 〈w − u, −v〉 ≥ 0. Since A is maximal, we have 0 ∈ Aw.
Therefore, w ∈ A−10 ∩ T −1(B−10). Since {zni} converges weakly to w ∈
A−10 ∩ T −1(B−10), it follows that

(3.27) l = lim
i→∞

〈f(z0) − z0, zni − z0〉 = 〈f(z0) − z0, w − z0〉 ≤ 0.

On the other hand, we see that

‖xn+1 − z0‖
2 = 〈xn+1 − z0, xn+1 − z0〉

= 〈αnf(xn) + βnxn + γnzn − z0, xn+1 − z0〉

= 〈αn(f(xn) − z0) + βn(xn − z0) + γn(zn − z0), xn+1 − z0〉

= αn〈f(xn) − f(z0) + f(z0) − z0, xn+1 − z0〉

+ βn〈xn − z0, xn+1 − z0〉 + γn〈zn − z0, xn+1 − z0〉

= αn〈f(xn) − f(z0), xn+1 − z0〉 + αn〈f(z0) − z0, xn+1 − z0〉

+ βn〈xn − z0, xn+1 − z0〉 + γn〈zn − z0, xn+1 − z0〉

≤ αnα‖xn − z0‖‖xn+1 − z0‖ + βn‖xn − z0‖‖xn+1 − z0‖

+ γn‖xn − z0‖‖xn+1 − z0‖ + αn〈f(z0) − z0, xn+1 − z0〉

= (αnα + βn + γn)‖|xn − z0‖‖xn+1 − z0‖

+ αn〈f(z0) − z0, xn+1 − z0〉

≤ (αnα + 1 − αn)
1

2
(‖xn − z0‖

2 + ‖xn+1 − z0‖
2)

+ αn〈f(z0) − z0, xn+1 − z0〉

=

(

ααn + 1 − αn
2

)

‖xn − z0‖
2

+

(

ααn + 1 − αn
2

)

‖xn+1 − z0‖
2

+ αn〈f(z0) − z0, xn+1 − z0〉

=

(

1 −
2(1 − α)αn

1 + (1 − α)αn

)

‖xn − z0‖
2

+

(

2(1 − α)αn
1 + (1 − α)αn

) (

1

1 − α

)

〈f(z0) − z0, xn+1 − z0〉.(3.28)

Also, we have

lim
n→∞

‖zn − xn+1‖ ≤ lim
n→∞

(‖zn − xn‖ + ‖xn+1 − xn‖) = 0.(3.29)

Then

lim sup
n→∞

〈f(z0) − z0, xn+1 − z0〉 ≤ 0.(3.30)

Since
∑

∞

n=1 αn = ∞, by Lemma 2.3 we conclude that xn → z0 as n → ∞.

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 354



Convergence theorems for finding the split common null point

Case 2. Suppose that there exists a subsequence {Γni} of the sequence
{Γni} such that Γni ≤ Γni+1 for all i ∈ N. In this case, we define τ : N → N by

τ(n) = max{k ≤ n : Γk < Γk+1}.(3.31)

Then by Lemma 2.6 we have Γτ(n) < Γτ(n)+1. Thus by (3.16) we have for all
n ∈ N,

γτ(n)(1 − γτ(n))‖zτ(n) − xτ(n)‖
2 ≤ α2τ(n)‖f(xτ(n)) − xτ(n)‖

2

+ 2ατ(n)γτ(n)(‖f(xτ(n)) − xτ(n)‖

× ‖zτ(n) − xτ(n)‖)

+ 2ατ(n)〈f(xτ(n)) − z0, xτ(n) − z0〉

− 2ατ(n)‖xτ(n) − z0‖
2
.(3.32)

Using limn→∞ αn = 0 and 0 < c ≤ γn ≤ d < 1, we have

lim
n→∞

‖zτ(n) − xτ(n)‖ = 0.(3.33)

As in the proof of Case 1, we can show that

lim
n→∞

‖xτ(n)+1 − xτ(n)‖ = 0.(3.34)

This gives

lim
n→∞

‖zτ(n) − xτ(n)+1‖ = 0.(3.35)

We next show that lim supn→∞〈f(z0) − z0, xτ(n)+1 − z0〉 ≤ 0. Put

l = lim sup
n→∞

〈f(z0) − z0, xτ(n)+1 − z0〉.(3.36)

So we have

l = lim sup
n→∞

〈f(z0) − z0, zτ(n) − z0〉.(3.37)

Without loss of generality, there exists a subsequence {zτ(ni)} of {zτ(n)} such
that

l = lim
i→∞

〈f(z0) − z0, zτ(ni) − z0〉(3.38)

and {zτ(ni)} converges weakly to some point w ∈ H. As in the proof of Case

1, we can show that w ∈ A−10 ∩ T −1(B−10). Then it follows that

l = lim
i→∞

〈f(z0) − z0, zτ(ni) − z0〉 = 〈f(z0) − z0, w − z0〉 ≤ 0.(3.39)

As in the proof of Case 1, we also obtain

‖xτ(n)+1 − z0‖
2

≤

(

1 −
2(1 − α)ατ(n)

1 + (1 − α)ατ(n)

)

‖xτ(n) − z0‖
2

+

(

2(1 − α)αn
1 + (1 − α)αn

) (

1

1 − α

)

〈f(z0) − z0, xτ(n)+1 − z0〉.(3.40)

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 355



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

Since Γτ(n) ≤ Γτ(n)+1,

(

2(1 − α)ατ(n)

1 + (1 − α)ατ(n)

)

‖xτ(n) − z0‖
2 ≤

(

2(1 − α)αn
1 + (1 − α)αn

)

×

(

1

1 − α

)

〈f(z0) − z0, xτ(n)+1 − z0〉.(3.41)

It is easily seen that
(

2(1−α)ατ(n)
1+(1−α)ατ(n)

)

> 0. Then we have

‖xτ(n) − z0‖
2 ≤

(

1

1 − α

)

〈f(z0) − z0, xτ(n)+1 − z0〉.(3.42)

This shows that

lim sup
n→∞

‖xτ(n) − z0‖
2 ≤ 0(3.43)

and hence ‖xτ(n) − z0‖ → 0 as n → ∞. Thus ‖xτ(n)+1 − z0‖ → 0 as n → ∞.
By Lemma 2.6, we obtain

‖xn − z0‖ ≤ ‖xτ(n)+1 − z0‖ → 0(3.44)

as n → ∞. This completes the proof. �

4. Examples and numerical results

In this section, we give examples including its numerical results for support-
ing our main theorem.

Example 4.1. Let H = R. For x ∈ R, we define G : R → R by

G(x) =

{

ωx if x ≥ 0,
+∞ otherwise.

Let F : R → R be defined by F(x) = ω|x| − ln(1 + ω|x|).

Choose x1 = 2, ω = 1, αn =
1

2n+1
, βn =

n
2n+1

, γn =
n

2n+1
for all n ∈ N.

Let f(x) = x
2
and T x = x . We aim to find the minimizers of F and G. Using

algorithm (3.1), we have the following numerical results:

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 356



Convergence theorems for finding the split common null point

n xn | xn+1 − xn |
1 1.471404251 5.2859547 × 10−1

2 1.265727840 7.8045565 × 10−1

3 0.527004926 7.8045565 × 10−1

4 0.250611252 1.3092799 × 10−1

5 0.160205803 7.9460312 × 10−2

6 0.114183169 5.3450423 × 10−2

7 0.083166153 3.7671482 × 10−2

8 0.061057394 2.7173989 × 10−2

9 0.045012810 1.9840027 × 10−2

10 0.033265124 1.4579710 × 10−2

11 0.024621382 1.0752746 × 10−2

12 0.018243176 7.9469695 × 10−3

...
...

...
50 0.000000260 1.1017075 × 10−7

Table 1 Numerical results of Example 4.1 for iteration process (3.1)

0 5 10 15 20 25 30 35 40 45 50
0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Number of iterations

X
n

Figure 1: Convergence behavior of {xn} in Table 1.

5 10 15 20 25 30 35 40 45 50
0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

Number of iterations

E
rr

o
r

Figure 2: Error plots for all sequences {xn} in Table 1.

Example 4.2. Let H = R3. For x ∈ R3, define G : R3 → R by G(x) =
‖Lx − y‖2, where

L =





1 2 2
3 0 1
2 1 −1



, y =





2
−3
1



 and F : R3 → R by F(x) = 5‖x‖2 +

(15, 6, −7)x + 10.

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 357



S. Suantai, K. Srisap, N. Naprang, M. Mamat, Y. Yundon and P. Cholamjiak

Let T =





1 0 1
0 −1 2
1 2 3



 . Find x ∈ R3 such that x is a minimizer of F and T x

also is a minimizer of G.

Choose x1=





1
−1
−1



, αn =
1

2n+1
, βn =

n
2n+1

, γn =
n

2n+1
for all n ∈ N and

let f(x) = x
2
. Using algorithm (3.1), we have the following numerical results:

n xn ‖xn+1 − xn‖
1 (0.1527,-0.6014,0.8512) 20.745517681470 ×10−1

2 (-0.5968,-0.6646,-0.0706) 11.896884171498 ×10−1

3 (-0.8523,-0.5312,0.7977) 9.149067074771 ×10−1

4 (-1.1354,-0.6011,0.3222) 5.577239966457 ×10−1

5 (-1.2027,-0.5415,0.7430) 4.302752764182 ×10−1

6 (-1.3156,-0.5838,0.4941) 2.766128927383 ×10−1

7 (-1.3323,-0.5554,0.7039) 2.124210254520 ×10−1

8 (-1.3822,-0.5792,0.5747) 1.406007674047 ×10−1

9 (-1.3869,-0.5656,0.6815) 1.078492388367 ×10−1

10 (-1.4115,-0.5788,0.6150) 7.214423578793 ×10−2

11 (-1.4139,-0.5724,0.6703) 5.574083184995 ×10−2

12 (-1.4273,-0.5798,0.6365) 3.720017649222 ×10−2

13 (-1.4295,-0.5769,0.6655) 2.926886211227 ×10−2

14 (-1.4375,-0.5811,0.6485) 1.930348053252 ×10−2

15 (-1.4397,-0.5800.0.6640) 1.568794063595 ×10−2

...
...

...
100 (-1.4790,-0.5913,0.6743) 8.809288634531 ×10−5

Table 2 Numerical results of Example 4.2 for iteration process (3.1)

0 10 20 30 40 50 60 70 80 90 100
−1.5

−1

−0.5

0

0.5

1

Number of iterations

X
n

Figure 3: Convergence behavior of {xn} in Table 2 .

c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 358



Convergence theorems for finding the split common null point

0 5 10 15 20 25
0

0.5

1

1.5

2

2.5

Number of iterations

E
rr

o
r

Figure 4: Error plots for all sequences {xn} in Table 2 .

From Table 2, we see that





−1.5
−0.6
0.7



 is a minimizer of F such that T





−1.5
−0.6
0.7





=





−0.8
2

−0.6



 is a minimizer of G.

Acknowledgements. S. Suantai wish to thank Chiang Mai University for

financial supports. P. Cholamjiak was supported by the Thailand Research

Fund and the Commission on Higher Education under Grant MRG5980248.

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