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Iterative algorithms with errors for zero points of $m$-accretive operators

Fixed Point Theory and Applications2013, 2013:148 doi:10.1186/1687-1812-2013-148

Sun Young Cho ([email protected])Xiaolong Qin ([email protected])Lin Wang ([email protected])

ISSN 1687-1812

Article type Research

Submission date 20 March 2013

Acceptance date 21 May 2013

Publication date 6 June 2013

Article URL http://www.fixedpointtheoryandapplications.com/content/2013/1/148

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ITERATIVE ALGORITHMS WITH ERRORS FOR ZERO POINTS OF

m-ACCRETIVE OPERATORS

XIAOLONG QIN1, SUN YOUNG CHO2, LIN WANG3

Abstract. In this paper, we study the convergence of paths for continuous pseudocontractions in a real

Banach space. As an application, we consider the problem of finding zeros ofm-accretive operators based

on an iterative algorithm with errors. Strong convergence theorems for zeros of m-accretive operators

are established in a real Banach space.

Keywords: accretive operator; fixed point; nonexpansive mapping; iterative algorithm; zero point.

2000 AMS Subject Classification: 47H05, 47H09, 47J25, 65J15.

1. Introduction and preliminaries

Let C be a nonempty closed convex subset of a real Banach space E and let E be the

dual space of E. Let , denote the pairing between E and E. The normalized duality

mapping J : E 2E

is defined by

J(x) = {f E : x, f = x2 = f2}

for all x E. In the sequel, we use j to denote the single-valued normalized duality

mapping. Let UE = {x E : x = 1}. E is said to be smooth or is said to have a

Gateaux differentiable norm if the limit

limt0

x + ty xt

exists for each x, y UE. E is said to have a uniformly Gateaux differentiable norm

if for each y UE, the limit is attained uniformly for all x UE. E is said to be

uniformly smooth or is said to have a uniformly Frechet differentiable norm if the limit is

attained uniformly for x, y UE. It is known that if the norm of E is uniformly Gateaux

differentiable, then the duality mapping J is single-valued and uniformly norm to weak

continuous on each bounded subset of E.1

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2 X. QIN, S.Y. CHO, L. WANG

Recall that a closed convex subset C of a Banach space E is said to have a normal

structure if for each bounded closed convex subset K of C which contains at least two

points, there exists an element x of K which is not a diametral point of K, i.e.,

sup{x y : y K} < d(K),

where d(K) is the diameter of K. It is well known that a closed convex subset of a

uniformly convex Banach space has the normal structure and a compact convex subset of

a Banach space has the normal structure; see [1] for more details.

Let T : C C be a nonlinear mapping. In this paper, we use F(T) to denote the

set of fixed points of T. Recall that T is said to be contractive if there exits a constant

(0, 1) such that

T x T y x y, x, y C.

For such a case, we also call T an -contraction. T is said to be a strong contraction if

there exits a constant (0, 1) and some j(x y) J(x y) such that

T x T y , j(x y) x y2, x, y C.

For such a case, we also call T an -strong pseudocontraction. T is said to be nonexpansive

if

T x T y x y, x, y C.

T is said to be pseudocontractive if there exists some j(x y) J(x y) such that

T x T y , j(x y) x y2, x, y C.

For the class of nonexpansive mappings, one classical way to study nonexpansive map-

pings is to use contractions to approximate a nonexpansive mapping ([2],[3]). More pre-

cisely, take t (0, 1) and define a contraction Tt : C C by

Ttx = tu + (1 t)T x, x C,

where u C is a fixed point. Banachs contraction mapping principle guarantees that Tt

has a unique fixed point xt in C. That is,

xt = tu + (1 t)T xt.

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ITERATIVE ALGORITHMS WITH ERRORS FOR m-ACCRETIVE OPERATORS 3

In the case that T enjoys a fixed point, Browder [2] proved that if E is a Hilbert space,

then xt converges strongly to a fixed point of T. Reich [3] extended Browders result to

the setting of Banach spaces and proved that if E is a uniformly smooth Banach space,then xt converges strongly to a fixed point of T.

Let D be a nonempty subset of C. Let Q : C D. Q is said to be contraction if

Q2 = Q; sunny if for each x C and t (0, 1), we have Q

tx + (1 t)Qx

= Qx; sunny

nonexpansive retraction if Q is sunny, nonexpansive, and contraction. K is said to be a

nonexpansive retract of C if there exists a nonexpansive retraction from C onto D.

The following result, which was established in [4,5], describes a characterization of

sunny nonexpansive retractions on a smooth Banach space.

Let E be a smooth Banach space and let C be a nonempty subset ofE. Let Q : E C

be a retraction and let j be the normalized duality mapping on E. Then the following

are equivalent:

(1) Q is sunny and nonexpansive;

(2) Qx Qy2 x y, j(Qx Qy), x, y E;

(3) x Qx,j(y Qx) 0, x E, y C.

Recently, Xu [6] improved Reichs results by considering the viscosity approximation

method which was first introduced by Moudafi [7]. To be more precise, he studied the

following continuous scheme:

xt = tf(xt) + (1 t)T xt, (X)

where f is an -contraction and T is a nonexpansive mapping. He showed that the

sequence {xt} converges strongly to a point in F(T) in a uniformly smooth Banach space.

If one defines Q : C F(T), where C denotes the set of -contractions, by

Q(f) := limt

xt,

then Q(f) solves the following variational inequality:

(I f)Q(f), j(Q(f) p) 0, p F(T).

The questions that naturally arise in connection with Xus results are as follows.

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(1) Can one extend the framework of the space from uniformly smooth Banach spaces

to a more general Banach space ?

(2) Can one extend the viscosity approximation method by considering strong pseu-docontractions instead of contractions ?

(3) Do Xus results still hold for a larger class of nonlinear mappings ?

In Section 2, we give an affirmative answer to the above questions.

Let I denote the identity operator on E. An operator A E E with domain

D(A) = {z E : Az = } and range R(A) = {Az : z D(A)} is said to be accretive if

for each xi D(A) and yi Axi, i = 1, 2, there exists j(x1 x2) J(x1 x2) such that

y1 y2, j(x1 x2) 0.

An accretive operator A is said to be m-accretiveifR(I+ rA) = E for all r > 0. In a real

Hilbert space, an operator A is m-accretive if and only if A is maximal monotone. In this

paper, we use A1(0) to denote the set of zeros of A. Interest in accretive operators, which

is an important class of nonlinear operators, stems mainly from their firm connection

with equations of evolution. It is known that many physically significant problems can

be modeled by initial value problems of the form

x(t) + Ax(t) = 0, x(0) = x0, (1.1)

where A is an accretive operator in an appropriate Banach space. Typical examples where

such evolution equations occur can be found in the heat, wave or Schrodinger equations.

If x(t) is dependent on t, then (1.1) is reduced to

Au = 0, (1.2)

whose solutions correspond to the equilibrium points of the system (1.1). Consequently,

considerable research efforts have been devoted, especially within the past 40 years or so,

to methods for finding approximate solutions (when they exist) of equation (1.2). An

early fundamental result in the theory of accretive operators, due to Browder [8], states

that the initial value problem (1.1) is solvable ifA is locally Lipschitz and accretive on E.

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For an accretive operator A, we can define a nonexpansive single-valued mapping Jr :

R(I+ rA) D(A) by Jr = (I+ rA)1 for each r > 0, which is called the resolvent of A.

One of classical methods for studying the problem 0 Ax, where A E E is anaccretive operator, is the following:

x0 E, xn+1 = Jrnxn, n 0,

where Jrn = (I + rnA)1 and {rn} is a sequence of positive real numbers.

As we know, many well-known problems arising in various branches of science can

be studied by using algorithms which are iterative in their nature. As an example, in

computer tomography with limited data, each piece of information implies the existence

of a convex set in which the required solution lies. The problem of finding a point in the

intersection of the sets is then of crucial interest, and it cannot be usually solved directly.

Therefore, an iterative algorithm must be used to approximate such a point.

Recall that the normal Mann iterative process was introduced by Mann [9] in 1953.

The normal Mann iterative process generates a sequence {xn} in the following manner:

x1 C, xn+1 = (1 n)xn + nT xn, n 1, (M)

where {n} is a sequence in the interval (0, 1). IfT is a nonexpansive mapping with a fixed

point and the control sequence {n} is chosen so that n=1

n(1 n) = , then the

sequence {xn} generated in the normal Mann iterative process converges weakly to a fixed

point of T. In an infinite-dimensional Hilbert space, the normal Mann iteration process

has only weak convergence, in general, even for nonexpansive mappings. Therefore, many

authors try to modify the normal Mann iterative process to have strong convergence for

nonexpansive mappings; see, e.g., [10-24] and the references therein.

Recently, Qin and Su [18] studied the problem of modifying the normal Mann iterative

process to have strong convergence for m-accretive operators. To be more precise, they

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considered the following iterative process:

x1 C,

yn = nxn + (1 n)Jrnxn,

xn+1 = u + (1 n)yn, n 1,

(QS)

where u is a fixed element in C and Jrn = (I + rnA)1. They proved that the sequence

{xn} generated in the above iterative process converges strongly to a zero of A.

In this paper, we study the convergence of paths for continuous pseudocontractions in

a real Banach space by viscosity approximation methods. As applications, we consider

the problem of finding zeros of m-accretive operators based on an iterative process with

errors. Strong convergence theorems of zeros are established in a real Banach space.

In order to prove our main results, we also need the following lemmas.

Lemma 1.1 [25] Let {an}, {bn}, and {cn} be three nonnegative real sequences satisfying

an+1 (1 tn)an + bn + cn, n 0,

where {tn} is a sequence in (0, 1). Assume that the following conditions are satisfied:(a)

n=0

tn = and bn = o(tn);

(b)

n=0cn < .

Then limn an = 0.

Lemma 1.2 [26] Let C be a nonempty, bounded, closed, and convex subset of a reflexive

Banach space E which also has the normal structure. Let T be a nonexpansive mapping

of C into itself. Then F(T) is nonempty.

Lemma 1.3 [27] Let C be a nonempty, closed, and convex subset of a Banach space E,

and let T : C C be a continuous and strong pseudocontraction. Then T has a unique

fixed point in C.

Lemma 1.4 [28] In a Banach space E, there holds the inequality

x + y2 x2 + 2y, j(x + y), x, y E,

where j(x + y) J(x + y).

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Set A = (0, 1), let B(A) denote the Banach space of all bounded real-valued functions

on A with a supremum norm and let X be a subspace of B(A).

Lemma 1.5 [1] Let C be a nonempty, closed, and convex subset of a Banach space E.

Suppose that the norm of E is uniformly Gateaux differentiable. Let {xt} be a bounded

set in E, and z C. Let t be a mean on X. Then txt z2 = minyCxt y if and

only if ty z, xt z 0 for all y C.

Lemma 1.6 [29] Let E be a Banach space and let A be an m-accretive operator. For

> 0, > 0, and x E, we have

Jx = J

x +

1

Jx

,

where J = (I + A)1 and J = (I + A)

1.

Lemma 1.7 [30] Let {xn} and {yn} be bounded sequences in a Banach space E, and let

{n} be a sequence in (0, 1) with

0 < lim infn

n lim supn

n < 1.

Suppose that xn+1 = (1 n)yn + nxn, n 1 and

lim supn

(yn+1 yn xn+1 xn) 0.

Then limn yn xn = 0.

2. Main results

Now, we are in a position to prove the strong convergence of paths for continuous

pseudocontractions.

Theorem 2.1. LetE be a real reflexive Banach space with the uniformly Gateaux differ-

entiable norm and the normal structure, and let C be a nonempty closed convex subset of

E. LetT : C C be a continuous pseudocontraction with a fixed point, and let f : C C

be a fixed continuous bounded and strong pseudocontraction with the coefficient (0, 1).

Let {xt} be a sequence generated by the following:

xt = tf(xt) + (1 t)T xt, (2.1)

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where t (0, 1). Then {xt} converges strongly as t 0 to a fixed point x of T, which is

the unique solution in F(T) to the following variational inequality:

f(x) x, j(x p) 0, p F(T).

Proof. For t (0, 1), define a mapping Tft : C C by

Tft x = tf(x) + (1 t)T x.

Then Tft : C C is a continuous, strong pseudocontraction for each t (0, 1). Indeed,

we have

T

f

t x T

f

t y, j(x y)

= tf(x) f(y), j(x y) + (1 t)T x T y , j(x y)

tx y2 + (1 t)x y2

=

1 t(1 )

x y2, x, y C.

In view of Lemma 1.3, we see that Tft has a unique fixed point xt in C for each t (0, 1).

Hence (2.1) is well defined. Next, we show that {xt} is bounded. For any p F(T) and

t (0, 1), we see that

xt p2 = xt p, j(xt p)

= tf(xt) p, j(xt p) + (1 t)T xt p, j(xt p)

= tf(xt) f(p), j(xt p) + tf(p) p, j(xt p)

+ (1 t)T xt p, j(xt p)

txt p2 + tf(p) p, j(xt p)

+ (1 t)xt p2

1 t(1 )

xt p2 + tf(p) p, j(xt p),

which implies that

xt p2

1

1 f(p) p, j(xt p). (2.2)

It follows that

xt p

1

1 f(p) p. (2.3)

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This shows that {xt} is bounded, so is {T xt}. On the other hand, we see from (2.1) that

xt T xt = tf(xt) T xt.Since f and {xt} are bounded, we obtain that f(xt) is bounded. It follows that

limt0

xt T xt = 0. (2.4)

Define g(x) = txt x2, x C, where t is a Banach limit. Then g(x) is continuous,

convex and g(x) as x . We see that g attains its infinimum over C (see, e.g.,

[1]). Set

D =

x C : g(x) = infyC

g(y)

.

Then D is a nonempty bounded closed convex subset of C. Next, we show that there

exits a point x D such that T x = x.

Indeed, for any y C, define a mapping S : C C by

Sx =1

2y +

1

2T x, x C.

It is easy to see that S is a continuous strong pseudocontraction. From Lemma 1.3, wesee that S has a unique fixed point x in C, that is, x = 1

2y + 1

2T x. This implies that

y = (2I T)x (2I T)(C).

This shows that C (2I T)(C). Define another mapping h : C C by

h(x) = (2I T)1x, x C.

We see that h is a nonexpansive mapping. Indeed, we have

h(x) h(y) = (2I T)1x (2I T)1y

I + (I T)1

x

I + (I T)1

y

x y, x, y C.

We also see that F(h) = F(T). Indeed,

x = h(x) 2x T x = x x = T x.

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On the other hand, we have

x h(x) = hh1(x) h(x)

h1(x) x

= (2I T)(x) x

= x T x.

It follows from (2.4) that

limt0

xt h(xt) = 0.

Fixing x D, we have

g(h(x)) = txt h(x)2

= th(xt) h(x)2

txt x2

= g(x).

This implies that h(x) D, that is, h(D) D. Since C is a nonempty closed convexsubset of E which has the normal structure, we see that h has a fixed point. We denote

it by x. It follows from F(h) = F(T) that T x = x. From Lemma 1.5, we see that

tp x, j(xt x

) 0, p C.

In particular, we have

tf(x

) x

, j(xt x

) 0. (2.5)

In view of (2.2), we arrive at

txt x2 0.

This implies that there exists a subnet {xt} of {xt} such that xt x.

On the other hand, we have

xt f(xt) = (1 t)

T xt f(xt)

.

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For any p F(T), we see that

xt f(xt), j(xt p)

= (1 t)T xt f(xt), j(xt p)

= (1 t)T xt xt, j(xt p) + (1 t)xt f(xt), j(xt p)

= (1 t)T xt p, j(xt p) + (1 t)p xt, j(xt p)

+ (1 t)xt f(xt), j(xt p)

(1 t)xt f(xt), j(xt p),

which implies that

xt f(xt), j(xt p) 0, p F(T). (2.6)

In particular, we have

xt f(xt), j(xt p) 0, p F(T). (2.7)

It follows that

x

f

(x

), j

(x

p

) 0,

p

F

(T

).

(2.8)

If there exits another subnet {xt} of {xt} such that xt x F(T). From (2.8), we

arrive at

x f(x), j(x x) 0. (2.9)

It is also from (2.6) that

xt f(xt), j(xt x) 0. (2.10)

It follows that

x f(x), j(x x) 0. (2.11)

Adding up (2.9) and (2.11), we obtain that

x f(x) x + f(x), j(x x) 0.

This implies that

x

x

2

x

x

2

.

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Note that (0, 1). We see that x = x. This shows that {xt} converges strongly to

x F(T), which is the unique solution to the variational inequality

(f I)x, p x 0, p F(T).

This completes the proof.

Remark 2.2. Note that uniformly smooth Banach spaces enjoy the uniformly Gateaux

differentiable norm and the uniform normal structure. Theorem 2.1, which gives an

affirmative answer to the questions presented in Section 1, improves Theorem 4.1 of Xu

[6] in the following aspects:

(1) improves the framework of spaces from uniformly smooth Banach spaces to the

Banach space with the uniformly Gateaux differentiable norm.

(2) improves the mapping f from the class of contractions to the class of strongly

pseudocontractions.

(3) improves the mapping T from the class of nonexpansive mappings to the class of

pseudocontractions.

Remark 2.3. Under the conditions of Theorem 2.1, we can define a mapping Q : C F(T), where C denotes the set of continuous bounded and strong pseudocontractions,

by

Q(f) := limt0

xt, f C.

From Proposition 1.1, we see that Q is a sunny nonexpansive retraction from C onto

F(T).

Remark 2.4. If T is a nonexpansive mapping and f(x) = u, a fixed element for every

x C in Theorem 2.1, then {xt} defined by

xt = tu + (1 t)T xt,

where t (0, 1), converges strongly as t 0 to a fixed point Q(u) of T, where Q is the

sunny nonexpansive retraction from C onto F(T). The point Q(u) is the unique solution

in F(T) to the following variational inequality:

u Q(u), j(Q(u) p) 0, p F(T).

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Next, we prove strong convergence of iterative processes with errors for m-accretive

operators.

Theorem 2.5. Let E be a real reflexive Banach space with the uniformly Gateaux dif-

ferentiable norm and let A be an m-accretive operator in E. Assume that C := D(A)

is convex and has the normal structure. Let {n}, {n}, {n}, and {n} be real number

sequences in (0, 1). Let QC be a sunny nonexpansive retraction from E onto C and let

{xn} be a sequence generated in the following manner:

x0 C,

yn = nxn + nJrn(xn + en+1) + nQCfn,

xn+1 = nu + (1 n)yn, n 0,

()

where {en} is a sequence in E, {fn} is a bounded sequence in E, {rn} is a positive

real numbers sequence, u is a fixed element in C and Jrn = (I + rnA)1. Assume that

A1(0) = and the above control sequences satisfy the following restrictions:

(a) n + n + n = 1 for each n 0;

(b) limn n = 0 and

n=1n = ;

(c) 0 < lim infn n lim supn n < 1;

(d)

n=1 en < and

n=0n < ;

(e) rn for each n 1 and limn |rn rn+1| = 0.

Then the sequence {xn} generated by () converges strongly to a zero Q(u) of A, which

is the unique solution to the following variational inequality:

(f I)Q(u), p Q(u) 0, p A1(0).

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Proof. First, we prove that {xn} is bounded. Fixing p A1(0), we see that

x1 p

= 0(u p) + (1 0)(y0 p)

0u p + (1 0)y0 p

= 0u p + (1 0)

0(x0 p) + 0

Jr0(x0 + e1) p

+ 0(QCf0 p)

0u p + (1 0)

0x0 p + 0Jr0(x0 + e1) p + 0f0 p

0u p + (1 0)

(1 0)x0 p + e1 + 0f0 p

K,

where K = u p + x0 p + e1 + f0 p < . Put

M1 = max{K, supn0

fn p}.

Next, we prove that

xn p M1 +

ni=1

ei, n 1. (2.12)

It is easy to see that (2.12) holds for n = 1. We assume that the result holds for some n.

It follows that

xn+1 p

nu p + (1 n)yn p

= nu p + (1 n)

n(xn p) + n

Jrn(xn + en+1) p

+ n(QCfn p)

nu p + (1 n)

nxn p + nJrn(xn + en+1) p + nfn p

nu p + (1 n)

(1 n)xn p + en+1 + nfn p

nM1 + (1 n)

(1 n)

M1 +

ni=1

ei

+ en+1 + nM1

= M1 +n+1

i=1

ei.

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This shows that (2.12) holds for all n 1. If rn+1 rn, we see from Lemma 1.6 that

Jrn(xn + en+1) Jrn+1(xn+1 + en+2)

rn

rn+1(xn + en+1) +

1

rn

rn+1

Jrn+1(xn + en+1) (xn+1 + en+2)

= rn

rn+1

(xn + en+1) (xn+1 + en+2)

+

rn+1 rnrn+1

Jrn+1(xn + en+1) (xn+1 + en+2)

xn xn+1 + en+1 + en+2 +

M2

(rn+1 rn),

(2.13)

where M2 is an appropriate constant such that

M2 supn1

{Jrn+1(xn + en+1) (xn+1 + en+2)}.

Put gn =xn+1nxn

1n. This implies that

xn+1 = (1 n)gn + nxn, n 0. (2.14)

Now, we compute gn+1 gn. Note that

gn+1 gn =n+1u + (1 n+1)yn+1 n+1xn+1

1 n+1

nu + (1 n)yn nxn1 n

= n+1(u yn+1) + n+1Jrn+1(xn+1 + en+2) + n+1QCfn+11 n+1

n(u yn) + nJrn(xn + en+1) + nQCfn

1 n

=n+1(u yn+1) + n+1

QCfn+1 Jrn+1(xn+1 + en+2)

1 n+1

n(u yn) + n

QCfn Jrn(xn + en+1)

1 n

+ Jrn+1(xn+1 + en+2) Jrn(xn + en+1).

(2.15)

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It follows that

gn+1 gn

n+11 n+1

u yn+1 + n1 n

yn u

+ n+1

1 n+1

QCfn+1 Jrn+1(xn+1 + en+2)

n

1 n

QCfn Jrn(xn + en+1)

+ Jrn+1(xn+1 + en+2) Jrn(xn + en+1)

n+1

1 n+1u yn+1 +

n

1 nyn u +

n+1

1 n+1QCfn+1 Jrn+1(xn+1 + en+2)

+n

1 nQCfn Jrn(xn + en+1) + Jrn+1(xn+1 + en+2) Jrn(xn + en+1).

(2.16)

Substituting (2.13) into (2.16), we arrive at

gn+1 gn xn xn+1

n+1

1 n+1u yn+1 +

n

1 nyn u + en+1 + en+2

+M2

|rn+1 rn| +

n

1 nyn u +

n+1

1 n+1QCfn+1 Jrn+1(xn+1 + en+2)

+ n1 n

QCfn Jrn(xn + en+1).

(2.17)

In a similar way, we can obtain (2.17) when rn rn+1. From the conditions (b)-(e), we

see that

lim supn

gn+1 gn xn xn+1

0.

It follows from Lemma 1.7 that

limn

gn xn = 0. (2.18)

In view of (2.14), we have

xn+1 xn = (1 n)(gn xn),

which combines with the condition (c) and (2.18) gives that

limnxn+1 xn = 0. (2.19)

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On the other hand, we see from the algorithm () that

xn Jrn(xn + en+1)

xn xn+1 + xn+1 yn + yn Jrn(xn + en+1)

xn xn+1 + nu yn + nxn Jrn(xn + en+1) + nQCfn xn.

This implies that

(1 n)xn Jrn(xn + en+1) xn xn+1 + nu yn + nQCfn xn.

From the conditions (b)-(d) and (2.19), we conclude that

limn

xn Jrn(xn + en+1) = 0. (2.20)

Since

n=1 en < , we see that

limn

xn Jrnxn = 0. (2.21)

Take a fixed number r such that > r > 0. From Lemma 1.6, we obtain that

Jrnxn Jrxn = Jr r

rnxn + (1

r

rn)Jrnxn

Jrxn

(1 rrn

)(Jrnxn xn)

Jrnxn xn.

(2.22)

Note that

xn Jrxn = xn Jrnxn + Jrnxn Jrxn

xn Jrnxn + Jrnxn Jrxn

2xn Jrnxn.

From (2.21), we see that

limn

xn Jrxn = 0. (2.23)

Next, we claim that

lim supn

u Q(u), j(xn Q(u)) 0, (2.24)

where

Qu = limt0 zt, u C

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and zt solves the fixed point equation

zt = tu + (1 t)Jrzt, t (0, 1),

from which it follows that

zt xn = (1 t)(Jrzt xn) + t(u xn).

For any t (0, 1), we see that

zt xn2 = (1 t)Jrzt xn, j(zt xn) + tu xn, j(zt xn)

= (1 t)

Jrzt Jrxn, j(zt xn) + Jrxn xn, j(zt xn)

+ tu zt, j(zt xn) + tzt xn, j(zt xn)

(1 t)

zt xn2 + Jrxn xnzt xn

+ tu zt, j(zt xn) + tzt xn2

zt xn2 + Jrxn xnzt xn + tu zt, j(zt xn).

It follows that

zt u, j(zt xn) 1

tJrxn xnzt xn t (0, 1).

In view of (2.23), we see that

lim supn

zt u, j(zt xn) 0. (2.25)

Since zt Q(u) as t 0 and the fact that j is strong to weak uniformly continuous on

bounded subsets of E, we see that

|u Q(u), j(xn Q(u)) zt u, j(zt xn)|

|u Q(u), j(xn Q(u)) u Q(u), j(xn zt)|

+ |u Q(u), j(xn zt) zt u, j(zt xn)|

|u Q(u), j(xn Q(u)) j(xn zt)| + |zt Q(u), J(xn zt)|

u Q(u)j(xn Q(u)) j(xn zt) + zt Q(u)xn zt 0 as t 0.

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Hence, for any > 0, there exists > 0 such that t (0, ) the following inequality

holds:

u Q(u), j(xn Q(u)) zt u, j(zt xn) + .

This implies that

lim supn

u Q(u), j(xn Q(u)) lim supn

zt u, j(zt xn) + .

Since is arbitrary and (2.25), we see that lim supnu Q(u), j(xn Q(u)) 0. That

is,

lim supn

u Q(u), j(xn+1 Q(u)) 0. (2.26)

Finally, we show that xn Q(u) as n . Note that

yn Q(u) = nxn + nJrn(xn + en+1) + nQCfn Q(u)

nxn Q(u) + nJrn(xn + en+1) Q(u) + nQCfn Q(u)

(n + n)xn Q(u) + en+1 + nfn Q(u)

xn Q(u) + n,

where n = en+1 + nM1. From the condition (d), we see that

n=0n < . It follows

from Lemma 1.5 that

xn+1 Q(u)2

= (1 n)(yn Q(u)) + n(u Q(u))2

(1 n)2yn Q(u)

2 + 2nu Q(u), J(xn+1 Q(u))

(1 n)(xn Q(u) + n)2 + 2nu Q(u), J(xn+1 Q(u))

= (1 n)(xn Q(u)2 + 2nxn Q(u) +

2n)

+ 2nu Q(u), J(xn+1 Q(u))

(1 n)xn Q(u)2 + nM3 + 2nu Q(u), J(xn+1 Q(u)),

(2.27)

where M3 is an appropriate constant such that M3 supn0{2xn Q(u) + n}. Let

n+1 = max{uQ(u), J(xn+1Q(u)), 0}. Next, we show that limn n+1 = 0. Indeed,

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from (2.26), for any given > 0, there exists a positive integer n1 such that

u Q(u), J(xn+1 Q(u)) < , n n1.

This implies that 0 n+1 < n n1. Since > 0 is arbitrary, we see that

limn n+1 = 0. It follows from (2.27) that

xn+1 Q(u)2 (1 n)xn Q(u)

2 + nM3 + 2nn+1.

Put an = xn Q(u)2, tn = n, bn = 2nn+1 and cn = nM3 for every n 0. In view

of Lemma 1.1, we can obtain the desired conclusion easily.

As an application of Theorem 2.5, we have the following results.

Corollary 2.6. Let E be a real reflexive Banach space with the uniformly Gateaux dif-

ferentiable norm and let A be an m-accretive operator in E. Assume that C := D(A)

is convex and has the normal structure. Let {n} and {n} be real number sequences in

(0, 1). Let {xn} be a sequence generated by (QS), where {rn} is a positive real numbers

sequence, u is a fixed element in C and Jrn = (I + rnA)1. Assume that A1(0) = and

the above control sequences satisfy the following restrictions:(a) limn n = 0 and

n=1

n = ;

(b) 0 < lim infn n lim supn n < 1;

(c) rn for each n 1 and limn |rn rn+1| = 0.

Then the sequence {xn} converges strongly to a zero of A.

Remark 2.7. Corollary 2.6 improves Theorem 2.1 of Qin and Su [18]. To be more

precise, we partially relax the restrictions on the parameters and extend the framework

of the space; see [18] for more details.

Remark 2.8. It is of interest to design an explicit iterative process to approximate zeros of

accretive operators by Moudafis viscosity approximation method with continuous strong

pseudocontractions.

Competing interests: The authors declare that they have no competing interests.

Authors contributions: All authors contributed equally and significantly in writing this

paper. All authors read and approved the final manuscript.

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Author details: 1Department of Mathematics, Hangzhou Normal University, Hangzhou

310036, China; 2Department of Mathematics, Gyeongsang National University, Jinju 660-

701, Korea;3

College of Statistics and Mathematics, Yunnan University of Finance andEconomics, Kunming 650221, China

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