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# Strong convergence of an new iterative method for a zero of accretive operator and nonexpansive mapping

## Abstract

Let E be a Banach space and A an m-accretive operator with a zero. Consider the iterative method that generates the sequence {x n } by the algorithm $x n + 1 = α n γϕ ( x n ) + ( I - α n F ) J r n x n$, where {a n } and {r n } are two sequences satisfying certain conditions, $J r n$ denotes the resolvent (I + r n A)-1 for r n > 0, F be a strongly positive bounded linear operator on E is $0 < γ < γ ̄$, and ϕ be a MKC on E. Strong convergence of the algorithm {x n } is proved assuming E either has a weakly continuous duality map or is uniformly smooth.

MSC: 47H09; 47H10

## 1 Introduction

Let E be a real Banach space, C a nonempty closed convex subset of E, and T : CC a mapping. Recall that T is nonexpansive if Tx - Tyx - y for all x, y C. A point x C is a fixed point of T provided Tx = x. Denote by F(T) the set of fixed points of T, that is, F(T) = {x C, Tx = x}.

It is assumed throughout the paper that T is a nonexpansive mapping such that $F ( t ) ≠∅$. The normalized duality mapping J from a Banach space E into $2 E *$ is given by J(x) = {f E* : 〈x, f〉 = x2 = f2}, x E, where E* denotes the dual space of E and 〈.,.〉 denotes the generalized duality pairing.

Theorem 1.1. (Banach [1]). Let (X, d) be a complete metric space and let f be a contraction on X, that is, there exists r (0, 1) such that d(f(x), f(y)) ≤ rd(x, y) for all x, y X. Then f has a unique fixed point.

Theorem 1.2. (Meir and Keeler [2]). Let (X, d) be a complete metric space and let ϕ be a Meir-Keeler contraction (MKC, for short) on X, that is, for every ε > 0, there exists δ > 0 such that d(x, y) < ε + δ implies d(ϕ(x), ϕ(y)) < ε for all x, y X. Then ϕ has a unique fixed point.

This theorem is one of generalizations of Theorem 1.1, because contractions are Meir-Keeler contractions.

Let F be a strongly positive bounded linear operator on E, that is, there exists a constant $γ ̃ >0$ such that

$F x , J ( x ) ≥ γ ̃ x 2 , a I - b F = sup x ≤ 1 ( a I - b F ) x , J ( x ) : a ∈ [ 0 , 1 ] , b ∈ [ 0 , 1 ] ,$

where I is the identity mapping and J is the normalized duality mapping.

Let D be a subset of C. Then Q : CD is called a retraction from C onto D if Q(x) = x for all x D. A retraction Q : CD is said to be sunny if Q(x + t(x - Q(x))) = Q(x) for all x C and t ≥ 0 whenever x + t(x - Q(x)) C. A subset D of C is said to be a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction of C onto D. In a smooth Banach space E, it is known (cf. [[3], p. 48]) that Q : CD is a sunny nonexpansive retraction if and only if the following condition holds:

$x - Q ( x ) , J ( z - Q ( x ) ) ≤ 0 , x ∈ C , z ∈ D .$
(1.1)

Recall that an operator A with domain D(A) and range R(A) in E is said to be accretive, if for each x i D(A) and y i Ax i , i = 1, 2, there is a j J(x2 - x1) such that

$y 2 - y 1 , j ≥ 0 .$

An accretive operator A is m-accretive if R(I + λA) = E for all λ > 0. Denote by N(A) the zero set of A; i.e.,

$N ( A ) : = A - 1 0 = { x ∈ D ( A ) : A x = 0 } .$

Throughout the rest of this paper it is always assumed that A is m-accretive and N(A) is nonempty. Denote by J r the resolvent of A for r > 0:

$J r = ( I + r A ) - 1 .$

Note that if A is m-accretive, then J r : EE is nonexpansive and F(J r ) = N(A) for all r > 0. We also denote by A r the Yosida approximation of A, i.e., $A r = 1 r ( I - J r )$. It is well known that J r is a nonexpansive mapping from E to C := D(A).

Recall that a gauge is a continuous strictly increasing function φ : [0, ∞) → [0, ∞) such that φ(0) = 0 and φ(t) → ∞ as t → ∞. Associated to a gauge φ is the duality mapping J φ : EE* defined by

$J φ ( x ) = x * ∈ E * : ⟨ x , x * ⟩ = x φ x , x * = φ x , x ∈ E .$

Following Browder [4], we say that a Banach space E has a weakly continuous duality map if there exists a gauge φ for which the duality map J φ is single-valued and weak-to-weak* sequentially continuous(i.e., if {x n } is a sequence in E weakly convergent to a point x, then the sequence J φ (x n ) converges weakly* to J φ (x)). It is known that lphas a weakly continuous duality map for all 1 < p < ∞, with gauge φ(t) = tp-1. Set

$Φ ( t ) = ∫ 0 t φ ( τ ) d τ , t ≥ 0 .$
(1.2)

Then

$J φ ( x ) = ∂ Φ x , x ∈ E ,$

where denotes the subdifferential in the sense of convex analysis.

Recently, Hong-Kun Xu [5] introduced the following iterative scheme: for x1 = x C,

$x n + 1 = α n u + ( 1 + α n ) J r n x n , ∀ n ≥ 1 ,$
(1.3)

where {a n } and {r n } are two sequences satisfying certain conditions, and $J r n$ denotes the resolvent (I + r n A)-1 for r n > 0. He proved the strong convergence of the algorithm {x n } assuming E either has a weakly continuous duality map or is uniformly smooth.

Motivated and inspired by the results of Hong-Kun Xu, we introduce the following iterative scheme: for any x0 E,

$x n + 1 = α n γ ϕ ( x n ) + ( I - α n F ) J r n x n , ∀ n ≥ 0 ,$
(1.4)

where {a n } and {r n } are two sequences satisfying certain conditions, $J r n$ denotes the resolvent (I + r n A)-1 for r n > 0, F be a strongly positive bounded linear operator on E is $0 < γ < γ ̄$, and ϕ be a MKC on E. Strong convergence of the algorithm {x n } is proved assuming E either has a weakly continuous duality map or is uniformly smooth. Our results extend and improve the corresponding results of Hong-Kun Xu [5] and many others.

## 2 Preliminaries

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

Lemma 2.1. [5]. Assume that E has a weakly continuous duality map J φ with gauge φ,

1. (i)

For all x, y E, there holds the inequality

$Φ x + y ≤ Φ x + y , J φ ( x + y ) .$
2. (ii)

Assume a sequence {x n } in E is weakly convergent to a point x, then there holds the equality

$lim sup n → ∞ Φ x n - y = lim sup n → ∞ Φ x n - x + Φ y - x , x , y ∈ E .$

Lemma 2.2. [6, 7]. Let {s n } be a sequence of nonnegative real numbers satisfying

$s n + 1 ≤ ( 1 - λ n ) s n + λ n δ n + γ n , n ≥ 0 ,$

where {λ n }, {δ n } and {γ n } satisfy the following conditions:

1. (i)

{λ n } [0,1] and $∑ n = 0 ∞ λ n = ∞$,

2. (ii)

lim supn→∞δ n ≤ 0 or $∑ n = 0 ∞ λ n δ n <∞$ (iii) $γ n ≥0 ( n ≥ 0 ) , ∑ n = 0 ∞ γ n <∞$. Then limn→∞s n = 0.

Lemma 2.3. (The Resolvent Identity [8, 9]). For λ > 0 and ν > 0 and x E,

$J λ x = J ν ν λ + 1 - ν λ J λ x .$

Lemma 2.4. (see [ [10], Lemma 2.3]). Assume that F is a strongly positive linear bounded operator on a smooth Banach space E with coefficient $γ ̄ >0$ and 0 < ρF-1. Then,

$I - ρ F ≤ 1 - ρ γ ̄ .$

Lemma 2.5. (see [ [11], Lemma 2.3]). Let ϕ be a MKC on a convex subset C of a Banach space E. Then for each ε > 0, there exists r (0,1) such that

$x - y ≥ ε implies ϕ x - ϕ y ≤ r x - y ∀ x , y ∈ C .$

Lemma 2.6. Let E be a reflexive Banach space which admits a weakly continuous duality map J φ with gauge φ. Let T : EE be a nonexpansive mapping. Now given ϕ : EE be a MKC, F be a strongly positive linear bounded operator with coefficient $γ ̄ >0$. Assume that $0<γ< γ ̄$, the sequence {x t } defined by x t = tγϕ(x t ) + (I - tF)Tx t . Then T has a fixed point if and only if {x t } remains bounded as t → 0+, and in this case, {x t } converges as t → 0+ strongly to a fixed point of T. If $x ̃ := lim t → 0 x t$, then $x ̃$ uniquely solves the variational inequality

$( F - γ ϕ ) x ̃ , J ( x ̃ - p ) ≤ 0 , P ∈ F ( T ) .$

Proof. The definition of {x t } is well defined. Indeed, from the definition of MKC, we can see MKC is also a nonexpansive mapping. Consider a mapping S t on E defined by

$S t ( x ) = t γ ϕ ( x ) + ( I - t F ) T x , x ∈ E .$

It is easy to see that S t is a contraction. Indeed, by Lemma 2.4, we have

$S t x - S t y ≤ t γ ϕ ( x ) - ϕ ( y ) + ( I - t F ) ( T x - T y ) ≤ t γ x - y + ( 1 - t γ ̄ ) T x - T y ≤ [ 1 - t ( γ ̄ - γ ) ] x - y ,$

for all x, y E. Hence S t has a unique fixed point, denoted as x t , which uniquely solves the fixed point equation

$x t = t γ ϕ ( x t ) + ( I - t F ) T x t , x t ∈ E .$
(2.1)

We next show the sequence {x t } is bounded. Indeed, we may assume $F ( t ) ≠∅$ and with no loss of generality t < F-1. Take p F(T) to deduce that, for t (0, 1),

$x t - p = t γ ϕ ( x t ) + ( I - t F ) T x t - p = t ( γ ϕ ( x t ) - F p ) + ( I - t F ) ( T x t - p ) ≤ ( 1 - t γ ̄ ) x t - p + t γ x t - p + t γ ϕ ( p ) - F p ≤ [ 1 - t ( γ ̄ - γ ) ] x t - p + t γ ϕ ( p ) - F p .$

Hence

$x t - p ≤ 1 γ ̄ - γ γ ϕ ( p ) - F p$

and {x t } is bounded.

Next assume that {x t } is bounded as t → 0+. Assume t n → 0+ and ${ x t n }$ is bounded. Since E is reflexive, we may assume that $x t n ⇀z$ for some z E. Since J φ is weakly continuous, we have by Lemma 2.1,

$lim sup n → ∞ Φ x t n - x = lim sup n → ∞ Φ x t n - z + Φ x - z , ∀ x ∈ E .$

Put

$f ( x ) = lim sup n → ∞ Φ x t n - x , x ∈ E .$

It follows that

$f ( x ) = f ( z ) + Φ x - z , x ∈ E .$

Since

$x t n - T x t n = t n γ ϕ ( x t n ) - F T x t n → 0 ,$

we obtain

(2.2)

On the other hand, however,

$f ( T z ) = f ( z ) + Φ T z - z$
(2.3)

Combining Equations (2.2) and (2.3) yields

$Φ T z - z ≤ 0 .$

Hence, Tz = z and z F(T).

Finally, we prove that {x t } converges strongly to a fixed point of T provided it remains bounded when t → 0.

Let {t n } be a sequence in (0, 1) such that t n → 0 and $x t n ⇀z$ as n → ∞. Then the argument above shows that z F(T). We next show that $x t n →z$. By contradiction, there is a number ε0 > 0 such that $x t n - z ≥ ε 0$. Then by Lemma 2.8, there is a number r (0, 1) such that

$ϕ ( x t n ) - ϕ ( z ) ≤ r x t n - z , x t n - z φ x t n - z = x t n - z , J φ ( x t n - z ) = t n ( γ ϕ ( x t n ) - F z ) + ( I - t n F ) ( T x t n - z ) , J φ ( x t n - z ) ≤ t n γ ϕ ( x t n ) - F z , J φ ( x n - z ) + ( I - t n F ) ( T x t n - z ) φ x t n - z ≤ ( 1 - t n γ ̄ ) x t n - z φ x t n - z + t n γ ϕ ( x t n ) - F z , J φ ( x n - z ) .$

It follows that

Therefore,

$x t n - z φ x t n - z ≤ 1 γ ̄ - γ r γ ϕ ( z ) - F z , J ϕ ( x t n - z ) .$

Now observing that $x t n ⇀z$ implies $J φ ( x t n - z ) ⇀0$, we conclude from the last inequality that

$lim n → ∞ x t n - z φ x t n - z = 0 .$

It contradicts $x t n - z φ x t n - z ≥ ε 0 φ ( ε 0 ) >0$. Hence $x t n →z$.

We finally prove that the entire net {x t } converges strongly. Towards this end, we assume that two null sequences {t n } and {s n } in (0, 1) are such that

$t n → 0 , x t n → z and s n → 0 , x s n → ź .$

We have to show $z=ź$. Indeed, for p F(T). Since

$x t = t γ ϕ ( x t ) + ( I - t f ) T x t ,$

we derive that

$( F - γ ϕ ) x t = - 1 t ( I - t F ) ( I - T ) x t .$
(2.4)

Notice

$( I - T ) x t - ( I - T ) p , J φ ( x t - p ) = x t - p φ x t - p + T p - T x t , J φ ( x t - p ) ≥ x t - p φ x t - p - T p - T x t J φ x t - p ≥ x t - p φ x t - p - φ x t - p = 0 .$

It follows that,

$〈 ( F − γ ϕ ) x t , J φ ( x t − p ) 〉 = − 1 t 〈 ( I − t F ) ( I − T ) x t , J φ ( x t − p ) 〉 ≤ 〈 ( F ( I − T ) x t , J φ ( x t − p ) 〉 .$
(2.5)

Now replacing t in (2.5) with t n and letting n → ∞, noticing $( I - T ) x t n → ( I - T ) z=0$ for z F(T), we obtain 〈(F - γϕ)z, J φ (z - p)〉 ≤ 0. In the same way, we have $( F - γ ϕ ) ź , J φ ( ź - p ) ≤0$.

Thus, we have

$( F - γ ϕ ) z , J φ ( z - ź ) ≤ 0 and ( F - γ ϕ ) ź , J φ ( ź - z ) ≤ 0 .$
(2.6)

$( F - γ ϕ ) z - ( F - γ ϕ ) ź , J φ ( z - ź ) ≤ 0 .$

On the other hand, without loss of generality, we may assume there is a number ε such that $z - ź ≥ε$, then by Lemma 2.5 there is a number r1 such that $ϕ ( z ) - ϕ ( ź ) ≤ r 1 z - ź$. Noticing that

$J φ = φ x / x J ( x ) x ≠ 0 , ( F - γ ϕ ) z - ( F - γ ϕ ) ź , J φ ( z - ź ) = F ( z - ź ) , J φ ( z - ź ) - γ ϕ z - γ ϕ ź , J φ ( z - ź ) ≥ γ ̄ z - ź J φ ( z - ź ) - γ r 1 z - ź J φ ( z - ź ) ≥ ( γ ̄ - γ r 1 ) z - ź φ z - ź > 0 .$

Hence $z=ź$ and {x t } converges strongly. Thus we may assume $x t → x ̃$. Since we have proved that, for all t (0, 1) and p F(T),

$〈 ( F − γ ϕ ) x t , J φ ( x t − p ) 〉 ≤ 〈 ( F ( I − T ) x t , J φ ( x t − p ) 〉 ,$

letting t → 0, we obtain that

$( F - γ ϕ ) x ̃ , J φ ( x ̃ - p ) ≤ 0 .$

This implies that

$( F - γ ϕ ) x ̃ , J ( x ̃ - p ) ≤ 0 .$

Lemma 2.7. (see [12]). Assume that C2C1 > 0. Then $J C 1 x - x ≤2 J C 2 x - x$ for all x E.

Lemma 2.8. [13]. Let C be a nonempty closed convex subset of a reflexive Banach space E which satisfies Opial's condition, and suppose T : CE is a nonexpansive mapping. Then the mapping I - T is demiclosed at zero, that is x n x and x n - Tx n → 0, then x = Tx.

Lemma 2.9. In a smooth Banach space E there holds the inequality

$x + y 2 ≤ x 2 + 2 y , J ( x + y ) , x , y ∈ E .$

## 3 Main result

Theorem 3.1. Suppose that E is reflexive which admits a weakly continuous duality map J φ with gauge φ and A is an m-accretive operator in E such that $F * =N ( A ) ≠∅$. Now given ϕ : EE be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient $γ ̄ >0,0<γ< γ ̄$. Assume

1. (i)

$lim n → ∞ α n =0, ∑ n = 0 ∞ α n =∞$;

2. (ii)

r n → ∞.

Then {x n } defined by (1.4) converges strongly to a point in F*.

Proof. First notice that {x n } is bounded. Indeed, take p F* to get

$x n + 1 - p = α n ϕ ( x n ) + ( I - α n F ) J r n x n - p = α n γ ϕ ( x n ) - α n F p + α n F p + ( I - α n F ) J r n x n - p = α n ( γ ϕ ( x n ) - F p ) + ( I - α n F ) ( J r n x n - p ) ≤ ( 1 - α n γ ̄ ) x n - p + α n γ x n - p + α n γ ϕ ( p ) - F p ≤ [ 1 - α n ( γ ̄ - γ ) ] x n - p + α n γ ϕ ( p ) - F p ≤ [ 1 - α n ( γ ̄ - γ ) ] x n - p + α n ( γ ̄ - γ ) γ ϕ ( p ) - F p γ ̄ - γ ≤ max x n - p , γ ϕ ( p ) - F p γ ̄ - γ , n ≥ 0 .$

By induction, we have

$x n - p ≤ max x 0 - p , γ ϕ ( p ) - F p γ ̄ - γ , n ≥ 0 .$

This implies that {x n } is bounded and hence

$x n + 1 - J r n x n = α n γ ϕ ( x n ) + ( I - α n F ) J r n x n - J r n x n = α n γ ϕ ( x n ) - F J r n x n → 0 .$

We next prove that

lim supn→∞γϕ(p) - Fp, J φ (x n - p)〉 ≤ 0, where p = limt→0x t with $x t =tγϕ ( x t ) + ( I - t F ) J r n x t$.

Since {x n } is bounded, take a subsequence ${ x n k }$ of {x n } such that

$lim sup n → ∞ γ ϕ ( p ) - F p , J ϕ ( x n - p ) = lim k → ∞ γ ϕ ( p ) - F p , J φ ( x n k - p ) .$
(3.1)

Since E is reflexive, we may further assume that $x n k ⇀ x ̃$. Moreover, since

$x n + 1 - J r n x n → 0 ,$

we obtain

$J r n k - 1 x n k - 1 ⇀ x ̃ .$

Taking the limit as k → ∞ in the relation

$J r n k - 1 x n k - 1 , A r n k - 1 x n k - 1 ∈ A ,$

we get $[ x ̃ , 0 ] ∈A$. That is, $x ̃ ∈ F *$. Hence by (3.1) and Lemma 2.6 we have

$lim sup n → ∞ γ ϕ ( p ) - F p , J φ ( x n - p ) = lim k → ∞ γ ϕ ( p ) - F p , J φ ( x n k - p ) = γ ϕ ( p ) - F p , J φ ( x ̃ - p ) ≤ 0 .$

Finally to prove that x n p, we apply Lemma 2.1 to get

An application of Lemma 2.2 yields that Φ(x n - p) → 0. That is, x n - p → 0, i.e., x n p. The proof is complete.

Theorem 3.2. Suppose that E is reflexive which admits a weakly continuous duality map J φ with gauge φ and A is an m-accretive operator in E such that $F * =N ( A ) ≠∅$. Now given ϕ : EE be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient $γ ̄ >0,0<γ< γ ̄$. Assume

1. (i)

$lim n → ∞ α n =0, ∑ n = 0 ∞ α n =∞$, and $∑ n = 1 ∞ α n + 1 - α n <∞ ( e . g . , α n = 1 n )$;

2. (ii)

r n ε for all n and $∑ n = 1 ∞ r n + 1 - r n <∞ ( e . g . , r n = 1 + 1 n )$.

Then {x n } defined by (1.4) converges strongly to a point in F*.

Proof. We only include the differences. We have

$x n + 1 = α n γ ϕ ( x n ) + ( I - α n F ) J r n x n , x n = α n - 1 γ ϕ ( x n - 1 ) + ( I - α n - 1 F ) J r n - 1 x n - 1 .$

Thus,

$x n + 1 - x n = ( I - α n F ) ( J r n x n - J r n - 1 x n - 1 ) + α n γ ϕ ( x n ) + α n - 1 γ ϕ ( x n - 1 ) + ( α n - α n - 1 ) F J r n - 1 x n - 1 .$
(3.2)

If rn-1r n , using the resolvent identity

$J r n x n = J r n - 1 r n - 1 r n x n + 1 - r n - 1 r n J r n x n ,$

we obtain

$J r n x n - J r n - 1 x n - 1 ≤ r n - 1 r n x n - x n - 1 + 1 - r n - 1 r n J r n x n - x n - 1 ≤ x n - x n - 1 + r n - r n - 1 r n J r n x n - x n - 1 ≤ x n - x n - 1 + 1 ε r n - 1 - r n J r n x n - x n - 1 .$
(3.2a)

It follows from (3.2) that

$x n + 1 - x n ≤ ( 1 - α n ( γ ̄ - γ ) ) x n - x n - 1 + M α n - α n - 1 + r n - 1 - r n ,$
(3.3)

where M > 0 is some appropriate constant. Similarly we can prove (3.3) if rn-1r n . By assumptions (i) and (ii) and Lemma 2.2, we conclude that

$x n + 1 - x n → 0 .$

This implies that

$x n - J r n x n ≤ x n + 1 - x n + x n + 1 - J r n x n → 0 ,$
(3.4)

since $x n + 1 - J r n x n = α n γ ϕ ( x n ) - F J r n x n →0$. It follows that

$A r n x n = 1 r n x n - J r n x n ≤ 1 ε x n - J r n x n → 0 .$

Now if ${ x n k }$ is a subsequence of {x n } converging weakly to a point $x ̃$, then taking the limit as k → ∞ in the relation

$[ J r n k x n k , A r n k x n k ] ∈ A ,$

we get $[ x ̃ , 0 ] ∈A$; i.e., $x ̃ ∈ F *$. We therefore conclude that all weak limit points of {x n } are zeros of A.

The rest of the proof follows that of Theorem 3.1.

Finally, we consider the framework of uniformly smooth Banach spaces. Assume r n ε for some ε > 0 (not necessarily r n → ∞), A is an m-accretive operator in E. Moreover let ϕ : EE be a MKC and F be a strongly positive linear bounded operator on E. Since $J r n$ is nonexpansive, the map $S:x∈E↦tγϕ ( x ) + ( I - t F ) J r n x$ is a contraction and for each integer n ≥ 1 it has a unique fixed z t,n E. Hence the scheme

$z t , n = t γ ϕ ( z t , n ) + ( I - t F ) J r n z t , n$
(3.5)

is well defined.

Note that {z t,n } is uniformly bounded; indeed, $z t , n - p ≤ 1 γ ̄ - γ γ ϕ ( p ) - F p$ for all t (0, 1), n ≥ 1 and p F*. A key component of the proof of the next theorem is the following lemma.

Lemma 3.1. The limit $ẑ= lim t → 0 z t , n$ is uniform for all n ≥ 1.

Proof. It suffices to show that for any positive integer n t (which may depend on t (0, 1)), if $z t , n t ∈E$ is the unique point in E that satisfies the property

$z t , n t = t γ ϕ ( z t , n t ) + ( I - t F ) J r n t z t , n t ,$
(3.6)

then ${ z t , n t }$ converges as t → 0 to a point in F*. For simplicity put

$w t = z t , n t and V t = J r n t .$

It follows that

$w t = t γ ϕ ( w t ) + ( I - t F ) V t w t .$
(3.7)

Note that Fix(V t ) = F* for all t. Note also that {w t } is bounded; indeed, we have $w t - p ≤ 1 γ ̄ - γ γ ϕ ( p ) - F p$ for all t (0, 1) and p F*. Since {V t w t } is bounded, it is easy to see that

$w t - V t w t = t γ ϕ ( w t ) - F V t w t → 0 , as t → 0 .$

Since r n ε for all n, by Lemma 2.7, we have

$w t - J ε w t ≤ 2 w t - J r n t w t = 2 w t - V t w t → 0 .$
(3.8)

Let {t k } be a sequence in (0,1) such that t k → 0 as k → ∞. Define a function f on E by

$f ( w ) = LI M k 1 2 w t k - w 2 , w ∈ E ,$

where LIM denotes a Banach limit on l. Let

$K : = w ∈ E : f ( w ) = min { f ( y ) : y ∈ E } .$

Then K is a nonempty closed convex bounded subset of E. We claim that K is also invariant under the nonexpansive mapping J ε . Indeed, noting (3.8), we have for w K,

$f ( J ε w ) = LI M k 1 2 w t k - J ε w 2 = LI M k 1 2 J ε w t k - J ε w 2 ≤ LI M k 1 2 w t k - w 2 = f ( w ) .$

Since a uniformly smooth Banach space has the fixed point property for nonexpansive mappings and since J ε is a nonexpansive self-mapping of E, J ε has a fixed point in K, say w'. Now since w' is also a minimizer of f over E, it follows that, for w E,

$0 ≤ f ( w ′ + λ ( w - w ′ ) ) - f ( w ′ ) λ = LI M k 1 2 w t k - w ′ + λ ( w ′ - w ) 2 - 1 2 w t k - w ′ 2 λ .$

Since E is uniformly smooth, the duality map J is uniformly continuous on bounded sets, letting λ → 0+ in the last equation yields

$0 ≤ LI M k w ′ - w , J ( w t k - w ′ ) , w ∈ E .$
(3.9)

Since

$w t k - w ′ = ( I - t k F ) ( V t k w t k - w ′ ) + t k ( γ ϕ ( w t k ) - F w ′ ) ,$

we obtain

$w t n - w ′ 2 = t k γ ϕ ( w t k ) - F w ′ , J ( w t k - w ′ ) + ( I - t k F ) ( V t k w t k - w ′ ) , J ( w t k - w ′ ) ≤ t k γ ϕ ( w t k ) - F w ′ , J ( w t k - w ′ ) + ( 1 - t k γ ̄ ) w t k - w ′ 2 ≤ t k γ ϕ ( w t k ) - γ ϕ ( w ′ ) , J ( w t k - w ′ ) + t k γ ϕ ( w ′ ) - F w ′ , J ( w t k - w ′ ) + ( 1 + t k γ ̄ ) w t k - w ′ 2 ≤ [ 1 - t k ( γ ̄ - γ ) ] w t k - w ′ 2 + t k γ ϕ ( w ′ ) - F w ′ , J ( w t k - w ′ ) .$

It follows that

$w t k - w ′ 2 ≤ 1 γ ̄ - γ γ ϕ ( w ′ ) - F w ′ , J ( w t k - w ′ ) .$
(3.10)

Upon letting w = γϕ(w') - Fw' + w' in (3.9), we see that the last equation implies

$LI M k w t k - w ′ 2 ≤ 0 .$
(3.11)

Therefore, ${ w t k }$ contains a subsequence, still denoted ${ w t k }$, converging strongly to w1 (say). By virtue of (3.8), w1 is a fixed point of J ε ; i.e., a point in F*.

To prove that the entire net {w t } converges strongly, assume {s k } is another null subsequence in (0, 1) such that $w s k → w 2$ strongly. Then w2 F*.

Repeating the argument of (3.10) we obtain

$w t - w ′ 2 ≤ 1 γ ̄ - γ γ ϕ ( w ′ ) - F w ′ , J ( w t - w ′ ) , ∀ w ′ ∈ F * .$

In particular,

$w 1 - w 2 2 ≤ 1 γ ̄ - γ γ ϕ ( w 1 ) - F w 1 , J ( w 2 - w 1 )$
(3.12)

and

$w 2 - w 1 2 ≤ 1 γ ̄ - γ γ ϕ ( w 2 ) - F w 2 , J ( w 1 - w 2 ) .$
(3.13)

Adding up the last two equations gives

$w 1 - w 2 2 ≤ 0 .$

That is, w1 = w2. This concludes the proof.

Theorem 3.3. Suppose that E is a uniformly smooth Banach space and A is an m-accretive operator in E such that $F * =N ( A ) ≠∅$. Now given ϕ : EE be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient $γ ̄ >0,0<γ< γ ̄$. Assume

1. (i)

$lim n → ∞ α n =0, ∑ n = 0 ∞ α n =∞$, and $∑ n = 1 ∞ α n + 1 - α n <∞ ( e . g . , α n = 1 n )$;

2. (ii)

limn→∞= r n = r,r R+, r n ε for all n and $∑ n = 1 ∞ r n + 1 - r n < ∞ ( e . g . , r n = 1 + 1 n )$.

Then {x n } defined by (1.4) converges strongly to a point in F*.

Proof. Since

$J r n x n - J r x n ≤ J r r r n x n + 1 - r r n J r n x n - J r x n ≤ r r n x n + 1 - r r n J r n x n - x n ≤ 1 - r r n J r n x n - x n → 0 as ( n → ∞ ) .$
(3.14)

Thus

$x n - J r x n ≤ x n - J r n x n + J r n x n - J r x n → 0 .$
(3.15)

We next claim that , where $ẑ= lim t → 0 z t , n$ with z t,n = tγϕ(z t,n ) + (I - tF)J r z t,n .

For this purpose, let ${ x n k }$ be a subsequence chosen in such a way that $lim sup n → ∞ γ ϕ ( ẑ ) - F ẑ , J ( x n - ẑ ) = lim k → ∞ γ ϕ ( ẑ ) - F ẑ , J ( x n k - ẑ )$ and $x n k ⇀ x ̃$. Moreover, since x n - J r x n → 0, using Lemma 2.8, we know $x ̃ ∈F ( J r )$. Hence by Lemma 2.6, we have

$lim sup n → ∞ 〈 γ ϕ ( z ^ ) − F z ^ , J ( x n − z ^ ) 〉 = lim k → ∞ 〈 γ ϕ ( z ^ ) − F z ^ , J ( x n k − z ^ ) 〉 = 〈 γ ϕ ( z ^ ) − F z ^ , x ˜ − z ^ ) 〉 ≤ 0.$
(3.16)

Finally to prove that $x n →ẑ$ strongly, we write

$x n + 1 - ẑ = ( I - α n F ) ( J r n x n - ẑ ) + α n ( γ ϕ ( x n ) - F ẑ ) .$

Apply Lemma 2.9 to get

$x n + 1 - ẑ 2 ≤ ( 1 - α n γ ̄ ) 2 x n - ẑ 2 + 2 α n γ ϕ ( x n ) - F ẑ , J ( x n + 1 - ẑ ) ≤ ( 1 - α n γ ̄ ) 2 x n - ẑ 2 + 2 α n γ ϕ ( x n ) - γ ϕ ( ẑ ) , J ( x n + 1 - ẑ ) + 2 α n γ ϕ ( ẑ ) - F ẑ , J ( x n + 1 - ẑ ) ≤ ( 1 - α n γ ̄ ) 2 x n - ẑ 2 + α n γ x n - ẑ 2 + x n + 1 - ẑ 2 + 2 α n γ ϕ ( ẑ ) - F ẑ , J ( x n + 1 - ẑ )$

It follows that

where $M 1 = sup n ≥ 1 x n - ẑ 2$. By Lemma 2.2 and (3.16), we see that $x n →ẑ$.

Remark 3.4. If γ = 1, F is the identity operator and ϕ(x n ) = u in our results, we can obtain Theorems 3.1, 4.1, 4.2, 4.4 and Lemma 4.3 of Hong-Kun Xu [5].

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## Author information

Correspondence to Changsong Hu.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

The main idea of this paper is proposed by Meng Wen. All authors read and approved the final manuscript.

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