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

Meng Wen and Changsong Hu*

Author Affiliations

Department of Mathematics, Hubei Normal University, Huangshi 435002, P. R. China

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Fixed Point Theory and Applications 2012, 2012:98 doi:10.1186/1687-1812-2012-98

 Received: 28 November 2011 Accepted: 15 June 2012 Published: 15 June 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

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

MSC: 47H09; 47H10

##### Keywords:
MKC; accretive operators; the resolvent operator; iterative method; weakly continuous duality map

### 1 Introduction

Let E be a real Banach space, C a nonempty closed convex subset of E, and T : C C a mapping. Recall that T is nonexpansive if ∥Tx - Ty∥ ≤ ∥x - 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 . The normalized duality mapping J from a Banach space E into 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 such that

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

Let D be a subset of C. Then Q : C D is called a retraction from C onto D if Q(x) = x for all x D. A retraction Q : C D 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 : C D is a sunny nonexpansive retraction if and only if the following condition holds:

(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 xi D(A) and yi Axi, i = 1, 2, there is a j J(x2 - x1) such that

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.,

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

Note that if A is m-accretive, then Jr : E E is nonexpansive and F(Jr) = N(A) for all r > 0. We also denote by Ar the Yosida approximation of A, i.e., . It is well known that Jr 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 Jφ : E E* defined by

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 Jφ is single-valued and weak-to-weak* sequentially continuous(i.e., if {xn} is a sequence in E weakly convergent to a point x, then the sequence Jφ(xn) converges weakly* to Jφ(x)). It is known that lp has a weakly continuous duality map for all 1 < p < ∞, with gauge φ(t) = tp-1. Set

(1.2)

Then

where denotes the subdifferential in the sense of convex analysis.

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

(1.3)

where {an} and {rn} are two sequences satisfying certain conditions, and denotes the resolvent (I + rnA)-1 for rn > 0. He proved the strong convergence of the algorithm {xn} 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,

(1.4)

where {an} and {rn} are two sequences satisfying certain conditions, denotes the resolvent (I + rnA)-1 for rn > 0, F be a strongly positive bounded linear operator on E is , and ϕ be a MKC on E. Strong convergence of the algorithm {xn} 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 Jφ with gauge φ,

(i) For all x, y E, there holds the inequality

(ii) Assume a sequence {xn} in E is weakly convergent to a point x, then there holds the equality

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

where {λn}, {δn} and {γn} satisfy the following conditions:

(i) {λn} ⊂ [0,1] and ,

(ii) lim supn→∞ δn ≤ 0 or (iii) . Then limn→∞ sn = 0.

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

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 and 0 < ρ ≤ ∥F-1. Then,

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

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

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

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

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

(2.1)

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

Hence

and {xt} is bounded.

Next assume that {xt} is bounded as t → 0+. Assume tn → 0+ and is bounded. Since E is reflexive, we may assume that for some z E. Since Jφ is weakly continuous, we have by Lemma 2.1,

Put

It follows that

Since

we obtain

(2.2)

On the other hand, however,

(2.3)

Combining Equations (2.2) and (2.3) yields

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

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

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

It follows that

Therefore,

Now observing that implies , we conclude from the last inequality that

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

We have to show . Indeed, for p F(T). Since

we derive that

(2.4)

Notice

It follows that,

(2.5)

Now replacing t in (2.5) with tn and letting n → ∞, noticing for z F(T), we obtain 〈(F - γϕ)z, Jφ(z - p)〉 ≤ 0. In the same way, we have .

Thus, we have

(2.6)

On the other hand, without loss of generality, we may assume there is a number ε such that , then by Lemma 2.5 there is a number r1 such that . Noticing that

Hence and {xt} converges strongly. Thus we may assume . Since we have proved that, for all t ∈ (0, 1) and p F(T),

letting t → 0, we obtain that

This implies that

Lemma 2.7. (see [12]). Assume that C2 C1 > 0. Then 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 : C E is a nonexpansive mapping. Then the mapping I - T is demiclosed at zero, that is xn x and xn - Txn∥ → 0, then x = Tx.

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

### 3 Main result

Theorem 3.1. Suppose that E is reflexive which admits a weakly continuous duality map Jφ with gauge φ and A is an m-accretive operator in E such that . Now given ϕ : E E be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient . Assume

(i) ;

(ii) rn → ∞.

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

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

By induction, we have

This implies that {xn} is bounded and hence

We next prove that

lim supn→∞γϕ(p) - Fp, Jφ(xn - p)〉 ≤ 0, where p = limt→0 xt with .

Since {xn} is bounded, take a subsequence of {xn} such that

(3.1)

Since E is reflexive, we may further assume that . Moreover, since

we obtain

Taking the limit as k → ∞ in the relation

we get . That is, . Hence by (3.1) and Lemma 2.6 we have

Finally to prove that xn p, we apply Lemma 2.1 to get

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

Theorem 3.2. Suppose that E is reflexive which admits a weakly continuous duality map Jφ with gauge φ and A is an m-accretive operator in E such that . Now given ϕ : E E be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient . Assume

(i) , and ;

(ii) rn ε for all n and .

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

Proof. We only include the differences. We have

Thus,

(3.2)

If rn-1 rn, using the resolvent identity

we obtain

(3.2a)

It follows from (3.2) that

(3.3)

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

This implies that

(3.4)

since . It follows that

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

we get ; i.e., . We therefore conclude that all weak limit points of {xn} 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 rn ε for some ε > 0 (not necessarily rn → ∞), A is an m-accretive operator in E. Moreover let ϕ : E E be a MKC and F be a strongly positive linear bounded operator on E. Since is nonexpansive, the map is a contraction and for each integer n ≥ 1 it has a unique fixed zt,n E. Hence the scheme

(3.5)

is well defined.

Note that {zt,n} is uniformly bounded; indeed, 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 is uniform for all n ≥ 1.

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

(3.6)

then converges as t → 0 to a point in F*. For simplicity put

It follows that

(3.7)

Note that Fix(Vt) = F* for all t. Note also that {wt} is bounded; indeed, we have for all t ∈ (0, 1) and p F*. Since {Vt wt} is bounded, it is easy to see that

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

(3.8)

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

where LIM denotes a Banach limit on l. Let

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

Since a uniformly smooth Banach space has the fixed point property for nonexpansive mappings and since Jε is a nonexpansive self-mapping of E, Jε 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,

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

(3.9)

Since

we obtain

It follows that

(3.10)

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

(3.11)

Therefore, contains a subsequence, still denoted , converging strongly to w1 (say). By virtue of (3.8), w1 is a fixed point of Jε; i.e., a point in F*.

To prove that the entire net {wt} converges strongly, assume {sk} is another null subsequence in (0, 1) such that strongly. Then w2 F*.

Repeating the argument of (3.10) we obtain

In particular,

(3.12)

and

(3.13)

Adding up the last two equations gives

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 . Now given ϕ : E E be a MKC, and let F be a strongly positive linear bounded operator on E with coefficient . Assume

(i) , and ;

(ii) limn→∞ = rn = r,r R+, rn ε for all n and .

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

Proof. Since

(3.14)

Thus

(3.15)

We next claim that , where with zt,n = tγϕ(zt,n) + (I - tF)Jrzt,n.

For this purpose, let be a subsequence chosen in such a way that and . Moreover, since ∥xn - Jrxn∥ → 0, using Lemma 2.8, we know . Hence by Lemma 2.6, we have

(3.16)

Finally to prove that strongly, we write

Apply Lemma 2.9 to get

It follows that

where . By Lemma 2.2 and (3.16), we see that .

Remark 3.4. If γ = 1, F is the identity operator and ϕ(xn) = 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].

### 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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