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Strong convergence of an new iterative method for a zero of accretive operator and nonexpansive mapping
Fixed Point Theory and Applications volume 2012, Article number: 98 (2012)
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 , where {a n } and {r n } are two sequences satisfying certain conditions, denotes the resolvent (I + r n A)-1 for r n > 0, F be a strongly positive bounded linear operator on E is , 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 : 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〉 = ∥x∥2 = ∥f∥2}, 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:
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
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 J r the resolvent of A for r > 0:
Note that if A is m-accretive, then J r : E → E 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., . 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 : 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 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
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,
where {a n } and {r n } are two sequences satisfying certain conditions, and 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,
where {a n } and {r n } are two sequences satisfying certain conditions, denotes the resolvent (I + r n A)-1 for r n > 0, F be a strongly positive bounded linear operator on E is , 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,
-
(i)
For all x, y ∈ E, there holds the inequality
-
(ii)
Assume a sequence {x n } in E is weakly convergent to a point x, then there holds the equality
Lemma 2.2. [6, 7]. Let {s n } be a sequence of nonnegative real numbers satisfying
where {λ n }, {δ n } and {γ n } satisfy the following conditions:
-
(i)
{λ n } ⊂ [0,1] and ,
-
(ii)
lim supn→∞δ n ≤ 0 or (iii) . Then limn→∞s n = 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 J φ 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 {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 , then uniquely solves the variational inequality
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
It is easy to see that S t is a contraction. Indeed, by Lemma 2.4, we have
for all x, y ∈ E. Hence S t has a unique fixed point, denoted as x t , which uniquely solves the fixed point equation
We next show the sequence {x t } 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 {x t } is bounded.
Next assume that {x t } is bounded as t → 0+. Assume t n → 0+ and is bounded. Since E is reflexive, we may assume that for some z ∈ E. Since J φ is weakly continuous, we have by Lemma 2.1,
Put
It follows that
Since
we obtain
On the other hand, however,
Combining Equations (2.2) and (2.3) yields
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 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
It contradicts . Hence .
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
We have to show . Indeed, for p ∈ F(T). Since
we derive that
Notice
It follows that,
Now replacing t in (2.5) with t n and letting n → ∞, noticing for z ∈ F(T), we obtain 〈(F - γϕ)z, J φ (z - p)〉 ≤ 0. In the same way, we have .
Thus, we have
Adding up (2.6) gets
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 {x t } 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 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
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 . 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)
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
By induction, we have
This implies that {x n } is bounded and hence
We next prove that
lim supn→∞〈γϕ(p) - Fp, J φ (x n - p)〉 ≤ 0, where p = limt→0x t with .
Since {x n } is bounded, take a subsequence of {x n } such that
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 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 . 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)
r n ≥ ε for all n and .
Then {x n } defined by (1.4) converges strongly to a point in F*.
Proof. We only include the differences. We have
Thus,
If rn-1≤ r n , using the resolvent identity
we obtain
It follows from (3.2) that
where M > 0 is some appropriate constant. Similarly we can prove (3.3) if rn-1≥ r n . By assumptions (i) and (ii) and Lemma 2.2, we conclude that
This implies that
since . It follows that
Now if is a subsequence of {x n } 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 {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 : 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 z t,n ∈ E. Hence the scheme
is well defined.
Note that {z t,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 n t (which may depend on t ∈ (0, 1)), if is the unique point in E that satisfies the property
then converges as t → 0 to a point in F*. For simplicity put
It follows that
Note that Fix(V t ) = F* for all t. Note also that {w t } is bounded; indeed, we have for all t ∈ (0, 1) and p ∈ F*. Since {V t w t } is bounded, it is easy to see that
Since r n ≥ ε for all n, by Lemma 2.7, we have
Let {t k } be a sequence in (0,1) such that t k → 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 J ε . 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 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,
Since E is uniformly smooth, the duality map J is uniformly continuous on bounded sets, letting λ → 0+ in the last equation yields
Since
we obtain
It follows that
Upon letting w = γϕ(w') - Fw' + w' in (3.9), we see that the last equation implies
Therefore, contains a subsequence, still denoted , 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 strongly. Then w2 ∈ F*.
Repeating the argument of (3.10) we obtain
In particular,
and
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→∞= r n = r,r ∈ R+, r n ≥ ε for all n and .
Then {x n } defined by (1.4) converges strongly to a point in F*.
Proof. Since
Thus
We next claim that , where with z t,n = tγϕ(z t,n ) + (I - tF)J r z t,n .
For this purpose, let be a subsequence chosen in such a way that and . Moreover, since ∥x n - J r x n ∥ → 0, using Lemma 2.8, we know . Hence by Lemma 2.6, we have
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(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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The main idea of this paper is proposed by Meng Wen. All authors read and approved the final manuscript.
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Wen, M., Hu, C. Strong convergence of an new iterative method for a zero of accretive operator and nonexpansive mapping. Fixed Point Theory Appl 2012, 98 (2012). https://doi.org/10.1186/1687-1812-2012-98
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DOI: https://doi.org/10.1186/1687-1812-2012-98