Strong Convergence of Two Iterative Algorithms for

International Mathematical Forum, 5, 2010, no. 44, 2165 - 2172

Strong Convergence of Two Iterative Algorithms for a Countable Family of Nonexpansive Mappings in Hilbert Spaces Jintana Joomwong Division of Mathematics , Faculty of Science Maejo University, Chiang Mai, Thailand, 50290 [email protected] Abstract In this paper, we study two iterative algorithms for a countable family of nonexpansive mappings in Hilbert Spaces.We prove that the proposed algorithms converge strongly to a fixed point of nonexpansive mappings {Tn }.The results of this paper extend and improve the results of Yonghong Yoa, et al. [14].

Mathematics Subject Classification: 47H05, 47H10 Keywords: Nonexpansive mapping; Fixed Point; Two iterative algorithms; Hilbert space

1

Introduction

Let H be a real Hilbert space with norm · and inner product ·, ·. And let C be a nonempty closed convex subset of H. A mapping T of C into itself is said to be nonexpansive if T x − T y ≤ x − y for each x, y ∈ C. We denote by F (T ) the set of fixed points of T . In 2003, for finding an element of F (S) ∩ V I(A, C), Takahashi and Toyoda [10] introduced the following iterative scheme: xn+1 = αn xn + (1 − αn )SP c(xn − λn Axn ) for every n = 0, 1, 2, ..., where x0 = x ∈ C,{αn }is a sequence in (0, 1), and {λn } is a sequence in (0, 2α). Let A is a strongly positive bounded linear operator on H. That is, there is a constant γ¯ > 0 with property Ax, x ≥ γ¯ x2 for all x ∈ H.

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A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbeert space H. 1 min Ax, x − x, b x∈C 2 where C is the fixed point set of a nonexpansive mapping T on H and b is a given point in H. Recently, Xu [12] proved that the sequence {xn } defined by the iterative method below, with the initial guess x0 ∈ H chosen arbitrarily: xn+1 = (I − αn A)T xn + αn u, n ≥ 0,

(1.1)

converges strongly to the unique solution of the minimization problem provided the sequence {αn } satisfies certain conditions. On the other hand, Aoyama, et al.,[1] introduce a Halpern type iterative sequence for finding a common fixed point of a countable family of nonexpansive mappings. Let x1 = x ∈ C and xn+1 = αn x + (1 − αn )Tn xn

(1.2)

for all n ∈ N, where C is a nonempty closed convex subset of a Banach space, {αn } is a sequence in [0, 1] and {Tn } is a sequence of nonexpansive mappings with some condition. They proved that {xn } defined by (1.2) converges strongly to a common fixed point of {Tn }. Very recently, Yonghong Yoa,et al.[14] introduced two iterative algorithms defined by, for given x0 ∈ C arbitrarily and let the sequence {xn }, n ≥ 0 be generated by yn = PC [(1 − αn )xn ], (1.3) xn+1 = (1 − βn )xn + βn T yn , and they prove that the algorithms strongly converge to a fixed point of nonexpansive mappings T . The research in this field, iterative algorithms for finding fixed points of nonlinear mappings, is important and find applications in a variety of applied areas of inverse problem, partial differential equations,image recovery and signal processing, see [2] [3] [4] [6]. In this paper motivate by result of Yonghong Yoa,et al.,and the ongoing research in this field, we introduced the two iterative algorithms in Hilbert space defined by, for given x0 ∈ C arbitrarily and let the sequence {xn }, n ≥ 0 be generated by yn = PC [(1 − αn )xn ], (1.4) xn+1 = (1 − βn )xn + βn Tn yn ,

Strong convergence for a countable family of nonexpansive mappings

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where {αn } and {βn } are real in [0,1] and {Tn } is a sequence of nonexpansive mappings with some conditions. Then we prove that the sequence {xn } defined by (1.4) converges strongly to a fixed point of {Tn }. The result of this result extends and improves the corresponding results of Yonghong Yoa,et al.,[14].

2

Preliminary Notes

Let H be a real Hilbert space and let C be a closed convex subset of H.Then, for any x ∈ H, there exists a unique nearest point u ∈ C such that x − u ≤ x − y, ∀y ∈ C. We denote u by PC (x), where PC is called the metric projection of H onto C. It is well known that PC is nonexpansive. Furthermore, for x ∈ H and u ∈ C, u = PC (x) ⇔ x − u, u − y ≥ 0, ∀y ∈ C Lemma 2.1. ([12]) Let C be a nonempty closed convex of a real Hilbert space H. Let T : C −→ C be a nonexpansive mapping. Then I − T is demi-closed at zero, i.e., if xn x ∈ C and xn − T xn → 0, then x = T x Lemma 2.2. ([8]) Let {xn } and {yn } be bounded sequences in a Banach space X and let {βn } be a sequence in [0, 1] with 0 < lim inf n−→∞ βn lim supn−→∞ βn < 1. Suppose that xn+1 = (1−βn )yn +βn xn for all integer n ≥ 0 and lim supn−→∞ (yn+1 − yn − xn+1 − xn ) 0. Then limn−→∞ yn − xn = 0. Lemma 2.3. ([12]).Let {an } be a sequence of nonnegative real numbers such that an+1 ≤ (1 − γn )an + γn δn , for all n ≥ 0 where {γn } be a sequence in (0,1) and {δn } is a sequence in R such that (i)

∞

n=10

γn = ∞,

(ii) lim supn−→∞ δn ≤ 0 or

∞

n=10

|δn γn | < ∞.

Then limn−→∞ an = 0. Lemma 2.4. .Let H be a real Hilbert space.Then for allx, y ∈ H,the following hold; (i) x + y2 = x2 + 2y, x + y, (ii) x + y2 = x2 + 2y, x

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Throughout the rest of this paper, Let each t ∈ (0, 1), we consider the following mapping Tt given by Tt x = T PC [(1 − t)x], ∀x ∈ C. It is easy to check that Tt x − Tt y ≤ (1 − t)x − y which implies that Tt is a contraction. Using the Banach contraction principle, there exists a unique fixed point xt of Tt in C, i.e., xt = T PC [(1 − t)xt ].

(2.1)

Lemma 2.5. ([14]). Let C be a nonempty bounded closed convex subset of Hilbert space H. Let T : C −→ C be a nonexpansive mapping with F (T ) = ∅. For each t ∈ (0, 1), let the net {xt } be generated by (2.1). Then, as t → 0, the net {xt } converges strongly to a fixed point of T . Lemma 2.6. ([11]). Let C be a nonempty bounded closed convex subset of Hilbert space H and {Tn } be a sequence of mappings of C into itself. Suppose that lim ρkl = 0,

k,l→∞

(2.2)

where ρkl = sup{Tk z − Tl z : z ∈ C} < ∞, for all k, l ∈ N. Then for each x ∈ C, {Tn x} converges strongly to some point of C. Moreover, let T be a mapping from C into itself defined by T x = limn→∞ Tn x, for all x ∈ C. Then lim supn→∞ {T z − Tn z : z ∈ C} = 0

3

Main results

In this section, we prove the strong convergence theorems for a countable family of nonexpansive mappings in a real Hilbert space. Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let {Tn } be a sequence of nonexpansive mapping of H into itself such that ∞ n=1 F (Tn ) is nonempty. Let {αn }, {βn } are the sequences in (0, 1). For given x0 ∈ C arbitrarily, let the sequence {xn }, n ≥ 0. be generated by (1.4). Suppose the following conditions are satisfied: (i) limn→∞ αn = 0 and ∞ n=10 αn = ∞; ∞ ∞ (ii) n=10 |αn+1 − αn | < ∞ and n=10 |βn+1 − βn | < ∞; (iii) 0 < lim inf n−→∞ βn ≤ lim supn−→∞ βn < 1.


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Suppose that for any Cof H, the sequence {Tn } satisfied condition (2.2) in lemma (2.6). Let T be a mapping of H into itself defined by T y = limn→∞ Tn y ∞ for all y ∈ H. If F (T ) = n=1 F (Tn ) then {xn } converges strongly to a fixed point z in F (T ). Proof. First, we observed ∞ that {xn } is bounded. Indeed, pick any p ∈ n=1 F (Tn ) = F (T ) to obtain, xn+1 − p = ≤ ≤ ≤ ≤ ≤ ≤

(1 − βn )xn + βn Tn yn − p (1 − βn )xn − p + βn Tn yn − p (1 − βn )xn − p + βn yn − p (1 − βn )xn − p + βn (1 − αn )xn − p (1 − βn )xn − p + βn [(1 − αn )xn − p + αn p] (1 − αn βn )xn − p + αn βn p max{xn − p, p}.

Hence, {xn } is bounded and so is {Tn xn }. Let B = {y ∈ H : y − p| ≤ K} where K = max{xn − p, p}, n ≥ 0. Clearly, B is bounded closed convex subset of H, T (B) ⊆ B, {xn } ⊆ B, and {Tn xn } ⊆ B. Set zn = Tn yn , n ≥ 0. It follows that zn+1 − zn ≤ Tn+1 yn+1 − Tn+1 yn + Tn+1 yn − Tn yn ≤ (1 − αn+1 )xn+1 − (1 − αn )xn + Tn+1 yn − Tn yn ≤ xn+1 − xn + αn+1 xn+1 + αn xn +Tn+1 (1 − αn+1 )xn+1 − Tn (1 − αn )xn ≤ xn+1 − xn + αn+1 xn+1 + αn xn + sup Tn+1 s − Tn s. s∈B

Then, we obtained zn+1 − zn − xn+1 − xn = αn+1 xn+1 + αn xn + sup Tn+1 s − Tn s. s∈B

By our assumptions, we get lim sup(zn+1 − zn − xn+1 − xn ) ≤ 0. n→∞

This together with lemma (2.2) imply that Therefore, lim xn+1 − xn =

n→∞

=

limn→∞ zn − xn = 0.

lim (1 − βn )xn + βn zn − xn

n→∞

lim βn zn − xn = 0

n→∞

J. Joomwong

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xn − xn+1 + xn+1 − Tn xn xn − xn+1 + (1 − βn )xn − βn Tn yn − Tn xn xn − xn+1 + (1 − βn )xn − Tn xn + βn yn − xn xn − xn+1 + (1 − βn )xn − Tn xn + βn (1 − αn )xn − xn xn − xn+1 + (1 − βn )xn − Tn xn + αn xn

That is xn − Tn xn ≤ β1n {xn − xn+1 + αn xn } → 0 as n → ∞. Let the net {xt } defined by (2.1). By lemma (2.5), we have xt → z as t → 0 Next,we prove that lim supn→∞ z, z − xn ≤ 0. Indeed, we calculate xt − xn 2 = (xt − Tn xn ) + (Tn xn − xn )2 = xt − Tn xn 2 + 2xt − Tn xn , Tn xn − xn + Tn xn − xn 2 ≤ xt − Tn xn 2 + 2xt − xn , Tn xn − xn −2Tn xn − xn , Tn xn − xn + Tn xn − xn 2 ≤ Tn PC [(1 − t)xt − Tn xn 2 + 2xt − xn Tn xn − xn −Tn xn − xn 2 ≤ (1 − t)xt − xn 2 + 2xt − xn Tn xn − xn ≤ xt − xn 2 − 2txt , xt − xn + t2 xt 2 +2xt − xn Tn xn − xn ≤ xt − xn 2 − 2txt , xt − xn + t2 M + MTn xn − xn where M > 0 such that sup{xt 2 , 2xt − xn , t ∈ (0, 1), n ≥ 1} ≤ M. xn − Tn xn . It follows that xt , xt − xn ≤ 2t M + M 2t Therefore, lim sup lim supxt , xt − xn ≤ 0. t→0

n→∞

We note that z, z − xn = z, z − xt + z − xt , xt − xn + xt , xt − xn ≤ z, z − xt + z − xt xt − xn + xt , xt − xn ≤ z, z − xt + z − xt M + xt , xt − xn . This together with xt → z and (3.1) implies that lim supz, z − xn ≤ 0 n→∞

(3.1)


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Finally, we show that xn → z. From (1.4), we have xn+1 − z2 = ≤ ≤ ≤ ≤

(1 − βn )xn + βn Tn yn − z2 (1 − βn )xn − z2 + βn Tn yn − z2 (1 − βn )xn − z2 + βn yn − z2 (1 − βn )xn − z2 + βn (1 − αn )(xn − z) − αn z2 (1 − βn )xn − z2 + βn [(1 − αn )xn − z2 −2αn (1 − αn )z, xn − z + αn2 z2 ]

≤ (1 − αn βn )xn − z2 + αn βn [2(1 − αn )z, z − xn + By lemma (2.3), we obtained xn → z proof.

as

αn z2 ]. βn

n → ∞. This completes the

Setting Tn ≡ T in Theorem 3.1, we have the following result. Corollary 3.2. [14, Theorem 3.2] Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a nonexpansive mapping such that F (T ) is nonempty. Let {αn }, {βn } are the sequences in (0, 1). For given x0 ∈ C arbitrarily, let the sequence {xn }, n ≥ 0. be generated by (1.3). Suppose the following conditions are satisfied: (i) limn→∞ αn = 0 and ∞ n=10 αn = ∞; (ii) 0 < lim inf n−→∞ βn ≤ lim supn−→∞ βn < 1. Then {xn } converges strongly to a fixed point of T . ACKNOWLEDGEMENTS. The author like to thank the referees for Their helpful comments and suggestions, which improve the presentation of this paper.

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