New strong convergence theorems for nonexpansive nonself-mappings ... methods for a nonexpansive nonself-mapping without
Computers and Mathematics with Applications 56 (2008) 1473–1478
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New strong convergence theorems for nonexpansive nonself-mappings without boundary conditions Yisheng Song College of Mathematics and Information Science, Henan Normal University, 453007, PR China
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Article history: Received 1 March 2007 Received in revised form 23 February 2008 Accepted 5 March 2008 Keywords: Nonexpansive nonself-mappings Explicit viscosity approximation Reflexive and strictly convex Banach space
a b s t r a c t The aim of this work is to establish the strong convergence of explicit viscosity-like methods for a nonexpansive nonself-mapping without boundary conditions defined in a Banach space. This gets rid of the restriction and dependence on the implicit anchor-like continuous path xt in the existing literature. © 2008 Elsevier Ltd. All rights reserved.
1. Introduction Let K be a nonempty closed convex subset of a Banach space E, and T be a nonexpansive mapping from K into E. Recently, the problem of approximating fixed points of nonexpansive nonself-mappings has been studied by many researchers. For example, for any given contraction f : K → K and the projector (or nonexpansive retraction) P from E onto K , the mapping Tft = P (tf + (1 − t)T ) obviously is a contraction from K to K for each fixed t ∈ (0, 1). Therefore, the Banach Contraction Principle guarantees the existence and the uniqueness of the fixed point for Tft in K ; i.e. ∃xt ∈ K such that xt = P (tf (xt ) + (1 − t)Txt ).
(1.1)
As t → 0, the strong convergence of the path xt have recently been studied by Song–Chen [1] in a reflexive Banach spaces E with a weakly sequentially continuous duality mapping; also by Song–Li [2] in a reflexive and strictly convex Banach spaces E with a uniformly Gâteaux differentiable norm; and by Matsushita–Takahashi [3] in a uniformly convex Banach space with a uniformly Gâteaux differentiable norm. In Refs. [1–3], they also investigated the following explicit iteration: xn+1 = P (αn f (xn ) + (1 − αn )Txn )
or xn+1 = αn f (xn ) + (1 − αn )PTxn .
(1.2)
Special cases of the path (1.1), in which f is a constant mapping (f (x) = u, for some u ∈ K ) and T is a nonexpansive nonselfmapping satisfying the weakly inward condition have been studied by Xu–Yin [4] in a Hilbert space; and by Xu [5] in a uniformly smooth Banach space; by Jung–Kim [6] in a uniformly convex Banach space with uniformly Gâteaux differentiable norm; also by Takahashi–Kim [7] in the frame of a reflexive Banach space with a uniformly Gâteaux differentiable norm and having the fixed point property for nonexpansive self-mapping. Other special cases of the path (1.1), in which T is a nonexpansive self-mapping (P = I, the identity operator) and f is a constant mapping (f (x) = u, for some u ∈ K ) have been studied by Browder [8] and Halpern [9] in a Hilbert space; and by Reich [10] in a uniformly smooth Banach space (also see Reich [11] and Goebel–Reich [12]); also by Takahashi–Ueda [13] in a uniformly convex Banach space with a uniformly Gâteaux differentiable norm; and by Xu [14] in a reflexive Banach space which has a weakly continuous duality mapping Jϕ . In the above results, the following explicit iteration is also investigated: xn+1 =
αn u + (1 − αn )Txn .
E-mail addresses:
[email protected],
[email protected]. 0898-1221/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2008.03.004
(1.3)
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In a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, the special cases of the path (1.1), in which P = I, the identity operator have been studied by Song–Chen [15] for a finite family of nonexpansive self-mappings {Tn }; by Song–Chen [16] for a continuous pseudocontractive self-mappings T . But the convergence of the explicit iteration (1.2) and (1.3) depends on the convergence of the path xt in the results mentioned above and in most other existing literatures. Reich [17] (Banach spaces) and Wittmann [18] (Hilbert space) respectively showed the convergence of the explicit iteration (1.3) without using the the analogous continuous scheme xt (P = I, f (x) ≡ u in (1.1)). In this paper, we shall get rid of the restriction and dependence on the path xt to prove that {xn } defined by (1.2) and (1.3) strongly converge to some fixed point of T . 2. Preliminaries Let E be a real Banach space with dual E∗ and K a nonempty closed convex subset of E. Let J denote the normalized duality ∗ mapping from E into 2E given by J(x) = {f ∈ E∗ , hx, f i = kxkkf k, kxk = kf k}, ∀ x ∈ E, where h·, ·i denotes the generalized duality pairing. We say F (T ) = {x ∈ D(T ); x = Tx}, the set of all fixed point for a mapping T , where D(T ) is domain of T . We ∗
write xn * x (respectively xn * x) to indicate that the sequence {xn } converges weakly (respectively weak∗ ) to x; as usual xn → x will symbolize (strong) convergence. Let T be a mapping with domain D(T ) and range R(T ) in Banach space E. T is said to be contractive if there exists β ∈ (0, 1) for any x, y ∈ D(T ) such that kTx − Tyk ≤ βkx − yk; T is called nonexpansive if the above inequality holds for β = 1. The norm of Banach space E is said to be Gâteaux differentiable (or E is said to be smooth) if the limit limt→0 kx+tytk−kxk exists for each x, y on the unit sphere S(E) of E. Moreover, if for each y in S(E) the limit above is uniformly attained for x in S(E), we say that the norm of E is uniformly Gâteaux differentiable. A Banach space E is said to be strictly convex if kxk = kyk = 1, x 6= y implies kx+2 yk < 1. A Banach space E is said to be uniformly convex if for all ε ∈ [0, 2], ∃δε > 0 such that kxk = kyk = 1 with kx − yk ≥ ε implies kx+2 yk < 1 − δε . If C and D are nonempty subsets of a Banach space E such that C is nonempty closed convex and D ⊂ C , then a mapping P : C → D is called a retraction from C to D if P is continuous with F (P ) = D. A mapping P : C → D is called sunny if P (Px + t(x − Px)) = Px, ∀x ∈ C whenever Px + t(x − Px) ∈ C and t > 0. 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. The term “sunny nonexpansive retraction” was coined by Reich in [19]. For more details, see Refs.[20,3,21,19,5,7,22]. Lemma 2.1 ([3, Lemma 3.1, 3.3]). Let E be a real smooth and strictly convex Banach space, and C be a nonempty closed convex subset of E which is also a sunny nonexpansive retract of E. Assume that T : C → E is a nonexpansive mapping and P is a sunny nonexpansive retraction of E onto C , then F (T ) = F (PT ). Let µ be a continuous linear functional on l∞ satisfying kµk = 1 = µ(1). Then we know that µ is a mean on N if and only if inf {an ; n ∈ N} ≤ µ(a) ≤ sup{an ; n ∈ N} for every a = (a1 , a2 , . . .) ∈ l∞ . According to time and circumstances, we µn (an ) instead of µ(a). A mean µ on N is called a Banach limit if µn (an ) = µn (an+1 ) for every a = (a1 , a2 , . . .) ∈ l∞ . Using the Hahn–Banach theorem, or the Tychonoff fixed point theorem, we can prove the existence of a Banach limit. The following results can be founded in Refs. [23–27]. Lemma 2.2 ([23, Proposition 2.2] and [24, Proposition 2]). Let α be a real number and (x0 , x1 , . . .) ∈ l∞ such that µn xn ≤ α for all Banach limits. If lim supn→∞ (xn+1 − xn ) ≤ 0, then lim supn→∞ xn ≤ α. Lemma 2.3 ([26, Lemma 2.5]). Let {an } be a sequence of nonnegative real numbers satisfying the property an+1 ≤ (1 − γn )an + γn βn ,
n ≥ 0,
where {γn } ⊂ (0, 1) and {βn } ⊂ R such that P P∞ (i) limn→∞ γn = 0 and ∞ n=0 γn = ∞; (ii) either n=0 |γn βn | < +∞ or lim supn→∞ βn ≤ 0. Then {an } converges to zero, as n → ∞. 3. Main results In this section, we will study the following iteration for a nonexpansive nonself-mapping T and a contractive self-mapping f: xn+1 = P (αn f (xn ) + (1 − αn )Txn ),
(3.1)
where P is a sunny nonexpansive retraction of E onto D(T ). In order to prove our main results, we need to show the following result by means of Lemma 2.2.
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Proposition 3.1. Let K be a nonempty closed convex subset of a Banach space E with a uniformly Gâteaux differentiable norm. Suppose that {xn } is a bounded sequence of E such that limn→∞ kxn+1 − xn k = 0, and µn is Banach limit. If z ∈ K such that µn kxn − zk2 = inf y∈K µn kxn − yk2 , then lim suphy − z, J(xn − z)i ≤ 0, n→∞
∀y ∈ K .
Proof. Let t ∈ (0, 1) and y ∈ K , then z+t(y−z) = (1−t)z+ty ∈ K by the convexity of K . Then µn kxn −zk2 ≤ µn kxn −z−t(y−z)k2 . Since
kxn − z − t(y − z)k2 = hxn − z, J(xn − z − t(y − z))i − thy − z, J(xn − z − t(y − z))i ≤ kxn − zkkxn − z − t(y − z)k − thy − z, J(xn − z − t(y − z))i kxn − zk2 + kxn − z − t(y − z)k2 ≤ − thy − z, J(xn − z − t(y − z))i, 2
then
kxn − z − t(y − z)k2 ≤ kxn − zk2 − 2thy − z, J(xn − z − t(y − z))i. Thus, taking the Banach limit over n ≥ 1 gives
µn kxn − z − t(y − z)k2 ≤ µn kxn − zk2 − 2tµn hy − z, J(xn − z − t(y − z))i. This implies 2tµn hy − z, J(xn − z − t(y − z))i ≤ µn kxn − zk2 − µn kxn − z − t(y − z)k2 ≤ 0. Therefore,
µn hy − z, J(xn − z − t(y − z))i ≤ 0. Since the normalized duality mapping J is single-valued and norm-weak∗ uniformly continuous on bounded subset of E, so we obtain that as t → 0,
hy − z, J(xn − z)i − hy − z, J(xn − z − t(y − z))i → 0 uniformly. Hence, for all ε > 0, there exists δ > 0 such that ∀ ∈ (0, δ) and for all n ≥ 1,
hy − z, J(xn − z)i < hy − z, J(xn − z − t(y − z))i + ε. Consequently,
µn hy − z, J(xn − z)i ≤ µn hy − z, J(xn − z − t(y − z))i + ε ≤ ε. Since ε is arbitrary, then
µn hy − z, J(xn − z)i ≤ 0. On the other hand, as limn→∞ kxn+1 − xn k = 0, then it follows from the norm-weak∗ uniform continuity of the duality mapping J that lim (hy − z, J(xn+1 − z)i − hy − z, J(xn − z)i) = 0.
n→∞
Hence, the sequence {hy − z, J(xn − z)i} satisfies the conditions of Lemma 2.2. As a result, lim supn→∞ hy − z, J(xn − z)i ≤ 0. This finishes the proof. The following lemma is proved by Song–Chen [1]. Lemma 3.2 ([1, Theorem 2.4]). Suppose K is a nonempty closed convex subset of a Banach space E which is also a sunny nonexpansive retract of E, and T : K → E is a nonexpansive mapping and F (T ) 6= ∅. Let f : K → K be a fixed contractive mapping with the contractive coefficient β ∈ (0, 1), and {xn } be defined by (3.1), and {αn } ⊂ (0, 1). Then (i) {xn } is bounded. P P∞ αn = 1, If {αn } ⊂ (0, 1) satisfies the conditions (C2) ∞ n=0 αn = ∞; (C3) either n=0 |αn+1 − αn | < +∞ or limn→∞ α then (ii) limn→∞ kxn − xn+1 k = 0. If, in addition, {αn } satisfies (C1) limn→∞ αn = 0, then (iii) limn→∞ kxn − PTxn k = 0.
n+1
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Let {xn } be defined by (3.1) and αn ∈ (0, 1). It follows from Lemma 3.2 (i) that {xn } is bounded. Let
ϕ(y) = µn kxn − yk2 for all y ∈ K , then ϕ(y) is convex and continuous, and ϕ(y) → ∞ as kyk → ∞. If E is reflexive, there exists z ∈ K such that ϕ(z) = inf y∈K ϕ(y) (see [22, Theorem 1.3.11]). So the set Kmin = {z ∈ K ; ϕ(z) = inf ϕ(y)} 6= ∅. y∈K
Clearly, Kmin is closed convex by the convexity and continuity of ϕ(y). Next, we show our main results. Theorem 3.3. Let E be a Banach space with a uniformly Gâteaux differentiable norm. Suppose K is a nonempty closed convex subset of E which is also a sunny nonexpansive retract of E, and T : K → E is a nonexpansive mapping with F (T ) 6= ∅. Assumed that f : K → K is a fixed contractive mapping with the contractive coefficient β ∈ (0, 1) and {αn } satisfies the conditions (C2) and (C3). T If Kmin F (T ) 6= ∅, then as n → ∞, {xn }, is defined by (3.1) converges strongly to some fixed point x∗ of T . T Proof. Take x∗ ∈ Kmin F (T ). It follows from Lemma 3.2 that {xn } is bounded and limn→∞ kxn − xn+1 k = 0. Since x∗ ∈ T Kmin F (T ), then
µn kxn − x∗ k2 = inf µn kxn − yk2 and x∗ = Tx∗ .
(3.2)
y∈K
Therefore, Proposition 3.1 assures that for f (x∗ ) ∈ K , lim suphf (x∗ ) − x∗ , J(xn+1 − x∗ )i ≤ 0.
(3.3)
n→∞
Next we show that xn → x∗ . Since xn+1 − (αn f (xn ) + (1 − αn )x∗ ) = (xn+1 − x∗ ) − αn (f (xn ) − x∗ )
and
kxn+1 − P(αn f (xn ) + (1 − αn )x∗ )k ≤ (1 − αn )kxn − x∗ k, then
kxn+1 − x∗ k2 = hxn+1 − (αn f (xn ) + (1 − αn )x∗ ), J(xn+1 − x∗ )i + αn hf (xn ) − x∗ , J(xn+1 − x∗ )i ≤ xn+1 − P(αn f (xn ) + (1 − αn )x∗ )kxn+1 − x∗ k + αn (hf (xn ) − f (x∗ ), J(xn+1 − x∗ )i + hf (x∗ ) − x∗ , J(xn+1 − x∗ )i) kf (xn ) − f (x∗ )k2 + kxn+1 − x∗ k2 + αn hf (x∗ ) − x∗ , J(xn+1 − x∗ )i ≤ (1 − αn )kxn − x∗ kkxn+1 − x∗ k + αn 2
≤ (1 − αn )
kxn − x∗ k2 + kxn+1 − x∗ k2 2
+ αn
β2 kxn − x∗ k2 + kxn+1 − x∗ k2 2
+ αn hf (x∗ ) − x∗ , J(xn+1 − x∗ )i. Therefore,
kxn+1 − x∗ k2 ≤ [1 − γn ]kxn − x∗ k2 + γn λn , where γn = αn (1 − β ) and λn = 2
lim sup λn = lim sup n→∞
n→∞
2
1−β2
2 1 − β2
(3.4)
hf (x ) − x , J(xn+1 − x )i. It is easily seen that ∗
∗
∗
P∞
n=1
γn = ∞ and
hf (x∗ ) − x∗ , J(xn+1 − x∗ )i ≤ 0.
Applying Lemma 2.3 onto (3.4), we conclude that xn → x∗ .
Theorem 3.4. Let E be a strictly convex Banach space with a uniformly Gâteaux differentiable norm. Suppose that K , T, f , {αn } are as in Theorem 3.3. Let {xn } be defined by the following equation, xn+1 =
αn f (xn ) + (1 − αn )PTxn ,
(3.5)
where P is a sunny nonexpansive retraction of E onto K . If Kmin F (T ) 6= ∅, then as n → ∞, {xn } converges strongly to some fixed point x∗ of T . T Proof. Take x∗ ∈ Kmin F (T ). It follows from Lemma 2.1 that F (T ) = F (PT ). Since the mapping A = PT is a nonexpansive mapping from K onto itself, then (3.5) can be rewritten as follows: T
xn+1 = P (αn f (xn ) + (1 − αn )Axn ),
(3.6)
Thus, from Theorem 3.3, we obtain xn → x ∈ F (T ) = F (PT ). ∗
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We now give an example in which our condition x∗ ∈ Kmin F (T ) is easily satisfied (more examples see Song [28]). Corollary 3.6 also improves and extends [3, Theorem 4.2]. Recall the set A of E is a Chebyshev set, if ∀x ∈ E, there exactly exists the unique element y0 ∈ A such that kx − y0 k = d(x, A), where d(x, A) = inf y∈A kx − yk. T
Lemma 3.5 (See [29, Theorem 5.1.18, Corollary 5.1.19]). E is a reflexive and strictly convex Banach space if and only if its each nonempty closed convex subsets is a Chebyshev set. Corollary 3.6. Let E be a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and {xn } be defined by the Eq. (3.1) (or (3.5)). Suppose K , T, f are as in Theorem 3.3 and {αn } satisfies the conditions (C1), (C2) and (C3). Then as n → ∞, {xn } converges strongly to some x∗ ∈ F (T ). Proof. It follows from the reflexivity of E that Kmin 6= ∅. Clearly, Kmin is closed convex by the convexity and continuity of ϕ(y). From Lemma 3.2, we have that {xn } is bounded and limn→∞ kxn − xn+1 k = 0 and limn→∞ kxn − PTxn k = 0. Hence, for ∀x ∈ Kmin , we get that
ϕ(PTx) = µn kxn − PTxk2 = µn kPTxn − PTxk2 ≤ µn kxn − xk2 = ϕ(x). Hence, PTx ∈ Kmin . As x is arbitrary, then PT (Kmin ) ⊂ Kmin . Since F (PT ) = F (T ) 6= ∅ by Lemma 2.1, then let p be one of those. It follows from Lemma 3.5 that there exists unique x∗ ∈ Kmin such that
kp − x∗ k = inf kp − xk. x∈Kmin
By p = PTp = Tp and PTx∗ ∈ Kmin , we have
kp − PTx∗ k = kPTp − PTx∗ k ≤ kp − x∗ k. Hence x∗ = PTx∗ = Tx∗ by the uniqueness of x∗ ∈ Kmin . Thus x∗ ∈ Kmin ∩ F (T ). This satisfies the conditions of Theorem 3.3. Consequently, {xn } converges strongly to x∗ as n → ∞. Corollary 3.7. Let E be a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of E. Suppose that T : K → K is a nonexpansive mapping with F (T ) 6= ∅, and f : K → K is a fixed contractive mapping with the contractive coefficient β ∈ (0, 1). Let {xn } be defined by xn+1 =
αn f (xn ) + (1 − αn )Txn .
(3.7)
If {αn } satisfies the conditions (C1), (C2) and (C3) as Lemma 3.2, then as n → ∞, {xn } converges strongly to some x ∈ Kmin ∩ F (T ). ∗
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