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Strong convergence of modified implicit iterative algorithms with perturbed mappings for continuous pseudocontractive mappings
Applied Mathematics and Computation 209 (2009) 162–176
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Strong convergence of modified implicit iterative algorithms with perturbed mappings for continuous pseudocontractive mappings Lu-Chuan Ceng a,1, Adrian Petrusßel b,*,2, Jen-Chih Yao c,3 a b c
Department of Mathematics, Shanghai Normal University, Shanghai 200234, China Department of Applied Mathematics, Babesß-Bolyai University Cluj-Napoca, 1 Koga˘lniceanu Street, 400084 Cluj-Napoca, Romania Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan
a r t i c l e
i n f o
Keywords: Pseudocontractive mapping Modified viscosity iterative process with perturbed mapping Modified implicit iterative scheme with perturbed mapping Strong convergence Reflexive and strictly convex Banach space Uniformly Gâteaux differentiable norm
a b s t r a c t Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm. First purpose of this paper is to introduce a modified viscosity iterative process with perturbation for a continuous pseudocontractive self-mapping T and prove that this iterative process converges strongly to x* 2 F(T) :¼ {x 2 Xjx = T(x)}, where x* is the unique solution in F(T) to the following variational inequality: hf ðx Þ x ; jðv x Þi 6 0 for all
v 2 FðTÞ:
Second aim of the paper is to propose two modified implicit iterative schemes with perturbation for a continuous pseudocontractive self-mapping T and prove that these iterative schemes strongly converge to the same point x* 2 F(T). Basically, we show that if the perturbation mapping is nonexpansive, then the convergence property of the iterative process holds. In this respect, the results presented here extend, improve and unify some very recent theorems in the literature, see [L.C. Zeng, J.C. Yao, Implicit iteration scheme with perturbed mapping for common fixed points of a finite family of nonexpansive mappings, Nonlinear Anal. 64 (2006) 2507–2515; H.K. Xu, Viscosity approximation methods for nonexpansive mappings, J. Math. Anal. Appl. 298 (2004) 279–291; Y.S. Song, R.D. Chen, Convergence theorems of iterative algorithms for continuous pseudocontractive mappings, Nonlinear Anal. (2006)]. Ó 2009 Published by Elsevier Inc.
1. Introduction and preliminaries Let X be a real Banach space with norm kk and let X* be its dual. The normalized duality mapping J from X into the family of nonempty (by Hahn–Banach theorem) weak* compact subsets of its dual X* is defined by
JðxÞ ¼ ff 2 X : hx; f i ¼ kxk2 ¼ kf k2 g
for all x 2 X;
* Corresponding author. E-mail addresses: [email protected] (L.-C. Ceng), [email protected] (A. Petrusßel), [email protected] (J.-C. Yao). 1 This research was partially supported by the Teaching and Research Award Fund for Outstanding Young Teachers in Higher Education Institutions of MOE, China and the Dawn Program Foundation in Shanghai. 2 This research was partially supported by a grant of the Romanian Research Council. 3 This research was partially supported by a grant from the National Science Council. 0096-3003/$ - see front matter Ó 2009 Published by Elsevier Inc. doi:10.1016/j.amc.2008.10.062
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where h, i denotes the generalized duality pairing. It is known (see [4]) that if X* is strictly convex or X is a Banach space with a uniformly Gâteaux differentiable norm (see the definitions below), then J is single-valued and, in this case, it will be denoted by j. When {xn} is a sequence in X, xn ? x (respectively, xn N x, xn N *x) will denote strong (respectively, weak, weak*) convergence of the sequence {xn} to x. The norm of X is called Gâteaux differentiable (and X is called smooth) if
lim t!0
kx þ tyk kxk t
ðÞ
exists for each x, y in its unit sphere U = {x 2 X: kxk = 1}. Recall that X is smooth if and only if J is single-valued. In this case, J : X ? X* is continuous from the strong topology to the weak* topology. The norm of X is called uniformly Gâteaux differentiable if for each y 2 U, this limit is attached uniformly for x 2 U. The norm of X is called Fréchet differentiable if for each x 2 U, the limit in (*) is attained uniformly for y 2 U. The norm of X is called uniformly Fréchet differentiable and X is called uniformly smooth, if the limit in (*) is attained uniformly for (x, y) 2 U U. If X is uniformly smooth, then J : X ? X* is single-valued and uniformly continuous on bounded sets of X from the strong topology of X to the strong topology of X*. Since the dual X* of X is uniformly convex if and only if the norm of X is uniformly Fréchet differentiable, every Banach space with a uniformly convex dual is reflexive and has a uniformly Gâteaux differentiable norm. Recall also that if X has a uniformly Gâteaux differentiable norm, then J is uniformly continuous on bounded sets of X from the strong topology of X to the weak* topology of X*. For every e with 0 6 e 6 2, the modulus dX(e) of convexity of X is defined by
kx þ yk dx ðeÞ ¼ inf 1 : kxk 6 1; kyk 6 1; kx yk P e : 2
A Banach space X is called uniformly convex if dX(e) > 0 for every e > 0. X is called strictly convex if kx + yk/2< 1 for every x, y 2 U with x – y. It is also well known that X is uniformly convex if and only if its dual X* is uniformly smooth. In this case, X is reflexive and strictly convex. The converse implication is false. A discussion on these and related concepts may be found in [4,10]. Let K be a nonempty closed convex subset of X. Recall that a self-mapping f : K ? K is called: (i) k-contraction if k 2 (0, 1) and
kf ðxÞ f ðyÞk 6 kkx yk for all x; y 2 k; (ii) nonexpansive if
kT x T y k 6 kx yk for all x; y 2 K; Furthermore, there holds the following result (see [4, Theorem 4.5.3]): If K is nonempty closed convex subset of a strictly convex Banach space X and T : K ? K is a nonexpansive mapping (i.e., kTx Tyk 6 kx yk, for all x, y 2 K), then the fixed point set F(T) of T is a closed convex subset of K. The self-mapping T : K ? K is called pseudocontractive (respectively, strongly pseudo-contractive), if for each x, y 2 K, there exists j(x y) 2 J(x y) such that
hT x T y ; jðx yÞi 6 kx yk2 (respectively, hTx Ty, j(x y)i 6 bkx yk2 for some b 2 (0, 1)). In 2003, Xu and Ori [2] introduced the following implicit iteration process for a finite family of nonexpansive selfmappings:
xn ¼ an xn1 þ ð1 an ÞT n xn
for all n P 0;
ð1:1Þ
where the initial point x0 2 K is chosen arbitrarily and Tn = TnmodN. They proved that {xn} converges weakly to a common fixed point of Tn, n = 1, 2, . . . , N in a Hilbert space with {an} (0, 1). Very recently, Zeng and Yao in [6] proposed an implicit iteration scheme with perturbed mapping for approximation of common fixed points of a finite family of nonexpansive self-mappings of a Hilbert space. They established some convergence theorems for this implicit iteration scheme. In particular, necessary and sufficient conditions for strong convergence of this implicit iteration scheme were obtained. Meantime, Chen et al. [3] extended the study of the iteration process (1.1) to a finite family of continuous pseudocontractive self-mappings. They established several strongly and weakly convergent theorems. On the other hand, in 2004, Xu [7] studied the following viscosity iteration process in a uniformly smooth Banach space X for a fixed contractive mapping f : K ? K and a nonexpansive mapping T : K ? K (where K is a nonempty closed convex subset of X),
xt ¼ tf ðxt Þ þ ð1 tÞTxt +
and proved that as t ? 0 , xt ? p 2 F(T).
ð1:2aÞ
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For a continuous pseudocontractive mapping T : K ? K, a nonexpansive mapping S : K ? K and a fixed Lipschitz strongly pseudocontractive mapping f : K ? K, we define the mapping T(S,f) : K ? K as follows:
TðS;f Þ ¼ tf þ r t S þ ðl t r t ÞT; where 0 < rt < 1 t. Then it is easy to see that the mapping T(S,f) is a continuous strongly pseudocontractive self-mapping of K. Therefore, T(S,f) has a unique fixed point in K [9, Corollary 2], i.e., for each given t 2 (0, 1) there exists xt 2 K such that
xt ¼ tf ðxt Þ þ rt Sxt þ ð1 t r t ÞTxt :
ð1:2bÞ
The above iterative process (1.2b) is called the modified viscosity iterative process with perturbed mapping, where S is called the perturbed mapping. We remark that if S = I and rt = 0 for all t 2 (0, 1), then (1.2b) reduces to (1.2a). Very recently, Song and Chen [11] proposed the following iteration process: Given any x0 2 K,
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ ð1 bn Þxn1 :
ð1:3Þ
They obtained strong convergence theorems for viscosity iterative process (1.2a) with continuous pseudocontractive mapping T and for the modified implicit iteration scheme (1.3) in a real reflexive and strictly convex Banach space X with a uniformly Gâteaux differentiable norm. Inspired by Song and Chen [11] and Zeng and Yao [6], we introduce and study, for a continuous pseudocontractive selfmapping T, a nonexpansive self-mapping S and a fixed contractive self-mapping f, the following modified implicit iterative algorithms with perturbed mappings: Algorithm 1.1. Let {an} (0, 1], {bn} (0, 1] and {cn} [0, 1] such that limn(cn/bn) = 0 and bn + cn < 1. Starting with an arbitrary initial point x0 2 K, generate a sequence {xn} via the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 bn cn Þxn :
ð1:4aÞ
Algorithm 1.2. Let {an}, {bn} and {cn} be as in Algorithm 1.1. Starting with an arbitrary initial point x0 2 K, generate a sequence {xn} via the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 :
ð1:4bÞ
We remark that if S = I and cn = 0, for all n > 0, then iterative scheme (1.4b) reduces to (1.3). In a real reflexive and strictly convex Banach space X with a uniformly Gâteaux differentiable norm, we not only prove that the modified viscosity iterative process (1.2b) with perturbed mapping for continuous pseudocontractive mapping T converges strongly to p 2 F(T) as t ? 0+, but also show that the modified implicit iterative schemes (1.4a) and (1.4b) with perturbed mappings converge strongly to the same fixed point p of T, which is the unique solution in F(T) to the following variational inequality:
hðf IÞp; jðu pÞi 6 0 for all u 2 FðTÞ: Recall that S : K ? K is called accretive if I S is pseudocontractive. We denote by Jr the resolvent of S, i.e., Jr = (I + rS)1. It is well known that Jr is nonexpansive, single-valued and F(Jr) = S1(0) = {z 2 D(S):0 = Sz} for all r > 0. For more details, see [1,4,9]. Let (E, d) be a metric space. Let x 2 E and C E. Denote
dðx; CÞ :¼ inf dðx; v Þ: m2C
Recall that C is called a Chebyshev set, if for each x 2 E, there exists a unique element y 2 C such that d(x, y) = d(x, C). It is well known that every nonempty closed convex subset of a strictly convex and reflexive Banach space X is a Chebyshev set [11, Corollary 5.1.19]. Let K be a nonempty closed convex subset of a real Banach space X and T : K ? K be a pseudocontractive map, then I T is accretive. We denote A = J1 = (2I T)1. Then F(A) = F(T) and the operator A : R(2I T) ? K is nonexpansive and single-valued. Next we state several lemmas which will be used in the sequel. The first lemma can be found in [11]. Lemma 1.1 [11, Lemma 1.1]. Let K be a nonempty closed convex subset of a real Banach space X and T : K ? K be a continuous pseudocontractive map. We denote A = (2I T)1. Then (i) [10, Theorem 6]. The map A is nonexpansive self-mapping on K, i.e., for all x, y 2 K, there hold
kAx Ayk 6 kx yk and Ax 2 K: (ii) If limn?1kxn Txnk = 0, then limn?1kxn Axnk = 0.
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The second lemma can be found in [7,8]. Lemma 1.2. Let {an} be a sequence of nonnegative real numbers satisfying the condition
anþ1 6 ð1 bn Þan þ bn cn
for all n P 0;
where {bn}, {cn} are sequences of real numbers such that P (i) fbn g ½0; 1; 1 n¼0 bn ¼ 1; (ii) lim supn?1 cn 6 0. Then, limn?1an = 0. Let l be a mean on positive integers N, i.e., a continuous linear functional l on l1 satisfying klk = 1 = l(1). Then we know that l is a mean on N if and only if
inffan : n 2 Ng 6 lðaÞ supfan : n 2 Ng for every a = (a1, a2, . . .) 2 l1. According to time and circumstances, we use ln(an) instead of l(a). A mean l on N is called a Banach limit if
ln ðan Þ ¼ ln ðanþ1 Þ for every a = (a1, a2, . . .) 2 l1. Using the Hahn–Banach theorem, we can prove the existence of a Banach limit. We know that if l is a Banach limit, then
lim inf n!1 an 6 ln ðan Þ 6 lim sup an n!1
1
for every a = (a1, a2, . . .) 2 l . Notice that the following lemma can be found in [5, Lemma 1,4, Lemma 4.5.4]. Lemma 1.3. Let C be a nonempty closed convex subset of a Banach space X with a uniformly Gâteaux differentiable norm. Suppose {xn} is a bounded sequence of X and l is a mean on N. Let z 2 C. Then
ln kxn zk2 ¼ min ln kxn yk2 y2C
if and only if
ln hy z; jðxn zÞi 6 0 for all y 2 C; where j is the normalized duality mapping of X.
2. Modified viscosity iterative algorithm with perturbed mapping Our first main result is the following. Theorem 2.1. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0, S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. Let {xt} be defined by
xt ¼ tf ðxt Þ þ rt Sxt þ ðl t rt ÞTxt
for all t 2 ð0; 1Þ;
ð2:1Þ +
where 0 6 rt < 1 t for all t 2 (0, 1) and limt?0 + (rt/t) = 0. Then as t ? 0 , xt converges strongly to some fixed point p of T such that p is the unique solution in F(T) to the following variational inequality:
hðf IÞp; jðu pÞi 6 0 for all u 2 FðTÞ:
ð2:2Þ
Proof. First, we show the uniqueness of a solution to the variational inequality (2.2). Indeed, suppose p, q 2 F(T) satisfy (2.2). Then we have
hðf IÞp; jðq pÞi 6 0
ð2:3Þ
hðf IÞq; jðp qÞi 6 0:
ð2:4Þ
and
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Adding up (2.3) and (2.4), we have
0 P hðI f Þp ððI f ÞqÞ; jðp qÞi ¼ kp qk2 hf ðpÞ f ðqÞ; jðp qÞi P ð1 bÞkp qk2 : Hence we have p = q. Below we use p 2 F(T) to denote the unique solution of (2.2). Taking a fixed u 2 F(T), we have
kxt uk2 ¼ htðf ðxt Þ uÞ þ r t ðSxt uÞ þ ð1 t r t ÞðTxt uÞ; jðxt uÞi ¼ thf ðxt Þ u; jðxt uÞi þ r t hSxt u; jðxt uÞi þ ð1 t r t ÞhTxt u; jðxt uÞi 6 thf ðxt Þ f ðuÞ; jðxt uÞi þ thf ðuÞ u; jðxt uÞi þ r t hSxt Su; jðxt uÞi þ r t hSu u; jðxt uÞi þ ð1 t r t Þkxt uk2 6 tbkxt uk2 þ thf ðuÞ u; jðxt uÞi þ rt kxt uk2 þ r t hSu u; jðxt uÞi þ ð1 t rt Þkxt uk2 ¼ btkxt uk2 þ thf ðuÞ u; jðxt uÞi þ r t hSu u; jðxt uÞi þ ð1 tÞkxt uk2 : Hence
ð1 bÞkxt uk2 6 hf ðuÞ u; jðxt uÞi þ ðrt =tÞhSu u; jðxt uÞi 6 kf ðuÞ uk kxt uk þ ðr t =tÞkSu uk kxt uk;
ð2:5Þ
which implies that
kxt uk 6
1 ðkf ðuÞ uk þ ðr t =tÞkSu ukÞ: 1b
Since limt?0 + (rt/t) = 0 and limt?0 + (rt/t) = 0, there exists t0 2 (0, 1) such that 0 < rt/t < 1/2 and 0 < t + rt 6 1/2 for all t 2 (0, t0). So, {xt : t 2 (0, t0)} is bounded. h Since f is a Lipschitz mapping and S is a nonexpansive mapping, kf(xt) f(u)k 6 Lkxt uk and kSxt Suk 6 kxt uk Then the sets {f(xt) : t 2 (0, t0)} and {Sxt : t 2 (0, t0)} are bounded. Note that xt = tf(xt) + rtSxt + (1 t rt)Txt. So we have
Txt ¼
1 t rt xt f ðxt Þ Sxt ; 1 t rt 1 t rt 1 t rt
which implies that
kTxt k 6
1 t rt kxt k þ kf ðxt Þk þ kSxt k: 1 t rt 1 t rt 1 t rt
Thus we have
kTxt k 6 2kxt k þ 2tkf ðxt Þk þ 2rt kSxt k;
for all t 2 ð0; t 0 Þ
and so {Txt : t 2 (0, t0)} is bounded. This implies that
lim kxt Txt k 6 lim tkf ðxt Þ Txt k þ lim rt kSxt Txt k ¼ 0:
t!0þ
t!0þ
t!0þ
From Lemma 1.1 we know that the mapping A = (2I T)1 : K ? K is nonexpansive, and F(A) = F(T) and limt?0 + kxt Axtk = 0, where I denotes the identity operator. We claim that the set {xt : t 2 (0, t0)} is relatively compact. Indeed, suppose xn := xtn and
gðxÞ ¼ ln kxn xk2
for all x 2 K;
where {tn} is a sequence in (0, t0) that converges to 0(n ? 0) and ln is a Banach limit. Define the set
K 1 ¼ fx 2 K : gðxÞ ¼ inf gðyÞg: y2K
Since X is a reflexive Banach space, then K1 is a nonempty bounded closed convex subset of X (see [4, Theorem 1.3.11]). Since limn?akxn Axnk = 0, we deduce that for each x 2 K1
gðAxÞ ¼ ln kxn Axk2 6 ln ðkxn Axn k þ kAxn AxkÞ2 6 ln kxn xk2 ¼ gðxÞ: Hence, Ax 2 K1. This shows that K1 is invariant under A. Since F(A) = F(T) – ;, we can take u 2 F(A) = F(T) arbitrarily. Since every nonempty closed convex subset of a strictly convex and reflexive Banach space X is a Chebyshev set (according to [10, Corollary 5.1.19]), there exists a unique q 2 K1 such that
L.-C. Ceng et al. / Applied Mathematics and Computation 209 (2009) 162–176
167
ku qk ¼ inf ku xk: x2K 1
Since u = Tu = Au and Aq 2 K1, we have
ku Aqk ¼ kAu Aqk 6 ku qk: Hence q = Aq. By Lemma 1.3, we have
ln hx q; jðxn qÞi 6 0 for all x 2 K: Taking x = f(q) 2 K in the last inequality, we conclude from (2.5) that
ln kxn qk2 6
1 l hf ðqÞ q; jðxn qÞi 6 0: 1b n
This implies that
ln kxn qk2 ¼ 0: Hence, there exists a subsequence of {xn} which converges strongly to q 2 F(T). Without loss of generality, we may assume that {xn} converges strongly to q 2 F(T). Next we show that q is a solution in F(T) to the variational inequality (2.2). Since u is a fixed point of T and since xt = tf(xt) + rtSxt + (1 t rt)Txt, so we deduce from the accretivity of I T that
hxt f ðxt Þ; jðxt uÞi ¼ ð1 tÞhTxt f ðxt Þ; jðxt yÞi þ r t hsxt Txt ; jðxt uÞi ¼ ð1 tÞhðI TÞxt ðI TÞu; jðxt uÞi þ ð1 tÞhxt f ðxt Þ; jðxt uÞi þ r t hSxt Txt ; jðxt uÞi 6 ð1 tÞhxt f ðxt Þ; jðxt uÞi þ r t hSxt Txt ; jðxt uÞi; hxt f ðxt Þ; jðxt uÞi 6 ðr t =tÞhSxt Txt ; jðxt uÞi:
ð2:6Þ
Since the sets {xt u : t 2 (0, t0)},{Sxt : t 2 (0, t0)} and {Txt : t 2 (0, t0)} are bounded, it follows from limt?0+(rt/t) = 0 that
lim ðr t =tÞhSxt Txt ; jðxt uÞi ¼ 0:
t!0þ
Note that the duality map J is single-valued and norm-to-weak* uniformly continuous on bounded sets of a Banach space X with uniformly Gâteaux differentiable norm. Since xn ? q as n ? 1, we get that for each u 2 F(T), we have
kðI f Þxn ðI f Þqk ! 0 as n ! 1 and
jhðI f Þxn ; jðxn uÞi hðI f Þq; jðq uÞij ¼ jhðI f Þxn ðI f Þq; jðxn uÞi þ hðI f Þq; jðxn uÞ jðq uÞij 6 kðI f Þxn ðI f Þqkkxn uk þ jhðI f Þq; jðxn uÞ jðq uÞij ! 0 as n ! 1: Therefore, it follows immediately from (2.6) that for each u 2 F(T),
hðI f Þq; jðq uÞi ¼ lim hxn f ðxn Þ; jðxn uÞi 6 0: n!1
Consequently, q 2 F(T) is a solution of (2.2); hence q = p by uniqueness. All in all, we have proved that the set {xt} is relatively compact and every cluster point of {xt} (as t ? 0+) equals p. Therefore, xt ? p as t ? 0+. Using the above Theorem 2.1, we can easily obtain the following corollary. Corollary 2.2 [11, Theorem 2.1]. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Suppose that T : K ? K is a continuous pseudocontractive mapping with F(T) – ; and f : K ? K is a Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0. Let {xt} be defined by
xt ¼ tf ðxt Þ þ ð1 tÞTxt
for all t 2 ð0; 1Þ:
ð2:7Þ
+
Then as t ? 0 , xt converges strongly to some fixed point p of T such that p is the unique solution in F(T) to the variational inequality (2.2). Proof. Put S = I and rt = 0 for all t 2 (0, 1). Then the conclusion follows immediately by Theorem 2.1. h A second result is: Theorem 2.3. Let K be a nonempty closed convex subset of a real Banach space X with a uniformly Gâteaux differentiable norm. Let f : K ? K be a fixed Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0, S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. If there exists a bounded sequence {xn} such that limn?1kxn Txnk = 0 and p ¼ limt!0þ zt exists, where {zt} is defined by (2.1), then
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lim suphf ðpÞ p; jðxn pÞi 6 0: n!1
Proof. Using the equality
zt xn ¼ ð1 t rt ÞðTxt xn Þ þ tðf ðzt Þ xn Þ þ r t ðSxt xn Þ and the inequality
hTx Ty; jðx yÞi 6 kx yk2 ; we obtain that
kzt xn k2 ¼ ð1 t rt ÞhTzt xn ; jðzt xn Þi þ thf ðzt Þ xn ; jðzt xn Þi þ r t hSzt xn ; jðzt xn Þi ¼ ð1 t rt ÞðhTzt Txn ; jðzt xn Þi þ hTxn xn ; jðzt xn ÞiÞ þ thf ðzt Þ zt ; jðzt xn Þi þ tkzt xn k2 þ r t hSzt zt ; jðzt xn Þi þ r t kzt xn k2 6 kzt xn k2 þ kTxn xn kkzt xn k þ thf ðxt Þ zt ; jðzt xn Þi þ r t kSzt zt kkzt xn k and hence
hf ðzt Þ zt ; jðxn zt Þi 6
kTxn xn k rt kzt xn k þ kSzt zt kkzt xn k: t t
ð2:8Þ
Note that as in the proof of Theorem 2.1, there exists t0 2 (0, 1) such that {zt : t 2 (0, t0)} and {Szt : t 2 (0, t0)} are bounded. Since {xn} is bounded and limn?1kxn Txnk = 0, taking the upper limit in (2.8), we have
lim suphf ðzt Þ zt ; jðxn zt Þi 6
rt kSzt zt kðkzt k þ supnP0 kxn kÞ: t
ð2:9Þ
Meantime, we also have
lim
t!0þ
rt kSzt zt kðkzt k þ sup kxn kÞ ¼ 0: t nP0
ð2:10Þ
On the other hand, since the set {zt xn : t 2 (0, t0) and n P 0} is bounded and the duality map J is single-valued and norm-toweak* uniformly continuous on bounded sets in the Banach space X with uniformly Gâteaux differentiable norm, thus it follows from zt ? p(t ? 0+) that
kðf ðzt Þ zt Þ ðf ðpÞ pÞk 6 kðf ðzt Þ f ðpÞk þ kzt pk ! 0; jhf ðzt Þ zt ; jðxn zt Þi hf ðpÞ p; jðxn pÞij ¼ jhðf ðzt Þ zt Þ ðf ðpÞ pÞ; jðxn zt Þi þ hf ðpÞ p; jðxn zt Þ jðxn pÞij
ð2:11Þ
6 jhf ðpÞ p; jðxn zt Þ jðxn pÞij þ kðf ðzt Þ zt Þ ðf ðpÞ pÞkkxn zt k ! 0 as t ! 0þ : Thus, from (2.10) and (2.11) we conclude that for an arbitrary e > 0, there exists d 2 (0, t0) such that for all t 2 (0, d) and all n P 0,
rt e kSzt zt kðkzt k þ supnP0 kxn kÞ < t 2 and
jhf ðzt Þ zt ; jðxn zt Þi hf ðpÞ p; jðxn pÞij <
e 2
;
which implies that
e
hf ðpÞ p; jðxn pÞi < hf ðzt Þ zt ; jðxn zt Þi þ : 2
ð2:12Þ
Consequently, for all t 2 (0, d) we derive from (2.9) and (2.12)
lim suphf ðpÞ p; jðxn pÞi 6 lim suphf ðzt Þ zt ; jðxn zt Þi þ n!1
n!1
Thus, since e is arbitrary, we infer that
lim suphf ðpÞ p; jðxn pÞi 6 0: n!1
The proof is complete.
h
e 2
6
rt e e e kSzt zt kðkzt k þ sup kxn kÞ þ < þ ¼ e: t 2 2 2 nP0
L.-C. Ceng et al. / Applied Mathematics and Computation 209 (2009) 162–176
169
3. Modified implicit iteration scheme with perturbed mapping We consider now the implicit iteration scheme with perturbed mapping given in Algorithm 1.1. Theorem 3.1. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed contractive mapping with contractive coefficient K 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 bn cn Þyn ;
ð3:1Þ
where {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; limn!1 bn ¼ 0; 1 n¼0 bn ¼ 1; (iii) {cn} (0, 1], limn?1(cn/bn) = 0, bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T) where x* is the unique solution in F(T) to the variational inequality (2.2). Proof. At first, we show that {xn} is bounded. Taking a fixed p 2 F(T), we have
kyn pk2 ¼ han xn þ ð1 an ÞTyn p; jðyn pÞi ¼ ð1 an ÞhTyn p; jðyn pÞi þ an hxn p; jðyn pÞi 6 ð1 an Þkyn pk2 þ an kxn pkkjðyn pÞk 6 ð1 an Þkyn pk2 þ an kxn pkkðyn pÞk and hence
kyn pk2 6 kxn pkkyn pk: So, kyn pk 6 kxn pk for all n P 0. Thus we have
xnþ1 pk 6 bn kf ðyn Þ pk þ cn kSyn pk þ ð1 bn cn Þkyn pk 6 bn ðkf ðyn Þ f ðpÞk þ kf ðpÞ pkÞ þ cn ðkSyn Spk þ kSp pkÞ þ ð1 bn cn Þkxn pk 6 bn kkyn pk þ bn kf ðpÞ pk þ cn kyn pk þ cn kSp pk þ ð1 bn cn Þkxn pk 6 bn kkxn pk þ bn kf ðpÞ pk þ cn kxn pk þ cn kSp pk þ ð1 bn cn Þkxn pk ¼ ð1 ð1 kÞbn Þkxn pk þ bn kf ðpÞ pk þ cn kSp pk: Since limn?1(cn/bn) = 0 we may assume, without loss of generality, that cn 6 bn for all n P 0. This implies that
1 kxnþ1 pk 6 ð1 ð1 kÞbn Þkxn pk þ ð1 kÞbn ðkf ðpÞ pk þ kSp pkÞ 1k 1 ðkf ðpÞ pk þ kSp pkÞ : 6 max kxn pk; 1k By induction, we drive
kxn pk 6 max kxo pk;
1 ðkf ðpÞ pk þ kSp pkÞ for all n P 0 1k
This shows that {xn} is bounded, and so is {yn}. Observe that
kf ðyn Þk 6 kf ðyn Þ f ðpÞk þ kf ðpÞk 6 kkyn pk þ kf ðpÞk and
kSyn k 6 kSyn Spk þ kSpk 6 kyn pk þ kSpk: Thus, {f(yn)} and {Syn} are bounded. Since limn?1an = 0, there exist n0 P 0 and a 2 (0, 1), such that an 6 a a for all n P n0. Note that yn = anxn + (1 an)Tyn. Hence we have
Tyn ¼
1 an y xn 1 an n 1 an
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and so
kTyn k 6
1 an 1 a ky k þ kxn k 6 ky k þ kxn k: 1 an n 1a n 1a 1 an
Consequently, the sequence {Tyn} is also bounded. From condition (i), we obtain
kyn Tyn k ¼ an kxn Tyn k ! 0 ðn ! 1Þ:
ð3:2Þ +
Put zt = tf(zt) + rtSzt + (1 t rt)Tzt. Then it follows from Theorem 2.1that as t ? 0 , zt converges strongly to some fixed point x* of T such that x* is the unique solution in F(T) to the variational inequality (2.2). It follows from (3.2) and Theorem 2.4 that
lim suphf ðx Þ x ; jðyn x Þi 6 0:
ð3:3aÞ
n!1
Furthermore, by (3.1) and (3.2) we have
kxn þ 1 yn k ¼ kbn f ðyn Þ þ cn Syn þ ð1 b cn Þyn ðan xn þ ð1 an ÞTyn Þk 6 an kxn Tyn k þ bn kf ðyn Þ yn k þ cn kSyn yn k þ kyn Tyn k ! 0 ðn ! 1Þ: Since the duality map J is single-valued and norm-to-weak* uniformly continuous on bounded sets in Banach space X with uniformly Gâteaux differentiable norm, we have
lim hf ðx Þ x ; jðxnþ1 x Þ jðyn x Þi ¼ 0:
n!1
So we obtain from (3.3a)
lim suphf ðx Þ x ; jðxnþ1 x Þi 6 lim suphf ðx Þ x ; jðyn x Þi þ lim suphf ðx Þ x ; jðxnþ1 x Þ jðyn x Þi n!1
n!1
n!1
¼ lim suphf ðx Þ x ; jðyn x Þi 6 0:
ð3:3bÞ
n!1
Finally we show that xn ? x* as n ? 1. Indeed, using the following inequalities: kyn x*k 6 kxn x*k, kf(x) f(y)kk 6 kkx yk and kSx Syk 6 kx yk, we have
kxnþ1 x k2 ¼ bn hf ðyn Þ x ; jðxnþ1 x Þi þ cn hSyn x ; jðxnþ1 x Þi þ ð1 bn cn Þhyn x ; jðxnþ1 x Þi 6 bn ðhf ðyn Þ f ðx Þ; jðxnþ1 x Þi þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðhSyn Sx ; jðxnþ1 x Þi þ hSx x ; jðxnþ1 x ÞiÞ þ ð1 bn cn Þkyn x kkjðxnþ1 x Þk 6 bn ðkf ðyn Þ f ðx Þkkjðxnþ1 x Þk þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðkSyn Sx kkjðxnþ1 x Þk þ kSx x kkjðxnþ1 x ÞkÞ þ ð1 bn cn Þkxn x kkjðxnþ1 x Þk 6 bn ðkkyn x kkxnþ1 x k þ hf ðx Þ x ; jðxnþl x ÞiÞ þ cn ðkyn x kkxnþi x k þ kSx x kkxnþ1 x kÞ þ ð1 bn cn Þkxn x kkxnþ1 x k 6 bn ðkkxn x kkxnþ1 x k þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðkxn x kkxnþ1 x k þ kSx x kkxnþ1 x kÞ þ ð1 bn cn Þkxn x kkxnþ1 x k ¼ ð1 ð1 kÞbn Þkxn x kkxnþ1 x k þ bn hf ðx Þ x ; jðxnþ1 x Þi þ cn kSx x kkxnþ1 x k ¼ ð1 ð1 kÞbn
kxn x k2 kxnþ1 x k2 þ bn hf ðx Þ x ; jðxnþ1 x Þi þ cn kSx x kkxnþ1 x k; 2
which hence implies that
1 ð1 kÞbn 2bn 2cn kxn x k2 þ hf ðx Þ x ; jðxnþ1 x Þi þ kSx x kkxnþ1 x k 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 þ ð1 kÞbn 2ð1 kÞbn 2ð1 kÞbn 1 c ¼ 1 ½hf ðx Þ x ; jðxnþ1 x Þi þ n kSx x kkxnþ1 x k: kxn x k2 þ 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 k bn
kxnþ1 x k2 6
2ð1kÞbn Put bn ¼ 1þð1kÞb and n
cn ¼
1 c ½hf ðx Þ x ; jðxnþ1 x i þ n kSx x kkxnþ1 x k: 1k bn
Then the last inequality can be rewritten as
kxnþ1 x k2 6 ð1 bn Þkxn x k2 þ bn cn :
ð3:4Þ
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Since *
P1
n¼o bn *
¼ 1, we have
P1
bn n¼0 1þð1kÞbn
¼ 1 and hence
P1
n¼0 bn
¼ 1. Note that limn?1(cn/bn = 0 and lim supn?1h f(x*)
x ,j(xn+1 x )i 6 0 due to (3.3b). Thus, according to the boundedness of {xn x*}, we have lim supn?1cn 6 0. Therefore, applying Lemma 1.2 to (3.4), we infer that limn?1kxn x*k = 0. The proof is complete. h Example 1. Let X = R2 with inner product h, i and norm k k defined by
hx; yi ¼ ac þ bd and kxk ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a2 þ b
2 for all x, y 2 R2 with x = (a, b) and y= (c, d). Let K = {x 2 R : kxk 6 1}. Let A be a 2 2 positively definite matrix such that 2 1 pffiffiffi kAk = 1, for instance, putting A ¼ 41 42 and u = (k, k) with jkj 6 2=2, we know that kAk = 1, u 2 K and Au = u. Given 3
3
j 2 (0, 1) we define the mappings T, S, f : K ? K as follows: T ¼ A; S ¼ A2
and f ¼ jA:
It is easy to see that T(0) = S(0) = f(0) = 0. Pick three sequences {an}, {bn}, {cn} in [0, 1] satisfying conditions (i)–(iii) in Theorem 3.1, where
bn ¼
^lð^l þ nÞ
8n P 0
ð^l þ n þ 1Þ2
and ^l ¼ 2=ð1 jÞ. For x0 2 K, let {xn} be given by the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 b n cn Þyn :
First, observe that
kyn 0k 6 an kxn 0k þ ð1 an ÞkTyn 0k 6 an kxn 0k þ ð1 an Þkyn 0k and hence
kyn 0k 6 kxn 0k 8 P 0: Second, observe that
kxnþ1 0k 6 bn kf ðyn Þ 0k þ cn kSyn 0k þ ð1 b cn Þkyn 0k 6 bn jkyn 0k þ cn kyn 0k þ ð1 bn cn Þkyn 0k ¼ ð1 ð1 jÞbn Þkyn 0k 6 ð1 ð1 jÞbn Þkxn 0k 6 ð1 ð1 jÞbn Þ ð1 ð1 jÞb0 Þkx0 0k ¼
n Y
ð1 ð1 jÞbj Þkx0 0k:
j¼0
Since 0 6 j 6 1 and
P1
n¼0 bn
¼ 1, we have
1 Y
ð1 ð1 jÞbn Þ ¼ lim
n!1
n¼0
n Y
ð1 ð1 jÞbj Þ ¼ 0;
j¼0
which, hence, implies that {xn} converges to 0 2 F(T). On the other hand, we claim that there holds the convergence rate estimate kxm 0k = O(1/m) for bn ¼ ^lð^l þ nÞ=ð^l þ n þ 1Þ2 ; 8n P 0 where ^l ¼ 2=ð1 jÞ. Indeed, by making use of the last estimation we derive for all n 6 0
kxnþ1 0k 6 ð1 ð1 jÞbn Þkxn 0k ¼ ¼
! ð^l þ nÞ2 þ 1 kxn 0k ð^l þ n þ 1Þ2
! ^l þ n 2 kxn 0k ¼ 1 ð1 jÞ 1 j ð^l þ n þ 1Þ2
and hence
ð^l þ n þ 1Þ2 kxnþ1 0k ð^l þ nÞ2 kxn 0k 6 kxn 0k: Summing this inequality from n = 0 to m 1(m P 1), we obtain
ð^l þ mÞ2 kxm 0k ^l2 kx0 0k 6 mkx0 0k:
1
ð2^l þ nÞ ð^l þ n þ 1Þ2
!
kxn 0k
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Thus,
kxm 0k 6
m þ ^l2 kx0 0k ðm þ ^lÞ2
from which it follows that
1 : kxm 0k ¼ O m
n pffiffiffi o Finally, it is easy to see that FðTÞ ¼ u 2 K : u ¼ ðk; kÞ with jkj 6 2=2 . Moreover, it is clear that p = 0 is the unique solution in F(T) to the following variational inequality:
hðf IÞp; u pÞ 6 0 for all u 2 FðTÞ: The second main result of this section deals with the scheme introduced in Algorithm 1.2. Theorem 3.2. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed contractive mapping with contractive coefficient k 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 ;
ð3:5Þ
where {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0,1], limn!1 an ¼ 0; P (ii) fbn g ð0; 1; 1 n¼0 bn ¼ 1; (iii) limn?1(cn/bn) = 0, bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T), where x* is the unique solution in F(T) of the variational inequality (2.2). Proof. At first, we show that {xn} is bounded. Taking a fixed p 2 F(T), we have
kxn pk2 ¼ han yn þ ð1 an ÞTxn p; jðxn pÞi ¼ ð1 an ÞhTxn p; jðxn pÞi þ an hyn p; jðxn pÞi 6 ð1 an Þkxn pk2 þ an kyn pkkjðxn pÞk 6 ð1 an Þkxn pk2 þ an kyn pkkxn pk and hence
kxn pk2 6 kyn pkkxn pk: So kxn p 6 kyn pk for all n P 0. Thus we have
kxn pk 6 kyn pk 6 bn kf ðxn1 Þ pk þ cn kSxn1 pk þ ð1 bn cn Þkxn1 pk 6 bn ðkf ðxn1 Þ f ðpÞk þ kf ðpÞ pkÞ þ cn ðkSxn1 Spk þ kSp pkÞ þ ð1 bn cn Þkxn1 pk 6 ð1 ð1 kÞbn Þkxn1 pk þ bn kf ðpÞ pk þ cn kSp pk: Since limn?1(cn/bn) = 0, we may assume, without loss of generality, that cn 6 bn for all n P 0. This implies that
1 kxn pk 6 ð1 ð1 kÞbn Þkxn1 pk þ ð1 kÞbn ðkf ðpÞ pk þ kSp pkÞ 1k 1 ðkf ðpÞ pk þ kSp pkÞ : 6 max kxn1 pk; 1k Further, we derive
kxn pk 6 max kx0 pk;
1 ðk=f ðpÞ pk þ kSp pkÞ for all n P 0; 1k
kf ðxn Þk 6 kf ðxn Þ f ðpÞk þ kf ðpÞk 6 kkxn pk þ kf ðpÞk; and
kSxn k 6 kSxn Spk þ kSpk 6 kxn pk þ kSpk: Thus, {xn}, {yn}, {f(xn)} and {Sxn} are bounded. Since limn?1an = 0, there exist n0 P 0 and a 2 (0, 1), such that an 6 a for all n P n0. Note that xn = anyn + (1 an)Txn. Hence we have
Txn ¼
1 an xn y 1 an 1 an n
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and so
kTxn k 6
1 an 1 a kxn k þ ky k 6 kxn k þ ky k: 1 an 1a 1a n 1 an n
Consequently, the sequence {Txn} is also bounded. From condition (i), we obtain
kxn Txn k ¼ an kyn Txn k ! 0 ðn ! 1Þ:
ð3:6Þ +
Put zt = tf(zt) + rtSzt + (1 t rt)Tzt. Then it follows from Theorem 2.1 that as t ? 0 , zt converges strongly to some fixed point x* of T such that x* is the unique solution in F(T) to the variational inequality (2.2). It follows from (3.6) and Theorem 2.4 that
lim suphf ðx Þ x ; jðxn x Þi 6 0:
ð3:7Þ
n!1
Finally we show that xn ? x* (n ? 1). Indeed, using inequalities: kSx Syk 6 kx yk,
hTx Ty; jðx yÞi 6 kx yk2
and kf ðxÞ f ðyÞk 6 kkx yk;
we get that
kxn x k2 ¼ ð1 an ÞhTxn x ; jðxn x Þi þ an hyn x ; jðxn x Þi 6 ð1 an Þkxn x k2 þ an ð1 bn cn Þhxn1 x ; jðxn x Þi þ an bn hf ðxn1 Þ x ; jðxn x Þi þ an cn hSxn1 x ; jðxn x Þi: Thus
kxn x k2 6 ð1 bn cn Þhxn1 x ; jðxn x Þi þ bn hf ðxn1 Þ x ; jðxn x Þi þ cn hSxn1 x ; jðxn x Þi 6 ð1 bn cn Þkxn1 x kkjðxn x Þk þ bn ðkkxn1 x kkjðxn x Þk þ hf ðx Þ x ; jðxn x ÞiÞ þ cn ðkxn1 x kkjðxn x Þk þ hSx x ; jðxn x ÞiÞ ¼ ð1 ð1 kÞbn Þkxn1 x kkjðxn x Þk þ bn hf ðx Þ x ; jðxn x Þi þ cn hSx x ; jðxn x Þi 6 ð1 ð1 kÞbn Þ
kxn1 x k2 þ kxn x k2 þ bn hf ðx Þ x ; jðxn x Þi þ cn hSx x ; jðxn x Þi; 2
which hence implies that
1 ð1 kÞbn 2bn 2cn kxn1 x k2 þ hf ðx Þ x ; jðxn x Þi þ kSx x kkxn x k 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 þ ð1 kÞbn 2ð1 kÞbn 2ð1 kÞbn 1 c ¼ 1 hf ðx Þ x ; jðxn x Þi þ n kSx x kkxn x k : kxn1 x k2 þ 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 k bn
kxn x g2 6
2ð1kÞbn Put bn ¼ 1þð1kÞb and n
cn ¼
1 c hf ðx Þ x ; jðxnþ1 x Þi þ n kSx x kkxn x k : 1k bn
Then the last inequality can be rewritten as
kxn x k2 6 ð1 bn Þkxn1 x k2 þ bn cn : Since
P1
n¼0 bn
¼ 1, we have
P1
bn n¼0 1þð1kÞbn
¼ 1 and hence
ð3:8Þ P1
n¼0 bn
*
¼ 1. Note that limn?1(cn/bn) = 0 and lim supn?1hf(x ) x*,
*
j(xn x )i 6 0 due to (3.7). Thus, according to the boundedness of {xn x*}, we have lim supn?1cn 6 0. Therefore, applying Lemma 1.2 to (3.8), we infer that limn?1kxn x*k = 0. The proof is complete. h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Example 2. Let X = R2 with inner product h, i and norm k k defined by hx, yi = ac + bd and kxk ¼ a2 þ b for all x, y 2 R2 with x = (a, b) and y = (c, d). Let {K = x 2 R2 : kxk 6 1}. Let A be a 2 2 positively definite matrix such that kAk = 1, for instance, 2 1 pffiffiffi putting A ¼ 31 32 and u = (k, k) with jkj 6 2=2, we know that kAk = 1, u 2 K and Au = u. Given j 2 (0, 1) we define the map3
3
pings T, S, f : K ? K as follows:
T ¼ A; S ¼ A2
and f ¼ jA:
It is easy to see that T(0) = S(0) = f(0) = 0. Pick three sequences {an}, {bn}, {cn}, in [0, 1] satisfying conditions (i)–(iii) in Theorem 3.1, where
bn ¼
^lð^l þ nÞ ð^l þ n þ 1Þ2
8n P 0
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and ^l ¼ 2=ð1 jÞ. For x0 2 K, let {xn} be given by the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 :
First, observe that
kxn 0k 6 an kyn 0k þ ð1 an ÞkTxn 0k 6 an kyn 0k þ ð1 an Þkxn 0k and hence
kxn 0k 6 kyn 0k 8n P 0: Second, observe that
kxn 0k 6 kyn 0k 6 bn kf ðxn1 Þ 0k þ cn kSxn1 0k þ ð1 bn cn Þkxn1 0k 6 bn jkxn1 0k þ cn kxn1 0k þ ð1 bn cn Þkxn1 0k ¼ ð1 ð1 jÞbn Þkxn1 0k 6 ð1 ð1 jÞbn Þ ð1 ð1 jÞb1 Þkx0 0k n Y kx0 0k ð1 ð1 jÞbj Þ ¼ : ð1 ð1 jÞb0 Þ j¼0 Since 0 < j < 1 and
P1
n¼0 bn
¼ 1, we have
1 n Y Y ð1 ð1 jÞbn Þ ¼ lim ð1 ð1 jÞbj Þ ¼ 0; n!1
n¼0
j¼0
which, hence, implies that {xn} converges to 0 2 F(T). On the other hand, we claim that there holds the convergence rate estimate kxm 0k = O(1/m) for bn ¼ ^lð^l þ nÞ=ð^l þ n þ 1Þ2 ; 8n P 0 where ^l ¼ 2=ð1 jÞ. Indeed, by making use of the last estimation we derive for all n P 0
! ^l þ n 2 kxn1 0k kxn 0k 6 ð1 ð1 jÞbn Þkxn1 0k ¼ 1 ð1 jÞ 1 j ð^l þ n þ 1Þ2 ! ! ^l þ nÞ2 þ 1 2ð^l þ nÞ kx kxn1 0k 0k ¼ ¼ 1 n1 ð^l þ n þ 1Þ2 ð^l þ n þ 1Þ2
and hence,
ð^l þ n þ 1Þ2 kxn 0k ð^l þ nÞ2 kxn1 0k 6 kxn1 0k: Summing this inequality from n = 1 to m(m P 1), we obtain
ð^l þ m þ 1Þ2 kxm 0k ð^l þ 1Þ2 kx0 0k 6 mkx0 0k: Thus,
kxm 0k 6
m þ ð^l þ 1Þ2 kx0 0k ðm þ ^l þ 1Þ2
from which it follows that
1 : kxm 0k ¼ O m
n pffiffiffi o Finally, it is easy to see that FðTÞ ¼ u 2 K : u ¼ ðk; kÞ with jkj 6 2=2 . Moreover, it is clear that p = 0 is the unique solution in F(T) to the following variational inequality:
hðf IÞp; u pi 6 0 for all u 2 FðtÞ: Remark 3.3. Whenever S = I and cn = 0, for all n P 0, it is easily seen that Theorem 3.2 reduces to Theorem 3.1 in Song and Chen [11]. Theorem 3.4. Suppose that K is a nonempty closed convex subset of real uniformly smooth Banach space X. Let f : K ? K be a contractive mapping with contractive coefficient k 2 (0,1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by (3.1). Suppose that {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions:
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175
(i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; limn!1 bn ¼ 0; 1 n¼0 bn ¼ 1; (iii) {cn} [0, 1], limn?1(cn/bn) = 0, (bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T), where x* is the unique solution in F(T) to the variational inequality (2.2). Proof. As in the proof of Theorem 3.1, we prove: 1. The sequences {xn}, {yn}, {Syn}, {Tyn} and {f(yn)} are bounded; 2. limn?1kyn Tynk = 0. Define A = (2I T)1. Then it follows from Lemma 1.1 that F(T) = F(A) and A is a nonexpansive self-mapping on K and
lim kyn Ayn k ¼ 0:
n!1
Applying Theorem 4.2 in Xu [7] for the nonexpansive self-mapping A, we obtain
lim suphf ðx Þ x ; jðyn x Þi 6 0;
ð3:9Þ
n!1
where x* 2 F(A) = F(T) is a unique solution to the variational inequality (2.2) in F(A) = F(T). Finally, we have to prove that xn ? x* as n ? 1. The method is similar to that given in the last part of the proof of Theorem 3.1, see (3.4) and Lemma 1.2. h Theorem 3.5. Suppose that K is a nonempty closed convex subset of real uniformly smooth Banach space X. Let f : K ? K be a fixed contractive mapping with contractive coefficient k 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by (3.5). Suppose that {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; 1 n¼0 bn ¼ 1; (iii) limn?1(cn/bn) = 0, bn + cn 6 1, n P 0. Then {xn} converges strongly to x* 2 F(T), where x* s the unique solution in F(T) of the variational inequality (2.2). Proof. As in the proof of Theorem 3.2, we can reach the following objectives: 1. The sequences {xn}, {Sxn}, {Txn} and {f(xn)} are bounded; 2. limn?1kxn Txnk = 0. We define A = (2I T)1. Then it follows from Lemma 1.1 that F(T) = F(A), and A is a nonexpansive self-mapping on K and
lim kxn Axn k ¼ 0:
n!1
Applying Theorem 4.2 in Xu [7] for the nonexpansive self-mapping A, we obtain
lim suphf ðx Þ x ; jðxn x Þi 6 0;
ð3:10Þ
n!1
where x* 2 F(A) = F(T) is the unique solution to the variational inequality (2.2) in F(T). By a similar approach to the proof of Theorem 3.2 (see the relation (3.8) and Lemma 1.2) we obtain that xn ? x* as n ? 1. h Remark 3.6. Whenever S = I and cn = 0, for all n P 0, it is easily seen that Theorem 3.5 reduces to Theorem 3.4 in Song and Chen [11]. Acknowledgements The authors are thankful to anonymous reviewers, for remarks and suggestions that improved the quality of the paper. References [1] S. Reich, An iterative procedure for constructing zeros of accretive sets in Banach spaces, Nonlinear Anal. 2 (1978) 85–92. [2] H.K. Xu, R.G. Ori, An implicit iteration process for nonexpansive mappings, Numer. Funct. Anal. Optim. 22 (2001) 767–773.
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[3] R.D. Chen, Y.S. Song, H.Y. Zhou, Convergence theorems for implicit iteration process for a finite family of continuous pseudocontractive mappings, J. Math. Anal. Appl. 314 (2006) 701–709. [4] W. Takahashi, Nonlinear Functional Analysis: Fixed Point Theory and its Applications, Yokohama Publishers Inc., Yokohama, 2002. [5] W. Takahashi, Y. Ueda, On Reich’s strong convergence for resolvents of accretive operators, J. Math. Anal. Appl. 104 (1984) 546–553. [6] L.C. Zeng, J.C. Yao, Implicit iteration scheme with perturbed mapping for common fixed points of a finite family of nonexpansive mappings, Nonlinear Anal. 64 (2006) 2507–2515. [7] H.K. Xu, Viscosity approximation methods for nonexpansive mappings, J. Math. Anal. Appl. 298 (2004) 279–291. [8] H.K. Xu, T.H. Kim, Convergence of hybrid steepest–descent methods for variational inequalities, J. Optim. Theory Appl. 119 (2003) 185–201. [9] K. Deimling, Zero of accretive operators, Manuscript Math. 13 (1974) 365–374. [10] R.E. Megginson, An Introduction to Banach Space Theory, Springer-Verlag, New York, Inc., 1998. [11] Y.S. Song, R.D. Chen, Convergence theorems of iterative algorithms for continuous pseudocontractive mappings, Nonlinear Anal. (2006), doi:10.1016/ j.na.2006.06.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Strong convergence of modified implicit iterative algorithms with perturbed mappings for continuous pseudocontractive mappings Lu-Chuan Ceng a,1, Adrian Petrusßel b,*,2, Jen-Chih Yao c,3 a b c
Department of Mathematics, Shanghai Normal University, Shanghai 200234, China Department of Applied Mathematics, Babesß-Bolyai University Cluj-Napoca, 1 Koga˘lniceanu Street, 400084 Cluj-Napoca, Romania Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan
a r t i c l e
i n f o
Keywords: Pseudocontractive mapping Modified viscosity iterative process with perturbed mapping Modified implicit iterative scheme with perturbed mapping Strong convergence Reflexive and strictly convex Banach space Uniformly Gâteaux differentiable norm
a b s t r a c t Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm. First purpose of this paper is to introduce a modified viscosity iterative process with perturbation for a continuous pseudocontractive self-mapping T and prove that this iterative process converges strongly to x* 2 F(T) :¼ {x 2 Xjx = T(x)}, where x* is the unique solution in F(T) to the following variational inequality: hf ðx Þ x ; jðv x Þi 6 0 for all
v 2 FðTÞ:
Second aim of the paper is to propose two modified implicit iterative schemes with perturbation for a continuous pseudocontractive self-mapping T and prove that these iterative schemes strongly converge to the same point x* 2 F(T). Basically, we show that if the perturbation mapping is nonexpansive, then the convergence property of the iterative process holds. In this respect, the results presented here extend, improve and unify some very recent theorems in the literature, see [L.C. Zeng, J.C. Yao, Implicit iteration scheme with perturbed mapping for common fixed points of a finite family of nonexpansive mappings, Nonlinear Anal. 64 (2006) 2507–2515; H.K. Xu, Viscosity approximation methods for nonexpansive mappings, J. Math. Anal. Appl. 298 (2004) 279–291; Y.S. Song, R.D. Chen, Convergence theorems of iterative algorithms for continuous pseudocontractive mappings, Nonlinear Anal. (2006)]. Ó 2009 Published by Elsevier Inc.
1. Introduction and preliminaries Let X be a real Banach space with norm kk and let X* be its dual. The normalized duality mapping J from X into the family of nonempty (by Hahn–Banach theorem) weak* compact subsets of its dual X* is defined by
JðxÞ ¼ ff 2 X : hx; f i ¼ kxk2 ¼ kf k2 g
for all x 2 X;
* Corresponding author. E-mail addresses: [email protected] (L.-C. Ceng), [email protected] (A. Petrusßel), [email protected] (J.-C. Yao). 1 This research was partially supported by the Teaching and Research Award Fund for Outstanding Young Teachers in Higher Education Institutions of MOE, China and the Dawn Program Foundation in Shanghai. 2 This research was partially supported by a grant of the Romanian Research Council. 3 This research was partially supported by a grant from the National Science Council. 0096-3003/$ - see front matter Ó 2009 Published by Elsevier Inc. doi:10.1016/j.amc.2008.10.062
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163
where h, i denotes the generalized duality pairing. It is known (see [4]) that if X* is strictly convex or X is a Banach space with a uniformly Gâteaux differentiable norm (see the definitions below), then J is single-valued and, in this case, it will be denoted by j. When {xn} is a sequence in X, xn ? x (respectively, xn N x, xn N *x) will denote strong (respectively, weak, weak*) convergence of the sequence {xn} to x. The norm of X is called Gâteaux differentiable (and X is called smooth) if
lim t!0
kx þ tyk kxk t
ðÞ
exists for each x, y in its unit sphere U = {x 2 X: kxk = 1}. Recall that X is smooth if and only if J is single-valued. In this case, J : X ? X* is continuous from the strong topology to the weak* topology. The norm of X is called uniformly Gâteaux differentiable if for each y 2 U, this limit is attached uniformly for x 2 U. The norm of X is called Fréchet differentiable if for each x 2 U, the limit in (*) is attained uniformly for y 2 U. The norm of X is called uniformly Fréchet differentiable and X is called uniformly smooth, if the limit in (*) is attained uniformly for (x, y) 2 U U. If X is uniformly smooth, then J : X ? X* is single-valued and uniformly continuous on bounded sets of X from the strong topology of X to the strong topology of X*. Since the dual X* of X is uniformly convex if and only if the norm of X is uniformly Fréchet differentiable, every Banach space with a uniformly convex dual is reflexive and has a uniformly Gâteaux differentiable norm. Recall also that if X has a uniformly Gâteaux differentiable norm, then J is uniformly continuous on bounded sets of X from the strong topology of X to the weak* topology of X*. For every e with 0 6 e 6 2, the modulus dX(e) of convexity of X is defined by
kx þ yk dx ðeÞ ¼ inf 1 : kxk 6 1; kyk 6 1; kx yk P e : 2
A Banach space X is called uniformly convex if dX(e) > 0 for every e > 0. X is called strictly convex if kx + yk/2< 1 for every x, y 2 U with x – y. It is also well known that X is uniformly convex if and only if its dual X* is uniformly smooth. In this case, X is reflexive and strictly convex. The converse implication is false. A discussion on these and related concepts may be found in [4,10]. Let K be a nonempty closed convex subset of X. Recall that a self-mapping f : K ? K is called: (i) k-contraction if k 2 (0, 1) and
kf ðxÞ f ðyÞk 6 kkx yk for all x; y 2 k; (ii) nonexpansive if
kT x T y k 6 kx yk for all x; y 2 K; Furthermore, there holds the following result (see [4, Theorem 4.5.3]): If K is nonempty closed convex subset of a strictly convex Banach space X and T : K ? K is a nonexpansive mapping (i.e., kTx Tyk 6 kx yk, for all x, y 2 K), then the fixed point set F(T) of T is a closed convex subset of K. The self-mapping T : K ? K is called pseudocontractive (respectively, strongly pseudo-contractive), if for each x, y 2 K, there exists j(x y) 2 J(x y) such that
hT x T y ; jðx yÞi 6 kx yk2 (respectively, hTx Ty, j(x y)i 6 bkx yk2 for some b 2 (0, 1)). In 2003, Xu and Ori [2] introduced the following implicit iteration process for a finite family of nonexpansive selfmappings:
xn ¼ an xn1 þ ð1 an ÞT n xn
for all n P 0;
ð1:1Þ
where the initial point x0 2 K is chosen arbitrarily and Tn = TnmodN. They proved that {xn} converges weakly to a common fixed point of Tn, n = 1, 2, . . . , N in a Hilbert space with {an} (0, 1). Very recently, Zeng and Yao in [6] proposed an implicit iteration scheme with perturbed mapping for approximation of common fixed points of a finite family of nonexpansive self-mappings of a Hilbert space. They established some convergence theorems for this implicit iteration scheme. In particular, necessary and sufficient conditions for strong convergence of this implicit iteration scheme were obtained. Meantime, Chen et al. [3] extended the study of the iteration process (1.1) to a finite family of continuous pseudocontractive self-mappings. They established several strongly and weakly convergent theorems. On the other hand, in 2004, Xu [7] studied the following viscosity iteration process in a uniformly smooth Banach space X for a fixed contractive mapping f : K ? K and a nonexpansive mapping T : K ? K (where K is a nonempty closed convex subset of X),
xt ¼ tf ðxt Þ þ ð1 tÞTxt +
and proved that as t ? 0 , xt ? p 2 F(T).
ð1:2aÞ
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For a continuous pseudocontractive mapping T : K ? K, a nonexpansive mapping S : K ? K and a fixed Lipschitz strongly pseudocontractive mapping f : K ? K, we define the mapping T(S,f) : K ? K as follows:
TðS;f Þ ¼ tf þ r t S þ ðl t r t ÞT; where 0 < rt < 1 t. Then it is easy to see that the mapping T(S,f) is a continuous strongly pseudocontractive self-mapping of K. Therefore, T(S,f) has a unique fixed point in K [9, Corollary 2], i.e., for each given t 2 (0, 1) there exists xt 2 K such that
xt ¼ tf ðxt Þ þ rt Sxt þ ð1 t r t ÞTxt :
ð1:2bÞ
The above iterative process (1.2b) is called the modified viscosity iterative process with perturbed mapping, where S is called the perturbed mapping. We remark that if S = I and rt = 0 for all t 2 (0, 1), then (1.2b) reduces to (1.2a). Very recently, Song and Chen [11] proposed the following iteration process: Given any x0 2 K,
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ ð1 bn Þxn1 :
ð1:3Þ
They obtained strong convergence theorems for viscosity iterative process (1.2a) with continuous pseudocontractive mapping T and for the modified implicit iteration scheme (1.3) in a real reflexive and strictly convex Banach space X with a uniformly Gâteaux differentiable norm. Inspired by Song and Chen [11] and Zeng and Yao [6], we introduce and study, for a continuous pseudocontractive selfmapping T, a nonexpansive self-mapping S and a fixed contractive self-mapping f, the following modified implicit iterative algorithms with perturbed mappings: Algorithm 1.1. Let {an} (0, 1], {bn} (0, 1] and {cn} [0, 1] such that limn(cn/bn) = 0 and bn + cn < 1. Starting with an arbitrary initial point x0 2 K, generate a sequence {xn} via the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 bn cn Þxn :
ð1:4aÞ
Algorithm 1.2. Let {an}, {bn} and {cn} be as in Algorithm 1.1. Starting with an arbitrary initial point x0 2 K, generate a sequence {xn} via the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 :
ð1:4bÞ
We remark that if S = I and cn = 0, for all n > 0, then iterative scheme (1.4b) reduces to (1.3). In a real reflexive and strictly convex Banach space X with a uniformly Gâteaux differentiable norm, we not only prove that the modified viscosity iterative process (1.2b) with perturbed mapping for continuous pseudocontractive mapping T converges strongly to p 2 F(T) as t ? 0+, but also show that the modified implicit iterative schemes (1.4a) and (1.4b) with perturbed mappings converge strongly to the same fixed point p of T, which is the unique solution in F(T) to the following variational inequality:
hðf IÞp; jðu pÞi 6 0 for all u 2 FðTÞ: Recall that S : K ? K is called accretive if I S is pseudocontractive. We denote by Jr the resolvent of S, i.e., Jr = (I + rS)1. It is well known that Jr is nonexpansive, single-valued and F(Jr) = S1(0) = {z 2 D(S):0 = Sz} for all r > 0. For more details, see [1,4,9]. Let (E, d) be a metric space. Let x 2 E and C E. Denote
dðx; CÞ :¼ inf dðx; v Þ: m2C
Recall that C is called a Chebyshev set, if for each x 2 E, there exists a unique element y 2 C such that d(x, y) = d(x, C). It is well known that every nonempty closed convex subset of a strictly convex and reflexive Banach space X is a Chebyshev set [11, Corollary 5.1.19]. Let K be a nonempty closed convex subset of a real Banach space X and T : K ? K be a pseudocontractive map, then I T is accretive. We denote A = J1 = (2I T)1. Then F(A) = F(T) and the operator A : R(2I T) ? K is nonexpansive and single-valued. Next we state several lemmas which will be used in the sequel. The first lemma can be found in [11]. Lemma 1.1 [11, Lemma 1.1]. Let K be a nonempty closed convex subset of a real Banach space X and T : K ? K be a continuous pseudocontractive map. We denote A = (2I T)1. Then (i) [10, Theorem 6]. The map A is nonexpansive self-mapping on K, i.e., for all x, y 2 K, there hold
kAx Ayk 6 kx yk and Ax 2 K: (ii) If limn?1kxn Txnk = 0, then limn?1kxn Axnk = 0.
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The second lemma can be found in [7,8]. Lemma 1.2. Let {an} be a sequence of nonnegative real numbers satisfying the condition
anþ1 6 ð1 bn Þan þ bn cn
for all n P 0;
where {bn}, {cn} are sequences of real numbers such that P (i) fbn g ½0; 1; 1 n¼0 bn ¼ 1; (ii) lim supn?1 cn 6 0. Then, limn?1an = 0. Let l be a mean on positive integers N, i.e., a continuous linear functional l on l1 satisfying klk = 1 = l(1). Then we know that l is a mean on N if and only if
inffan : n 2 Ng 6 lðaÞ supfan : n 2 Ng for every a = (a1, a2, . . .) 2 l1. According to time and circumstances, we use ln(an) instead of l(a). A mean l on N is called a Banach limit if
ln ðan Þ ¼ ln ðanþ1 Þ for every a = (a1, a2, . . .) 2 l1. Using the Hahn–Banach theorem, we can prove the existence of a Banach limit. We know that if l is a Banach limit, then
lim inf n!1 an 6 ln ðan Þ 6 lim sup an n!1
1
for every a = (a1, a2, . . .) 2 l . Notice that the following lemma can be found in [5, Lemma 1,4, Lemma 4.5.4]. Lemma 1.3. Let C be a nonempty closed convex subset of a Banach space X with a uniformly Gâteaux differentiable norm. Suppose {xn} is a bounded sequence of X and l is a mean on N. Let z 2 C. Then
ln kxn zk2 ¼ min ln kxn yk2 y2C
if and only if
ln hy z; jðxn zÞi 6 0 for all y 2 C; where j is the normalized duality mapping of X.
2. Modified viscosity iterative algorithm with perturbed mapping Our first main result is the following. Theorem 2.1. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0, S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. Let {xt} be defined by
xt ¼ tf ðxt Þ þ rt Sxt þ ðl t rt ÞTxt
for all t 2 ð0; 1Þ;
ð2:1Þ +
where 0 6 rt < 1 t for all t 2 (0, 1) and limt?0 + (rt/t) = 0. Then as t ? 0 , xt converges strongly to some fixed point p of T such that p is the unique solution in F(T) to the following variational inequality:
hðf IÞp; jðu pÞi 6 0 for all u 2 FðTÞ:
ð2:2Þ
Proof. First, we show the uniqueness of a solution to the variational inequality (2.2). Indeed, suppose p, q 2 F(T) satisfy (2.2). Then we have
hðf IÞp; jðq pÞi 6 0
ð2:3Þ
hðf IÞq; jðp qÞi 6 0:
ð2:4Þ
and
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Adding up (2.3) and (2.4), we have
0 P hðI f Þp ððI f ÞqÞ; jðp qÞi ¼ kp qk2 hf ðpÞ f ðqÞ; jðp qÞi P ð1 bÞkp qk2 : Hence we have p = q. Below we use p 2 F(T) to denote the unique solution of (2.2). Taking a fixed u 2 F(T), we have
kxt uk2 ¼ htðf ðxt Þ uÞ þ r t ðSxt uÞ þ ð1 t r t ÞðTxt uÞ; jðxt uÞi ¼ thf ðxt Þ u; jðxt uÞi þ r t hSxt u; jðxt uÞi þ ð1 t r t ÞhTxt u; jðxt uÞi 6 thf ðxt Þ f ðuÞ; jðxt uÞi þ thf ðuÞ u; jðxt uÞi þ r t hSxt Su; jðxt uÞi þ r t hSu u; jðxt uÞi þ ð1 t r t Þkxt uk2 6 tbkxt uk2 þ thf ðuÞ u; jðxt uÞi þ rt kxt uk2 þ r t hSu u; jðxt uÞi þ ð1 t rt Þkxt uk2 ¼ btkxt uk2 þ thf ðuÞ u; jðxt uÞi þ r t hSu u; jðxt uÞi þ ð1 tÞkxt uk2 : Hence
ð1 bÞkxt uk2 6 hf ðuÞ u; jðxt uÞi þ ðrt =tÞhSu u; jðxt uÞi 6 kf ðuÞ uk kxt uk þ ðr t =tÞkSu uk kxt uk;
ð2:5Þ
which implies that
kxt uk 6
1 ðkf ðuÞ uk þ ðr t =tÞkSu ukÞ: 1b
Since limt?0 + (rt/t) = 0 and limt?0 + (rt/t) = 0, there exists t0 2 (0, 1) such that 0 < rt/t < 1/2 and 0 < t + rt 6 1/2 for all t 2 (0, t0). So, {xt : t 2 (0, t0)} is bounded. h Since f is a Lipschitz mapping and S is a nonexpansive mapping, kf(xt) f(u)k 6 Lkxt uk and kSxt Suk 6 kxt uk Then the sets {f(xt) : t 2 (0, t0)} and {Sxt : t 2 (0, t0)} are bounded. Note that xt = tf(xt) + rtSxt + (1 t rt)Txt. So we have
Txt ¼
1 t rt xt f ðxt Þ Sxt ; 1 t rt 1 t rt 1 t rt
which implies that
kTxt k 6
1 t rt kxt k þ kf ðxt Þk þ kSxt k: 1 t rt 1 t rt 1 t rt
Thus we have
kTxt k 6 2kxt k þ 2tkf ðxt Þk þ 2rt kSxt k;
for all t 2 ð0; t 0 Þ
and so {Txt : t 2 (0, t0)} is bounded. This implies that
lim kxt Txt k 6 lim tkf ðxt Þ Txt k þ lim rt kSxt Txt k ¼ 0:
t!0þ
t!0þ
t!0þ
From Lemma 1.1 we know that the mapping A = (2I T)1 : K ? K is nonexpansive, and F(A) = F(T) and limt?0 + kxt Axtk = 0, where I denotes the identity operator. We claim that the set {xt : t 2 (0, t0)} is relatively compact. Indeed, suppose xn := xtn and
gðxÞ ¼ ln kxn xk2
for all x 2 K;
where {tn} is a sequence in (0, t0) that converges to 0(n ? 0) and ln is a Banach limit. Define the set
K 1 ¼ fx 2 K : gðxÞ ¼ inf gðyÞg: y2K
Since X is a reflexive Banach space, then K1 is a nonempty bounded closed convex subset of X (see [4, Theorem 1.3.11]). Since limn?akxn Axnk = 0, we deduce that for each x 2 K1
gðAxÞ ¼ ln kxn Axk2 6 ln ðkxn Axn k þ kAxn AxkÞ2 6 ln kxn xk2 ¼ gðxÞ: Hence, Ax 2 K1. This shows that K1 is invariant under A. Since F(A) = F(T) – ;, we can take u 2 F(A) = F(T) arbitrarily. Since every nonempty closed convex subset of a strictly convex and reflexive Banach space X is a Chebyshev set (according to [10, Corollary 5.1.19]), there exists a unique q 2 K1 such that
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167
ku qk ¼ inf ku xk: x2K 1
Since u = Tu = Au and Aq 2 K1, we have
ku Aqk ¼ kAu Aqk 6 ku qk: Hence q = Aq. By Lemma 1.3, we have
ln hx q; jðxn qÞi 6 0 for all x 2 K: Taking x = f(q) 2 K in the last inequality, we conclude from (2.5) that
ln kxn qk2 6
1 l hf ðqÞ q; jðxn qÞi 6 0: 1b n
This implies that
ln kxn qk2 ¼ 0: Hence, there exists a subsequence of {xn} which converges strongly to q 2 F(T). Without loss of generality, we may assume that {xn} converges strongly to q 2 F(T). Next we show that q is a solution in F(T) to the variational inequality (2.2). Since u is a fixed point of T and since xt = tf(xt) + rtSxt + (1 t rt)Txt, so we deduce from the accretivity of I T that
hxt f ðxt Þ; jðxt uÞi ¼ ð1 tÞhTxt f ðxt Þ; jðxt yÞi þ r t hsxt Txt ; jðxt uÞi ¼ ð1 tÞhðI TÞxt ðI TÞu; jðxt uÞi þ ð1 tÞhxt f ðxt Þ; jðxt uÞi þ r t hSxt Txt ; jðxt uÞi 6 ð1 tÞhxt f ðxt Þ; jðxt uÞi þ r t hSxt Txt ; jðxt uÞi; hxt f ðxt Þ; jðxt uÞi 6 ðr t =tÞhSxt Txt ; jðxt uÞi:
ð2:6Þ
Since the sets {xt u : t 2 (0, t0)},{Sxt : t 2 (0, t0)} and {Txt : t 2 (0, t0)} are bounded, it follows from limt?0+(rt/t) = 0 that
lim ðr t =tÞhSxt Txt ; jðxt uÞi ¼ 0:
t!0þ
Note that the duality map J is single-valued and norm-to-weak* uniformly continuous on bounded sets of a Banach space X with uniformly Gâteaux differentiable norm. Since xn ? q as n ? 1, we get that for each u 2 F(T), we have
kðI f Þxn ðI f Þqk ! 0 as n ! 1 and
jhðI f Þxn ; jðxn uÞi hðI f Þq; jðq uÞij ¼ jhðI f Þxn ðI f Þq; jðxn uÞi þ hðI f Þq; jðxn uÞ jðq uÞij 6 kðI f Þxn ðI f Þqkkxn uk þ jhðI f Þq; jðxn uÞ jðq uÞij ! 0 as n ! 1: Therefore, it follows immediately from (2.6) that for each u 2 F(T),
hðI f Þq; jðq uÞi ¼ lim hxn f ðxn Þ; jðxn uÞi 6 0: n!1
Consequently, q 2 F(T) is a solution of (2.2); hence q = p by uniqueness. All in all, we have proved that the set {xt} is relatively compact and every cluster point of {xt} (as t ? 0+) equals p. Therefore, xt ? p as t ? 0+. Using the above Theorem 2.1, we can easily obtain the following corollary. Corollary 2.2 [11, Theorem 2.1]. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Suppose that T : K ? K is a continuous pseudocontractive mapping with F(T) – ; and f : K ? K is a Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0. Let {xt} be defined by
xt ¼ tf ðxt Þ þ ð1 tÞTxt
for all t 2 ð0; 1Þ:
ð2:7Þ
+
Then as t ? 0 , xt converges strongly to some fixed point p of T such that p is the unique solution in F(T) to the variational inequality (2.2). Proof. Put S = I and rt = 0 for all t 2 (0, 1). Then the conclusion follows immediately by Theorem 2.1. h A second result is: Theorem 2.3. Let K be a nonempty closed convex subset of a real Banach space X with a uniformly Gâteaux differentiable norm. Let f : K ? K be a fixed Lipschitz strongly pseudocontractive mapping with pseudocontractive coefficient b 2 (0, 1) and the Lipschitz constant L > 0, S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. If there exists a bounded sequence {xn} such that limn?1kxn Txnk = 0 and p ¼ limt!0þ zt exists, where {zt} is defined by (2.1), then
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lim suphf ðpÞ p; jðxn pÞi 6 0: n!1
Proof. Using the equality
zt xn ¼ ð1 t rt ÞðTxt xn Þ þ tðf ðzt Þ xn Þ þ r t ðSxt xn Þ and the inequality
hTx Ty; jðx yÞi 6 kx yk2 ; we obtain that
kzt xn k2 ¼ ð1 t rt ÞhTzt xn ; jðzt xn Þi þ thf ðzt Þ xn ; jðzt xn Þi þ r t hSzt xn ; jðzt xn Þi ¼ ð1 t rt ÞðhTzt Txn ; jðzt xn Þi þ hTxn xn ; jðzt xn ÞiÞ þ thf ðzt Þ zt ; jðzt xn Þi þ tkzt xn k2 þ r t hSzt zt ; jðzt xn Þi þ r t kzt xn k2 6 kzt xn k2 þ kTxn xn kkzt xn k þ thf ðxt Þ zt ; jðzt xn Þi þ r t kSzt zt kkzt xn k and hence
hf ðzt Þ zt ; jðxn zt Þi 6
kTxn xn k rt kzt xn k þ kSzt zt kkzt xn k: t t
ð2:8Þ
Note that as in the proof of Theorem 2.1, there exists t0 2 (0, 1) such that {zt : t 2 (0, t0)} and {Szt : t 2 (0, t0)} are bounded. Since {xn} is bounded and limn?1kxn Txnk = 0, taking the upper limit in (2.8), we have
lim suphf ðzt Þ zt ; jðxn zt Þi 6
rt kSzt zt kðkzt k þ supnP0 kxn kÞ: t
ð2:9Þ
Meantime, we also have
lim
t!0þ
rt kSzt zt kðkzt k þ sup kxn kÞ ¼ 0: t nP0
ð2:10Þ
On the other hand, since the set {zt xn : t 2 (0, t0) and n P 0} is bounded and the duality map J is single-valued and norm-toweak* uniformly continuous on bounded sets in the Banach space X with uniformly Gâteaux differentiable norm, thus it follows from zt ? p(t ? 0+) that
kðf ðzt Þ zt Þ ðf ðpÞ pÞk 6 kðf ðzt Þ f ðpÞk þ kzt pk ! 0; jhf ðzt Þ zt ; jðxn zt Þi hf ðpÞ p; jðxn pÞij ¼ jhðf ðzt Þ zt Þ ðf ðpÞ pÞ; jðxn zt Þi þ hf ðpÞ p; jðxn zt Þ jðxn pÞij
ð2:11Þ
6 jhf ðpÞ p; jðxn zt Þ jðxn pÞij þ kðf ðzt Þ zt Þ ðf ðpÞ pÞkkxn zt k ! 0 as t ! 0þ : Thus, from (2.10) and (2.11) we conclude that for an arbitrary e > 0, there exists d 2 (0, t0) such that for all t 2 (0, d) and all n P 0,
rt e kSzt zt kðkzt k þ supnP0 kxn kÞ < t 2 and
jhf ðzt Þ zt ; jðxn zt Þi hf ðpÞ p; jðxn pÞij <
e 2
;
which implies that
e
hf ðpÞ p; jðxn pÞi < hf ðzt Þ zt ; jðxn zt Þi þ : 2
ð2:12Þ
Consequently, for all t 2 (0, d) we derive from (2.9) and (2.12)
lim suphf ðpÞ p; jðxn pÞi 6 lim suphf ðzt Þ zt ; jðxn zt Þi þ n!1
n!1
Thus, since e is arbitrary, we infer that
lim suphf ðpÞ p; jðxn pÞi 6 0: n!1
The proof is complete.
h
e 2
6
rt e e e kSzt zt kðkzt k þ sup kxn kÞ þ < þ ¼ e: t 2 2 2 nP0
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169
3. Modified implicit iteration scheme with perturbed mapping We consider now the implicit iteration scheme with perturbed mapping given in Algorithm 1.1. Theorem 3.1. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed contractive mapping with contractive coefficient K 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 bn cn Þyn ;
ð3:1Þ
where {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; limn!1 bn ¼ 0; 1 n¼0 bn ¼ 1; (iii) {cn} (0, 1], limn?1(cn/bn) = 0, bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T) where x* is the unique solution in F(T) to the variational inequality (2.2). Proof. At first, we show that {xn} is bounded. Taking a fixed p 2 F(T), we have
kyn pk2 ¼ han xn þ ð1 an ÞTyn p; jðyn pÞi ¼ ð1 an ÞhTyn p; jðyn pÞi þ an hxn p; jðyn pÞi 6 ð1 an Þkyn pk2 þ an kxn pkkjðyn pÞk 6 ð1 an Þkyn pk2 þ an kxn pkkðyn pÞk and hence
kyn pk2 6 kxn pkkyn pk: So, kyn pk 6 kxn pk for all n P 0. Thus we have
xnþ1 pk 6 bn kf ðyn Þ pk þ cn kSyn pk þ ð1 bn cn Þkyn pk 6 bn ðkf ðyn Þ f ðpÞk þ kf ðpÞ pkÞ þ cn ðkSyn Spk þ kSp pkÞ þ ð1 bn cn Þkxn pk 6 bn kkyn pk þ bn kf ðpÞ pk þ cn kyn pk þ cn kSp pk þ ð1 bn cn Þkxn pk 6 bn kkxn pk þ bn kf ðpÞ pk þ cn kxn pk þ cn kSp pk þ ð1 bn cn Þkxn pk ¼ ð1 ð1 kÞbn Þkxn pk þ bn kf ðpÞ pk þ cn kSp pk: Since limn?1(cn/bn) = 0 we may assume, without loss of generality, that cn 6 bn for all n P 0. This implies that
1 kxnþ1 pk 6 ð1 ð1 kÞbn Þkxn pk þ ð1 kÞbn ðkf ðpÞ pk þ kSp pkÞ 1k 1 ðkf ðpÞ pk þ kSp pkÞ : 6 max kxn pk; 1k By induction, we drive
kxn pk 6 max kxo pk;
1 ðkf ðpÞ pk þ kSp pkÞ for all n P 0 1k
This shows that {xn} is bounded, and so is {yn}. Observe that
kf ðyn Þk 6 kf ðyn Þ f ðpÞk þ kf ðpÞk 6 kkyn pk þ kf ðpÞk and
kSyn k 6 kSyn Spk þ kSpk 6 kyn pk þ kSpk: Thus, {f(yn)} and {Syn} are bounded. Since limn?1an = 0, there exist n0 P 0 and a 2 (0, 1), such that an 6 a a for all n P n0. Note that yn = anxn + (1 an)Tyn. Hence we have
Tyn ¼
1 an y xn 1 an n 1 an
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and so
kTyn k 6
1 an 1 a ky k þ kxn k 6 ky k þ kxn k: 1 an n 1a n 1a 1 an
Consequently, the sequence {Tyn} is also bounded. From condition (i), we obtain
kyn Tyn k ¼ an kxn Tyn k ! 0 ðn ! 1Þ:
ð3:2Þ +
Put zt = tf(zt) + rtSzt + (1 t rt)Tzt. Then it follows from Theorem 2.1that as t ? 0 , zt converges strongly to some fixed point x* of T such that x* is the unique solution in F(T) to the variational inequality (2.2). It follows from (3.2) and Theorem 2.4 that
lim suphf ðx Þ x ; jðyn x Þi 6 0:
ð3:3aÞ
n!1
Furthermore, by (3.1) and (3.2) we have
kxn þ 1 yn k ¼ kbn f ðyn Þ þ cn Syn þ ð1 b cn Þyn ðan xn þ ð1 an ÞTyn Þk 6 an kxn Tyn k þ bn kf ðyn Þ yn k þ cn kSyn yn k þ kyn Tyn k ! 0 ðn ! 1Þ: Since the duality map J is single-valued and norm-to-weak* uniformly continuous on bounded sets in Banach space X with uniformly Gâteaux differentiable norm, we have
lim hf ðx Þ x ; jðxnþ1 x Þ jðyn x Þi ¼ 0:
n!1
So we obtain from (3.3a)
lim suphf ðx Þ x ; jðxnþ1 x Þi 6 lim suphf ðx Þ x ; jðyn x Þi þ lim suphf ðx Þ x ; jðxnþ1 x Þ jðyn x Þi n!1
n!1
n!1
¼ lim suphf ðx Þ x ; jðyn x Þi 6 0:
ð3:3bÞ
n!1
Finally we show that xn ? x* as n ? 1. Indeed, using the following inequalities: kyn x*k 6 kxn x*k, kf(x) f(y)kk 6 kkx yk and kSx Syk 6 kx yk, we have
kxnþ1 x k2 ¼ bn hf ðyn Þ x ; jðxnþ1 x Þi þ cn hSyn x ; jðxnþ1 x Þi þ ð1 bn cn Þhyn x ; jðxnþ1 x Þi 6 bn ðhf ðyn Þ f ðx Þ; jðxnþ1 x Þi þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðhSyn Sx ; jðxnþ1 x Þi þ hSx x ; jðxnþ1 x ÞiÞ þ ð1 bn cn Þkyn x kkjðxnþ1 x Þk 6 bn ðkf ðyn Þ f ðx Þkkjðxnþ1 x Þk þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðkSyn Sx kkjðxnþ1 x Þk þ kSx x kkjðxnþ1 x ÞkÞ þ ð1 bn cn Þkxn x kkjðxnþ1 x Þk 6 bn ðkkyn x kkxnþ1 x k þ hf ðx Þ x ; jðxnþl x ÞiÞ þ cn ðkyn x kkxnþi x k þ kSx x kkxnþ1 x kÞ þ ð1 bn cn Þkxn x kkxnþ1 x k 6 bn ðkkxn x kkxnþ1 x k þ hf ðx Þ x ; jðxnþ1 x ÞiÞ þ cn ðkxn x kkxnþ1 x k þ kSx x kkxnþ1 x kÞ þ ð1 bn cn Þkxn x kkxnþ1 x k ¼ ð1 ð1 kÞbn Þkxn x kkxnþ1 x k þ bn hf ðx Þ x ; jðxnþ1 x Þi þ cn kSx x kkxnþ1 x k ¼ ð1 ð1 kÞbn
kxn x k2 kxnþ1 x k2 þ bn hf ðx Þ x ; jðxnþ1 x Þi þ cn kSx x kkxnþ1 x k; 2
which hence implies that
1 ð1 kÞbn 2bn 2cn kxn x k2 þ hf ðx Þ x ; jðxnþ1 x Þi þ kSx x kkxnþ1 x k 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 þ ð1 kÞbn 2ð1 kÞbn 2ð1 kÞbn 1 c ¼ 1 ½hf ðx Þ x ; jðxnþ1 x Þi þ n kSx x kkxnþ1 x k: kxn x k2 þ 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 k bn
kxnþ1 x k2 6
2ð1kÞbn Put bn ¼ 1þð1kÞb and n
cn ¼
1 c ½hf ðx Þ x ; jðxnþ1 x i þ n kSx x kkxnþ1 x k: 1k bn
Then the last inequality can be rewritten as
kxnþ1 x k2 6 ð1 bn Þkxn x k2 þ bn cn :
ð3:4Þ
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Since *
P1
n¼o bn *
¼ 1, we have
P1
bn n¼0 1þð1kÞbn
¼ 1 and hence
P1
n¼0 bn
¼ 1. Note that limn?1(cn/bn = 0 and lim supn?1h f(x*)
x ,j(xn+1 x )i 6 0 due to (3.3b). Thus, according to the boundedness of {xn x*}, we have lim supn?1cn 6 0. Therefore, applying Lemma 1.2 to (3.4), we infer that limn?1kxn x*k = 0. The proof is complete. h Example 1. Let X = R2 with inner product h, i and norm k k defined by
hx; yi ¼ ac þ bd and kxk ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a2 þ b
2 for all x, y 2 R2 with x = (a, b) and y= (c, d). Let K = {x 2 R : kxk 6 1}. Let A be a 2 2 positively definite matrix such that 2 1 pffiffiffi kAk = 1, for instance, putting A ¼ 41 42 and u = (k, k) with jkj 6 2=2, we know that kAk = 1, u 2 K and Au = u. Given 3
3
j 2 (0, 1) we define the mappings T, S, f : K ? K as follows: T ¼ A; S ¼ A2
and f ¼ jA:
It is easy to see that T(0) = S(0) = f(0) = 0. Pick three sequences {an}, {bn}, {cn} in [0, 1] satisfying conditions (i)–(iii) in Theorem 3.1, where
bn ¼
^lð^l þ nÞ
8n P 0
ð^l þ n þ 1Þ2
and ^l ¼ 2=ð1 jÞ. For x0 2 K, let {xn} be given by the following iterative scheme:
yn ¼ an xn þ ð1 an ÞTyn ; xnþ1 ¼ bn f ðyn Þ þ cn Syn þ ð1 b n cn Þyn :
First, observe that
kyn 0k 6 an kxn 0k þ ð1 an ÞkTyn 0k 6 an kxn 0k þ ð1 an Þkyn 0k and hence
kyn 0k 6 kxn 0k 8 P 0: Second, observe that
kxnþ1 0k 6 bn kf ðyn Þ 0k þ cn kSyn 0k þ ð1 b cn Þkyn 0k 6 bn jkyn 0k þ cn kyn 0k þ ð1 bn cn Þkyn 0k ¼ ð1 ð1 jÞbn Þkyn 0k 6 ð1 ð1 jÞbn Þkxn 0k 6 ð1 ð1 jÞbn Þ ð1 ð1 jÞb0 Þkx0 0k ¼
n Y
ð1 ð1 jÞbj Þkx0 0k:
j¼0
Since 0 6 j 6 1 and
P1
n¼0 bn
¼ 1, we have
1 Y
ð1 ð1 jÞbn Þ ¼ lim
n!1
n¼0
n Y
ð1 ð1 jÞbj Þ ¼ 0;
j¼0
which, hence, implies that {xn} converges to 0 2 F(T). On the other hand, we claim that there holds the convergence rate estimate kxm 0k = O(1/m) for bn ¼ ^lð^l þ nÞ=ð^l þ n þ 1Þ2 ; 8n P 0 where ^l ¼ 2=ð1 jÞ. Indeed, by making use of the last estimation we derive for all n 6 0
kxnþ1 0k 6 ð1 ð1 jÞbn Þkxn 0k ¼ ¼
! ð^l þ nÞ2 þ 1 kxn 0k ð^l þ n þ 1Þ2
! ^l þ n 2 kxn 0k ¼ 1 ð1 jÞ 1 j ð^l þ n þ 1Þ2
and hence
ð^l þ n þ 1Þ2 kxnþ1 0k ð^l þ nÞ2 kxn 0k 6 kxn 0k: Summing this inequality from n = 0 to m 1(m P 1), we obtain
ð^l þ mÞ2 kxm 0k ^l2 kx0 0k 6 mkx0 0k:
1
ð2^l þ nÞ ð^l þ n þ 1Þ2
!
kxn 0k
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Thus,
kxm 0k 6
m þ ^l2 kx0 0k ðm þ ^lÞ2
from which it follows that
1 : kxm 0k ¼ O m
n pffiffiffi o Finally, it is easy to see that FðTÞ ¼ u 2 K : u ¼ ðk; kÞ with jkj 6 2=2 . Moreover, it is clear that p = 0 is the unique solution in F(T) to the following variational inequality:
hðf IÞp; u pÞ 6 0 for all u 2 FðTÞ: The second main result of this section deals with the scheme introduced in Algorithm 1.2. Theorem 3.2. Let X be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, and K be a nonempty closed convex subset of X. Let f : K ? K be a fixed contractive mapping with contractive coefficient k 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 ;
ð3:5Þ
where {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0,1], limn!1 an ¼ 0; P (ii) fbn g ð0; 1; 1 n¼0 bn ¼ 1; (iii) limn?1(cn/bn) = 0, bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T), where x* is the unique solution in F(T) of the variational inequality (2.2). Proof. At first, we show that {xn} is bounded. Taking a fixed p 2 F(T), we have
kxn pk2 ¼ han yn þ ð1 an ÞTxn p; jðxn pÞi ¼ ð1 an ÞhTxn p; jðxn pÞi þ an hyn p; jðxn pÞi 6 ð1 an Þkxn pk2 þ an kyn pkkjðxn pÞk 6 ð1 an Þkxn pk2 þ an kyn pkkxn pk and hence
kxn pk2 6 kyn pkkxn pk: So kxn p 6 kyn pk for all n P 0. Thus we have
kxn pk 6 kyn pk 6 bn kf ðxn1 Þ pk þ cn kSxn1 pk þ ð1 bn cn Þkxn1 pk 6 bn ðkf ðxn1 Þ f ðpÞk þ kf ðpÞ pkÞ þ cn ðkSxn1 Spk þ kSp pkÞ þ ð1 bn cn Þkxn1 pk 6 ð1 ð1 kÞbn Þkxn1 pk þ bn kf ðpÞ pk þ cn kSp pk: Since limn?1(cn/bn) = 0, we may assume, without loss of generality, that cn 6 bn for all n P 0. This implies that
1 kxn pk 6 ð1 ð1 kÞbn Þkxn1 pk þ ð1 kÞbn ðkf ðpÞ pk þ kSp pkÞ 1k 1 ðkf ðpÞ pk þ kSp pkÞ : 6 max kxn1 pk; 1k Further, we derive
kxn pk 6 max kx0 pk;
1 ðk=f ðpÞ pk þ kSp pkÞ for all n P 0; 1k
kf ðxn Þk 6 kf ðxn Þ f ðpÞk þ kf ðpÞk 6 kkxn pk þ kf ðpÞk; and
kSxn k 6 kSxn Spk þ kSpk 6 kxn pk þ kSpk: Thus, {xn}, {yn}, {f(xn)} and {Sxn} are bounded. Since limn?1an = 0, there exist n0 P 0 and a 2 (0, 1), such that an 6 a for all n P n0. Note that xn = anyn + (1 an)Txn. Hence we have
Txn ¼
1 an xn y 1 an 1 an n
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and so
kTxn k 6
1 an 1 a kxn k þ ky k 6 kxn k þ ky k: 1 an 1a 1a n 1 an n
Consequently, the sequence {Txn} is also bounded. From condition (i), we obtain
kxn Txn k ¼ an kyn Txn k ! 0 ðn ! 1Þ:
ð3:6Þ +
Put zt = tf(zt) + rtSzt + (1 t rt)Tzt. Then it follows from Theorem 2.1 that as t ? 0 , zt converges strongly to some fixed point x* of T such that x* is the unique solution in F(T) to the variational inequality (2.2). It follows from (3.6) and Theorem 2.4 that
lim suphf ðx Þ x ; jðxn x Þi 6 0:
ð3:7Þ
n!1
Finally we show that xn ? x* (n ? 1). Indeed, using inequalities: kSx Syk 6 kx yk,
hTx Ty; jðx yÞi 6 kx yk2
and kf ðxÞ f ðyÞk 6 kkx yk;
we get that
kxn x k2 ¼ ð1 an ÞhTxn x ; jðxn x Þi þ an hyn x ; jðxn x Þi 6 ð1 an Þkxn x k2 þ an ð1 bn cn Þhxn1 x ; jðxn x Þi þ an bn hf ðxn1 Þ x ; jðxn x Þi þ an cn hSxn1 x ; jðxn x Þi: Thus
kxn x k2 6 ð1 bn cn Þhxn1 x ; jðxn x Þi þ bn hf ðxn1 Þ x ; jðxn x Þi þ cn hSxn1 x ; jðxn x Þi 6 ð1 bn cn Þkxn1 x kkjðxn x Þk þ bn ðkkxn1 x kkjðxn x Þk þ hf ðx Þ x ; jðxn x ÞiÞ þ cn ðkxn1 x kkjðxn x Þk þ hSx x ; jðxn x ÞiÞ ¼ ð1 ð1 kÞbn Þkxn1 x kkjðxn x Þk þ bn hf ðx Þ x ; jðxn x Þi þ cn hSx x ; jðxn x Þi 6 ð1 ð1 kÞbn Þ
kxn1 x k2 þ kxn x k2 þ bn hf ðx Þ x ; jðxn x Þi þ cn hSx x ; jðxn x Þi; 2
which hence implies that
1 ð1 kÞbn 2bn 2cn kxn1 x k2 þ hf ðx Þ x ; jðxn x Þi þ kSx x kkxn x k 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 þ ð1 kÞbn 2ð1 kÞbn 2ð1 kÞbn 1 c ¼ 1 hf ðx Þ x ; jðxn x Þi þ n kSx x kkxn x k : kxn1 x k2 þ 1 þ ð1 kÞbn 1 þ ð1 kÞbn 1 k bn
kxn x g2 6
2ð1kÞbn Put bn ¼ 1þð1kÞb and n
cn ¼
1 c hf ðx Þ x ; jðxnþ1 x Þi þ n kSx x kkxn x k : 1k bn
Then the last inequality can be rewritten as
kxn x k2 6 ð1 bn Þkxn1 x k2 þ bn cn : Since
P1
n¼0 bn
¼ 1, we have
P1
bn n¼0 1þð1kÞbn
¼ 1 and hence
ð3:8Þ P1
n¼0 bn
*
¼ 1. Note that limn?1(cn/bn) = 0 and lim supn?1hf(x ) x*,
*
j(xn x )i 6 0 due to (3.7). Thus, according to the boundedness of {xn x*}, we have lim supn?1cn 6 0. Therefore, applying Lemma 1.2 to (3.8), we infer that limn?1kxn x*k = 0. The proof is complete. h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Example 2. Let X = R2 with inner product h, i and norm k k defined by hx, yi = ac + bd and kxk ¼ a2 þ b for all x, y 2 R2 with x = (a, b) and y = (c, d). Let {K = x 2 R2 : kxk 6 1}. Let A be a 2 2 positively definite matrix such that kAk = 1, for instance, 2 1 pffiffiffi putting A ¼ 31 32 and u = (k, k) with jkj 6 2=2, we know that kAk = 1, u 2 K and Au = u. Given j 2 (0, 1) we define the map3
3
pings T, S, f : K ? K as follows:
T ¼ A; S ¼ A2
and f ¼ jA:
It is easy to see that T(0) = S(0) = f(0) = 0. Pick three sequences {an}, {bn}, {cn}, in [0, 1] satisfying conditions (i)–(iii) in Theorem 3.1, where
bn ¼
^lð^l þ nÞ ð^l þ n þ 1Þ2
8n P 0
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and ^l ¼ 2=ð1 jÞ. For x0 2 K, let {xn} be given by the following iterative scheme:
xn ¼ an yn þ ð1 an ÞTxn ; yn ¼ bn f ðxn1 Þ þ cn Sxn1 þ ð1 bn cn Þxn1 :
First, observe that
kxn 0k 6 an kyn 0k þ ð1 an ÞkTxn 0k 6 an kyn 0k þ ð1 an Þkxn 0k and hence
kxn 0k 6 kyn 0k 8n P 0: Second, observe that
kxn 0k 6 kyn 0k 6 bn kf ðxn1 Þ 0k þ cn kSxn1 0k þ ð1 bn cn Þkxn1 0k 6 bn jkxn1 0k þ cn kxn1 0k þ ð1 bn cn Þkxn1 0k ¼ ð1 ð1 jÞbn Þkxn1 0k 6 ð1 ð1 jÞbn Þ ð1 ð1 jÞb1 Þkx0 0k n Y kx0 0k ð1 ð1 jÞbj Þ ¼ : ð1 ð1 jÞb0 Þ j¼0 Since 0 < j < 1 and
P1
n¼0 bn
¼ 1, we have
1 n Y Y ð1 ð1 jÞbn Þ ¼ lim ð1 ð1 jÞbj Þ ¼ 0; n!1
n¼0
j¼0
which, hence, implies that {xn} converges to 0 2 F(T). On the other hand, we claim that there holds the convergence rate estimate kxm 0k = O(1/m) for bn ¼ ^lð^l þ nÞ=ð^l þ n þ 1Þ2 ; 8n P 0 where ^l ¼ 2=ð1 jÞ. Indeed, by making use of the last estimation we derive for all n P 0
! ^l þ n 2 kxn1 0k kxn 0k 6 ð1 ð1 jÞbn Þkxn1 0k ¼ 1 ð1 jÞ 1 j ð^l þ n þ 1Þ2 ! ! ^l þ nÞ2 þ 1 2ð^l þ nÞ kx kxn1 0k 0k ¼ ¼ 1 n1 ð^l þ n þ 1Þ2 ð^l þ n þ 1Þ2
and hence,
ð^l þ n þ 1Þ2 kxn 0k ð^l þ nÞ2 kxn1 0k 6 kxn1 0k: Summing this inequality from n = 1 to m(m P 1), we obtain
ð^l þ m þ 1Þ2 kxm 0k ð^l þ 1Þ2 kx0 0k 6 mkx0 0k: Thus,
kxm 0k 6
m þ ð^l þ 1Þ2 kx0 0k ðm þ ^l þ 1Þ2
from which it follows that
1 : kxm 0k ¼ O m
n pffiffiffi o Finally, it is easy to see that FðTÞ ¼ u 2 K : u ¼ ðk; kÞ with jkj 6 2=2 . Moreover, it is clear that p = 0 is the unique solution in F(T) to the following variational inequality:
hðf IÞp; u pi 6 0 for all u 2 FðtÞ: Remark 3.3. Whenever S = I and cn = 0, for all n P 0, it is easily seen that Theorem 3.2 reduces to Theorem 3.1 in Song and Chen [11]. Theorem 3.4. Suppose that K is a nonempty closed convex subset of real uniformly smooth Banach space X. Let f : K ? K be a contractive mapping with contractive coefficient k 2 (0,1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by (3.1). Suppose that {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions:
L.-C. Ceng et al. / Applied Mathematics and Computation 209 (2009) 162–176
175
(i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; limn!1 bn ¼ 0; 1 n¼0 bn ¼ 1; (iii) {cn} [0, 1], limn?1(cn/bn) = 0, (bn + cn 6 1, for all n P 0. Then {xn} converges strongly to x* 2 F(T), where x* is the unique solution in F(T) to the variational inequality (2.2). Proof. As in the proof of Theorem 3.1, we prove: 1. The sequences {xn}, {yn}, {Syn}, {Tyn} and {f(yn)} are bounded; 2. limn?1kyn Tynk = 0. Define A = (2I T)1. Then it follows from Lemma 1.1 that F(T) = F(A) and A is a nonexpansive self-mapping on K and
lim kyn Ayn k ¼ 0:
n!1
Applying Theorem 4.2 in Xu [7] for the nonexpansive self-mapping A, we obtain
lim suphf ðx Þ x ; jðyn x Þi 6 0;
ð3:9Þ
n!1
where x* 2 F(A) = F(T) is a unique solution to the variational inequality (2.2) in F(A) = F(T). Finally, we have to prove that xn ? x* as n ? 1. The method is similar to that given in the last part of the proof of Theorem 3.1, see (3.4) and Lemma 1.2. h Theorem 3.5. Suppose that K is a nonempty closed convex subset of real uniformly smooth Banach space X. Let f : K ? K be a fixed contractive mapping with contractive coefficient k 2 (0, 1), S : K ? K be a nonexpansive mapping and T : K ? K be a continuous pseudocontractive mapping with F(T) – ;. For x0 2 K, let {xn} be given by (3.5). Suppose that {an}, {bn} and {cn} are three sequences of nonnegative real numbers satisfying the conditions: (i) {an} (0, 1], limn?1an = 0; P (ii) fbn g ð0; 1; 1 n¼0 bn ¼ 1; (iii) limn?1(cn/bn) = 0, bn + cn 6 1, n P 0. Then {xn} converges strongly to x* 2 F(T), where x* s the unique solution in F(T) of the variational inequality (2.2). Proof. As in the proof of Theorem 3.2, we can reach the following objectives: 1. The sequences {xn}, {Sxn}, {Txn} and {f(xn)} are bounded; 2. limn?1kxn Txnk = 0. We define A = (2I T)1. Then it follows from Lemma 1.1 that F(T) = F(A), and A is a nonexpansive self-mapping on K and
lim kxn Axn k ¼ 0:
n!1
Applying Theorem 4.2 in Xu [7] for the nonexpansive self-mapping A, we obtain
lim suphf ðx Þ x ; jðxn x Þi 6 0;
ð3:10Þ
n!1
where x* 2 F(A) = F(T) is the unique solution to the variational inequality (2.2) in F(T). By a similar approach to the proof of Theorem 3.2 (see the relation (3.8) and Lemma 1.2) we obtain that xn ? x* as n ? 1. h Remark 3.6. Whenever S = I and cn = 0, for all n P 0, it is easily seen that Theorem 3.5 reduces to Theorem 3.4 in Song and Chen [11]. Acknowledgements The authors are thankful to anonymous reviewers, for remarks and suggestions that improved the quality of the paper. References [1] S. Reich, An iterative procedure for constructing zeros of accretive sets in Banach spaces, Nonlinear Anal. 2 (1978) 85–92. [2] H.K. Xu, R.G. Ori, An implicit iteration process for nonexpansive mappings, Numer. Funct. Anal. Optim. 22 (2001) 767–773.
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