Strong convergence of modified Mann iterations

Nonlinear Analysis 61 (2005) 51 – 60 www.elsevier.com/locate/na

Strong convergence of modified Mann iterations Tae-Hwa Kima,1 , Hong-Kun Xub,∗,2 a Division of Mathematical Sciences, Pukyong National University, Pusan 608-737, Korea b School of Mathematical Sciences, University of KwaZulu-Natal, Private Bag X54001, Durban 4000,

South Africa Received 30 March 2004; accepted 15 November 2004

Abstract The Mann iterations for nonexpansive mappings have only weak convergence even in a Hilbert space. We propose two modifications of the Mann iterations in a uniformly smooth Banach space, one for nonexpansive mappings and the other for the resolvent of accretive operators. The two modified Mann iterations are proved to have strong convergence. 䉷 2004 Elsevier Ltd. All rights reserved. MSC: Primary 47H09; Secondary 46B25 Keywords: Strong convergence; Modified Mann iteration; Uniformly smooth Banach space; Nonexpansive mapping; Accretive operator

1. Introduction Let X be a real Banach space, C a nonempty closed convex subset of X, and T : C → C a mapping. Recall that T is nonexpansive if T x − T y x − y for all x, y ∈ C. A point x ∈ C is a fixed point of T provided T x = x. Denote by F ix(T ) the set of fixed points of T; that is, F ix(T ) = {x ∈ C : T x = x}. It is assumed throughout that T is a nonexpansive mapping such that F ix(T ) = ∅. ∗ Corresponding author.

E-mail address: [email protected] (H.-K. Xu). 1 Supported by Pukyong National University Abroad Fund in 2004. 2 Supported in part by the National Research Foundation of South Africa.

0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.11.011

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T.-H. Kim, H.-K. Xu / Nonlinear Analysis 61 (2005) 51 – 60

Construction of fixed points of nonexpansive mappings is an important subject in the theory of nonexpansive mappings and its applications in a number of applied areas, in particular, in image recovery and signal processing (see, e.g., [2,4,11,15,19,20]). However, the sequence {T n x}∞ n=0 of iterates of the mapping T at a point x ∈ C may, in general, not behave well. This means that it may not converge (even in the weak topology). One way to overcome this difficulty is to use Mann’s iteration method that produces a sequence {xn } via the recursive manner: xn+1 = n xn + (1 − n )T x n ,

n 0,

(1)

where the initial guess x0 ∈ C is chosen arbitrarily. For example, Reich [13] proved that if X is a uniformly convex Banach space with a Fréchet differentiable norm and if {n } is chosen such that ∞ n=1 n (1 − n ) = ∞, then the sequence {xn } defined by (1) converges weakly to a fixed point of T. However, this scheme has only weak convergence even in a Hilbert space [7]. Some attempts to modify the Mann iteration method (1) so that strong convergence is guaranteed have recently been made. Nakajo and Takahashi [10] proposed the following modification of the Mann iteration method (1) in a Hilbert space H:  x0 = x ∈ C,     yn = n xn + (1 − n )T x n , (2) Cn = {z ∈ C : yn − z xn − z},   0}, − z, x − x = {z ∈ C : x Q  n 0 n n  xn+1 = PCn ∩Qn (x0 ), where PK denotes the metric projection from H onto a closed convex subset K of H. They proved that if the sequence {n } is bounded above from one, then {xn } defined by (2) converges strongly to PF ix(T ) (x0 ). Their argument does not work outside the Hilbert space setting. Also, at each iteration step, an additional projection is needed to calculate. Some related work can also be found in Shioji and Takahashi [16] and Kamimura and Takahashi [9]. Without knowing the rate of convergence of (2), we propose here a simpler modification of Mann’s iteration method (1) inspired by the ideas in [17] (see also [18]). Our modified Mann’s iteration scheme is a convex combination of a fixed point in C and the Mann’s iteration method (1) and works in a uniformly smooth Banach space. There is no additional projection involved in our new scheme (see (3) below). Let C be a closed convex subset of a Banach space and T : C → C a nonexpansive mapping such that F ix(T ) = ∅. Define {xn } in the following way: x0 = x ∈ C, (3) yn = n xn + (1 − n )T x n , xn+1 = n u + (1 − n )yn , where u ∈ C is an arbitrary (but fixed) element in C, and {n } and {n } are two sequences in (0, 1). We prove, under certain appropriate assumptions on the sequences {n } and {n }, which will be made precise in Section 3, that {xn } defined by (3) converges to a fixed point of T.


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Our second modification of Mann’s iteration method (1) is an adaption to (3) for finding a zero of an m-accretive operator A, for which we assume that the zero set A−1 (0) = ∅. Our iteration process {xn } is given by x0 = x ∈ X, (4) yn = Jrn xn , xn+1 = n u + (1 − n )yn , where for each r > 0, Jr = (I + rA)−1 is the resolvent of A. Again we prove, in a uniformly smooth Banach space and under certain appropriate assumptions on the sequences {n } and {rn } which will be made precise in Section 4, that {xn } defined by (4) converges strongly to a zero of A. 2. Preliminaries Let X be a real Banach space. Recall the (normalized) duality map J from X into X∗ , the dual space of X, is given by J (x) = {x ∗ ∈ X∗ : x, x ∗ = x2 = x ∗ 2 },

x ∈ X.

We are going to work in uniformly smooth Banach spaces that can be characterized by duality mappings as follows (see [5] for more details): Lemma 1. A Banach space X is uniformly smooth if and only if the duality map J is singlevalued and norm-to-norm uniformly continuous on bounded sets of X. In our convergence results in the next sections, we need to estimate the square-norm xn+1 − p2 in terms of the square-norm xn − p2 , where xi is the ith iterate for i 1, and p is a fixed point of our mapping T. To do this, we need the following well-known (subdifferential) inequality: Lemma 2. In a Banach space X, there holds the inequality x + y2 x2 + 2y, j (x + y),

x, y ∈ X,

where j (x + y) ∈ J (x + y). Recall that if C and D are nonempty subsets of a Banach space X such that C is nonempty closed convex and D ⊂ C, then a map Q : C → D is called a retraction from C onto D provided Q(x) = x for all x ∈ D. A retraction Q : C → D is sunny [3,12] provided Q(x + t (x − Q(x))) = Q(x) for all x ∈ C and t 0 whenever x + t (x − Q(x)) ∈ C. A sunny nonexpansive retraction is a sunny retraction, which is also nonexpansive. Sunny nonexpansive retractions play an important role in our argument. They are characterized as follows [3,8,12]: If X is a smooth Banach space, then Q : C → D is a sunny nonexpansive retraction if and only if there holds the inequality x − Qx, J (y − Qx) 0

for all x ∈ C

and y ∈ D.

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Reich [14] showed that if X is uniformly smooth and if D is the fixed point set of a nonexpansive mapping from C into itself, then there is a sunny nonexpansive retraction from C onto D and it can be constructed as follows. Lemma 3 (Reich [14]). Let X be a uniformly smooth Banach space and let T : C → C be a nonexpansive mapping with a fixed point. For each fixed u ∈ C and every t ∈ (0, 1), the unique fixed point xt ∈ C of the contraction C x → tu + (1 − t)T x converges strongly as t → 0 to a fixed point of T. Define Q : C → F ix(T ) by Qu = s − limt→0 xt . Then Q is the unique sunny nonexpansive retract from C onto F ix(T ); that is, Q satisfies the property: u − Qu, J (z − Qu) 0,

u ∈ C,

z ∈ F ix(T ).

We also need the following lemma that can be found in the existing literature (see e.g. [17,18]): Lemma 4. Let {an }∞ n=0 be a sequence of nonnegative real numbers satisfying the property an+1 (1 − n )an + n n ,

n 0,

∞ where {n }∞ n=0 ⊂ (0, 1) and {n }n=0 are such that (i) limn→∞ n = 0 and ∞ = ∞, n=0 n (ii) either lim supn→∞ n 0 or ∞ n=0 |n n | < ∞.

Then {an }∞ n=0 converges to zero.

3. Convergence to a fixed point of nonexpansive mappings In this section, we propose a modification of the Mann’s iteration method (1) to have strong convergence. The modified Mann’s iteration process is a convex combination of a fixed point in C and the Mann’s iteration process (1). The advantage of this modification is that not only strong convergence is guaranteed, but also computations of iterates are not substantially increased. Also our new iteration process works in a Banach space setting as opposed to [10], iteration process that works only in the framework of Hilbert spaces. Theorem 1. Let C be a closed convex subset of a uniformly smooth Banach space X and let T : C → C be a nonexpansive mapping such that F ix(T ) = ∅. Given a point u ∈ C ∞ and given sequences {n }∞ n=0 and {n }n=0 in (0, 1), the following conditions are satisfied: (i) 0; n → 0, n → ∞ (ii) ∞ = ∞, = ∞; n=0 n n=0 n ∞ ∞ | − | < ∞, (iii) n n=0 n+1 n=0 |n+1 − n | < ∞.


Define a sequence {xn }∞ n=0 in C by x0 = x ∈ C arbitrarily, yn = n xn + (1 − n )T x n , n 0, xn+1 = n u + (1 − n )yn , n 0.

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(5)

Then {xn }∞ n=0 strongly converges to a fixed point of T. Proof. First we observe that {xn }∞ n=0 is bounded. Indeed, if we take a fixed point p of T, noting that yn − p n xn − p + (1 − n )T x n − p xn − p, we have xn+1 − p n u − p + (1 − n )yn − p n u − p + (1 − n )xn − p max{xn − p, u − p}. Now, an induction yields xn − p max{x0 − p, u − p}, n 0.

(6)

Hence, {xn } is bounded, so is {yn }. As a result, we obtain by condition (i), xn+1 − yn = n u − yn → 0.

(7)

We next show that xn − T x n → 0.

(8)

It suffices to show that xn+1 − xn → 0. Indeed, if (9) holds, then noting (7), we obtain xn − T x n xn − xn+1 + xn+1 − yn + yn − T x n = xn − xn+1 + xn+1 − yn + n xn − T x n → 0, i.e., (8) holds. In order to prove (9), we calculate xn+1 − xn . After some manipulations we get xn+1 − xn = (n − n−1 )(u − T x n−1 ) + (1 − n )n (xn − xn−1 ) + [(n − n−1 )(1 − n ) − (n − n−1 )n−1 ](xn−1 − T x n−1 ) + (1 − n )(1 − n )(T x n − T x n−1 ). It follows that xn+1 − xn (1 − n )(1 − n )T x n − T x n−1 + (1 − n )n xn − xn−1 + |(n − n−1 )(1 − n ) − (n − n−1 )n−1 |xn−1 − T x n−1 + |n − n−1 |u − T x n−1 .

(9)

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Hence, xn+1 − xn (1 − n )xn − xn−1 + (|n − n−1 | + 2|n − n−1 |),

(10)

where > 0 is a constant such that max{u − Tx n−1 , xn−1 − T x n−1 } for all n. By ∞ assumptions (i)–(iii), we have that limn→∞ n =0, ∞ n=1 (|n − n−1 |+ n=1 n =∞, and 2|n − n−1 |) < ∞. Hence, Lemma 4 is applicable to (10) and we obtain xn+1 − xn → 0. Next, we claim that lim sup u − q, J (xn − q) 0,

(11)

n→∞

where q = Q(u) = s − limt→0 zt with zt being the fixed point of the contraction z → tu + (1 − t)T z (see Lemma 3). In order to prove (11), we need some more information on q, which is obtained from that of zt (cf. [16]). Indeed, zt solves the fixed point equation zt = tu + (1 − t)T zt . Thus we have zt − xn = (1 − t)(T zt − xn ) + t (u − xn ). We apply Lemma 2 to get zt − xn 2 (1 − t)2 T zt − xn 2 + 2tu − xn , J (zt − xn )

(1 − 2t + t 2 )zt − xn 2 + an (t) + 2tu − zt , J (zt − xn ) + 2tzt − xn 2 , where an (t) = (2zt − xn + xn − T x n )xn − T x n → 0

as n → ∞.

(12)

It follows that t 1 zt − u, J (zt − xn ) zt − xn 2 + an (t). 2 2t Letting n → ∞ in (13) and noting (12) yields t lim sup zt − u, J (zt − xn ) M, 2 n→∞

(13)

(14)

where M > 0 is a constant such that M zt − xn 2 for all t ∈ (0, 1) and n 1. Since the set {zt − xn } is bounded, the duality map J is norm-to-norm uniformly continuous on bounded sets of X (Lemma 1), and zt strongly converges to q. By letting t → 0 in (14), it is not hard to find that the two limits can be interchanged and (11) is thus proven. Finally, we show that xn → q strongly and this concludes the proof. Indeed, using Lemma 2 again we obtain xn+1 − q2 = (1 − n )(yn − q) + n (u − q)2

(1 − n )2 yn − q2 + 2n u − q, J (xn+1 − q) (1 − n )xn − q2 + 2n u − q, J (xn+1 − q). Now we apply Lemma 4 and use (11) to see that xn − q → 0.


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4. Convergence to a zero of accretive operators Let X be a real Banach space. Recall that a (possibly multivalued) operator A with domain D(A) and range R(A) in X is accretive if, for each xi ∈ D(A) and yi ∈ Ax i (i = 1, 2), there exists a j ∈ J (x2 − x1 ) such that y2 − y1 , j 0. (Here J is the duality map.) An accretive operator A is m-accretive if R(I + rA) = X for each r > 0. Throughout this section we always assume that A is m-accretive and has a zero (i.e., the inclusion 0 ∈ Az is solvable). The set of zeros of A is denoted by F. Hence, F = {z ∈ D(A) : 0 ∈ A(z)} = A−1 (0). For each r > 0, we denote by Jr the resolvent of A, i.e., Jr = (I + rA)−1 . Note that if A is m-accretive, then Jr : X → X is nonexpansive and F ix(Jr ) = F for all r > 0. We need the resolvent identity (see [1], where more details on accretive operators can be found). Lemma 5 (The Resolvent Identity). For > 0 and > 0 and x ∈ X, J x = J J x . x+ 1−

(15)

Theorem 2. Assume that X is a uniformly smooth Banach space and A is an m-accretive operator in X such that A−1 (0) = ∅. Let {xn } be defined by x0 = x ∈ X, (16) yn = Jrn xn , n 0, xn+1 = n u + (1 − n )yn , n 0. Suppose {n } and {rn } satisfy the conditions: (i) limn→∞ n = 0 and ∞ n=0 n = ∞; ∞ | − | < ∞; (ii) n n=0 n+1 (iii) rn for some > 0 and for all n 1. Also assume

∞

1 − rn−1 < ∞.

rn

n=1

Then {xn } converges strongly to a zero of A. Proof. First of all we show that {xn } is bounded. Take a p ∈ F = A−1 (0). It follows that xn+1 − p = n (u − p) + (1 − n )(Jrn xn − p) n u − p + (1 − n )xn − p. By induction, we get xn − p max{x0 − p, u − p},

n 0.

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This implies that {xn } is bounded. It then follows that xn+1 − Jrn xn → 0.

(17)

Simple calculations show that xn+1 − xn = (n − n−1 )(u − yn−1 ) + (1 − n )(yn − yn−1 ).

(18)

The resolvent identity (15) implies that rn−1 rn−1 yn = Jrn−1 Jrn xn , xn + 1 − rn rn which in turn implies that

rn−1

yn − yn−1 xn − xn−1 + 1 − Jr xn − xn . rn n

(19)

Combining (18) and (19) gives

rn−1

xn+1 − xn (1 − n )xn − xn−1 + M |n − n−1 | + 1 − , rn

(20)

where M is a constant such that M max{u − yn , Jr xn − xn } for all n 0 and r > 0. ∞ = 0, By assumptions (i)–(iii) in the theorem, we have that lim n n→∞ n=0 n = ∞,

∞

rn−1 and n=1 |n − n−1 | + 1 − rn < ∞. Hence, Lemma 4 is applicable to (20) and we conclude that xn+1 − xn → 0. Take a fixed number r such that > r > 0. Again from the resolvent identity (15) we find

r r

Jrn xn − Jr xn = Jr Jrn xn − Jr xn xn + 1 −

rn rn r xn − Jrn xn 1− rn xn − xn+1 + xn+1 − Jrn xn → 0. It follows that xn+1 − Jr xn+1 xn+1 − Jrn xn + Jrn xn − Jr xn + Jr xn − Jr xn+1 xn+1 − Jrn xn + Jrn xn − Jr xn + xn − xn+1 . Hence, xn − Jr xn → 0.

(21)

Since in a uniformly smooth Banach space, the sunny nonexpansive retract Q from X onto the fixed point set F ix(Jr ) (=F =A−1 (0)) of Jr is unique, it must be obtained from Reich’s theorem (Lemma 3). Namely, Q(u) = s − lim zt , t→0

u ∈ X,

(22)


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where t ∈ (0, 1) and zt ∈ X solves the fixed point equation zt = tu + (1 − t)Jr zt .

(23)

Since we can derive zt − xn = t (u − xn ) + (1 − t)(Jr xt − xn ), applying Lemma 2 we get zt − xn (1 − t)2 Jr zt − xn 2 + 2tu − xn , J (zt − xn )

(1 − t)2 zt − xn 2 + an (t) + 2tu − zt , J (zt − xn ) + 2tzt − xn 2 , where an (t) = 2zt − xn · Jr xn − xn + Jr xn − xn 2 → 0 by (21). It follows that t 1 zt − u, J (zt − xn ) zt − xn 2 + an (t). 2 2t

(24)

Therefore, letting n → ∞ in (24) we get t lim sup zt − u, J (zt − xn ) M, 2 n→∞

(25)

where M is a constant such that M zt −xn 2 for all t ∈ (0, 1) and n 1. Since zt → Q(u) strongly and the duality map J is norm-to-norm uniformly continuous on bounded sets of X, it follows that (by letting t → 0 in (25)) lim sup u − Q(u), J (xn − Q(u)) 0.

(26)

n→∞

Finally, we prove that {xn } strongly converges to Q(u). Indeed we have xn+1 − Q(u)2 = n (u − Q(u)) + (1−n )(Jrn xn −Q(u))2

(1−n )2 Jrn xn −Q(u))2 +2n u − Q(u), J (xn+1 − Q(u)) (1 − n )xn − Q(u)2 + 2n u − Q(u), J (xn+1 − Q(u)). By Lemma 4 and (26), we obtain that xn − Q(u) → 0.

Remark. Theorem 2 removes the assumption that the space X has a weakly continuous duality map in [6, Theorem 2.5].

Acknowledgements The authors thank the referee for his comments and suggestions, which improved the presentation of this manuscript.

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