EAST JOURNAL ON APPROXIMATIONS Volume 13, Number 2 (2007), 211-221

A MODIFICATION OF MANN’S ITERATION SCHEME Yu Miao

In this paper, we present a modifications on Mann’s iteration process with continuous pseudo-contractive mapping in Banach space and prove weak and strong convergence theorems for the scheme with errors which overcome the gaps in [16], but also complement, improve and extend further the corresponding ones of [16, 20]. Mathematics Subject Classification (2000): 47H05, 47H10. Key words and phrases: Continuous pseudo-contractive mappings, strong and weak convergence, Mann’s iteration, errors.

1. Introduction Let E be a real Banach space with dual E ∗ and K a non-empty closed ∗ convex subset of E. Let J : E → 2E denote the normalized duality mapping defined by J(x) := {f ∈ E ∗ : hx, f i = kxk2 , kxk = kf k}, ∀x ∈ E, where h·, ·i denotes the generalized duality pairing. We denote by I the identity operator. Definition 1.1. Following Morales [11], we call a mapping T with domain D(T ) and range R(T ) in E (i) strongly pseudo-contractive if for some constant k < 1 and ∀x, y ∈ D(T ) (1.1)

(λ − k)kx − yk ≤ k(λI − T )x − (λI − T )yk

for all λ > k, which is equivalent to the condition that there exists j(x − y) ∈ J(x − y) such that (1.2)

hT x − T y, j(x − y)i ≤ βkx − yk2 , 0 < β < 1

(see Kato [8]);

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Mann’s iteration scheme

(ii) pseudo-contractive if (1.1) holds for k = 1, which is equivalent to the condition that there exists j(x − y) ∈ J(x − y) such that hT x − T y, j(x − y)i ≤ kx − yk2 ,

(1.3) (see Kato [8]);

(iii) non-expansive if for ∀x, y ∈ D(T ), the following inequality holds: kT x − T yk ≤ kx − yk; (iv) hemicontractive if for any x ∈ D(T ) and y ∈ F (T ), there exists j(x − y) in J(x − y) such that (1.4)

hT x − y, j(x − y)i ≤ kx − yk2 .

Here F (T ) denotes the set of fixed points of T , that is, F (T ) , {x ∈ D(T ) : T x = x}. Every non-expansive mapping is a pseudo-contraction. The converse is not true as seen from the example: T (x) = 1 − x2/3 , 0 ≤ x ≤ 1, a continuous pseudo-contraction which is not non-expansive. For the importance of the fixed points of pseudo-contractions the reader may consult [1]. If E is a Hilbert space (J = I), it is easy to see that (1.2), (1.3) and (1.4) are equivalent, respectively, to the following inequalities (cf. Osilike [13] and Osilike and Udomene [14]): (1.5)

kT x − T yk2 ≤ kx − yk2 + βk(x − T x) − (y − T y)k2 , ∀x, y ∈ D(T ),

(1.6)

kT x − T yk2 ≤ kx − yk2 + k(x − T x) − (y − T y)k2 , ∀x, y ∈ D(T ),

and (1.7)

kT x − yk2 ≤ kx − yk2 + kx − T xk2 , ∀x ∈ D(T ), y ∈ F (T ).

Construction of fixed points of non-expansive mappings is an important subject in the theory of non-expansive mappings and its applications in a number of applied areas, in particular, in image recovery and signal processing (see, e.g., [2, 3, 15, 19, 23, 24, 25]). However, the sequence {T n x}∞ n=0 of iterates of the mapping T at a point x ∈ K 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: (1.8)

xn+1 = an xn + (1 − an )T xn , x0 ∈ K, n ≥ 0,

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where T is a non-expansive mapping. For example, Reich [17] proved that if E is a uniformly convex Banach P∞space with a Fr´echet differentiable norm and if {an } is chosen such that n=1 an (1 − an ) = ∞, then the sequence {xn } defined by (1.8) converges weakly to a fixed point of T . However, this scheme has only weak convergence even in a Hilbert space [7]. In connection with the iterative approximation of fixed points of pseudocontractions, Chidume and Moore (1999) [4] gave the following remarks (the following questions are still open): Does the Mann iteration process always converge for continuous pseudo-contractions, or for even Lipschitz pseudocontractions? Let E be a Banach space and K be a non-empty compact convex subset of E. Let T : K → K be a Lipschitz pseudo-contractive map. Under this setting, even for E = H, a Hilbert space, the answer to the above question is not known. Furthermore, Chidume and Mutangadura [5] provide an example of a Lipschitz pseudo-contractive map with a unique fixed point for which the Mann iteration sequence failed to converge and they stated there ”This resolves a long standing open problem”. Recently Rafiq [16] studied the strong convergence of the following Mann type implicit process for hemicontractive mapping T with an ∈ [δ, 1 − δ] for some δ ∈ (0, 1/2): (1.9)

xn+1 = an xn + (1 − an )T xn+1 .

However, Rafiq’s control conditions exclude the natural choice of an = 1/n. In [20], Song discusses the iteration process (1.9) for continuous pseudocontractive mapping which overcomes the restriction in Rafiq [16] (includes the case an = 1/n). In the present paper, we consider the following Mann type implicit iteration process with errors associated with pseudo-contractive mapping: Let K be a non-empty closed convex subset of the Banach space E and T be a continuous pseudo-contractive mappings from K into itself. Assume that the sequence {xn } is defined by (1.10)

x0 ∈K; xn+1 =an xn + bn T xn+1 + cn un ,

where {an }, {bn }, {cn } are sequences in (0, 1) with an + bn + cn = 1 and {un } is a bounded sequence in K. Note that if cn ≡ 0, then the iteration process (1.10) induces (1.9). We will show several weak and strong convergence results for the iterative scheme (1.10) in Hilbert space and Banach space, respectively (it includes the natural choice of an = 1/n ). For continuous pseudo-contractive mappings, the results presented in this paper not only overcome the gaps in [16] but also complement, improve and extend further the corresponding ones of [16, 20].

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Mann’s iteration scheme Next we state the following useful lemma.

Lemma 1.1. (Tan and Xu [22]). Let {sn }, {tn } be two non-negative sequences satisfying sn+1 ≤ sn + tn ∀n ≥ 1. P∞ If n=1 tn < ∞, then limn→∞ sn exists.

2. Weak convergence theorems Lemma 2.1. Let K be a non-empty closed convex subset of the Banach space E and F (T ) 6= ∅. In addition, assume that {an }, {bn }, {cn } are sequences in (0, 1) with an + bn + cn = 1, bn ∈ (β, 1) for some β ∈ (0, 1) and ∞ X

cn < ∞. 1 − bn n=1 Then, for the iteration process {xn } which is defined in (1.10), we have (i) for any p ∈ F (T ), limn→∞ kxn − pk exists; (ii) both {xn } and {T xn } are bounded. Proof. Let p ∈ F (T ). From the equivalent definition (1.6) of pseudocontractive mapping and the well-known Cauchy-Schwarz inequality (see also [10]), it is easy to see that

(2.1)

kxn+1 − pk2 = ≤ + + ≤

han xn + bn T xn+1 + cn un − p, j(xn+1 − p)i an hxn − p, j(xn+1 − p)i bn hT (xn+1 − p), j(xn+1 − p)i cn hun − p, j(xn+1 − p)i an kxn − pk kxn+1 − pk

+ bn kxn+1 − pk2 + cn kun − pk kxn+1 − pk. If kxn+1 −pk = 0, the result is obvious. So, we suppose that kxn+1 −pk ≥ 0, and then (2.1) implies an cn kxn − pk + kun − pk 1 − bn 1 − bn cn cn kxn − pk + kun − pk =kxn − pk − 1 − bn 1 − bn cn kun − pk. ≤kxn − pk + 1 − bn

kxn+1 − pk ≤ (2.2)

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Since {un } is a bounded sequence in K, by the assumption ∞ X

cn <∞ 1 − bn n=1 and Lemma 1.1, limn→∞ kxn+1 − pk exists and (i) holds. From (i) in Lemma 2.1, we know that {xn } is bounded. By the iteration process (1.10) we obtain 1 kxn+1 − an xn − cn un k bn 1 an cn ≤ kxn+1 k + kxn k + kun k β β β

kT xn+1 k =

which means that {T xn } is bounded.

2

Definition 2.1. A Banach space E is said to satisfy Opial’s condition (Opial [12])if it possesses the property: if a sequence {xn } in E converges weakly to x ∈ E and x 6= y, then lim inf kxn − yk > lim inf kxn − xk. n→∞

n→∞

Banach spaces satisfying Opial’s condition include Hilbert spaces and lp (1 < p < ∞) spaces while the Lp spaces (for p 6= 2) are not Opial spaces. Theorem 2.1. Suppose that the assumptions of Lemma 2.1 are fulfilled and limn→∞ an = 0. Then (i) limn→∞ kxn+1 − T xn+1 k = 0; (ii) in addition, if E is both reflexive and satisfies Opial’s condition, then {xn } weakly converges to a fixed point of T . Proof. (i) From the iteration process (1.10) and (ii) in Lemma 2.1, it is easy to verify that lim kxn+1 −T xn+1 k

n→∞

= lim kan xn + (bn − 1)T xn+1 + cn un k (2.3)

n→∞

≤ lim an kxn − T xn+1 k n→∞

+ lim |an + bn − 1|kT xn k + lim cn kun k. n→∞

Since the assumption

n→∞

∞ X

cn <∞ 1 − bn n=1

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Mann’s iteration scheme

implies cn → 0, by the fact that {xn }, {T xn } and {un } are bounded and limn→∞ an → 0, we have lim kxn+1 − T xn+1 k = 0.

n→∞

(ii) Put A = (2I −T )−1 , from Theorem 6 in Martin [9] we know that A is a non-expansive self-mapping on K and F (T ) = F (A) (see also [6, 18, 20, 21]). From (2.3) and A = (2I − T )−1 , we have kxn+1 − Axn+1 k

= kAA−1 xn+1 − Axn+1 k ≤ kA−1 xn+1 − xn+1 k = kxn+1 − T xn+1 k,

and lim kxn+1 − Axn+1 k = 0.

n→∞

Since E is a reflexive Banach space and {xn } is bounded, then there exits subsequence {xnk } of {xn } such that {xnk } converges weakly to some point x∗ ∈ K, and lim kxnk − Axnk k = 0. k→∞

By the Opial’s condition and A being non-expansive, if x∗ 6= Ax∗ , we have lim inf kxnk − x∗ k < lim inf kxnk − Ax∗ k ≤ lim inf kxnk − x∗ k. k→∞

k→∞

k→∞

The contradiction implies x∗ ∈ F (A) = F (T ). If there exists another subsequence xnj such that xnj converges weakly to some point y ∈ K and y 6= x∗ (we also have y ∈ F (T )), then by (i) of Lemma 2.1 and the Opial’s condition, we obtain lim kxn − x∗ k = lim kxnk − x∗ k < lim kxnk − yk

n→∞

k→∞

k→∞

= lim kxnj − yk < lim kxnj − x∗ k j→∞

j→∞

∗

= lim kxnk − x k. n→∞

This contradiction means y = x∗ . So the desired results are obtained.

2

Theorem 2.2. Let K be a non-empty closed convex subset of the Hilbert space E, other assumptions being the same as in Lemma 2.1. Then (i) limn→∞ kxn+1 − T xn+1 k = 0; (ii) {xn } weakly converges to a fixed point of T .

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Proof. At first we recall the well-known identity in Hilbert space E: for any x, y ∈ E and t ∈ [0, 1], (2.4)

ktx + (1 − t)yk2 = tkxk2 + (1 − t)kyk2 − t(1 − t)kx − yk2 .

Given p ∈ F (T ), we have kxn+1 − pk2 = kbn (T xn+1 − p + cn (un − xn )) (2.5)

+ (1 − bn )(xn − p + cn (un − xn ))k2 =: kXn k2 + 2hXn , Yn i + kYn k2 ,

where

Xn = bn ((1 − cn )(T xn+1 − p) + cn (un − xn )) + (1 − bn )((1 − cn )(xn − p) + cn (un − xn ))

and Yn = bn cn (T xn+1 − p) + (1 − bn )cn (xn − p). By (2.4) and simple calculations, we have kXn k2 = bn k(1 − cn )(T xn+1 − p) + cn (un − xn )k2 + (1 − bn )k(1 − cn )(xn − p) + cn (un − xn )k2

(2.6)

− bn (1 − bn )(1 − cn )2 kT xn+1 − xn k2 n = bn (1 − cn )kT xn+1 − pk2 + cn kun − xn k2 o − cn (1 − cn )kT xn+1 − p − (un − xn )k2 n + (1 − bn ) (1 − cn )kxn − pk2 + cn kun − xn k2 o − cn (1 − cn )k2xn − p − un k2 − bn (1 − bn )(1 − cn )2 kT xn+1 − xn k2 ,

(2.7)

kYn k2 = c2n kbn (T xn+1 − p) + (1 − bn )(xn − p)k2 n = c2n bn kT xn+1 − pk2 + (1 − bn )kxn − pk2 o − bn (1 − bn )kT xn+1 − xn k2 ,

and (2.8)

hXn , Yn i = cn hXn , bn (T xn+1 − p) + (1 − bn )(xn − p)i.

From Lemma 2.1 and our assumptions we know that {xn }, {T xn } and {un } are bounded. Thus, from (2.6) to (2.8), it is easy to check that kxn+1 − pk2 = bn (1 − cn )kT xn+1 − pk2 (2.9)

+ (1 − bn )(1 − cn )kxn − pk2 − bn (1 − bn )(1 − cn )2 kT xn+1 − xn k2 + O(cn ).

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Mann’s iteration scheme

Furthermore, from (1.6) and the iteration process (1.10), we have kT xn+1 − pk2 ≤kxn+1 − pk2 + kxn+1 − T xn+1 k2 ≤ kxn+1 − pk2 © ª2 + an kxn − T xn+1 k + cn kT xn+1 k + cn kun k

(2.10)

= kxn+1 − pk2 + a2n kxn − T xn+1 k2 + O(cn ). From (2.9) and (2.10), it follows that kxn+1 − pk2 ≤ bn (1 − cn )kxn+1 − pk2 + (1 − bn )(1 − cn )kxn − pk2

(2.11)

+ bn (1 − cn )(a2n − (1 − bn )(1 − cn ))kxn − T xn+1 k2 + O(cn ),

which implies bn (1 − cn )((1 − bn )(1 − cn )− a2n )kxn − T xn+1 k2 ≤ (1 − bn )(1 − cn )kxn − pk2

(2.12)

− (1 − bn (1 − cn ))kxn+1 − pk2 . Since an + bn < 1 and bn ∈ (β, 1) for some β ∈ (0, 1), then (1 − bn )(1 − cn ) − a2n > (1 − bn )bn , 1 − bn (1 − cn ) > (1 − cn )(1 − bn ). By the fact that limn→∞ kxn − pk exists (from (i) in Lemma 2.1) and (2.12), we have (1 − bn )(1 − cn )b2n kxn − T xn+1 k2 ≤ bn (1 − cn )((1 − bn )(1 − cn ) − a2n )kxn − T xn+1 k2 ≤ (1 − bn )(1 − cn )kxn − pk2 − (1 − bn )(1 − cn )kxn+1 − pk2 which implies lim sup β 2 kxn − T xn+1 k2 ≤ lim {kxn − pk2 − kxn+1 − pk2 } = 0. n→∞

n→∞

Thus lim kxn − T xn+1 k = 0.

n→∞

Combining the iteration process (1.10), we obtain lim kxn+1 − T xn+1 k = lim {an kxn − T xn+1 k + cn kT xn+1 k + cn kun k} = 0.

n→∞

n→∞

Putting A = (2I − T )−1 , since every Hilbert space is a reflexive Banach space and satisfies Opial’s condition, the remainder of the proof is the same as the proof of Theorem 2.1. 2

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3. Strong convergence theorems Theorem 3.1. Assume that K is a compact convex subset of a real Banach space E and the other assumptions are the same as in Lemma 2.1. In addition, suppose that limn→∞ an = 0. Then {xn } strongly converges to a fixed point of T . Proof. Since K is a compact convex subset of E and {xn } is bounded, there exists a subsequence {xnk } of {xn } such that xnk strongly converges to x∗ ∈ K. By Theorem 2.1 and the continuity of the mapping T , we have kx∗ − T x∗ k = lim kxnk − T xnk k = 0, k→∞

which implies x∗ ∈ F (T ). From (i) of Lemma 2.1, for all p ∈ F (T ), limn→∞ kxn −pk exists. Therefore, we have lim kxn − x∗ k = 0. n→∞

2 With the same proof, we obtain Theorem 3.2. Assume that K is a compact convex subset of a real Hilbert space E and the other assumptions are the same as in Theorem 2.2. Then {xn } strongly converges to a fixed point of T . From Theorem 3.1 and Theorem 3.2, we have the following sufficient and necessary conditions. Theorem 1. Suppose that K is a non-empty closed convex subset of E and the other assumptions are the same as in Theorem 3.1 or Theorem 3.2. Then {xn } strongly converges to a fixed point of T if and only if {xn } exists as a strongly convergent subsequence.

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Received December 21, 2006

Yu Miao College of Mathematics and Information Science Henan Normal University 453007 Henan, CHINA E-mail: [email protected]