Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 817392, 6 pages http://dx.doi.org/10.1155/2013/817392 Research Article The Problem of Image Recovery by the Metric Projections in Banach Spaces Yasunori Kimura1 and Kazuhide Nakajo2 1 2 Department of Information Science, Toho University, Miyama, Funabashi, Chiba 274-8510, Japan Sundai Preparatory School, Kanda-Surugadai, Chiyoda-ku, Tokyo 101-8313, Japan Correspondence should be addressed to Yasunori Kimura; yasunori@is.sci.toho-u.ac.jp Received 24 September 2012; Accepted 28 December 2012 Academic Editor: Jaan Janno Copyright © 2013 Y. Kimura and K. Nakajo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We consider the problem of image recovery by the metric projections in a real Banach space. For a countable family of nonempty closed convex subsets, we generate an iterative sequence converging weakly to a point in the intersection of these subsets. Our convergence theorems extend the results proved by Bregman and Crombez. 1. Introduction Let πΆ1 , πΆ2 , . . . , πΆπ be nonempty closed convex subsets of a real Hilbert space π» such that βππ=1 πΆπ =ΜΈ 0. Then, the problem of image recovery may be stated as follows: the original unknown image π§ is known a priori to belong to the intersection of {πΆπ }ππ=1 ; given only the metric projections ππΆπ of π» onto πΆπ for π = 1, 2, . . . , π, recover π§ by an iterative scheme. Such a problem is connected with the convex feasibility problem and has been investigated by a large number of researchers. Bregman [1] considered a sequence {π₯π } generated by cyclic projections, that is, π₯0 = π₯ ∈ π», π₯1 = ππΆ1 π₯, π₯2 = ππΆ2 π₯1 , π₯3 = ππΆ3 π₯2 , . . . , π₯π = ππΆπ π₯π−1 , π₯π+1 = ππΆ1 π₯π , π₯π+2 = ππΆ2 π₯π+1 , . . .. It was proved that {π₯π } converges weakly to an element of βππ=1 πΆπ for an arbitrary initial point π₯ ∈ π». Crombez [2] proposed a sequence {π¦π } generated by π¦0 = π¦ ∈ π», π¦π+1 = πΌ0 π¦π + ∑ππ=1 πΌπ (π¦π + π π (ππΆπ π¦π − π¦π )) for π = 0, 1, 2, . . ., where 0 < πΌπ < 1 for all π = 0, 1, 2, . . . , π with ∑ππ=0 πΌπ = 1 and 0 < π π < 2 for every π = 1, 2, . . . , π. It was proved that {π¦π } converges weakly to an element of βππ=1 πΆπ for an arbitrary initial point π¦ ∈ π». Later, Kitahara and Takahashi [3] and Takahashi and Tamura [4] dealt with the problem of image recovery by convex combinations of nonexpansive retractions in a uniformly convex Banach space πΈ. Alber [5] took up the problem of image recovery by the products of generalized projections in a uniformly convex and uniformly smooth Banach space πΈ whose duality mapping is weakly sequentially continuous (see also [6, 7]). On the other hand, using the hybrid projection method proposed by Haugazeau [8] (see also [9–11] and references therein) and the shrinking projection method proposed by Takahashi et al. [12] (see also [13]), Nakajo et al. [14] and Kimura et al. [15] considered this problem by the metric projections and proved convergence of the iterative sequence to a common point of countable nonempty closed convex subsets in a uniformly convex and smooth Banach space πΈ and in a strictly convex, smooth, and reflexive Banach space πΈ having the Kadec-Klee property, respectively. Kohsaka and Takahashi [16] took up this problem by the generalized projections and obtained the strong convergence to a common point of a countable family of nonempty closed convex subsets in a uniformly convex Banach space whose norm is uniformly GaΜteaux differentiable (see also [17, 18]). Although these results guarantee the strong convergence, they need to use metric or generalized projections onto different subsets for each step, which are not given in advance. In this paper, we consider this problem by the metric projections, which are one of the most familiar projections to deal with. The advantage of our results is that we use projections onto the given family of subsets only, to generate 2 Abstract and Applied Analysis the iterative scheme. Our convergence theorems extend the results of [1, 2] to a Banach space πΈ, and they deduce the weak convergence to a common point of a countable family of nonempty closed convex subsets of πΈ. There are a number of results dealing with the image recovery problem from the aspects of engineering using nonlinear functional analysis (see, e.g., [19]). Comparing with these researches, we may say that our approach is more abstract and theoretical; we adopt a general Banach space with several conditions for an underlying space, and therefore, the technique of the proofs can be applied to various mathematical results related to nonlinear problems defined on Banach spaces. for every π¦∗ ∈ πΈ∗ , where π ∈ R satisfies 1/π + 1/π = 1. For π > 1, the duality mapping π½π of a smooth Banach space πΈ is said to be weakly sequentially continuous if π₯π β π₯ implies ∗ that π½π π₯π β π½π π₯ (see [20, 21] for details). The following is proved by Xu [22] (see also [23]). Theorem 1 (Xu [22]). Let πΈ be a smooth Banach space and π > 1. Then, πΈ is π-uniformly convex if and only if there exists a constant π > 0 such that βπ₯ + π¦βπ ≥ βπ₯βπ +πβ¨π¦, π½π π₯β©+πβπ¦βπ holds for every π₯, π¦ ∈ πΈ. Remark 2. For a π-uniformly convex and smooth Banach space πΈ, we have that the constant π in the theorem above satisfies π ≤ 1. Let 2. Preliminaries Throughout this paper, let N be the set of all positive integers, R the set of all real numbers, πΈ a real Banach space with norm β⋅β, and πΈ∗ the dual of πΈ. For π₯ ∈ πΈ and π₯∗ ∈ πΈ∗ , we denote by β¨π₯, π₯∗ β© the value of π₯∗ at π₯. We write π₯π → π₯ to indicate that a sequence {π₯π } converges strongly to π₯. Similarly, π₯π β π₯ ∗ and π₯π β π₯ will symbolize weak and weak∗ convergence, respectively. We define the modulus πΏπΈ of convexity of πΈ as follows: πΏπΈ is a function of [0, 2] into [0, 1] such that σ΅©σ΅©σ΅©π₯ + π¦σ΅©σ΅©σ΅© σ΅© : π₯, π¦ ∈ πΈ, βπ₯β = 1, πΏπΈ (π) = inf {1 − σ΅© 2 (1) σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π¦σ΅©σ΅© = 1, σ΅©σ΅©π₯ − π¦σ΅©σ΅© ≥ π} for every π ∈ [0, 2]. πΈ is called uniformly convex if πΏπΈ (π) > 0 for each π > 0. Let π > 1. πΈ is said to be π-uniformly convex if there exists a constant π > 0 such that πΏπΈ (π) ≥ πππ for every π ∈ [0, 2]. It is obvious that a π-uniformly convex Banach space is uniformly convex. πΈ is said to be strictly convex if βπ₯ + π¦β/2 < 1 for all π₯, π¦ ∈ πΈ with βπ₯β = βπ¦β = 1 and π₯ =ΜΈ π¦. We know that a uniformly convex Banach space is strictly convex and reflexive. For every π > 1, the (generalized) ∗ duality mapping π½π : πΈ → 2πΈ of πΈ is defined by σ΅© σ΅© π½π π₯ = {π¦∗ ∈ πΈ∗ : β¨π₯, π¦∗ β© = βπ₯βπ , σ΅©σ΅©σ΅©π¦∗ σ΅©σ΅©σ΅© = βπ₯βπ−1 } (2) for all π₯ ∈ πΈ. When π = 2, π½2 is called the normalized duality mapping. We have that for π, π > 1, βπ₯βπ π½π π₯ = βπ₯βπ π½π π₯ for all π₯ ∈ πΈ. πΈ is said to be smooth if the limit σ΅© σ΅©σ΅© σ΅©π₯ + π‘π¦σ΅©σ΅©σ΅© − βπ₯β (3) lim σ΅© π‘→0 π‘ exists for every π₯, π¦ ∈ πΈ with βπ₯β = βπ¦β = 1. We know that the duality mapping π½π of πΈ is single valued for each π > 1 if πΈ is smooth. It is also known that if πΈ is strictly convex, then the duality mapping π½π of πΈ is one to one in the sense that π₯ =ΜΈ π¦ implies that π½π π₯ ∩ π½π π¦ = 0 for all π > 1. If πΈ is reflexive, then π½π is surjective, and π½π−1 is identical to the duality mapping π½π∗ : πΈ∗ → 2πΈ defined by σ΅© σ΅©π σ΅© σ΅©π−1 π½π∗ π¦∗ = {π₯ ∈ πΈ : β¨π₯, π¦∗ β© = σ΅©σ΅©σ΅©π¦∗ σ΅©σ΅©σ΅© , βπ₯β = σ΅©σ΅©σ΅©π¦∗ σ΅©σ΅©σ΅© } (4) σ΅©π σ΅© π0 = sup {π > 0 : σ΅©σ΅©σ΅©π₯ + π¦σ΅©σ΅©σ΅© ≥ βπ₯βπ + π β¨π¦, π½π π₯β© σ΅© σ΅©π + πσ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© ∀π₯, π¦ ∈ πΈ} . (5) Then, there exists a positive real sequence {ππ } such that limπ → ∞ ππ = π0 and βπ₯ + π¦βπ ≥ βπ₯βπ + πβ¨π¦, π½π π₯β© + ππ βπ¦βπ for all π₯, π¦ ∈ πΈ and π ∈ N. So, we get βπ₯ + π¦βπ ≥ βπ₯βπ + πβ¨π¦, π½π π₯β© + π0 βπ¦βπ for every π₯, π¦ ∈ πΈ. Therefore, π0 is the maximum of constants. In the case of Hilbert spaces, the normalized duality mapping π½2 is the identity mapping and π0 = 1. Let πΈ be a smooth Banach space and π > 1. The function ππ : πΈ × πΈ → R is defined by σ΅© σ΅©π ππ (π¦, π₯) = σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© − π β¨π¦, π½π π₯β© + (π − 1) βπ₯βπ (6) for every π₯, π¦ ∈ πΈ. We have ππ (π₯, π¦) ≥ 0 for all π₯, π¦ ∈ πΈ and ππ (π§, π₯) + ππ (π₯, π¦) = ππ (π§, π¦) + πβ¨π₯ − π§, π½π π₯ − π½π π¦β© for every π₯, π¦, π§ ∈ πΈ. It is known that if πΈ is strictly convex and smooth, then, for π₯, π¦ ∈ πΈ, ππ (π¦, π₯) = ππ (π₯, π¦) = 0 if and only if π₯ = π¦. Indeed, suppose that ππ (π¦, π₯) = ππ (π₯, π¦) = 0. Then, since 0 = ππ (π¦, π₯) + ππ (π₯, π¦) σ΅© σ΅©π = π (βπ₯βπ + σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© − β¨π₯, π½π π¦β© − β¨π¦, π½π π₯β©) σ΅© σ΅©π σ΅© σ΅©π−1 σ΅© σ΅© ≥ π (βπ₯βπ + σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© − βπ₯β σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© − σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© βπ₯βπ−1 ) (7) σ΅© σ΅©π−1 σ΅© σ΅© = π (βπ₯β − σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅©) (βπ₯βπ−1 − σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© ) ≥ 0, we have βπ₯β = βπ¦β. It follows that β¨π¦, π½π π₯β© = π−1 (βπ¦βπ + (π − 1)βπ₯βπ − ππ (π¦, π₯)) = βπ¦βπ and βπ½π π₯β = βπ₯βπ−1 = βπ¦βπ−1 , which implies that π½π π¦ = π½π π₯. Since π½π is one to one, we have π₯ = π¦ (see also [17]). We have the following result from Theorem 1. Abstract and Applied Analysis 3 Lemma 3. Let π > 1 and πΈ be a π-uniformly convex and smooth Banach space. Then, for each π₯, π¦ ∈ πΈ, σ΅©π σ΅© ππ (π₯, π¦) ≥ π0 σ΅©σ΅©σ΅©π₯ − π¦σ΅©σ΅©σ΅© (8) = −ππ (π¦π,π , π₯π ) − ππ π,π β¨π¦π,π − π§, π½π (π₯π − ππΆπ π₯π )β© Proof. Let π₯, π¦ ∈ πΈ. By Theorem 1, we have (9) (12) σ΅©σ΅©π σ΅©π σ΅© ≥ π0 σ΅©σ΅©σ΅©π₯ − π¦σ΅©σ΅©σ΅© , = −ππ (π¦π,π , π₯π ) − ππ π,π β¨π¦π,π − π₯π , π½π (π₯π − ππΆπ π₯π )β© − ππ π,π β¨π₯π − π§, π½π (π₯π − ππΆπ π₯π )β© where π0 is maximum in Remark 2. Hence, we get σ΅© ππ (π₯, π¦) = βπ₯βπ − σ΅©σ΅©σ΅©π¦σ΅©σ΅© − π β¨π₯ − π¦, π½π π¦β© ππ (π§, π¦π,π ) − ππ (π§, π₯π ) = −ππ (π¦π,π , π₯π ) + π β¨π¦π,π − π§, π½π π¦π,π − π½π π₯π β© holds, where π0 is maximum in Remark 2. σ΅©π σ΅© σ΅© σ΅©π βπ₯βπ ≥ σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© + π β¨π₯ − π¦, π½π π¦β© + π0 σ΅©σ΅©σ΅©π₯ − π¦σ΅©σ΅©σ΅© , Proof. Let π¦π,π = π½π∗ (π½π π₯π − π π,π π½π (π₯π − ππΆπ π₯π )) for π ∈ N and π = 1, 2, . . . , π. Then, for π§ ∈ βπ∈N πΆπ , we obtain (10) for all π ∈ N and π = 1, 2, . . . , π. Using Remark 5 with that π§ ∈ πΆπ , we get β¨π₯π − π§, π½π (π₯π − ππΆπ π₯π )β© = β¨π₯π − ππΆπ π₯π , π½π (π₯π − ππΆπ π₯π )β© which is the desired result. Let πΆ be a nonempty closed convex subset of a strictly convex and reflexive Banach space πΈ, and let π₯ ∈ πΈ. Then, there exists a unique element π₯0 ∈ πΆ such that βπ₯0 − π₯β = inf π¦∈πΆβπ¦ − π₯β. Putting π₯0 = ππΆπ₯, we call ππΆ the metric projection onto πΆ (see [24]). We have the following result [25, p. 196] for the metric projection. Lemma 4. Let πΆ be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space πΈ, and let π₯ ∈ πΈ. Then, π¦ = ππΆπ₯ if and only if β¨π¦ − π§, π½2 (π₯ − π¦)β© ≥ 0 for all π§ ∈ πΆ, where ππΆ is the metric projection onto πΆ. Remark 5. For π > 1, it holds that βπ₯βπ½π π₯ = βπ₯βπ−1 π½2 π₯ for every π₯ ∈ πΈ. Therefore, under the same assumption as Lemma 4, we have that π¦ = ππΆπ₯ if and only if β¨π¦ − π§, π½π (π₯ − π¦)β© ≥ 0 for all π§ ∈ πΆ. 3. Main Results + β¨ππΆπ π₯π − π§, π½π (π₯π − ππΆπ π₯π )β© σ΅©π σ΅© ≥ σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© for every π ∈ N and π = 1, 2 . . . , π. Thus, by Lemma 3 we have ππ (π§, π¦π,π ) − ππ (π§, π₯π ) σ΅© σ΅©π ≤ −π0 σ΅©σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© − ππ π,π β¨π¦π,π − π₯π , π½π (π₯π − ππΆπ π₯π )β© σ΅© σ΅©π − ππ π,π σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© σ΅©π−1 σ΅© σ΅©π σ΅© σ΅©σ΅© ≤ −π0 σ΅©σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© + ππ π,π σ΅©σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© σ΅© σ΅©π − ππ π,π σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© Theorem 6. Let π, π > 1 be such that 1/π + 1/π = 1. Let {πΆπ }π∈N be a family of nonempty closed convex subsets of a πuniformly convex and smooth Banach space πΈ whose duality mapping π½π is weakly sequentially continuous. Suppose that βπ∈N πΆπ =ΜΈ 0. Let π π,π ∈]0, (1+1/(π−1))π−1 π0 [ and πΌπ,π ∈ [0, 1] for all π ∈ N and π = 1, 2, . . . , π with ∑ππ=1 πΌπ,π = 1 for every π ∈ N, where π0 is maximum in Remark 2. Let {π₯π } be a sequence generated by π₯1 = π₯ ∈ πΈ and π₯π+1 = π ( ∑ πΌπ,π (π½π π₯π − π π,π π½π (π₯π − ππΆπ π₯π ))) (14) for each π ∈ N and π = 1, 2, . . . , π. Since it holds that Firstly, we consider the iteration of Crombez’s type and get the following result. π½π∗ (13) (11) π π‘ ≤ (15) for π , π‘ ≥ 0, π, π > 1 with 1/π + 1/π = 1, and π½ > 0, we have σ΅©π−1 σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© ≤ 1 σ΅©σ΅© σ΅©π σ΅©π¦ − π₯π σ΅©σ΅©σ΅© π½π π σ΅© π,π 1/(π−1) π + π½π (16) − 1 σ΅©σ΅© σ΅©π σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅© σ΅© σ΅© π for every π ∈ N, π½π > 0 and π ≥ π. Therefore, it follows that ππ (π§, π¦π,π ) − ππ (π§, π₯π ) π=1 for every π ∈ N. If 0 < lim inf π → ∞ π π,π ≤ lim supπ → ∞ π π,π < (1 + 1/(π − 1))π−1 π0 and lim inf π → ∞ πΌπ,π > 0 for each π ∈ N, then {π₯π } converges weakly to a point in β∞ π=1 πΆπ . π‘π 1 π π + π½π−1 π½π π ≤( π π,π σ΅©π σ΅© − π0 ) σ΅©σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© π½π 1/(π−1) + π π,π ((π − 1) π½π σ΅©π σ΅© − π) σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© (17) 4 Abstract and Applied Analysis for every π ∈ N, π = 1, 2, . . . , π, and π½π > 0. Since π’2 ∈ πΆπ for every π ∈ N. Let limπ → ∞ ππ (π’1 , π₯π ) = π1 and limπ → ∞ ππ (π’2 , π₯π ) = π2 . Since π ππ (π§, π₯π+1 ) = βπ§βπ − π β¨π§, ∑ πΌπ,π π½π π¦π,π β© π1 − π2 = lim (ππ (π’1 , π₯ππ ) − ππ (π’2 , π₯ππ )) π→∞ π=1 σ΅© σ΅©π σ΅© σ΅©π = σ΅©σ΅©σ΅©π’1 σ΅©σ΅©σ΅© − σ΅©σ΅©σ΅©π’2 σ΅©σ΅©σ΅© + π lim β¨π’2 − π’1 , π½π π₯ππ β© σ΅©σ΅©π/(π−1) σ΅©σ΅© π σ΅©σ΅© σ΅©σ΅© σ΅© + (π − 1) σ΅©σ΅© ∑ πΌπ,π π½π π¦π,π σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π=1 σ΅© σ΅© π→∞ and π½π is weakly sequentially continuous, we have π ≤ βπ§βπ − π ∑ πΌπ,π β¨π§, π½π π¦π,π β© (18) π=1 σ΅© σ΅©π σ΅© σ΅©π π1 − π2 = σ΅©σ΅©σ΅©π’1 σ΅©σ΅©σ΅© − σ΅©σ΅©σ΅©π’2 σ΅©σ΅©σ΅© + π β¨π’2 − π’1 , π½π π’1 β© = −ππ (π’2 , π’1 ) . π σ΅© σ΅©π + (π − 1) ∑ πΌπ,π σ΅©σ΅©σ΅©π¦π,π σ΅©σ΅©σ΅© π = ∑ πΌπ,π ππ (π§, π¦π,π ) π=1 Using the idea of [9, p. 256], we also have the following result by the iteration of Bregman’s type. for every π ∈ N, we have ππ (π§, π₯π+1 ) − ππ (π§, π₯π ) π π=1 π π,π σ΅©π σ΅© − π0 ) σ΅©σ΅©σ΅©π¦π,π − π₯π σ΅©σ΅©σ΅© π½π π 1/(π−1) + ∑ πΌπ,π π π,π ((π − 1) π½π π=1 σ΅©π σ΅© − π) σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© (19) for all π ∈ N and π½1 , π½2 , . . . , π½π > 0. Since π π,π ∈]0, (1 + 1/(π − 1))π−1 π0 [, πΌπ,π ∈ [0, 1] for all π ∈ N and π = 1, 2, . . . , π, π−1 0 < lim inf π π,π ≤ lim sup π π,π < (1 + π→∞ π→∞ 1 ) (π − 1) π0 , (20) π→∞ for each π ∈ N, we can choose π½π > 0 for every π ∈ N such 1/(π−1) − π) ≤ 0 that πΌπ,π (π π,π /π½π − π0 ) ≤ 0, πΌπ,π π π,π ((π − 1)π½π for all π ≥ π and π→∞ π π,π − π0 ) < 0, π½π lim sup πΌπ,π π π,π ((π − π→∞ 1/(π−1) 1) π½π σ΅©σ΅© π₯π+1 = π½π∗ (π½π π₯π − π π π½π (π₯π − ππΆπ(π) π₯π )) (25) for every π ∈ N, where the index mapping π : N → πΌ satisfies that, for every π ∈ πΌ, there exists ππ ∈ N such that π ∈ {π(π), . . . , π(π + ππ − 1)} for each π ∈ N. If 0 < lim inf π → ∞ π π ≤ lim supπ → ∞ π π < (1 + 1/(π − 1))π−1 π0 , then, {π₯π } converges weakly to a point in βπ∈πΌ πΆπ . (21) − π) < 0 σ΅© σ΅© σ΅© − π₯π σ΅©σ΅©σ΅© = lim σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ π₯π σ΅©σ΅©σ΅©σ΅© = 0 π→∞ ππ (π§, π₯π+1 ) − ππ (π§, π₯π ) ≤( ππ σ΅©π σ΅© − π0 ) σ΅©σ΅©σ΅©π₯π+1 − π₯π σ΅©σ΅©σ΅© π½ (26) σ΅©π σ΅© + π π ((π − 1) π½1/(π−1) − π) σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ(π) π₯π σ΅©σ΅©σ΅©σ΅© for each π ∈ N. Hence, there exists limπ → ∞ ππ (π§, π₯π ) for every π§ ∈ βπ∈N πΆπ and lim σ΅©π¦ π → ∞ σ΅© π,π Theorem 7. Let π, π > 1 be such that 1/π + 1/π = 1. Let πΌ be a countable set and {πΆπ }π∈πΌ a family of nonempty closed convex subsets of a π-uniformly convex and smooth Banach space πΈ whose duality mapping π½π is weakly sequentially continuous. Suppose that βπ∈πΌ πΆπ =ΜΈ 0. Let π π ∈]0, (1 + 1/(π − 1))π−1 π0 [ for all π ∈ N, where π0 is maximum in Remark 2, and let {π₯π } be a sequence generated by π₯1 = π₯ ∈ πΈ and Proof. Let π§ ∈ βπ∈πΌ πΆπ . As in the proof of Theorem 6, we have lim inf πΌπ,π > 0 lim sup πΌπ,π ( (24) Similarly, we obtain π2 − π1 = −ππ (π’1 , π’2 ). So, we get ππ (π’1 , π’2 ) + ππ (π’2 , π’1 ) = 0, that is, π’1 = π’2 . Therefore, {π₯π } converges weakly to a point in βπ∈N πΆπ . π=1 ≤ ∑ πΌπ,π ( (23) (22) for all π ∈ N. It follows from Lemma 3 that {π₯π } is bounded. Let {π₯ππ } and {π₯ππ } be subsequences of {π₯π } such that π₯ππ β π’1 and π₯ππ β π’2 . Then, we get βπ₯ππ −ππΆπ π₯ππ β → 0 which implies that π’1 ∈ πΆπ for every π ∈ N. In the same way, we also have for all π ∈ N and π½ > 0. Since π π ∈]0, (1 + 1/(π − 1))π−1 π0 [ for all π ∈ N and 0 < lim inf π → ∞ π π ≤ lim supπ → ∞ π π < (1 + 1/(π − 1))π−1 π0 , we can find that π½ > 0 such that lim sup ( π→∞ ππ − π0 ) < 0, π½ (27) lim sup π π ((π − 1) π½1/(π−1) − π) < 0. π→∞ Then, there exists limπ → ∞ ππ (π§, π₯π ) for every π§ ∈ βπ∈πΌ πΆπ and σ΅©σ΅© lim σ΅©π₯ π → ∞ σ΅© π+1 σ΅© σ΅© σ΅© − π₯π σ΅©σ΅©σ΅© = lim σ΅©σ΅©σ΅©σ΅©π₯π − ππΆπ(π) π₯π σ΅©σ΅©σ΅©σ΅© = 0. π→∞ (28) Abstract and Applied Analysis 5 So, we have that {π₯π } is bounded from Lemma 3. Let {π₯ππ } be a subsequence of {π₯π } such that π₯ππ β π’. For fixed π ∈ πΌ, there exists a strictly increasing sequence {ππ } ⊂ N such that ππ ≤ ππ ≤ ππ + ππ − 1 and π(ππ ) = π for every π ∈ N. It follows that σ΅©σ΅© σ΅© σ΅©σ΅©π₯ππ − π₯ππ σ΅©σ΅©σ΅© ≤ σ΅© σ΅© ππ +ππ −1 ∑ π=ππ σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 − π₯π σ΅©σ΅©σ΅© (29) π π ∈]0, 2[ for all π ∈ N, and let {π₯π } be a sequence generated by π₯1 = π₯ ∈ π» and π₯π+1 = π₯π − π π (π₯π − ππΆπ(π) π₯π ) for every π ∈ N, where the index mapping π : N → πΌ satisfies that, for every π ∈ πΌ, there exists ππ ∈ N such that π ∈ {π(π), . . . , π(π + ππ − 1)} for each π ∈ N. If 0 < lim inf π → ∞ π π ≤ lim supπ → ∞ π π < 2, then, {π₯π } converges weakly to a point in βπ∈πΌ πΆπ . for all π ∈ N which implies that π₯ππ β π’. Since limπ → ∞ βπ₯ππ − ππΆπ π₯ππ β = 0, π’ ∈ πΆπ for every π ∈ πΌ. So, we get π’ ∈ βπ∈πΌ πΆπ . As in the proof of Theorem 6, using that π½π is weakly sequentially continuous, we get that {π₯π } converges weakly to a point in βπ∈πΌ πΆπ . Acknowledgment Suppose that the index set πΌ is a finite set {0, 1, 2, . . . , π − 1}. For the cyclic iteration, the index mapping π is defined by π(π) = π mod π for each π ∈ πΌ. Clearly it satisfies the assumption in Theorem 7. In the case where the index set πΌ is countably infinite, that is, πΌ = N, one of the simplest examples of π : N → N can be defined as follows: References 1 { { { { 2 { { { { {3 π (π) = { {. . . , { { { { π { { { {. . . . (π = 2π − 1 for some π ∈ N) , (π = 2 (2π − 1) for some π ∈ N) , (π = 4 (2π − 1) for some π ∈ N) , (π = 2π−1 (2π − 1) for some π ∈ N) , (30) Then, the assumption in Theorem 7 is satisfied by letting ππ = 2π for each π ∈ πΌ = N. 4. Deduced Results Since a real Hilbert space π» is 2-uniformly convex and the maximum π0 in Remark 2 is equal to 1, we get the following results. At first, we have the following theorem which generalizes the results of [2] by Theorem 6. Theorem 8. Let {πΆπ }π∈N be a family of nonempty closed convex subsets of π» such that βπ∈N πΆπ =ΜΈ 0. Let π π,π ∈]0, 2[ and πΌπ,π ∈ [0, 1] for all π ∈ N and π = 1, 2, . . . , π with ∑ππ=1 πΌπ,π = 1 for every π ∈ N. Let {π₯π } be a sequence generated by π₯1 = π₯ ∈ π» and π π₯π+1 = ∑ πΌπ,π (π₯π − π π,π (π₯π − ππΆπ π₯π )) (31) π=1 for every π ∈ N. If it holds that 0 < lim inf π → ∞ π π,π ≤ lim supπ → ∞ π π,π < 2 and lim inf π → ∞ πΌπ,π > 0 for each π ∈ N, then, {π₯π } converges weakly to a point in β∞ π=1 πΆπ . Next, we have the following theorem which extends the result of [1] by Theorem 7. Theorem 9. Let πΌ be a countable set and {πΆπ }π∈πΌ a family of nonempty closed convex subsets of π» such that βπ∈πΌ πΆπ =ΜΈ 0. Let (32) The first author was supported by the Grant-in-Aid for Scientific Research no. 22540175 from the Japan Society for the Promotion of Science. [1] L. M. 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