Volume 2012, Issue 1 575014
Research Article
Open Access

Strong Convergence Theorems for the Generalized Split Common Fixed Point Problem

Cuijie Zhang

Corresponding Author

Cuijie Zhang

College of Science, Civil Aviation University of China, Tianjin 300300, China

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First published: 08 May 2012
Citations: 1
Academic Editor: Rudong Chen

Abstract

We introduce the generalized split common fixed point problem (GSCFPP) and show that the GSCFPP for nonexpansive operators is equivalent to the common fixed point problem. Moreover, we introduce a new iterative algorithm for finding a solution of the GSCFPP and obtain some strong convergence theorems under suitable assumptions.

1. Introduction

Let H1 and H2 be real Hilbert spaces and let A : H1H2 be a bounded linear operator. Given intergers p, r ≥ 1, let us recall that the multiple-set split feasibility problem (MSSFP) was recently introduced [1] and is to find a point:
(1.1)
where and are nonempty closed convex subsets of H1 and H2, respectively. If p = r = 1, the MSSFP (1.1) becomes the so-called split feasibility problem (SFP) [2] which is to find a point:
(1.2)
where C and Q are nonempty closed convex subsets of H1 and H2, respectively. Recently, the SFP (1.2) and MSSFP (1.1) have been investigated by many researchers; see, [310].
Since every closed convex subset in a Hilbert space is looked as the fixed point set of its associating projection, the MSSFP (1.1) becomes a special case of the split common fixed point problem (SCFPP), which is to find a point:
(1.3)
where Ui : H1H1  (i = 1,2, …, p) and Tj : H2H2  (j = 1,2, …, r) are nonlinear operators. If p = r = 1, the problem (1.3) reduces to the so-called two-set SCFPP, which is to find a point:
(1.4)
Censor and Segal in [11] firstly introduced the concept of SCFPP in finite-dimensional Hilbert spaces and considered the following iterative algorithm for the two-set SCFPP (1.4) for Class- operators:
(1.5)
where x0H1, 0 < γ < 2/∥A2 and I is the identity operator. They proved the convergence of the algorithm (1.5) to a solution of problem (1.4). Moreover, they introduced a parallel iterative algorithm, which converges to a solution of the SCFPP (1.3). However, the parallel iterative algorithm does not include the algorithm (1.5) as a special case.
Very recently, Wang and Xu in [12] considered the SCFPP (1.3) for Class- operators and introduced the following iterative algorithm for solving the SCFPP (1.3):
(1.6)
Under some mild conditions, they proved some weak and strong convergence theorems. Their iterative algorithm (1.6) includes Censor and Segal’s algorithm (1.5) as a special case for the two-set SCFPP (1.4). Moreover, they prove that the SCFPP (1.3) for the Class- operators is equivalent to a common fixed point problem. This is also a classical method. Many problems eventually converted to a common fixed point problem; see [1315].
Motivated and inspired by the aforementioned research works, we introduce a generalized split common fixed point problem (GSCFPP) which is to find a point:
(1.7)
Then, we show that the GSCFPP (1.7) for nonexpansive operators is equivalent to the following common fixed point problem:
(1.8)
where Vj = IγA*(ITj)A  (0 < γ ≤ 1/∥A2) for every j. Moreover, we give a new iterative algorithm for solving the GSCFPP (1.7) for nonexpansive operators and obtain some strong convergence theorems.

2. Preliminaries

Throughout this paper, we write xnx and xnx to indicate that {xn} converges weakly to x and converges strongly to x, respectively.

An operator T : HH is said to be nonexpansive if ∥TxTy∥≤∥xy∥ for all x, yH. The set of fixed points of T is denoted by F(T). It is known that F(T) is closed and convex. An operator f : HH is called contraction if there exists a constant ρ ∈ [0,1) such that ∥f(x) − f(y)∥≤ρxy∥ for all x, yH. Let C be a nonempty closed convex subset of H. For each xH, there exists a unique nearest point in C, denoted by PCx, such that ∥xPCx∥≤∥xy∥ for every yC. PC is called a metric projection of H onto C. It is known that for each xH,
(2.1)
for all yC.
Let {Tn} be a sequence of operators of H into itself. The set of common fixed points of {Tn} is denoted by F({Tn}), that is, . A sequence {Tn} is said to be strongly nonexpansive if each {Tn} is nonexpansive and
(2.2)
whenever {xn} and {yn} are sequences in C such that {xnyn} is bounded and ∥xnyn∥−∥TnxnTnyn∥→0; see [16, 17]. A sequence {zn} in H is said to be an approximate fixed point sequence of {Tn} if znTnzn → 0. The set of all bounded approximate fixed point sequences of {Tn} is denoted by ; see [16, 17]. We know that if {Tn} has a common fixed point, then is nonempty; that is, every bounded sequence in the common fixed point set is an approximate fixed point sequence. A sequence {Tn} with a common fixed point is said to satisfy the condition (Z) if every weak cluster point of {xn} is a common fixed point whenever . A sequence {Tn} of nonexpansive mappings of H into itself is said to satisfy the condition (R) if
(2.3)
for every nonempty bounded subset D of H; see [18].

In order to prove our main results, we collect the following lemmas in this section.

Lemma 2.1 (see [16].)Let C be a nonempty subset of a Hilbert space H. Let {Tn} be a sequence of nonexpansive mappings of C into H. Let {λn} be a sequence in [0,1] such that lim  inf nλn > 0. Let {Un} be a sequence of mappings of C into H defined by Un = λnI + (1 − λn)Tn for n, where I is the identity mapping on C. Then {Un} is a strongly nonexpansive sequence.

Lemma 2.2 (see [16].)Let H be a Hilbert space, C a nonempty subset of H, and {Sn} and {Tn} sequences of nonexpansive self-mappings of C. Suppose that {Sn} or {Tn} is a strongly nonexpansive sequence and is nonempty. Then .

Lemma 2.3 (see [17].)Let H be a Hilbert space, and C a nonempty subset of H. Both {Sn} and {Tn} satisfy the condition (R) and {Tny : n,   yD} is bounded for any bounded subset D of C. Then {SnTn} satisfies the condition (R).

Lemma 2.4 (see [19].)Let {xn} and {yn} be bounded sequences in a Banach space X and let {βn} be a sequence in [0,1] with 0 < lim  inf nβn ≤ limsup nβn < 1. Suppose xn+1 = (1 − βn)  yn + βn  xn  for all integers n    0 and

(2.4)
Then lim nynxn∥ = 0.

Lemma 2.5 (see [20].)Assume that {an} is a sequence of nonnegative real numbers such that

(2.5)
where {γn} is a sequence in (0,1) and {δn} is a sequence such that
  • (i)

    ,

  • (ii)

    limsup nδn/γn ≤ 0 or .

Then lim nan = 0.

3. Main Results

Now we state and prove our main results of this paper.

Lemma 3.1. Let A : H1H2 be a given bounded linear operator and let Tn : H2H2 be a sequence of nonexpansive operators. Assume

(3.1)
For each constant γ > 0, Vn is defined by the following:
(3.2)
Then Fix ({Vn}) = A−1(Fix ({Tn})). Moreover, for 0 < γ ≤ 1/∥A2, Vn is nonexpansive on H1 for n.

Proof. Since the inclusion A−1(Fix ({Tn}))⊆Fix ({Vn}) is evident, now we only need to show the converse inclusion. If z ∈ Fix ({Vn}), then we have A*(ITn)Az = 0. Since A−1(Fix ({Tn})) ≠ , we take an arbitrary pA−1(Fix ({Tn})). Hence

(3.3)
It follows that (1/2)∥AzTnAz2 ≤ 0, then Az = TnAz for every n, hence zA−1(Fix ({Tn})). Next we turn to show that Vn is a nonexpansive operator for n. Since Tn is nonexpansive, we have
(3.4)
Hence
(3.5)
For 0 < γ ≤ 1/∥A2, we can immediately obtain that Vn is a nonexpansive operator for every n.

From Lemma 3.1, we can obtain that the solution set of GSCFPP (1.7) is identical to the solution set of problem (1.8).

Theorem 3.2. Let {Un} and {Vn} be sequences of nonexpansive operators on Hilbert space H1. Both {Un} and {Vn} satisfy the conditions (R) and (Z). Let f : H1H1 be a contraction with coefficient ρ ∈ [0,1). Suppose Ω = Fix (Un)⋂Fix (Vn) ≠ . Take an initial guess x1H1 and define a sequence {xn} by the following algorithm:

(3.6)
where {αn}, {βn}, {γn}, and {λn} are sequences in [0,1]. If the following conditions are satisfied:
  • (i)

    αn + βn + γn = 1,   for  all  n ≥ 1;

  • (ii)

    lim nαn = 0  and   ;

  • (iii)

    0 < lim  inf nβn ≤ limsup nβn < 1;

  • (iv)

    0 < lim  inf nλn ≤ limsup nλn < 1;

  • (v)

    lim n | λn+1λn | = 0,

then {xn} converges strongly to w ∈ Ω where w = PΩf(w).

Proof. We proceed with the following steps.

Step 1. First show that there exists w ∈ Ω such that w = PΩf(w).

In fact, since f is a contraction with coefficient ρ, we have

(3.7)
for every x, y. Hence PΩf is also a contraction. Therefore, there exists a unique w ∈ Ω such that w = PΩf(w).

Step 2. Now we show that {xn} is bounded.

Let p ∈ Ω, then p ∈ Fix ({Un}) and p ∈ Fix ({Vn}). Hence

(3.8)
Then
(3.9)
By induction on n,
(3.10)
for every n. This shows that {xn} and {Vnxn} are bounded, and hence, {Unyn}, {yn}, and {f(xn)} are also bounded.

Step 3. We claim that and , where An = λnI + (1 − λn)Vn.

We first show the former equality. Let {zn} be a bounded sequence in H1. If , then

(3.11)
Hence . On the other hand, if , combining (3.11) and limsup nλn < 1, we obtain that ∥Vnznzn∥→0. Hence . Therefore, .

Next, we show the latter equality. Using Lemma 2.1, we know that {An} is a strongly nonexpansive sequence. Thus, since , from Lemma 2.2 we have

(3.12)

Step 4. {Sn} satisfies the condition (R), where Sn = UnAn.

Let D be a nonempty bounded subset of H1. From the definition of {An}, we have, for all yD,

(3.13)
It follows that
(3.14)
Since {Vn} satisfies the condition (R) and lim n | λn+1λn | = 0, we have
(3.15)
that is, {An} satisfies the condition (R). Since {Any : n,     yD} is bounded for any bounded subset D of H1, by using Lemma 2.3, we have that {VnAn} satisfies the condition (R), that is, {Sn} satisfies the condition (R).

Step 5. We show ∥xn+1xn∥→0.

We can write (3.6) as xn+1 = βnxn + (1 − βn)zn where zn = (αnf(xn) + γnSnxn)/1 − βn. It follows that

(3.16)
From Step 2, we may assume that {xn} ⊂ D, where D is a bounded set of H1. Then from (3.16), we obtain
(3.17)
It follows that
(3.18)
Since {Sn} satisfies the condition (R), combining αn → 0 as n, we have
(3.19)
Hence by Lemma 2.4, we get ∥znxn∥→0 as n. Consequently,
(3.20)

Step 6. We claim that .

From (3.6), we have

(3.21)
and hence
(3.22)
Since ∥xn+1xn∥→0, αn → 0 and limsup nβn < 1, we derive
(3.23)
Thus (3.23) and Steps 2 and 3 imply that
(3.24)

Step 7. Show limsup nf(w) − w, xnw〉≤0, where w = PΩf(w).

Since {xn} is bounded, there exist a point vH1 and a subsequence of {xn} such that

(3.25)
and . Since {Un} and {Vn} satisfy the condition (Z), from Step 6, we have vF({Un})∩F({Vn}). Using (2.1), we get
(3.26)

Step 8. Show xnw = PΩf(w).

Since w ∈ Ω, using (3.8), we have

(3.27)
which implies that
(3.28)
for every n. Consequently, according to Step 7, ρ ∈ [0,1), and Lemma 2.5, we deduce that {xn} converges strongly to w = PΩ(w). This completes the proof.

Combining Lemma 3.1 and Theorem 3.2, we can obtain the following strong convergence theorem for solving the GSCFPP (1.7).

Theorem 3.3. Let {Un} and {Tn} be sequences of nonexpansive operators on Hilbert space H1 and H2, respectively. Both {Un} and {Tn} satisfy the conditions (R) and (Z). Let f : H1H1 be a contraction with coefficient ρ ∈ [0,1). Suppose that the solution set Ω of GSCFPP (1.7) is nonempty. Take an initial guess x1H1 and define a sequence {xn} by the following algorithm:

(3.29)
where γ ∈ (0, 1/∥A2), and {αn}, {βn}, {γn}, {λn} are sequences in [0,1]. If the following conditions are satisfied:
  • (i)

    αn + βn + γn = 1,   for  all  n ≥ 1;

  • (ii)

    lim nαn = 0  and  ;

  • (iii)

    0 < liminf nβn ≤ limsup nβn < 1;

  • (iv)

    0 < liminf nλn ≤ limsup nλn < 1;

  • (v)

    lim n | λn+1λn | = 0,

then {xn} converges strongly to w ∈ Ω where w = PΩf(w).

Proof. Set Vn = IγA*(ITn)A. By Lemma 3.1, Vn is a nonexpansive operator for every n. We can rewrite (3.29) as

(3.30)

We only need to prove that {Vn} satisfies the conditions (R) and (Z). Assume that D is a nonempty bounded subset of H1. For every yD, we have

(3.31)
Since {Tn} satisfies the condition (R), and D = {Ay : yD} is bounded, it follows from (3.31) that
(3.32)
Therefore, {Vn} satisfies the condition (R).

Assume that xnz and xnVnxn → 0; we next show that Vnz = z. By using xnVnxn → 0, we have A*(ITn)Axn → 0. Since A−1(Fix ({Tn})) ≠ , we choose an arbitrary point pA−1(Fix ({Tn})); then for every n,

(3.33)
Hence
(3.34)
Then we get . Since {Tn} satisfies the condition (Z) and AxnAz, we have AzF({Tn}). From Lemma 3.1, we have z ∈ Fix ({Vn}).

Let T : HH be a nonexpansive mapping with a fixed point, and define Tn = T for all n. Then {Tn} satisfies the conditions (R) and (Z). Thus, one obtains the algorithm for solving the two-set SCFPP (1.4).

Corollary 3.4. Let U and T be nonexpansive operators on Hilbert space H1 and H2, respectively. Let f : H1H1 be a contraction with coefficient ρ ∈ [0,1). Suppose that the solution set Ω of SCFPP (1.4) is nonempty. Take an initial guess x1H1 and define a sequence {xn} by the following algorithm in (3.29), where γ ∈ (0, 1/∥A2), and {αn}, {βn}, {γn}, {λn} are sequences in [0,1]. If the following conditions are satisfied:

  • (i)

    αn + βn + γn = 1,   for  all  n ≥ 1;

  • (ii)

    lim nαn = 0 and ;

  • (iii)

    0 < lim  inf nβn ≤ limsup nβn < 1;

  • (iv)

    0 < lim  inf nλn ≤ limsup nλn < 1;

  • (v)

    lim n | λn+1λn | = 0.

Then {xn} converges strongly to w ∈ Ω where w = PΩf(w).

Remark 3.5. By adding more operators to the families {Un} and {Tn} by setting Ui = I for ip + 1 and Tj = I for jr + 1, the SCFPP (1.3) can be viewed as a special case of the GSCFPP (1.7).

Acknowledgment

This research is supported by the science research foundation program in Civil Aviation University of China (07kys09), the Fundamental Research Funds for the Central Universities (Program No. ZXH2011D005), and the NSFC Tianyuan Youth Foundation of Mathematics of China (No. 11126136).

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