Strong Convergence Theorems for the Generalized Split Common Fixed Point Problem
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
2. Preliminaries
Throughout this paper, we write xn⇀x and xn → x to indicate that {xn} converges weakly to x and converges strongly to x, respectively.
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 n→∞λn > 0. Let {Un} be a sequence of mappings of C into H defined by Un = λnI + (1 − λn)Tn for n ∈ 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 ∈ ℕ, y ∈ D} 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 n→∞βn ≤ limsup n→∞βn < 1. Suppose xn+1 = (1 − βn) yn + βn xn for all integers n ≥ 0 and
Lemma 2.5 (see [20].)Assume that {an} is a sequence of nonnegative real numbers such that
- (i)
,
- (ii)
limsup n→∞δn/γn ≤ 0 or .
3. Main Results
Now we state and prove our main results of this paper.
Lemma 3.1. Let A : H1 → H2 be a given bounded linear operator and let Tn : H2 → H2 be a sequence of nonexpansive operators. Assume
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*(I − Tn)Az = 0. Since A−1(Fix ({Tn})) ≠ ∅, we take an arbitrary p ∈ A−1(Fix ({Tn})). Hence
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 : H1 → H1 be a contraction with coefficient ρ ∈ [0,1). Suppose Ω = Fix (Un)⋂ Fix (Vn) ≠ ∅. Take an initial guess x1 ∈ H1 and define a sequence {xn} by the following algorithm:
- (i)
αn + βn + γn = 1, for all n ≥ 1;
- (ii)
lim n→∞αn = 0 and ;
- (iii)
0 < lim inf n→∞βn ≤ limsup n→∞βn < 1;
- (iv)
0 < lim inf n→∞λn ≤ limsup n→∞λn < 1;
- (v)
lim n→∞ | λn+1 − λn | = 0,
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
Step 2. Now we show that {xn} is bounded.
Let p ∈ Ω, then p ∈ Fix ({Un}) and p ∈ Fix ({Vn}). Hence
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
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
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 y ∈ D,
Step 5. We show ∥xn+1 − xn∥→0.
We can write (3.6) as xn+1 = βnxn + (1 − βn)zn where zn = (αnf(xn) + γnSnxn)/1 − βn. It follows that
Step 6. We claim that .
From (3.6), we have
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 : H1 → H1 be a contraction with coefficient ρ ∈ [0,1). Suppose that the solution set Ω of GSCFPP (1.7) is nonempty. Take an initial guess x1 ∈ H1 and define a sequence {xn} by the following algorithm:
- (i)
αn + βn + γn = 1, for all n ≥ 1;
- (ii)
lim n→∞αn = 0 and ;
- (iii)
0 < liminf n→∞βn ≤ limsup n→∞βn < 1;
- (iv)
0 < liminf n→∞λn ≤ limsup n→∞λn < 1;
- (v)
lim n→∞ | λn+1 − λn | = 0,
Proof. Set Vn = I − γA*(I − Tn)A. By Lemma 3.1, Vn is a nonexpansive operator for every n ∈ ℕ. We can rewrite (3.29) as
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 y ∈ D, we have
Assume that xn⇀z and xn − Vnxn → 0; we next show that Vnz = z. By using xn − Vnxn → 0, we have A*(I − Tn)Axn → 0. Since A−1(Fix ({Tn})) ≠ ∅, we choose an arbitrary point p ∈ A−1(Fix ({Tn})); then for every n ∈ ℕ,
Let T : H → H be a nonexpansive mapping with a fixed point, and define Tn = T for all n ∈ 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 : H1 → H1 be a contraction with coefficient ρ ∈ [0,1). Suppose that the solution set Ω of SCFPP (1.4) is nonempty. Take an initial guess x1 ∈ H1 and define a sequence {xn} by the following algorithm in (3.29), where γ ∈ (0, 1/∥A∥2), 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 n→∞αn = 0 and ;
- (iii)
0 < lim inf n→∞βn ≤ limsup n→∞βn < 1;
- (iv)
0 < lim inf n→∞λn ≤ limsup n→∞λn < 1;
- (v)
lim n→∞ | λn+1 − λn | = 0.
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).