# A Weak Convergence Theorem for Common Fixed Points of Two Nonlinear Mappings in Hilbert Spaces

Authors: Takanori Ibaraki; Shunsuke Kajiba; Yukio Takeuchi
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

In this paper, by using properties of attractive points, we study an iteration scheme combining simplified Baillon type and Mann type to find a common fixed point of commutative two nonlinear mappings in Hilbert spaces. Then, we apply the obtained results to prove a new weak convergence theorem.


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## Body

## 1. Introduction

In 1963, DeMarr [1] proved a common fixed point theorem for a family of commuting nonexpansive mappings in a Banach space. After DeMarr, many researchers studied this subject (see [2–6] and others).On the other hand, in 1975, Baillon [7] proved a mean convergence theorem known as the first nonlinear ergodic theorem in a Hilbert space. After Baillon, many researchers have studied topics related to his mean convergence theorem. In 1997, Shimizu and Takahashi [8] introduced the iteration scheme that combines Baillon type and Halpern type [9]. Then, they proved a strong convergence theorem to a common fixed point of a finite family of commutative nonexpansive mappings in Hilbert spaces. In 1998, Atsushiba and Takahashi [10] introduced the iteration scheme that combines Baillon type and Mann type [11] and proved a weak convergence theorem to a common fixed point of commutative two nonexpansive mappings in uniformly convex Banach spaces. In 2002, Suzuki [12] studied for common fixed points of commutative two nonexpansive mappings in general Banach spaces. Then, he proved a strong convergence theorem using Atsushiba and Takahashi’s iteration scheme. Stimulated by Suzuki [12], Takeuchi [13] introduced a new iteration scheme combining simplified Baillon type and Mann type and proved the following strong convergence theorem in general Banach spaces.Theorem 1 (see [13]).
LetE be a Banach space and let C be a compact convex subset of E. Let S and T be nonexpansive self-mappings on C with ST=TS. Let an be a sequence in 0,1 satisfying 0<liminfn⟶∞an≤limsupn⟶∞an<1. Let x1∈C and define a sequence xn in C by
(1)xn+1=an2n∑i=1n∑j=ii+1SiTjxn+1−anxn,for each n∈ℕ. Then, xn converges strongly to some common fixed point z of S and T.Also, some researchers studied topics related to common fixed points of various nonlinear mappings or semigroups of nonlinear mappings, and some convergence theorems were proved; for example, see [14–16] and others. In addition, Baillon’s theorem [7] evolved as follows. In 2011, Takahashi and Takeuchi [17] proposed the notion of an attractive point of a mapping T. They denote by AT the set of all attractive points of T and by PAT the metric projection from H onto the closed convex set AT. An attractive point is an important notion related to fixed points (see [18–21] and therein). Then, Takahashi and Takeuchi [17] proved the following Baillon type mean convergence theorem finding an attractive point for a wide class of nonlinear mappings called generalized hybrid [22].Theorem 2 (see [17]).
LetH be a Hilbert space and let C be a nonempty subset of H. Let T be a generalized hybrid mapping from C into itself. Let vn and bn be sequences defined by
(2)v1∈C,vn+1=Tvn,bn=1n∑k=1nvk,for each n∈ℕ. Suppose vn is bounded. Then, the following hold:
(1)
AT is nonempty, closed, and convex(2)
bn converges weakly to u∈AT such that u=limn⟶∞PATvnMotivated by the works as above, considering properties of attractive points, we study the iteration scheme proposed by Takeuchi [13]. Then, using the obtained results, we prove a new weak convergence theorem for common fixed points of commutative two nonlinear mappings in Hilbert spaces.


## 2. Preliminaries

We present some of fundamental concepts and some symbols used throughout this paper. We denote byℝ the set of all real numbers, by ℕ the set of all positive integers, and by ℕ0 the set of all nonnegative integers. Also, we denote by ℕi the set k∈ℕ0:i≤k for each i∈ℕ0 and by ℕi,j the set k∈ℕ0:i≤k≤j for each i,j∈ℕ0 with i≤j. Obviously, ℕ1=ℕ.H always denotes a real Hilbert space with inner product ·,· and induced norm ·. Let xn be a sequence in H. Sometimes the strong convergence and weak convergence of xn to a point x∈H are denoted by xn⟶x and xn⇀x, respectively. So, xn⟶x implies limn⟶∞xn−x=0, and xn⇀x implies limn⟶∞xn−x,y=0 for each y∈H. Then, we know the following basic facts:
(i)
A nonempty closed convex subsetC of H is weakly closed(ii)
A bounded sequence ofH has a weakly convergent subsequence(iii)
xn⇀z∈H if every weak cluster point of xn and z are the sameA Hilbert spaceH has the Opial property; that is, if a sequence un in H converges weakly to a point u∈H, then
(3)liminfn⟶∞un−u<liminfn⟶∞un−v,for all v∈Hwith u≠v.LetC be a nonempty subset of H and let T be a mapping from C into H. Sometimes, we denote by T0 the identity mapping I on C. Then, we denote by FT the set of all fixed points of T and by AT the set of all attractive points of T, that is,
(4)FT=x∈C:x=Tx,AT=y∈H:Tx−y≤x−yforallx∈C.T is called nonexpansive if Tx−Ty≤x−y for all x,y∈C. We say that I−T is demiclosed at 0 if z∈FT holds whenever xn is a sequence in C such that xn⇀z for some z∈C and xn−Txn⟶0. Then, I−T is demiclosed at 0 if T is nonexpansive. T is called quasi-nonexpansive if ∅≠FT⊂AT. It is easy to see that a nonexpansive mapping T with FT≠∅ is quasi-nonexpansive. Aoyama et al. [23] proposed λ-hybrid mappings for λ∈ℝ. T is called λ–hybrid if
(5)Tx−Ty2≤x−y2+21−λx−Tx,y−Ty,for all x,y∈C. We can easily verify that a mapping of several important classes of nonlinear mappings is λ-hybrid for some λ∈ℝ. For example, a nonspreading mapping [24] is 0-hybrid; a hybrid mapping [25] is 1/2-hybrid and a nonexpansive mapping is 1-hybrid. T satisfies FT⊂AT if T is λ-hybrid. So a λ-hybrid mapping T is quasi-nonexpansive if FT≠∅. According to Falset et al. [26], T is said to satisfy the condition (E) if there is s∈0,∞ such that
(6)x−Ty≤x−y+sx−Tx,for all x,y∈C. If T satisfies the condition (E), then FT⊂AT. So T is quasi-nonexpansive if T satisfies the condition (E) and FT≠∅.LetC be a subset of H and let T1 and T2 be self-mappings on C. We denote by F the common fixed point set FT1∩FT2 and by A the common attractive point set AT1∩AT2. We place importance on the condition F⊂A. Therefore, we present some facts relevant to this condition. It is easy to see the following:
(a)
F⊂A implies neither FT1⊂AT1 nor FT2⊂AT2(b)
F≠∅ does not imply A≠∅ without the assumption F⊂A(c)
In the case whenC is closed and convex, A≠∅ implies F≠∅. However, A≠∅ does not imply F⊂A(d)
IfT1 and T2 are quasi-nonexpansive mappings, then F⊂A(e)
Even if∅≠F⊂A, neither T1 nor T2 need be quasi-nonexpansiveTo clearly understand such situations, a specific example from [20] is given below.Example 1 (see [20]).
LetD=x=s,t∈ℝ2:s∈0,1,t∈s/2,2s. Then, D is compact and convex. Let T1 and T2 be self-mappings on D defined by
(7)T1x=T1s,t=12s,t+12t,t=12s+14t,t,T2x=T2s,t=12s,t+s,12s=s,14s+12t,for each x=s,t∈D.In this example, we easily see(8)FT1=x1,x2∈D:x2=2x1,FT2=x1,x2∈D:x2=x12,AT1=x1,x2∈ℝ2:x1≤0,AT2=x1,x2∈ℝ2:x2≤0,F=0,0,A=x1,x2∈ℝ2:x1≤0,x2≤0.From this, we can easily confirm the following:(i)
FT1∩FT2⊂AT1∩AT2 holds(ii)
NeitherFT1⊂AT1 nor FT2⊂AT2 holds(iii)
NeitherT1 nor T2 is quasi-nonexpansiveRefer to [17–21] for more details of attractive points.


## 3. Lemmas

We begin this section with preparing the required symbols for the iteration scheme we are dealing with and then present some lemmas which are needed to prove our main result.LetC be a nonempty subset of a Hilbert space H. Let T1 and T2 be self-mappings on C. For each n∈ℕ, define mappings Men, Mon, and Mn from C into H, respectively, by
(9)Menx=1n∑i=0n−1T1iT2ix,Monx=1n∑i=0n−1T1iT2i+1x,Mnx=12n∑i=0n−1∑j=ii+1T1iT2jx=12Menx+12Monx,for eachx∈C. Then, for each n∈ℕ, v∈FMen∩FMon∩FMn holds if v∈F, and u∈AMen∩AMon∩AMn holds if u∈A. In the case of T1T2=T2T1, the following holds:
(10)T1iT2ix=T1T2ix,T1iT2i+1x=T1T2iT2x=T2T1T2ix,for all i∈ℕ0, x∈C.Here are some lemmas which are needed to get our main result. First, since a Hilbert space has the Opial property, the following lemma is easily obtained; for example, see Atsushiba et al. [18].Lemma 3.
LetH be a Hilbert space and let C be a nonempty subset of H. Let un be a sequence in H such that un−w converges for each w∈C. Suppose uni and unj are subsequences of un which converge weakly to u,v∈C, respectively. Then, u=v.The following lemma is due to Ibaraki and Takeuchi [20].Lemma 4 (see [20]).
LetH be a Hilbert space and let C be a nonempty subset of H. Let T be a mapping from C into H. Let a∈0,1, x∈C, and w=ax+1−aTx. Suppose v∈AT. Then, the following holds:
(11)a1−aTx−x2≤x−v2−w−v2.
Suppose further thatC is bounded. Let r>supx∈Cx−v. Then,
(12)a1−a2rTx−x2≤x−v−w−v.We need the following trivial lemma to prove Lemma6.Lemma 5.
Letann∈ℕ0 be a sequence in 0,∞. Then, for each n∈ℕ2, there exists in∈ℕ0,n−2 satisfying mini∈ℕ0,n−2ai+ai+1=ain+ain+1, and therefore, the following holds:
(13)∑i=0n−1ai≥n−12ain+ain+1.Proof.
Fix anyn∈ℕ2. It is trivial that there exists in∈ℕ0,n−2 such that
(14)mini∈ℕ0,n−2ai+ai+1=ain+ain+1.
In the case whenn is odd, n∈ℕ3 and in satisfies
(15)∑i=0n−1ai≥∑i=0n−2ai=∑k=0n−3/2a2k+a2k+1≥∑k=0n−3/2ain+ain+1=n−12ain+ain+1.
In the case whenn is even, in satisfies
(16)∑i=0n−1ai=∑k=0n−2/2a2k+a2k+1≥∑k=0n−2/2ain+ain+1=n2ain+ain+1.
Thus, we see thatin satisfies (13).Lemma 6.
LetH be a Hilbert space and let C be a nonempty subset of H. Let T1 and T2 be self-mappings on C satisfying ∅≠A=AT1∩AT2. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let xn be a bounded sequence in C and define yn by
(17)yn=anMnxn+1−anxn,
for eachn∈ℕ. Suppose limn⟶∞xn−yn=0. Then, there is a sequence inn∈ℕ2 such that in∈ℕ0,n−2⊂ℕ0 for each n∈ℕ2 and
(18)limn⟶∞T1inT2inxn−xn=0,limn⟶∞T1inT2in+1xn−xn=0,limn⟶∞T1in+1T2in+1xn−xn=0.Proof.
For eachn∈ℕ, let Men and Mon be as in (9). Note that 0<a1−b≤an1−an for all n∈ℕ. Fix any u∈A. Since xn is a bounded sequence in C, there is an r∈0,∞ satisfying r>supn∈ℕxn−u. For each n∈ℕ, by (12) in Lemma 4, we see that
(19)anMenxn+1−anxn−u=ann∑i=0n−1T1iT2ixn+1−anxn−u=an1n∑i=0n−1T1iT2ixn−u+1−anxn−u=1n∑i=0n−1anT1iT2ixn−u+1−anxn−u≤1n∑i=0n−1anT1iT2ixn−u+1−anxn−u≤1n∑i=0n−1xn−u−an1−an2rT1iT2ixn−xn2≤xn−u−a1−b2r1n∑i=0n−1T1iT2ixn−xn2.
Then, we have(20)a1−b2r1n∑i=0n−1T1iT2ixn−xn2≤xn−u−anMenxn+1−anxn−u.
In the same way, we also have(21)a1−b2r1n∑i=0n−1T1iT2i+1xn−xn2≤xn−u−anMonxn+1−anxn−u.
By (20) and (21), the following holds:
(22)a1−b4r1n∑i=0n−1T1iT2ixn−xn2+T1iT2i+1xn−xn2≤12xn−u−anMenxn+1−anxn−u+12xn−u−anMonxn+1−anxn−u≤xn−u−12anMenxn+12anMonxn+1−anxn−u=xn−u−anMnxn+1−anxn−u=xn−u−yn−u≤xn−yn.
For eachi∈ℕ0, we regard T1iT2ixn−xn2+T1iT2i+1xn−xn2 as ai in Lemma 5. From Lemma 5, for each n∈ℕ2, there exists in∈ℕ0,n−2⊂ℕ0 satisfying (13). So, we have the sequence inn∈ℕ2 which consists of such in. Then, for each n∈ℕ2, it follows from (13) that
(23)n−12T1inT2inxn−xn2+T1inT2in+1xn−xn2+T1in+1T2in+1xn−xn2+T1in+1T2in+2xn−xn2≤∑i=0n−1T1iT2ixn−xn2+T1iT2i+1xn−xn2.
From this, by (22), we see
(24)a1−b8rn−1nT1inT2inxn−xn2≤a1−b4r1nn−12T1inT2inxn−xn2+T1inT2in+1xn−xn2+T1in+1T2in+1xn−xn2+T1in+1T2in+2xn−xn2≤a1−b4r1n∑i=0n−1T1iT2ixn−xn2+T1iT2i+1xn−xn2≤xn−yn.
Then, bylimn⟶∞xn−yn=0, we immediately see
(25)limsupn⟶∞a1−b8rn−1nT1inT2inxn−xn2≤limn⟶∞xn−yn=0.
Thus, bya1−b/8r>0 and limn⟶∞n−1/n=1, we obtain
(26)limn⟶∞T1inT2inxn−xn2=0.
In the same way, we also have the following:(27)limn⟶∞T1inT2in+1xn−xn2=0,limn⟶∞T1in+1T2in+1xn−xn2=0.The following lemmas for aλ-hybrid mapping and a mapping satisfying the condition (E) were known; for example, see Ibaraki and Takeuchi [20]. Of course, these are extensions of the demiclosed principle in the Hilbert space setting.Lemma 7.
LetH be a Hilbert space and let C be a subset of H. Let T be a λ-hybrid mapping from C into H. Suppose xn is a sequence in C which converges weakly to some u∈C and satisfies limn⟶∞Txn−xn=0. Then, u∈FT.Lemma 8.
LetH be a Hilbert space and let C be a subset of H. Let T be a mapping from C into H which satisfies the condition E. Suppose xn is a sequence in C which converges weakly to some u∈C and satisfies limn⟶∞Txn−xn=0. Then, u∈FT.


## 4. Main Result and Applications

We present a weak convergence theorem which is our main result.Theorem 9.
LetH be a Hilbert space and let C be a nonempty closed convex subset of H. Let T1 and T2 be self-mappings on C satisfying T1T2=T2T1. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let x1∈C and define a sequence xn in C by
(28)xn+1=an2n∑i=0n−1∑j=ii+1T1iT2jxn+1−anxn=anMnxn+1−anxn,for each n∈ℕ. Set A=AT1∩AT2 and F=FT1∩FT2. Suppose A≠∅ and I−Tj is demiclosed at 0 for each j∈1,2. Then, the following hold:
(1)
A weakly convergent subsequence ofxn exists, and every weakly convergent subsequence of xn converges weakly to a point of F(2)
In the case ofF⊂A, xn converges weakly to some z∈FProof.
Fix anyv∈A. It is trivial that A⊂AMn for all n∈ℕ. Then, by (11) in Lemma 4, we see
(29)0≤an1−anMnxn−xn2≤xn−v2−xn+1−v2,
for eachn∈ℕ. From this, we see that xn+1−v≤xn−v for all n∈ℕ; that is, xn−v converges. So, since 0<a1−b≤an1−an for all n∈ℕ, we have
(30)limn⟶∞Mnxn−xn2=0.
Furthermore, we know that for eachn∈ℕ,
(31)xn+1−xn=anMnxn+1−anxn−xn=anMnxn−xn.
By (30) and (31), limn⟶∞xn+1−xn=0 holds. For each n∈ℕ, we may regard xn+1 as yn in Lemma 6. Thus, since xn is bounded, by Lemma 6, there is a sequence inn∈ℕ2 such that in∈ℕ0,n−2⊂ℕ0 for each n∈ℕ2 and
(32)limn⟶∞T1inT2inxn−xn=0,limn⟶∞T1inT2in+1xn−xn=0,limn⟶∞T1in+1T2in+1xn−xn=0.
We show that (1) holds. Sincexn is bounded, xn has a weakly convergent subsequence. Let xnl be a subsequence of xn which converges weakly to some z∈H. Since C is weakly closed and xnl is a sequence in C, we see z∈C.
We showz∈FT2. By T1T2=T2T1 and (32), we see
(33)liml⟶∞T1inlT2inlxnl−xnl=0,limsupl⟶∞T2T1inlT2inlxnl−T1inlT2inlxnl≤limsupl⟶∞T1inlT2inl+1xnl−xnl+limsupl⟶∞xnl−T1inlT2inlxnl=liml⟶∞T1inlT2inl+1xnl−xnl+liml⟶∞xnl−T1inlT2inlxnl=0.
From the latter, we seeliml⟶∞T2T1inlT2inlxnl−T1inlT2inlxnl=0. Since xnl converges weakly to z∈C, by liml⟶∞T1inlT2inlxnl−xnl=0, T1inlT2inlxnl also converges weakly to z. So, since I−T2 is demiclosed at 0, we see z∈FT2.
We showz∈FT1. Similarly to the discussion above, we have the following:
(34)liml⟶∞T1inlT2inl+1xnl−xnl=0,liml⟶∞T1T1inlT2inl+1xnl−T1inlT2inl+1xnl=0.
So,T1inlT2inl+1xnl converges weakly to z and
(35)liml⟶∞T1T1inlT2inl+1xnl−T1inlT2inl+1xnl=0.
SinceI−T1 is demiclosed at 0, we see z∈FT1. Thus, z∈F=FT1∩FT2.
By the argument so far, we see that a weakly convergent subsequence ofxn exists and any weakly convergent subsequence of xn has to converge weakly to a point of F. So, we confirmed that (1) holds.
We show that (2) holds. We already know thatxn−v converges for each v∈A. Suppose F⊂A. Then, by (1), F⊂A, and Lemma 3, all weak cluster points of xn coincide. That is, xn itself converges weakly to a point z∈F.Next, we present some results derived from Theorem9.Theorem 10.
LetH be a Hilbert space and let C be a nonempty closed convex subset of H. Let T1 and T2 be quasi-nonexpansive self-mappings on C satisfying T1T2=T2T1. Assume that F=FT1∩FT2≠∅ and I−Tj is demiclosed at 0 for each j∈1,2. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let x1∈C and define a sequence xn in C by
(36)xn+1=anMnxn+1−anxn,for each n∈ℕ. Then, xn converges weakly to some z∈F.Proof.
SinceT1 and T2 are quasi-nonexpansive, we know that ∅≠F=FT1∩FT2⊂AT1∩AT2=A holds. Thus, by Theorem 9, we have the result.Theorem 11.
LetH be a Hilbert space and let C be a nonempty closed convex subset of H. Let T1 and T2 be nonexpansive self-mappings on C satisfying T1T2=T2T1. Assume F=FT1∩FT2≠∅. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let x1∈C and define a sequence xn in C by
(37)xn+1=anMnxn+1−anxn,for each n∈ℕ. Then, xn converges weakly to some z∈F.Proof.
If a mappingT is nonexpansive and FT≠∅, then T is quasi-nonexpansive. We also know that I−T is demiclosed at 0 if T is nonexpansive. Thus, by Theorem 10, we have the result.By considering Lemmas7 and 8, we have the following theorems.Theorem 12.
LetH be a Hilbert space and let C be a nonempty closed convex subset of H. Let T1 and T2 be self-mappings on C satisfying T1T2=T2T1. Assume that T1 is λ-hybrid, T2 is μ-hybrid, and F=FT1∩FT2≠∅. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let x1∈C and define a sequence xn in C by
(38)xn+1=anMnxn+1−anxn,for each n∈ℕ. Then, xn converges weakly to some z∈F.Proof.
SinceT1 is λ-hybrid and T2 is μ-hybrid, by F=FT1∩FT2≠∅, we know that T1 and T2 are quasi-nonexpansive. By Lemma 7, we also know that I−T1 and I−T2 are demiclosed at 0. Thus, by Theorem 10, we have the result.Theorem 13.
LetH be a Hilbert space and let C be a nonempty closed convex subset of H. Let T1 and T2 be self-mappings on C satisfying T1T2=T2T1. Assume that T1 and T2 satisfy the condition E and F=FT1∩FT2≠∅. Let a,b∈0,1 satisfy a≤b and let an be a sequence in a,b. For each n∈ℕ, let Mn be as in (9). Let x1∈C and define a sequence xn in C by
(39)xn+1=anMnxn+1−anxn,for each n∈ℕ. Then, xn converges weakly to some z∈F.Proof.
SinceT1 and T2 satisfy the condition (E), by F=FT1∩FT2≠∅, we know that T1 and T2 are quasi-nonexpansive. By Lemma 8, we also know that I−T1 and I−T2 are demiclosed at 0. Thus, by Theorem 10, we have the result.


## 5. Supplement

In this section, we present some examples that complement the argument so far. In advance, recall that weak and strong topologies on a Euclidean space coincide.In the previous section, Theorem9, neither T1 nor T2 need to be quasi-nonexpansive. However, in Theorems 10–13, we deal only with quasi-nonexpansive mappings T1 and T2. We therefore give an example where all conditions of Theorem 9 are satisfied and T1 is not quasi-nonexpansive.Example 2.
LetD=x=s,t∈ℝ2:s∈0,1,t∈−s,s. Let T1 and T2 be continuous self-mappings on D defined, respectively, by
(40)T1x=T1s,t=12s+t,t,T2x=T2s,t=s,−t,for each x=s,t∈D.In this example, notingF=FT1∩FT2 and A=AT1∩AT2, we see
(41)FT1=x1,x2∈D:x1=x2,FT2=x1,x2∈D:x2=0,AT1=x1,x2∈ℝ2:x1≤0,AT2=x1,x2∈ℝ2:x2=0,F=0,0,A=x1,x2∈ℝ2:x1≤0,x2=0.From these, we can easily verify the following:(i)
D is a nonempty compact convex subset of ℝ2(ii)
FT1⊄AT1, that is, T1 is not quasi-nonexpansive(iii)
T2 is nonexpansive and FT2≠∅; that is, T2 is quasi-nonexpansive(iv)
I−T1 and I−T2 are demiclosed at 0, since T1 and T2 are continuous(v)
T1T2=T2T1 and ∅≠F⊂ASo, all conditions of Theorem9 are satisfied and T1 is not quasi-nonexpansive. This implies that the sequence yn generated by the procedure in Theorem 9 converges strongly to 0,0∈F⊂A.Recall that the conditionF⊂A is unnecessary to prove (1) of Theorem 9. That is, without the condition F⊂A, we proved that the sequence xn in Theorem 9 has a subsequence which converges weakly to a point z∈F. So, to gain a better understanding the contents of this paper, we are interested in what happens when the condition F⊂A is missing from Theorem 9. Then, from this point of view, we give the following example.Example 3.
LetD=0,1. Let T1 and T2 be continuous self-mappings on D defined, respectively, by
(42)T1s=2s2whens∈0,12,T1s=2s−122+12whens∈12,1,T2s=4s3whens∈0,12,T2s=4s−123+12whens∈12,1.For this example, we easily see(43)FT1=FT2=0,12,1,AT1=AT2=−∞,0,F=FT1∩FT2=0,12,1,A=AT1∩AT2=−∞,0,T1T2s=24s32=32s6=42s23=T2T1swhens∈0,12,T1T2s=24s−123+12−122+12=32s−126+12,T2T1s=42s−122+12−123+12=32s−126+12whens∈12,1.From these, we see the following:(i)
D is a nonempty compact convex subset of ℝ(ii)
NeitherT1 nor T2 is quasi–nonexpansive(iii)
I−T1 and I−T2 are demiclosed at 0, since T1 and T2 are continuous(iv)
T1T2=T2T1 and A≠∅So, withoutF⊂A, all conditions of Theorem 9 are satisfied.In this example, for ease of verification, we have chosenT1 and T2 which are special in the following sense:
(i)
We can regardT1 and T2 as self-mappings on 0,1/2(ii)
We can regardT1 and T2 as self-mappings on 1/2,1Accordingly, the following should be noted:(i)
F=0,1/2 and A=−∞,0 if T1 and T2 are considered like the former(ii)
F=1/2,1 and A=−∞,1/2 if T1 and T2 are considered like the latterWe know that the sequencexn generated by the procedure in Theorem 9 has a subsequence which converges strongly to a point z of F. Especially, in this example, we can confirm that the sequence xn itself converges strongly to z. That is, by the argument so far, we can easily verify the following:
(i)
xn converges strongly to 0∈F∩A if x1∈0,1/2(ii)
xn converges strongly to 1/2∈F if x1∈1/2,1(iii)
xn converges strongly to 1∈F if x1=1From this, we simultaneously see the following: Example3 is illustrative example such that z∈A may not hold when the condition F⊂A is missing.



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*Source: 9568060-2022-08-26.xml*