# Fixed Point and Acute Point Theorems for New Mappings in a Banach Space

Authors: Toshiharu Kawasaki
Journal: Journal of Mathematics
 (2019)

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## Abstract

In a Hilbert space, concepts of attractive point and acute point are studied by many researchers. Moreover, these concepts extended to Banach space. In this paper, we introduce a new class of mappings on Banach space corresponding to the class of all widely more generalized hybrid mappings on Hilbert space. Moreover, we introduce some extensions of acute point and prove some acute point theorems.


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## Body

## 1. Introduction

In [1] Takahashi and Takeuchi introduced a concept of attractive point in a Hilbert space. Let H be a real Hilbert space, let C be a nonempty subset of H, and let T be a mapping from C into H. x∈H is called an attractive point of T if (1)x-Ty≤x-yfor any y∈C. Let(2)AT=x∈H∣x-Ty≤x-yforanyy∈C.Moreover, they proved that the Baillon type ergodic theorem [2] for generalized hybrid mappings [3] without convexity of C. A mapping T from C into H is said to be generalized hybrid if there exist α,β∈R such that (3)αTx-Ty2+1-αx-Ty2≤βTx-y2+1-βx-y2for any x,y∈C. Such a mapping is said to be (α,β)-generalized hybrid. The class of all generalized hybrid mappings is a new class of nonlinear mappings including nonexpansive mappings, nonspreading mappings [4], and hybrid mappings [5]. A mapping T from C into H is said to be nonexpansive if (4)Tx-Ty≤x-yfor any x,y∈C; nonspreading if (5)2Tx-Ty2≤Tx-y2+Ty-x2for any x,y∈C; hybrid if (6)3Tx-Ty2≤x-y2+Tx-y2+Ty-x2for any x,y∈C. Any nonexpansive mapping is (1,0)-generalized hybrid; any nonspreading mapping is (2,1)-generalized hybrid; any hybrid mapping is 3/2,1/2-generalized hybrid.Motivated these mappings, in [6] Kawasaki and Takahashi introduced a new very wider class of mappings, called widely more generalized hybrid mappings, than the class of all generalized hybrid mappings. A mapping T from C into H is widely more generalized hybrid if there exist α,β,γ,δ,ε,ζ,η∈R such that (7)αTx-Ty2+βx-Ty2+γTx-y2+δx-y2+εx-Tx2+ζy-Ty2+ηx-Tx-y-Ty2≤0for any x,y∈C. Such a mapping is said to be (α,β,γ,δ,ε,ζ,η)-widely more generalized hybrid. This class includes the class of all generalized hybrid mappings and also the class of all k-pseudocontractions [7] for k∈[0,1]. A mapping T from C into H is called a k-pseudocontraction if (8)Tx-Ty2≤x-y2+kx-Tx-y-Ty2for any x,y∈C. Any (α,β)-generalized hybrid mapping is (α,1-α,-β,β-1,0,0,0)-widely more generalized hybrid; any k-pseudocontraction is (1,0,0,-1,0,0,-k)-widely more generalized hybrid. Moreover, they proved some fixed point theorems and some ergodic theorems; for instance, see [6, 8–10].There are some studies on Banach space related to these results. In [11] Takahashi, Wong and Yao introduced the generalized nonspreading mapping and the skew-generalized nonspreading mapping in a Banach space. Let E be a smooth Banach space and let C be a nonempty subset of E. A mapping T from C into E is said to be generalized nonspreading if there exist α,β,γ,δ,ε,ζ∈R such that (9)αϕTx,Ty+βϕx,Ty+γϕTx,y+δϕx,y≤εϕTy,Tx-ϕTy,x+ζϕy,Tx-ϕy,xfor any x,y∈C, where J is the duality mapping on E and (10)ϕu,v=u2-2u,Jv+v2.Such a mapping is said to be (α,β,γ,δ,ε,ζ)-generalized nonspreading. A mapping T from C into E is said to be skew-generalized nonspreading if there exist α,β,γ,δ,ε,ζ∈R such that (11)αϕTx,Ty+βϕx,Ty+γϕTx,y+δϕx,y≤εϕTy,Tx-ϕy,Tx+ζϕTy,x-ϕy,xfor any x,y∈C. Such a mapping is said to be (α,β,γ,δ,ε,ζ)-skew-generalized nonspreading. These classes include the class of generalized hybrid mappings in a Hilbert space; however, it does not include the class of widely more generalized hybrid mappings. Moreover, they introduced some extensions of attractive point and proved some attractive point theorems. x∈E is an attractive point of T if (12)ϕx,Ty≤ϕx,yfor any y∈C; x∈E is a skew-attractive point of T if (13)ϕTy,x≤ϕy,xfor any y∈C. Let (14)AT=x∈E∣ϕx,Ty≤ϕx,yforanyy∈C;BT=x∈E∣ϕTy,x≤ϕy,xforanyy∈C.They also proved the following.Theorem 1.
LetE be a reflexive and smooth Banach space, let C be a nonempty subset of E, and let T be an (α,β,γ,δ,ε,ζ)-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+β>0. Then T has an attractive point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.
Additionally, ifE is strictly convex and C is closed and convex, then T has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Theorem 2.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty subset of E, and let T be an (α,β,γ,δ,ε,ζ)-skew-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+γ>0. Then T has a skew-attractive point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.
Additionally, ifE is strictly convex and J(C) is closed and convex, then T has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.On the other hand, in [12] Atsushiba, Iemoto, Kubota, and Takeuchi introduced a concept of acute point as an extension of attractive point in a Hilbert space. Let H be a real Hilbert space, let C be a nonempty subset of H, and let T be a mapping from C into H and k∈[0,1]. x∈H is called a k-acute point of T if (15)x-Ty2≤x-y2+ky-Ty2for any y∈C. Let (16)AkT=x∈H∣x-Ty2≤x-y2+ky-Ty2foranyy∈C.Moreover, using a concept of acute point, they proved convergence theorems without convexity of C.Motivated by these results, in this paper, we introduce a new class of mappings on Banach space corresponding to the class of all widely more generalized hybrid mappings on Hilbert space. Moreover, we introduce some extensions of acute point and prove some acute point theorems. Moreover, in the next paper [13], we show some mean convergence theorems for the new class of mappings.


## 2. Preliminaries

We know that the following hold; for instance, see [14].(T1)
LetE be a normed space, let C be a nonempty, closed, and convex subset in E, and let D be a nonempty, compact, and convex subset in E. Suppose that C∩D=∅. Then (17)dC,D≝inf⁡x-y∣x∈C,y∈D>0.(T2)
LetE be a normed space and let C and D be nonempty and convex subsets with d(C,D)>0. Then there exists x∗∈E∗, where E∗ is the topological dual space of E, such that x∗=1 and (18)inf⁡x,x∗∣x∈C=dC,D+sup⁡y,x∗∣y∈D.(T3)
LetE be a Banach space, let E∗ be the topological dual space of E, and let J be the duality mapping on E defined by (19)Jx=x∗∈E∗∣x2=x,x∗=x∗2for anyx∈E. Then E is strictly convex if and only if J is injective; that is, x≠y implies J(x)∩J(y)=∅.(T4)
LetE be a Banach space, let E∗ be the topological dual space of E, and let J be the duality mapping on E. Then E is reflexive if and only if J is surjective; that is, ⋃x∈EJ(x)=E∗.(T5)
LetE be a Banach space and let J be the duality mapping on E. Then E is smooth if and only if J is single-valued.(T6)
LetE be a Banach space and let J be the duality mapping on E. If J is single-valued, then J is norm-to-weak∗ continuous.(T7)
LetE be a Banach space and let J be the duality mapping on E. Then E is strictly convex if and only if (20)1-x,y∗>0for anyx,y∈E with x=y and x=y=1 and for any y∗∈J(y).(T8)
LetE be a Banach space and let E∗ be the topological dual space of E. Then E is reflexive if and only if E∗ is reflexive.(T9)
LetE be a Banach space and let E∗ be the topological dual space of E. If E∗ is strictly convex, then E is smooth. Conversely, E is reflexive and smooth, then E∗ is strictly convex.(T10)
LetE be a Banach space and let E∗ be the topological dual space of E. If E∗ is smooth, then E is strictly convex. Conversely, E is reflexive and strictly convex, then E∗ is smooth.LetE be a smooth Banach space, let J be the duality mapping on E, and let ϕ be the mapping from E×E into [0,∞) defined by (21)ϕx,y=x2-2x,Jy+y2for any x,y∈E. Since by (T5) J is single-valued, ϕ is well-defined. It is obvious that x=y implies ϕ(x,y)=0. Conversely, by (T7)(T11)
IfE is also strictly convex, then ϕ(x,y)=0 implies x=y.LetE be a strictly convex and smooth Banach space. By (T3) an (T5) J is a bijective mapping from E onto J(E). In particular, if E is also reflective, then by (T4) J is a bijective mapping from E onto E∗. Suppose that E is strictly convex, reflective, and smooth. Let ϕ∗ be the mapping from E∗×E∗ into [0,∞) defined by (22)ϕ∗x∗,y∗=x∗2-2J-1y∗,x∗+y∗2for any x∗,y∗∈E∗. Then(23)ϕ∗x∗,y∗=ϕJ-1y∗,J-1x∗holds. ThereforeT11∗
ϕ∗x∗,y∗=0 if and only if x∗=y∗.Letl∞ be the Banach space consists of all bounded sequences and μ∈l∞∗. Sometimes we denote by μnxn the value μ(xnn=1∞). If μ∈l∞∗ satisfies μ(e)=μ=1, where e=1n=1∞, then μ is called a mean. If a mean μ satisfies μnxn+1=μnxn, then μ is called a Banach limit. We know that there exists some Banach limits. If xnn=1∞∈l∞ and μ is a mean, then the following holds:(T12)
inf⁡{xn∣n∈N}≤μnxn≤sup⁡{xn∣n∈N}.The following lemma is introduced in [11]. For completeness we write its proof.Lemma 3.
LetE be a Banach space, let E∗ be the dual space of E, let {xn∣n∈N} be a bounded sequence in E, and let μ be a mean on l∞. Then there exists a unique z0∈co¯{xn∣n∈N} such that μnxn,x∗=z0,x∗ for any x∗∈E∗.Proof.
Define a mappingg from E∗ into R by g(x∗)=μnxn,x∗ for any x∗∈E∗. Then g is linear and bounded clearly and hence g∈E∗∗. Since E is reflexive, we can put z0=g.
Lastly, we show thatz0∈co¯{xn∣n∈N}. Assume that z0∉co¯{xn∣n∈N}. Since co¯{xn∣n∈N} is nonempty, close, and convex and z0 is nonempty, compact, and convex, by (T1) and (T2) there exists x∗∈E∗ such that x∗=1 and (24)inf⁡x,x∗∣x∈co¯xn∣n∈N=dco¯xn∣n∈N,z0+sup⁡y,x∗∣y∈z0>z0,x∗.On the other hand, by (T12) we obtain (25)inf⁡x,x∗∣x∈co¯xn∣n∈N≤inf⁡xn,x∗∣n∈N≤μnxn,x∗=z0,x∗.It is a contradiction. Therefore z0∈co¯{xn∣n∈N}.


## 3. Fixed Point Theorems

LetE be a smooth Banach space and let C be a nonempty subset of E. A mapping T from C into E is called a generalized pseudocontraction if there exist α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2∈R such that(26)α1ϕTx,Ty+α2ϕTy,Tx+β1ϕx,Ty+β2ϕTy,x+γ1ϕTx,y+γ2ϕy,Tx+δ1ϕx,y+δ2ϕy,x+ε1ϕTx,x+ε2ϕx,Tx+ζ1ϕy,Ty+ζ2ϕTy,y≤0for any x,y∈C. Such a mapping is called an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction.If all parameters are negative, of course, all mappings satisfy the above inequality. However, in the following argument, such cases are properly excluded.Theorem 4.
LetE be a strictly convex, reflexive and smooth Banach space, let C be a nonempty, closed, and convex subset of E and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Then the following hold. (1)
Suppose thatT satisfies α1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, and one of the following
(1-1)
α1+β1+ζ1>0 and ζ2≥0;(1-2)
α1+β1+ζ1≥0 and ζ2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(2)
Suppose thatT satisfies α2+β2+γ2+δ2≥0, α1+γ1≥0, β1+δ1≥0, ζ1≥0, ζ2≥0, and one of the following:
(2-1)
α2+γ2+ε2>0 and ε1≥0;(2-2)
α2+γ2+ε2≥0 and ε1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(3)
Suppose thatT satisfies α1+α2+β2+γ1≥0, β1+γ2+δ1+δ2≥0, ε2+ζ1≥0, and one of the following:
(3-1)
α1+α2+β1+γ2+ε2+ζ1>0 and ε1+ζ2≥0;(3-2)
α1+α2+β1+γ2+ε2+ζ1≥0 and ε1+ζ2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+α2+β1+β2+γ1+γ2+δ1+δ2>0, then the fixed point is unique.Proof.
IfT has a fixed point z, then {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. Moreover, we suppose that (1) holds. Since(27)ϕu,v=ϕu,w+ϕw,v+2u-w,Jw-Jvfor any u,v,w∈E, we obtain (28)α1ϕTx,Ty+α2ϕTy,Tx+β1ϕx,Ty+β2ϕTy,x+γ1ϕTx,y+γ2ϕy,Tx+δ1ϕx,y+δ2ϕy,x+ε1ϕTx,x+ε2ϕx,Tx+ζ1ϕy,Ty+ζ2ϕTy,y=α1ϕTx,Ty+α2ϕTy,Tx-α1ϕx,Ty+α1+β1ϕx,y+ϕy,Ty+2x-y,Jy-JTy+β2ϕTy,x+γ1ϕTx,y+γ2ϕy,Tx+δ1ϕx,y+δ2ϕy,x+ε1ϕTx,x+ε2ϕx,Tx+ζ1ϕy,Ty+ζ2ϕTy,y=α1ϕTx,Ty+α2ϕTy,Tx-α1ϕx,Ty+β2ϕTy,x+γ1ϕTx,y+γ2ϕy,Tx+α1+β1+δ1ϕx,y+δ2ϕy,x+ε1ϕTx,x+ε2ϕx,Tx+α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1x-y,Jy-JTy.Since α1+β1+δ1≥-γ1, β2≥-α2, δ2≥-γ2, ε1≥0, and ε2≥0, we obtain (29)α1ϕTx,Ty+α2ϕTy,Tx-α1ϕx,Ty+β2ϕTy,x+γ1ϕTx,y+γ2ϕy,Tx+α1+β1+δ1ϕx,y+δ2ϕy,x+ε1ϕTx,x+ε2ϕx,Tx+α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1x-y,Jy-JTy≥α1ϕTx,Ty-ϕx,Ty+α2ϕTy,Tx-ϕTy,x+γ1ϕTx,y-ϕx,y+γ2ϕy,Tx-ϕy,x+α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1x-y,Jy-JTy.Therefore (30)α1ϕTx,Ty-ϕx,Ty+α2ϕTy,Tx-ϕTy,x+γ1ϕTx,y-ϕx,y+γ2ϕy,Tx-ϕy,x+α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1x-y,Jy-JTy≤0.Replacing x by Tnz, we obtain (31)α1ϕTn+1z,Ty-ϕTnz,Ty+α2ϕTy,Tn+1z-ϕTy,Tnz+γ1ϕTn+1z,y-ϕTnz,y+γ2ϕy,Tn+1z-ϕy,Tnz+α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1Tnz-y,Jy-JTy≤0.Applying a Banach limit μ to both sides of this inequality, we obtain (32)α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1μnTnz-y,Jy-JTy≤0.By Lemma 3 there exists a unique z0∈co¯{Tnz∣n∈N}⊂C such that(33)μnTnz-y,Jy-JTy=z0-y,Jy-JTy.Therefore we obtain (34)α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1z0-y,Jy-JTy≤0.for any y∈C. Putting y=z0, we obtain (35)α1+β1+ζ1ϕz0,Tz0+ζ2ϕTz0,z0≤0.In the case of (1-1), since α1+β1+ζ1>0, ζ2≥0 and E is strictly convex, by (T11) we obtain z0=Tz0. In the case of (1-2), since α1+β1+ζ1≥0, ζ2>0, and E is strictly convex, by (T11), we obtain z0=Tz0.
In the case of (2), changing the variablesx and y in (26), we obtain(36)α2ϕTx,Ty+α1ϕTy,Tx+γ2ϕx,Ty+γ1ϕTy,x+β2ϕTx,y+β1ϕy,Tx+δ2ϕx,y+δ1ϕy,x+ζ2ϕTx,x+ζ1ϕx,Tx+ε2ϕy,Ty+ε1ϕTy,y≤0.Suppose that (2) holds. Using the result in the case of (1), we can show that T has a fixed point.
In the case of (3), adding (26) and (36), we obtain(37)α1+α2ϕTx,Ty+α1+α2ϕTy,Tx+β1+γ2ϕx,Ty+β2+γ1ϕTy,x+β2+γ1ϕTx,y+β1+γ2ϕy,Tx+δ1+δ2ϕx,y+δ1+δ2ϕy,x+ε1+ζ2ϕTx,x+ε2+ζ1ϕx,Tx+ε2+ζ1ϕy,Ty+ε1+ζ2ϕTy,y≤0Suppose that (3) holds. Using the result in the case of (1), we can show that T has a fixed point.
Lastly, suppose thatα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0 in the case of (1). Let z1 and z2 be fixed points of T. Then (38)0≥α1ϕTz1,Tz2+α2ϕTz2,Tz1+β1ϕz1,Tz2+β2ϕTz2,z1+γ1ϕTz1,z2+γ2ϕz2,Tz1+δ1ϕz1,z2+δ2ϕz2,z1+ε1ϕTz1,z1+ε2ϕz1,Tz1+ζ1ϕz2,Tz2+ζ2ϕTz2,z2=α1+β1+γ1+δ1ϕz1,z2+α2+β2+γ2+δ2ϕz2,z1.By (T11) we obtain z1=z2 holds. The cases of (2) and (3) are similar.LetE∗ be the dual space of a strictly convex, reflexive, and smooth Banach space E and let C∗ be a nonempty subset of E∗. A mapping T∗ from C∗ into E∗ is called a ∗-generalized pseudocontraction if there exist α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2∈R such that (39)α1ϕ∗T∗x∗,T∗y∗+α2ϕ∗T∗y∗,T∗x∗+β1ϕ∗x∗,T∗y∗+β2ϕ∗T∗y∗,x∗+γ1ϕ∗T∗x∗,y∗+γ2ϕ∗y∗,T∗x∗+δ1ϕ∗x∗,y∗+δ2ϕ∗y∗,x∗+ε1ϕ∗T∗x∗,x∗+ε2ϕ∗x∗,T∗x∗+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗≤0for any x∗,y∗∈C∗. Such a mapping is called an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-∗-generalized pseudocontraction.Theorem 5.
LetE∗ be the dual space of a strictly convex, reflexive, and smooth Banach space E, let C∗ be a nonempty, closed, and convex subset of E∗, and let T∗ be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-∗-generalized pseudocontraction from C∗ into itself. Then the following hold.(1)
Suppose thatT∗ satisfies α1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, and one of the following:
(1-1)
α1+β1+ζ1>0 and ζ2≥0;(1-2)
α1+β1+ζ1≥0 and ζ2>0.ThenT∗ has a fixed point if and only if there exists z∗∈C∗ such that {(T∗)nz∗∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(2)
Suppose thatT∗ satisfies α2+β2+γ2+δ2≥0, α1+γ1≥0, β1+δ1≥0, ζ1≥0, ζ2≥0, and one of the following:
(2-1)
α2+γ2+ε2>0 and ε1≥0;(2-2)
α2+γ2+ε2≥0 and ε1>0.ThenT∗ has a fixed point if and only if there exists z∗∈C∗ such that {(T∗)nz∗∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(3)
Suppose thatT∗ satisfies α1+α2+β2+γ1≥0, β1+γ2+δ1+δ2≥0, ε2+ζ1≥0, and one of the following:
(3-1)
α1+α2+β1+γ2+ε2+ζ1>0 and ε1+ζ2≥0;(3-2)
α1+α2+β1+γ2+ε2+ζ1≥0 and ε1+ζ2>0.ThenT∗ has a fixed point if and only if there exists z∗∈C∗ such that {(T∗)nz∣n∈N∪{0}} is bounded.Moreover, ifα1+α2+β1+β2+γ1+γ2+δ1+δ2>0, then the fixed point is unique.Proof.
IfT∗ has a fixed point z∗, then {(T∗)nz∗∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∗∈C∗ such that {(T∗)nz∗∣n∈N∪{0}} is bounded. Moreover, we suppose that (1) holds. From (23) and (27) we obtain (40)ϕ∗u∗,v∗=ϕ∗u∗,w∗+ϕ∗w∗,v∗+2J-1w∗-J-1v∗,u∗-w∗for any u∗,v∗,w∗∈E∗. Similarly to the proof of Theorem 4 we obtain(41)α1ϕ∗T∗x∗,T∗y∗-ϕ∗x∗,T∗y∗+α2ϕ∗T∗y∗,T∗x∗-ϕ∗T∗y∗,x∗+γ1ϕ∗T∗x∗,y∗-ϕ∗x∗,y∗+γ2ϕ∗y∗,T∗x∗-ϕ∗y∗,x∗+α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1J-1y∗-J-1T∗y∗,x∗-y∗≤0.Replacing x∗ by T∗nz∗, we obtain (42)α1ϕ∗T∗n+1z∗,T∗y∗-ϕ∗T∗nz∗,T∗y∗+α2ϕ∗T∗y∗,T∗n+1z∗-ϕ∗T∗y∗,T∗nz∗+γ1ϕ∗T∗n+1z∗,y∗-ϕ∗T∗nz∗,y∗+γ2ϕ∗y∗,T∗n+1z∗-ϕ∗y∗,T∗nz∗+α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1J-1y∗-J-1T∗y∗,T∗nz∗-y∗≤0.Applying a Banach limit μ to both sides of this inequality, we obtain (43)α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1μnJ-1y∗-J-1T∗y∗,T∗nz∗-y∗≤0.By Lemma 3 there exists a unique z0∗∈co¯{T∗nz∗∣n∈N}⊂C∗ such that (44)μnJ-1y∗-J-1T∗y∗,T∗nz∗-y∗=J-1y∗-J-1T∗y∗,z0∗-y∗.Therefore we obtain (45)α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1J-1y∗-J-1T∗y∗,z0∗-y∗≤0.for any y∗∈C∗. Putting y∗=z0∗, we obtain (46)α1+β1+ζ1ϕ∗z0∗,T∗z0∗+ζ2ϕ∗T∗z0∗,z0∗≤0.In the case of (1-1), since α1+β1+ζ1>0, ζ2≥0, and E∗ is strictly convex, by T11∗ we obtain z0∗=T∗z0∗. In the case of (1-2), since α1+β1+ζ1≥0, ζ2>0, and E∗ is strictly convex, by (T11)∗ we obtain z0∗=T∗z0∗.
In the cases of (2) and (3) we can show similarly.By Theorem5 we obtain the following.Theorem 6.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty subset of E satisfying J(C) is closed and convex, and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Then the following hold. (1)
Suppose thatT satisfies α2+β2+γ2+δ2≥0, α1+β1≥0, γ1+δ1≥0, ε1≥0, ε2≥0, and one of the following:
(1-1)
α2+β2+ζ2>0 and ζ1≥0;(1-2)
α2+β2+ζ2≥0 and ζ1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(2)
Suppose thatT satisfies α1+β1+γ1+δ1≥0, α2+γ2≥0, β2+δ2≥0, ζ1≥0, ζ2≥0, and one of the following:
(2-1)
α1+γ1+ε1>0 and ε2≥0;(2-2)
α1+γ1+ε1≥0 and ε2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(3)
Suppose thatT satisfies α1+α2+β1+γ2≥0, β2+γ1+δ1+δ2≥0, ε1+ζ2≥0, and one of the following:
(3-1)
α1+α2+β2+γ1+ε1+ζ2>0 and ε2+ζ1≥0;(3-2)
α1+α2+β2+γ1+ε1+ζ2≥0 and ε2+ζ1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+α2+β1+β2+γ1+γ2+δ1+δ2>0, then the fixed point is unique.Proof.
IfT has a fixed point z, then {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. Since E is strictly convex, reflexive, and smooth, by (T8), (T9), and (T10) E∗ is also so. By assumption J(C) is nonempty, close, and convex. Let T∗=JTJ-1. Then T∗ is a mapping from J(C) into itself. Putting z∗=Jz, since T∗nz∗=JTnz, T∗nz∗=JTnz=Tnz, and hence {T∗nz∗∣n∈N∪{0}} is bounded. Putting x∗=Jx and y∗=Jy, (26) is equivalent to (47)α2ϕ∗T∗x∗,T∗y∗+α1ϕ∗T∗y∗,T∗x∗+β2ϕ∗x∗,T∗y∗+β1ϕ∗T∗y∗,x∗+γ2ϕ∗T∗x∗,y∗+γ1ϕ∗y∗,T∗x∗)+δ2ϕ∗x∗,y∗+δ1ϕ∗(y∗,x∗+ε2ϕ∗T∗x∗,x∗+ε1ϕ∗x∗,T∗x∗+ζ2ϕ∗y∗,T∗y∗+ζ1ϕ∗T∗y∗,y∗≤0from (23). Therefore T∗ is an (α2,α1,β2,β1,γ2,γ1,δ2,δ1,ε2,ε1,ζ2,ζ1)-∗-generalized pseudocontraction from J(C) into itself. By Theorem 5 we have that T∗ has a fixed point z0∗. Then J-1z0∗ is a fixed point of T.


## 4. Acute Point and Skew-Acute Point

LetE be a smooth Banach space, let C be a nonempty subset of E, let T be a mapping from C into E, and let k,l∈R. x∈E is called a (k,l)-acute point of T if(48)ϕx,Ty≤ϕx,y+kϕy,Ty+lϕTy,yfor any y∈C. x∈E is called a (k,l)-skew-acute point of T if(49)ϕTy,x≤ϕy,x+kϕy,Ty+lϕTy,yfor any y∈C. Let (50)Ak,lT=x∈E∣ϕx,Ty≤ϕx,y+kϕy,Ty+lϕTy,yforanyy∈C;Bk,lT=x∈E∣ϕTy,x≤ϕy,x+kϕy,Ty+lϕTy,yforanyy∈C.It is obvious that(51)Ak1,l1T⊂Ak2,l2T,Bk1,l1T⊂Bk2,l2Tfor any k1,k2,l1,l2∈R with k1≤k2 and l1≤l2.The following lemmas are important property characterizing them.Lemma 7.
LetE be a smooth Banach space, let C be a nonempty subset of E, let T be a mapping from C into E, and let k,l∈R. Then Ak,l(T) is closed and convex.Proof.
Since (48) is equivalent to (52)2x,Jy-JTy≤k-1ϕy,Ty+lϕTy,y+2y,Jy-JTyfrom (27), Ak,l(T) is closed and convex.Lemma 8.
LetE be a smooth Banach space, let C be a nonempty subset of E, let T be a mapping from C into E, and let k,l∈R. Then Bk,l(T) is closed.Proof.
Inequality (49) is equivalent to (53)2y-Ty,Jx≤kϕy,Ty+l-1ϕTy,y+2y-Ty,Jyfrom (27). Moreover, by (T6) J is norm-to-weak∗ continuous. Therefore Bk,l(T) is closed.LetE∗ be the dual space of a strictly convex, reflexive, and smooth Banach space E, let C∗ be a nonempty subset of E∗, let T∗ be a mapping from C∗ into E∗, and let k,l∈R. x∗∈E∗ is called a (k,l)-∗-acute point of T∗ if(54)ϕ∗x∗,T∗y∗≤ϕ∗x∗,y∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗for any y∗∈C∗. x∗∈E∗ is called a (k,l)-∗-skew-acute point of T∗ if(55)ϕ∗T∗y∗,x∗≤ϕ∗y∗,x∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗for any y∗∈C∗. Let (56)Ak,l∗T∗=x∗∈E∗∣ϕ∗x∗,T∗y∗≤ϕ∗x∗,y∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗foranyy∗∈C∗;Bk,l∗T∗=x∗∈E∗∣ϕ∗T∗y∗,x∗≤ϕ∗y∗,x∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗foranyy∗∈C∗.Lemma 9.
LetE∗ be the dual space of a strictly convex, reflective, and smooth Banach space E, let C∗ be a nonempty subset of E∗, let T∗ be a mapping from C∗ into E∗, and let k,l∈R. Then Ak,l∗(T∗) is closed and convex.Proof.
Since(57)ϕ∗x∗,T∗y∗≤ϕ∗x∗,y∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗is equivalent to (58)2J-1y∗-J-1T∗y∗,x∗≤k-1ϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗+2J-1y∗-J-1T∗y∗,y∗from (27) and (23), Ak,l∗(T∗) is closed and convex.Lemma 10.
LetE∗ be the dual space of a strictly convex, reflexive, and smooth Banach space E, let C∗ be a nonempty subset of E∗, let T∗ be a mapping from C∗ into E∗, and let k,l∈R. Then Bk,l∗(T∗) is closed.Proof.
Since(59)ϕ∗T∗y∗,x∗≤ϕ∗y∗,x∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗is equivalent to (60)2J-1x∗,y∗-T∗y∗≤kϕ∗y∗,T∗y∗+l-1ϕ∗T∗y∗,y∗+2J-1y∗,y∗-T∗y∗from (27) and (23), moreover, by (T6) J-1 is norm-to-weak∗ continuous. Therefore Bk,l∗(T∗) is closed.Lemma 11.
LetE be a strictly convex, reflective, and smooth Banach space, let C be a nonempty subset of E, let T be a mapping from C into E, let T∗=JTJ-1, and let k,l∈R. Then (61)Ak,l∗T∗=JBl,kT,Bk,l∗T∗=JAl,kT.In particular, JBk,lT is closed and convex and JAk,lT is closed.Proof.
Letx∗∈Ak,l∗(T∗). Then (62)ϕ∗x∗,T∗y∗≤ϕ∗x∗,y∗+kϕ∗y∗,T∗y∗+lϕ∗T∗y∗,y∗for any y∗∈J(C). From (23) (63)ϕJ-1T∗y∗,J-1x∗≤ϕJ-1y∗,J-1x∗+kϕJ-1T∗y∗,J-1y∗+lϕJ-1y∗,J-1T∗y∗for any y∗∈J(C). Since J-1T∗=TJ-1, putting y=J-1y∗, we obtain (64)ϕTy,J-1x∗≤ϕy,J-1x∗+lϕy,Ty+kϕTy,y.Therefore J-1x∗∈Bl,k(T) and hence Ak,l∗(T∗)=J(Bl,k(T)).
B k , l ∗ ( T ∗ ) = J A l , k T can be shown similarly.
Moreover, by Lemma9J(Bk,l(T)) is closed and convex and by Lemma 10J(Ak,l(T)) is closed.Lemma 12.
LetE be a strictly convex and smooth Banach space, let C be a nonempty subset of E, let T be a mapping from C into E, and let k,l∈R. Then the following hold.(1)
Ifk,l∈-∞,1×-∞,0∖{(1,0)}, then C∩Ak,l(T) is included in the set of all fixed points of T.(2)
If(k,l)∈(-∞,0]×(-∞,1]∖{(0,1)}, then C∩Bk,l(T) is included in the set of all fixed points of T.Proof.
Letx∈C∩Ak,l(T). Then (48) holds for any y∈C. Putting y=x, by (T11) we obtain (1-k)ϕ(x,Tx)-lϕ(Tx,x)≤0. If (k,l)∈(-∞,1]×(-∞,0]∖{(1,0)}, then by (T11) we obtain x=Tx.
Letx∈C∩Bk,l(T). Then (49) holds for any y∈C. Putting y=x, by (T11) we obtain -kϕ(x,Tx)+(1-l)ϕ(Tx,x)≤0. If (k,l)∈(-∞,0]×(-∞,1]∖{(0,1)}, then by (T11) we obtain x=Tx.


## 5. Acute Point and Skew-Acute Point Theorems

Theorem 13.
LetE be a reflexive and smooth Banach space, let C be a nonempty subset of E, and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Suppose that there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded. Then the following hold.(1)
Ifα1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, and α1+β1>0, then there exists a -ζ1/α1+β1,-ζ2/α1+β1-acute point.(2)
Ifα2+β2+γ2+δ2≥0, α1+γ1≥0, β1+δ1≥0, ζ1≥0, ζ2≥0, and α2+γ2>0, then there exists a -ε2/α2+γ2,-ε1/α2+γ2-acute point.(3)
Ifα1+α2+β2+γ1≥0, β1+γ2+δ1+δ2≥0, ε1+ζ2≥0, ε2+ζ1≥0, and α1+α2+β1+γ2>0, then there exists a -ε2+ζ1/α1+α2+β1+γ2,-ε1+ζ2/α1+α2+β1+γ2-acute point.Proof.
Suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. In the case of (1), similarly to the proof of Theorem 4, we have that there exists a unique z0∈co¯{Tnz∣n∈N} such that (65)α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+2α1+β1z0-y,Jy-JTy≤0.for any y∈C. Since (66)2p-q,Jr-Js=ϕp,s+ϕq,r-ϕp,r-ϕq,sfor any p,q,r,s∈E, we obtain (67)α1+β1+ζ1ϕy,Ty+ζ2ϕTy,y+α1+β1ϕz0,Ty-ϕz0,y-ϕy,Ty=α1+β1ϕz0,Ty-α1+β1ϕz0,y+ζ1ϕy,Ty+ζ2ϕTy,y≤0.Since α1+β1>0, we obtain (68)ϕz0,Ty≤ϕz0,y-ζ1α1+β1ϕy,Ty-ζ2α1+β1ϕTy,y.Therefore z0 is a -ζ1/α1+β1,-ζ2/α1+β1-acute point.
In the cases of (2), changing the variablesx and y in (26), we have that there exists a unique z0∈co¯{Tnz∣n∈N} such that (69)α2+γ2ϕz0,Ty-α2+γ2ϕz0,y+ε2ϕy,Ty+ε1ϕTy,y≤0.Since α2+γ2>0, we obtain (70)ϕz0,Ty≤ϕz0,y-ε2α2+γ2ϕy,Ty-ε1α2+γ2ϕTy,y.Therefore z0 is a -ε2/α2+γ2,-ε1/α2+γ2-acute point.
In the case of (3), adding (26) and (36), we have that there exists a unique z0∈co¯{Tnz∣n∈N} such that (71)α1+α2+β1+γ2ϕz0,Ty-α1+α2+β1+γ2ϕz0,y+ε2+ζ1ϕy,Ty+ε1+ζ2ϕTy,y≤0.Since α1+α2+β1+γ2>0, we obtain (72)ϕz0,Ty≤ϕz0,y-ε2+ζ1α1+α2+β1+γ2ϕy,Ty-ε1+ζ2α1+α2+β1+γ2ϕTy,y.Therefore z0 is a -ε2+ζ1/α1+α2+β1+γ2,-ε1+ζ2/α1+α2+β1+γ2-acute point.Theorem 14.
LetE∗ be the dual space of a strictly convex, reflexive, and smooth Banach space E, let C∗ be a nonempty subset of E∗, and let T∗ be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-∗-generalized pseudocontraction from C∗ into itself. Suppose that there exists z∗∈C∗ such that {(T∗)nz∗∣n∈N∪{0}} is bounded. Then the following hold.(1)
Ifα1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, and α1+β1>0, then there exists a -ζ1/α1+β1,-ζ2/α1+β1-∗-acute point.(2)
Ifα2+β2+γ2+δ2≥0, α1+γ1≥0, β1+δ1≥0, ζ1≥0, ζ2≥0, and α2+γ2>0, then there exists a -ε2/α2+γ2,-ε1/α2+γ2-∗-acute point.(3)
Ifα1+α2+β2+γ1≥0, β1+γ2+δ1+δ2≥0, ε1+ζ2≥0, ε2+ζ1≥0, and α1+α2+β1+γ2>0, then there exists a -ε2+ζ1/α1+α2+β1+γ2,-ε1+ζ2/α1+α2+β1+γ2-∗-acute point.Proof.
Similar to the proof of Theorem5 we obtain (73)α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1J-1y∗-J-1T∗y∗,z0∗-y∗≤0.for any y∗∈C∗. Since (74)2J-1r∗-J-1s∗,p∗-q∗=ϕ∗p∗,s∗+ϕ∗q∗,r∗-ϕ∗p∗,r∗-ϕ∗q∗,s∗for any p∗,q∗,r∗,s∗∈E∗, we obtain (75)α1+β1+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗+2α1+β1J-1y∗-J-1T∗y∗,z0∗-y∗=α1+β1ϕ∗z0∗,T∗y∗-α1+β1ϕ∗z0∗,y∗+ζ1ϕ∗y∗,T∗y∗+ζ2ϕ∗T∗y∗,y∗≤0.Since α1+β1>0, we obtain (76)ϕ∗z0∗,T∗y∗≤ϕ∗z0∗,y∗-ζ1α1+β1ϕ∗y∗,T∗y∗-ζ2α1+β1ϕ∗T∗y∗,y∗.Therefore z0∗ is a -ζ1/α1+β1,-ζ2/α1+β1-∗-acute point.
In the cases of (2) and (3) we can show similarly.By Theorem14 we obtain the following.Theorem 15.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty subset of E, and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Suppose that there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded. Then the following hold.(1)
Ifα2+β2+γ2+δ2≥0, α1+β1≥0, γ1+δ1≥0, ε1≥0, ε2≥0, and α2+β2>0, then there exists a -ζ2/α2+β2,-ζ1/α2+β2-skew-acute point.(2)
Ifα1+β1+γ1+δ1≥0, α2+γ2≥0, β2+δ2≥0, ζ1≥0, ζ2≥0, and α1+γ1>0, then there exists a -ε1/α1+γ1,-ε2/α1+γ1-skew-acute point.(3)
Ifα1+α2+β1+γ2≥0, β2+γ1+δ1+δ2≥0, ε1+ζ2≥0, ε2+ζ1≥0, and α1+α2+β2+γ1>0, then there exists a -ε1+ζ2/α1+α2+β2+γ1,-ε2+ζ1/α1+α2+β2+γ1-skew-acute point.Proof.
By Theorem14, T∗=JTJ-1 has a ∗-acute point z0∗∈co¯{T∗nJz∣n∈N}. By Lemma 11J-1z0∗ is a skew-acute point of T.


## 6. Applications

By Theorem13 we can show easily the first part of Theorem 1. Note that A(T)=A0,0(T).Proof of the First Part of of Theorem1.
LetE be a reflexive and smooth Banach space, let C be a nonempty subset of E, and let T be an (α,β,γ,δ,ε,ζ)-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+β>0.
Suppose thatT has an attractive point z. Then it is clear that {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. T is an (α,-ε,β,ε,γ,-ζ,δ,ζ,0,0,0,0)-generalized pseudocontraction. These parameters satisfy the conditions of Theorem 13 (1). Therefore there exists a (0,0)-acute point, that is, an attractive point.By Theorem15 we can show easily the first part of Theorem 2. Note that B(T)=B0,0(T).Proof of the First Part of Theorem2.
LetE be a strictly convex, reflexive and smooth Banach space, let C be a nonempty subset of E, and let T be an (α,β,γ,δ,ε,ζ)-skew-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+γ>0.
Suppose thatT has a skew-attractive point z. Then it is clear that {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. T is an (α,-ε,β,-ζ,γ,ε,δ,ζ,0,0,0,0)-generalized pseudocontraction. These parameters satisfy the conditions of Theorem 15 (2). Therefore there exists a (0,0)-skew-acute point, that is, a skew-attractive point.By Theorem13 and Lemma 12 (1) we obtain the following.Theorem 16.
LetE be a strictly convex, reflexive and smooth Banach space, let C be a nonempty, closed, and convex subset of E, and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Then the following hold.(1)
Suppose thatT satisfies α1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, α1+β1>0, and one of the following:
(1-1)
α1+β1+ζ1>0 and ζ2≥0;(1-2)
α1+β1+ζ1≥0 and ζ2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(2)
Suppose thatT satisfies α2+β2+γ2+δ2≥0, α1+γ1≥0, β1+δ1≥0, ζ1≥0, ζ2≥0, α2+γ2>0, and one of the following:
(2-1)
α2+γ2+ε2>0 and ε1≥0;(2-2)
α2+γ2+ε2≥0 and ε1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(3)
Suppose thatT satisfies α1+α2+β2+γ1≥0, β1+γ2+δ1+δ2≥0, ε2+ζ1≥0, α1+α2+β1+γ2>0, and one of the following:
(3-1)
α1+α2+β1+γ2+ε2+ζ1>0 and ε1+ζ2≥0;(3-2)
α1+α2+β1+γ2+ε2+ζ1≥0 and ε1+ζ2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+α2+β1+β2+γ1+γ2+δ1+δ2>0, then the fixed point is unique.Proof.
IfT has a fixed point z, then {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. If T satisfies α1+β1+γ1+δ1≥0, α2+β2≥0, γ2+δ2≥0, ε1≥0, ε2≥0, and α1+β1>0, then by Theorem 13 there exits z0∈co¯{Tnz∣n∈N}⊂C such that it is a -ζ1/α1+β1,-ζ2/α1+β1-acute point. Moreover, If (1-1) holds, then-ζ1/α1+β1<1 and -ζ2/α1+β1≤0;If (1-2) holds, then-ζ1/α1+β1≤1 and -ζ2/α1+β1<0.
By Lemma12 (1) z0 is a fixed point of T.
In the cases of (2) and (3) we can show similarly.Remark 17.
Theorem16 is the same as Theorem 4. The only difference is that Theorem 16 needs more assumptions than Theorem 4.By Theorem16 we can show easily the second part of Theorem 1.Proof of the Second Part of Theorem1.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty, closed, and convex subset of E, and let T be an (α,β,γ,δ,ε,ζ)-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+β>0.
Suppose thatT has a fixed point z. Then it is clear that {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. T is an (α,-ε,β,ε,γ,-ζ,δ,ζ,0,0,0,0)-generalized pseudocontraction. These parameters satisfy the conditions of Theorem 16 (1). Therefore there exists a fixed point.By Theorem15 and Lemma 12 (2) we obtain the following.Theorem 18.
LetE be a strictly convex, reflexive, and smooth Banach space, let C be a nonempty subset of E satisfying J(C) is closed and convex, and let T be an (α1,α2,β1,β2,γ1,γ2,δ1,δ2,ε1,ε2,ζ1,ζ2)-generalized pseudocontraction from C into itself. Then the following hold. (1)
Suppose thatT satisfies α2+β2+γ2+δ2≥0, α1+β1≥0, γ1+δ1≥0, ε1≥0, ε2≥0, α2+β2>0, and one of the following:
(1-1)
α2+β2+ζ2>0 and ζ1≥0;(1-2)
α2+β2+ζ2≥0 and ζ1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(2)
Suppose thatT satisfies α1+β1+γ1+δ1≥0, α2+γ2≥0, β2+δ2≥0, ζ1≥0, ζ2≥0, α1+γ1>0, and one of the following:
(2-1)
α1+γ1+ε1>0 and ε2≥0;(2-2)
α1+γ1+ε1≥0 and ε2>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+β1+γ1+δ1>0 or α2+β2+γ2+δ2>0, then the fixed point is unique.(3)
Suppose thatT satisfies α1+α2+β1+γ2≥0, β2+γ1+δ1+δ2≥0, ε1+ζ2≥0, α1+α2+β2+γ1>0, and one of the following:
(3-1)
α1+α2+β2+γ1+ε1+ζ2>0 and ε2+ζ1≥0;(3-2)
α1+α2+β2+γ1+ε1+ζ2≥0 and ε2+ζ1>0.ThenT has a fixed point if and only if there exists z∈C such that {Tnz∣n∈N∪{0}} is bounded.Moreover, ifα1+α2+β1+β2+γ1+γ2+δ1+δ2>0, then the fixed point is unique.Proof.
IfT has a fixed point z, then {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. If T satisfies α2+β2+γ2+δ2≥0, α1+β1≥0, γ1+δ1≥0, ε1≥0, ε2≥0, and α2+β2>0, then by Theorem 15 there exits z0∈J-1(co¯{T∗nJz∣n∈N})⊂C such that it is a -ζ1/α2+β2,-ζ2/α2+β2-skew-acute point. Moreover,If (1-1) holds, then-ζ1/α2+β2≤0 and -ζ2/α2+β2<1;If (1-2) holds, then-ζ1/α2+β2<0 and -ζ2/α2+β2≤1.
By Lemma12 (2) z0 is a fixed point of T.
In the cases of (2) and (3) we can show similarly.Remark 19.
Theorem18 is the same as Theorem 6. The only difference is that Theorem 18 needs more assumptions than Theorem 6.
By Theorem18 we can show easily the second part of Theorem 2.Proof of the Second Part of Theorem2.
LetE be a strictly convex, reflexive and smooth Banach space, let C be a nonempty subset of E satisfying J(C) is closed and convex, and let T be an (α,β,γ,δ,ε,ζ)-skew-generalized nonspreading mapping from C into itself satisfying α+β+γ+δ≥0 and α+γ>0.
Suppose thatT has a fixed point z. Then it is clear that {Tnz∣n∈N∪{0}} is bounded.
Conversely, suppose that there existsz∈C such that {Tnz∣n∈N∪{0}} is bounded. T is an (α,-ε,β,-ζ,γ,ε,δ,ζ,0,0,0,0)-generalized pseudocontraction. These parameters satisfy the conditions of Theorem 18 (2). Therefore there exists a fixed point.



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*Source: 5874305-2019-06-09.xml*