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# Fixed Point Theorems of a New Generalized Nonexpansive Mapping

Written By

Shi Jie

Submitted: April 2nd, 2019 Reviewed: July 5th, 2019 Published: September 25th, 2019

DOI: 10.5772/intechopen.88421

From the Edited Volume

## Functional Calculus

Edited by Kamal Shah and Baver Okutmuştur

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## Abstract

This paper introduces a T−Da mapping that is weaker than the nonexpansive mapping. It introduces several iterations for the fixed point of the T−Da mapping. It gives fixed point theorems and convergence theorems for the T−Da mapping in Banach space, instead of uniformly convex Banach space. This paper gives some basic properties on the T−Da mapping and gives the example to show the existence of T−Da mapping. The results of this paper are obtained in general Banach spaces. It considers some sufficient conditions for convergence of fixed points of mappings in general Banach spaces under higher iterations.

### Keywords

• iteration
• convergence theorems
• nonexpansive mapping
• fixed point
• 2010 MSC: 47H09
• 47H10

## 1. Introduction

In this paper, E is a Banach space, C is a nonempty closed convex subset of E, and FixT=xC:Tx=x.

Definition 1. T is contraction mapping if there is r01

TxTyrxyforallxyC.

Definition 2. T is nonexpansive mapping if

TxTyxyforallxyC.

Definition 3. T is quasinonexpansive mapping if

TxTyxyforallxC,yFT.

Definition 4.T:CC is a TDa mapping on a subset C, if there is a121, TxTyxy for all αa1,xC,yCTxα, where CTxα=yCy=1αp+αTppCTppTxx.

In 2008 Suzuki [1] defined a mapping T in Banach space: 12TxTyxy implies TxTyxy. And T is said to satisfy condition (C). Suzuki [1] showed that the mapping satisfying condition (C) is weaker than nonexpansive mapping and stronger than quasinonexpansive mapping.

Suzuki [1] proved the theorem T is a mapping in Banach space, T satisfies condition (C), and {xn} is the sequence defined by the iteration process:

x1=xC,xn+1=1αnxn+αnTxn,E1

then {xn} converges to a fixed point of T.

Suzuki [1] gave this convergence theorem in an ordinary Banach space, and the mapping satisfying condition (C) is weaker than nonexpansive mapping.

In 2016, Thakur [2] proved the theorem T is a mapping in uniformly convex Banach space, T satisfies condition (C), and xn is the sequence defined by iteration process:

x1=xC,xn+1=Tyn,yn=Tzn,zn=1αnxn+αnTxn,E2

then {xn} converges to a fixed point of T.

Thakur [2] claimed that the rate of iteration is fastest of known iterations. However, the disadvantage is that their results must be in uniformly convex Banach space, instead of the ordinary Banach space.

The aim of this article is there exists a generalized nonexpansive mapping, which makes the sequence generated by Thakur’s iteration converge to a fixed point in a general Banach space.

The following propositions are obvious:

Proposition 1. If T is nonexpansive, then T satisfies condition (Da).

Proposition 2. If T is TDa mapping, then T is quasinonexpansive.

Proposition 3. Suppose T:CC is a TDa mapping. Then, for x,yC:

1. T2xTxTxxforallxC.

2. T2xTyTxyorT2yTxTyxforallx,yC.

Proof:

1. Since TxxTxx,TxCTx1, we have T2xTxTxx.

2. For all x,yC, TxxTyy or TyyTxx.

Then TxCTyα or TyCTxα.

It follows that T2xTyTxy or T2yTxTyx.

Example 1

Tx=1.1x2x3x4,x1=3,0x2x3x4,x13,

where

x=x1x2x3x4,x103,x20,0.01,x30,0.01,x40,0.01.
x1=maxx1+x3x2+x4

Set

x=3000

and

y=2.5000

We see that

TxTy1=1.1>xy1.

Hence, T is not a nonexpansive mapping.

To verify that T is a TDa mapping, consider the following cases:

Case 1:

α11191,x=x1x2x3x4,x13.y=y1y2y3y4CTxα,

then y13. We have

TyTx=0y2x2y3x3y4x4y1x1y2x2y3x3y4x4=yx

Case 2:

α11191,x=x1x2x3x4,x1=3.y=y1y2y3y4CTxα,

then y10,1.9. We have

TyTx=1.1y2x2y3x3y4x41.11yx

Hence, T is a TDa mapping, and T is not nonexpansive.

## 2. Fixed point

In this section, we prove convergence theorems for fixed point of the TDa mapping in Banach space.

Lemma 1. Let C be bounded convex subset of a Banach space B. Assume that T:CC is TDa mapping and xn,yn,zn are sequences generated by iteration:

x1=xC,xn+1=Tyn,yn=Tzn,zn=1αnxn+αnTxn,E3

where 12<aαnb<1. Then

1. Txn+1xn+1TynynTznznTxnxn.

2. limnTxnxn=limnTynyn=limnTznzn=r0.

Proof: (1) From Proposition 3 and zn=1αnxn+αnTxn, we have

Txn+1xn+1T2ynTynTynyn=T2znTznTznzn=TznTxn+1αnTxnxnznxn+1αnTxnxn=Txnxn.

(2) From (1), we have 0Txn+1xn+1Txnxn. So limnTxnxn=r0. Now, we have limnTxnxn=limnTynyn=limnTznzn=r0.

Lemma 2. Assume that T:CC is a TDa mapping and xn,yn,zn are sequences generated by iteration (3). 12<aαnb<1. Let um satisfy u3n2=xn,u3n1=zn,u3n=yn. Then, for all n1,p1

1+k=3n23n+p3βkTu3n2u3n2Tu3n2+pu3n2+k=nn+p121αkTu3n2u3n2Tu3n2+3pu3n2+3p,E4

where

βk=αnk=3n21k3n2

Proof: From Lemma 1, we have

Txn+1xn+1Tznzn=Tzn1αnxnαnTxn1αnTznxn+αnTznTxn1αnTznxn+αnznxn=1αnTznxn+αn2Txnxn.

So, for p=1 and all n1

1+β3n2Tu3n2u3n2=1+αnTxnxnTxnxn+11αnTxnxnTxn+1xn+1=Tu3n1u3n2+11αnTu3n2u3n2Tu3n+1u3n+1Tu3n1u3n2+21αnTu3n2u3n2Tu3n+1u3n+1.

(4) holds.

We make the inductive assumption that (4) holds for a given p>1 and all n>0 and obtain, upon replacing n with n+1

(1+k=3n+13n+pβk) Tu3n+1u3n+1 Tu3n+1+pu3n+1 +(k=n+1n+p21αk)( Tu3n+1u3n+1 Tu3n+1+3pu3n+1+3p ).E5

And obviously

k3n2,Tuk+1TukβkTu3n2u3n2,E6
k>t,TukTutukut.E7

Case 1: p=3m,m1. From (6) and (7)

Tu3n+1+pu3n+1=Txn+m+1xn+1=Txn+m+1Tynxn+m+1yn=Tyn+mTznyn+mzn=Tzn+m1αnxnαnTxn1αnTzn+mxn+αnTzn+mTxn1αnTzn+mxn+αnTzn+mTxn+m+Txn+mTyn+m1++TynTzn+TznTxn=1αnTu3n1+pu3n2+αnk=3n23n2+pTuk+1Tuk1αnTu3n1+pu3n2+αnk=3n23n2+pβkTu3n2u3n2.

It follows that

Tu3n+1+pu3n+11αnTu3n1+pu3n2+αnk=3n23n2+pβkTu3n2u3n2E8

Using (5) and (8), we have

1+k=3n+13n+pβkTu3n+1u3n+11αnTu3n1+pu3n2+αnk=3n23n2+pβkTu3n2u3n2+k=n+1n+p21αkTu3n+1u3n+1Tu3n+1+3pu3n+1+3p.

From 1+k=3n+13n+pβkk=n+1n+p11αk and Tu3n+1u3n+1Tu3n2u3n2, we have

1+k=3n+13n+pβkTu3n2u3n21αnTu3n1+pu3n2+αnk=3n23n2+pβkTu3n2u3n2+k=n+1n+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Then

1+k=3n+13n+pβkαnk=3n23n2+pβk1αnTu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

It follows that

1+k=3n23n2+pβkTu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Thus, for n,p+1, (4) holds.

Case 2: p=3m+1,m0. From (6) and (7), we have

Tu3n+1+pu3n+1=Tzn+m+1xn+1=Tzn+m+1Tynzn+m+1yn=1αm+n+1xm+n+1+αm+n+1Txm+n+1Tzn
1αm+n+1xm+n+1Tzn+αm+n+1Txm+n+1Tzn1αm+n+1Tym+nTzn+αm+n+1xm+n+1zn1αm+n+1(Tym+nTzm+n+Tzm+nTxn+m++Txn+1Tyn+TynTzn)+αm+n+1xm+n+1zn=1αm+n+1k=3n13n2+pTuk+1Tuk+αm+n+1Tym+n1αnxnαnTxn1αm+n+1k=3n13n2+pβkTu3n2u3n2+αm+n+11αnTym+nxn+αnTym+nTxn1αm+n+1k=3n13n2+pβkTu3n2u3n2+αm+n+11αnTym+nxn+αm+n+1αnTym+nTzm+n+Tzm+nTxm+n++TynTzn+TznTxn=1αm+n+1k=3n13n2+pβkTu3n2u3n2+αm+n+11αnTym+nxn+αm+n+1αnk=3n23n2+pTuk+1Tuk1αm+n+1k=3n13n2+pβkTu3n2u3n2+αm+n+11αnTym+nxn+αm+n+1αnk=3n23n2+pβkTukuk=1αm+n+1k=3n13n2+pβkTu3n2u3n2+αm+n+11αnTu3n1+pu3n2+αm+n+1αnk=3n23n2+pβkTukuk=1αm+n+1+αm+n+1αnk=3n23n2+pβkTu3n2u3n2αn1αm+n+1Tu3n2u3n2+αm+n+11αnTu3n1+pu3n2.

It follows that

Tu3n+1+pu3n+11αm+n+1+αm+n+1αnk=3n23n2+pβkTu3n2u3n2αn1αm+n+1Tu3n2u3n2+αm+n+11αnTu3n1+pu3n2E9

Using (5) and (9), we have

1+k=3n+13n+pβkTu3n+1u3n+11αm+n+1+αm+n+1αnk=3n23n2+pβkTu3n2u3n2
αn1αm+n+1Tu3n2u3n2+αm+n+11αnTu3n1+pu3n2+k=n+1n+p21αkTu3n+1u3n+1Tu3n+1+3pu3n+1+3p.

From 1+k=3n+13n+pβkk=n+1n+p11αk and Tu3n+1u3n+1Tu3n2u3n2, we have

1+k=3n+13n+pβkTu3n2u3n21αm+n+1+αm+n+1αnk=3n23n2+pβkTu3n2u3n2αn1αm+n+1Tu3n2u3n2+αm+n+11αnTu3n1+pu3n2+k=n+1n+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Then

1+k=3n+13n+pβk+αn1αm+n+11αm+n+1+αm+n+1αnk=3n23n2+pβkαm+n+11αnTu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

It follows that

1+k=3n23n2+pβkTu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Thus, for n,p+1, (4) holds.

Case 3: p=3m+2,m0. From (6) and (7), we have

Tu3n+1+pu3n+1=Tyn+m+1Tynyn+m+1yn=Tzn+m+1Tznzn+m+1znzn+m+11αnxnαnTxn1αnzn+m+1xn+αnzn+m+1Txn=1αn1αn+m+1xn+m+1+αn+m+1Txn+m+1xn+αn1αn+m+1xn+m+1+αn+m+1Txn+m+1Txn
1αn1αm+n+1xn+m+1xn+αm+n+1Txn+m+1xn+αn1αm+n+1xn+m+1Txn+αm+n+1Txn+m+1Txn1αnαm+n+1Txn+m+1xn+1αm+n+11αn+αm+n+1αnxn+m+1xn+αn1αm+n+1Tyn+m+1Txn1αnαm+n+1Txn+m+1xn+1αm+n+11αn+αm+n+1αnTyn+mTzn+m++TznTxn+Txnxn+αn1αm+n+1Tyn+m+1Tzn+m+1++TynTzn+TznTxn1αnαm+n+1Txn+m+1xn+1αm+n+11αn+αm+n+1αnk=3n23n3+pTuk+1Tuk+Tu3n2x3n2+1αm+n+1αnk=3n23n3+pTuk+1Tuk1αnαm+n+1Txn+m+1xn+1αm+n+11αn+αm+n+1αnk=3n23n2+pβkTxnxn+1αm+n+1αnk=3n23n3+pβkTxnxn1αnαm+n+1Tu3n1+pu3n2+1αm+n+1+αnαm+n+1k=3n23n2+pβkαn1αm+n+1Tu3n2u3n2.

It follows that

Tu3n+1+pu3n+11αnαm+n+1Tu3n1+pu3n2+1αm+n+1+αnαm+n+1k=3n23n2+pβkαn1αm+n+1Tu3n2u3n2E10

Using (5) and (10), we have

1+k=3n+13n+pβkTu3n+1u3n+11αnαm+n+1Tu3n1+pu3n2+1αm+n+1+αnαm+n+1k=3n23n2+pβkαn1αm+n+1Tu3n2u3n2+k=n+1n+p21αkTu3n+1u3n+1Tu3n+1+3pu3n+1+3p.

From 1+k=3n+13n+pβkk=n+1n+p21αk and Tu3n+1u3n+1Tu3n2u3n2, we have

1+k=3n+13n+pβkTu3n2u3n21αnαm+n+1Tu3n1+pu3n2+1αm+n+1+αnαm+n+1k=3n23n2+pβkαn1αm+n+1Tu3n2u3n2+k=n+1n+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Then

(1+k=3n+13n+pβk1αm+n+1+αnαm+n+1k=3n23n2+pβkαn1αm+n+11αnαm+n+1Tu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

It follows that

1+k=3n23n2+pβkTu3n2u3n2Tu3n1+pu3n2+k=nn+p21αkTu3n2u3n2Tu3n+1+3pu3n+1+3p.

Thus, for n,p+1, (4) holds. This completes the induction.

Lemma 3.T:CC is a TDa mapping, TxxTyy. Then

xTy3Txx+xy.

Proof: Since TxxTyy, we have TxCTyα. Then

T2xTyTxy.

It follows that

xTyxTx+T2xTx+T2xTy.

From Proposition 3, we have

xTy2Txx+Txy2Txx+Txx+xy=3Txx+xy.

Theorem 1. Assume that T:CC is a TDa mapping and xn,yn,zn are sequences generated by iteration (3), 12<aαnb<1. Then limnTxnxn=0.

Proof: Since C is bounded, there must exists d>0, for every xC,xd. Let um satisfy u3n2=xn,u3n1=zn,u3n=yn. From Lemma 1, limkTukuk=r0. Assume r>0. Let ε satisfy

e61bdr+1ε<r

and choose n so that for every p>0

Tu3n2u3n2Tu3n2+3pu3n2+3p<ε.

Now choose p so that rk=3n23n+p4βkdrk=3n23n+p3βk.

Since 12<aαnb<1, for every k,t, we have 1+αk<3αk,αt<2αk. From Lemma 2 and rTu3n2u3n2, we have

d+rr1+k=3n23n+p3βk1+k=3n23n+p3βkTu3n2u3n2Tu3n2+pu3n2+k=nn+p121αkTu3n2u3n2Tu3n2+3pu3n2+3p<d+k=nn+p121αkε=d+eΣk=nn+p1ln1+1+αk1αkεd+eΣk=nn+p11+αk1αkεd+e31bΣk=nn+p1αkεd+e61bΣk=3n23n+p3βkεd+e61bΣk=3n23n+p4βk+1ε<d+e61bdr+1ε<d+r.

This is a contradiction. So limkTukuk=0. That is to say, limnTxnxn=0. This completes the proof.

Theorem 2. Assume that T:CC is a TDa mapping and xn is generated by iteration (3), 12<aαnb<1. Then the sequence xn converges to a fixed point of T.

Proof: Since C is compact, there exists a subsequence xnkxn which converges to some zC. By Lemma 3, we have xnkTz3Txnkxnk+xnkz. Since limnkTxnkxnk=0 and limnkxnkz=0, we have limnkxnkTz=0. This implies that z=Tz. On the other hand, from Proposition 3

xn+1zynzznzαnTxnz+1αnxnzxnz.

So, limnxnz exists. Therefore, limnxnz=0. This completes the proof.

## References

1. 1. Suzuki T. Fixed point theorems and convergence theorems for some generalized nonexpansive mappings. Journal of Mathematical Analysis and Applications. 2008;340:1088-1095
2. 2. Thakur BS, Thakur D, Postolache M. A new iterative scheme for numerical reckoning fixed points of Suzuki’s generalized nonexpansive mappings. Journal of Applied Mathematics and Computing. 2016;275:147-155

Written By

Shi Jie

Submitted: April 2nd, 2019 Reviewed: July 5th, 2019 Published: September 25th, 2019