A Suzuki Type Fixed-Point Theorem

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International Journal of Mathematics and Mathematical Sciences Volume 2011, Article ID 736063,9pages

doi:10.1155/2011/736063

Research Article

A Suzuki Type Fixed-Point Theorem

Ishak Altun and Ali Erduran

Department of Mathematics, Faculty of Science and Arts, Kirikkale University, Yahsihan, 71450 Kirikkale, Turkey

Correspondence should be addressed to Ishak Altun,[email protected] Received 16 December 2010; Accepted 7 February 2011

Academic Editor: Genaro Lopez

Copyrightq2011 I. Altun and A. Erduran. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We present a fixed-point theorem for a single-valued map in a complete metric space using implicit relation, which is a generalization of several previously stated results including that of Suziki 2008.

1. Introduction

There are a lot of generalizations of Banach fixed-point principle in the literature. See 1–

5. One of the most interesting generalizations is that given by Suzuki 6. This interesting fixed-point result is as follows.

Theorem 1.1. LetX, dbe a complete metric space, and letT be a mapping onX. Define a non- increasing functionθfrom0,1into1/2,1by

θr

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

1, 0≤r≤

√5−1 2 , 1−r

r² ,

√5−1

2 ≤r≤ 1

√2, 1

1r, 1

√2 ≤r <1.

1.1

Assume that there existsr∈0,1, such that θrdx, Tx≤d

x, y

impliesd

Tx, Ty

≤rd x, y

, 1.2

for allx, y∈X, then there exists a unique fixed-pointzofT. Moreover, limnTⁿxzfor allx∈X.

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Like other generalizations mentioned above in this paper, the Banach contraction principle does not characterize the metric completeness of X. However,Theorem 1.1does characterize the metric completeness as follows.

Theorem 1.2. Define a nonincreasing functionθas inTheorem 1.1, then for a metric spaceX, d the following are equivalent:

iXis complete,

iiEvery mappingTonXsatisfying1.2has a fixed point.

In addition to the above results, Kikkawa and Suzuki 7 provide a Kannan type version of the theorems mentioned before. In8, it is provided a Chatterjea type version.

Popescu 9 gives a Ciric type version. Recently, Kikkawa and Suzuki also provide multivalued versions which can be found in10,11. Some fixed-point theorems related to Theorems1.1and1.2have also been proven in12,13.

The aim of this paper is to generalize the above results using the implicit relation technique in such a way that

F d

Tx, Ty , d

x, y

, dx, Tx, d y, Ty

, d x, Ty

, d y, Tx

≤0, 1.3

forx, y∈X, whereF:0,∞⁶ → Ris a function as given inSection 2.

2. Implicit Relation

Implicit relations on metric spaces have been used in many papers. See1,14–16.

LetR denote the nonnegative real numbers, and let Ψbe the set of all continuous functionsF :0,∞⁶ → Rsatisfying the following conditions:

F1:Ft1, . . . , t6is nonincreasing in variablest2, . . . , t6, F2: there existsr∈0,1, such that

Fu, v, v, u, uv,0≤0 2.1

or

Fu, v,0, uv, u, v≤0 2.2

or

Fu, v, v, v, v, v≤0 2.3

impliesu≤rv,

F3:Fu,0,0, u, u,0>0, for allu >0.

Example 2.1. Ft1, . . . , t6 t1−rt2, wherer∈0,1. It is clear thatF∈Ψ.

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Example 2.2. Ft1, . . . , t6 t1−αt3t4, whereα∈0,1/2.

LetFu, v, v, u, uv,0 u−αuv≤0, then we haveu≤α/1−αv. Similarly, let Fu, v,0, uv, u, v≤0, then we haveu≤α/1−αv. Again, letFu, v, v, v, v, v≤0, then u≤2αv. Sinceα/1−α≤2α <1,F2is satisfied withr 2α. AlsoFu,0,0, u, u,0 1−αu >

0, for allu >0. Therefore,F∈Ψ.

Example 2.3. Ft1, . . . , t6 t1−αmax{t3, t4}, whereα∈0,1/2.

Let Fu, v, v, u, uv,0 u−αmax{u, v} ≤ 0, then we have u ≤ αv ≤ α/1 − αv. Similarly, let Fu, v,0, u v, u, v ≤ 0, then we have u ≤ α/1 −αv. Again, let Fu, v, v, v, v, v ≤ 0, thenu ≤ αv ≤ α/1−αv. Thus,F2 is satisfied withr α/1−α.

AlsoFu,0,0, u, u,0 1−αu >0, for allu >0. Therefore,F ∈Ψ.

Example 2.4. Ft1, . . . , t6 t1−αt5t6, whereα∈0,1/2.

LetFu, v, v, u, uv,0 u−αuv≤0, then we haveu≤α/1−αv. Similarly, let Fu, v,0, uv, u, v≤0, then we haveu≤α/1−αv. Again, letFu, v, v, v, v, v≤0, then u≤2αv. Sinceα/1−α≤2α <1,F2is satisfied withr 2α. AlsoFu,0,0, u, u,0 1−αu >

0, for allu >0. Therefore,F∈Ψ.

Example 2.5. Ft1, . . . , t6 t1−at3−bt4, wherea, b∈0,1/2.

LetFu, v, v, u, uv,0 u−av−bu≤ 0, then we haveu≤ a/1−bv. Similarly, let Fu, v,0, uv, u, v ≤ 0, then we have u ≤ b/1−bv. Again, let Fu, v, v, v, v, v ≤ 0, then u ≤ abv. Thus,F2 is satisfied withr max{a/1−b, b/1−b, ab}. Also Fu,0,0, u, u,0 1−bu >0, for allu >0. Therefore,F∈Ψ.

3. Main Result

Theorem 3.1. Let X, d be a complete metric space, and let T be a mapping on X. Define a nonincreasing functionθfrom0,1into1/2,1as inTheorem 1.1. Assume that there existsF∈Ψ, such thatθrdx, Tx≤dx, yimplies

F d

Tx, Ty , d

x, y

, dx, Tx, d y, Ty

, d x, Ty

, d y, Tx

≤0, 3.1

for allx, y∈X, thenT has a unique fixed-pointzand lim_nTⁿxzholds for everyx∈X.

Proof. Sinceθr≤1,θrdx, Tx≤dx, Txholds for everyx∈X, by hypotheses, we have

F d

Tx, T²x , dx, Tx, dx, Tx, d

Tx, T²x , d

x, T²x ,0 ≤0, 3.2

and so fromF1, F

d

Tx, T²x , dx, Tx, dx, Tx, d

Tx, T²x , dx, Tx d

Tx, T²x ,0 ≤0. 3.3

ByF2, we have

d

Tx, T²x ≤rdx, Tx, 3.4

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for allx∈X. Now fixu∈Xand define a sequence{un}inXbyun Tⁿu. Then from3.4, we have

dun, u_n1 d

Tu_n−1, T²u_n−1 ≤rdu_n−1, Tu_n−1≤ · · · ≤rⁿdu, Tu. 3.5

This shows that_∞

n1dun, un1<∞, that is,{un}is Cauchy sequence. SinceX is complete, {un}converges to some pointz∈X. Now, we show that

dTx, z≤rdx, z ∀x∈X\ {z}. 3.6

Forx ∈X\ {z}, there existsn0 ∈N, such thatdun, z ≤ dx, z/3 for alln ≥ n0. Then, we have

θrdun, Tun≤dun, Tun dun, un1

≤dun, z dz, u_n1

≤ 2

3dx, z dx, z−dx, z 3

≤dx, z−dun, z≤dun, x.

3.7

Hence, by hypotheses, we have

FdTun, Tx, dun, x, dun, Tun, dx, Tx, dun, Tx, dx, Tun≤0, 3.8

and so

Fdun1, Tx, dun, x, dun, un1, dx, Tx, dun, Tx, dx, un1≤0. 3.9

Lettingn → ∞, we have

Fdz, Tx, dz, x,0, dx, Tx, dz, Tx, dx, z≤0, 3.10

and so

Fdz, Tx, dz, x,0, dx, z dz, Tx, dz, Tx, dx, z≤0. 3.11

ByF2, we have

dz, Tx≤rdx, z, 3.12

and this shows that3.6is true.

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Now, we assume thatT^mz /zfor allm∈N, then from3.6, we have

d

T^m1z, z ≤r^mdTz, z, 3.13

for allm∈N.

Case 1. Let 0≤r≤√

5−1/2. In this case,θr 1. Now, we show by induction that

dTⁿz, Tz≤rdz, Tz, 3.14

forn≥2. From3.4,3.14holds forn2. Assume that3.14holds for somenwithn≥2.

Since

dz, Tz≤dz, Tⁿz dTⁿz, Tz

≤dz, Tⁿz rdz, Tz, 3.15

we have

dz, Tz≤ 1

1−rdz, Tⁿz, 3.16

and so

θrd

Tⁿz, Tⁿ¹z d

Tⁿz, Tⁿ¹z ≤rⁿdz, Tz

≤ rⁿ

1−rdz, Tⁿz≤ r²

1−rdz, Tⁿz

≤dz, Tⁿz.

3.17

Therefore, by hypotheses, we have

F d

Tⁿ¹z, Tz , dTⁿz, z, d

Tⁿz, Tⁿ¹z , dz, Tz, dTⁿz, Tz, d

z, Tⁿ¹z

≤0, 3.18

and so

F d

Tⁿ¹z, Tz , rⁿ⁻¹dTz, z, rⁿdz, Tz, dz, Tz, rdz, Tz, rⁿdz, Tz ≤0, 3.19

then

F d

Tⁿ¹z, Tz , dTz, z, dz, Tz, dz, Tz, dz, Tz, dz, Tz ≤0, 3.20

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and byF2, we have

d

Tⁿ¹z, Tz ≤rdTz, z. 3.21

Therefore,3.14holds.

Now, from3.6, we have d

Tⁿ¹z, z ≤rdTⁿz, z≤rⁿdTz, z. 3.22

This shows thatTⁿz → z, which contradicts3.14.

Case 2. Let√

5−1/2≤r ≤√

2/2. In this case,θr 1−r/r². Again we want to show that 3.14is true forn≥2. From3.4,3.14holds forn2. Assume that3.14holds for some nwithn≥2. Since

dz, Tz≤dz, Tⁿz dTⁿz, Tz

≤dz, Tⁿz rdz, Tz, 3.23

we have

dz, Tz≤ 1

1−rdz, Tⁿz, 3.24

and so

θrd

Tⁿz, Tⁿ¹z 1−r r² d

Tⁿz, Tⁿ¹z ≤ 1−r rⁿ d

Tⁿz, Tⁿ¹z

≤1−rdz, Tz≤dz, Tⁿz.

3.25

Therefore, as in the previous case, we can prove that3.14is true forn≥2. Again from3.6, we have

d

Tⁿ¹z, z ≤rdTⁿz, z≤rⁿdTz, z. 3.26

This shows thatTⁿz → z, which contradicts3.14.

Case 3. Let√

2/2≤r <1. In this case,θr 1/1r. Note that forx, y∈X, either θrdx, Tx≤d

x, y

3.27

or

θrd

Tx, T²x ≤d Tx, y

3.28

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holds. Indeed, if

θrdx, Tx> d x, y

,

θrd

Tx, T²x > d Tx, y

, 3.29

then we have

dx, Tx≤d x, y

d Tx, y

< θr

dx, Tx d Tx, T²x

≤θrdx, Tx rdx, Tx dx, Tx,

3.30

which is a contradiction. Therefore, either

θrdu2n, Tu2n≤du2n, z 3.31

or

θrdu_2n1, Tu_2n1≤du_2n1, z 3.32

holds for everyn∈N. If

θrdu2n, Tu2n≤du2n, z 3.33

holds, then by hypotheses we have

FdTu2n, Tz, du2n, z, du2n, Tu2n, dz, Tz, du2n, Tz, dz, Tu2n≤0, 3.34

and so

Fdu_2n1, Tz, du2n, z, du2n, u_2n1, dz, Tz, du2n, Tz, dz, u_2n1≤0. 3.35

Fdz, Tz,0,0, dz, Tz, dz, Tz,0≤0, 3.36

which contradictsF3. If

θrdu_2n1, Tu_2n1≤du_2n1, z 3.37

holds, then by hypotheses we have

FdTu_2n1, Tz, du_2n1, z, du_2n1, Tu_2n1, dz, Tz, du_2n1, Tz, dz, Tu_2n1≤0, 3.38

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and so

Fdu2n2, Tz, du2n1, z, du2n1, u2n2, dz, Tz, du2n1, Tz, dz, u2n2≤0. 3.39

Fdz, Tz,0,0, dz, Tz, dz, Tz,0≤0, 3.40

which contradictsF3.

Therefore, in all the cases, there existsm∈N, such thatT^mzz. Since{Tⁿz}is Cauchy sequence, we obtainTz z. That is,zis a fixed point of T. The uniqueness of fixed point follows easily from3.6.

Remark 3.2. If we combineTheorem 3.1with Examples2.1,2.2,2.3, and2.4, we have Theorem 2 of6, Theorem 2.2 of7, Theorem 3.1 of7, and Theorem 4 of8, respectively.

UsingExample 2.5, we obtain the following result.

Corollary 3.3. Let X, d be a complete metric space, and let T be a mapping on X. Define a nonincreasing functionθfrom0,1into1/2,1as inTheorem 1.1. Assume that

θrdx, Tx≤d x, y

3.41

implies

d

Tx, Ty

≤adx, Tx bd y, Ty

, 3.42

for allx, y∈X, wherea, b∈0,1/2, then there exists a unique fixed point ofT.

Remark 3.4. We obtain some new results, if we combine Theorem 3.1with some examples ofF.

References

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727–730, 1972.

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71–76, 1968.

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2949, 2008.

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A Suzuki Type Fixed-Point Theorem

Research Article

A Suzuki Type Fixed-Point Theorem

Ishak Altun and Ali Erduran

1. Introduction

2. Implicit Relation

3. Main Result

References

Submit your manuscripts at http://www.hindawi.com

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