A Generalization of Kannan's Fixed Point Theorem

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Volume 2009, Article ID 192872,10pages doi:10.1155/2009/192872

Research Article

A Generalization of Kannan’s Fixed Point Theorem

Yusuke Enjouji, Masato Nakanishi, and Tomonari Suzuki

Department of Mathematics, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan

Correspondence should be addressed to Tomonari Suzuki,suzuki-t@mns.kyutech.ac.jp

Received 22 December 2008; Accepted 23 March 2009

Recommended by Jerzy Jezierski

In order to observe the condition of Kannan mappings, we prove a generalization of Kannan’s fixed point theorem. Our theorem involves constants and we obtain the best constants to ensure a fixed point.

Copyrightq2009 Yusuke Enjouji et al. 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.

1. Introduction

A mappingTon a metric spaceX, dis calledKannanif there existsα∈0,1/2such that

dTx, Ty≤αdx, Tx αdy, Ty 1.1

for allx, y ∈X. Kannan1proved that ifXis complete, then every Kannan mapping has a fixed point. It is interesting that Kannan’s theorem is independent of the Banach contraction principle2. Also, Kannan’s fixed point theorem is very important because Subrahmanyam

3 proved that Kannan’s theorem characterizes the metric completeness. That is, a metric spaceX is complete if and only if every Kannan mapping onXhas a fixed point. Recently, Kikkawa and Suzuki proved a generalization of Kannan’s fixed point theorem. See also4–8.

Theorem 1.1see9. Define a nonincreasing functionϕfrom0,1/2into1/2,1by

ϕα

⎧ ⎪ ⎨ ⎪ ⎩

1 if0≤α <√2−1,

1−α if√2−1≤α < 1

2.

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LetTbe a mapping on a complete metric spaceX, d. Assume that there existsα∈0,1/2such that

ϕαdx, Tx≤dx, yimplies dTx, Ty≤αdx, Tx αdy, Ty 1.3

for allx, y∈X. ThenT has a unique fixed pointz. MoreoverlimnTnxzholds for everyx∈X. Remark 1.2. ϕαis the best constant for everyα∈0,1/2.

From this theorem, we can tell that a Kannan mapping withα <√2−1 is much stronger than a Kannan mapping withα≥√2−1.

Whilexandyplay the same role in1.1,xandydo not play the same role in1.3. So we can consider “αdx, Tx βdy, Ty” instead of “αdx, Tx αdy, Ty.” And it is a quite natural question of what is the best constant for each pairα, β. In this paper, we give the complete answer to this question.

2. Preliminaries

Throughout this paper we denote byNthe set of all positive integers and byRthe set of all real numbers.

We use two lemmas. The first lemma is essentially proved in5.

Lemma 2.1see5,9. LetX, dbe a metric space and letT be a mapping onX. Letx∈Xsatisfy dTx, T2_x_≤_{rdx, Tx}_{for some}_r _∈₀_,₁_{. Then for}_y_∈_{X, either}

1r−1dx, Tx≤dx, y or 1r−1dTx, T2_x _≤_d_{Tx, y} _2.1

holds.

The second lemma is obvious. We use this lemma several times in the proof of Theorem 4.1.

Lemma 2.2. Leta,A,b, andBbe four real numbers such thata≤Aandb≤B. ThenaBAb≤ abABholds.

3. Fixed Point Theorem

In this section, we prove a fixed point theorem. We first putΔandΔjj1, . . . ,4by

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β0 β1

α0 α1

1

2 3 4

Figure 1:Δjj1, . . . ,4

Δ2α, β∈Δ:α≥β, αββ2<1,

Δ3α, β∈Δ:α≥β, αββ2≥1,

Δ4α, β∈Δ:α≤β, αβα2≥1.

3.1

SeeFigure 1.

Theorem 3.1. Define a nonincreasing functionψfromΔinto1/2,1by

ψα, β

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

1 ifα, β∈Δ1,

1 ifα, β∈Δ2,

1−β ifα, β∈Δ3,

1−β

1−βα if

α, β∈Δ4.

3.2

LetTbe a mapping on a complete metric spaceX, d. Assume that there existsα, β∈Δsuch that

ψα, βdx, Tx≤dx, yimpliesdTx, Ty≤αdx, Tx βdy, Ty 3.3

for allx, y∈X. ThenThas a unique fixed pointz. MoreoverlimnTnxzholds for everyx∈X. Proof. We put

q: β

1−α∈0,1, r: α

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Sinceψα, β≤1,ψα, βdx, Tx≤dx, Txholds. From the assumption, we have

dTx, T2x ≤αdx, Tx βdTx, T2x 3.5

and hence

dTx, T2x ≤rdx, Tx 3.6

for allx∈X. Since

ψα, βdTx, T2x ≤dTx, T2x ≤rdx, Tx≤dTx, x, 3.7

we have

dT2x, Tx ≤αdTx, T2x βdx, Tx 3.8

and hence

dTx, T2x ≤qdx, Tx 3.9

for allx∈X.

Fixu∈Xand putunTnuforn∈N. From3.6, we have

∞

n1

dun, un1≤ ∞

n1

rndu, Tu<∞. 3.10

So{un}is a Cauchy sequence inX. SinceXis complete,{un}converges to some pointz∈X.

We next show

dz, Tx≤βdx, Tx ∀x∈X\ {z}. 3.11

Since{un}converges, for suﬃciently largen∈N, we have

ψα, βdun, Tun≤dun, un1≤dun, x 3.12

and hence

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Therefore we obtain

dz, Tx lim

n→ ∞dun1, Tx nlim→ ∞dTun, Tx ≤ lim

n→ ∞

αdun, Tun βdx, Tx

βdx, Tx 3.14

for allx∈X\ {z}. By3.11, we have

dx, Tx≤dx, z dz, Tx≤dx, z βdx, Tx 3.15

and hence

1−βdx, Tx≤dx, z ∀x∈X\ {z}. 3.16

Let us prove thatz is a fixed point ofT. In the case whereα, β ∈ Δ1, arguing by contradiction, we assumeTz /z. Then we have

dTz, T2z ≤rdz, Tz< dz, Tz lim

n→ ∞dTz, un. 3.17

So for suﬃciently largen∈N,

ψα, βdTz, T2z dTz, T2z ≤dTz, un 3.18

holds and hence

dT2z, z lim

n→ ∞d

T2z, Tun

≤ lim

n→ ∞

αdTz, T2z βdun, Tun αd

Tz, T2z .

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Thus we obtain

dz, Tz≤dz, T2z dTz, T2z ≤1αdTz, T2z

≤1αrdz, Tz αα

2

1−β dz, Tz < dz, Tz,

3.20

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In the case whereα, β∈Δ2, if we assumeTz /z, then we have

dz, Tz≤dz, T2z dTz, T2z ≤1βdTz, T2z

≤1βqdz, Tz ββ 2

1−αdz, Tz < dz, Tz,

3.21

which is a contradiction. ThereforeTzzholds.

In the case whereα, β∈Δ3, we consider the following two cases.

iThere exist at least two natural numbersnsatisfyingunz. iiun/zfor suﬃciently largen∈N.

In the first case, if we assumeTz /z, then{un}cannot be Cauchy. ThereforeTz z. In the

second case, we have by3.16,ψα, βdun, Tun≤dun, zfor suﬃciently largen∈N. From

the assumption,

dz, Tz lim

n→ ∞dTun, Tz≤nlim→ ∞

αdun, Tun βdz, Tz

βdz, Tz. 3.22

Sinceβ <1, we obtainTzz.

In the case whereα, β∈Δ4, we note thatψα, β 1r−1. ByLemma 2.1, either

ψα, βdun, Tun≤dun, z or ψ

α, βdTun, T2un ≤dTun, z 3.23

holds for everyn∈N. Thus there exists a subsequence{nj}of{n}such that

ψα, βdunj, Tunj ≤d

unj, z 3.24

forj∈N. From the assumption, we have

dz, Tz lim

j→ ∞d

Tunj, Tz ≤ lim j→ ∞

αdunj, Tunj βdz, Tz βdz, Tz. 3.25

Sinceβ <1, we obtainTzz. Therefore we have shownTzzin all cases. From3.11, the fixed pointzis unique.

Remark 3.2. We have shownTzz, dividing four cases. It is interesting that the four methods are all diﬀerent. We can rewriteψby

ψα, β

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

1 ifαβmin{α, β}2<1,

1−β

1−βminα, β ifαβmin{α, β}

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4. The Best Constants

In this section, we prove the following theorem, which informs thatψα, βis the best constant for everyα, β∈Δ.

Theorem 4.1. Define a functionψ as inTheorem 3.1. For everyα, β ∈ Δ, there exist a complete metric spaceX, dand a mappingT onXsuch thatT has no fixed points and

ψα, βdx, Tx< dx, yimpliesdTx, Ty≤αdx, Tx βdy, Ty 4.1

for allx, y∈X.

Proof. We putqandrby3.4.

In the case whereα, β∈Δ1∪Δ2, define a complete subsetXof the Euclidean space

RbyX {−1,1}. We also define a mappingT onXbyTx −xforx∈X. ThenT does not have any fixed points and

ψα, βdx, Tx 2≥dx, y 4.2

for allx, y∈X.

In the case whereα, β∈Δ3, we put

p: β

1−β ∈0,1. 4.3

We note thatψα, β1p 1. Define a complete subsetXof the Euclidean spaceRby

X{0,1} ∪ {xn:n∈N∪ {0}}, 4.4

wherexn 1−q−pn forn∈N∪ {0}. Define a mappingT onX byT0 1,T1 x0,and Txnxn1forn∈N∪ {0}. Then we have

dT1, T0 qαd1, T1 βd0, T0≤αd0, T0 βd1, T1,

ψα, βd0, T0> ψα, βdxn, Txn

1−qpnd0, xn

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forn∈N∪ {0}. Since

dTxn, T1−

αdxn, Txn βd1, T1

1−q

1−−pn1−α

βp

n1₋ β2

1−α−β

≤1−q

1− β

2

1−α−β

1−qpn1

1−α

β

≤0,

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we have

dTxn, T1≤αdxn, Txn βd1, T1≤αd1, T1 βdxn, Txn 4.7

forn∈N∪ {0}. Form, n∈N∪ {0}withm < n, since

dTxn, Txm−

αdxn, Txn βdxm, Txm

1−q−pn1−−pm1−α βp

n1₋_pm1

≤1−qpn1pm1−α βp

n1₋_pm1

≤0,

4.8

we have

dTxn, Txm≤αdxn, Txn βdxm, Txm≤αdxm, Txm βdxn, Txn. 4.9

In the case whereα, β ∈ Δ4, we note thatψα, β1r 1. We also note thatr ≥

2−1/2>1/2. Let∞be the Banach space consisting of all functionsffromNintoRi.e.,fis a

real sequencesuch thatf:sup_n|fn|<∞. Let{en}be the canonical basis of∞. Define

a complete subsetXof∞by

X{0, e1} ∪ {xn:n∈N∪ {0}}, 4.10

where

xn 1−rrnen1−1−rrnen2 4.11

forn∈N∪ {0}. We note that

dxm, xn

⎧ ⎨ ⎩

1−r2_rm _if_m₁_n,

1−rrm _if_m₁_{< n,} 4.12

form, n∈Nwithm < n. Define a mappingTonX byT0 e1,Te1 x0, andTxn xn1for n∈N∪ {0}. Then we have

dT0, Te1 rαd0, T0 βde1, Te1≤αde1, Te1 βd0, T0, ψα, βd0, T0> ψα, βdxn, Txn 1−rrnd0, xn

4.13

forn∈N∪ {0}. Since

dTe1, Tx0−

αde1, Te1 βdx0, Tx0

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we have

dTe1, Tx0≤αde1, Te1 βdx0, Tx0≤αdx0, Tx0 βde1, Te1. 4.15

Sinceαβα2 _≥_{1, we have}

dTe1, Txn 1−r ≤αrαde1, Te1

< αde1, Te1 βdxn, Txn≤αdxn, Txn βde1, Te1

4.16

forn∈N. We have

dTxn, Txn1

1−r2 rn1αdxn, Txn βdxn1, Txn1 ≤αdxn1, Txn1 βdxn, Txn

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forn∈N∪ {0}. Form, n∈N∪ {0}withm1< n, we have

ψα, βdxm, Txm 1−rrmdxm, xn, dTxn, Txm−

αdxn, Txn βdxm, Txm

< dTxn, Txm−βdxm, Txm rm11−r−βrm1−r2 rm₁₋_r_α₋_β_≤₀_.

4.18

This completes the proof.

Acknowledgment

T. Suzuki is supported in part by Grants-in-Aid for Scientific Research from the Japanese Ministry of Education, Culture, Sports, Science and Technology.

References

1 R. Kannan, “Some results on fixed points. II,”American Mathematical Monthly, vol. 76, pp. 405–408, 1969.

2 S. Banach, “Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales,”Fundamenta Mathematicae, vol. 3, pp. 133–181, 1922.

3 P. V. Subrahmanyam, “Completeness and fixed-points,”Monatshefte f ¨ur Mathematik, vol. 80, no. 4, pp. 325–330, 1975.

4 M. Kikkawa and T. Suzuki, “Three fixed point theorems for generalized contractions with constants in complete metric spaces,”Nonlinear Analysis: Theory, Methods & Applications, vol. 69, no. 9, pp. 2942– 2949, 2008.

5 T. Suzuki, “A generalized Banach contraction principle that characterizes metric completeness,”

Proceedings of the American Mathematical Society, vol. 136, no. 5, pp. 1861–1869, 2008.

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7 T. Suzuki and M. Kikkawa, “Some remarks on a recent generalization of the Banach contraction principle,” inProceedings of the 8th International Conference on Fixed Point Theory and Its Applications (ICFPTA ’08), S. Dhompongsa, K. Goebel, W. A. Kirk, S. Plubtieng, B. Sims, and S. Suantai, Eds., pp. 151–161, Yokohama, Chiang Mai, Thailand, July 2008.

8 T. Suzuki and C. Vetro, “Three existence theorems for weak contractions of Matkowski type,”

International Journal of Mathematics and Statistics, vol. 6, supplement 10, pp. 110–120, 2010.