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,[email protected] 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 called Kannan if there existsα∈0,1/2such that
d
Tx, Ty
≤αdx, Tx αd y, 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 if 0≤α <√ 2−1, 1−α if√
2−1≤α < 1
2. 1.2
LetTbe a mapping on a complete metric spaceX, d. Assume that there existsα∈0,1/2such that
ϕαdx, Tx≤d x, y
implies d
Tx, Ty
≤αdx, Tx αd y, Ty
1.3
for allx, y∈X. ThenT has a unique fixed pointz. Moreover limnTnxzholds 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, T2x≤rdx, Txfor somer ∈0,1. Then fory∈X, either
1r−1dx, Tx≤d x, y
or 1r−1d
Tx, T2x ≤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
Δ α, β
:α≥0, β≥0, αβ <1 , Δ1
α, β
∈Δ:α≤β, αβα2<1 ,
β0 β1
α0 α1
1
2 3 4
Figure 1:Δjj1, . . . ,4
Δ2 α, β
∈Δ:α≥β, αββ2<1 , Δ3
α, β
∈Δ:α≥β, αββ2≥1 , Δ4
α, β
∈Δ:α≤β, αβα2≥1 .
3.1
SeeFigure 1.
Theorem 3.1. Define a nonincreasing functionψfromΔinto1/2,1by
ψ α, β
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
1 if
α, β
∈Δ1,
1 if
α, β
∈Δ2, 1−β if
α, 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
ψ α, β
dx, Tx≤d x, y
impliesd
Tx, Ty
≤αdx, Tx βd y, Ty
3.3
for allx, y∈X. ThenThas a unique fixed pointz. Moreover limnTnxzholds for everyx∈X.
Proof. We put
q: β
1−α∈0,1, r: α
1−β ∈0,1. 3.4
Sinceψα, β≤1,ψα, βdx, Tx≤dx, Txholds. From the assumption, we have
d
Tx, T2x ≤αdx, Tx βd
Tx, T2x 3.5
and hence
d
Tx, T2x ≤rdx, Tx 3.6
for allx∈X. Since
ψ α, β
d
Tx, T2x ≤d
Tx, T2x ≤rdx, Tx≤dTx, x, 3.7
we have
d
T2x, Tx ≤αd
Tx, T2x βdx, Tx 3.8
and hence
d
Tx, 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
ψ α, β
dun, Tun≤dun, un1≤dun, x 3.12
and hence
dTun, Tx≤αdun, Tun βdx, Tx. 3.13
Therefore we obtain
dz, Tx lim
n→ ∞dun1, Tx lim
n→ ∞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−β
dx, 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
d
Tz, T2z ≤rdz, Tz< dz, Tz lim
n→ ∞dTz, un. 3.17
So for sufficiently largen∈N,
ψ α, β
d
Tz, T2z d
Tz, T2z ≤dTz, un 3.18
holds and hence
d
T2z, z lim
n→ ∞d
T2z, Tun
≤ lim
n→ ∞
αd
Tz, T2z βdun, Tun αd
Tz, T2z .
3.19
Thus we obtain
dz, Tz≤d
z, T2z d
Tz, T2z ≤1αd Tz, T2z
≤1αrdz, Tz αα2
1−β dz, Tz
< dz, Tz,
3.20
which is a contradiction. Therefore we obtainTzz.
In the case whereα, β∈Δ2, if we assumeTz /z, then we have dz, Tz≤d
z, T2z d
Tz, T2z ≤ 1β
d Tz, T2z
≤ 1β
qdz, Tz ββ2
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≤ lim
n→ ∞
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
ψ α, β
dun, Tun≤dun, z or ψ α, β
d
Tun, T2un ≤dTun, z 3.23
holds for everyn∈N. Thus there exists a subsequence{nj}of{n}such that
ψ α, β
d
unj, Tunj ≤d
unj, z 3.24
forj∈N. From the assumption, we have dz, Tz lim
j→ ∞d
Tunj, Tz ≤ lim
j→ ∞
αd
unj, 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
ψ α, β
⎧⎪
⎪⎨
⎪⎪
⎩
1 ifαβmin{α, β}2<1, 1−β
1−βmin
α, β ifαβmin{α, β}2≥1. 3.26
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
ψ α, β
dx, Tx< d x, y
impliesd
Tx, Ty
≤αdx, Tx βd y, 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
ψ α, β
dx, Tx 2≥d x, 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, ψ
α, β
d0, T0> ψ α, β
dxn, Txn 1−q
pnd0, xn 4.5
forn∈N∪ {0}. Since
dTxn, T1−
αdxn, Txn βd1, T1
1−q
1−−pn1−α
βpn1− β2 1−α−β
≤
1−q
1− β2 1−α−β
1−q pn1
1−α
β
≤0,
4.6
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
βpn1−pm1
≤ 1−q
pn1pm1−α
βpn1−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:supn|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 ifm1n,
1−rrm ifm1< 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, ψ
α, β
d0, T0> ψ α, β
dxn, Txn 1−rrnd0, xn 4.13
forn∈N∪ {0}. Since
dTe1, Tx0−
αde1, Te1 βdx0, Tx0
1−β
1−2r2 ≤0, 4.14
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 4.17
forn∈N∪ {0}. Form, n∈N∪ {0}withm1< n, we have ψ
α, β
dxm, Txm 1−rrmdxm, xn, dTxn, Txm−
αdxn, Txn βdxm, Txm
< dTxn, Txm−βdxm, Txm rm11−r−βrm
1−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.
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