Fixed Point Theory and Applications Volume 2008, Article ID 649749,8pages doi:10.1155/2008/649749
Research Article
Some Similarity between Contractions and Kannan Mappings
Misako Kikkawa1and Tomonari Suzuki2
1Department of Mathematics, Faculty of Science, Saitama University, 255 Shimo-Okubo, Sakura, Saitama 338-8570, Japan
2Department of Mathematics, Kyushu Institute of Technology, Sensuicho, Tobata, Kitakyushu 804-8550, Japan
Correspondence should be addressed to Tomonari Suzuki,[email protected] Received 11 October 2007; Accepted 13 November 2007
Recommended by J. R. L. Webb
Contractions are always continuous and Kannan mappings are not necessarily continuous. This is a very big difference between both mappings. However, we know that relaxed both mappings are quite similar. In this paper, we discuss both mappings from a new point of view.
Copyrightq2008 M. Kikkawa and T. Suzuki. 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
LetX, dbe a metric space and letT be a mapping onX. ThenTis called a contraction if there existsr∈0,1such that
dTx, Ty≤rdx, y 1.1
for allx, y∈X.Tis called Kannan if there existsα∈0,1/2such that
dTx, Ty≤αdx, Tx αdy, Ty 1.2
for all x, y ∈ X. We know that if X is complete, then every contraction and every Kannan mapping have a unique fixed point, see1,2. We know that both conditions are independent, that is, there exist a contraction, which is not Kannan, and a Kannan mapping, which is not a contraction. Thus we cannot compare both conditions directly. So we compare both indirectly.
Fact 1
Banach fixed-point theorem, which is often called the Banach contraction principle, is very im- portant because it is a very forceful tool in nonlinear analysis. We think that Kannan fixed-point theorem is also very important because Subrahmanyam3proved that Kannan theorem char- acterizes the metric completeness of underlying spaces, that is, a metric spaceX is complete if and only if every Kannan mapping onXhas a fixed point. On the other hand, Connell4 gave an example of a metric spaceXsuch thatXis not complete and every contraction onX has a fixed point. Thus the Banach theorem cannot characterize the metric completeness ofX.
Therefore, we consider that the notion of contractions is stronger from this point of view.
Fact 2
Using the notion ofτ-distances, Suzuki5considered some weaker contractions and Kannan mappings and proved the following.
iIf T is a contraction with respect to aτ-distance, thenT is Kannan with respect to anotherτ-distance.
iiIf T is Kannan with respect to aτ-distance, thenT is a contraction with respect to anotherτ-distance.
That is, both conditions are completely the same.
Recently, Suzuki6proved the following theorem, see also7.
Theorem 1.1see6. Define a nonincreasing functionθfrom0,1onto1/2,1by
θr
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
1 if 0≤r≤ 1 2
√5−1 , 1−r
r2 if 1 2
√5−1
≤r≤ 1
√2, 1
1r if 1
√2 ≤r <1.
1.3
Then for a metric spaceX, d, the following are equivalent:
iXis complete,
iievery mappingTonX, satisfying the following, has a fixed point: there existsr∈0,1such thatθrdx, Tx≤dx, yimpliesdTx, Ty≤rdx, yfor allx, y∈X.
Remark 1.2.θris the best constant for everyr.
The purpose of this paper is to prove a Kannan version ofTheorem 1.1. Then we compare the theoremTheorem 2.2withTheorem 1.1and attempt to judge which is stronger from our new point of view.
2. Kannan mappings
Throughout this paper we denote byNthe set of all positive integers and byRthe set of all real numbers.
In this section, we prove our main result. We begin with the following lemma.
Lemma 2.1. LetX, dbe a metric space and letTbe a mapping onX.Letx∈XsatisfydTx, T2x≤ rdx, Txfor somer∈0,1.Then fory∈X,either
1
1rdx, Tx≤dx, y or 1 1rd
Tx, T2x
≤dTx, y 2.1
holds.
Proof. We assume 1
1rdx, Tx> dx, y, 1 1rd
Tx, T2x
> dTx, y. 2.2
Then we have
dx, Tx≤dx, y dy, Tx
< 1 1r
dx, Tx d
Tx, T2x
≤ 1 1r
dx, Tx rdx, Tx
dx, Tx.
2.3
This is a contradiction.
The following theorem is a Kannan version ofTheorem 1.1.
Theorem 2.2. Define a nonincreasing functionϕfrom0,1into1/2,1by
ϕr
⎧⎪
⎪⎨
⎪⎪
⎩
1 if 0≤r < 1
√2, 1
1r if 1
√2 ≤r <1. 2.4
LetX, dbe a complete metric space and letT be a mapping onX. Letα ∈ 0,1/2and putr : α/1−α∈0,1. Assume that
ϕrdx, Tx≤dx, y impliesdTx, Ty≤αdx, Tx αdy, Ty 2.5
for allx, y∈X, thenThas a unique fixed pointzand limnTnxzholds for everyx∈X.
Proof. Sinceϕr≤1,ϕrdx, Tx≤dx, Txholds. From the assumption, we have d
Tx, T2x
≤αdx, Tx αd
Tx, T2x
, 2.6
and hence
d
Tx, T2x
≤rdx, Tx 2.7
forx∈X. Letu∈X. Putu0uandunTnufor alln∈N. From2.7, we have ∞
n1
d
un, un1
≤∞
n1
rnd u0, u1
<∞. 2.8
1
0.5 r0 r1
The graph ofθ
a
r0 r1
The graph ofϕ
b Figure 1
So{un}is a Cauchy sequence inXand by the completeness ofX, there exists a pointzsuch thatun→z.
We next show
dz, Tx≤αdx, Tx, ∀x∈X withx / z. 2.9
Sinceun →z, there existsn0 ∈ Nsuch thatdun, z≤ 1/3dx, zfor alln∈Nwithn≥ n0. Then we have
ϕrd
un, Tun
≤d
un, Tun d
un, un1
≤d un, z
d un1, z
≤2
3dx, z dx, z−1 3dx, z
≤dx, z−d un, z
≤dun, x ,
2.10
and hence
d
Tun, Tx
≤αd un, Tun
αdx, Tx forn∈Nwithn≥n0. 2.11
Therefore, we obtain
dz, Tx lim
n→∞d
un1, Tx lim
n→∞d
Tun, Tx
≤ lim
n→∞
αd
un, Tun
αdx, Tx αdx, Tx
2.12
forx∈Xwithx / z.
Let us prove thatzis a fixed point ofT. In the case where 0 ≤ r < 1/√
2, arguing by contradiction, we assume thatTz / z. Then we have, from2.9,
d z, T2z
≤αdTz, T2z
≤αrdz, Tz, 2.13
and hence
dz, Tz≤d z, T2z
d
Tz, T2z
≤αrdz, Tz rdz, Tz r2r2
1r dz, Tz
< r1
1rdz, Tz dz, Tz.
2.14
This is a contradiction. Therefore, we obtainTz z. In the case where 1/√
2 ≤ r < 1, from Lemma 2.1, either
ϕrd
u2n, u2n1
≤d u2n, z
or ϕrd
u2n1, u2n2
≤d
u2n1, z
2.15 holds forn∈N. Thus there exists a subsequence{nj}of{n}such that
ϕrd
unj, unj1
≤d unj, z
2.16 forj ∈N. From the assumption, we have
dz, Tz lim
j→∞d
unj1, Tz
≤lim
j→∞
αd
unj, unj1
αdz, Tz
αdz, Tz. 2.17
Sinceα <1/2, we haveTzz. Therefore, we have shownTzzin both cases.
From2.9, we obtain that the fixed pointzis unique.
Remark 2.3. Sinceθr ≤ ϕrfor everyr, we can consider that Kannan is stronger from our new point of view. Thoughθandϕare different, we remark that the graphs ofθandϕare quite similar.
The following theorem shows thatϕris the best constant for everyr.
Theorem 2.4. Define a functionϕas inTheorem 2.2. For everyα∈0,1/2,puttingrα/1−α, there exist a complete metric spaceX, dand a mappingTonXsuch thatT has no fixed points and
ϕrdx, Tx< dx, y implies dTx, Ty≤αdx, Tx αdy, Ty 2.18 for allx, y∈X.
Proof. In the case where 0≤r <1/√
2, define a complete subsetXof the Euclidean spaceRby X {−1,1}. We also define a mappingT onXbyTx−xforx∈X. ThenT dose not have a fixed point and
ϕrdx, Tx 2≥dx, y 2.19
for allx, y∈X. In the case where 1/√
2≤r < 1, define a complete subsetXof the Euclidean spaceRby
X{0,1} ∪ xn:n∈N∪ {0}
, 2.20
wherexn 1−r−rnfor alln∈N∪ {0}. Define a mappingTonXbyT01,T11−r, and Txnxn1forn∈N∪ {0}. Then the following are obvious:
idT0, T1 rαd0, T0 αd1, T1,
iiϕrd0, T0≥ϕrdxn, Txn d0, xnforn∈N∪ {0}.
Also, we have d
Txm, Txn
≤d 0, Txm
d 0, Txn
αd
xm, Txm
αd xn, Txn
, d
T1, Txn
−
αd1, T1 αd
xn, Txn
≤d0, T1 d 0, Txn
−
αd1, T1 αd
xn, Txn d0, T1−αd1, T1 1−2r2
1r ≤0
2.21 form, n∈N∪ {0}.
3. Generalized Kannan mappings
It is a very natural question of whether or not another fixed-point theorem withθexists. In this section, we give a positive answer to this problem.
Theorem 3.1. Define a nonincreasing functionθas inTheorem 1.1. LetX, dbe a complete metric space and letTbe a mapping onX. Suppose that there existsr∈0,1such that
θrdx, Tx≤dx, y impliesdTx, Ty≤rmax dx, Tx, dy, Ty
3.1
for allx, y∈X. ThenT has a unique fixed pointzand limnTnxzholds for everyx∈X.
Proof. Sinceθrdx, Tx≤dx, Tx, we have, from the assumption, d
Tx, T2x
≤rmax dx, Tx, d
Tx, T2x
3.2
and hence
d
Tx, T2x
≤rdx, Tx 3.3
forx∈X. Letu∈X. Putu0uandunTnufor alln∈N. As in the proof ofTheorem 2.2, we can prove that{un}converges to somez∈X.
We next show
dz, Tx≤rdx, Tx for allx∈X withx / z. 3.4
Sinceun→z, we haveθrdun, Tun≤dun, xfor sufficiently largen∈N. Hence we obtain, from the assumption,
dz, Tx lim
n→∞d
un1, Tx lim
n→∞d
Tun, Tx
≤ lim
n→∞rmax{d
un, Tun
, dx, Tx}rdx, Tx 3.5
forx∈Xwithx / z.
Let us prove thatzis a fixed point ofT. In the case where 0≤r <1/√
2, we note
θr≤1−r
r2 . 3.6
We will show, by induction,
d
Tnz, Tz
≤rdz, Tz 3.7
forn∈Nwithn≥2. Whenn2,3.7becomes3.3, thus3.7holds. We assumedTnz, Tz≤ rdz, Tzfor somen∈Nwithn≥2. Since
dz, Tz≤d z, Tnz
d
Tnz, Tz
≤d z, Tnz
rdz, Tz, 3.8
we havedz, Tz≤1/1−rdz, Tnz, and hence θrd
Tnz, Tn1z
≤1−r r2 d
Tnz, Tn1z
≤ 1−r rn d
Tnz, Tn1z
≤1−rdz, Tz≤d z, Tnz
.
3.9
Therefore, by the assumption, we have d
Tn1z, Tz
≤rmax d
Tnz, Tn1z
, dz, Tz
rdz, Tz. 3.10
By induction,3.7holds forn∈Nwithn≥2. Arguing, by contradiction, we assumeTz / z.
Then from3.7,Tnz / zholds for alln∈N. Then by3.4, we have d
Tn1z, z
≤rd
Tnz, Tn1z
≤rn1dz, Tz. 3.11
This impliesTnz→z, which contradicts3.7. Therefore, we obtainTzz. In the case where 1/√
2≤r <1, as in the proof ofTheorem 2.2, we can show that there exists a subsequence{nj} of{n}such thatϕrdunj, unj1≤dunj, zforj∈N. From the assumption, we have
dz, Tz lim
j→∞d
unj1, Tz
≤lim
j→∞rmax d
unj, unj1
, dz, Tz
rdz, Tz. 3.12
Sincer <1, the above inequality implies thatTzz. Therefore, we have shown thatTzzin both cases.
From3.4, we obtain that the fixed pointzis unique.
Remark 3.2. When the second author was provingTheorem 1.1, he did not feel thatθrwas natural. However, since the above proof is easier to understand howθrworks, the authors can faintly feel thatθris natural.
The following theorem shows thatθris the best constant for everyr.
Theorem 3.3. Define a functionθas inTheorem 1.1. Then for anyr ∈0,1,there exist a complete metric spaceX, dand a mappingTonXsuch thatT has no fixed points and
θrdx, Tx< dx, y impliesdTx, Ty≤rmax dx, Tx, dy, Ty
3.13 for allx, y∈X.
Proof. We have already shown the conclusion in the case where 0≤r≤1/2√
5−1or 1/√ 2≤ r <1 becauseϕr θrholds. So let us consider the case where1/2√
5−1< r <1/√ 2.
Define a complete subsetXof the Euclidean spaceRbyX{xn:n∈N},wherex00,x11, x2 1−r,andxn 1−r−r2−rn−3forn≥ 3. Define a mappingT onX byTxn xn1for n∈N. Then the following are obvious:
idTx0, Tx1 rrdx0, Tx0 rmax{dx0, Tx0, dx1, Tx1}, iiθrdx0, Tx0≥θrdx2, Tx2 1−rdx0, x2,
iiiθrdx0, Tx0≥θrdxn, Txn 1−r2/r2dx0, xn≥dx0, xnforn≥3, ivdTx1, Tx2 r2rdx1, Tx1.
Since
x3< x5< x7<· · ·< x0<· · ·< x8< x6< x4< x2< x1, 3.14
we have the following:
idTx1, Txn< dx2, x3 r2rdx1, Tx1forn≥3,
iidTx2, Txn−rdx2, Tx2≤dx3, x4−r32r2−1≤0 forn≥3, iiidTxm, Txn≤dTxm, Txm1 rdxm, Txmfor 3≤m < n.
This completes the proof.
Acknowledgment
The second author 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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