Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2011, Article ID 736063, 9 pages
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, ishakaltun@yahoo.com
Received 16 December 2010; Accepted 7 February 2011
Academic Editor: Genaro Lopez
Copyright q 2011 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. Let X, d be a complete metric space, and let T be a mapping on X. Define a nonincreasing function θ from 0, 1 into 1/2, 1 by
√
⎧
5−1
⎪
⎪
⎪
,
1,
0≤r≤
⎪
⎪
2
⎪
⎪
⎪
√
⎨
1−r
5−1
1
θr ≤r≤ √ ,
,
2
⎪
2
⎪ r
2
⎪
⎪
⎪
⎪
1
1
⎪
⎪
⎩
, √ ≤ r < 1.
1r
2
1.1
Assume that there exists r ∈ 0, 1, such that
θrdx, T x ≤ d x, y
implies d T x, T y ≤ rd x, y ,
1.2
for all x, y ∈ X, then there exists a unique fixed-point z of T . Moreover, limn T n x z for all x ∈ X.
2
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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 1.1 does
characterize the metric completeness as follows.
Theorem 1.2. Define a nonincreasing function θ as in Theorem 1.1, then for a metric space X, d
the following are equivalent:
i X is complete,
ii Every mapping T on X satisfying 1.2 has a fixed point.
In addition to the above results, Kikkawa and Suzuki 7 provide a Kannan type
version of the theorems mentioned before. In 8, 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 in 10, 11. Some fixed-point theorems related to
Theorems 1.1 and 1.2 have also been proven in 12, 13.
The aim of this paper is to generalize the above results using the implicit relation
technique in such a way that
F d T x, T y , d x, y , dx, T x, d y, T y , d x, T y , d y, T x ≤ 0,
1.3
for x, y ∈ X, where F : 0, ∞6 → R is a function as given in Section 2.
2. Implicit Relation
Implicit relations on metric spaces have been used in many papers. See 1, 14–16.
Let R denote the nonnegative real numbers, and let Ψ be the set of all continuous
functions F : 0, ∞6 → R satisfying the following conditions:
F1 : Ft1 , . . . , t6 is nonincreasing in variables t2 , . . . , t6 ,
F2 : there exists r ∈ 0, 1, such that
Fu, v, v, u, u v, 0 ≤ 0
2.1
Fu, v, 0, u v, u, v ≤ 0
2.2
Fu, v, v, v, v, v ≤ 0
2.3
or
or
implies u ≤ rv,
F3 : Fu, 0, 0, u, u, 0 > 0, for all u > 0.
Example 2.1. Ft1 , . . . , t6 t1 − rt2 , where r ∈ 0, 1. It is clear that F ∈ Ψ.
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3
Example 2.2. Ft1 , . . . , t6 t1 − αt3 t4 , where α ∈ 0, 1/2.
Let Fu, v, v, u, u v, 0 u − αu v ≤ 0, then we have u ≤ 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, then
u ≤ 2αv. Since α/1 − α ≤ 2α < 1, F2 is satisfied with r 2α. Also Fu, 0, 0, u, u, 0 1 − αu >
0, for all u > 0. Therefore, F ∈ Ψ.
Example 2.3. Ft1 , . . . , t6 t1 − α max{t3 , t4 }, where α ∈ 0, 1/2.
Let Fu, v, v, u, u v, 0 u − α max{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, then u ≤ αv ≤ α/1 − αv. Thus, F2 is satisfied with r α/1 − α.
Also Fu, 0, 0, u, u, 0 1 − αu > 0, for all u > 0. Therefore, F ∈ Ψ.
Example 2.4. Ft1 , . . . , t6 t1 − αt5 t6 , where α ∈ 0, 1/2.
Let Fu, v, v, u, u v, 0 u − αu v ≤ 0, then we have u ≤ 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, then
u ≤ 2αv. Since α/1 − α ≤ 2α < 1, F2 is satisfied with r 2α. Also Fu, 0, 0, u, u, 0 1 − αu >
0, for all u > 0. Therefore, F ∈ Ψ.
Example 2.5. Ft1 , . . . , t6 t1 − at3 − bt4 , where a, b ∈ 0, 1/2.
Let Fu, v, v, u, u v, 0 u − av − bu ≤ 0, then we have u ≤ a/1 − bv. Similarly,
let Fu, v, 0, u v, u, v ≤ 0, then we have u ≤ b/1 − bv. Again, let Fu, v, v, v, v, v ≤
0, then u ≤ a bv. Thus, F2 is satisfied with r max{a/1 − b, b/1 − b, a b}. Also
Fu, 0, 0, u, u, 0 1 − bu > 0, for all u > 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 function θ from 0, 1 into 1/2, 1 as in Theorem 1.1. Assume that there exists F ∈ Ψ,
such that θrdx, T x ≤ dx, y implies
F d T x, T y , d x, y , dx, T x, d y, T y , d x, T y , d y, T x ≤ 0,
3.1
for all x, y ∈ X, then T has a unique fixed-point z and limn T n x z holds for every x ∈ X.
Proof. Since θr ≤ 1, θrdx, T x ≤ dx, T x holds for every x ∈ X, by hypotheses, we have
F d T x, T 2 x , dx, T x, dx, T x, d T x, T 2 x , d x, T 2 x , 0 ≤ 0,
3.2
and so from F1 ,
F d T x, T 2 x , dx, T x, dx, T x, d T x, T 2 x , dx, T x d T x, T 2 x , 0 ≤ 0.
3.3
By F2 , we have
d T x, T 2 x ≤ rdx, T x,
3.4
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International Journal of Mathematics and Mathematical Sciences
for all x ∈ X. Now fix u ∈ X and define a sequence {un } in X by un T n u. Then from 3.4,
we have
dun , un1 d T un−1 , T 2 un−1 ≤ rdun−1 , T un−1 ≤ · · · ≤ r n du, T u.
3.5
This shows that ∞
n1 dun , un1 < ∞, that is, {un } is Cauchy sequence. Since X is complete,
{un } converges to some point z ∈ X. Now, we show that
dT x, z ≤ rdx, z
∀x ∈ X \ {z}.
3.6
For x ∈ X \ {z}, there exists n0 ∈ N, such that dun , z ≤ dx, z/3 for all n ≥ n0 . Then, we
have
θrdun , T un ≤ dun , T un dun , un1 ≤ dun , z dz, un1 ≤
dx, z
2
dx, z dx, z −
3
3
3.7
≤ dx, z − dun , z ≤ dun , x.
Hence, by hypotheses, we have
FdT un , T x, dun , x, dun , T un , dx, T x, dun , T x, dx, T un ≤ 0,
3.8
Fdun1 , T x, dun , x, dun , un1 , dx, T x, dun , T x, dx, un1 ≤ 0.
3.9
and so
Letting n → ∞, we have
Fdz, T x, dz, x, 0, dx, T x, dz, T x, dx, z ≤ 0,
3.10
Fdz, T x, dz, x, 0, dx, z dz, T x, dz, T x, dx, z ≤ 0.
3.11
and so
By F2 , we have
dz, T x ≤ rdx, z,
and this shows that 3.6 is true.
3.12
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5
z for all m ∈ N, then from 3.6, we have
Now, we assume that T m z /
d T m1 z, z ≤ r m dT z, z,
3.13
for all m ∈ N.
√
Case 1. Let 0 ≤ r ≤ 5 − 1/2. In this case, θr 1. Now, we show by induction that
dT n z, T z ≤ rdz, T z,
3.14
for n ≥ 2. From 3.4, 3.14 holds for n 2. Assume that 3.14 holds for some n with n ≥ 2.
Since
dz, T z ≤ dz, T n z dT n z, T z
≤ dz, T n z rdz, T z,
3.15
we have
dz, T z ≤
1
dz, T n z,
1−r
3.16
and so
θrd T n z, T n1 z d T n z, T n1 z ≤ r n dz, T z
≤
rn
r2
dz, T n z ≤
dz, T n z
1−r
1−r
3.17
≤ dz, T n z.
Therefore, by hypotheses, we have
F d T n1 z, T z , dT n z, z, d T n z, T n1 z , dz, T z, dT n z, T z, d z, T n1 z ≤ 0,
3.18
and so
F d T n1 z, T z , r n−1 dT z, z, r n dz, T z, dz, T z, rdz, T z, r n dz, T z ≤ 0,
3.19
then
F d T n1 z, T z , dT z, z, dz, T z, dz, T z, dz, T z, dz, T z ≤ 0,
3.20
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International Journal of Mathematics and Mathematical Sciences
and by F2 , we have
d T n1 z, T z ≤ rdT z, z.
3.21
d T n1 z, z ≤ rdT n z, z ≤ r n dT z, z.
3.22
Therefore, 3.14 holds.
Now, from 3.6, we have
This shows that T n z → z, which contradicts 3.14.
√
√
Case 2. Let 5 − 1/2 ≤ r ≤ 2/2. In this case, θr 1 − r/r 2 . Again we want to show that
3.14 is true for n ≥ 2. From 3.4, 3.14 holds for n 2. Assume that 3.14 holds for some
n with n ≥ 2. Since
dz, T z ≤ dz, T n z dT n z, T z
≤ dz, T n z rdz, T z,
3.23
we have
dz, T z ≤
1
dz, T n z,
1−r
3.24
and so
1−r 1−r θrd T n z, T n1 z 2 d T n z, T n1 z ≤ n d T n z, T n1 z
r
r
3.25
≤ 1 − rdz, T z ≤ dz, T z.
n
Therefore, as in the previous case, we can prove that 3.14 is true for n ≥ 2. Again from 3.6,
we have
d T n1 z, z ≤ rdT n z, z ≤ r n dT z, z.
3.26
This shows that T n z → z, which contradicts 3.14.
√
Case 3. Let 2/2 ≤ r < 1. In this case, θr 1/1 r. Note that for x, y ∈ X, either
θrdx, T x ≤ d x, y
3.27
θrd T x, T 2 x ≤ d T x, y
3.28
or
International Journal of Mathematics and Mathematical Sciences
7
holds. Indeed, if
θrdx, T x > d x, y ,
θrd T x, T 2 x > d T x, y ,
3.29
then we have
dx, T x ≤ d x, y d T x, y < θr dx, T x d T x, T 2 x
3.30
≤ θrdx, T x rdx, T x dx, T x,
which is a contradiction. Therefore, either
θrdu2n , T u2n ≤ du2n , z
3.31
θrdu2n1 , T u2n1 ≤ du2n1 , z
3.32
θrdu2n , T u2n ≤ du2n , z
3.33
or
holds for every n ∈ N. If
holds, then by hypotheses we have
FdT u2n , T z, du2n , z, du2n , T u2n , dz, T z, du2n , T z, dz, T u2n ≤ 0,
3.34
Fdu2n1 , T z, du2n , z, du2n , u2n1 , dz, T z, du2n , T z, dz, u2n1 ≤ 0.
3.35
and so
Letting n → ∞, we have
Fdz, T z, 0, 0, dz, T z, dz, T z, 0 ≤ 0,
3.36
θrdu2n1 , T u2n1 ≤ du2n1 , z
3.37
which contradicts F3 . If
holds, then by hypotheses we have
FdT u2n1 , T z, du2n1 , z, du2n1 , T u2n1 , dz, T z, du2n1 , T z, dz, T u2n1 ≤ 0,
3.38
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International Journal of Mathematics and Mathematical Sciences
and so
Fdu2n2 , T z, du2n1 , z, du2n1 , u2n2 , dz, T z, du2n1 , T z, dz, u2n2 ≤ 0.
3.39
Letting n → ∞, we have
Fdz, T z, 0, 0, dz, T z, dz, T z, 0 ≤ 0,
3.40
which contradicts F3 .
Therefore, in all the cases, there exists m ∈ N, such that T m z z. Since {T n z} is Cauchy
sequence, we obtain T z z. That is, z is a fixed point of T . The uniqueness of fixed point
follows easily from 3.6.
Remark 3.2. If we combine Theorem 3.1 with Examples 2.1, 2.2, 2.3, and 2.4, we have Theorem
2 of 6, Theorem 2.2 of 7, Theorem 3.1 of 7, and Theorem 4 of 8, respectively.
Using Example 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 function θ from 0, 1 into 1/2, 1 as in Theorem 1.1. Assume that
θrdx, T x ≤ d x, y
3.41
d T x, T y ≤ adx, T x bd y, T y ,
3.42
implies
for all x, y ∈ X, where a, b ∈ 0, 1/2, then there exists a unique fixed point of T .
Remark 3.4. We obtain some new results, if we combine Theorem 3.1 with some examples
of F.
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