1. Introduction
Banach’s contraction principle is a magnific tool in many fields of nonlinear analysis and in mathematical analysis. Applications in these fields are very interests and promise to other new applications. Banach’s contraction principle has been generalized and extended in many directions. The authors [1] have given an important result about contractions in multi-valued complete metric spaces. We have given the generalization of this result which is a particular case of author’s [2] paper. In this paper, we prove a new fixed point theorem for multi-valued mapping defined on complete metric spaces. To realize this result we give the proof of an intuitive lemma which is used to complete the proof of the main result.
Theorem 1.1 (Đorić and Lazović) Define a nonincreasing function
from
into
by
(1)
Let
be a complete metric space and T be a mapping from X into
. Assuming that there exists
such that
implies
(2)
for all
. Then, there exists
, such that
.
This theorem is a particular case of our main result when
.
Theorem 1.2 (Kaliaj) Let
be a complete metric space and let
be a multi-valued mapping, assuming that T is
-Suzuki integral contraction. Then, for any
we have
(3)
Our main result is a particular case of this theorem when
is absolutely continuous.
Definition 1.3 Let
be a complete metric space and let
be the family of all nonempty closed bounded subsets of X. Define the Hausdorff metric
(4)
for all
.
It is well-known that, if
is a complete metric space, then
is also a complete metric space.
Definition 1.4 Let
be a multi-valued mapping. We say that T is a
-Suzuki contraction with
(5)
if there exists
such that, the implication
(6)
holds whenever
, where
(7)
Definition 1.5 The multi-valued mapping T is an
-Suzuki integral contraction if
is defined in conditions of Theorem 1.1 and there exist
and
, such that the implication
(8)
holds whenever
, where
(9)
Definition 1.6 The multi-valued mapping T is said to be a
-Suzuki contraction with
(10)
if there exists
and
such that, the implication
(11)
holds whenever
, where
(12)
Definition 1.7 A function
is said to be absolutely continuous in
if, given
, there exists some
such that
(13)
whenever
is a finite collection of mutually disjoint sub-intervals of
with
.
Lemma 1.8 (Kaliaj) Let
be a complete metric space and let
be a
-Suzuki integral contraction and
with
(14)
and
for all
.
Then, exists a
such that
(15)
for all
.
When
is absolutely continuous, we have this
Corollary 1.9 Let
be a complete metric space and let
be a
-Suzuki contraction and
with
(16)
and
for all
.
Then, exists a
such that
(17)
for all
.
The main result is Theorem 2.1. First, we give the proof of this lemma:
Lemma 1.10 Let
be a complete metric space and
be a
-Suzuki contraction. Then, the implication
holds for all
.
Proof: First, since
, it follows that
. By the definition of distance of a point to the set we can obtain this inequality:
Because the function
takes values in
, it implies
Since
is increase monotonic we have
and from the fact that T is a
-Suzuki contraction, it follows that
where,
Substituting, the last inequality will be transformed in
which implies
(18)
Remember that, from the definition of Hausdorff distance, we can write:
which implies
Using the fact that
is monotone increasing, it follows that
But
and since
, it follows that
Indeed, using the fact that H is a Hausdorff distance, we can write:
and so,
If the maximum element of the set
it was
then,
which is a contradiction, because
. The last result with the result of inequality (18) implies
Since y was arbitrary, the last result holds
.
Since
is absolutely continuous, by the Corollary 1.9, we can write:
(19)
2. The Main Result
Theorem 2.1 Let
be a complete metric space and let the mapping
be a
-Suzuki contraction. Then, T has a fixed point.
Proof. Let
be an arbitrary fixed point in X. Choose a real number
. If
, then
. Hence
is a fixed point of T and the proof has finished.
Assume that
. Then, there exists
with
Since
, using Lemma 1.10 we obtain:
(20)
We assume that
since:
(21)
As before, if
, for similarity,
is a fixed point for the mapping T and the proof is done.
Assume that
. Since
is continuous at
, given
there exists
, such that
and, since there exists
such that
it follows that
(22)
Inductively, assume now that we chose
. The, by Lemma 1.10 we have
If
, then
is a fixed point of
-Suzuki contraction T, and the proof is done.
Assume that
. Since
is continuous at
, given
there exists
, such that
and, since there exists
such that
it follows that
(23)
Since
, by Lemma 1.10, we have
By above construction, we obtain a sequence
with terms in X such that
(24)
Hence, we get
and since
it follows that
Summing side by side for
to
it follows that
Since
the last result yields that
is a Cauchy sequence in X, and by completeness of X it follows that
converges to a point
. We are going to prove that z is a fixed point of T. Suppose the contrary, i.e.,
. It follows that two cases are possibles:
1)
2)
We study each case as follows:
1) For an fixed arbitrary
, since
we obtain
and since by Corollary 1.9 and Lemma 1.10 we have also
and
it follows that
(25)
Since
was arbitrary, the last results yield
for all
. From equality
it follows that there exists sequence
such that
Then, by inequality (25) we obtain
and since
it follows that
or
which is a contradiction and as consequence,
.
2) Let be an arbitrary
. Since
we get
(26)
Since
, by Corollary 1.9, we have
(27)
If
then
(28)
Otherwise, if
then we obtain by inequality (26) that
and using inequality (26) again, we get
Since x was arbitrary, combining last result with inequality (28) yields
Then, by hypothesis, it follow that
whenever
. Clearly, if
then, the last inequality also holds. Thus, the last inequality holds for all
. In particular, for
, we have
Hence, by Theorem IX.4.1 in [3], it follows that
This contradiction shows that
and the proof is done.
3. Conclusion
In this paper, we studied the
-Suzuki contraction for multi-valued mappings in the complete metric spaces generated by the family of all nonempty closed bounded subsets of a set X, refereed as
. We proved that this contraction has a fixed point.