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Fixed Points for Multivalued Mappings and the Metric Completeness
Fixed Point Theory and Applications volume 2009, Article number: 972395 (2009)
Abstract
We consider the equivalence of the existence of fixed points of single-valued mappings and multivalued mappings for some classes of mappings by proving some equivalence theorems for the completeness of metric spaces.
1. Introduction
The Banach contraction principle [1] states that for a complete metric space , every contraction
on
, that is, for some
,
for all
, has a (unique) fixed point.
Connell [2] gave an example of a noncomplete metric space on which every contraction on
has a fixed point. Thus contractions cannot characterize the metric completeness of
Theorem 1.1 (see [3, Kannan]).
Let be a complete metric space. Let
be a Kannan mapping on
, that is, for some
,
for all
Then
has a (unique) fixed point.
Subrahmanyam [4] proved that Kannan mappings can be used to characterize the completeness of the metric. That is, a metric space is complete if and only if every Kannan mapping on
has a fixed point.
In 2008 Suzuki [5] introduced a new type of mappings and presented a generalization of the Banach contraction principle in which the completeness can also be characterized by the existence of fixed points of these mappings. Define a nonincreasing function from
onto
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ1_HTML.gif)
Theorem 1.2 (see [5]).
For a metric space , the following are equivalent:
(i) is complete;
(ii)every mapping on
such that there exists
,
implies
for all
has a fixed point.
In 2008, Kikkawa and Suzuki [6] partially extended Theorem 1.2 to multivalued mappings.
Theorem 1.3 (see [6]).
Define a strictly decreasing function from
onto
by
. Let
be a complete metric space and let
be a multivalued mapping with bounded and closed values. Assume that there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ2_HTML.gif)
for all , then there exists
such that
Obviously, the converse of Theorem 1.3 is valid since for all
Moţ and Petruşel [7] proved the following theorem which is a generalization of Kikkawa and Suzuki Theorem.
Theorem 1.4 (see [7]).
Let be a complete metric space and let
be a multivalued mapping with closed values and satisfies the following: if for nonnegative numbers
with
and for each
, one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ3_HTML.gif)
Then has a fixed point.
In this paper, we will characterize the completeness of a metric space by the existence of fixed points for both single-valued and multivalued mappings. We first aim to extend, in Section 3, the Suzuki's result (Theorem 1.2) to more general classes of mappings. We then consider multivalued mappings in Section 4. We also show in this section that the converse of Theorem 1.4 is true.
2. Preliminaries
Let be a complete metric space and let
be a mapping. We say that
is a Caristi mapping if there exists a lower semicontinuous function
such that
is bounded below and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ4_HTML.gif)
Recall that a mapping is lower semicontinuous if for each
and for every
, there exists a neighborhood
of
such that
for all
For a metric space , let
and
denote, respectively, a collection of all nonempty closed subsets of
and a collection of all nonempty bounded closed subsets of
. Let
be the Hausdorff metric on
. That is, for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ5_HTML.gif)
where is the distance from a point
in
to a subset
of
.
The next theorem plays important roles in this paper.
Theorem 2.1 (see cf. [8]).
If is a mapping of a complete metric space
into the family of all nonempty closed subsets of
and
is a lower semicontinuous function such that the following condition holds:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ6_HTML.gif)
then has at least one fixed point.
3. Completeness and Single-valued Mappings
In 2008, Kikkawa and Suzuki [9] proved fixed point theorems for some generalized Kannan mappings. Let be a nonincreasing function defined from [0,1) onto (1/2,1] by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ7_HTML.gif)
Theorem 3.1 (see [9]).
Let be a complete metric space and let
be a mapping on
. Let
and put
Assume that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ8_HTML.gif)
for all , then
has a unique fixed point
and
holds for every
Theorem 3.2 (see [9]).
Let be a complete metric space and let
be a mapping on
. Suppose that there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ9_HTML.gif)
for all Then
has a unique fixed point
and
holds for every
The above theorems inspire us to present another version of Theorem 1.2. Before doing that we present first the following theorem. The proof of which is a mild modification of the proofs in [5, 9].
Theorem 3.3.
Let be a complete metric space and let
be a mapping on
such that there exists
,
implies
for all
, then
has a fixed point.
Proof.
Since ,
holds for every
, and thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ10_HTML.gif)
If for some
, then
and we get a fixed point
of
Suppose now that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ11_HTML.gif)
We fix and define a sequence
in
by
.
Then , and so
. Thus
is a Cauchy sequence. Since
is complete,
converges to some point
We show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ12_HTML.gif)
Suppose Since
as
, there exists
such that
for each
Observe that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ13_HTML.gif)
Hence for each
Letting
we get
for all
and we obtain (3.6).
As in the proof of [5, Theorem  1.2], we show that for some
from which it is proved that
is a fixed point of
For this purpose, we assume
for all
and find a contradiction. We show, by induction, that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ14_HTML.gif)
From (3.6) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ15_HTML.gif)
Suppose . Thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ16_HTML.gif)
Thus (3.8) holds and now we find a contradiction in each of the following cases.
Case 1 ().
We have .
Assume then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ17_HTML.gif)
which is a contradiction. So
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ18_HTML.gif)
Hence , which is a contradiction.
Case 2 ().
We have .
We show, by induction, that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ19_HTML.gif)
If then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ20_HTML.gif)
which is a contradiction. Therefore
Suppose . Thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ21_HTML.gif)
If then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ22_HTML.gif)
which is a contradiction. Hence and thus (*) holds.
For ,
We have
, which is a contradiction.
Case 3 ().
We claim that or
Suppose not,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ23_HTML.gif)
which is a contradiction. So there exists a subsequence of
such that
:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ24_HTML.gif)
Thus , which is a contradiction.
In fact the following theorem shows that the converse of Theorem 3.3 is valid.
Theorem 3.4.
Let be a metric space. Then the following are equivalent:
(i) is complete;
(ii)for each , every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ25_HTML.gif)
for all has a fixed point;
(iii)for each , every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ26_HTML.gif)
for all has a fixed point;
(iv)for each , every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ27_HTML.gif)
for all has a fixed point;
(v)For nonnegative numbers with
, every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ28_HTML.gif)
for all has a fixed point;
(vi)for each , every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ29_HTML.gif)
for all has a fixed point;
(vii)for each , every mapping
on
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ30_HTML.gif)
for all has a fixed point.
Proof.
The implication (i)(vii) is exactly Theorem 3.3.
(vii)(vi). Let
satisfy (3.22). We show that
satisfies (3.23) to obtain a fixed point for
. Let
,
. Thus
, and (3.23) holds.
(vi)(v). Let
satisfy (3.21). To show
satisfies (3.22), let
,
,
and
. Notice that
Thus
So we get
, and (3.22) holds.
(v)(ii). Let
satisfy (3.18). To show
satisfies (3.21), let
Thus
, and so
and (3.21) holds.
(ii)(i). Follows the same proof of Theorem 1.2. Notice that, for
(vii)(iv). Let
satisfy (3.20). To show
satisfies (3.23), let
. Thus
.
(iv)(iii). Let
satisfy (3.19). We show
satisfies (3.20). Let
,
. Thus
(iii)(i). We know that every Kannan mapping belongs to the class of mappings in (iii). Thus
is complete by Subrahmanyam [4].
4. Completeness and Multivalued Mappings
Inspired by Theorem 1.2 and Theorem 1.3, we prove the following theorem for a larger class of mappings under some certain assumptions.
Theorem 4.1.
Let be a metric space. Then the following are equivalent:
(i) is complete;
(ii)for each , every mapping
such that
implies
,
and the function
is lower semicontinuous has a fixed point.
Observe that Theorem 4.1 is not covered by Theorem 3.4 when considering as single-valued mappings.
Proof of Theorem 4.1.
(i)(ii). Let
be small enough so that
and define
. For any
, we can find some
satisfying
. To apply Theorem 2.1, it remains to show that
. We have
Thus
Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ31_HTML.gif)
Let .
Case :
.
Case :
.
Case :
which is impossible.
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ32_HTML.gif)
Thus has a fixed point by Theorem 2.1.
(ii)(i). Suppose
is not complete.
Define a function as in the proof of Theorem 1.2 and a mapping
as follows:
for each since
and
, there exists
satisfying
We put and write
It is obvious that for all
Since
for all
, for all
, thus
That is,
does not have a fixed point. Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ33_HTML.gif)
We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ34_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ35_HTML.gif)
Fix with
To show that the mapping
satisfies the condition in (ii), that is, for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ36_HTML.gif)
Observe that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ37_HTML.gif)
Case 1 ().
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ38_HTML.gif)
Case 2 ().
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ39_HTML.gif)
Therefore (4.6) holds.
It remains to show that the mapping is lower semicontinuous, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ40_HTML.gif)
Suppose not, then there exists such that
for all
for each
. Since
such that
We have
, for all large
. Thus for each
,
, for all large
So for those
. Consequently,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ41_HTML.gif)
which impliest that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ42_HTML.gif)
a contradiction. Thus the mapping is lower semicontinuous.
The converse of Theorem 1.4 is also valid by following the same proof of Theorem 1.2. Assuming that is not complete, we find a fixed point free mapping
satisfying the condition in Theorem 1.4. Following the same proof of Theorem 1.2 by replacing
by
where
, we obtain
for all
and
is fixed point free. We now verify the condition in Theorem 1.4 for
Fix with
We show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ43_HTML.gif)
Observe that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ44_HTML.gif)
Case 1 ().
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ45_HTML.gif)
Case 2 ().
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ46_HTML.gif)
Therefore (4.13) holds, and the proof of the converse of Theorem 1.4 is complete.
Moreover, by following the proof of Theorem 1.4, we can partially extend the class of mappings and still obtain their fixed points. Notice that
Theorem 4.2.
Let be a metric space. Then the following are equivalent:
(i) is complete.
(ii)every mapping such that for each
with
and for each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ47_HTML.gif)
has a fixed point.
Proof.
(i)(ii). Following the same proof of Theorem 1.4 by replacing
in its proof by
Thus we obtain a sequence
such that
(1), for each
and;
(2) for
Choose so that
and therefore
. We see that the sequence
is Cauchy in
, and so
converges to some
We show
for each
Suppose Since
as
, there exists
such that
for each
We have
Hence
for each
Letting
, we get
for all
as desired.
Next, we show for all
For
we obtain for each
such that
Clearly
, for all
Hence, as
we get
and so
implying that
for
Finally, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ48_HTML.gif)
Thus and
has a fixed point.
(ii)(i). Let
and
, we have
implying
. Hence
is complete by the converse of Theorem 1.4.
5. Caristi Set-Valued Mappings
In 2008, Ćirić [10] proved the following fixed point theorems.
Theorem 5.1 ([10]).
Let be a complete metric space and let
If there exist constants
such that for any
there is
satisfying the following two conditions:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ49_HTML.gif)
Then has a fixed point in
provided a function
is lower semicontinuous.
Theorem 5.2 ([10]).
Let be a complete metric space and
If there exists a function
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ50_HTML.gif)
and such that for any there is
satisfying the following two conditions:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ51_HTML.gif)
Then has a fixed point in
provided a function
is lower semicontinuous.
We give a simple proof of each of these theorems.
Proof of Theorem 5.1.
Define a lower semi-continuous function by
For any
we can find some
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ52_HTML.gif)
We show that . Let
. Clearly,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ53_HTML.gif)
Hence has a fixed point by Theorem 2.1.
Proof of Theorem 5.2.
Let and
. For each
there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ54_HTML.gif)
Furthermore, . Indeed,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F972395/MediaObjects/13663_2008_Article_1196_Equ55_HTML.gif)
Thus has a fixed point by Theorem 2.1.
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Acknowledgment
The authors would like to thank the Thailand Research Fund (grant BRG4780016) for its support.
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Dhompongsa, S., Yingtaweesittikul, H. Fixed Points for Multivalued Mappings and the Metric Completeness. Fixed Point Theory Appl 2009, 972395 (2009). https://doi.org/10.1155/2009/972395
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DOI: https://doi.org/10.1155/2009/972395