Common Fixed Points for Two Contractive Mappings of Integral Type in Metric Spaces ()
1. Introduction and Preliminaries
Throughout this paper, we assume that
and
satisfying that
is Lebesgue
integral, summable on each compact subset of
and
for each
.
The famous Banach’s contraction principle is as follows:
Theorem 1.1 ([1] ). Let f be a self mapping on a complete metric space
satisfying
(1.1)
where
is a constant. Then f has a unique fixed point
such that
for each
.
It is known that the Banach contraction principle has a lot of generalizations and various applications in many directions; see, for examples, [2] - [15] and the references cited therein. In 1962, Rakotch [11] extended the Banach contraction principle with replacing the contraction constant c in (1.1) by a contraction function
and obtained the next theorem.
Theorem 1.2 ( [11] ). Let f be a self-mapping on a complete metric space
satisfying
(1.2)
where
is a monotonically decreasing function. Then f has a unique fixed point
such that
for each
.
In 2002, Branciari [12] gave an integral version of Theorem 1.1 as follows.
Theorem 1.3 ( [12] ). Let f be a self-mapping on a complete metric space
satisfying
(1.3)
where
is a constant and
. Then f has a unique fixed point
such that
for each.
In 2011, Liu and Li [13] modified the method of Rakotch to generalize the Branciari’s fixed point theorem with replacing the contraction constant c in (1.3) by contraction functions
and
and established the following fixed point theorem:
Theorem 1.4 ([13] ). Let f be a self-mapping on a complete metric space
satisfying
(1.4)
where
and
are two functions with
for all
;
;
for all
.
Then f has a unique fixed point
such that
for each
.
Here, we will use the methods in [3] [9] [13] to discuss the unique existence problems of common fixed points for two self-mappings satisfying two different contractive conditions of integral type in a complete metric space.
2. Common Fixed Point Theorems
Lemma 2.1 ([13] ). Let
and
be a nonnegative sequence with
. Then
.
Lemma 2.2([13] ). Let
and
be a nonnegative sequence. Then
![]()
Now, we will give the first main result in this paper.
Theorem 2.1. Let
be a complete metric space,
two mappings. If for each
,
(2.1)
where
and
are three functions satisfying the following conditions
(2.2)
(2.3)
Then f and g have a unique common fixed point u, and the sequence
defined by
,
for any
converges to u.
Proof.
Let
. We construct a sequence
satisfying the following conditions
,
for all
. Let
for all
.
For
, by (2.1),
![]()
hence by (2.3),
(2.4)
Similarly, by (2.1),
![]()
hence by (2.3),
(2.5)
Combining (2.4) and (2.5), we have
(2.6)
Now, we prove that
(2.7)
Otherwise, there exists
such that
(2.8)
Obviously,
. If
, then by (2.3), (2.4), (2.6) and (2.8),
![]()
which is a contradiction. Similarly, if
, then by (2.3), (2.5), (2.6) and (2.8),
![]()
which is also a contradiction. Hence (2.7) holds. Therefore there exists
such that
. If
, then by Lemma 2.1, (2.3) and (2.4),
![]()
which is a contradiction. Therefore
, that is,
.
We claim that
is a Cauchy sequence. Otherwise, there
such that for
, there exist
with
such that the parity of
and
is different and
![]()
For k, let
denotes the least integer exceeding
and satisfying the above, then
(2.9)
hence
(2.10)
Let
, then we obtain
. But
![]()
hence we obtain
(2.11)
If
is even and
is odd, then by Lemma 2.1, (2.11) and (2.1),
![]()
which is a contradiction. Similarly, we obtain the same contradiction for the case that
is odd and
is even. Hence
is a Cauchy sequence, therefore
for some
by the completeness of X.
If
, then
, hence by (2.1) and Lemma 2.1,
![]()
which is a contradiction, hence
. Similarly, we obtain
. Therefore
is a common fixed point of f and g.
If
is another common fixed point of f and g, then
, hence by (2.1),
![]()
which is a contradiction, hence
, i.e.,
is the unique common fixed point of f and g.
From Theorem 2.1, we obtain the next more general common fixed point theorem.
Theorem 2.2. Let
be a complete metric space,
and
two mappings. If for each
,
(2.12)
where
,
are three functions satisfying (2.2) and (2.3). Then f and g have a unique common fixed point u, and the sequence
defined by
for any
converges to u.
Proof.
Let
and
, then F and G satisfy all of the conditions of Theorem 2.1, hence there exists an unique element
such that
. If
, then
, hence by (2.12),
![]()
which is a contradiction, hence
. Similarly,
. So u is a common fixed point of f and g. The uniqueness is obvious.
From now on, we will discuss the second common fixed point problem for two mappings with implicit contraction of integral type.
Let
if and only if
is a continuous and non-decreasing function about the 4th and 5th variables and satisfying the following conditions:
(i) There exists
such that
implies
;
(ii) There exists
such that
implies
;
(iii)
for all
.
Example 2.1. Define
as follows
![]()
where
for all
and
. Then
.
The function
is called to be sub-additive if and only if for all
,
![]()
Example 2.2. Let
for each
. Then obviously
and for all
,
![]()
Hence
is a sub-additive function.
Theorem 2.3. Let
be a complete metric space,
two mappings. If for each
,
(2.13)
where
is sub-additive and
. Then f and g have a unique common fixed point.
Proof.
We take any element
and consider the sequence
constructed by
and
for all
. Let
for all
.
Since
![]()
So by (i),
(2.14)
Similarly,
![]()
So by (ii),
(2.15)
Combining (2.14) and (2.15), we have
(2.16)
Obviously,
for all
. If there exits
such that
, then
. If
, then by (2.14) and (2.16)
![]()
which is a contradiction. Similarly, if
, then by (2.15) and (2.16)
![]()
which is also a contradiction. Hence we have
![]()
Therefore there exists
. If
, then
![]()
which is a contradiction. Therefore,
, i.e., ![]()
We claim that
is Cauchy. Otherwise, just as the line of proof of Theorem 2.1, there exists
such that for
there exist
with
such that the parity of
and
is different and (2.11) holds.
If
is even and
is odd, then by Lemma 2.1, (2.11), (2.13) and (iii),
![]()
This is a contradiction. Similarly, we obtain the same contradiction for the case that
is odd and
is even. Therefore,
is a Cauchy sequence. Let
.
If
, then
, hence by Lemma 2.1 and (2.13) and (iii),
![]()
which is a contradiction, hence
. Similarly, we obtain
. Therefore,
is a common fixed point of f and g.
If
is another common fixed point of f and g, then
, hence by (2.13) and (iii),
![]()
This is a contradiction. Hence
is the unique common fixed point of f and g.
Using Theorem 2.3 and the Example 2.2, we have the next result.
Theorem 2.4. Let
be a complete metric space,
two mappings. If
![]()
where
. Then f and g have a unique common fixed point u.
Combining Theorem 2.4 and Example 2.1, we obtain the following result.
Theorem 2.5. Let
be a complete metric space,
two mappings. If for each
,
![]()
where
for all
and
. Then f and g have a unique common.
Acknowledgements
The research is partially supported by the National Natural Science of Foundation of China (No. 11361064).
NOTES
*Corresponding author.