We start this section with our main result.
Theorem 8 Let be a complete metric space. Suppose that is a pair of nonempty closed subsets of X and is nonempty. Suppose also that the pair has the P-property. If a non-self-mapping is a generalized Geraghty-contraction satisfying , then there exists a unique best proximity point, that is, there exists in A such that .
Let us fix an element
, we can find
. Further, as
, there is an element
. Recursively, we obtain a sequence
with the following property:
Due to the fact that the pair
has the P
-property, we derive that
From (2.1), we get
On the other hand, by (2.1) and (2.2) we obtain that
Consequently, we have
If there exists
, then the proof is completed. In fact, due to (2.2), we have
which yields that
. Hence, equation (2
.1) implies that
For the rest of the proof, we suppose that
. Owing to the fact T
is a generalized Geraghty-contraction, we derive that
Then, by (2.3) and (2.6), we deduce that
. Then we get that
a contradiction. As a result, we conclude that
By (2.6), we get
is a nonincreasing sequence and bounded below. Thus, there exists
. We shall show that
. Suppose, on the contrary,
. Then, by (2.8), we have
. In what follows,
On the other hand, since
, we conclude
, that is,
holds for all
satisfies the P
-property, then, for all
, we can write,
. We also have
. It follows that
Taking (2.9) into consideration, we find
We shall show that
is a Cauchy sequence. Suppose, on the contrary, that we have
Due to the triangular inequality, we have
Regarding (1.6) and (2.12), we have
Taking (2.10), (2.13) and (2.9) into account, we derive that
Owing to (2.11), we get
. By the property of β
, we have
. Consequently, we have
, a contradiction. Hence, we conclude that the sequence
is Cauchy. Since A
is a closed subset of the complete metric space
, and we can find
. We assert that
. Suppose, on the contrary, that
. First, we obtain the following inequalities:
in the inequalities above, we conclude that
On the other hand, we obtain
Taking limit as
in the inequality above, we find
So, we deduce that
. As a consequence, we derive
Combining (1.6) and (2.14), we find
together with (2.15), we get
. Hence, we have
As a result, we deduce that , a contradiction. So, and hence , is a best proximity point of T. Hence, we conclude that T has a best proximity point.
We claim that the best proximity point of T is unique.
Suppose, on the contrary, that
are two distinct best proximity points of T
. Thus, we have
By using the P
-property, we find
Due to the fact that T
is a generalized Geraghty-contraction, we have
a contradiction. This completes the proof. □
Remark 9 Let be a metric space and A be any nonempty subset of X. It is evident that a pair satisfies the P-property.
Corollary 10 Suppose that is a complete metric space and A is a nonempty closed subset of X. If a self-mapping is a generalized Geraghty-contraction, then it has a unique fixed point.
Proof Taking Remark 9 into consideration, we conclude the desired result by applying Theorem 8 with . □
In order to illustrate our main result, we present the following example.
with the metric
and consider the closed subsets
be the mapping defined by
Since , the pair has the P-property.
Notice that and and .
, we have
where is defined as .
Notice that β is nondecreasing since .
and it is easily seen that the function belongs to F.
Therefore, since the assumptions of Theorem 8 are satisfied, by Theorem 8 there exists a unique
The point is .