In order to illustrate our results, we present some examples.

**Example 4.1** Consider
with the usual metric.

Let

*A* and

*B* be the subsets of

*X* defined by

Obviously,
and *A*, *B* are nonempty closed subsets of *X*.

Moreover, it is easily seen that
and
.

Let

be the mapping defined as

In the sequel, we check that *T* is a Geraghty-contraction.

In fact, for

with

, we have

Suppose that
(the same reasoning works for
).

Then, since

is strictly increasing in

, we have

This proves (5).

Taking into account (4) and (5), we have

where
for
, and
for
and
.

Obviously, when
, the inequality (6) is satisfied.

It is easily seen that
by using elemental calculus.

Therefore, *T* is a Geraghty-contraction.

Notice that the pair
satisfies the *P*-property.

then

and

and consequently,

By Theorem 3.1, *T* has a unique best proximity point.

Obviously, this point is
.

The condition *A* and *B* are nonempty closed subsets of the metric space
is not a necessary condition for the existence of a unique best proximity point for a Geraghty-contraction
as it is proved with the following example.

**Example 4.2** Consider

with the usual metric and the subsets of

*X* given by

Obviously,
and *B* is not a closed subset of *X*.

Note that
and
.

We consider the mapping

defined as

Now, we check that *T* is a Geraghty-contraction.

In fact, for

with

, we have

In what follows, we need to prove that

In fact, suppose that
(the same argument works for
).

Put
and
(notice that
since the function
for
is strictly increasing).

and since

, we have that

, and consequently, from the last inequality it follows that

Applying

*ϕ* (notice that

) to the last inequality and taking into account the increasing character of

*ϕ*, we have

and this proves (8).

where
for
and
. Obviously, the inequality (9) is satisfied for
with
.

Now, we prove that
.

In fact, if
, then the sequence
is a bounded sequence since in the contrary case,
and thus
. Suppose that
. This means that there exists
such that, for each
, there exists
with
. The bounded character of
gives us the existence of a subsequence
of
with
convergent. Suppose that
. From
, we obtain
and, as the unique solution of
is
, we obtain
.

Thus,
and this contradicts the fact that
for any
.

Therefore,
and this proves that
.

A similar argument to the one used in Example 4.1 proves that the pair
has the *P*-property.

On the other hand, the point

is a best proximity point for

*T* since

Moreover,
is the unique best proximity point for *T*.

Indeed, if

is a best proximity point for

*T*, then

Taking into account that the unique solution of this equation is
, we have proved that *T* has a unique best proximity point which is
.

Notice that in this case *B* is not closed.

Since for any nonempty subset *A* of *X*, the pair
satisfies the *P*-property, we have the following corollary.

**Corollary 4.1**
*Let*
*be a complete metric space and*
*A*
*be a nonempty closed subset of X*. *Let*
*be a Geraghty*-*contraction*. *Then*
*T*
*has a unique fixed point*.

*Proof* Using Theorem 3.1 when
, the desired result follows. □

Notice that when
, Corollary 4.1 is Theorem 1.1 due to Gerahty [1].