We start this section by defining some sets of auxiliary functions. Let ℱ denote all functions

such that

if and only if

. Let Ψ and Φ be the subsets of ℱ such that

**Lemma 2.1**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a sequence in*
*X*
*such that*
*is nonincreasing*,

*Proof*

Due to Lemma 1.1, we have

Letting

regarding the assumption of the lemma, we derive that

If

is not

*G*-Cauchy, then, due to Proposition 1.2, there exist

and corresponding subsequences

and

of ℕ satisfying

for which

where

is chosen as the smallest integer satisfying (13), that is,

By (13), (14) and the rectangle inequality (G5), we easily derive that

Letting

in (15) and using (10), we get

Passing to the limit when

and using (10) and (16), we obtain that

Passing to the limit when

and using (10) and (16), we obtain that

Passing to the limit when

and using (10) and (16), we obtain that

By regarding the assumptions (G3) and (G5) together with the expression (13), we derive the following:

Letting

in the inequality above and using (12) and (16), we conclude that

□

**Theorem 2.1**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*Suppose that there exist functions*
*and*
*such that the map*
*T*
*satisfies the inequality*
*Then*
*T*
*has a unique fixed point in*
.

*Proof* First we show the existence of a fixed point of the map

*T*. For this purpose, we take an arbitrary

and define a sequence

in the following way:

We have

,

,

, … since

*T* is a cyclic mapping. If

for some

, then, clearly, the fixed point of the map

*T* is

. From now on, assume that

for all

. Consider the inequality (29) by letting

and

,

If

, then the expression (32) implies that

So, the inequality (34) yields

. Thus, we conclude that

This contradicts the assumption

for all

. So, we derive that

Hence the inequality (32) turns into

Thus,

is a nonnegative, nonincreasing sequence that converges to

. We will show that

. Suppose, on the contrary, that

. Taking

in (36), we derive that

By the continuity of

*ψ* and the lower semi-continuity of

*ϕ*, we get

Then it follows that

. Therefore, we get

, that is,

Lemma 1.1 with

and

implies that

Next, we will show that

is a

*G*-Cauchy sequence in

. Suppose, on the contrary, that

is not

*G*-Cauchy. Then, due to Proposition 1.2, there exist

and corresponding subsequences

and

of ℕ satisfying

for which

where

is chosen as the smallest integer satisfying (42), that is,

By (42), (43) and the rectangle inequality (G5), we easily derive that

Letting

in (44) and using (39), we get

Notice that for every

there exists

satisfying

such that

Thus, for large enough values of

*k*, we have

, and

and

lie in the adjacent sets

and

respectively for some

. When we substitute

and

in the expression (29), we get that

By using Lemma 2.1, we obtain that

So, we have

. We deduce that

. This contradicts the assumption that

is not

*G*-Cauchy. As a result, the sequence

is

*G*-Cauchy. Since

is

*G*-complete, it is

*G*-convergent to a limit, say

. It easy to see that

. Since

, then the subsequence

, the subsequence

and, continuing in this way, the subsequence

. All the m subsequences are

*G*-convergent in the

*G*-closed sets

and hence they all converge to the same limit

. To show that the limit

*w* is the fixed point of

*T*, that is,

, we employ (29) with

,

. This leads to

Passing to limsup as

, we get

Thus,
and hence
, that is,
.

Finally, we prove that the fixed point is unique. Assume that

is another fixed point of

*T* such that

. Then, since both

*v* and

*w* belong to

, we set

and

in (29), which yields

On the other hand, by setting

and

in (29), we obtain that

If

, then

. Indeed, by definition, we get that

. Hence

. If

, then by (56)

and by (55),

and, clearly,

. So, we conclude that

. Otherwise,

. Then by (58),

and by (57),

and, clearly,
. So, we conclude that
. Hence the fixed point of *T* is unique. □

**Remark 2.1** We notice that some fixed point result in the context of *G*-metric can be obtained by usual (well-known) fixed point theorems (see, *e.g.*, [50, 51]). In fact, this is not a surprising result due to strong relationship between the usual metric and *G*-metric space (see, *e.g.*, [2, 3, 5]). Note that a *G*-metric space tells about the distance of three points instead of two points, which makes it original. We also emphasize that the techniques used in [50, 51] are not applicable to our main theorem.

To illustrate Theorem 2.1, we give the following example.

**Example 2.1** Let

and let

be given as

. Let

and

. Define the function

as

Clearly, the function *G* is a *G*-metric on *X*. Define also
as
and
as
. Obviously, the map *T* has a unique fixed point
.

It can be easily shown that the map

*T* satisfies the condition (29). Indeed,

On the other hand, we have the following inequality:

By elementary calculation, we conclude from (65) and (64) that

Combining the expressions (62) and (65), we obtain that

Hence, all conditions of Theorem 2.1 are satisfied. Notice that 0 is the unique fixed point of *T*.

For particular choices of the functions *ϕ*, *ψ*, we obtain the following corollaries.

**Corollary 2.1**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*Suppose that there exists a constant*
*such that the map*
*T*
*satisfies*
*Then*
*T*
*has a unique fixed point in*
.

*Proof* The proof is obvious by choosing the functions *ϕ*, *ψ* in Theorem 2.1 as
and
. □

**Corollary 2.2**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*Suppose that there exist constants*
*a*,

*b*,

*c*,

*d*
*and*
*e*
*with*
*and there exists a function*
*such that the map*
*T*
*satisfies the inequality*
*for all*
*and*
,
. *Then*
*T*
*has a unique fixed point in*
.

*Proof*

By Corollary 2.1, the map *T* has a unique fixed point. □

**Corollary 2.3**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*Suppose that there exist functions*
*and*
*such that the map*
*T*
*satisfies the inequality*
*Then*
*T*
*has a unique fixed point in*
.

*Proof* The expression (75) coincides with the expression (30). Following the lines in the proof of Theorem 2.1, by letting
and
, we get the desired result. □

Cyclic maps satisfying integral type contractive conditions are amongst common applications of fixed point theorems. In this context, we consider the following applications.

**Corollary 2.4**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*Suppose also that there exist functions*
*and*
*such that the map*
*T*
*satisfies*
*for all*
*and*
,
. *Then*
*T*
*has a unique fixed point in*
.

**Corollary 2.5**
*Let*
*be a*
*G*-

*complete*
*G*-

*metric space and*
*be a family of nonempty*
*G*-

*closed subsets of*
*X*
*with*
.

*Let*
*be a map satisfying*
*where*
*and*
*for all*
*and*
,
. *Then*
*T*
*has a unique fixed point in*
.

*Proof* The proof is obvious by choosing the function *ϕ*, *ψ* in Corollary 2.4 as
and
. □