Some remarks on b v ( s ) -metric spaces and ﬁxed point results with an application ∗

. We compare the newly deﬁned b v ( s ) -metric spaces with several other abstract spaces like metric spaces, b -metric spaces and show that some well-known results, which hold in the latter class of spaces, may not hold in b v ( s ) -metric spaces. Besides, we introduce the notions of sequential compactness and bounded compactness in the framework of b v ( s ) -metric spaces. Using these notions, we prove some ﬁxed point results involving Nemytzki–Edelstein type mappings in this setting, from which several comparable ﬁxed point results can be deduced. In addition to these, we ﬁnd some existence and uniqueness criteria for the solution to a certain type of mixed Fredholm– Volterra integral equations.


Introduction
There are many interesting extensions of the notion of metric spaces available in the literature where several classical fixed point results have been studied. One of such extensions is the concept of b-metric spaces, which was introduced by Bakhtin [3] in 1989, and later on, in 1993, it had been further investigated by Czerwik [5]. Afterward, in the year of 2000, Branciari [4] coined the notion of rectangular metric spaces or generalized metric spaces by modifying the triangle inequality of the usual metric spaces. As a generalization of b-metric spaces and rectangular metric spaces, George et al. [11] introduced the concept of rectangular b-metric spaces. Following this direction, in [4], Branciari introduced the concept of v-generalized metric spaces. Taking into account all these concepts, many mathematicians have elaborated several fixed point results in these settings, some of which meliorated and improved the original fixed point theories in usual metric spaces and some others give new results in the literature. In the literature, there is a huge amount of relevant texts available for intent readers (see [2,7,11,13,18] and the references therein).
In an attempt to extend all kinds of above mentioned generalizations of metric spaces, Mitrović and Radenović [16] introduced the concept of b v (s)-metric spaces. Before going further, we recall the definition of such abstract spaces. [16,Def. 1.8].) Let X be a nonempty set, v ∈ N and s 1 a real number. A function d : X ×X → R is said to be a b v (s)-metric if for all x, y ∈ X and for all distinct points u 1 , u 2 , . . . , u v ∈ X, each of them different from x and y, the following conditions hold: If d is a b v (s)-metric on X, then the pair (X, d) is said to be a b v (s)-metric space. It is easy to observe that the class of b v (s)-metric spaces is larger than that of all other metric spaces. Thus, it is a natural question to ask whether all properties of the above mentioned metric spaces remain invariant in case of b v (s)-metric spaces or not. So in this paper, one of our main motivations is to compare some properties of sequences, the metric function d in b v (s)-metric spaces with that of usual metric spaces, b-metric spaces.
To be specific, we show that in a b v (s)-metric space (X, d), a convergent sequence may not be Cauchy, the metric function d need not be continuous. Moreover, we show that a sufficient condition for a sequence to be Cauchy in the standard metric space, as well as in a b-metric space, may not work in this structure.
On the other hand, many mathematicians have achieved some interesting fixed point results in the setting of b v (s)-metric spaces, such as the authors of [16] proved fixed point results associated to Banach and Reich contractions, the authors of [1] obtained a common fixed point theorem due to Jungck, the authors of [15] achieved fixed point result due to Sehgal-Guseman. It may be observed that the existing results in this structure are concerned with a different type of contraction mappings only. But at the same time, there are some important results regarding contractive type mappings in the standard metric spaces; see [8][9][10]12]. So it is a natural question to ask whether the fixed point results related to different contractive conditions can be proved in the setting of b v (s)-metric spaces or not. In fact, we are able to find some fixed point results concerning contractive type maps in this setting. We first recall the definition of a contractive mapping in metric spaces and then highlight some salient points regarding the existence of fixed points. A self-mapping T on a metric space (X, d) is said to be a contractive mapping if d(T x, T y) < d(x, y) holds for all x, y ∈ X with x = y. It may be noted that completeness of the underlying space X does not give the guaranty of the existence of a fixed point of this map, but if X is compact, it is guaranteed; see [8]. In this article, we also try to find a (mild) additional criteria on the underlying b v (s)-metric space X, which confirms the existence of a fixed point for a contractive mapping. To proceed in this direction, we introduce the notions of sequential compactness and bounded compactness of b v (s)-metric spaces and establish correlations between them. In such spaces, we establish some fixed point theorems related to contractive mapping, which improve and generalize some standard fixed point results due to Nemytzki [17], Edelstein [8] and Suzuki [19].
Another importance of the (metric) fixed point theory is that it is an invaluable tool for finding existence and/or uniqueness criteria of solution(s) of several types of differential equations, integral equations, fractional integral equations, matrix equations, etc. In most of the cases, the fixed point results of usual metric spaces are applied to find the criteria as mentioned above. But the fixed point results of b v (s)-metric spaces are yet to be employed to investigate for such tools. So at the end of this paper, we utilize one of our obtained results to find some criteria for the existence and uniqueness of solution of a special type of integral equation.

Preliminaries
It is well known that Nemytzki's result in [17], regarding contractive mappings, was first in the literature. Later on, Edelstein proved in [8] that a contractive self-mapping on a compact metric space has a unique fixed point. One can easily verify that, in this result, the compactness of X cannot be replaced by completeness. Therefore, some additional conditions have to be imposed on X or on T together with the completeness of X to ensure the existence of fixed point of T . Many researchers tried to find such additional conditions. One of such conditions is given by the next theorem, which was proved bý Cirić.
Theorem 1. (See [6].) Let (X, d) be a complete metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y, and let for any > 0, there exists δ > 0 such that for any x, y ∈ X. Then T has a unique fixed point z, and for any x ∈ X, the sequence of iterates (T n x) converges to z.
Suzuki introduced another additional weaker assumption with the completeness of (X, d) to assure fixed point of T in the following theorem. [19,Thm. 5].) Let (X, d) be a complete metric space, T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y, and let the following hold: For any x ∈ X and for any > 0, there exists δ > 0 such that for any i, j ∈ N ∪ {0}. Then T has a unique fixed point z, and for any x ∈ X, the sequence of iterates (T n x) converges to z.
On the other hand, Mitrović and Radenović introduced the notions of Cauchy sequences and completeness in b v (s)-metric space. [16,Def. 1.9].) Let (X, d) be a b v (s)-metric space, (x n ) be a sequence in X and x ∈ X.

Definition 2. (See
(i) The sequence (x n ) is said to be a Cauchy sequence if for any > 0, there exists N ∈ N such that d(x n , x n+p ) < for all n > N and for all p ∈ N. (ii) The sequence (x n ) is said to be convergent to x if for any > 0, there exists N ∈ N such that d(x n , x) < for all n > N , and this fact is represented by In this manuscript, we now introduce the concepts of sequential compactness and bounded compactness of a b v (s)-metric space.  4. Let (X, d) be a b v (s)-metric space. Then X is said to be sequentially compact if every sequence (x n ) in X has a subsequence, which converges to some point of X. Again, a subset A of X is said to be sequentially compact if every sequence (x n ) in A has a subsequence, which converges to some point of A.
Definition 5. Let (X, d) be a b v (s)-metric space. Then X is said to be boundedly compact if every bounded sequence (x n ) in X has a subsequence, which converges to some point of X. Again, a subset A of X is said to be boundedly compact if every bounded sequence (x n ) in A has a subsequence, which converges to some point of A.
Clearly, it follows from the definition that every sequentially compact b v (s)-metric space is boundedly compact but not conversely. To investigate this, we frame the following example. Example 1. Consider the set X = X 1 ∪ X 2 , where X 1 = {1/n: n ∈ N and n 2} and n , y = 1 m and |n − m| = 1, 3; if x, y ∈ X 1 , x = y, x = 1 n , y = 1 m and |n − m| = 1, 3; n if x ∈ X 1 , y ∈ X 2 and x = 1 n or y ∈ X 1 , x ∈ X 2 and y = 1 n ; 5 if x, y ∈ X 2 and x = y; 0 if x, y ∈ X and x = y.
Then it is an easy task to verify that (X, d) is a b 3 (2)-metric space.
http://www.journals.vu.lt/nonlinear-analysis Note that the sequence (1/(n+2)) has no subsequence, which converges to some point of X. So, (X, d) is not sequentially compact. But one can easily check that a sequence (x n ) in X is bounded if and only if the range of the sequence (x n ) is finite. Thus every bounded sequence in X has a subsequence, which converges to some point of X, i.e, (X, d) is boundedly compact.

Comparison of b v (s)-metric space with other spaces
In this section, we point out some properties of usual metric spaces and b-metric spaces, which are not true in case of b v (s)-metric spaces. The first one is given in the following remark.
Remark 1. We know that a convergent sequence in usual metric spaces is always a Cauchy sequence. But this fact is not true in case of b v (s)-metric spaces. To show this, we illustrate the following example in the line of [7, Ex. 1].
Example 2. Let us take X = {0, 1/n: n ∈ N}. We define a function d : Then it is easy to check that (X, d) is a b 2 (1)-metric space. Now we consider the sequence (x n ) in X where x n = 1/n for all n ∈ N. Then we have Therefore, (x n ) is convergent in X but not a Cauchy sequence.
In the next part of this section, we state a lemma in b v (s)-metric space, which also occurs in usual metric spaces.
Proof. The proof of this lemma is similar to that of usual metric spaces.

Remark 2.
If (X, d) is a metric space, then we know that the function d is continuous on X × X with respect to the product metric on X × X, but this is not true in case of b v (s)-metric spaces. This fact can be substantiated from [7, Ex. 1] in case of rectangular metric spaces of Branciari, i.e., in b 2 (1)-metric spaces.
Remark 3. We know that in a metric space (X, d) for a sequence (x n ), if there exists a real number µ such that 0 µ < 1 and d(x n , x n+1 ) µ · d(x n−1 , x n ) holds for all n ∈ N, then (x n ) is a Cauchy sequence in X. This result also holds in b-metric spaces, i.e., in b 1 (s)-metric spaces, which is proved in [14] by Miculescu and Mihail. Now in the next example, we show that this result may not hold in arbitrary b v (s)-metric spaces.
Example 3. We consider the set X = N. Define a function d : if n = m and n, m both are even or both are odd.
Note that for any natural number n, holds for µ = 1/2. But (x n ) is not a Cauchy sequence, since d(x n , x n+2 ) = 1 for all natural numbers n.

Fixed point results
To begin with, we prove a fixed point theorem related to contractive mappings in the structure of sequentially compact b v (s)-metric spaces.
for all x, y ∈ X with x = y. Assume that the function d is continuous on X × X. Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point.
Proof. At first, we fixed an element x 0 in X and then we consider a sequence (x n ), which is defined by x n = T n x 0 for every natural number n. Now we show that the sequence of real numbers (s n ), defined by s n = d(x n , x n+1 ), converges to 0. If x n = x n+1 for some n ∈ N, then clearly (s n ) converges to 0. So we now assume that x n = x n+1 for all n ∈ N. Then we have This proves that (s n ) is a decreasing sequence of nonnegative real numbers and hence convergent to some a 0. Again, by the sequential compactness of (X, d), there exists a convergent subsequence of (x n ), say, (x n k ). Further, let this subsequence converges to z ∈ X. From the contractivity condition of T , it is continuous on X, and hence the subsequences (x n k +1 ) and (x n k +2 ) converge to T z and T 2 z respectively. Then we have Again, we have Now if a > 0, then z = T z and so we have d(T z, T 2 z) < d(z, T z), i.e., a < a, which is a contradiction. So we must have a = 0, i.e., (s n ) converges to 0. Further, since a = 0, we have T z = z. So z is a fixed point of T . Now we examine the uniqueness of this fixed point. To do this, let z 1 be another fixed point of T . Then we have which is a contradiction. Hence, z is the only fixed point of T .
Finally, we prove that (x n ) converges to z. If x n = z for some n ∈ N, then clearly (x n ) converges to z. Let us now, assume that x n = z for all n ∈ N. Since, (x n ) contains a subsequence (x n k ), which converges to z, z is a cluster point of the sequence (x n ). Let z 1 be another cluster point of (x n ), then (x n ) contains a subsequence, which converges to z 1 . Then by similar arguments as above we can prove that z 1 is a fixed point of T and this will again lead to a contradiction. Henceforth, z is the only cluster point of (x n ). Next, we consider the sequence (t n ) of real numbers given by t n = d(x n , z) for all n ∈ N. Therefore, This shows that the sequence (t n ) contains the subsequence (t n k ), which converges to 0 and so 0 is a cluster point of {t n }.
Consequently, (t n ) is a decreasing sequence of nonnegative real numbers and hence convergent. But 0 is a cluster point of the convergent sequence (t n ), so (t n ) must converge to 0. Therefore, t n → 0 as n → ∞, i.e., d(x n , z) → 0 as n → ∞. Thus, (x n ) converges to z, i.e., (T n x 0 ) converges to z. Since x 0 ∈ X was arbitrary, it follows that (T n x) converges to the fixed point z for any x ∈ X.
From Theorem 3, we can deduce the following corollary by taking s = 1 and v = 1.

Corollary 1.
Let (X, d) be a compact metric space and T : X → X a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Then T has a unique fixed point z, and for any x ∈ X, the sequence (T n x) converges to z.
In the above theorem sequential compactness condition cannot be replaced by bounded compactness of the space, which follows from the following example. Thus all the conditions of Theorem 3 are satisfied but T is fixed point free.
Therefore, we are in search of an additional condition either on X or on T with the bounded compactness of X to get a unique fixed point of T. Here in the next theorem we deal with one such additional condition.
Theorem 4. Let (X, d) be a boundedly compact b v (s)-metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Assume that the function d is continuous on X × X. Further, assume that, for any x ∈ X and for any k ∈ N with k v, there exists a real number M > 0 (depending on x) such that d(x, T k−v x) M . Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point.
Proof. We choose an element x 0 from X and then we consider the sequence of iterates (x n ) where x n = T x n−1 for all n 1. If x n = x n+1 for some natural number n, then it is easy to notice that T has a fixed point, the fixed point is unique and the sequence (x n ) converges to that fixed point. So now we assume that x n = x n+1 for all natural numbers n. Then we claim that all terms of (x n ) are distinct. To prove our claim, we presume that x n = x m for some natural numbers n, m with m > n. Then T x n = T x m i.e. x n+1 = x m+1 . Therefore, which is a contradiction. So our claim is correct. Now for any n ∈ N, we have Let k, n ∈ N be arbitrary but fixed. First suppose that k v. Then by hypothesis we get a real number Let M 2 = max k<v {d(x v , x k )}. Then, clearly, M 2 is finite. Thus by using equation (1), we get d(x n , x n+k ) < s vd(x 0 , x 1 ) + M 2 = M 3 , say.
Therefore, from our above discussions, we see that for all n, k ∈ N, which shows that the sequence (x n ) is bounded. So by bounded compactness of X, there exists a convergent subsequence of (x n ), let it be (x n k ). Let lim k→∞ x n k = z. Then proceeding as Theorem 3, we can show that z is the unique fixed point of T and the sequence of iterates (x n ) converges to z.
As a consequence of Theorem 4, we obtain the following corollary.
Corollary 2. Let (X, d) be a boundedly compact b v (s)-metric space such that d is continuous on X × X, and let T : X → X be a bounded mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Then T has a unique fixed point, and for any x ∈ X, the sequence (T n x) converges to that fixed point.
Taking s = 1 and v = 1 in Theorem 4, we get the following corollary.
Corollary 3. Let (X, d) be a boundedly compact metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X. Further, assume that, for any x ∈ X and for any k ∈ N with k 1 there exists a real number M > 0 (depending on x) such that d(x, T k−1 x) M . Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point.
Now we cite the following example, which supports the above theorem.
Example 5. Consider the set X = {1/n: n ∈ N, n 2}. We define a mapping d : Then it is an easy task to verify that (X, d) is a b 3 (3)-metric space. Also, it is simply noticeable that (X, d) is boundedly compact but not sequentially compact. Now we define a mapping T : X → X by Then, clearly, d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Further, for any x ∈ X and for any k ∈ N with k 3, there exists a real number M > 0 (depending on x) such that d(x, T k−3 x) < M (here M = 1/x + 4). Then by Theorem 4, T has a unique fixed point. Note that x = 1/4 is the unique fixed point of T .
From the definitions of bounded compactness and completeness of b v (s) metric spaces we see that every boundedly compact b v (s)-metric space is complete. Thus Example 4 also shows that if (X, d) is a complete b v (s)-complete metric space and T : X → X is a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ with x = y, then T may not have a fixed point. So we again need some additional assumption with the completeness of X to warrant the fixed point of T .
Next, we consider the following example.
Example 6. Let (x i n ) be a sequence whose ith term is 1 and all other terms are 0. Consider the set X = {(x i n ): i ∈ N}. Now we define a function d : Then it is trivial to check that (X, d) is a b 2 (10)-metric space. Further, (X, d) is complete but not boundedly compact. We now define T : X → X by Let (x i n ), (x j n ) ∈ X be arbitrary with i = j. Then the following three cases may arise.
n , x j+11 n = 10 i + j + 22 Case 2. Let i, j > 10. Then n , x j+11 n = 10 i + j + 22 Case 3. Let exactly any one of i and j is greater than 10. Then Thus, d(T x, T y) < d(x, y) for all x, y ∈ X with x = y.
Note that for any x ∈ X and for any k ∈ N with k v (here v = 2), there exists a real number M > 0 (here we may take M = 200) such that d(x, T k−v x) M but still T has no fixed point.
Thus we see that the additional condition, which is considered in Theorem 4 together with the completeness of X, does not deliver fixed point of T . It means that we have to find out some different additional condition with the completeness of X so as to ensure the existence of fixed point of T . In the following theorem, we consider such an additional condition in case of b v (1)-metric spaces, which was firstly given by Suzuki [19].
Theorem 5. Let (X, d) be a complete b v (1)-metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) holds for all x, y ∈ X with x = y. Further, assume that for any x ∈ X and for any > 0, there exists δ > 0 such that for any i, j ∈ N ∪ {0}. Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point.
Proof. First, we choose an arbitrary but fixed element x 0 in X, and then we consider a sequence (x n ), which is defined by x n = T n x 0 for every natural number n. If x n = x n+1 for some natural number n, then x n is the unique fixed point of T . So, now we assume that x n = x n+1 for every natural numbers n. Under this assumption, following Theorem 4, we can show that all terms of (x n ) are distinct.
Next, we consider the sequence of real numbers (s n ), where s n = d(x n , x n+1 ) for all n ∈ N. Then s n+1 = d(x n+1 , x n+2 ) < d(x n , x n+1 ) = s n for all n ∈ N. Therefore, (s n ) is a decreasing sequence of nonnegative real numbers and hence convergent to some a 0. If a > 0, then by given condition there exists δ > 0 such that for all n ∈ N. Again, by definition of a, for the above δ > 0, there exists n ∈ N such that d(x n , x n+1 ) < a + δ. Therefore, d(x n+1 , x n+2 ) a, and this leads to a contradiction. So we must have a = 0, i.e., lim In a similar manner, we can show that Now let > 0 be arbitrary. Then by given condition we get a δ > 0 such that On the other hand, by equations (2) and (3), for the above δ > 0, there exists a natural number N such that for all n N . Now for each n N , we show by mathematical induction on j that d(x n+2v+1 , x n+2v+j ) 2 holds for all j ∈ N. The result is obviously true for j = 1. Assume that the result is true for some j ∈ N. So, d(x n+2v+1 , x n+2v+j ) /2, which implies Then Using equations (4) and (5) in equation (6), we get for all n N . This implies that for all n N . Thus by induction it follows that for all n N and for all j ∈ N, which proves that (x n ) is a Cauchy sequence. So, by the completeness of X, (x n ) converges to some z ∈ X. Again, since T is contractive, the sequence (x n ) converges to T z also. So z is a fixed point of T . The uniqueness of the fixed point can be analogously proved from Theorem 3.
Since we choose x 0 ∈ X arbitrarily, it follows that (T n x) converges to the unique fixed point z for all x ∈ X.
It is very interesting to verify whether the additional criteria, used in the above theorem, will work for s > 1 or not. If not, then it is also necessary to find an additional assumption, which will ensure the existence of a fixed point in the complete b v (s)-metric structure. In this respect, we pose the following open problem.
Open question. Let (X, d) be a complete b v (s)-metric space, and let T be a self-mapping on X such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. If s > 1, then find out a weaker additional assumption on T , which will ensure that T has a fixed point.
From Theorem 5 we can derive several important corollaries. We present a number of selected ones, which extend several well-known results in the literature.  [19,Thm. 5].) Let (X, d) be a complete metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Further, assume that for any x ∈ X and for any > 0, there exists δ > 0 such that for any i, j ∈ N ∪ {0}. Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point.

Corollary 5.
Let (X, d) be a complete rectangular metric space, and let T : X → X be a mapping such that d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Further, assume that for any x ∈ X and for any > 0, there exists δ > 0 such that for any i, j ∈ N ∪ {0}. Then T possesses a unique fixed point, and for any x ∈ X, the Picard's iterative sequence (T n x) converges to that fixed point. Now we present an example in order to endorse Theorem 5.
if any one of x and y lies in X and the other in X \ X 1 ; 0 elsewhere.
Then it is easy to check that (X, d) is a b 2 (1)-metric space. Next, we define a mapping T : X → X by Let x, y ∈ X be arbitrary with x = y. Then the following three cases may arise. Case 1. Let x, y ∈ X 1 . Then Case 2. Let x, y ∈ X \ X 1 . Then Case 3. Let x ∈ X 1 and y ∈ X \ X 1 . Then Thus d(T x, T y) < d(x, y) for all x, y ∈ X with x = y. Next, assume that x ∈ X and > 0 be arbitrary. Here we choose δ = . Therefore, if x ∈ X \ X 1 , then, clearly, x, T j+1 x http://www.journals.vu.lt/nonlinear-analysis for any i, j ∈ N ∪ {0}. Now we assume that x ∈ X 1 and d(T i x, T j x) < + δ for some i, j ∈ N ∪ {0}. Therefore, Thus we see that all conditions of Theorem 5 hold good. So by this theorem, T has a unique fixed point in X. Indeed, x : [0, 1] → R, defined by x(t) = 0 for all t ∈ [0, 1], is the fixed point of T in X.
Also, it can be easily verified that T y(t) 0 for all t ∈ [0, 1]. Therefore, T y ∈ A, and hence T is a self-map on A.
The above relation is true for all t ∈ [0, 1], and hence we have d(T y 1 , T y 2 ) (α 1 β 1 + α 2 β 2 )d(y 1 , y 2 ) =⇒ d(T y 1 , T y 2 ) < d(y 1 , y 2 ). http://www.journals.vu.lt/nonlinear-analysis Thus we see that all the conditions of Theorem 3 hold good here, and so we can assure that T has a unique fixed point in A, say, y. By the formulation T and A we see that y is the unique solution of equation (7), and the solution y is nonnegative and satisfies the condition | 1 0 y(t) dt| 1.
It is well known that the mixed Fredholm-Volterra integral equations arise from the mathematical model of the spatio-temporal developments of an epidemic model and also from several parabolic boundary value problems. It may be noted that these are all associated to some physical problems. Here we demonstrate a specific example, which validates the aforementioned result.