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(iii) Sp(x, x, x) ≤ Sp(x, y, z),

(iv) Sp(x, x, y)= Sp(y,y,x).

The pair (X,Sp) is called a partial S-.

Next, we give the definition of an Ms-metric space, but first we introduce the following notation.

Notation.

1. msx,y,z := min{ms(x, x, x), ms(y,y,y), ms(z,z,z)}

2. Msx,y,z := max{ms(x, x, x), ms(y,y,y), ms(z,z,z)}

3 + Definition 2. An Ms-metric on a nonempty set X is a ms : X → R such that for all x,y,z,t ∈ X, the following conditions are satisfied:

1. ms(x, x, x)= ms(y,y,y)= ms(x, x, y) if and only if x = y,

2. msx,y,z ≤ ms(x, y, z),

3. ms(x, x, y)= ms(y,y,x),

4. (ms(x, y, z) − msx,y,z ) ≤ (ms(x, x, t) − msx,x,t )+(ms(y,y,t) − msy,y,t )+(ms(z,z,t) −

msz,z,t ).

The pair (X, ms) is called an Ms-metric space. Notice that the condition ms(x, x, x) = ms(y,y,y)= ms(z,z,z)= ms(x, y, z) ⇔ x = y = z implies (1) above.

It is a straightforward to verify that every partial S-metric space is an Ms-metric space but the converse is not true. The following is an example of an Ms-metric which is not a partial S-metric space.

Example 1.1. Let X = {1, 2, 3} and define the Ms-metric space ms on X by ms(1, 2, 3)=6, ms(1, 1, 2) = ms(2, 2, 1) = ms(1, 1, 1)=8, ms(1, 1, 3) = ms(3, 3, 1) = ms(3, 3, 2) = ms(2, 2, 3)=7, ms(2, 2, 2)=9, and ms(3, 3, 3)=5. It is not difficult to see that (X, ms) is an Ms-metric space, but since ms(1, 1, 1) 6≤ ms(1, 2, 3) we deduce that ms is not a partial S-metric space.

Definition 3. Let (X, ms) be an Ms-metric space. Then:

1. A {xn} in X converges to a point x if and only if

lim (ms(xn, xn, x) − msx ,x ,x)=0. n→∞ n n

2 2. A sequence {xn} in X is said to be Ms- if and only if

lim (ms(xn, xn, xm) − msx ,x ,x ), and lim (Msx ,x ,x − msx ,x ,x ) n,m→∞ n n m n→∞ n n m n n m

exist and finite.

3. An Ms-metric space is said to be complete if every Ms-Cauchy sequence {xn} converges to a point x such that

lim (ms(xn, xn, x) − msx ,x ,x)=0 and lim (Msx ,x ,x − msx ,x ,x)=0. n→∞ n n n→∞ n n n n

A in the Ms−metric (X, ms) space with center x ∈ X and radius η > 0 is defined by Bs[x, η]= {y ∈ X | ms(x, x, y) − msx,x,y ≤ η}.

The of (X, Ms) is generated by means of the basis β = {Bs[x, η]: η > 0}.

Lemma 1.2. Assume xn → x and yn → y as n → ∞ in an Ms−metric space (X, ms). Then, lim (ms(xn, xn,yn) − msx ,x ,y )= ms(x, x, y) − msx,x,y n→∞ n n n Proof. The proof follows by the inequality (4) in Definition 2. Indeed, we have

|(ms(xn, xn,yn) − msxn,xn,yn ) − (ms(x, x, y) − msx,x,y)| ≤ 2[(ms(xn, xn, x) − msxn,xn,x)

+ (ms(yn,yn,y) − msyn,yn,y)] (1)

2 Fixed Point Theorems

In this section, we consider some results about the existence and the uniqueness of fixed point for self mappings on an Ms-metric space, under different contractions principles.

Theorem 1. Let (X, ms) be a complete Ms-metric space and T be a self mapping on X satisfying the following condition:

ms(Tx,Tx,Ty) ≤ kms(x, x, y), (2)

for all x, y ∈ X, where k ∈ [0, 1). Then T has a unique fixed point u. Moreover, ms(u,u,u)= 0.

Proof. Since k ∈ [0, 1), we can choose a natural number n0 such that for a given 0 <ǫ< 1, ǫ n0 n0 i we have k < . Let T ≡ F and F x0 = x for all natural numbers i, where x0 is arbitrary. 8 i Hence, for all x, y ∈ X, we have

n0 n0 n0 n0 ms(Fx,Fx,Fy)= ms(T x, T x, T y) ≤ k ms(x, x, y).

3 For any i, we have

ms(xi+1, xi+1, xi)= ms(F xi, F xi, F xi−1) n0 ≤ k ms(xi, xi, xi−1)

n0+i ≤ k ms(x1, x1, x0) → 0 as i →∞.

Similarly, by (2) we have ms(xi, xi, xi) → 0 as i →∞. Thus, we choose l such that ǫ ǫ m (x +1, x +1, x ) < and m (x , x , x ) < . s l l l 8 s l l l 4 ǫ Now, let η = 2 + ms(xl, xl, xl). Define the set

Bs[xl, η]= {y ∈ X | ms(xl, xl,y) − msxl,xl,y ≤ η}.

Note that, xl ∈ Bs[xl, η]. Therefore Bs[xl, η] =6 ∅. Let z ∈ Bs[xl, η] be arbitrary. Hence,

n0 ms(Fz,Fz,Fxl) ≤ k ms(z,z,xl) ǫ ǫ ≤ [ + m + m (x , x , x )] 8 2 sz,z,xl s l l l ǫ < [1 + 2m (x , x , x )]. 8 s l l l ǫ Also, we know that m (F x , F x , x )= m (x +1, x +1, x ) < . Therefore, s l l l s l l l 8

ms(Fz,Fz,xl) − msFz,Fz,xl ≤ 2[(ms(Fz,Fz,Fxl) − msFz,Fz,Fxl)]+(ms(F xl, F xl, xl) − msFxl,Fxl,xl )

≤ 2ms(Fz,Fz,Fxl)+ ms(F xl, F xl, xl)] ǫ ǫ ≤ 2 (1+2m (x , x , x )) + 8 s l l l 8 ǫ ǫ ǫ = + + m (x , x , x ) 4 8 2 s l l l ǫ < + m (x , x , x ). 2 s l l l

Thus, F z ∈ Bb[xl, η] which implies that F maps Bb[xl, η] into itself. Thus, by repeating n this process we deduce that for all n ≥ 1 we have F xl ∈ Bb[xl, η] and that is xm ∈ Bb[xl, η] for all m ≥ l. Therefore, for all m > n ≥ l where n = l + i for some i

ms(xn, xn, xm)= ms(F xn−1, F xn−1, F xm−1) n0 ≤ k ms(xn−1, xn−1, xm−1) 2n0 ≤ k ms(xn−2, xn−2, xm−2) . .

in0 ≤ k ms(xl, xl, xm−i)

≤ ms(xl, xl, xm−i) ǫ ≤ + m − + m (x , x , x ) 2 sxl,xl,xm i s l l l ǫ ≤ +2m (x , x , x ). 2 s l l l

4 ǫ Also, we have ms(xl, xl, xl) < 4 , which implies that ms(xn, xn, xm) < ǫ for all m > n > l, and thus ms(xn, xn, xm) − msxn,xn,xm < ǫ for all m > n > l. By the contraction condition (2) we see that the sequence {ms(xn, xn, xn)} is decreasing and hence, for all m>n>l, we have

Msxn,xn,xm − msxn,xn,xm ≤ Msxn,xn,xm

= ms(xn, xn, xn)

≤ kms(xn−1, xn−1, xn−1) . . n ≤ k ms(x0, x0, x0) → 0 as n →∞.

Thus, we deduce that

lim (ms(xn, xn, xm) − msx ,x ,x )=0, and lim (Msx ,x ,x − msx ,x ,x )=0 n,m→∞ n n m n→∞ n n m n n m

Hence, the sequence {xn} is an Ms-Cauchy. Since X is complete, there exists u ∈ X such that lim ms(xn, xn,u) − msx ,x ,u =0, lim Msx ,x ,u − msx ,x ,u =0 n→∞ n n n→∞ n n n n

The contraction condition (2) implies that ms(xn, xn, xn) → 0 as n →∞. Moreover, notice that lim Msx ,x ,u − msx ,x ,u = lim |ms(xn, xn, xn) − ms(u,u,u)| =0, n→∞ n n n n n→∞ and hence ms(u,u,u) = 0. Since xn → u, ms(u,u,u) = 0 and ms(xn, xn, xn) → 0 as n → ∞ then limn→∞ ms(xn, xn,u) = limn→∞ msxn,xn,u = 0. Since ms(T xn, T xn,Tu) ≤ kms(xn, xn,u) → 0 as n →∞, then T xn → Tu. Now, we show that Tu = u. By Lemma 1.2 and that T xn → Tu and xn+1 = T xn → u, we have

lim ms(xn, xn,u) − msx ,x ,u =0 n→∞ n n

= lim ms(xn+1, xn+1,u) − msx +1,x +1,u n→∞ n n

= lim ms(T xn, T xn,u) − msTx ,Tx ,u n→∞ n n

= ms(u,u,u) − msTu,Tu,u

= ms(Tu,Tu,u) − msTu,Tu,u.

Hence, ms(Tu,Tu,u) = msTu,Tu,u = ms(u,u,u) , but also by the contraction condition (2) we see that msTu,Tu,u = ms(Tu,Tu,Tu) . Therefore, (2) in Definition 2 implies that Tu = u.

To prove the uniqueness of the fixed point u, assume that T has two fixed points u, v ∈ X; that is, Tu = u and T v = v. Thus,

ms(u,u,v)= ms(Tu,Tu,Tv) ≤ kms(u,u,v) < ms(u,u,v),

5 ms(u,u,u)= ms(Tu,Tu,Tu) ≤ kms(u,u,u) < ms(u,u,u), and ms(v,v,v)= ms(Tv,Tv,Tv) ≤ kms(v,v,v) < ms(v,v,v),

which implies that ms(u,u,v)=0= ms(u,u,u) = ms(v,v,v), and hence u = v as desired. Finally,assume that u is a fixed point of T. Then applying the contraction condition (2) with k ∈ [0, 1), implies that

ms(u,u,u)= ms(Tu,Tu,Tu)

≤ kms(u,u,u) . . n ≤ k ms(u,u,u)

Taking the as n tends to infinity, implies that is ms(u,u,u)=0. In the following result, we prove the existence and uniqueness of a fixed point for a self mapping in Ms-metric space, but under a more general contraction.

Theorem 2. Let (X, ms) be a complete Ms-metric space and T be a self mapping on X satisfying the following condition:

ms(Tx,Tx,Ty) ≤ λ[ms(x, x, T x)+ ms(y,y,Ty)], (3)

1 for all x, y ∈ X, where λ ∈ [0, 2 ). Then T has a unique fixed point u, where ms(u,u,u)=0. n Proof. Let x0 ∈ X be arbitrary. Consider the sequence {xn} defined by xn = T x0 and msn = ms(xn, xn, xn+1). Note that if there exists a natural number n such that msn = 0,

then xn is a fixed point of T and we are done. So, we may assume that msn > 0, for n ≥ 0. By (3), we obtain for any n ≥ 0,

msn = ms(xn, xn, xn+1)= ms(T xn−1, T xn−1, T xn)

≤ λ[ms(xn−1, xn−1, T xn−1)+ ms(xn, xn, T xn)]

= λ[ms(xn−1, xn−1, xn)+ ms(xn, xn, xn+1)]

= λ[msn−1 + msn ].

λ 1 Hence, msn ≤ λmsn−1 +λmsn , which implies msn ≤ µmsn−1 , where µ = 1−λ < 1 as λ ∈ [0, 2 ). By repeating this process we get n msn ≤ µ ms0 .

Thus, limn→∞ msn =0. By (3), for all natural numbers n, m we have

n n m ms(xn, xn, xm)= ms(T x0, T x0, T x0)= ms(T xn−1, T xn−1, T xm−1)

≤ λ[ms(xn−1, xn−1, T xn−1)+ ms(xm−1, xm−1, T xm−1)]

= λ[ms(xn−1, xn−1, xn)+ ms(xm−1, xm−1, xm)]

= λ[msn−1 + msm−1 ].

6 ǫ Since limn→∞ msn =0, for every ǫ> 0, we can find a natural number n0 such that msn < 2 ǫ and msm < 2 for all m, n > n0. Therefore, it follows that ǫ ǫ ǫ ǫ m (x , x , x ) ≤ λ[m − + m − ] <λ[ + ] < + = ǫ for all n, m > n0. s n n m sn 1 sm 1 2 2 2 2 This implies that

ms(xn, xn, xm) − msxn,xn,xm < ǫ for all n, m > n0.

Now, for all natural numbers n, m we have

Msxn,xn,xm = ms(T xn−1, T xn−1, T xn−1)

≤ λ[ms(xn−1, xn−1, T xn−1)+ ms(xn−1, xn−1, T xn−1)]

= λ[ms(xn−1, xn−1, xn)+ ms(xn−1, xn−1, xn)]

= λ[msn−1 + msn−1 ]

=2λmsn−1 .

ǫ As limn→∞ msn−1 = 0, for every ǫ > 0 we can find a natural number n0 such that msn < 2 and for all m, n > n0. Therefore, it follows that ǫ ǫ ǫ ǫ M ≤ λ[m − + m − ] <λ[ + ] < + = ǫ for all n, m > n0, sxn,xn,xm sn 1 sn 1 2 2 2 2 which implies that

Msxn,xn,xm − msxn,xn,xm < ǫ for all n, m > n0.

Thus, {xn} is an Ms-Cauchy sequence in X. Since X is complete, there exists u ∈ X such that lim ms(xn, xn,u) − msx ,x ,u =0. n→∞ n n Now, we show that u is a fixed point of T in X. For any natural number n we have,

lim ms(xn, xn,u) − msx ,x ,u =0 n→∞ n n

= lim ms(xn+1, xn+1,u) − msx +1,x +1,u n→∞ n n

= lim ms(T xn, T xn,u) − msTx ,Tx ,u n→∞ n n

= ms(Tu,Tu,u) − msTu,Tu,u.

This implies that ms(Tu,Tu,u) − msu,u,Tu = 0, and that is ms(Tu,Tu,u) = msu,u,Tu. Now, assume that

ms(Tu,Tu,u)= ms(Tu,Tu,Tu) ≤ 2λms(u,u,Tu)=2λms(Tu,Tu,u) < ms(u,u,Tu).

Thus,

ms(Tu,Tu,u)= ms(u,u,u) ≤ ms(Tu,Tu,Tu) ≤ 2λms(u,u,Tu) < ms(u,u,Tu)

7 Therefore, Tu = u and thus u is a fixed point of T.

Next, we show that if u is a fixed point, then ms(u,u,u)=0. Assume that u is a fixed point of T, then using the contraction (3), we have

ms(u,u,u)= ms(Tu,Tu,Tu)

≤ λ[ms(u,u,Tu)+ ms(u,u,Tu)]

=2λms(u,u,Tu)

=2λms(u,u,u) 1 < m (u,u,u) since λ ∈ [0, ); s 2

that is, ms(u,u,u)=0. Finally, To prove uniqueness, assume that T has two fixed points say u, v ∈ X. Hence,

ms(u,u,v)= ms(Tu,Tu,Tv) ≤ λ[ms(u,u,Tu)+ms(v,v,Tv)] = λ[ms(u,u,u)+ms(v,v,v)]=0,

which implies that ms(u,u,v)=0= ms(u,u,v)= ms(u,u,v), and hence u = v as required.

In closing, the authors would like to bring to the reader’s attention that in this interesting Ms-metric space it is possible to add some control functions in both contractions of Theorems 1, and 2.

Theorem 3. Let (X, ms) be a complete Ms-metric space and T be a self mapping on X satisfying the following condition: for all x, y, z ∈ X

ms(Tx,Ty,Tz) ≤ ms(x, y, z) − φ(ms(x, y, z)), (4)

where φ : [0, ∞) → [0, ∞) is a continuous and non-decreasing function and φ−1(0) = 0 and φ(t) > 0 for all t> 0. Then T has a unique fixed point in X.

n−1 Proof. Let x0 ∈ X. Define the sequence {xn} in X such that xn = T x0 = T xn−1, for all n ∈ IN. Note that if there exists an n ∈ IN such that xn+1 = xn, then xn is a fixed point for T . Without lost of generality, assume that xn+1 =6 xn, for all n ∈ IN. Now

ms(xn, xn+1, xn+1) = ms(T xn−1, T xn, T xn)

≤ ms(xn−1, xn, xn) − φ(ms(xn−1, xn, xn))

≤ ms(xn−1, xn, xn). (5)

Similarly, we can prove that ms(xn−1, xn, xn) ≤ ms(xn−2, xn−1, xn−1). Hence, ms(xn, xn+1, xn+1) is a monotone decreasing sequence. Hence there exists r ≥ 0 such that

lim ms(xn, xn+1, xn+1)= r. n→∞

8 Now, by taking the limit as n →∞ in the inequality (5), we get r ≤ r − φ(r) which leads to a contradiction unless r = 0. Therefore,

lim ms(xn, xn+1, xn+1)=0. n→∞

Suppose that {xn} is not an Ms-Cauchy sequence. Then there exists an ǫ> 0 such that we can find subsequences xmk and xnk of {xn} such that m (x , x , x ) − m ≥ ǫ. (6) s nk mk mk sxnk ,xmk ,xmk

Choose nk to be the smallest integer with nk > mk and satisfies the inequality (6). Hence, m (x , x , x ) − m < ǫ. s nk mk−1 mk−1 sxnk ,xmk−1 ,xmk−1 Now, ǫ ≤ m (x , x , x ) − m s mk nk nk sxmk ,xnk ,xnk ≤ m (x , x , x )+2m (x , x , x ) − m s mk nk−1 nk−1 s nk−1 nk−1 nk−1 sxmk ,xnk−1 ,xnk−1

≤ ǫ +2ms(xnk−1 , xnk−1 , xnk−1 ) < ǫ, as n →∞. Hence, we have a contradiction.

Without lost of generality, assume that msxn,xn,xm = ms(xn, xn, xn). Then we have

0 ≤ msxn,xn,xm − msxn,xn,xm ≤ Msxn,xn,xm

= ms(xn, xn, xn)

= ms(T xn−1, T xn−1, T xn−1)

≤ ms(xn−1, xn−1, xn−1) − φ(ms(xn−1, xn−1, xn−1))

≤ ms(xn−1, xn−1, xn−1) . .

≤ ms(x0, x0, x0)

Hence, lim msx ,x ,x − msx ,x ,x exists and finite. Therefore, {xn} is an Ms-Cauchy se- n→∞ n n m n n m quence. Since X is complete, the sequence {xn} converges to an element x ∈ X; that is,

0 = lim ms(xn, xn, x) − msx ,x ,x n→∞ n n

= lim ms(xn+1, xn+1, x) − msx +1,x +1,x n→∞ n n

= lim ms(T xn, T xn, x) − msTx ,Tx ,x n→∞ n n

= ms(T x, T x, x) − msTx,Tx,x. Similarly to the proof of Theorem 2, it is not difficult to show that this implies that, T x = x and so x is a fixed point. Finally, we show that T has a unique fixed point. Assume that there are two fixed points u, v ∈ X of T . If we have ms(u,u,v) > 0, then Condition (4) implies that

ms(u,u,v)= ms(Tu,Tu,Tv) ≤ ms(u,u,v) − φ(ms(u,u,v)) < ms(u,u,v),

9 and that is a contradiction. Therefore, ms(u,u,v) = 0 and similarly ms(u,u,u)= Ms(v,v,v)= 0 and thus u = v as desired. In closing, is it possible to define the same space but without the condition, (i.e. ms(x, x, y) =6 ms(y,y,x)?) If possible, what kind of results can be obtained in such space?

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