We establish some unique fixed point theorems in complete partial metric spaces for generalized weakly S-contractive mappings, containing two altering distance functions under certain assumptions. Also, we discuss some examples in support of our main results.

1. Introduction and Preliminaries

An abstract metric space was first introduced and studied by the French mathematician Frechet [1] in 1906. Many researchers have generalized the concept of metric space as cone metric space, semimetric space, quasimetric space, and so forth, along with the generalization of contraction mappings with applications (see [2–7]). The best approximations of functions in locally convex spaces were discussed by Mishra et al. [8] and Mishra [9]. The degree of approximation of signals in Lp-space is established in [10].

Matthews [11, 12] initiated the concept of partial metric space as another generalization of metric space to study the denotational semantics of dataflow networks. Also, Matthews [11] generalized the Banach contraction principle to the class of partial metric spaces as follows: let (X,p) be a complete partial metric space, and then a self-mapping T on X, satisfying (1)pTx,Ty≤kpx,y∀x,y∈X,where 0≤k<1, has a unique fixed point.

After the Matthews [11] historical contribution, several researchers have established some more fixed point theorems in partial metric spaces and also discussed its topological properties (see [13–15] and references therein).

First, we recall some useful definitions and results, which is useful throughout the paper.

Definition 1 (see [<xref ref-type="bibr" rid="B13">11</xref>, <xref ref-type="bibr" rid="B14">12</xref>]).

Let X be a nonempty set, and a mapping p:X×X→[0,∞) satisfying the following conditions is called a partial metric space on X:

p(x,y)=p(y,x),

p(x,x)=p(x,y)=p(y,y)⇔x=y,

p(x,x)≤p(x,y),

p(x,y)≤p(x,z)+p(z,y)-p(z,z),

for all x,y,z∈X, and the pair (X,p) is called a partial metric space. In the rest of the paper, (X,p) represents a partial metric space equipped with a partial metric p, unless otherwise stated. Let (X,p) be a partial metric space and then let a function dp:X×X→[0,∞) be defined as (2)dp(x,y)=2p(x,y)-p(y,y)-p(x,x)which is a metric on X. Consider the function dm:X×X→[0,∞) such that (3)dmx,y=maxpx,y-px,x,px,y-py,y=p(x,y)-minpx,x,py,y;then dm is a metric on X, and both of the above metrics dp and dm are equivalent [16].
Remark 2 (see [<xref ref-type="bibr" rid="B6">17</xref>]).

In a partial metric space (X,p),

p(x,y)=0⇒x=y but if x=y, then p(x,y) may not be zero,

p(x,y)>0 for all x≠y,

for all x,y∈X.
Example 3 (see [<xref ref-type="bibr" rid="B8">16</xref>]).

Consider a mapping p:[0,∞)×[0,∞)→[0,∞) such that p(x,y)=max{x,y} for all x,y∈[0,∞). Then p will satisfy all the property of partial metric, and hence ([0,∞),p) is a partial metric space but fails to be the condition of p(x,x)=0 for all nonzero x∈[0,∞). Therefore ([0,∞),p) is not a metric space.

Example 4 (see [<xref ref-type="bibr" rid="B8">16</xref>, <xref ref-type="bibr" rid="B12">18</xref>]).

Let pi:X×X→[0,∞)(i=1,2,3) be three mappings and for any arbitrary mapping f:X→[0,∞) such that

p1x,y=d(x,y)+p(x,y),

p2x,y=d(x,y)+max{f(x),f(y)},

p3x,y=d(x,y)+r,

for all r≥0, where (X,d) and (X,p) are a metric space and a partial metric space, respectively. Then each pi is the partial metric on X.
Definition 5 (see [<xref ref-type="bibr" rid="B7">19</xref>]).

In a partial metric space (X,p), (1) a sequence {xn} is said to be convergent to a point x∈X if and only if limn→∞p(xn,x)=p(x,x).

(2) A sequence {xn} is called Cauchy sequence if and only if limn,m→∞p(xn,xm) is finite.

(3) If every Cauchy sequence {xn} converges to a point x∈X such that (4)limn,m→∞p(xn,xm)=p(x,x),then (X,p) is known as complete partial metric space.

Definition 6 (see [<xref ref-type="bibr" rid="B11">20</xref>, <xref ref-type="bibr" rid="B21">21</xref>]).

A self-mapping ψ on a positive real number is said to be an altering distance function, if it holds for all t∈[0,∞) such that

ψ is continuous and nondecreasing,

ψ(t)=0⇔t=0.

The generalization of contractive mappings into C-contractive mappings has been introduced by Chatterjea [6].

Definition 7 (see [<xref ref-type="bibr" rid="B1">2</xref>, <xref ref-type="bibr" rid="B21">21</xref>]).

A self-mapping T on a metric space (X,d), satisfying (5)dTx,Ty≤12dx,Ty+dTx,y-ϕdx,Ty,dTx,y,for all x,y∈X and ϕ:[0,∞)2→[0,∞) is a continuous mapping with ϕ(x,y)=0 if and only if x=y=0 is called weakly C-contractive mapping or a weak C-contraction.

Shukla and Tiwari [3] have introduced the concept of weakly S-contractive mappings.

Definition 8 (see [<xref ref-type="bibr" rid="B3">3</xref>]).

A self-mapping T on a complete metric space (X,d) is said to be weakly S-contractive mapping or a weak S-contraction, if the following inequality holds: (6)dTx,Ty≤13dx,Ty+dTx,y+dx,y-ϕdx,Ty,dTx,y,dx,y,for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function with ϕ(x,y,z)=0 if and only if x=y=z=0.

Lemma 9 (see [<xref ref-type="bibr" rid="B16">7</xref>, <xref ref-type="bibr" rid="B9">14</xref>]).

In a partial metric space (X,p), if a sequence {xn} is convergent to a point x∈X, then limn→∞p(xn,x)≤p(x,z) for all z∈X. Also, if p(x,x)=0, then (7)limn→∞p(xn,z)=p(x,z)∀z∈X.

Lemma 10 (see [<xref ref-type="bibr" rid="B2">13</xref>]).

If {x2n} is not a Cauchy sequence in (X,p) and two sequences {m(k)} and {n(k)} of positive integers such that n(k)>m(k)>k, then the four sequences (8)px2mk,x2nk+1,px2mk,x2nk,px2mk-1,x2nk+1,px2mk-1,x2nktend to ε>0, when k→∞.

Lemma 11 (see [<xref ref-type="bibr" rid="B2">13</xref>, <xref ref-type="bibr" rid="B8">16</xref>]).

In a partial metric space (X,p):

a sequence {xn} is a Cauchy if and only if it is a Cauchy in (X,dp),

X is complete if and only if it is complete in (X,dp).

In addition, limn→∞dp(xn,x)=0 if and only if (9)limn,m→∞p(xn,xm)=limn→∞p(xn,x)=p(x,x).

If {xn} is a Cauchy sequence in the metric space (X,dp), we have(10)limn,m→∞dp(xn,xm)=0and therefore, by definition of dp, we have (11)limn,m→∞p(xn,xm)=0.

2. Main ResultsTheorem 12.

Let (X,p) be a complete partial metric space and ψ and φ be two altering distance functions such that ψ(t)-φ(t)≥0∀t≥0. Then the self-continuous nondecreasing mapping T on X, satisfying the condition(12)ψpTx,Ty≤φpx,Ty+pTx,y+px,y3-ϕpx,Ty,pTx,y,px,y,for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function such that ϕ(x,y,z)=0 if and only if x=y=z=0, has a unique fixed point in X.

Proof.

First we prove that if fixed point of T exists, then it will be unique. On the contrary, we consider two fixed points z,u∈X of T such that z≠u. Then by (12), we have (13)ψpz,u=ψpTz,Tu≤φpz,Tu+pTz,u+pz,u3-ϕpz,Tu,pTz,u,pz,u⟹0≤(ψ-φ)(p(z,u))≤-ϕ(p(z,u),p(z,u),p(z,u)).By the property of ϕ, we obtain (14)ϕpz,u,pz,u,pz,u=0⟹pz,u=0.Using Remark 2, we obtain z=u, which is a contradiction with respect to z≠u. Thus, we conclude that T has a unique fixed point in X.

Next, we show that the mappings T, satisfying (12), have a fixed point. We choose an arbitrary point x0 in X. If x0=Tx0, then the theorem follows trivially. Now, we suppose that x0≤Tx0 and we choose x1∈X such that Tx0=x1. Since T is a nondecreasing function, then we have x0≤x1=Tx0≤Tx1. Again, let x2=Tx1. Then we get (15)x0≤x1=Tx0≤Tx1=x2≤Tx2.Proceeding with this work, we obtained a sequence {xn} in X such that xn+1=Txn and (16)x0≤x1≤x2≤x3≤⋯≤xn≤xn+1⋯.Supposing that p(xn0,xn0+1)=0 for some n0≥0, then by Remark 2 we have (17)xn0=xn0+1=Txn0,thatis,xn0isafixedpointofT.Again, we suppose that p(x2n,x2n+1)>0∀n≥0. Firstly, we prove that the sequence {p(x2n,x2n+1)} is nonincreasing. Suppose this is not true, and then(18)p(x2n,x2n+1)≥p(x2n-1,x2n)∀n≥0.Putting x=x2n-1 and y=x2n in (12) and using (P4), we have (19)ψpx2n,x2n+1=ψpTx2n-1,Tx2n≤φpx2n-1,Tx2n+pTx2n-1,x2n+px2n-1,x2n3-ϕpx2n-1,Tx2n,pTx2n-1,x2n,px2n-1,x2n.Using (P4) above, we get (20)ψpx2n,x2n+1≤φ2px2n-1,x2n+px2n,x2n+13-ϕpx2n-1,x2n+1,px2n,x2n,px2n-1,x2n.Using (18) above, we have (21)0≤ψ-φpx2n,x2n+1≤-ϕ(p(x2n-1,x2n+1),p(x2n,x2n),p(x2n-1,x2n)).⟹ϕ(p(x2n-1,x2n+1),p(x2n,x2n),p(x2n-1,x2n))=0⟹p(x2n-1,x2n+1)=0,p(x2n,x2n)=0,p(x2n-1,x2n)=0llllllllllllllllllllll∀n≥0⟹p(x2n,x2n+1)=0∀n≥0,which contradicts our assumption that p(x2n-1,x2n+1)>0 for all n≥0. Thus, we deduce that {p(x2n,x2n+1)} is a nonincreasing sequence. Therefore(22)p(x2n,x2n+1)≤p(x2n-1,x2n)∀n≥0.Since {p(x2n,x2n+1)} is a monotonically decreasing and bounded below sequence in X, then there exists r≥0 such that(23)limn→∞p(x2n,x2n+1)=r.Using (23) and letting n→∞ in (20), we get (24)0≤ψ-φr≤-ϕlimn→∞px2n-1,x2n+1,limn→∞px2n,x2n,r⟹ϕlimn→∞px2n-1,x2n+1,limn→∞px2n,x2n,r=0⟹limn→∞p(x2n-1,x2n+1)=0,limn→∞px2n,x2n=0,r=0.Then (23) reduces to(25)limn→∞p(x2n,x2n+1)=0∀n≥0.Now, we have required proving that the sequence {xn} is a Cauchy sequence in the metric space (X,dp) and so in (X,p) by Lemma 11. On the contrary, that is, the sequence {x2n} not being a Cauchy sequence in (X,dp), sequences in Lemma 10 tend to ε, when k→∞. Now, we put x=x2n(k)+1 and y=x2m(k) in (12). We have (26)ψpx2nk+1,x2mk=ψpTx2nk,Tx2mk-1≤φpx2nk,Tx2mk-1+pTx2nk,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,Tx2mk-1,pTx2nk,x2mk-1,px2nk,x2mk-1=φpx2nk,x2mk+px2nk+1,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,x2mk,px2nk+1,x2mk-1,px2nk,x2mk-1.Taking k→∞ and applying Lemma 10 in the above inequality, we have (27)0≤ψ-φε≤-ϕε,ε,ε⟹ϕε,ε,ε=0⟹ε=0,which is a contradiction with respect to ε>0. Thus {x2n} is a Cauchy sequence in (X,dp) and so in (X,p). Since (X,p) is complete, (X,dp) is also complete (by Lemma 11). Therefore, the Cauchy sequence {xn} converges in (X,dp); that is, limn→∞dp(xn,z)=0; then by Lemma 11, we have(28)p(z,z)=limn→∞p(xn,z)=limn,m→∞p(xn,xm).By Lemma 11, we get limn,m→∞dp(xn,xm)=0. So, by definition of dp, we get (29)dp(xn,xm)=2p(xn,xm)-p(xn,xn)-p(xm,xm).Using (24) and taking n,m→∞ in the above inequality, we obtain(30)limn,m→∞p(xn,xm)=0.From (28) and (30), we get(31)p(z,z)=limn→∞p(xn,z)=0.By (P4), we obtain (32)p(z,Tz)≤p(z,xn)+p(xn,Tz)-p(xn,xn).Taking n→∞ and using (31), (24), and Lemma 9 in the above inequality, we have(33)p(z,Tz)≤p(Tz,Tz).From (P2), we have(34)p(Tz,Tz)≤p(z,Tz).By (33) and (34), we get(35)p(z,Tz)=p(Tz,Tz).From (35) and (12), we obtain (36)ψpz,Tz=ψpTz,Tz≤φpz,Tz+pTz,z+pz,z3-ϕpz,Tz,pTz,z,pz,z.Using (31) and property of φ in the above inequality, we obtain (37)0≤ψ-φpz,Tz≤-ϕpz,Tz,pTz,z,0⟹ϕ(p(z,Tz),p(Tz,z),0)=0⟹p(z,Tz)=0⟹Tz=z.Thus, z is a unique fixed point of T in X.

Example 13.

Let ([0,1],p) be a complete partial metric space defined by p(x,y)=max{x,y}∀x,y∈[0,1]. Consider a self-map T on [0,1] such that Tx=3x2+2x3. Also, we define ψ,φ:[0,∞)→[0,∞) such that ψ(t)=t+t2/2, φ(t)=3t2+t3, respectively, and ϕ:[0,∞)3→[0,∞) such that ϕ(p,q,r)=(p+q+r)2/54.

If x≥y, then (38)pTx,Ty=maxTx,Ty=Tx=3x2+2x3,ψpTx,Ty=3x2+2x3+92x4+6x5+2x6,φpx,y+pTx,y+px,Ty3=2x+3x2+2x323+2x+3x2+2x3,ϕpx,y,pTx,y,px,Ty=2x+3x2+2x3254.We observe that, for all x,y∈[0,1], (39)ψpTx,Ty≤φpx,y+pTx,y+px,Ty3-ϕpx,y,pTx,y,px,Ty.Similarly, we can show the result for y≥x. Thus, (12) holds for all x,y∈[0,1] and satisfies all the requirements of Theorem 12. So, 0 is the unique fixed point of T.

Corollary 14.

Let (X,p) be a complete partial metric space. Then the self-continuous nondecreasing mapping T on X, satisfying the condition(40)ψpTx,Ty≤ψpx,Ty+pTx,y+px,y3-ϕpx,Ty,pTx,y,px,yfor all x,y∈X and ψ and ϕ which are the same as in Theorem 12, has a unique fixed point in X.

Corollary 15.

In Corollary 14, if partial metric space (X,p) is replaced by usual metric space (X,d), then it reduces to the result of [21].

Corollary 16.

In Theorem 12, if we take ψ(t)=ϕ(t)=t and partial metric space (X,p) is replaced by usual metric space (X,d), then we obtain the main result of [3], which unifies the main result of [2].

Corollary 17.

If we put p(x,y)=0 in (12) and let ϕ:[0,∞)×[0,∞)→[0,∞) be a function, such that ϕ(x,y)=0 if and only if x=y=0, then Theorem 12 reduces to Theorem 2.1 of [13].

Theorem 18.

Let (X,p) be a complete partial metric space and ψ and φ be two altering distance functions such that ψ(t)-φ(t)≥0∀t≥0. Then the two self-continuous nondecreasing mappings S and T on X, satisfying the condition(41)ψpTx,Sy≤φpx,Sy+pTx,y+px,y3-ϕ(p(x,Sy),p(Tx,y),p(x,y))for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function such that ϕ(x,y,z)=0 if and only if x=y=z=0, having a unique common fixed point in X.

Proof.

First, we show that the common fixed point of T and S is unique, if it exists. On the contrary, we assume two common fixed points z,u∈X of T and S such that z≠u. Then by (41), we get (42)ψpz,u=ψpTz,Su≤φ(pz,Su+pTz,u+pz,u3)-ϕ(p(z,Su),p(Tz,u),p(z,u))⟹0≤(ψ-φ)(p(z,u))≤-ϕ(p(z,u),p(z,u),p(z,u)).Property of ϕ implies that (43)ϕ(p(z,u),p(z,u),p(z,u))=0⟹p(z,u)=0⟹z=u,which contradicts our assumption that u≠z. Therefore, we conclude that T and S have a unique common fixed point in X.

Now, we prove that the mappings S and T, satisfying (41), have a common fixed point in X. We choose an arbitrary point x0 in X. If x0=Sx0 and x0=Tx0, then theorem follows trivially. So, we suppose that x0≠Sx0 and x0≠Tx0. Then we construct a sequence {xn} in X, in such a way that Sx2n+1=x2n+2 and Tx2n=x2n+1∀n≥0.

Let us assume that p(x2n,x2n+1)>0 and p(x2n,x2n+2)>0∀n≥0. Then, we can prove that S and T have a common fixed point in X. Firstly, we show that {p(x2n,x2n+1)} is nonincreasing sequence. Suppose this is not true, and then(44)p(x2n,x2n+1)≥p(x2n-1,x2n)∀n≥0.Putting x=x2n and y=x2n+1 in (41) and using (P4), we get (45)ψpx2n+1,x2n+2=ψpTx2n,Sx2n+1≤φpx2n,Sx2n+1+pTx2n,x2n+1+px2n,x2n+13-ϕpx2n,Sx2n+1,pTx2n,x2n+1,px2n,x2n+1.Using (P4) above, we get (46)ψpx2n+1,x2n+2≤φ2px2n,x2n+1+px2n+1,x2n+23-ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1.By (44) and (46), we obtain (47)0≤ψ-φpx2n+1,x2n+2≤-ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1⟹ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1=0⟹p(x2n+1,x2n+1)=0,px2n,x2n+1=0,p(x2n,x2n+2)=0llllllllllllllllllllll∀n≥0,which is a contradiction with respect to p(x2n,x2n+1)>0 and p(x2n,x2n+2)>0∀n≥0. Therefore {p(x2n,x2n+1)} is a nonincreasing sequence in X. Thus, we have(48)p(x2n,x2n+1)≤p(x2n-1,x2n)∀n≥0.Since {p(x2n,x2n+1)} is a monotonically decreasing sequence in X, then there exists r≥0 such that(49)limn→∞p(x2n,x2n+1)=r.Letting n→∞ in (46) and using (49), consequently we get (50)0≤(ψ-φ)(r)≤-ϕ(limn→∞p(x2n,x2n+2),limn→∞p(x2n+1,x2n+1),r)⟹ϕ(limn→∞p(x2n,x2n+2),limn→∞p(x2n+1,x2n+1),r)=0⟹limn→∞p(x2n,x2n+2)=0,limn→∞p(x2n+1,x2n+1)=0,r=0.Then (49) will get reduced to(51)limn→∞p(x2n,x2n+1)=r=0∀n≥0.

Now, we have to show that {xn} is a Cauchy sequence in the partial metric space (X,p). By similar arguments as used in case of proving Theorem 12 we find that the sequence {x2n} is a Cauchy sequence. Putting x=x2n(k) and y=x2m(k)-1 in (41), we have (52)ψpx2nk+1,x2mk=ψ(p(Tx2nk,Sx2mk-1))≤φpx2nk,Sx2mk-1+pTx2nk,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,Sx2mk-1,pTx2nk,x2mk-1,px2nk,x2mk-1.Taking k→∞ and using Lemma 10 in the above inequality, we obtain (53)0≤(ψ-φ)(ε)≤-ϕ(ε,ε,ε)⟹ϕ(ε,ε,ε)=0⟹ε=0,which contradicts our assumption that ε>0. Thus {x2n} is a Cauchy sequence in (X,dp) and so in (X,p). Further, by similar arguments of Theorem 12, we obtain(54)p(z,z)=limn→∞p(xn,z)=limn,m→∞p(xn,xm)=0.By substituting x=z, y=x2m(k)-1 in (41), we obtain(55)ψpTz,x2mk=ψ(p(Tz,Sx2mk-1))≤φpz,Sx2mk-1+pTz,x2mk-13-1+pz,x2mk-13-1-ϕpz,Sx2mk-1,pTz,x2mk-1,pz,x2mk-1.Letting k→∞ and using (54) with property of nondecreasing function φ in the above inequality, we obtain (56)0≤(ψ-φ)p(Tz,z)≤-ϕp0,pTz,z,0⟹ϕ(0,p(Tz,z),0)=0⟹p(z,Tz)=0⟹Tz=z.Hence z is a fixed point of T. Similarly, if we take x=x2n(k)+1 and y=z in (41) and use (54), we obtain Sz=z. By uniqueness of the fixed point, z is a unique common fixed point of S and T.

Again, if p(x2n,x2n+1)=0 or p(x2n,x2n+2)=0∀n≥0, then we will show that the mappings S and T have a common fixed point in X.

Here, we suppose that p(x2n,x2n+2)=0∀n≥0. Then by Remark 2, x2n=x2n+2, for all n≥0. Let n=k, and then(57)x2k=x2k+2∀k≥0.From (41), we get (58)ψpx2k+1,x2k+2=ψ(p(Tx2k,Sx2k+1))≤φpx2k,Sx2k+1+pTx2k,x2k+13-1+px2k,x2k+13-1-ϕpx2k,Sx2k+1,pTx2k,x2k+1,px2k,x2k+1.Using (P4), (P1), and (57) above, we obtain (59)0≤ψ-φpx2k+1,x2k+2≤-ϕ(p(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1))⟹ϕp(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1)=0⟹p(x2k,x2k+2)=0,p(x2k+1,x2k+1)=0,p(x2k,x2k+1)=0⟹x2k=x2k+1=x2k+2∀k≥0.Similarly, we can show that (60)x2k=x2k+1=x2k+2=x2k+3=⋯∀k≥0.Thus {xn} becomes a constant sequence. So xn=Txn=Sxn for all n≥0. Hence xn is a common fixed point of T and S.

Finally, we assume that p(x2n,x2n+1)=0∀n≥0. Then by Remark 2, we have x2n=x2n+1∀n≥0. Let n=k, and then(61)x2k=x2k+1∀k≥0.Using (58), (61), and (P4) with property of nondecreasing function φ, we have (62)0≤ψ-φpx2k+1,x2k+2≤-ϕ(p(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1)).Using similar property of ϕ, as used in first case, we have (63)x2k=x2k+1=x2k+2=x2k+3=⋯∀k≥0.Thus, {xn} becomes a constant sequence. So xn=Txn=Sxn. Hence xn is a common fixed point of T and S.

Example 19.

Let T, p, ψ, φ, and ϕ all be the same as in Example 13 and a self-mapping S on [0,1] defined as Sx=x2/2+x3/3. Then 0 is a unique common fixed point of S and T. One can compute the solution similarly as done in Example 13.

Corollary 20.

Two self-continuous nondecreasing mappings S and T on a complete partial metric space (X,p), satisfying the condition(64)ψpTx,Sy≤ψpx,Sy+pTx,y+px,y3-ϕ(p(x,Sy),p(Tx,y),p(x,y))for all x,y∈X and ψ and ϕ, are the same as in Theorem 18, having a unique common fixed point in X.

Corollary 21.

In Corollary 20, if partial metric space (X,p) is replaced by usual metric space (X,d), then one gets Theorem 2.3 of [21].

Corollary 22.

If one puts p(x,y)=0 in (41) and lets ϕ:[0,∞)×[0,∞)→[0,∞) be a function, such that ϕ(x,y)=0 if and only if x=y=0, then Theorem 18 reduces to Theorem 2.3 of [13].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to express their deep gratitude to the anonymous learned referee(s) and the editor for their valuable suggestions and constructive comments, which resulted in the subsequent improvement of this research article. Special thanks are due to our great Master and friend academician Professor M. Mursaleen, Editor of the Journal of Function Spaces, for his efforts to send the reports of the paper timely. The authors are also grateful to all the editorial board members and reviewers of esteemed journal, that is, Journal of Function Spaces. The first author Lakshmi Narayan Mishra acknowledges the Ministry of Human Resource Development, New Delhi, India, for supporting this research article. All the authors carried out the proof of theorems. Each author contributed equally in the development of the paper. Vishnu Narayan Mishra conceived of the study and participated in its design and coordination. The second author Shiv Kant Tiwari is grateful to Ms. Jagrati Bilthare for her valuable suggestions during the preparation of this paper. All the authors read and approved the final version of paper.

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