Sciresol Sciresol https://indjst.org/author-guidelines Indian Journal of Science and Technology 0974-5645 10.17485/IJST/v13i39.1629 Some results in multiplicative metric space using absorbing mappings Vijayabaskerreddy B basker.bonuga@gmail.com 1 Srinivas V 2 Mathematics Section, Department S & H, Sreenidhi Institute of Science and Technology (SNIST) Hyderabad, Telangana, 501301, +91 9505742650 India Maths Department, PG Science college Saifabad, Osmania University Hyderabad, Telangana, 500004 India 13 39 2020 Abstract

Objective/Aim: To generate three common fixed point results for four self mappings in complete multiplicative metric space (MMS). Method: The method involves applying of point wise absorbing mappings with different combinations such as complete subspace, reciprocally continuous and compatible mappings and semi-compatible mappings. Findings: All the results are supported by the provision of valid examples. Novelty/Improvement: The concept of reciprocally continuity along with semi compatible mappings is used which is weaker than the concept of continuity and compatibility.

Keywords Fixed point absorbing maps compatible mappings semi-compatible mappings reciprocally continuous mappings None
Introduction

The extraction of fixed point theorems has been fascinating area to the researchers due to its remarkable applications in many areas of mathematics and other allied subjects. The notion of multiplicative distance is initiated in multiplicative calculus 1. Subsequently, the multiplicative metric space (MMS) has been introduced 2. Proved fixed point theorem in multiplicative metric space using weak commuting mappings 3. Introduced the notion of Absorbing mappings in fuzzy metric space and prove common fixed point theorems 4. The notion of semi-compatible mappings is introduced in d-topological space 5. The notion of reciprocally continuous mappings became instrumental in proving some common fixed point theorems in metric space 6. Recently some results in multiplicative space are seen in 7 and8 using the concept of semi compatible mappings.

In this study, absorbing mappings notion is initiated in MMS to generate some common fixed point theorems under different conditions.

Preliminaries

Before establishing our Theorems we present some definitions and results that they are needed.

Definition 2.1. 1: For a non empty set X ,a MMS (X, d*) is defined as function d*:X×X(0,) holding the conditions:

MMS (i) d*α,β1 for all α,βX  and d*α,β=1 α=β

MMS (ii) d*α,β=d*β,α for all α,βX

MMS (iii) d*α,βd*α,γ.d*γ,β for all α,β,γX.

Definition 2.2. 2: In a MMS a sequence {αη} converges to α in X if d*αη,α<ε for each >1 and for all ηη0  and η0N.

Definition 2.3. 2: Multiplicative Cauchy sequence in MMS is one which holds d*αη, βm< for all m,η>N and for all >1.

Definition 2.4. 2: A complete MMS is one in which every Cauchy sequence is converges in it.

Definition 2.5 : In a MMS having two maps A and S then S said to be A-absorbing if d*(Aα,ASα)d*R(Aα,Sα) for some real number R>0 and for all αX.

Example 2.5.1 : Let X=(0,).Define d*:X×X(0,) by d*(α,β)=eα-β then (X,d*) is multiplicative metric space. Defined A,S:X×X(0,) as

A(α)=12 for α120 for α=12 and S(α)=12 for all αX

Then A is S-absorbing for R1.

Definition 2.6: A condition d*Ax,ASxd*RAx,Sx for some R>0 and for given xX holds then the map S in a MMS is said to be point wise A-absorbing.

Example 2.6.1: Let X=[0,10]. Define d*:X×XR+ by d*=eα-β then (X,d*) is MMS. The mappings Aα=1  and  Sα=3α2α+1 for all αX then A is point wise S-absorbing.

Definition 2.7: We define the pair (S,A) in MMS as compatible or asymptotically commuting if for some  tX,  limηd*SAαη,ASαη=1 whenever { αη} is a sequence in X such that limηAxη=limηSxη=t .

Example 2.7.1: Let X=1,10 Define d*:X×XR+ by d*(α,β)=eα-β then (X,d*) is MMS. The mappings A and S are defined as

A(α)=1 for 1α2 and α=34 for α>33α-15 for α(2,3) and S(α)=2 for 1α22α+15 for α>2

Let αη=2+1η for η >0.  Then it is easy to see that the pair ( A,S) is not compatible but A is S-absorbing.

Definition 2.8: Mappings S and A of multiplicative metric space (X,d*) are said to be semi compatible if limηd*SAαη,Aζ=1  for all ζ>0  whenever {αη} is a sequence in X such that limηSαη=limηAαη=ζ for some ζX.

Example 2.8.1: Let X,d* be a multiplicative metric space where X=[0,1] and d*α,β=eα-β. We define the functions S and A by

S(α)=16-α if 0α112112 if 112<α1  A(α)=112 if 0α112110 if 112<α1

Let αη=112-1η for η1, then the pair ( S,A) is semi- compatible.

Definition 2.9: A pair of self maps  (S,A) of multiplicative metric space (X,d*) is said to be reciprocally continuous if limηd*ASαη,Aζ=1 and limηd*SAαη,Sζ=1whenever there exists a sequence { αη} in X such that limηSαη=t,limnAαη=t for some tX.

Example 2.9.1: Let (X,d*) be a multiplicative metric space where X=[-1,1] and d*α,β=eα-β.We define mappings Aand S by

A(α)=15 if -1α<1616 if 16α1 and S(α)=15 if -1α<166α+536 if 16α1

Let the sequence αη=16+1η for η1 then the pair ( A,S) is reciprocally continuous.

Now some common fixed point theorems are to be established using different conditions in multiplicative metric space.

Main Results

3.1 Theorem: The mappings S, T, A and B defined in a complete MMS holding the conditions:

3.1.1 S(X)B(X),T(X)A(X)

3.1.2 d*(Sα,Tβ)d*(Aα,Bβ)·d*(Aα,Sα)·d*(Bβ,Tβ)·d*(Sα,Bβ)·d*(Aα,Tβ)λ6 where λ0.12 for all α,βX

3.1.3 If the mappings S and Tare point wise A-absorbing and point wise B-absorbing respectively.

3.1.4 If the range of one of the mappings S,T,A or B is a complete subspace of X  then A,B,S and T have a unique common fixed point in  X.

Following discussion is useful in proving of Theorem 3.1

Since  SXBX, consider a point  α0X, there exists α1X such that Sα0=Bα1=β0. For this α1 there exists  α2X such that Tα1=Aα2=β1. Continuing this process we get  Sα2η=Bα2η+1=β2η(say) and  Tα2η+1=Aα2η+2=β2η+1say.

Now we can define βη in X, we obtain

d*β2η,β2η+1d*Aα2η,Bα2η+1·d*Aα2η,Sα2η·d*Bα2η+1,Tα2η+1d*Sα2η,Bα2η+1·d*Aα2η,Tα2η+1λ6                    d*β2η-1,β2η.d*β2η-1,β2η.d*β2η,β2η+1.d*β2η,β2η.d*β2η-1,β2η+1λ6                    d*3β2η-1,β2η.d*2β2η,βy2η+1λ6

this implies that

d*β2η,β2η+1d*32λ3-λβ2η-1,β2η.

Let 32λ1-λ=h, then

d*β2η,β2η+1d*hβ2η-1,β2η.

We also obtain d*β2η+1,β2η+2d*hβ2η,β2η+1.

Therefore d*βη,βη+1d*hβη-1,βηd*hηβ1,β0 for all η2.

Let m,ηN such that mη, then we get

d*βm,βηd*βm,βm-1.d*βm-1,βm-2d*βη+1,βηd*hm-1β1,β0.d*hm-2β1,β0d*hηβ1,β0d*hη1-hβ1,β0

this implies d*βm,βη1 as η.

Hence βη is multiplicative Cauchy sequence.

By the completeness of X, there exists ζX such that βηζ as η. Moreover, Sα2η=Bα2η+1=β2n and Tα2η+1=Aα2η+2=β2η+1 are sub sequences of βη, consequently    Sα2η, Bα2η+1 , Tα2η+1  Aα2η+2 convege to ζ as η.

Proof of Theorem 3.1

Let A(X) be the range of X being a complete subspace, then there exists a point Au such that limηAα2η=Au . By condition 3.1.4 we get Tα2η+1Au, Sα2η-2Au,Bα2ηAu as ηαin view of discussion Au=ζ.

Put α=u ,β=α2η+1 in 3.1.2 we have

d*Su,Tα2η+1     d*Au,Bα2η+1.d*Au,Su.d*Bα2η+1,Tα2η+1.d*Su,Bα2η+1.d*(Au,Tα2η+1)λ6

letting ηwe obtain

d*(Su,Au)d*Au,Au.d*Au,Su.d*Au,Au.d*Su,Au.d*(Au,Au)λ6

this gives

d*(Au,Su)d*λ3(Au,Su) , a contradiction to the definition of 2.1

which implies Au=Su.

Since S(X)B(X) then there exists w𝕏 such that Au=Bw.

Put α=u, β=w  in the inequality we get

d*Su,Twd*Au,Bw.d*Au,Su.d*Bw,Tw.d*Su,Bw.d*Au,Twλ6d*(Au,Tw)d*λ3(Au,Tw)

this implies Au=Tw.

Thus we have Su=Au=Tw=Bw.

Since S is a point wise A-absorbing this makes

d*Au,ASud*RAu,Su

this implies Au=ASu=AAu.

By putting α=Au,β=w in condition 3.1.2

d*SAu,Twd*AAu,Bw.d*AAu,SAu.d*Bw,Tw.d*SAu,Bw.d*(AAu,Tw)λ6d*SAu,Aud*λ3SAu,Au

this implies SAu=Au.

Therefore ASu=SAu=Au.

Similarly T is a point wise B-absorbing

implies d*(Bw,BTw)d*R(Bw,Tw)

and implies Bw=BTw.

This gives Bw=BTw=BBw=Tw=Au.

Thus we have BAu=Au.

Now we claim TTw=Tw for this put α=u, β=Tw in condition then we get

d*Su,TTwd*Au,BTw.d*Au,Su.d*BTw,TTw.d*Su,Bw.d*Au,TTwλ6

this gives d*Au,TTwd*λ3Au,TTw

this gives Au=TTw=TTw=TAu

and  this gives  TAu=Au.

Therefore AAu=SAu=BAu=TAu=Au.

Thus Au is a common fixed point of  S,T,A, and B and ζ is common fixed point of the mappings  S,T, A and B. The uniqueness of the fixed point can be easily verified. The proof is similar when T(X) ,or B(X) or S(X)  is assumed to be complete subspace of X.

The following example satisfies all the properties of Theorem 3.1.

3.1.6 Example: Let 𝕏,d* be MMS where 𝕏=[0,8] and d*=e|α-β| where α,β𝕏 .

The self mappings A,B,S and T defined as

\begin{aligned} \mathrm {A}(\alpha) &=\mathrm {B}(\alpha)=\alpha \text { if } \alpha \in[0,8] \\ \\ \mathrm {S}(\alpha) &=\left\{\begin{array}{ll} \alpha \text { if } \alpha \in(01) \\ 2 \text { if } \alpha \in[1,8] \end{array} \quad \mathrm {T}(\alpha)=\left\{\begin{array}{l} 1 \text { if } \alpha \in(0,1) \\ 2 \text { if } \alpha \in[1,8] \end{array}\right.\right. \end{aligned}

where\;\alpha=2\;common\;\;fixed\;point for the four mappings S,T,A,\;and\;B

3.2 Theorem: Let S,T,\;A\;and\;B\;be self mappings of a complete multiplicative metric space, they satisfying the following conditions 3.1.1 and 3.1.2 together with

3.2.1 T be a point wise \text{B -absorbing }

3.2.2 $$\text { the pair of mappings }(\mathrm{S}, \mathrm{A}) \text { is reciprocally continuous and compatible }$$ then the mappings S,T,A\;and\;B have a unique common fixed point in X.

Proof

Using the condition 3.2.2 we have {{\underset{\eta\rightarrow\infty}{lim}}{SA}}\alpha_{2\eta}=S\zeta\; and {{\underset{\eta\rightarrow\infty}{lim}}{AS\alpha_{2\eta}}}=A\zeta and {{\underset{\eta\rightarrow\infty}{lim}}{d^\ast\left(SA\alpha_{2\eta},AS\alpha_{2\eta}\right)=1}}.

This gives S\zeta=A\zeta.

Put \alpha=\zeta\;and\;\beta=\alpha_{2\eta+1} in condition 3.1.2 then

d^\ast\left(S\zeta,T\alpha_{2\eta+1}\right)\leq\left(d^\ast\left(A\zeta,B\alpha_{2\eta+1}\right).d^\ast\left(A\zeta,S\zeta\right).d^\ast\left(B\alpha_{2\eta+1},T\alpha_{2\eta+1}\right).d^\ast\left(S\zeta,B\alpha_{2\eta+1}\right).d^\ast\left(A\zeta,T\alpha_{2\eta+1}\right)\right\}^\frac\lambda6

letting \eta\rightarrow\infty we get,

d^\ast\left(A\zeta,\zeta\right)\leq\left(d^\ast\left(A\zeta,\zeta\right).d^\ast\left(A\zeta,A\zeta\right).d^\ast\left(\zeta,\zeta\right),d^\ast\left(A\zeta,\zeta\right).d^\ast\left(A\zeta,\zeta\right)\right\}^\frac\lambda6

this gives

This gives A\zeta=\zeta.

Therefore\;A\zeta=S\zeta=\zeta.

Since S(X)B(X) then there exists a point u in X such that {S\alpha}_{2\eta}=Bu

letting \eta\rightarrow\infty this gives

\;\zeta={{\underset{\eta\rightarrow\infty}{lim}}{{S\alpha}_{2\eta}}}=Bu,\;this gives \zeta=Bu.

Now put \;\alpha=\zeta\;,\;\beta=u in condition 3.1.2 then we get

d^\ast\left(S\zeta,Tu\right)\leq\left(d^\ast\left(A\zeta,Bu\right),d^\ast\left(A\zeta,S\zeta\right),d^\ast\left(Bu,Tu\right),d^\ast\left(S\zeta,Bu\right),d^\ast\left(A\zeta,Tu\right)\right\}^\frac\lambda6.

On using A\zeta=S\zeta=\zeta, we get

d^\ast(\zeta,Tu)\leq\left\{d^\ast(\zeta,Tu)\right\}^\frac\lambda3

implies \zeta=Tu.

Therefore S\zeta=A\zeta=Bu=Tu=\zeta.

Again on using B-absorbing nature of the pair \;(B,T) , we have

d^\ast(Bu,BTu)\leq{d^\ast}^R(Bu,Tu) , a contradiction in view of Bu=Tu=\zeta.

Therefore Bu=BTu=Tu and implies B\zeta=\zeta.

Put \alpha=\zeta,\;\beta=\zeta in the condition 3.1.2 and on simplification this leads to

d^\ast(\zeta,T\zeta)\leq{d^\ast}^{\frac\lambda3}(\zeta,T\zeta)

implies T\zeta=\zeta.

Consequently S\zeta=T\zeta=B\zeta=A\zeta=\;\zeta , proving that the point \zeta is a common fixed point of \;S,T,A\;and\;B.

The uniqueness of the fixed point can be easily proved.

The following example supports the conditions of Theorem 3.2.

3.2.3 Example: Let (X,d^\ast) be MMS where X=\lbrack0,2\rbrack and d^\ast=e^{\left|\alpha-\beta\right|} and four self mappings defined as

$$\mathrm {A}(\alpha)=\left\{\begin{array}{cc} 1 \text { if } \alpha \in[0,1) & \\ 6 / 4 \text { if } \alpha \in[1,2) \\ 5 / 4 \text { if } \alpha=2 \end{array} \quad, \quad \mathrm {B}(\alpha)=\left\{\begin{array}{c} 1 / 4 \text { if } \alpha \in[0,1) \\ 6 / 4 \text { if } \alpha \in[1,2) \\ 1 \text { if } \alpha=2 \end{array}\right.\right.$$

$$\mathrm {S}(\alpha)=\frac{6}{4} \text { if } \alpha \in[0,2] \quad \text { and } \quad \mathrm {T}(\alpha)=\left\{\begin{array}{l} 5 / 4 \text { if } \alpha \in[0,1) \\ 6 / 4 \text { if } \alpha \in[1,2] \end{array}\right.$$

After verifying the conditions of Theorem 3.2 in a routine manner \alpha=\frac64 is arrived as the unique common fixed point.

3.3 Theorem

Let S,T\;A\;and\;B be four self mappings of a complete multiplicative metric space \;(X,d^\ast) satisfying the conditions 3.1.1, 3.1.2 along with

3.3.1 The pair (A,S) reciprocally continuous with semi-compatible and T be Point wise B-absorbing or (\;B,T) is reciprocally continuous and semi compatible with S be point wise A-absorbing then A,B,S\;and\;T have unique common fixed point.

Proof: Since the pair of maps (A,S) is reciprocally continuous and semi compatible then we have {{\underset{\eta\rightarrow\infty}{lim}}{SA\alpha_\eta}}=S\zeta.{{\underset{\eta\rightarrow\infty}{lim}}{AS\alpha_\eta=A\zeta}} and {{\underset{\eta\rightarrow\infty}{lim}}{d^\ast\left(AS\alpha_\eta,S\zeta\right)=1.}}

Hence we get \;S\zeta=A\zeta.

Now we claim S\zeta=\zeta for this Put \alpha=\zeta\;,\beta=\alpha_{2\eta+1} in contraction condition 3.1.2.

letting \eta\rightarrow\alpha, we obtain

d^\ast(S\zeta,\zeta)\leq\left(d^\ast\left(S\zeta,\zeta\right).d^\ast\left(S\zeta,S\zeta\right).d^\ast\left(\zeta,\zeta\right).d^\ast\left(S\zeta,\zeta\right).d^\ast(S\zeta,\zeta)\right\}^\frac\lambda6

this gives

d^\ast(S\zeta,\zeta)\leq{d^\ast(S\zeta,\zeta)}^{\frac\lambda2} is a contradiction, implies S\zeta=\zeta.

Hence we get S\zeta=A\zeta=\zeta.

Since S(X)\subseteq B(X) there exists a point u\in X such that \;\;{S\alpha}_{2\eta}=Bu\;letting\;\eta\rightarrow\infty we get {{\underset{\eta\rightarrow\infty}{lim}}{S\alpha_{2\eta}}}=\zeta=Bu.

To claim Tu=\zeta substitute \alpha=\zeta\;,\beta=u in contraction condition 3.1.2

d^\ast(S\zeta,Tu)\leq\left(d^\ast\left(A\zeta,Bu\right).d^\ast\left(A\zeta,S\zeta\right).d^\ast\left(Bu,Tu\right).d^\ast\left(S\zeta,Bu\right).d^\ast\left(A\zeta,Tu\right)\right\}^\frac\lambda6

this gives

d^\ast(\zeta,Tu)\leq\left(Max\left(d^\ast\left(\zeta,\zeta\right).d^\ast\left(\zeta,\zeta\right).d^\ast\left(\zeta,Tu\right).d^\ast\left(\zeta,\zeta\right).d^\ast(\zeta,Tu)\right\}\right\}^\frac\lambda6

this implies

d^\ast(\zeta,Tu)\leq{d^\ast(\zeta,Tu)}^{\frac\lambda3} is a contradiction, Hence we get Tu=\zeta.

Therefore S\zeta=A\zeta=Bu=Tu=\zeta.

Again on using the point wise B-absorbing nature of T there\;exists\;a\;real\;number R>0 such that d^\ast(Bu,BTu)\leq{d^\ast}^R(Bu,Tu) implies \;Bu=BTu\;\;,that\;is\;\zeta=B\zeta.

Put \alpha=\zeta,\beta=\zeta in the condition 3.1.2 results T\zeta=\zeta.

Thus A\zeta=B\zeta=T\zeta=S\zeta=\zeta , giving \zeta is a common fixed point of S,T,A\;and\;B .The uniqueness of the common fixed can be easily proved.

The following example satisfies all the conditions of Theorem 3.3.

3.3.4 Example: Let $$\left(\mathbb{X}, d^{*}\right)$$ be MMS where $$\mathbb{X}=[1,18] \text { and } d^{*}=e^{|\alpha-\beta|}$$ . Define four self mappings $$\mathcal{A}, \mathcal{B}, \mathcal{S} \text { and } \mathcal{T}$$.

A(\alpha)=\left\{\begin{array}{c}1\text{ if }\alpha=1\\10\text{ if }1<\alpha\leq4\\\frac{\alpha+1}3\text{ if }\alpha\in(4,18\rbrack\end{array}\quad B(\alpha)=\left\{\begin{array}{c}1\text{ if }\alpha=1\\5\text{ if }\alpha\in(1,18\rbrack\end{array}\right.\right.

S(\alpha)=\left\{\begin{array}{c}1\text{ if }\alpha=1\\5\text{ if }\alpha\in(1,4\rbrack\\1\text{ if }\alpha\in(4,18\rbrack\end{array}\quad T(\alpha)=\left\{\begin{array}{c}1\text{ if }\alpha=1\\2\text{ if }\alpha\in(1,18\rbrack\end{array}\right.\right.

After simple verification of the conditions of Theorem 3.3, \alpha=1 is emerged as the unique common fixed point for the four self mappings.

Conclusion

This study is focussed on proving three common fixed point results: In Theorem 3.1 the concept of point wise absorbing mappings together with reciprocally continuous mappings is applied. In Theorem 3.2, the notion of point wise absorbing mapping is used along with reciprocally continuous and compatible mappings is discussed. Finally, in the last Theorem 3.3, the concept of absorbing mappings is applied along with reciprocally continuous and semi compatible mappings. Further, all the results are substantiated with suitable examples.

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