International Journal of Analysis and Applications
ISSN 2291-8639
Volume 10, Number 2 (2016), 90-94
http://www.etamaths.com

CHARACTERIZATION OF MULTIPLICATIVE METRIC COMPLETENESS

BADSHAH E ROME AND MUHAMMAD SARWAR∗

Abstract. We established fixed point theorems in multiplicative metric spaces. The obtained re-
sults generalize Banach contraction principle in multiplicative metric spaces and also characterize

completeness of the underlying multiplicative metric space.

1. Introduction and preliminaries

In 1970 Michael Grossman and Robert Katz [11] established a new calculus called multiplicative
calculus also termed as exponential calculus. Florack and Van Assen [10] used the idea of multiplicative
calculus in biomedical image analysis. Bashirov et al.[3] demonstrated the efficiency of multiplicative
calculus over the Newtonian calculus. They elaborated that multiplicative calculus is more effective
than Newtonian calculus for modeling various problems from different fields . Bashirov and Bashirova
[4] used the concept of multiplicative calculus for deriving function that shows dynamics of literary
text. Bashirov et al.[2] further demonstrated the usefulness of multiplicative calculus by proving the
fundamental theorem of multiplicative calculus. By defining multiplicative distance they provided foun-
dation for multiplicative metric spaces. Özavsar and Cevikel [13] presented the notion of multiplicative
contraction mapping. Besides some other results, they proved the well known Banach contraction prin-
ciple for such contraction in multiplicative metric spaces. HXiaoju et al.[12] established common fixed
point theorems for weak commutative mappings in the setting of multiplicative metric space. Abbas
et al.[1] established common fixed point results of quasi-weak commutative mappings on a closed ball
in the framework of multiplicative metric spaces. Banach contraction principle has been a very advan-
tageous and effectual means in nonlinear analysis. Generalization of the Banach contraction principle
has been one of the most enquired branch of research. Banach theorem has many generalizations;
(see [5, 6, 7, 8, 17]). Sarwar and Rome [16] established several generalizations of Banach contraction
principle and proved Cantor intersection theorem in the framework of Multiplicative metric spaces.
Tomonari Suzuki [18] proved a fixed point result which generalizes Banach theorem and characterizes
metric completeness.

In the current article we prove fixed point results in the set up of multiplicative metric spaces. The
derived results results generalized Banach contraction principle in multiplicative metric spaces and
characterize completeness of the underlying multiplicative metric space. For various definitions and
elements of multiplicative calculus we refer the reader to [1, 2, 3, 9, 11, 12, 13, 14, 15].

Definition 1.1. [2] Let M be a nonempty set. A mapping d : M × M → [1,∞) is said to be
multiplicative metric on M if the following condition are satisfied:

(1) d(x,y) ≥ 1 for all x,y ∈ M;
(2) d(x,y) = 1 if and only if x = y;
(3) d(x,y) = d(y,x) for all x,y ∈ M;
(4) d(x,z) ≤ d(x,y).d(y,z) for all x,y,z ∈ M. And the pair (M,d) is called multiplicative metric

space.

2010 Mathematics Subject Classification. Primary 47H10, 54H25; Secondary 55M20.
Key words and phrases. multiplicative metric space; multiplicative contraction mapping; multiplicative Cauchy se-

quence; fixed point.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

90



CHARACTERIZATION OF MULTIPLICATIVE METRIC COMPLETENESS 91

Example 1.1. [2] The mapping d∗ : (0,∞) × (0,∞) → [1,∞) defined as d∗(x,y) = |x
y
|∗, where

|a|∗ =

{
a if a ≥ 1
1
a

if a < 1
. is a multiplicative metric.

Definition 1.2. [1, 12, 13] A sequence {xn} in a multiplicative metric space (X,d) is said to be
multiplicative Cauchy sequence if for all � > 1 there exits a positive integer n0 such that d(xn,xm) <
� ∀n,m ≥ n0.

Definition 1.3. [1, 12, 13] A multiplicative metric space (X,d) is said to be complete if every multi-
plicative Cauchy sequence in Xconverges in X.

Definition 1.4. [1, 12, 13] Let (X,d) be a multiplicative metric space. A mapping f : X → X is
called multiplicative contraction if there exists a real constant λ ∈ [0, 1) such that d(f(x1),f(x2)) ≤
d(x1,x2)

λ ∀x,y ∈ X.

2. Main results

This section studied two fixed point theorems in the setting of multiplicative metric spaces. The first
result generalized the Banach contraction principal while the second one characterizes multiplicative
metric completeness.

Theorem 2.1. Let (M,d) be a complete multiplicative metric space. Let f be a mapping on M and
ϕ : [0, 1) → (1/2, 1] a non increasing function defined as follows

ϕ(γ) =




1 if 0 ≤ γ ≤ (
√

5 − 1)/2,
(1 −γ)γ−2 if (

√
5 − 1)/2 < γ < 1/

√
2,

(1 + γ)−1 if 1/
√

2 ≤ γ < 1.
Let there exists γ ∈ [0, 1) such that

(1) d(x,fx)ϕ(γ) ≤ d(x,y) ⇒ d(fx,fy) ≤ d(x,y)γ ∀ x,y ∈ M.
Then f has a unique fixed point z. Furthermore limnf

nx = z for all x ∈ M.

Proof. As ϕ(γ) ≤ 1 therefor d(x,fx)ϕ(γ) ≤ d(x,fx) ∀ x ∈ M. condition (1) implies
(2) d(fx,f2x) ≤ d(x,fx)γ ∀ x ∈ M.
Fix v ∈ M and define a sequence {vn} in M by vn = fnv. Relation (2) implies that d(vn,vn+1) ≤
d(v,fv)γ

n

. Therefor
∏∞
n=1 d(vn,vn+1) ≤ d(v,fv)

γ
1−γ < ∞. It means {vn} is a Cauchy sequence. As

M is complete so {vn} converges to some point z ∈ M. We show that
(3) d(fx,z) ≤ d(x,z)γ ∀ x ∈ M\{z}.
For x ∈ M\{z} there will be some positive integer m such that d(vn,z) ≤ d(x,z)1/3
∀ n ≥ m. We have

d(vn,fvn)
ϕ(γ) ≤ d(vn,fvn) = d(vn,vn+1) ≤ d(vn,z)d(z,vn+1)

≤ d(x,z)2/3 =
d(x,z)

d(x,z)1/3
≤

d(x,z)

d(vn,z)
≤ d(vn,x).

Using hypothesis of theorem, we get d(vn+1,fx) ≤ d(vn,x)γ for n ≥ m. Letting n → ∞, we get
d(z,fx) ≤ d(z,x)γ. Hence (3) is proved. Now let us suppose by the way of contradiction that fiz 6= z
for all i ∈ N. Then (3) gives

(4) d(fi+1z,z) ≤ d(fz,z)γ
i

∀ i ∈ N.
Now consider the following cases.
• 0 ≤ γ ≤ (

√
5 − 1)/2,

• (
√

5 − 1)/2 < γ < 1/
√

2,

• 1/
√

2 ≤ γ < 1.
When 0 ≤ γ ≤ (

√
5 − 1)/2, then γ2 + γ − 1 ≤ 0, also 2γ2 ≤ 3 −

√
5 < 1.

If d(f2z,z) < d(f2z,f3z), then

d(z,fz) ≤ d(z,f2z)d(f2z,fz) < d(f2z,f3z)d(f2z,fz) ≤ d(z,fz)γ
2+γ ≤ d(z,fz).



92 ROME AND SARWAR

Which is contradiction. Therefor d(f2z,z) ≥ d(f2z,f3z) = d(f2z,f ◦f2z)ϕ(γ).
Using hypothesis of the theorem and (4),we have

d(z,fz) ≤ d(z,f3z)d(f3z,fz) ≤ d(z,fz)γ
2

d(f2z,z)γ ≤ d(z,fz)γ
2

d(fz,z)γ
2

= d(fz,z)2γ
2

< d(fz,z).

Which is contraction. And when (
√

5 − 1)/2 < γ < 1/
√

2 then 2γ2 < 1. If we suppose d(f2z,z) <
d(f2z,f3z)ϕ(γ), then using (2) we have

d(z,fz) ≤ d(z,f2z)d(f2z,fz) < d(f2z,f3z)ϕ(γ)d(f2z,fz) ≤ d(z,fz)ϕ(γ)γ
2

d(z,fz)γ

= d(z,fz)ϕ(γ)γ
2+γ = d(z,fz)(1−γ)γ

−2γ2+γ = d(z,fz),

giving a contradiction. Hence d(f2z,z) ≥ d(f2z,f3z)ϕ(γ) = d(f2z,f ◦ f2z)ϕ(γ). And this, like the
preceding case, produces the following contradiction.

d(z,fz) ≤ d(z,fz)2γ
2

< d(z,fz).

Finally when 1/
√

2 ≤ γ < 1. Then for x,y ∈ M, either d(x,fx)ϕ(γ) ≤ d(x,y) or d(fx,f2x)ϕ(γ) ≤
d(fx,y). In case d(x,fx)ϕ(γ) > d(x,y) and d(fx,f2x)ϕ(γ) > d(fx,y), then using multiplicative trian-
gular inequality and (2), we have

d(x,fx) ≤ d(x,y)d(y,fx) < d(x,fx)ϕ(γ)d(fx,f2x)ϕ(γ) = (d(x,fx)d(fx,f2x))ϕ(γ)

= d(x,fx)
(1+γ)ϕ(γ)

= d(x,fx)
(1+γ)(1+γ)−1

= d(x,fx).

Which is again contradiction. Now since d(v2n,v2n+1)
ϕ(γ) ≤ d(v2n,z) or

d(v2n+1,v2n+2)
ϕ(γ) ≤ d(v2n+1,z) ∀n ∈ N. Therefore using hypothesis of the theorem, either

d(v2n+1,fz) ≤ d(v2n,z)γ ≤ d(v2n,z) or d(v2n+2,fz) ≤ d(v2n+1,z)γ ≤ d(v2n+1,z) ∀ n ∈ N. Now {vn}
converges to z, but the above inequalities indicate that there is a subsequence of {vn} which converges
to fz. Therefore fz = z. This contradicts the supposition. Hence in all the above cases, there will be
some i ∈ N such that fiz = z. As {fnz} is a Cauchy sequence, therefore fz = z. In order to show
uniqueness of the fixed point of f, let w ∈ M\{z} be another fixed point of f. Then using (3), we
have the contradiction, d(w,z) = d(fw,z) ≤ d(w,z)γ < d(w,z). Hence z is the only fixed point of f
in M. �

Theorem 2.2. Let (M,d) be a multiplicative metric space and ϕ be a mapping as defined in Theorem
2.1. For γ ∈ [0, 1) and β ∈ (0,ϕ(γ)], let Sγ,β be the family of mappings f on M satisfying the following:
(1) For x,y ∈ M, d(x,fx)β ≤ d(x,y) ⇒ d(fx,fy) ≤ d(x,y)γ.
Let Tγ,β be the family of mappings f on M satisfying (1) and the following:
(2) f(M) is countably infinite.
(3) Every subset of f(M) is closed.
Then the following are equivalent:
(a) M is complete.
(b) Every mapping f ∈ Sγ,β has a fixed point for all γ ∈ [0, 1).
(c) There exist γ ∈ (0, 1) and β ∈ (0,ϕ(γ)] such that every mapping f ∈ Tγ,β has a fixed point.

Proof. As β ≤ ϕ(γ), therefore using Theorem 2.1, (a) implies (b). And as Tγ,β ⊂ Sγ,β, therefore (b)
implies (c). Next we prove that (c) implies (a). Let (c) holds but M is not complete. It means there
exists a Cauchy sequence {vn} which doesn’t converge in M. Define a mapping g : M → [1,∞) by
g(x) = limnd(x,vn) for x ∈ M. With the properties of multiplicative metric, the following are obvious:
(i) g(x)/g(y) ≤ d(x,y) ≤ g(x)g(y) for all x,y ∈ M,
(ii) g(x) > 1 for all x ∈ M and
(iii) limng(vn) = 1.
Define a mapping f on M as follows: As for each x ∈ M, g(x) > 1 and limng(vn) = 1, therefore there
exists η ∈ N such that g(vη) ≤ g(x)

γβ
3+γβ . For f(x) = vη,

(5) obviously g(fx) ≤ g(x)
γβ

3+γβ and fx ∈{vn : n ∈ N} for all x ∈ M.
This implies that g(fx) < g(x) for all x ∈ M therefore fx 6= x for all x ∈ M. That is f has no fixed
point. Now since f(M) ⊂{vn : n ∈ N}, therefore condition (2) is satisfied. Obviously every subset of



CHARACTERIZATION OF MULTIPLICATIVE METRIC COMPLETENESS 93

f(M) is closed, that is (3) holds. In order to prove (1), fix x,y ∈ M such that d(x,fx)β ≤ d(x,y). In
case where g(y) > (g(x))2,

(i) and (5) imply that d(fx,fy) ≤ g(fx)g(fy) ≤ (g(x)g(y))
γβ

3+γβ ≤ (g(x)g(y))
γ
3

< (g(x)g(y))
γ
3 (

g(y)

(g(x))2
)

2γ
3 ≤

(g(y)
g(x)

)γ ≤ d(x,y)γ.
And when g(y) ≤ (g(x))2, then again using (i) and (5), we have

d(x,y) ≥ d(x,fx)β ≥
( g(x)
g(fx)

)β ≥ ( g(x)
g(x)

γβ
3+γβ

)β
= g(x)

3β
3+γβ .

And therefore d(fx,fy) ≤ g(fx)g(fy) ≤ (g(x)g(y))
γβ

3+γβ ≤ (g(x)(g(x))2)
γβ

3+γβ

≤ g(x)
3γβ

3+γβ =
(
g(x)

3β
3+γβ

)γ ≤ d(x,y)γ.
Therefore (1) is proved. Hence f ∈ Tγ,β. And by (c), f has a fixed point. which is contradiction.
Consequently M is complete. This completes the proof. �

We conclude with the following example which supports Theorem 2.1.

Example 2.1. Let M = R+,set of positive real numbers. Consider the multiplicative metric d :
M ×M → [1,∞) defined by d(x,y) = e|x−y|. Then (M,d) is complete multiplicative metric space. Let
ϕ be a mapping as defined in Theorem 2.1. T : M → M be mapping defined by T(x) = 1

5+x
, such that

d(x,fx)ϕ(γ) = d(x,fx)ϕ(
1
2
) = d(x,fx) = e|x−

1
5+x
| ≤ e|x−y| = d(x,y) then d(fx,fy) = e|x−y||

1
(5+x)(5+y)

| ≤
e

1
2
|x−y| = d(x,y)γ ∀x,y ∈ M.

Obviously T has unique fixed point 0.1925824036 ∈ M.

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94 ROME AND SARWAR

Department of Mathematics, University of Malakand, Chakdara Dir(L), Pakistan

∗Corresponding author: sarwarswati@gmail.com