International Journal of Analysis and Applications

Volume 16, Number 4 (2018), 472-483

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-16-2018-472

A NEW FIXED POINT THEOREM IN MODULAR METRIC SPACES

ALİ MUTLU1, KÜBRA ÖZKAN1,∗ AND UTKU GÜRDAL2

1Manisa Celal Bayar University, Faculty of Science and Arts, Department of Mathematics, 45140,

Manisa/TURKEY

2Mehmet Akif Ersoy University, Faculty of Science and Arts, Department of Mathematics,

Burdur/TURKEY

∗Corresponding author: kubra.ozkan@hotmail.com

Abstract. In this article, we first give a new fixed point theorem which is main theorem of our study in

modular metric spaces. After that, by using this theorem, we express some interesting results. Moreover,

we characterize completeness in modular metric spaces via this theorem. Finally, we use our main result to

show the existence of solution for a specific problem in dynamic programming.

1. Introduction

The fixed point theory is used in many different fields of mathematics such as topology, analysis, nonlinear

analysis and operator theory. Moreover, it can be applied to different disciplines such as statistics, economy,

engineering, etc. In literature, studies of fixed point theory cover a wide range. The most basic and famous

fixed point theorem is Banach fixed point theorem which was introduced in 1922 [6]. It guarantees the

existence and uniqueness of solution of a functional equation. Besides Banach, many different fixed point

theorems were introduced such as Kannan, Caristi, Coupled, Suzuki, etc [1, 2, 7, 8, 13–16, 19, 23, 24].

In 1950, Nakano introduced modular spaces [21]. Then Chistyakov introduced the concept of modular

metric spaces, which have a physical interpretation, via F-modulars [9] in 2008 and he further developed

Received 2018-03-20; accepted 2018-05-09; published 2018-07-02.

2010 Mathematics Subject Classification. 46A80, 47H10, 54H25.

Key words and phrases. modular metric space; fixed point theorem; complete modular metric.

c©2018 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

472

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-472


Int. J. Anal. Appl. 16 (4) (2018) 473

the theory of these spaces in 2010 [10]. Then many authors made various studies on this structures, e.g.

[3–5, 11, 12, 17, 18, 20].

In this paper, we first give a new fixed point theorem which is main theorem of our study. After that, by

using this theorem, we express some interesting results. Moreover, we characterize completeness in modular

metric spaces via this theorem. Finally, we use our main theorem to show the existence of solution for a

specific problem in dynamic programming.

2. Modular Metric Spaces

Here, we express a series of definitions of some basic concepts related to modular metric spaces.

Definition 2.1. [22] Let X be a linear space on R. If a functional ρ : X → [0,∞] satisfies the following

conditions, we call that ρ is a modular on X:

(1) ρ(0) = 0;

(2) If x ∈ X and ρ(αx) = 0 for all numbers α > 0, then x = 0;

(3) ρ(−x) = ρ(x), for all x ∈ X;

(4) ρ(αx + βy) ≤ ρ(x) + ρ(y) for all α,β ≥ 0 with α + β = 1 and x,y ∈ X.

Let X 6= ∅ and λ ∈ (0,∞). Generally, a function ω : (0,∞) ×X ×X → [0,∞] is denoted as ωλ(x,y) =

ω(λ,x,y) for all x,y ∈ X and λ > 0.

Definition 2.2. [10] Let X 6= ∅. A function ω : (0,∞) × X × X → [0,∞], which satisfies the following

conditions for all x,y,z ∈ X, is called a metric modular on X:

(m1) ωλ(x,y) = 0 for all λ > 0 ⇔ x = y;

(m2) ωλ(x,y) = ωλ(y,x) for all λ > 0;

(m3) ωλ+µ(x,y) ≤ ωλ(x,z) + ωµ(z,y) for all λ,µ > 0.

If 0 < µ < λ, from properties of metric modular, we obtain that

ωλ(x,y) ≤ ωλ−µ(x,x) + ωµ(x,y) = ωµ(x,y)

for all x,y ∈ X.

From [10, 11], we know that for a fixed x0 ∈ X, the two sets

Xω = Xω(x0) = {x ∈ X : ωλ(x,x0) → 0 as λ →∞}

and

X∗ω = X
∗
ω(x0) = {x ∈ X : ∃λ = λ(x) > 0 such that ωλ(x,x0) < ∞}

are said to be modular spaces.



Int. J. Anal. Appl. 16 (4) (2018) 474

It is known [10, 11] that if ω is a metric modular on a nonempty set X, then the modular space Xω can

be equipped with a metric, generated by ω and given by

dω(x,y) = inf{λ > 0 : ωλ(x,y) ≤ λ}

for all x,y ∈ Xω. The pair (Xω,dω) is called a modular metric space.

Definition 2.3. [18] Let Xω be a modular metric space, {xn}n∈N be a sequence in Xω and C ⊆ Xω. Then

(1) {xn}n∈N is called a modular convergent sequence such that xn → x, x ∈ Xω, if for λ > 0

ωλ(xn,x) → 0 as n →∞.

(2) {xn}n∈N is called a modular Cauchy sequence, if for λ > 0

ωλ(xn,xm) → 0 as m,n →∞.

(3) C is called closed, if the limit of a modular convergent sequence in C always belongs to C.

(4) C is called complete modular, if every modular Cauchy sequence {xn} in C is modular convergent in C.

(5) C is called bounded, if

δω(C) = sup{ωλ(x,y) : x,y ∈ C, λ > 0} < ∞.

3. Main Results

Let ω : (0,∞)×X ×X → [0,∞] be a metric modular on X, Xω be a modular metric space, C ⊆ Xω and

ψ : C → R+ be a function on C. ψ is called lower semi-continuous (l.s.c.) on C if

lim
n→∞

ωλ(xn,x) = 0 ⇒ ψ(x) ≤ lim
n→∞

inf(ψ(xn))

for all {xn}⊆ C and λ > 0.

Theorem 3.1. Let ω be a metric modular on X, Xω be a complete modular metric space, ψ : Xω → R+ be

a lower semi-continuous function on Xω and T : Xω → Xω be a mapping satisfying the inequality

ωλ(x,Tx) ≤ ψ(x) −ψ(Tx) (3.1)

for all x ∈ Xω and λ > 0. Then T has a fixed point in Xω.

Proof. For any x ∈ Xω denote,

P(x) = {y ∈ Xω : ωλ(x,y) ≤ ψ(x) −ψ(y) for all λ > 0},

α(x) = inf{ψ(y) : y ∈ P(x)}.



Int. J. Anal. Appl. 16 (4) (2018) 475

As x ∈ P(x), P(x) is not empty and 0 ≤ α(x) ≤ ψ(x). Let x ∈ Xω be an arbitrary point. Now, we construct

a sequence {xn} in Xω as follows: Let x1 = x and when x1,x2, ...,xn have been chosen, choose xn+1 ∈ P(xn)

such that ψ(xn+1) ≤ α(xn) + 1n, for all n ∈ N. By doing so, we get a sequence {xn} satisfying the conditions

ωλ(xn,xn+1) ≤ ψ(xn) −ψ(xn+1),

α(xn) ≤ ψ(xn+1) ≤ α(xn) + 1n
(3.2)

for all n ∈ N and λ > 0. Then {ψ(xn)} is a nonincreasing sequence in R and it is bounded from below by

zero. So, the sequence {ψ(xn)} is convergent to a number M ≥ 0. By virtue of (3.2), we get

M = lim
n→∞

ψ(xn) = lim
n→∞

α(xn). (3.3)

Now, let k ∈ N be arbitrary. From (3.2) and (3.3), there exists at least a number Nk such that ψ(xn) < M + 1k
for all n ≥ Nk. Since ψ(xn) is monotone, we get

M ≤ ψ(xm) ≤ ψ(xn) < M +
1

k

for m ≥ n ≥ Nk. It follows that

ψ(xn) −ψ(xm) <
1

k
for all m ≥ n ≥ Nk. (3.4)

Preserving the generality, suppose that m > n and m,n ∈ N. From (3.2), we get

ω λ
m−n

(xn,xn+1) ≤ ψ(xn) −ψ(xn+1)

for λ
m−n > 0. Now, we obtain that

ωλ(xn,xm) ≤ ω λ
m−n

(xn,xn+1) + ω λ
m−n

(xn+1,xn+2) + · · · + ω λ
m−n

(xm−1,xm)

≤ ψ(xn) −ψ(xn+1) + ψ(xn+1) −ψ(xn+2) + · · · + ψ(xm−1) −ψ(xm)

= ψ(xn) −ψ(xm)

(3.5)

for all m,n ≥ Nk. Then by (3.4),

ωλ(xn,xm) <
1

k
for all m ≥ n ≥ Nk. (3.6)

Letting k or m and n tend to infinity in (3.6), we conclude that

lim
m,n→∞

ωλ(xn,xm) = 0.

Then {xn}n∈N is a modular Cauchy sequence. Since Xω is complete modular, there exists a point u ∈ Xω

such that xn → u as n →∞. Since ψ is lower semi-continuous, using (3.5), we have

ψ(u) ≤ lim
m→∞

inf ψ(xm)

≤ lim
m→∞

inf(ψ(xn) −ωλ(xn,xm))

= ψ(xn) −ωλ(xn,u)



Int. J. Anal. Appl. 16 (4) (2018) 476

and hence

ωλ(xn,u) ≤ ψ(xn) −ψ(u).

Thus, u ∈ P(xn) for all n ∈ N and hence α(xn) ≤ ψ(u). Then by (3.3), we get M ≤ ψ(u). Moreover, using

lower semi-continuity of ψ and (3.3), we have

ψ(u) ≤ lim
n→∞

inf ψ(xn) = M.

So, ψ(u) = M. From (3.1), we know that Tu ∈ P(u). Since u ∈ P(u) for n ∈ N, we have

ωλ(xn,Tu) ≤ ωλ
2
(xn,u) + ωλ

2
(u,Tu)

≤ ψ(xn) −ψ(u) + ψ(u) −ψ(Tu)

= ψ(xn) −ψ(Tu).

Then Tu ∈ P(xn) and this implies α(xn) ≤ ψ(Tu). Hence, we obtain M ≤ ψ(Tu). From (3.1), we get

ψ(Tu) ≤ ψ(u). As ψ(u) = M, we have

ψ(u) = M ≤ ψ(Tu) ≤ ψ(u).

Therefore, ψ(Tu) = ψ(u). Then from (3.1), we get

ωλ(u,Tu) ≤ ψ(u) −ψ(Tu) = 0.

Thus, Tu = u. �

Theorem 3.2. Let ω be a metric modular on X and Xω be a complete modular metric space and ψ : Xω → R

be a lower semi-continuous function on Xω. If ψ is bounded below, then there exists a point u ∈ Xω such

that

ψ(u) < ψ(x) + ωλ(u,x)

for each x ∈ Xω, x 6= u and λ > 0.

Proof. Following the proof Theorem 3.1, we obtain a sequence {xn} that converges to some u ∈ Xω. Under

the same notations, for any u ∈ Xω, define

P(u) = {x ∈ Xω : ωλ(u,x) ≤ ψ(u) −ψ(x) for all λ > 0}

α(u) = inf{ψ(x) : x ∈ P(u)}.



Int. J. Anal. Appl. 16 (4) (2018) 477

We will show that u /∈ P(u) as x 6= u. Suppose the contrary, that is, we get v ∈ P(u) for some v 6= u. Then

0 < ωλ(u,v) ≤ ψ(u) −ψ(v) implies ψ(v) < ψ(u) = M. Since

ωλ(xn,v) ≤ ωλ
2
(xn,u) + ωλ

2
(u,v)

≤ ψ(xn) −ψ(u) + ψ(u) −ψ(v)

= ψ(xn) −ψ(v)

for all λ > 0, v ∈ P(xn). So,

α(xn) ≤ ψ(v) for all n ∈ N.

Letting n tends to infinity, we get M ≤ ψ(v). This equation contradicts with ψ(v) < M = ψ(u). Therefore,

for each x ∈ Xω, x 6= u implies x /∈ P(u), that is

x 6= u ⇒ ωλ(u,x) > ψ(u) −ψ(x).

�

Theorem 3.3. Let Xω and Yω be complete modular metric spaces and the mapping T : Xω → Xω be

arbitrary. Assume that there exists a closed mapping S : Xω → Yω, a lower semi-continuous mapping

ψ : S(Xω) → R+ and a constant c > 0 such that for any x ∈ Xω and λ > 0

ωλ(x,Tx) ≤ ψ(Sx) −ψ(STx) and

c ·ωλ(Sx,STx)} ≤ ψ(Sx) −ψ(STx).
(3.7)

Then the mapping T has a fixed point.

Proof. For any x ∈ Xω, we set

P(x) = {z ∈ Xω : ωλ(x,z) ≤ ψ(Sx) −ψ(Sz) and

c ·ωλ(Sx,Sz)}≤ ψ(Sx) −ψ(Sz) for all λ > 0}

α(x) = inf{ψ(Sz) : z ∈ P(x)}.

As x ∈ P(x), it is clear that P(x) 6= ∅ and 0 ≤ α(x) ≤ ψ(Sx). Similar to the proof of Theorem 3.1, choose

a sequence {xn} in Xω: x1 = x, xn+1 ∈ P(xn) such that

ψ(Sxn+1) ≤ α(xn) +
1

n

for all n ≥ 1. Thus we obtain that

ωλ(xn,xn+1) ≤ ψ(Sxn) −ψ(Sxn+1),

c ·ωλ(Sxn,Sxn+1)} ≤ ψ(Sxn) −ψ(Sxn+1)
(3.8)



Int. J. Anal. Appl. 16 (4) (2018) 478

and

ψ(Sxn+1) −
1

n
≤ α(xn) ≤ ψ(Sxn+1). (3.9)

From (3.8), {ψ(Sxn)} is a nonincreasing and bounded sequence on R. So, {ψ(Sxn)} is a modular convergent

sequence. Therefore, by (3.9) there is a number M ≥ 0 such that

M = lim
n→∞

α(xn) = lim
n→∞

ψ(Sxn). (3.10)

Now, let k ∈ N be an arbitrary point. From (3.10), there exists some Nk such that ψ(Sxn) < M + 1k for all

n ≥ Nk. Thus, by monotonicity of {ψ(Sxn)}, for all m ≥ n ≥ Nk we have

M ≤ ψ(Sxm) ≤ ψ(Sxn) ≤ M +
1

k
.

So,

ψ(Sxn) −ψ(Sxm) ≤
1

k
. (3.11)

Preserving the generality, suppose that m > n and m,n ∈ N. From (3.8), we easily obtain that

ω λ
m−n

(xn,xn+1) ≤ ψ(Sxn) −ψ(Sxn+1) and

c ·ω λ
m−n

(Sxn,STxn+1)} ≤ ψ(Sxn) −ψ(Sxn+1)
(3.12)

for λ
m−n > 0. From (3.8), (3.12) and condition (M3) of modular metric, we have

ωλ(xn,xm) ≤ ω λ
m−n

(xn,xn+1) + ω λ
m−n

(xn+1,xn+2) + · · · + ω λ
m−n

(xm−1,xm)

≤ ψ(Sxn) −ψ(Sxn+1) + ψ(Sxn+1) −ψ(Sxn+2)

+ · · · + ψ(Sxm−1) −ψ(Sxm)

= ψ(Sxn) −ψ(Sxm)

c ·ωλ(Sxn,Sxm) ≤ c ·ω λ
m−n

(Sxn,Sxn+1) + c ·ω λ
m−n

(Sxn+1,Sxn+2)

+ · · · + cω λ
m−n

(Sxm−1,Sxm)

≤ ψ(Sxn) −ψ(Sxn+1) + ψ(Sxn+1) −ψ(Sxn+2)

+ · · · + ψ(Sxm−1) −ψ(Sxm)

= ψ(Sxn) −ψ(Sxm).

(3.13)

From (3.11), we get

ωλ(xn,xm) <
1

k
and c ·ωλ(Sxn,Sxm) <

1

k

for all m ≤ n ≤ Nk and k ∈ N. Therefore, {xn}n∈N is a modular Cauchy sequence in Xω and {Sxn}n∈N is a

modular Cauchy sequence in Yω. By completeness of Xω and Yω, there exist p ∈ Xω and q ∈ Yω such that



Int. J. Anal. Appl. 16 (4) (2018) 479

xn → p and Sxn → q. The fact that, S is a closed mapping implies Sp = q. Since ψ is lower semi-continuous,

using equation (3.13), we have

ψ(q) = ψ(Sp) ≤ lim
m→∞

inf ψ(Sxm) ≤ lim
m→∞

inf(ψ(Sxn) −ωλ(xn,xm))

= ψ(Sxn) −ωλ(xn,p).

Then we obtain

ωλ(xn,p) ≤ ψ(Sxn) −ψ(Sp)

for λ > 0. Similarly, we get

c ·ωλ(xn,p) ≤ ψ(Sxn) −ψ(Sp).

Thus, p ∈ P(xn) for all n ∈ N. Then α(xn) ≤ ψ(Sp). So, by (3.10), we get M ≤ ψ(Sp). On the other hand,

using lower semi-continuity of ψ and (3.10), we have

ψ(q) = ψ(Sp) = lim
m→∞

α(xn) = M.

Therefore, ψ(Sp) = M. By benefiting from the proof of Theorem 3.2, we obtain that x 6= p implies x /∈ P(p).

From (3.7), it’s clear that Tp ∈ P(p), then we have Tp = p. �

Corollary 3.1. Theorem 3.3 holds with inequality

max{ωλ(x,Tx),c ·ωλ(Sx,STx)}≤ ψ(Sx) −ψ(STx)

in the place of inequality (3.7).

Example 3.1. Let X = R. We define the mapping ω : (0,∞) × R × R → [0,∞] by ωλ(x,y) =
|x−y|
1+λ

for all

x,y ∈ R and λ > 0. Then it is obvious that Rω is a complete modular metric space. Define T : Rω → Rω by

Tx = x
4

and ψ : Rω → [0,∞] by ψ(x) = 3|x|. Then for all x,y ∈ R and λ > 0, we have

ωλ(x,Tx) =
|x−Tx|

1 + λ
=
|x− x

4
|

1 + λ
=

3|x|
4(1 + λ)

≤
3

4
|x|

and

ψ(x) −ψ(Tx) = 3|x|−
3|x|

4
=

9

4
|x|.

Hence, ωλ(x,Tx) ≤ ψ(x) − ψ(Tx). From Theorem 3.1, the mapping T has a fixed point. Moreover, it is

0 ∈ Rω.



Int. J. Anal. Appl. 16 (4) (2018) 480

4. Characterization of Completeness

We now prove a new theorem, which together with Theorem 3.1 characterizes completeness in modular

metric spaces.

Theorem 4.1. Let Xω be a modular metric space which is not complete modular. Then there exists a fixed

point free function T : Xω → Xω and a lower semi-continuous mapping ψ : Xω → R+ such that

ωλ(x,Tx) ≤ ψ(x) −ψ(Tx)

for all x ∈ Xω and λ > 0.

Proof. Let {xn}⊂ Xω be a modular Cauchy sequence, which has no limit. We define a function φ : Xω → R+

by

φ(u) = lim
m→∞

ωλ(u,xm), u ∈ Xω for all λ > 0.

Given x ∈ Xω and let n denote the smallest positive integer such that

0 <
1

2
ωλ(x,xn) ≤ φ(x) −φ(xn) for all λ > 0. (4.1)

Note that φ(xn) → 0 as φ(x) > 0. With n so determined, we define function T : Xω → Xω as Tx = xn. Let

ψ(x) = 2φ(x). Then from (4.1), we obtain that

ωλ(x,Tx) ≤ ψ(x) −ψ(Tx).

�

5. Application

Let Xω be a complete modular metric space, Y be a Banach space, M ⊆ Xω, S ⊆ Y and θ : M ×S → M,

H : M ×S ×R → R be two functions. Using Theorem 3.1, we show the existence of a bounded solution for

the following problem in dynamic programming:

We take a g ∈ B(M) such that

g(t) = sup
s∈S
{H(t,s,g(θ(t,s)))} (5.1)

where t ∈ M and B(M) is a Banach space which consists of all bounded real functionals on M with the

norm ‖g‖ = supt∈M |g(t)|. We define a complete modular metric on B(M) with

ωλ(g,k) = sup
t∈Z

{∣∣∣∣g(t) −k(t)1 + λ
∣∣∣∣
}

(5.2)

for all g,k ∈ B(M) and λ > 0. If we take a Cauchy sequence {gn}n∈N in B(M), then from completeness of

Xω, there exists a function u ∈ B(M) such that the sequence {gn}n∈N is convergent to u.



Int. J. Anal. Appl. 16 (4) (2018) 481

Theorem 5.1. Let θ : M ×S → M, H : M ×S × R → R be bounded and ψ : B(M) → R+ be lower semi

continuous on Xω and define by ψ(g) = ‖g‖. We define a operator T : B(M) → B(M) by

T(g)(t) = sup
s∈S
{H(t,s,g(θ(t,s)))}

for all g ∈ B(M) and t ∈ M. If

sup
t∈M

∣∣∣∣g(t) −H(t,s,g(θ(t,s)))λ
∣∣∣∣ ≤ ψ(g) −ψ(T(g)) (5.3)

for all λ > 0, g,k ∈ B(M) and s ∈ S, then the functional equation (5.1) has a bounded solution.

Proof. Let t ∈ M and g ∈ B(M). Then there exists s ∈ S and � > 0 such that

T(g)(x) < H(t,s,g(θ(t,s))) + � (5.4)

and

T(g)(x) > H(t,s,g(θ(t,s))). (5.5)

On the other hand, it is obvious that

g(t) < g(t) + � (5.6)

and

g(t) > g(t) − �. (5.7)

for all � > 0. By using the inequalities (5.5) and (5.6) we obtain that

g(t) −T(g)(t) < g(t) −H(t,s,g(θ(t,s))) + �

≤ |g(t) −H(t,s,g(θ(t,s)))| + �.
(5.8)

Similarly, by using the inequalities (5.4) and (5.7) we obtain that

T(g)(t) −g(t) < H(t,s,g(θ(t,s))) −g(t) + 2�

≤ |H(t,s,g(θ(t,s))) −g(t)| + 2�.
(5.9)

Therefore, from the inequalities (5.8) and (5.9), we get

|g(t) −T(g)(t)| < |g(t) −H(t,s,g(θ(t,s)))| + 2�. (5.10)

If we divide both sides of the inequality (5.10) by 1 + λ, we get∣∣∣∣g(t) −T(g)(t)1 + λ
∣∣∣∣ <

∣∣∣∣g(t) −H(t,s,g(θ(t,s)))1 + λ
∣∣∣∣ + 2�1 + λ (5.11)

for all λ > 0. Since 2�
1+λ

> 0 in the inequality (5.11), we can ignore the contrary 2�
1+λ

. Then we have∣∣∣∣g(t) −T(g)(t)1 + λ
∣∣∣∣ <

∣∣∣∣g(t) −H(t,s,g(θ(t,s)))1 + λ
∣∣∣∣



Int. J. Anal. Appl. 16 (4) (2018) 482

for all λ > 0. Then from property of supremum, we get

sup
t∈Z

∣∣∣∣g(t) −T(g)(t)1 + λ
∣∣∣∣ < sup

t∈Z

∣∣∣∣g(t) −H(t,s,g(θ(t,s)))1 + λ
∣∣∣∣.

Then from inequalities (5.2) and (5.3) we obtain that

ωλ(g,T(g)) < ψ(g) −ψ(T(g)).

Therefore, from Theorem 3.1, T has a fixed point u ∈ B(Z). Then the functional equation (5.1) has a

bounded solution. �

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	1. Introduction
	2. Modular Metric Spaces
	3. Main Results
	4.  Characterization of Completeness
	5. Application
	References