Fixed Point Theorems for Generalized Set Valued F-contractions in Metric Spaces

Preprint

Article

Fixed Point Theorems for Generalized Set Valued F-contractions in Metric Spaces

Altmetrics

Downloads

156

Views

Comments

A peer-reviewed article of this preprint also exists.

Seong-Hoon Cho^*

This version is not peer-reviewed

Submitted:

28 December 2023

Posted:

03 January 2024

You are already at the latest version

Alerts

Abstract

In this paper, a positive answer to the open question of Fabino et al. is given, and the concept of I\c sik type set valued contractions is introduced and fixed point theorem for such contractions is established. Examples to support main theorems are given, and an application to integral inclusion is given.

Keywords:

Subject: Computer Science and Mathematics - Analysis

MSC: 47H10; 54H25

1. Introduction

Wardowski [21] introduced the notion of F-contraction mappings and generalized Banach contraction principle by proving that every F-contractions on complete metric spaces have only one fixed point, where

F : (0, \infty) \to (- \infty, \infty)

is a function such that

(F1): F is stlictly increasing;
(F2): $\forall {s_{n}} \subset (0, \infty)$ ,

$lim_{n \to \infty} s_{n} = 0 ⟺ lim_{n \to \infty} F (s_{n}) = - \infty;$
(F3): $\exists q \in (0, 1) : {lim}_{t \to 0^{+}} t^{q} F (t) = 0$ .

Among several results ([1,2,3,4,7,8,10,11,12,13,15,16,18,19,20,22,23]) generalizing Wardowski’s result, Piri and Kumam [17] introduced the concept of Suzuki type F-contractions and obtained related fixed point results in complete metric spaces, where

F : (0, \infty) \to (- \infty, \infty)

is a stlictly increasing function such that

(F4): $inf F = - \infty$ ;
(F5): F is continuous on $(0, \infty)$ .

Note that for a function

F : (0, \infty) \to (- \infty, \infty)

, the following are equivalent.

(1): (F2) is satisfied;
(2): (F4) is satisfied;
(3): ${lim}_{t \to 0^{+}} F (t) = - \infty$ .

Hence, we have that

lim_{n \to \infty} s_{n} = 0 \Rightarrow lim_{n \to \infty} F (s_{n}) = - \infty

whenever (F4) holds.

Very recently, Fabiano et al. [9] gave a generalization of Wardowski’s result [21] by reducing the condition on function

F : (0, \infty) \to (- \infty, \infty)

and by using the right limit of function

F : (0, \infty) \to (- \infty, \infty)

. They proved the following theorem.

Theorem 1.(Theorem 2.3 [9]) Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to E

is a map such that for all

x, y \in E with ρ (T x, T y) > 0

τ + F (ρ (T x, T y)) \leq F (ρ (x, y))

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function If

(F 1)

is satisfied, then T possesses only one fixed point.

In [9], Fabiano et al. asked the following question:

Question.(Question 4.3 [9]) Can conditions for the function F be reduced to (

F 1

) and (

F 2

), and can the proof be made simpler in some results for multivalued mappings in the same way as it was presented in [9] for single-valued mappings?

In this paper, we give a positive answer to the above question by extending above theorem to set-valued maps and obtain a fixed point result for Işik type set valued contractions. We give examples to interpret main results and an application to integral inclusion.

Let

(E, ρ)

be a metric space. We denote by

C L (E)

the family of all nonempty closed subsets of E.

Let

H (\cdot, \cdot)

be the generalized Pompeiu-Hausdorff distance[5] on

C L (E)

, i.e., for all

A, B \in C L (E)

H (A, B) = \{\begin{matrix} max {sup_{a \in A} ρ (a, B), sup_{b \in B} ρ (b, A)}, & if the maximum exists, \\ \infty, & otherwise, \end{matrix}

where

ρ (a, B) = inf {ρ (a, b) : b \in B}

is the distance from the point a to the subset B.

Let

δ (A, B) = sup {ρ (a, b) : a \in A, b \in B}

. When

A = {x}

, we denote

δ (A, B)

δ (x, B)

For

A, B \in C L (E)

, let

D (A, B) = {sup}_{x \in A} d (x, B) = {sup}_{x \in A} {inf}_{y \in B} d (x, y) .

Then, we have that for all

A, B \in C L (E)

D (A, B) \leq H (A, B) \leq δ (A, B) .

Lemma 1.

Let

l > 0

, and let

{t_{n}}, {s_{n}} \subset (l, \infty)

be non-increasing sequences such that

t_{n} < s_{n}, \forall n = 1, 2, 3, \dots and lim_{n \to \infty} t_{n} = lim_{n \to \infty} s_{n} = l .

F : (0, \infty) \to (- \infty, \infty)

is non-decreasing, then we have

lim_{n \to \infty} F (t_{n}) = lim_{n \to \infty} F (s_{n}) = F (l^{+}) > 0

where

F (l^{+})

denotes

{lim}_{t \to l^{+}} F (t)

Proof.

Since

{t_{n}}

and

{s_{n}}

are non-increasing, it follows from the nondecreasingness of F that

lim_{t \to l^{+}} F (t) = lim_{n \to \infty} F (t_{n + 1}) \leq lim_{n \to \infty} F (t_{n}) \leq lim_{n \to \infty} F (s_{n}) \leq lim_{n \to \infty} F (s_{n - 1}) \leq lim_{t \to l^{+}} F (t) .

Thus we have

lim_{n \to \infty} F (t_{n}) = lim_{n \to \infty} F (s_{n}) = lim_{t \to l^{+}} F (t) = F (l^{+}) > 0 .

□

Lemma 2.

[6] Let

(E, ρ)

be a metric space. If

{x_{n}}

is not a Cauchy sequence, then there exists

ϵ > 0

for which we can find subsequences

{x_{m (k)}}

and

{x_{n (k)}}

{x_{n}}

such that

m (k)

is the smallest index for which

m (k) > n (k) > k, ρ (x_{m (k)}, x_{n (k)}) \geq ϵ and ρ (x_{m (k) - 1}, x_{n (k)}) < ϵ . (1.1)

Further if

lim_{n \to \infty} ρ (x_{n}, x_{n + 1}) = 0,

then we have that

lim_{k \to \infty} ρ (x_{n (k)}, x_{m (k)}) = lim_{k \to \infty} ρ (x_{n (k) + 1}, x_{m (k)})

= lim_{k \to \infty} ρ (x_{n (k)}, x_{m (k) + 1}) = lim_{k \to \infty} ρ (x_{n (k) + 1}, x_{m (k) + 1}) = ϵ . (1.2)

Lemma 3.

Let

(E, ρ)

be a metric space, and let

A, B \in C L (E)

a \in A

and

ρ (a, B) < c

, then there exists

b \in B

such that

ρ (a, b) < c

Proof.

Let

ϵ = c - ρ (a, B)

It follows from definition of infimum that there exists

b \in B

such that

ρ (a, b) < ρ (a, B) + ϵ

. Hence,

ρ (a, b) < c

. □

Lemma 4.

Let

(E, ρ)

be a metric space, and let

A, B \in C L (E)

and

ϕ : [0, \infty) \to [0, \infty)

be a strictly increasing function. If

a \in A

and

ρ (a, B) + ϕ (ρ (a, B)) < c

, then there exists

b \in B

such that

ρ (a, b) + ϕ (ρ (a, b)) < c

Proof.

Since

ϕ

is strictly increasing,

ρ (a, B) < ϕ^{- 1} (c - ρ (a, B)) .

By Lemma 3, there exists

b^{'} \in B

such that

ρ (a, b^{'}) < ϕ^{- 1} (c - ρ (a, B))

which yields

ρ (a, B) < c - ϕ (ρ (a, b^{'})) .

Again, by applying Lemma 3, there exists

b^{''} \in B

such that

ρ (a, b^{''}) < c - ϕ (ρ (a, b^{'})) .

Let

min {ρ (a, b^{'}), ρ (a, b^{''})} = ρ (a, b) .

Then, we have that

ρ (a, b) + ϕ (ρ (a, b)) < c .

□

Lemma 5.

(E, ρ)

is a metric space, then

K (E) \subset C L (E)

, where

K (E)

is the family of nonempty compact subsets of E.

Proof.

Let

A \in K (E)

, and let

{x_{n}} \subset A

be a sequence such that

lim_{n \to \infty} ρ (x_{n}, x) = 0, where x \in E .

It follows from compactness of A that there exists a convergent subsequence

{x_{n (k)}}

{x_{n}}

Let

lim_{n \to \infty} ρ (x_{n (k)}, a) = 0, for a \in A .

Since

lim_{n \to \infty} ρ (x_{n (k)}, x) = 0,

x = a \in A .

Hence

A \in C L (E)

. □

2. Fixed point results

Let

(E, ρ)

be a metric space.

A set valued map

T : E \to C L (E)

is called generalized F-contraction if the following condition holds:

for all

x, y \in E with H (T x, T y) > 0

τ + F (H (T x, T y)) \leq F (m (x, y)) (2.1)

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function, and

\begin{matrix} m (x, y) = & max {ρ (x, y), ρ (x, T x), ρ (y, T y), \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]} . \end{matrix}

We now prove our main result.

Theorem 2.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a generalized F-contraction. If

(F 1)

is satisfied, then T possesses a fixed point.

Proof.

Let

x_{0} \in E

be a point, and let

x_{1} \in T x_{0}

x_{1} \in T x_{1}

, then the proof is completed.

Assume that

x_{1} \notin T x_{1}

Then,

ρ (x_{1}, T x_{1}) > 0

, because

T x_{1} \in C L (X)

. Hence,

H (T x_{0}, T x_{1}) \geq d (x_{1}, T x_{1}) > 0 .

From (2.1) we have that

τ + F (H (T x_{0}, T x_{1})) \leq F (m (x_{0}, x_{1})) . (2.2)

We infer that

\begin{matrix} m (x_{0}, x_{1}) = max {ρ (x_{0}, x_{1}), ρ (x_{0}, T x_{0}), ρ (x_{1}, T x_{1}), \frac{1}{2} [ρ (x_{0}, T x_{1}) + ρ (x_{1}, T x_{0})]} \\ = & max {ρ (x_{0}, x_{1}), ρ (x_{1}, T x_{1})}, because that ρ (x_{0}, T x_{0}) \leq ρ (x_{0}, x_{1}) and \\ \frac{1}{2} [ρ (x_{0}, T x_{1}) + ρ (x_{1}, T x_{0})] \leq \frac{1}{2} [ρ (x_{0}, x_{1}) + ρ (x_{1}, T x_{1})] . \end{matrix}

m (x_{0}, x_{1}) = ρ (x_{1}, T x_{1})

, then from (2.2) we obtain that

F (ρ (x_{1}, T x_{1})) < τ + F (H (T x_{0}, T x_{1})) \leq F (ρ (x_{1}, T x_{1}))

which is a contradiction.

Thus,

m (x_{0}, x_{1}) = ρ (x_{0}, x_{1})

. It follows from (2.2) that

\begin{matrix} \frac{1}{2} τ + F (ρ (x_{1}, T x_{1})) < τ + F (H (T x_{0}, T x_{1})) \leq F (ρ (x_{0}, x_{1})) . \end{matrix}

(2.3)

Since (F1) is satisfied, we obtain that

ρ (x_{1}, T x_{1}) < F^{- 1} (\frac{1}{2} τ + F (H (T x_{0}, T x_{1}))) .

Applying Lemma 3, there exists

x_{2} \in T x_{1}

such that

ρ (x_{1}, x_{2}) < F^{- 1} (\frac{1}{2} τ + F (H (T x_{0}, T x_{1})))

which implies

F (ρ (x_{1}, x_{2})) < \frac{1}{2} τ + F (H (T x_{0}, T x_{1})) \leq F (ρ (x_{0}, x_{1})) - \frac{1}{2} τ . (2.4)

Again from (2.1) we have that

\frac{1}{2} τ + F (ρ (x_{2}, T x_{2})) < τ + F (H (T x_{1}, T x_{2})) \leq F (ρ (x_{1}, x_{2})) (2.5)

which implies

ρ (x_{2}, T x_{2}) < F^{- 1} (\frac{1}{2} τ + F (H (T x_{1}, T x_{2}))) .

By Lemma 3, there exists

x_{3} \in T x_{2}

such that

ρ (x_{2}, x_{3}) < F^{- 1} (\frac{1}{2} τ + F (H (T x_{1}, T x_{2}))) .

Hence, we obtain that

F (ρ (x_{2}, x_{3})) < \frac{1}{2} τ + F (H (T x_{1}, T x_{2})) \leq F (ρ (x_{1}, x_{2})) - \frac{1}{2} τ . (2.6)

Inductively, we have that for all

n = 1, 2, 3, \dots,

x_{n} \in T x_{n - 1}

and

F (ρ (x_{n}, x_{n + 1})) < \frac{1}{2} τ + F (H (T x_{n - 1}, x_{n})) \leq F (ρ (x_{n - 1}, x_{n})) - \frac{1}{2} τ . (2.7)

Because (F1) is satisfied,

ρ (x_{n}, x_{n + 1}) < ρ (x_{n - 1}, x_{n}), \forall n = 1, 2, 3, \dots .

Hence, there exists

r \geq 0

such that

lim_{n \to \infty} ρ (x_{n}, x_{n + 1}) = r .

Assume that

r > 0

By Lemma 1, we have that

lim_{n \to \infty} F (ρ (x_{n}, x_{n + 1})) = lim_{n \to \infty} F (ρ (x_{n - 1}, x_{n})) = lim_{t \to r^{+}} F (t) = F (r^{+}) . (2.8)

Taking limit

n \to \infty

in (2.7) and using (2.8), we obtain that

F (r^{+}) \leq F (r^{+}) - \frac{1}{2} τ

which is a contradiction, because

τ > 0 .

Thus, we obtain that

lim_{n \to \infty} ρ (x_{n}, x_{n + 1}) = 0 .

(2.9)

Now, we show that

{x_{n}}

is a Cauchy sequence.

Assume that

{x_{n}}

is not a Cauchy sequence.

Then, there exists

ϵ > 0

for which we can find subsequences

{x_{m (k)}}

and

{x_{n (k)}}

{x_{n}}

such that

m (k)

is the smallest index for which (1.1) holds.

That is, the following are satisfied:

m (k) > n (k) > k, ρ (x_{m (k)}, x_{n (k)}) \geq ϵ and ρ (x_{m (k) - 1}, x_{n (k)}) < ϵ .

It follows from (2.1) that

\begin{matrix} F (ρ (x_{n (k) + 1}, T x_{m (k)}) < τ + F (ρ (x_{n (k) + 1}, T x_{m (k)}) \\ \leq & τ + F (H (T x_{n (k)}, T x_{m (k)}) \leq F (m (x_{n (k)}, x_{m (k)})) . \end{matrix}

(2.10)

We infer that

\begin{matrix} ϵ \leq ρ (x_{n (k)}, x_{m (k)}) \leq m (x_{n (k)}, x_{m (k)}) \\ = & max {ρ (x_{n (k)}, x_{m (k)}), ρ (x_{n (k)}, T x_{n (k)}), ρ (x_{m (k)}, T x_{m (k)}), \\ \frac{1}{2} [ρ (x_{n (k)}, T x_{m (k)}) + ρ (x_{m (k)}, T x_{n (k)})]} \\ \leq & max {ρ (x_{n (k)}, x_{m (k)}), ρ (x_{n (k)}, x_{n (k) + 1}), ρ (x_{m (k)}, x_{m (k) + 1}), \\ \frac{1}{2} [ρ (x_{n (k)}, x_{m (k) + 1}) + ρ (x_{m (k)}, x_{n (k) + 1})]} \end{matrix}

(2.11)

Taking limit as

k \to \infty

on both sides of (2.11) and using (1.2), we obtain that

lim_{k \to \infty} m (x_{n (k)}, x_{m (k)}) = ϵ . (2.12)

Since (F1) is satisfied, from (2.10) we have that

ρ (x_{n (k) + 1}, T x_{m (k)}) < F^{- 1} (τ + F (ρ (x_{n (k) + 1}, T x_{m (k)})) .

By applying Lemma 3, there exists

y_{m (k)} \in T x_{m (k)}

such that

ρ (x_{n (k) + 1}, y_{m (k)}) < F^{- 1} (τ + F (ρ (x_{n (k) + 1}, T x_{m (k)})) .

Hence,

F (ρ (x_{n (k) + 1}, y_{m (k)})) < τ + F (ρ (x_{n (k) + 1}, T x_{m (k)}) .

Thus, it follows from (2.10) that

\begin{matrix} F (ρ (x_{n (k) + 1}, y_{m (k)})) \\ < & τ + F (ρ (x_{n (k) + 1}, y_{m (k)})) < τ + F (ρ (x_{n (k) + 1}, T x_{m (k)}) \\ \leq & τ + F (H (T x_{n (k)}, T x_{m (k)}) \\ \leq & F (m (x_{n (k)}, x_{m (k)})) \end{matrix}

(2.13)

which leads to

ρ (x_{n (k) + 1}, y_{m (k)}) < m (x_{n (k)}, x_{m (k)}), \forall k = 1, 2, 3, \dots . (2.14)

By taking

lim sup

k \to \infty

in (2.14) and using (2.12), we have that

lim_{k \to \infty} sup ρ (x_{n (k) + 1}, y_{m (k)}) \leq ϵ . (2.15)

Since

ρ (x_{n (k) + 1}, T x_{m (k)}) \leq ρ (x_{n (k) + 1}, y_{m (k)}),

\begin{matrix} ρ (x_{n (k) + 1}, x_{m (k)}) \\ \leq & ρ (x_{n (k) + 1}, T x_{m (k)}) + ρ (T x_{m (k)}, x_{m (k)}) \\ \leq & ρ (x_{n (k) + 1}, y_{m (k)}) + ρ (x_{m (k) + 1}, x_{m (k)}) \end{matrix}

(2.16)

Taking

lim inf

k \to \infty

in (2.16) and using (1.2), we obtain that

ϵ \leq lim_{k \to \infty} inf ρ (x_{n (k) + 1}, y_{m (k)}) . (2.17)

It follows from (2.15) and (2.17) that

lim_{k \to \infty} ρ (x_{n (k) + 1}, y_{m (k)}) = ϵ . (2.18)

By applying Lemma 1 to (2.13) with (2.12), (2.14) and (2.18), we obtain that

F (ϵ^{+}) \leq τ + F (ϵ^{+}) \leq F (ϵ^{+})

which leads to a contradiction.

Hence,

{x_{n}}

is a Cauchy sequence.

From the completeness of E, there exists

x_{*} = lim_{n \to \infty} x_{n} \in E .

It follows from (2.1) that

\begin{matrix} F (ρ (x_{n + 1}, T x_{*})) < τ + F (ρ (x_{n + 1}, T x_{*})) \\ \leq & τ + F (H (T x_{n}, T x_{*})) \leq F (m (x_{n}, x_{*})), \end{matrix}

(2.19)

where

m (x_{n}, x_{*}) = max {ρ (x_{n}, x_{*}), ρ (x_{n}, x_{n + 1}), ρ (x_{*}, T x_{*}),

\frac{1}{2} [ρ (x_{*}, x_{n + 1}) + ρ (x_{n}, T x_{*})]}

Since (F1) is satisfied, from (2.19) we have that

ρ (x_{n + 1}, T x_{*}) < m (x_{n}, x_{*}) . (2.20)

We have that

lim_{n \to \infty} ρ (x_{n + 1}, T x_{*}) = lim_{n \to \infty} m (x_{n}, x_{*}) = ρ (x_{*}, T x_{*}) . (2.21)

Assume that

ρ (x_{*}, T x_{*}) > 0

By Lemma 1, we have that

\begin{matrix} lim_{n \to \infty} F (ρ (x_{n + 1}, T x_{*})) = lim_{n \to \infty} F (m (x_{n}, x_{*})) \\ = & lim_{t \to ρ {(x_{*}, T x_{*})}^{+}} F (t) = F (ρ {(x_{*}, T x_{*})}^{+}) . \end{matrix}

(2.22)

Applying (2.22) to (2.19), we obtain that

F (ρ {(x_{*}, T x_{*})}^{+}) \leq τ + F (ρ {(x_{*}, T x_{*})}^{+}) \leq F (ρ {(x_{*}, T x_{*})}^{+})

which leads to a contradiction.

Hence,

ρ (x_{*}, T x_{*}) = 0

, and hence

x_{*} \in T x_{*} .

□

The following example interprets Theorem 2.

Example 1.

Let

E = [0, 1]

and

ρ (x, y) = | x - y |, \forall x, y \in E

Then

(E, ρ)

is a complete metric space.

Define a set-valued map

T : E \to C L (E)

T x = \{\begin{matrix} {1}, & (x = 0) \\ {\frac{2}{5}, \frac{1}{2}}, & (0 < x \leq 1) . \end{matrix}

Let

τ = ln \frac{3}{2}

and

F (t) = ln t, \forall t > 0

We show that T is a generalized F-contraction.

We have that for all

x, y \in E

H (T x, T y) > 0 ⟺ (x = 0 and 0 < y \leq 1), (0 < x \leq 1 and y = 0) .

1^{\circ}

: Let

x = 0 and 0 < y \leq 1

Then, we obtain that

\begin{matrix} τ + F (H (T x, T y)) - F (ρ (x, T x)) \\ = & τ + F (\frac{3}{5}) - F (1) \\ = & ln \frac{3}{2} + ln \frac{3}{5} - ln 1 \\ = & ln 9 - ln 10 < 0 . \end{matrix}

Thus,

τ + F (H (T x, T y)) < F (ρ (x, T x))

which implies

τ + F (H (T x, T y)) < F (m (x, y)) .

2^{\circ}

: Let

0 \leq x < 1 and y = 1

. Then, we have that

\begin{matrix} τ + F (H (T x, T y)) - F (ρ (y, T y)) \\ = & τ + F (\frac{3}{5}) - F (1) \\ = & ln \frac{3}{2} + ln \frac{3}{5} - ln 1 \\ = & ln 9 - ln 10 < 0 . \end{matrix}

Thus,

τ + F (H (T x, T y)) < F (ρ (y, T y))

which leads to

τ + F (H (T x, T y)) < F (m (x, y)) .

Hence, F is generalized F-contraction. The assumptions of Theorem 2 are satisfied. By Theorem 2, T possesses two fixed points,

\frac{2}{5}

and

\frac{1}{2} .

Remark 1.

Theorem 2 is a positive answer to Question 4.3 of [9].

By Theorem 2, we have the following result.

Corollary 1.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

τ + F (H (T x, T y)) \leq F (l (x, y))

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function, and

\begin{matrix} l (x, y) = & max {ρ (x, y), \frac{1}{2} [ρ (x, T x) + ρ (y, T y)], \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]} . \end{matrix}

(F 1)

is satisfied, then T possesses a fixed point.

Corollary 2.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C B (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

τ + F (H (T x, T y)) \leq F (ρ (x, y))

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function. If

(F 1)

is satisfied, then T possesses a fixed point.

Corollary 3.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

\begin{matrix} τ + F (H (T x, T y)) \\ \leq & F (a ρ (x, y) + b ρ (x, T x) + c ρ (y, T y) + e [ρ (x, T y) + ρ (y, T x)]) \end{matrix}

(2.23)

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function, and

a, b, c, e \geq 0

and

a + b + c + 2 e = 1

. If

(F 1)

is satisfied, then T possesses a fixed point.

Proof.

It follows from (2.23) that

\begin{matrix} τ + F (H (T x, T y)) \\ \leq & F (a ρ (x, y) + b ρ (x, T x) + c ρ (y, T y) + e [ρ (x, T y) + ρ (y, T x)]) \\ = & F (a ρ (x, y) + b ρ (x, T x) + c ρ (y, T y)] + 2 e \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]) \\ \leq & F ((a + b + c + 2 e) max {ρ (x, y), ρ (x, T x), ρ (y, T y), \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]}) \\ = & F (m (x, y)) . \end{matrix}

By Theorem 2, T possesses a fixed point. □

Corollary 4.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

\begin{matrix} τ + F (H (T x, T y)) \\ \leq & F (a ρ (x, y) + b [ρ (x, T x) + ρ (y, T y)] + c [ρ (x, T y) + ρ (y, T x)]) \end{matrix}

(2.24)

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function, and

a, b, c \geq 0

and

a + 2 b + 2 c = 1

. If

(F 1)

is satisfied, then T possesses a fixed point.

Proof.

It follows from (2.24) that

\begin{matrix} τ + F (H (T x, T y)) \\ \leq & F (a ρ (x, y) + b [ρ (x, T x) + ρ (y, T y)] + c [ρ (x, T y) + ρ (y, T x)]) \\ = & F (a ρ (x, y) + 2 b \frac{1}{2} [ρ (x, T x) + ρ (y, T y)] + 2 c \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]) \\ \leq & F ((a + 2 b + 2 c) max {ρ (x, y), \frac{1}{2} [ρ (x, T x) + ρ (y, T y)], \frac{1}{2} [ρ (x, T y) + ρ (y, T x)]}) \\ = & F (l (x, y)) . \end{matrix}

By Corollary 2.2, T possesses a fixed point. □

Corollary 5.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

τ + F (H (T x, T y)) \leq F (\frac{1}{2} [ρ (x, T x) + ρ (y, T y)])

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function. If

(F 1)

is satisfied, then T possesses a fixed point.

Corollary 6.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for all

x, y \in E with H (T x, T y) > 0

τ + F (H (T x, T y)) \leq F (\frac{1}{2} [ρ (x, T y) + ρ (y, T x)])

where

τ > 0

and

F : (0, \infty) \to (- \infty, \infty)

is a function. If

(F 1)

is satisfied, then T possesses a fixed point.

Theorem 3.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is an Işik type set-valued contraction, i.e. for each

x, y \in E

and each

u \in T x

, there exists

v \in T y

such that

ρ (u, v) \leq ϕ (ρ (x, y)) - ϕ (ρ (u, v))

(2.25)

where

ϕ : [0, \infty) \to [0, \infty)

is a function such that

lim_{t \to 0^{+}} ϕ (t) = 0 .

(2.26)

Then, T possesses a fixed point.

Proof.

Let

x_{0} \in E

, and let

x_{1} \in T x_{0}

. Then there exits

x_{2} \in T x_{1}

such that

ρ (x_{1}, x_{2}) \leq ϕ (ρ (x_{0}, x_{1})) - ϕ (ρ (x_{1}, x_{2})) .

Again, there exists

x_{3} \in T x_{2}

such that

ρ (x_{2}, x_{3}) \leq ϕ (ρ (x_{1}, x_{2})) - ϕ (ρ (x_{2}, x_{3})) .

Inductively, we have a sequence

{x_{n}} \subset E

such that for all

n = 1, 2, 3, \dots

x_{n} \in T x_{n - 1} and ρ (x_{n}, x_{n + 1}) \leq ϕ (ρ (x_{n - 1}, x_{n})) - ϕ (ρ (x_{n}, x_{n + 1})) . (2.27)

From (2.27)

{ϕ (ρ (x_{n - 1}, x_{n}))}

is a non-increasing sequence and bounded below by 0. Hence, there exists

r \geq 0

such that

lim_{n \to \infty} ϕ (ρ (x_{n - 1}, x_{n})) = r .

We show that

{x_{n}}

is a Cauchy sequence.

Let

m, n

be any positive integers such that

m > n

Then, we have that

\begin{matrix} ρ (x_{n}, x_{m}) \\ \leq & ρ (x_{n}, x_{n + 1}) + ρ (x_{n + 1}, x_{n + 2}) + \dots + ρ (x_{m - 1}, x_{m}) \\ \leq & ϕ (ρ (x_{n - 1}, x_{n})) - ϕ (ρ (x_{m - 1}, x_{m})) \\ \leq & ϕ (ρ (x_{n - 1}, x_{n})) - r . \end{matrix}

(2.28)

Letting

m, n \to \infty

in (2.28), we obtain that

lim_{n, m \to \infty} ρ (x_{n}, x_{m}) = 0 .

Thus,

{x_{n}}

is a Cauchy sequence. It follows from the completeness of E that

x_{*} = lim_{n \to \infty} x_{n} exists .

(2.29)

Now, we show that

x_{*}

is a fixed point for T.

It follows from (2.25) that for

x_{n} \in T x_{n - 1}

, there exists

v \in T x_{*}

such that

ρ (x_{n}, v) \leq ϕ (ρ (x_{n - 1}, x_{*})) - ϕ (ρ (x_{n}, v)) \leq ϕ (ρ (x_{n - 1}, x_{*})) .

(2.30)

Taking limit

n \to \infty

in equation (2.30) and using (2.26), we infer that

lim_{n \to \infty} ρ (x_{n}, v) = 0

which implies

x_{*} = v \in T x_{*} .

□

Example 2.

Let

E = {x_{n} : x_{n} = 1 + 2 + 3 + \dots + n, n = 1, 2, 3, \dots}

and

ρ (x, y) = | x - y |, \forall x, y \in E

Then,

(E, ρ)

is a complete metric space.

Define a map

T : E \to C L (E)

T x = \{\begin{matrix} {x_{1}}, & (x = x_{1}) \\ {x_{1}, x_{2}, x_{3}, \dots x_{n - 1}}, & (x = x_{n}) . \end{matrix}

Let

ϕ (t) = \frac{1}{2} t, \forall t \geq 0

We show that condition (2.25) is satisfied.

Consider the following two cases.

1^{\circ}

: Let

x = x_{1} and y = x_{n}, n = 2, 3, 4, \dots

Then, for

u = x_{1} \in T x

, there exists

v = x_{1} \in T y

such that

ρ (u, v) = 0 < \frac{1}{2} ρ (x_{1}, x_{n}) = ϕ (ρ (x_{1}, x_{n})) = ϕ (ρ (x_{1}, x_{n})) - ϕ (ρ (u, v)) .

2^{\circ}

: Let

x = x_{n} and y = x_{m}, m > n, n = 2, 3, 4, \dots

For

u = x_{k} \in T x (k = 1, 2, 3, \dots, n - 1)

, there exists

v = x_{k} \in T y

such that

ρ (u, v) = 0 < \frac{1}{2} ρ (x_{n}, x_{m}) = ϕ (ρ (x_{n}, x_{m})) = ϕ (ρ (x_{n}, x_{m})) - ϕ (ρ (u, v)) .

This show that T satisfies condition (2.25). Thus, all conditions of Theorem 3 hold. From Theorem 3T possesses a fixed point,

x_{*} = x_{1}

Corollary 7.

Let

(E, ρ)

be a complete metric space. Suppose that

T : E \to C L (E)

is a set-valued map such that for each

x, y \in E

H (T x, T y) < ϕ (ρ (x, y)) - ϕ (H (T x, T y)),

where

ϕ : [0, \infty) \to [0, \infty)

is a strictly increasing function such that

lim_{t \to 0^{+}} ϕ (t) = 0 .

Then, T possesses a fixed point.

Proof.

Let

x, y \in E

and let

u \in T x

ϕ

is strictly increasing,

ρ (u, T y) + ϕ (ρ (u, T y)) < ϕ (ρ (x, y)) .

Applying Lemma 4, there exists

v \in T y

such that

ρ (u, v) + ϕ (ρ (u, v)) < ϕ (ρ (x, y)) .

By Theorem 3, T possesses a fixed point. □

From Theorem 3 we have the following result.

Corollary 8.

[14] Let

(E, ρ)

be a complete metric space. Suppose that

f : E \to E

is a map such that for each

x, y \in E

ρ (f x, f y) \leq ϕ (ρ (x, y)) - ϕ (ρ (f x, f y))

where

ϕ : [0, \infty) \to [0, \infty)

is a function such that

lim_{t \to 0^{+}} ϕ (t) = 0 .

Then, f possesses a fixed point.

3. Application

In this section, we give an application of our result to integral inclusion.

Let

[a, b] \subset R

be a closed interval, and let

C ([a, b], R)

be the family of continuous mapping from

[a, b] into R

Let

E = C ([a, b], R)

and

ρ (x, y) = {sup}_{t \in [a, b]} | x (t) - y (t) |

for all

x, y \in E

Then,

(E, ρ)

is a complete metric space.

Consider the Fredholm type integral inclusion:

x (t) \in \int_{a}^{b} K (t, s, x (s)) d s + f (t), t \in [a, b]

(3.1)

where

f \in E

K : [a, b] \times [a, b] \times R \to C B (R)

, and

x \in E

is the unknown function.

Suppose that the following conditions are satisfied:

( $1^{\circ}$ ): for each $x \in E, K (\cdot, \cdot, x (s)) = K_{x} (\cdot, \cdot)$ is continuous;
( $2^{\circ}$ ): there exists a continuous function $Z : [a, b] \times [a, b] \to [0, \infty)$ such that for all $t, s \in [a, b]$ and all $u, v \in E$ ,

$| k_{u} (t, s) - k_{v} (t, s) | \leq Z (t, s) ρ (u (s), v (s))$

where $k_{u} (t, s) \in K_{u} (t, s), k_{v} (t, s) \in K_{v} (t, s);$
( $3^{\circ}$ ): there exists $α > 1$ such that

$sup_{t \in [a, b]} \int_{a}^{b} Z (t, s) d s \leq \frac{1}{2 + α} .$

Theorem 4.

Let

(E, ρ)

be a complete metric space. If conditions

(1^{\circ})

(2^{\circ})

and

(3^{\circ})

are satisfied, then the integral inclusion (3.1) has a solution.

Proof.

Define a set-valued map

T : E \to C B (E)

T x = {y \in E : y (t) \in \int_{a}^{b} K (t, s, x (s)) d s + f (t), t \in [a, b]} .

Let

x \in E

be given.

For the set-valued map

K_{x} (t, s) : [a, b] \times [a, b] \to C B (R)

, by applying Michael’s selection theorem, there exists a continuous map

k_{x} (t, s) : [a, b] \times [a, b] \to R

such that

k_{x} (t, s) \in K_{x} (t, s), \forall t, s \in [a, b] .

Thus,

\int_{a}^{b} k_{x} (t, s) d s + f (t) \in T x,

and so

T x \neq \emptyset .

Since f and

k_{x}

are continuous,

T x \in C B (E)

for each

x \in E

Let

y_{1} \in T x_{1}

Then,

y_{1} (t) \in \int_{a}^{b} K (t, s, x_{1} (s)) d s + f (t), t \in [a, b] .

Hence, there exists

k_{x_{1}} (t, s) \in K_{x_{1}} (t, s), \forall t, s \in [a, b]

such that

y_{1} (t) = \int_{a}^{b} k_{x_{1}} (t, s) d s + f (t), \forall t, s \in [a, b] .

It follows from (

2^{\circ}

) that there exists

z (t, s) \in K_{x_{2}} (t, s)

such that

| k_{x_{1}} (t, s) - z (t, s) | \leq Z (t, s) ρ (x_{1} (s), x_{2} (s)), \forall t, s \in [a, b] .

Let

U : [a, b] \times [a, b] \to C B (R)

be defined by

U (t, s) = K_{x_{2}} (t, s) \cap {u \in R : ρ (k_{x_{1}} (t, s), u) \leq ρ (x_{1} (s), x_{2} (s))} .

From (

1^{\circ}

) U is continuous. Hence, it follows that there exists a continuous map

k_{x_{2}} : [a, b] \times [a, b] \to R

such that

k_{x_{2}} (t, s) \in U (t, s), \forall t, s \in [a, b] .

Let

y_{2} (t) = \int_{a}^{b} k_{x_{2}} (t, s) d s + f (t), \forall t, s \in [a, b] .

Then,

y_{2} (t) \in \int_{a}^{b} K_{x_{2}} (t, s) d s + f (t) = \int_{a}^{b} K (t, s, x_{2} (s)) d s + f (t), \forall t, s \in [a, b],

and so

y_{2} \in T x_{2} .

Thus, we obtain that

\begin{matrix} ρ (y_{1}, y_{2}) = | \int_{a}^{b} k_{x_{1}} (t, s) - k_{x_{2}} (t, s) d s | \\ \leq & sup_{t \in [a, b]} \int_{a}^{b} | k_{x_{1}} (t, s) - k_{x_{2}} (t, s) | d s \\ \leq & sup_{t \in [a, b]} \int_{a}^{b} Z (t, s) d s ρ (x_{1} (s), x_{2} (s)) \\ \leq & \frac{1}{2 + α} ρ (x_{1} (s), x_{2} (s)) . \end{matrix}

Thus, we have that

(1 + \frac{1}{2} α) δ (T x_{1}, T x_{2}) \leq \frac{1}{2} ρ (x_{1}, x_{2})

which implies

(1 + \frac{1}{2} α) H (T x_{1}, T x_{2}) \leq \frac{1}{2} ρ (x_{1}, x_{2}) .

Hence, we obtain that

\begin{matrix} H (T x_{1}, T x_{2})) \leq ϕ (ρ (x_{1}, x_{2})) - ϕ (α H (T x_{1}, T x_{2})) \\ < & ϕ (ρ (x_{1}, x_{2})) - ϕ (H (T x_{1}, T x_{2})) where ϕ (t) = \frac{1}{2} t, \forall t \geq 0 . \end{matrix}

By Corollary 7, T possesses a fixed point, and hence the integral inclusion (3.1) has a solution. □

4. Conclusions

Our results are generalizations and extensions of F-contractions and Işik contractions to set-valued maps on metric spaces. We give a positive answer to the Question 4.3 of [9] and an application to integral inclusion.

Data Availability Statement

Not applicable.

Acknowledgments

The author express his gratitude to the referees for careful reading and giving variable comments. This research was supported by Hanseo University.

Conflicts of Interest

The author declares no conflict of interest.

References

J. Ahmad, A. Al-Rawashdeh, A. Azam, Some new fixed point theorems for generalized F-contractions in complete metric spaces, Fixed Point Theory Appl. (2015), 2015:80.
W.M. Alfaqih, M. Imdad, R. Gubran, An observation on F-weak contractions and discontinuity at the fixed point with an applications, J. Fixed Point Theory Appl. (2020) 22:66.
M. Arshad, S. Khan and J. Ahmad, Fixed point results for F-contractions involving some new rational expressions, JP Journal of Fixed Point Theory and Applications, 11(1) (2016), 79-97.
S.V. Bedre, Remarks on F-weak contractions and discontinuity at the fixed point, Advances in theTheory of Nonlinear Analysis and its Applications 4(4)(2020), 260-265.
V. Berinde, M. Păcurar, The role of the Pompeiu-Hausdorff metric in fixed point theory, Creat. Math. Inform. 22(2)(2013), 143-150.
S.H. Cho, J.S. Bae, Fixed points of weak α-contraction type maps, Fixed Point Theory and Appl. 2014, 175(2014).
M. Cosentino, P. Vetro, Fixed point results for F-contractive mappings of Hardy Rogers-Type, Filomat 28(4) (2014), 715-722.
N.V. Dung, V.L. Hang, A fixed point theorem for generalized F-contractions on complete metric spaces, Vietnam J. Math. 43(2015), 743-753.
N. Fabiano, Z. Kadelburg, N. Mirkov, V. Š. Čavić, S. Radenović, On F-contractions: A Survey, Contemporary Mathematics (2022) 3:3, 327.
H. Huang, Z.D. Mitrović, K. Zoto, S. Radenović, On convex F-contraction in b-metric spaces, Axioms 2021, 10,71. [CrossRef]
A. Hussain, Ćirić type α-ψF-contraction involving fixed point on a closed ball, Honam Math. J. 41(2019), 19-34.
A. Hussain and M. Arshad, New type of multivalued F-Contraction involving fixed Point on Closed Ball, J. Math. Comp. Sci. 10 (2017), 246-254.
N. Hussain and P. Salimi, Suzuki-wardowski type fixed point theorems for α-GF-contractions, Taiwanese J. Math., 20 (2014). [CrossRef]
H. Işik, B. Mohammadi, M.R. Haddadi, V. Parvaneh, On a new generalization of Banach contraction principle with application, Mathematics 2019,7,862. [CrossRef]
N. Konwar, P. Debnath, Fixed point results for a family of Interpolative F-contractions in b-metric spaces, Axioms 2022, 11, 621. [CrossRef]
G. Minak, A. Halvaci and I. Altun, Ćirić type generalized F-contractions on complete metric spaces and fixed point results, Filomat, 28(6) (2014), 1143-1151.
H. Piri, P. Kumam, Some fixed point theorems concerning F-contraction in complete metric spaces, Fixed Point Theory and Appl 2014, Article ID 210(2014). [CrossRef]
H. Piri, P. Kumam, Fixed point theorems for generalized F-Suzuki-contraction mappings in complete b-metric spaces, Fixed Point Theory and Applications (2016) 2016:90. [CrossRef]
N.A. Secelean, Iterated function systems consisting of F-contractions, Fixed Point Theory Appl. (2013), Article ID 277 (2013). [CrossRef]
T. Stephen, Y. Rohen, M.K. Singh, K.S. Devi, Some rational F-contractions in b-metric spaces and fixed points, Nonlinear Functional Analysis and Applications Vol. 27, No. 2 (2022), pp. 309-322.
D. Wardowski, Fixed point theory of a new type of contractive mappings in complete metric spacs, Fixed Point Theory and Appl 2012, Article ID 94(2012). [CrossRef]
D. Wardowski, N.V. Dung, Fixed points f-weak contractions on complete metric spaces, Demonstr. Math. 1(2014), 146-155.
M. Younis, N. Mirkov, A. Savić, M. Pantović, S. Radenović, Some critical remarks on recent results concerning F-contractions in b-metric spaces, CUBO, A mathematical Journal, Accepted: 27 December 2022.

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

MDPI Initiatives

Important Links

Choose an area of interest and we will send you notifications of new preprints at your preferred frequency.

Disclaimer

Fixed Point Theorems for Generalized Set Valued F-contractions in Metric Spaces

Abstract

1. Introduction

2. Fixed point results

3. Application

4. Conclusions

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

MDPI Initiatives

Important Links

Subscribe