Contraction on -Metric Spaces

Hindawi Journal of Function Spaces Volume 2017, Article ID 9389768, 11 pages https://doi.org/10.1155/2017/9389768

Research Article Common Fixed Points of Four Maps Satisfying 𝐹-Contraction on 𝑏-Metric Spaces Muhammad Nazam,1 Ma Zhenhua,2,3 Sami Ullah Khan,1,4 and Muhammad Arshad1 1

Department of Mathematics and Statistics, International Islamic University, H-10, Islamabad, Pakistan Department of Mathematics and Physics, Hebei University of Architecture, Zhangjiakou 075024, China 3 School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 10001, China 4 Department of Mathematics, Gomal University D. I. Khan, Khyber Pakhtunkhwa 29050, Pakistan 2

Correspondence should be addressed to Ma Zhenhua; mazhenghua [email protected] Received 16 October 2017; Accepted 7 December 2017; Published 28 December 2017 Academic Editor: Ahmad S. Al-Rawashdeh Copyright © 2017 Muhammad Nazam et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We manifest some common fixed point theorems for four maps satisfying C´ır´ıc type 𝐹-contraction and Hardy-Rogers type 𝐹contraction defined on complete 𝑏-metric spaces. We apply these results to infer several new and old corresponding results. These results also generalize some results obtained previously. We dispense examples to validate our results.

1. Introduction and Preliminaries After the famous Banach Contraction Principle, a large number of researchers revealed many fruitful generalizations of Banach’s fixed point theorem. One of these generalizations is known as 𝐹-contraction presented by Wardowski [1]. Wardowski [1] evinced the fact that every 𝐹-contraction defined on complete 𝑏-metric space has a unique fixed point. The concept of 𝐹-contraction proved to be another milestone in fixed point theory and numerous research papers on 𝐹contraction have been published (see, e.g., [2–8]). Recently, Cosentino and Vetro [9] established a fixed point result for Hardy-Rogers type 𝐹-contraction and Mınak et al. [10] presented a fixed point result for C´ır´ıc type generalized 𝐹contraction. In 1989, Bakhtin [11] investigated the concept of 𝑏-metric spaces; however, Czerwik [12] initiated study of fixed point of self-mappings in 𝑏-metric spaces and proved an analogue of Banach’s fixed point theorem. Since then, numerous research articles have been published comprising fixed point theorems for various classes of single-valued and multivalued operators in 𝑏-metric spaces (see, e.g., [13–19]). We shall bring into use the idea of C´ır´ıc type 𝐹-contraction and Hardy-Rogers type 𝐹-contraction comprising four

self-mappings defined on 𝑏-metric space. We present common fixed point results for four self-maps satisfying C´ır´ıc type and Hardy-Rogers type 𝐹-contraction on 𝑏-metric space. We apply our results to infer several new and old results. We denote (0, ∞) by R+ , [0, ∞) by R+0 , (−∞, +∞) by R, and set of natural numbers by N. We bring back into reader’s mind some definitions and properties of 𝑏-metric. Definition 1 (see [12]). Let 𝑀 be a nonempty set and 𝑠 ≥ 1 be a real number. A function 𝑑𝑏 : 𝑀 × 𝑀 → [0, ∞) is said to be a 𝑏-metric if, for all 𝜃, 𝜌, 𝜎 ∈ 𝑀, one has (𝑑𝑏 1) 𝜃 = 𝜌 if and only if 𝑑𝑏 (𝜃, 𝜌) = 0, (𝑑𝑏 2) 𝑑𝑏 (𝜃, 𝜌) = 𝑑𝑏 (𝜌, 𝜃), (𝑑𝑏 3) 𝑑𝑏 (𝜃, 𝜌) ≤ 𝑠[𝑑𝑏 (𝜃, 𝜎) + 𝑑𝑏 (𝜎, 𝜌)]. In this case, the pair (𝑀, 𝑑𝑏 , 𝑠) is called a 𝑏-metric space (with coefficient 𝑠). Definition 1 allows us to remark that the class of 𝑏-metric spaces is effectually more general than that of metric spaces because a 𝑏-metric is a metric when 𝑠 = 1. The following example describes the significance of a 𝑏-metric.

2

Journal of Function Spaces

Example 2. Let (𝑀, 𝑑) be a metric space and 𝑑𝑏 (𝜃, 𝜌) = (𝑑(𝜃, 𝜌))𝑟 , 𝑟 > 1 is a real number. Then 𝑑𝑏 is 𝑏-metric space with 𝑠 = 2𝑟−1 . Apparently, (𝑑𝑏 1) and (𝑑𝑏 2) of Definition 1 are satisfied. If 1 < 𝑟 < ∞, then the convexity of the function 𝑓(𝜃) = 𝜃𝑟 (𝜃 > 0) implies (

𝑗+𝑙 1 ) ≤ (𝑗𝑟 + 𝑙𝑟 ) 2 2

(𝑟𝑛 , 𝑠𝑛 ) = 0, lim 𝑟𝑛 = 𝑡 for some 𝑡 ∈ 𝑀,

𝑟

𝑟

𝑟

≤ 2𝑟−1 [(𝑑 (𝜃, 𝜌)) + (𝑑 (𝜌, 𝜎)) ]

For the notions like convergence, completeness, and Cauchy sequence in the setting of 𝑏-metric spaces, the reader is referred to Aghajani et al. [20], Czerwik [12], AminiHarandi [21], Huang et al. [13], Hussain et al. [14], Radenović et al. [15], Khamsi and Hussain [16], Latif et al. [22], Parvaneh et al. [17], Roshan et al. [18], and Zabihi and Razani [23]. In line with Wardowski [1], Cosentino et al. [24] investigated a nonlinear function 𝐹 : R+ → R complying with the following axioms: (𝐹1 ) 𝐹 is strictly increasing. (𝐹2 ) For each sequence {𝑟𝑛 } of positive numbers, the following equation holds: lim𝑛→∞ 𝑟𝑛 = 0 if and only if lim𝑛→∞ 𝐹(𝑟𝑛 ) = −∞. (𝐹3 ) There exists 𝜃 ∈ (0, 1) such that lim𝑟𝑛 →0+ (𝑟𝑛 )𝜃 𝐹(𝑟𝑛 ) = 0. (𝐹4 ) 𝜏 + 𝐹(𝑠𝑟𝑛 ) ≤ 𝐹(𝑟𝑛−1 ) implies 𝜏 + 𝐹(𝑠𝑛 𝑟𝑛 ) ≤ 𝐹(𝑠𝑛−1 𝑟𝑛−1 ), for each 𝑛 ∈ N and some 𝜏 > 0. We denote by Δ 𝐹𝑠 the set of all functions satisfying the conditions (𝐹1 )–(𝐹4 ). Example 3. Let 𝐹 : R+ → R be defined by (a) 𝐹(𝑟) = ln(𝑟), (b) 𝐹(𝑟) = 𝑟 + ln(𝑟). It is easy to check that (a) and (b) are members of Δ 𝐹𝑠 . Definition 4 (see [25]). Let (𝑀, 𝑑𝑏 , 𝑠) be a 𝑏-metric space. The pair {𝑓, 𝑔} is said to be compatible if and only if (𝑓𝑔 (𝑟𝑛 ) , 𝑔𝑓 (𝑟𝑛 )) = 0,

𝑛→∞

for some 𝑡 ∈ 𝑀.

Proof. Due to the triangular inequality we have 𝑑𝑏 (𝑠𝑛 , 𝑡) ≤ 𝑠 [𝑑𝑏 (𝑠𝑛 , 𝑟𝑛 ) + 𝑑𝑏 (𝑟𝑛 , 𝑡)] ,

(5)

and the result follows after applying limit as 𝑛 → ∞.

Therefore, 𝑑𝑏 (𝜃, 𝜌) ≤ 𝑠[𝑑𝑏 (𝜃, 𝜌) + 𝑑𝑏 (𝜌, 𝜎)], where 𝑠 = 2𝑟−1 , which shows that (𝑀, 𝑑𝑏 , 𝑠) is a 𝑏-metric space. Nevertheless, if (𝑀, 𝑑) is a metric space, then (𝑀, 𝑑𝑏 , 𝑠) may not be a metric space. Indeed, if 𝑀 = R and 𝑑(𝜃, 𝜌) = |𝜃− 𝜌| (a usual metric), then 𝑑𝑏 (𝜃, 𝜌) = [𝑑(𝜃, 𝜌)]2 does not define a metric on 𝑀.

lim 𝑓 (𝑟𝑛 ) = lim 𝑔 (𝑟𝑛 ) = 𝑡

Then lim𝑛→∞ 𝑠𝑛 = 𝑡.

(2)

= 2𝑟−1 [𝑑𝑏 (𝜃, 𝜌) + 𝑑𝑏 (𝜌, 𝜎)] .

whenever {𝑟𝑛 } is a sequence in 𝑀 so that

(4)

𝑛→∞

(1)

𝑑𝑏 (𝜃, 𝜎) = (𝑑 (𝜃, 𝜎))𝑟 ≤ (𝑑 (𝜃, 𝜌) + 𝑑 (𝜌, 𝜎))

𝑛→∞

lim 𝑑 𝑛→∞ 𝑏

𝑟

that gives (𝑗 + 𝑙)𝑟 ≤ 2𝑟−1 (𝑗𝑟 + 𝑙𝑟 ). Thus, for all 𝜃, 𝜌, 𝜎 ∈ 𝑀, one has


Lemma 5. Let (𝑀, 𝑑𝑏 , 𝑠) be a 𝑏-metric space. If there exist two sequences {𝑟𝑛 }, {𝑠𝑛 } such that

(3)

2. Círíc Type Fixed Point Theorem In this section we set up a fixed point theorem for four selfmaps satisfying C´ır´ıc type 𝐹-contraction. We explain this theorem through an example and discuss its consequences. Theorem 6. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 and 𝜏 > 0, the inequality (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M (𝑟1 , 𝑟2 ))) ,

(6)

holds, where M (𝑟1 , 𝑟2 ) = max {𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) ,

(7)

𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 )) }, 2𝑠 then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible. Proof. Let 𝑟0 ∈ 𝑀. As 𝑓(𝑀) ⊆ 𝑇(𝑀), there exists 𝑟1 ∈ 𝑀 such that 𝑓(𝑟0 ) = 𝑇(𝑟1 ). Since 𝑔(𝑟1 ) ∈ 𝑆(𝑀), we can choose 𝑟2 ∈ 𝑀 such that 𝑔(𝑟1 ) = 𝑆(𝑟2 ). In general, 𝑟2𝑛+1 and 𝑟2𝑛+2 are chosen in 𝑀 such that 𝑓(𝑟2𝑛 ) = 𝑇(𝑟2𝑛+1 ) and 𝑔(𝑟2𝑛+1 ) = 𝑆(𝑟2𝑛+2 ). Define a sequence {𝑗𝑛 } in 𝑀 such that 𝑗2𝑛 = 𝑓(𝑟2𝑛 ) = 𝑇(𝑟2𝑛+1 ) and 𝑗2𝑛+1 = 𝑔(𝑟2𝑛+1 ) = 𝑆(𝑟2𝑛+2 ) for all 𝑛 ≥ 0; we show that {𝑗𝑛 } is a Cauchy sequence. Assume that 𝑑𝑏 (𝑓(𝑟2𝑛 ), 𝑔(𝑟2𝑛+1 )) > 0; then from contractive condition (6), we get 𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) = 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 ))) ≤ 𝐹 (M (𝑟2𝑛 , 𝑟2𝑛+1 )) − 𝜏,

(8)


3

for all 𝑛 ∈ N ∪ {0}, where

To prove {𝑗𝑛 } as a Cauchy sequence, we use (19) and, for 𝑚 > 𝑛 ≥ 𝑛1 , we consider

M (𝑟2𝑛 , 𝑟2𝑛+1 ) = max {𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) ,

𝑚−1

𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑆 (𝑟2𝑛 )) , 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 )) , 𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) } 2𝑠

𝑖=𝑛

(9)

= max {𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) , 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛−1 ) , 𝑑𝑏 (𝑗2𝑛+1 , 𝑗2𝑛 ) ,

𝑖=𝑛

𝑖=𝑛

1 𝑛1/𝜅

.

(20)

lim 𝑓 (𝑟2𝑛 ) = lim 𝑇 (𝑟2𝑛+1 ) = lim 𝑔 (𝑟2𝑛+1 )

𝑛→∞

𝑛→∞

𝑛→∞

(21)

= lim 𝑆 (𝑟2𝑛+2 ) = 𝜐. 𝑛→∞

= max {𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) , 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )} .

We show that 𝜐 is a common fixed point of 𝑓, 𝑔, 𝑆, and 𝑇. Since 𝑆 is continuous, therefore,

If M(𝑟2𝑛 , 𝑟2𝑛+1 ) = 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ), then 𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) ≤ 𝐹 (𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) − 𝜏,

𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) ≤ 𝐹 (𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 )) − 𝜏,

lim 𝑆𝑓 (𝑟2𝑛 ) = 𝑆 (𝜐) = lim 𝑆2 (𝑟2𝑛+2 ) .

𝑛→∞

(10)

(22)

𝑛→∞

Since, the pair {𝑓, 𝑆} is compatible, so,

which is a contradiction due to 𝐹1 . Therefore,


(11)

(𝑓𝑆 (𝑟2𝑛 ) , 𝑆𝑓 (𝑟2𝑛 )) = 0, (23)

by Lemma 5 lim 𝑓𝑆 (𝑟2𝑛 ) = 𝑆 (𝜐) .

for all 𝑛 ∈ N. Similarly,

𝑛→∞

𝐹 (𝑠𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 )) ≤ 𝐹 (𝑑𝑏 (𝑗2𝑛−2 , 𝑗2𝑛−1 )) − 𝜏,

(12)

for all 𝑛 ∈ N. Hence, from (11) and (12), we have 𝐹 (𝑠𝑑𝑏 (𝑗𝑛 , 𝑗𝑛+1 )) ≤ 𝐹 (𝑑𝑏 (𝑗𝑛−1 , 𝑗𝑛 )) − 𝜏,

𝜏 + 𝐹 (𝑠𝑛 𝑏𝑛 ) ≤ 𝐹 (𝑠𝑛−1 𝑏𝑛−1 ) , 𝑛 ∈ N.

(14)

Repeating the process, we obtain 𝐹 (𝑠𝑛 𝑏𝑛 ) ≤ 𝐹 (𝑏0 ) − 𝑛𝜏, 𝑛 ∈ N.

(15)

On taking limit 𝑛 → ∞ in (15), we have lim𝑛→∞ 𝐹(𝑠𝑛 𝑏𝑛 ) = −∞; due to axiom (𝐹2 ) we get lim𝑛→∞ 𝑠𝑛 𝑏𝑛 = 0 and (𝐹3 ) implies that there exists 𝜅 ∈ (0, 1) such that 𝜅

lim (𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑠𝑛 𝑏𝑛 ) = 0.

(16)

𝑛→∞

Now put 𝑟1 = 𝑆(𝑟2𝑛 ) and 𝑟2 = 𝑟2𝑛+1 in (6) and suppose on the contrary that 𝑑𝑏 (𝑆(𝜐), 𝜐) > 0; we obtain 𝐹 (𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 )))

(13)

for all 𝑛 ∈ N. Let 𝑏𝑛 = 𝑑𝑏 (𝑗𝑛 , 𝑗𝑛+1 ) for each 𝑛 ∈ N, from (13) and axiom (𝐹4 ) we have

𝜅

𝜅

(𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑠𝑛 𝑏𝑛 ) − (𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑏0 ) ≤ − (𝑠𝑛 𝑏𝑛 ) 𝑛𝜏 ≤ 0.

where M (𝑆 (𝑟2𝑛 ) , 𝑟2𝑛+1 ) = max {𝑑𝑏 (𝑆2 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) , 𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑆2 (𝑟2𝑛 )) , 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 )) , 𝑑𝑏 (𝑆2 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) 2𝑠

On taking limit 𝑛 → ∞ in (17), we have 𝜅

lim (𝑛 (𝑠𝑛 𝑏𝑛 ) ) = 0.

𝑛→∞

(18)

This implies there exists 𝑛1 ∈ N, such that 𝑛(𝑠𝑛 𝑏𝑛 )𝜅 ≤ 1 for all 𝑛 ≥ 𝑛1 or 1 𝑛1/𝜅

∀𝑛 ≥ 𝑛1 .

(19)

(25) }.

Taking the upper limit in (24), we have 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) ≤ 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) − 𝜏 < 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) ,

(17)

(24)

≤ 𝐹 (M (𝑆 (𝑟2𝑛 ) , 𝑟2𝑛+1 )) − 𝜏,

From (15), for all 𝑛 ∈ N, we obtain

𝑠𝑛 𝑏𝑛 ≤

∞

1/𝜅 ) entails The convergence of the series ∑∞ 𝑖=𝑛 (1/𝑖 lim𝑛,𝑚→∞ 𝑑𝑏 (𝑗𝑛 , 𝑗𝑚 ) = 0. Hence {𝑗𝑛 } is a Cauchy sequence in (𝑀, 𝑑𝑏 , 𝑠). Since (𝑀, 𝑑𝑏 , 𝑠) is a complete 𝑏-metric space, there exists 𝜐 ∈ 𝑀 such that lim𝑛→∞ 𝑑𝑏 (𝑗𝑛 , 𝜐) = 0. Also we have

𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛+1 ) + 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛 ) } 2𝑠

𝜅

∞

𝑑𝑏 (𝑗𝑛 , 𝑗𝑚 ) ≤ ∑ 𝑠𝑖 𝑏𝑖 ≤ ∑𝑠𝑖 𝑏𝑖 ≤ ∑

(26)

a contradiction; hence, 𝑑𝑏 (𝑆(𝜐), 𝜐) = 0 implies 𝑆(𝜐) = 𝜐. Again since 𝑇 is continuous, therefore, lim 𝑇𝑔 (𝑟2𝑛+1 ) = 𝑇 (𝜐) = lim 𝑇2 (𝑟2𝑛+1 ) .

𝑛→∞

𝑛→∞

(27)

Since the pair {𝑔, 𝑇} is compatible, so, lim 𝑑 𝑛→∞ 𝑏

(𝑔𝑇 (𝑟2𝑛+1 ) , 𝑇𝑔 (𝑟2𝑛+1 )) = 0,

by Lemma 5 lim 𝑔𝑇 (𝑟2𝑛+1 ) = 𝑇 (𝜐) . 𝑛→∞

(28)

4


Now put 𝑟2 = 𝑇(𝑟2𝑛+1 ) and 𝑟1 = 𝑟2𝑛 in (6) and suppose on the contrary that 𝑑𝑏 (𝜐, 𝑇(𝜐)) > 0; we obtain 𝐹 (𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑔𝑇 (𝑟2𝑛+1 )))

< 𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) ,

(37)

a contradiction; hence, 𝑑𝑏 (𝑔(𝜐), 𝜐) = 0 which then implies 𝑔(𝜐) = 𝜐. Hence, 𝜐 proves to be a common fixed point of the four maps 𝑓, 𝑔, 𝑆, and 𝑇. Also 𝜐 is unique; indeed if 𝜔 is another fixed point of 𝑓, 𝑔, 𝑆, and 𝑇, then from (6), we have

where M (𝑟2𝑛 , 𝑇 (𝑟2𝑛+1 )) = max {𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑇2 (𝑟2𝑛+1 )) ,

𝐹 (𝑑𝑏 (𝜐, 𝜔)) = 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜔)))

2

𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑆 (𝑟2𝑛 )) , 𝑑𝑏 (𝑔𝑇 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 )) , 2𝑠

𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) ≤ 𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) − 𝜏

(29)

≤ 𝐹 (M (𝑟2𝑛 , 𝑇 (𝑟2𝑛+1 ))) − 𝜏,

𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑔𝑇 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑇2 (𝑟2𝑛+1 ))

This implies that

(30)

≤ 𝐹 (M (𝜐, 𝜔)) − 𝜏,

(38)

where

}.

M (𝜐, 𝜔) = max {𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜔)) , 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐)) ,

Taking the upper limit in (29), we have 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) ≤ 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) − 𝜏 < 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) ,

𝑑𝑏 (𝑔 (𝜔) , 𝑇 (𝜔)) , (31)

a contradiction; hence, 𝑑𝑏 (𝜐, 𝑇(𝜐)) = 0 which then implies 𝑇(𝜐) = 𝜐. Now assuming 𝑑𝑏 (𝑓(𝜐), 𝜐) > 0 on the contrary, from (6) we get 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝑔 (𝑟2𝑛+1 ))) ≤ 𝐹 (M (𝜐, 𝑟2𝑛+1 )) − 𝜏,

(32)

where

𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝜔)) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝜔)) }. 2𝑠 Thus, from (38), we have 𝐹 (𝑑𝑏 (𝜐, 𝜔)) < 𝐹 (𝑑𝑏 (𝜐, 𝜔)) ,

(33)

𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝑟2𝑛+1 )) }. 2𝑠

which is a contradiction. Hence, 𝜐 = 𝜔 and 𝜐 is a unique common fixed point of the four maps 𝑓, 𝑔, 𝑆, and 𝑇.

Example 7. Let 𝑀 = [0, 1] and define 𝑑𝑏 : 𝑀 × 𝑀 → R+0 by 𝑑𝑏 (𝑟1 , 𝑟2 ) = |𝑟1 − 𝑟2 |2 , so (𝑀, 𝑑𝑏 , 2) is a complete 𝑏-metric space. Define the mappings 𝑓, 𝑔, 𝑆, 𝑇 : 𝑀 → 𝑀, for all 𝑟 ∈ 𝑀 by 𝑟 16 𝑓 (𝑟) = ( ) , 3

Taking the upper limit in (32) and using the fact that 𝑆(𝜐) = 𝑇(𝜐) = 𝜐, we have 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) ≤ 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) − 𝜏

𝑟 8 𝑔 (𝑟) = ( ) , 3

(34)

𝑟 8 𝑆 (𝑟) = ( ) , 3

a contradiction; hence, 𝑑𝑏 (𝑓(𝜐), 𝜐) = 0 which then implies 𝑓(𝜐) = 𝜐. Finally, assume on the contrary 𝑑𝑏 (𝜐, 𝑔(𝜐)) > 0. From (6) and the fact that 𝑆(𝜐) = 𝑇(𝜐) = 𝑓(𝜐) = 𝜐, we obtain

𝑟 4 𝑇 (𝑟) = ( ) . 3

< 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) ,

𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) = 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝑔 (𝜐))) ≤ 𝐹 (M (𝜐, 𝜐)) − 𝜏,

(35)

lim 𝑓 (𝑟𝑛 ) = lim 𝑆 (𝑟𝑛 ) = 𝑡,

𝑛→∞

𝑛→∞

for some 𝑡 ∈ 𝑀,

(42)

then

M (𝜐, 𝜐) = max {𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜐)) , 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐)) ,

𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝜐)) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝜐)) }. 2𝑠

(41)

Clearly, 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings complying with 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). We note that the pair {𝑓, 𝑆} is compatible. If {𝑟𝑛 } is a sequence in 𝑀 satisfying

where

𝑑𝑏 (𝑔 (𝜐) , 𝑇 (𝜐)) ,

(40)

The following example explains Theorem 6.

M (𝜐, 𝑟2𝑛+1 ) = max {𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝑟2𝑛+1 )) , 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐)) , 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 )) ,

(39)

󵄨 󵄨 󵄨2 󵄨2 lim 󵄨󵄨𝑓 (𝑟𝑛 ) − 𝑡󵄨󵄨󵄨 = lim 󵄨󵄨󵄨𝑆 (𝑟𝑛 ) − 𝑡󵄨󵄨󵄨 = 0, 𝑛→∞

𝑛→∞ 󵄨

(36)

(43)

and equivalently 󵄨󵄨 𝑟 16 󵄨󵄨2 󵄨󵄨 𝑟 8 󵄨󵄨2 󵄨 󵄨 󵄨 󵄨 lim 󵄨󵄨󵄨( 𝑛 ) − 𝑡󵄨󵄨󵄨 = lim 󵄨󵄨󵄨( 𝑛 ) − 𝑡󵄨󵄨󵄨 = 0 𝑛→∞ 󵄨 3 𝑛→∞ 󵄨 3 󵄨 󵄨󵄨 󵄨 󵄨 󵄨

(44)


5

implies

holds, where

󵄨2 󵄨2 󵄨 󵄨 16 8 lim 󵄨󵄨(𝑟𝑛 ) − 316 𝑡󵄨󵄨󵄨󵄨 = lim 󵄨󵄨󵄨󵄨(𝑟𝑛 ) − 38 𝑡󵄨󵄨󵄨󵄨 = 0. 𝑛→∞

𝑛→∞ 󵄨󵄨

1/16

(45)

󵄨 󵄨2 lim 𝑑 (𝑓𝑆 (𝑟𝑛 ) , 𝑆𝑓 (𝑟𝑛 )) = lim 󵄨󵄨󵄨𝑓𝑆 (𝑟𝑛 ) − 𝑆𝑓 (𝑟𝑛 )󵄨󵄨󵄨 𝑛→∞ 𝑏 𝑛→∞ 󵄨 󵄨2 = 󵄨󵄨󵄨𝑓 (𝑡) − 𝑆 (𝑡)󵄨󵄨󵄨 = |0 − 0|2 = 0

(46)

for 𝑡 = 0 ∈ 𝑀.

󵄨 󵄨2 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) = 󵄨󵄨󵄨𝑓 (𝑟1 ) − 𝑔 (𝑟2 )󵄨󵄨󵄨 2 󵄨󵄨 𝑟 16 𝑟 8 󵄨󵄨󵄨 󵄨 = 󵄨󵄨󵄨( 1 ) − ( 2 ) 󵄨󵄨󵄨 󵄨󵄨 3 3 󵄨󵄨 2

1 1 2 ≤( + ) 𝑑𝑏 (𝑇 (𝑟1 ) , 𝑆 (𝑟2 )) 6561 81 =

then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible.

Theorem 9 (Sehgal type). Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0, the inequality + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M4 (𝑟1 , 𝑟2 )))

(47)

(53)

with 𝑘 ∈ [0, 𝑠) , then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible.

The above inequality can be written as

Theorem 10 generalizes the result presented by C´ır´ıc [28].

(48)

≤ ln (M (𝑟1 , 𝑟2 )) . Define the function 𝐹 : R+ → R by 𝐹(𝑟) = ln(𝑟), for all 𝑟 ∈ 𝑅+ > 0. Hence, for all 𝑟1 , 𝑟2 ∈ 𝑀 such that 𝑑𝑏 (𝑓(𝑟1 ), 𝑔(𝑟2 )) > 0, 𝜏 = ln(43046721/6724) we obtain 𝜏 + 𝐹 (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M (𝑟1 , 𝑟2 )) .

M4 (𝑟1 , 𝑟2 ) = 𝑘 max {𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 ))} ,

6724 𝑑 (𝑇 (𝑟1 ) , 𝑆 (𝑟2 )) 43046721 𝑏

43046721 ) + ln (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) 6724

(52)

holds, where

6724 ≤ M (𝑟1 , 𝑟2 ) . 43046721

ln (

with 𝑘 ∈ [0, 𝑠) ,

(𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 2 󵄨󵄨 𝑟 8 𝑟 4 󵄨󵄨󵄨 󵄨󵄨 1 󵄨󵄨( ) − ( 2 ) 󵄨󵄨󵄨 󵄨󵄨 3 3 󵄨󵄨

(51)

Theorem 9 generalizes the result presented by Sehgal [27].

Similarly, the pair {𝑔, 𝑇} is compatible. Now for each 𝑟1 , 𝑟2 ∈ 𝑀, consider

𝑟1 8 𝑟 4 ) + ( 2) ] 3 3

= 𝑘 max {𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 ))} ,

1/8

Uniqueness of limit gives 𝑡 = 𝑡 which implies 𝑡 = 0, 1. Due to the continuity of 𝑓, 𝑆 we have

= [(

M3 (𝑟1 , 𝑟2 )

(49)

Thus, contractive condition (6) is satisfied for all 𝑟1 , 𝑟2 ∈ 𝑀. Hence, all the hypotheses of Theorem 6 are satisfied; note that 𝑓, 𝑔, 𝑆, 𝑇 have a unique common fixed point 𝑟 = 0.

Theorem 10. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0, the inequality (𝐹 (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M5 (𝑟1 , 𝑟2 )))

(54)

holds, where for all 𝑘 ∈ [0, 𝑠) M5 (𝑟1 , 𝑟2 ) = 𝑘 max {𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) ,

(55)

𝑑𝑏 (𝑔 (𝑟2 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 ))} ,

The following theorems can easily be obtained by chasing the proof of Theorem 6. Theorem 8 generalizes the result presented by Bianchini [26].


Theorem 8 (Bianchini type). Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏metric space and 𝑓, 𝑔, 𝑆, 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0, the inequality

Theorem 11. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0, the inequality

(𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M3 (𝑟1 , 𝑟2 )))

(50)

(𝐹 (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M6 (𝑟1 , 𝑟2 )))

(56)

6


holds, where for all 𝑘 ∈ [0, 𝑠)

holds, where M1 (𝑟1 , 𝑟2 ) = 𝑎1 𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) + 𝑎2 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 ))

M6 (𝑟1 , 𝑟2 ) = 𝑘 max {𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) + 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) , 2

+ 𝑎3 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 ))

(57)

then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible. Theorem 12. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If for all 𝑟1 , 𝑟2 ∈ 𝑀, 𝑘 ∈ [0, 𝑠) the inequality (58)

holds, where M (𝑟1 , 𝑟2 ) = max {𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) , 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) , 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) ,

(64)

M1 (𝑟2𝑛 , 𝑟2𝑛+1 ) = 𝑎1 𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) + 𝑎2 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑆 (𝑟2𝑛 )) + 𝑎3 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 ))

(60)

thus, if 𝑑𝑏 (𝑓(𝑟1 ), 𝑔(𝑟2 )) > 0, we have (61)

where 𝜏 = ln(𝑠/𝑘) > 0; thus, the contractive condition (58) reduces to (6) with 𝐹(𝑟) = ln(𝑟). Hence, Theorem 6 is a generalization of [29, Theorem 3.1].

3. Hardy and Rogers Type Fixed Point Theorem

+ 𝑎4 [𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 ))] = 𝑎1 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 )

(65)

+ 𝑎2 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛−1 ) + 𝑎3 𝑑𝑏 (𝑗2𝑛+1 , 𝑗2𝑛 ) + 𝑎4 [𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛+1 ) + 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛 )] = (𝑎1 + 𝑎2 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) + (𝑎3 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) .

In this section, we set up a fixed point theorem for four self-maps satisfying Hardy and Rogers type 𝐹-contraction. We give an example to validate this theorem and discuss its consequences. Theorem 13. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0, the inequality + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M1 (𝑟1 , 𝑟2 )))

≤ 𝐹 (M1 (𝑟2𝑛 , 𝑟2𝑛+1 )) − 𝜏, for all 𝑛 ∈ N ∪ {0}, where

Proof. Since, for all 𝑟1 , 𝑟2 ∈ 𝑀, we have

(𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏

Proof. Let 𝑟0 ∈ 𝑀. As 𝑓(𝑀) ⊆ 𝑇(𝑀), there exists 𝑟1 ∈ 𝑀 such that 𝑓(𝑟0 ) = 𝑇(𝑟1 ). Since 𝑔(𝑟1 ) ∈ 𝑆(𝑀), we can choose 𝑟2 ∈ 𝑀 such that 𝑔(𝑟1 ) = 𝑆(𝑟2 ). In general 𝑟2𝑛+1 and 𝑟2𝑛+2 are chosen in 𝑀 such that 𝑓(𝑟2𝑛 ) = 𝑇(𝑟2𝑛+1 ) and 𝑔(𝑟2𝑛+1 ) = 𝑆(𝑟2𝑛+2 ). Define a sequence {𝑗𝑛 } in 𝑀 such that 𝑗2𝑛 = 𝑓(𝑟2𝑛 ) = 𝑇(𝑟2𝑛+1 ) and 𝑗2𝑛+1 = 𝑔(𝑟2𝑛+1 ) = 𝑆(𝑟2𝑛+2 ) for all 𝑛 ≥ 0; we show that {𝑗𝑛 } is a Cauchy sequence. Assume that 𝑑𝑏 (𝑓(𝑟2𝑛 ), 𝑔(𝑟2𝑛+1 )) > 0; then from contractive condition (62), we get

(59)


𝜏 + ln (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ ln (M (𝑟1 , 𝑟2 )) ,


𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) = 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 )))

𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 )) }, 2𝑠

𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) ≤ 𝑘M (𝑟1 , 𝑟2 ) ,

(63)

𝑤𝑖𝑡ℎ 𝑎𝑖 ≥ 0 (𝑖 = 1, 2, 3, 4) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎1 + 𝑎2 + 𝑎3 + 2𝑠𝑎4 = 1,

𝑑𝑏 (𝑔 (𝑟2 ) , 𝑆 (𝑟1 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 )) }, 2𝑠

𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) ≤ 𝑘M (𝑟1 , 𝑟2 )

+ 𝑎4 [𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 ))]

(62)

Now from (64) we have 𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) ≤ 𝐹 ((𝑎1 + 𝑎2 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) + (𝑎3 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) − 𝜏. Since 𝐹 is strictly increasing, (66) implies 𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) ≤ (𝑎1 + 𝑎2 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) + (𝑎3 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) ,

(66)


7 1/𝜅 ) entails The convergence of the series ∑∞ 𝑖=𝑛 (1/𝑖 lim𝑛,𝑚→∞ 𝑑𝑏 (𝑗𝑛 , 𝑗𝑚 ) = 0. Hence {𝑗𝑛 } is a Cauchy sequence in (𝑀, 𝑑𝑏 , 𝑠). Since (𝑀, 𝑑𝑏 , 𝑠) is a complete 𝑏-metric space, there exists 𝜐 ∈ 𝑀 such that lim𝑛→∞ 𝑑𝑏 (𝑗𝑛 , 𝜐) = 0. Also we have

(1 − 𝑎3 − 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) < (𝑠 − 𝑎3 − 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) ≤ (𝑎1 + 𝑎2 + 𝑠𝑎4 ) 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) ,

lim 𝑓 (𝑟2𝑛 ) = lim 𝑇 (𝑟2𝑛+1 ) = lim 𝑔 (𝑟2𝑛+1 )

𝑎1 + 𝑎2 + 𝑠𝑎4 𝑑 (𝑗 , 𝑗 ) . 1 − 𝑎3 − 𝑠𝑎4 𝑏 2𝑛−1 2𝑛

𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) ≤

𝑛→∞

𝑛→∞

= lim 𝑆 (𝑟2𝑛+2 ) = 𝜐.

We show that 𝜐 is a common fixed point of 𝑓, 𝑔, 𝑆, and 𝑇. Since 𝑆 is continuous, therefore,

Since 𝑎1 + 𝑎2 + 𝑎3 + 2𝑠𝑎4 = 1, therefore 𝑎1 + 𝑎2 + 𝑠𝑎4 𝑑 (𝑗 , 𝑗 ) 1 − 𝑎3 − 𝑠𝑎4 𝑏 2𝑛−1 2𝑛

(68)

= 𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 ) .

lim 𝑆𝑓 (𝑟2𝑛 ) = 𝑆 (𝜐) = lim 𝑆2 (𝑟2𝑛+2 ) .

𝑛→∞


𝐹 (𝑠𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 )) ≤ 𝐹 (𝑑𝑏 (𝑗2𝑛−2 , 𝑗2𝑛−1 )) − 𝜏,

(70)

(71)

for all 𝑛 ∈ N. Let 𝑏𝑛 = 𝑑𝑏 (𝑗𝑛 , 𝑗𝑛+1 ) for each 𝑛 ∈ N; from (71) and axiom (𝐹4 ) we have 𝜏 + 𝐹 (𝑠𝑛 𝑏𝑛 ) ≤ 𝐹 (𝑠𝑛−1 𝑏𝑛−1 ) , 𝑛 ∈ N.

(72)

Now put 𝑟1 = 𝑆(𝑟2𝑛 ) and 𝑟2 = 𝑟2𝑛+1 in (62) and suppose on the contrary that 𝑑𝑏 (𝑆(𝜐), 𝜐) > 0; we obtain 𝐹 (𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 ))) ≤ 𝐹 (M1 (𝑆 (𝑟2𝑛 ) , 𝑟2𝑛+1 )) − 𝜏,

𝐹 (𝑠𝑛 𝑏𝑛 ) ≤ 𝐹 (𝑏0 ) − 𝑛𝜏, 𝑛 ∈ N.

(73)

On taking limit 𝑛 → ∞ in (73), we have lim𝑛→∞ 𝐹(𝑠𝑛 𝑏𝑛 ) = −∞; due to axiom (𝐹2 ) we get lim𝑛→∞ 𝑠𝑛 𝑏𝑛 = 0 and (𝐹3 ) implies that there exists 𝜅 ∈ (0, 1) such that 𝜅

lim (𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑠𝑛 𝑏𝑛 ) = 0.

(74)

𝑛→∞

M1 (𝑆 (𝑟2𝑛 ) , 𝑟2𝑛+1 ) = 𝑎1 𝑑𝑏 (𝑆2 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 )) + 𝑎2 𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑆2 (𝑟2𝑛 ))

𝜅

𝜅

+ 𝑑𝑏 (𝑓𝑆 (𝑟2𝑛 ) , 𝑇 (𝑟2𝑛+1 ))] . Taking the upper limit in (82) and considering the fact that 𝑎1 + 2𝑎4 < 1, we have 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) ≤ 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) − 𝜏 < 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝜐)) ,

(75)

On taking limit 𝑛 → ∞ in (75), we have 𝜅

lim (𝑛 (𝑠𝑛 𝑏𝑛 ) ) = 0.

(76)

𝑛→∞

This implies there exists 𝑛1 ∈ N, such that 𝑛(𝑠𝑛 𝑏𝑛 )𝜅 ≤ 1 for all 𝑛 ≥ 𝑛1 or 𝑠𝑛 𝑏𝑛 ≤

1 𝑛1/𝜅

∀𝑛 ≥ 𝑛1 .

(77)

To prove {𝑗𝑛 } as a Cauchy sequence, we use (77) and for 𝑚 > 𝑛 ≥ 𝑛1 ; we consider 𝑚−1

∞

∞

𝑑𝑏 (𝑗𝑛 , 𝑗𝑚 ) ≤ ∑ 𝑠𝑖 𝑏𝑖 ≤ ∑𝑠𝑖 𝑏𝑖 ≤ ∑ 𝑖=𝑛

𝑖=𝑛

𝑖=𝑛 𝑛

1

. 1/𝜅

(84)

a contradiction; hence, 𝑑𝑏 (𝑆(𝜐), 𝜐) = 0 implies 𝑆(𝜐) = 𝜐. Again since 𝑇 is continuous, therefore, lim 𝑇𝑔 (𝑟2𝑛+1 ) = 𝑇 (𝜐) = lim 𝑇2 (𝑟2𝑛+1 ) .

𝑛→∞

𝑛→∞

(85)

Since, the pair {𝑔, 𝑇} is compatible, lim 𝑑 𝑛→∞ 𝑏

(𝑔𝑇 (𝑟2𝑛+1 ) , 𝑇𝑔 (𝑟2𝑛+1 )) = 0,

by Lemma 5 lim 𝑔𝑇 (𝑟2𝑛+1 ) = 𝑇 (𝜐) .

(86)

𝑛→∞

Now put 𝑟2 = 𝑇(𝑟2𝑛+1 ) and 𝑟1 = 𝑟2𝑛 in (62) and suppose on the contrary that 𝑑𝑏 (𝜐, 𝑇(𝜐)) > 0; we obtain 𝐹 (𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑔𝑇 (𝑟2𝑛+1 )))

(78)

(83)

+ 𝑎4 [𝑑𝑏 (𝑆2 (𝑟2𝑛 ) , 𝑔 (𝑟2𝑛+1 ))

From (73), for all 𝑛 ∈ N, we obtain (𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑠𝑛 𝑏𝑛 ) − (𝑠𝑛 𝑏𝑛 ) 𝐹 (𝑏0 ) ≤ − (𝑠𝑛 𝑏𝑛 ) 𝑛𝜏 ≤ 0.

(82)

where

+ 𝑎3 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 ))

Repeating the process, we obtain

(81)

𝑛→∞

for all 𝑛 ∈ N. Hence, from (69) and (70), we have 𝐹 (𝑠𝑑𝑏 (𝑗𝑛 , 𝑗𝑛+1 )) ≤ 𝐹 (𝑑𝑏 (𝑗𝑛−1 , 𝑗𝑛 )) − 𝜏,

(80)

(𝑓𝑆 (𝑟2𝑛 ) , 𝑆𝑓 (𝑟2𝑛 )) = 0,

by Lemma 5 lim 𝑓𝑆 (𝑟2𝑛 ) = 𝑆 (𝜐) .

(69)

for all 𝑛 ∈ N. Similarly,

𝜅

𝑛→∞

Since, the pair {𝑓, 𝑆} is compatible, so,

Thus, from (66) we obtain 𝐹 (𝑠𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 )) ≤ 𝐹 (𝑑𝑏 (𝑗2𝑛−1 , 𝑗2𝑛 )) − 𝜏,

(79)

𝑛→∞

(67)

𝑑𝑏 (𝑗2𝑛 , 𝑗2𝑛+1 ) ≤

𝑛→∞

≤ 𝐹 (M1 (𝑟2𝑛 , 𝑇 (𝑟2𝑛+1 ))) − 𝜏,

(87)

8

Journal of Function Spaces a contradiction; hence, 𝑑𝑏 (𝑔(𝜐), 𝜐) = 0 which then implies 𝑔(𝜐) = 𝜐. Hence, 𝜐 proves to be a common fixed point of the four maps 𝑓, 𝑔, 𝑆, and 𝑇. Also 𝜐 is unique. Indeed if 𝜔 is another fixed point of 𝑓, 𝑔, 𝑆, and 𝑇, then, from (62), we have

where M1 (𝑟2𝑛 , 𝑇 (𝑟2𝑛+1 )) = 𝑎1 𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑇2 (𝑟2𝑛+1 )) + 𝑎2 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑆 (𝑟2𝑛 )) + 𝑎3 𝑑𝑏 (𝑔𝑇 (𝑟2𝑛+1 ) , 𝑇2 (𝑟2𝑛+1 ))

𝐹 (𝑑𝑏 (𝜐, 𝜔)) = 𝐹 (𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜔)))

(88)

≤ 𝐹 (M1 (𝜐, 𝜔)) − 𝜏,

+ 𝑎4 [𝑑𝑏 (𝑆 (𝑟2𝑛 ) , 𝑔𝑇 (𝑟2𝑛+1 )) where

+ 𝑑𝑏 (𝑓 (𝑟2𝑛 ) , 𝑇2 (𝑟2𝑛+1 ))] .

M1 (𝜐, 𝜔)

Taking the upper limit in (87), we have 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) ≤ 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) − 𝜏 < 𝐹 (𝑑𝑏 (𝜐, 𝑇 (𝜐))) ,

= 𝑎1 𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜔)) + 𝑎2 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐)) + 𝑎3 𝑑𝑏 (𝑔 (𝜔) , 𝑇 (𝜔))

(89)

a contradiction; hence, 𝑑𝑏 (𝜐, 𝑇(𝜐)) = 0 which then implies 𝑇(𝜐) = 𝜐. Assume that 𝑑𝑏 (𝑓(𝜐), 𝜐) > 0 and from (62) we get 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝑔 (𝑟2𝑛+1 ))) ≤ 𝐹 (M1 (𝜐, 𝑟2𝑛+1 )) − 𝜏,

(96)

(97)

+ 𝑎4 [𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝜔)) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝜔))] . Thus, from (96), we have 𝐹 (𝑑𝑏 (𝜐, 𝜔)) < 𝐹 (𝑑𝑏 (𝜐, 𝜔)) ,

(90)

(98)

which is a contradiction. Hence, 𝜐 = 𝜔 and 𝜐 is a unique common fixed point of the four maps 𝑓, 𝑔, 𝑆, and 𝑇.

where M1 (𝜐, 𝑟2𝑛+1 )

The following example explains Theorem 13.

= 𝑎1 𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝑟2𝑛+1 )) + 𝑎2 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐))

+ 𝑎4 [𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝑟2𝑛+1 )) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝑟2𝑛+1 ))] .

Example 14. Let 𝑀 = [0, ∞) and define 𝑑𝑏 : 𝑀 × 𝑀 → R+0 by 𝑑𝑏 (𝑟1 , 𝑟2 ) = |𝑟1 − 𝑟2 |2 , so, (𝑀, 𝑑𝑏 , 2) is a complete 𝑏-metric space. Define the mappings 𝑓, 𝑔, 𝑆, 𝑇 : 𝑀 → 𝑀, for all 𝑟 ∈ 𝑀, by

Taking the upper limit in (90) and using the fact that 𝑆(𝜐) = 𝑇(𝜐) = 𝜐, we have

𝑟 𝑓 (𝑟) = ln (1 + ) , 6

𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) ≤ 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) − 𝜏

𝑟 𝑔 (𝑟) = ln (1 + ) , 7

+ 𝑎3 𝑑𝑏 (𝑔 (𝑟2𝑛+1 ) , 𝑇 (𝑟2𝑛+1 ))

< 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝜐)) ,

(91)

(92)

𝑆 (𝑟) = 𝑒 − 1,

a contradiction; hence, 𝑑𝑏 (𝑓(𝜐), 𝜐) = 0 which then implies 𝑓(𝜐) = 𝜐. Finally, assume that 𝑑𝑏 (𝜐, 𝑔(𝜐)) > 0. From (62) and the equation 𝑆(𝜐) = 𝑇(𝜐) = 𝑓(𝜐) = 𝜐, we obtain 𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) = 𝐹 (𝑑𝑏 (𝑓 (𝜐) , 𝑔 (𝜐))) ≤ 𝐹 (M1 (𝜐, 𝜐)) − 𝜏,

(99)

7𝑟

(93)

𝑇 (𝑟) = 𝑒6𝑟 − 1. Clearly, 𝑓, 𝑔, 𝑆, and 𝑇 are continuous self-mappings complying with 𝑓(𝑀) = 𝑇(𝑀) = 𝑔(𝑀) = 𝑆(𝑀). We note that the pair {𝑓, 𝑆} is compatible. Indeed, let {𝑟𝑛 } be a sequence in 𝑀 satisfying lim 𝑓 (𝑟𝑛 ) = lim 𝑆 (𝑟𝑛 ) = 𝑡, for some 𝑡 ∈ 𝑀.

(100)

󵄨 󵄨 󵄨2 󵄨2 lim 󵄨󵄨𝑓 (𝑟𝑛 ) − 𝑡󵄨󵄨󵄨 = lim 󵄨󵄨󵄨𝑆 (𝑟𝑛 ) − 𝑡󵄨󵄨󵄨 = 0, 𝑛→∞

(101)

𝑛→∞

where M1 (𝜐, 𝜐)

𝑛→∞

Then

= 𝑎1 𝑑𝑏 (𝑆 (𝜐) , 𝑇 (𝜐)) + 𝑎2 𝑑𝑏 (𝑓 (𝜐) , 𝑆 (𝜐)) + 𝑎3 𝑑𝑏 (𝑔 (𝜐) , 𝑇 (𝜐))

𝑛→∞ 󵄨

(94)

and equivalently 󵄨󵄨 󵄨󵄨2 𝑟 󵄨2 󵄨 lim 󵄨󵄨󵄨󵄨ln (1 + 𝑛 ) − 𝑡󵄨󵄨󵄨󵄨 = lim 󵄨󵄨󵄨󵄨𝑒7𝑟𝑛 − 1 − 𝑡󵄨󵄨󵄨󵄨 = 0 𝑛→∞ 󵄨 𝑛→∞ 6 󵄨

+ 𝑎4 [𝑑𝑏 (𝑆 (𝜐) , 𝑔 (𝜐)) + 𝑑𝑏 (𝑓 (𝜐) , 𝑇 (𝜐))] . This implies that

(102)

implies

𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) ≤ 𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) − 𝜏 < 𝐹 (𝑑𝑏 (𝜐, 𝑔 (𝜐))) ,

(95)

󵄨󵄨 ln (𝑡 + 1) 󵄨󵄨󵄨󵄨2 󵄨2 󵄨 󵄨 lim 󵄨󵄨󵄨󵄨𝑟𝑛 − (6𝑒𝑡 − 6)󵄨󵄨󵄨󵄨 = lim 󵄨󵄨󵄨𝑟𝑛 − 󵄨󵄨 = 0. 𝑛→∞ 𝑛→∞ 󵄨 󵄨󵄨 7 󵄨

(103)


9

The uniqueness of limit gives 6𝑒𝑡 − 6 = ln(𝑡 + 1)/7; thus, 𝑡 = 0 is only possible solution. Due to the continuity of 𝑓, 𝑆 we have 󵄨 󵄨2 lim 𝑑 (𝑓𝑆 (𝑟𝑛 ) , 𝑆𝑓 (𝑟𝑛 )) = lim 󵄨󵄨󵄨𝑓𝑆 (𝑟𝑛 ) − 𝑆𝑓 (𝑟𝑛 )󵄨󵄨󵄨 𝑛→∞ 𝑏 𝑛→∞ (104) 󵄨 󵄨2 = 󵄨󵄨󵄨𝑓 (𝑡) − 𝑆 (𝑡)󵄨󵄨󵄨 = |0 − 0|2 = 0 for 𝑡 = 0 ∈ 𝑀. Similarly, the pair {𝑔, 𝑇} is compatible. Now for each 𝑟1 , 𝑟2 ∈ 𝑀, consider 󵄨 󵄨2 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) = 󵄨󵄨󵄨𝑓 (𝑟1 ) − 𝑔 (𝑟2 )󵄨󵄨󵄨 󵄨󵄨 𝑟 𝑟 󵄨󵄨2 = 󵄨󵄨󵄨󵄨ln (1 + ) − ln (1 + )󵄨󵄨󵄨󵄨 6 7 󵄨 󵄨 𝑟 𝑟 2 1 2 ≤ ( − ) ≤ ( ) |7𝑟 − 6𝑟|2 6 7 42 ≤( =(

1 2 󵄨󵄨 7𝑟 6𝑟 󵄨󵄨2 ) 󵄨󵄨𝑒 − 𝑒 󵄨󵄨󵄨 42 󵄨

(105)

1 ) 𝑑𝑏 (𝑇 (𝑟1 ) , 𝑆 (𝑟2 )) 42

≤ ln (M1 (𝑟1 , 𝑟2 )) .

(106)

Define the function 𝐹 : R+ → R by 𝐹(𝑟) = ln(𝑟), for all 𝑟 ∈ 𝑅+ > 0. Hence, for all 𝑟1 , 𝑟2 ∈ 𝑀, such that 𝑑𝑏 (𝑓(𝑟1 ), 𝑔(𝑟2 )) > 0, 𝜏 = 2 ln(42) we obtain 𝜏 + 𝐹 (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M (𝑟1 , 𝑟2 )) .

(107)

Thus, contractive condition (62) is satisfied for all 𝑟1 , 𝑟2 ∈ 𝑀. Hence, all the hypotheses of Theorem 13 are satisfied. Note that 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝑟 = 0. Theorem 15 generalizes the result presented by Reich [30] and can be proved by following the proof of Theorem 13. Theorem 15 (Reich type). Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If, for all 𝑟1 , 𝑟2 ∈ 𝑀, for some 𝐹 ∈ Δ 𝐹𝑠 , 𝜏 > 0, the inequality (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M2 (𝑟1 , 𝑟2 )))

(108)

holds, where for all 𝑎𝑖 ≥ 0 (𝑖 = 1, 2, 3) such that 𝑎1 + 𝑎2 + 𝑎3 = 1 M2 (𝑟1 , 𝑟2 ) = 𝑎1 𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) + 𝑎2 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 ))

(𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )))

(110)

≤ 𝐹 (𝑎 [𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 ))])) holds for all 𝑎 ≥ 0 such that 2𝑎 = 1, then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible.

Theorem 17. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If for all 𝑟1 , 𝑟2 ∈ 𝑀 for some 𝐹 ∈ Δ 𝐹𝑠 , 𝜏 > 0 the inequality

2

2 ln (42) + ln (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )))

Theorem 16 (Chatterjea type). Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If for all 𝑟1 , 𝑟2 ∈ 𝑀 for some 𝐹 ∈ Δ 𝐹𝑠 𝜏 > 0 the inequality

Similarly, Theorems 17 and 18 can be proved by following the proof of Theorem 13.

2

1 ) M1 (𝑟1 , 𝑟2 ) . 42 The above inequality can be written as ≤(

Theorem 16 generalizes the result presented by Chatterjea [31] and can be proved by following the proof of Theorem 13.

(109)

+ 𝑎3 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) , then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible.

(𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M7 (𝑟1 , 𝑟2 )))

(111)

holds, where M7 (𝑟1 , 𝑟2 ) = 𝑎1 (𝑟1 , 𝑟2 ) 𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 )) + 𝑎2 (𝑟1 , 𝑟2 ) 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) + 𝑎3 (𝑟1 , 𝑟2 ) 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 )) + 𝑎4 (𝑟1 , 𝑟2 )

(112)

⋅ [𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 ))] with 𝑎𝑖 (𝑟1 , 𝑟2 ) (𝑖 = 1, 2, 3, 4) are nonnegative functions such that sup {𝑎1 (𝑟1 , 𝑟2 ) + 𝑎2 (𝑟1 , 𝑟2 ) + 𝑎3 (𝑟1 , 𝑟2 )

𝑟1 ,𝑟2 ∈𝑀

(113)

+ 2𝑠𝑎4 (𝑟1 , 𝑟2 )} = 1, then 𝑓, 𝑔, 𝑆, and 𝑇 have a unique common fixed point 𝜐 provided 𝑆, 𝑇 are continuous and pairs {𝑓, 𝑆}, {𝑔, 𝑇} are compatible. Theorem 18. Let (𝑀, 𝑑𝑏 , 𝑠) be a complete 𝑏-metric space and 𝑓, 𝑔, 𝑆, and 𝑇 are self-mappings on (𝑀, 𝑑𝑏 , 𝑠) such that 𝑓(𝑀) ⊆ 𝑇(𝑀), 𝑔(𝑀) ⊆ 𝑆(𝑀). If for all 𝑟1 , 𝑟2 ∈ 𝑀 for some 𝐹 ∈ Δ 𝐹𝑠 , 𝜏 > 0 the inequality (𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 )) > 0 implies 𝜏 + 𝐹 (𝑠𝑑𝑏 (𝑓 (𝑟1 ) , 𝑔 (𝑟2 ))) ≤ 𝐹 (M8 (𝑟1 , 𝑟2 )))

(114)

10

Journal of Function Spaces (ii) There exists 𝜏 > 0 and, for each 𝑟 ∈ 𝑀, one has

holds, where M8 (𝑟1 , 𝑟2 ) = 𝑎1 𝑑𝑏 (𝑆 (𝑟1 ) , 𝑇 (𝑟2 ))

𝑡

∫ 𝐻 (𝑟) 𝑑𝑟 ≤ √

𝑎 + 𝑎3 + 2 [𝑑𝑏 (𝑓 (𝑟1 ) , 𝑆 (𝑟1 )) 2

𝑎

+ 𝑑𝑏 (𝑔 (𝑟2 ) , 𝑇 (𝑟2 ))]

(115)

𝑎 + 𝑎5 [𝑑𝑏 (𝑆 (𝑟1 ) , 𝑔 (𝑟2 )) + 4 2𝑠

(120)

𝑡 ∈ [𝑎, 𝑏] .

(iii) There exists a sequence {𝑟𝑛 } in = that lim𝑛→∞ 𝑑𝑏 (𝑓𝑆(𝑟𝑛 ), 𝑆𝑓(𝑟𝑛 )) lim𝑛→∞ 𝑑𝑏 (𝑔𝑇(𝑟𝑛 ), 𝑇𝑔(𝑟𝑛 )) = 0, whenever

𝑀 such 0 and

lim 𝑓 (𝑟𝑛 ) = lim 𝑆 (𝑟𝑛 ) = 𝑡,

𝑛→∞ 5

+ 𝑑𝑏 (𝑓 (𝑟1 ) , 𝑇 (𝑟2 ))]

𝑒−𝜏 , 𝑠

lim 𝑔 (𝑟𝑛 ) = lim 𝑇 (𝑟𝑛 ) = 𝑡

with 𝑎𝑖 ≥ 0, ∑𝑎𝑖 = 1,

𝑛→∞

𝑖=1


Then the system of integral equations given in (117) has a solution.

󵄨 󵄨 𝑑𝑏 (𝑓𝑢 (𝑡) , 𝑔V (𝑡)) = ( sup 󵄨󵄨󵄨𝑓𝑢 (𝑡) − 𝑔V (𝑡)󵄨󵄨󵄨) 𝑡∈[𝑎,𝑏]

2

𝑡

= ( sup ∫ |𝐾 (𝑡, 𝑟, 𝑆 (𝑢 (𝑡)) − 𝐽 (𝑡, 𝑟, 𝑇 (V (𝑡))))| 𝑑𝑟) 𝑡∈[𝑎,𝑏] 𝑎

2

(116)

𝑡∈[𝑎,𝑏]

2

𝑡

≤ ( sup ∫ 𝐻 (𝑟) |𝑆 (𝑢 (𝑡)) − 𝑇 (V (𝑡))| 𝑑𝑟)

for all 𝑢, V ∈ 𝐶([𝑎, 𝑏], R). Obviously, (𝑀, 𝑑𝑏 , 2) is a complete 𝑏-metric space. We shall apply Theorem 6 to show the existence of a common solution of the system of Volterra type integral equations given by 𝑡

𝑡∈[𝑎,𝑏] 𝑎

(122) 𝑡

≤ (√ sup |𝑆 (𝑢 (𝑡)) − 𝑇 (V (𝑡))|2 ∫ 𝐻 (𝑟) 𝑑𝑟)

2

𝑎

𝑡∈[𝑎,𝑏]

𝑡

𝑢 (𝑡) = 𝑝 (𝑡) + ∫ 𝐾 (𝑡, 𝑟, 𝑆 (𝑢 (𝑡))) 𝑑𝑟, 𝑡

for some 𝑡 ∈ 𝑀.

2

Let 𝑀 = 𝐶([𝑎, 𝑏], R) be the space of all continuous real valued functions defined on [𝑎, 𝑏], 𝑎 > 0. Let the function 𝑑𝑏 : 𝑀 × 𝑀 → [0, ∞) be defined by

𝑎

(121)

𝑛→∞

Proof. Following assumptions (i) and (ii), we have

4. Application of Theorem 6 to a System of Integral Equations

𝑑𝑏 (𝑢, V) = ( sup |𝑢 (𝑡) − V (𝑡)|) ,

𝑛→∞

2

= 𝑑𝑏 (𝑆 (𝑢 (𝑡)) , 𝑇 (V (𝑡))) (∫ 𝐻 (𝑟) 𝑑𝑟) 𝑎

(117) ≤ 𝑑𝑏 (𝑆 (𝑢 (𝑡)) , 𝑇 (V (𝑡)))

𝑤 (𝑡) = 𝑝 (𝑡) + ∫ 𝐽 (𝑡, 𝑟, 𝑇 (V (𝑡))) 𝑑𝑟, 𝑎

for all 𝑡, where 𝑝 : 𝑀 → R is a continuous function and 𝐾, 𝐽 : [𝑎, 𝑏] × [𝑎, 𝑏] × 𝑀 → R are lower semicontinuous operators. Now we prove the following theorem to ensure the existence of a solution of the system of integral equations (117). Theorem 19. Let 𝑀 = 𝐶([𝑎, 𝑏], R) and define the mappings 𝑓, 𝑔 : 𝑀 → 𝑀 by

−𝜏

𝑒

𝑠

≤ M1 (𝑢 (𝑡) , V (𝑡))

𝑒−𝜏 . 𝑠

Consequently, we have 𝑠𝑑𝑏 (𝑓𝑢 (𝑡) , 𝑔V (𝑡)) ≤ 𝑒−𝜏 M1 (𝑢 (𝑡) , V (𝑡)) .

(123)

As the natural logarithm belongs to Δ 𝐹𝑠 , applying it on above inequality and after some simplification, we get 𝜏 + ln (𝑠𝑑𝑏 (𝑓𝑢 (𝑡) , 𝑔V (𝑡))) ≤ ln (M1 (𝑢 (𝑡) , V (𝑡))) . (124)

𝑡

𝑓𝑢 (𝑡) = 𝑝 (𝑡) + ∫ 𝐾 (𝑡, 𝑟, 𝑆 (𝑢 (𝑡))) 𝑑𝑟, 𝑎

𝑡

(118)

𝑔𝑢 (𝑡) = 𝑝 (𝑡) + ∫ 𝐽 (𝑡, 𝑟, 𝑇 (V (𝑡))) 𝑑𝑟,

So taking 𝐹(𝑟) = ln(𝑟), all hypotheses of Theorem 6 are satisfied. Hence the system of integral equations given in (117) has a unique common solution.

𝑎

where 𝑝 : 𝑀 → R is a continuous function and 𝐾, 𝐽 : [𝑎, 𝑏] × [𝑎, 𝑏] × 𝑀 → R are lower semicontinuous operators. Assume the following conditions are satisfied: (i) There exists a continuous function 𝐻 : 𝑀 → [0, ∞) such that |𝐾 (𝑡, 𝑟, 𝑆) − 𝐽 (𝑡, 𝑟, 𝑇)| ≤ 𝐻 (𝑟) |𝑆 (𝑢 (𝑡)) − 𝑇 (V (𝑡))| for each 𝑡, 𝑟 ∈ [𝑎, 𝑏] and 𝑆, 𝑇 ∈ 𝑀.

(119)

Conflicts of Interest The authors declare that they have no conflicts of interest.

Acknowledgments Ma Zhenhua acknowledges the support of Scientific Research Foundations of Education Burrean of Hebei Province, Grant no. QN2016191.


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Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

International Journal of Mathematics and Mathematical Sciences

Mathematical Problems in Engineering

Journal of

Mathematics Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014


Volume 2014

Volume 2014


Volume 2014

#HRBQDSDĮ,@SGDL@SHBR

Journal of

Volume 201


Discrete Dynamics in Nature and Society

Journal of

Function Spaces Hindawi Publishing Corporation http://www.hindawi.com

Abstract and Applied Analysis

Volume 2014


Volume 2014


Volume 2014

International Journal of

Journal of

Stochastic Analysis

Optimization



Volume 2014

Volume 2014