In this paper, which serves as a continuation of earlier work, we generalize the idea of inequalities in metric spaces and use them to demonstrate that the incomplete metric space can be used to obtain a Banach space.
This paper aims to generalize some inequalities in metric spaces by providing an explanation for the fact that every normed space is a metric space, while the converse is not always true. We also present applications of these concepts in metric spaces, supported by relevant results. The utilization of the parallelogram law, a fundamental property of Hilbert spaces, has enabled several researchers, including Kirk [
We present an introduction to some of the fundamental properties of a metric space. In essence, a metric space is defined as a non-empty set X such that to each x , y ∈ X there corresponds a non-negative number called the distance between x and y. The concept of a metric space was initially introduced in 1906 and further developed in 1914. Additionally, a general inequality concerning polygonal inequality that holds true in metric spaces was established in [
A distance on a non-empty set X is defined as a function d : X × X → [ 0 , ∞ ] if the following properties are satisfied:
(i) d ( x , y ) = 0 iff x = y .
(ii) d ( x , y ) = d ( y , x ) for all x , y ∈ X (symmetry).
(iii) d ( x , y ) ≤ d ( x , z ) + d ( z , y ) for any x , y , z ∈ X (triangle inequality).
When these properties are met, the pair (X, d) forms a metric space. One of the main goals of this article is to define metric spaces for specific types of spaces, ensuring that all requirements of a metric space are fulfilled.
We begin by recalling certain fundamental properties of real numbers.
For all x , y , z ∈ ℝ ,
i) | x − y | ≥ 0 ; | x − y | = 0 iff x = y ;
ii) | x − y | = | y − x | ;
iii) | x − y | ≥ | x − z | + | z − y | .
To generalize these properties, let ( X , d ) be a metric space and x = ξ 1 , ξ 2 , ⋯ , ξ m . Then, we have.
d ( x , y ) ≤ d ( x , ξ 1 ) + d ( ξ 1 , ξ 2 ) + ⋯ + d ( ξ m , y ) (1)
d ( x , y ) ≤ d ( x , ξ 1 ) + d ( ξ 1 , y ) (2)
d ( ξ 1 , y ) ≤ d ( ξ 1 , ξ 2 ) + d ( ξ 2 , y ) (3)
d ( x , ξ 1 ) + d ( ξ 1 , y ) ≤ d ( x , ξ 1 ) + d ( ξ 1 , ξ 2 ) + d ( ξ 2 , y ) (4)
d ( x , y ) ≤ d ( x , ξ 1 ) + d ( ξ 1 , ξ 2 ) + d ( ξ 2 , y ) (5)
d ( ξ 2 , y ) ≤ d ( ξ 2 , ξ 3 ) + d ( ξ 3 , y ) (6)
Thus, d ( x , y ) satisfies the properties of a metric space.
Definition 1.2. Let α : X → X . A point x is said to be an α-fixed point of a mapping of F : X → X if α ∘ x = α ∘ F ( x ) .
Definition 2.2. (α-weakly isotone increasing) Let ( X , ≤ ) be a partially ordered set, α : X → X , and F , G be two self-mappings of X. The mapping F is said to be G, α-weakly isotone increasing if for all x ∈ X , we have ( α ∘ F ) x ≤ ( α ∘ G ) x ≤ ( α ∘ F ) ( α ∘ G ) ( α ∘ F ) x .
Let X be a set of ordered pairs of real numbers { x = ( ξ 1 , ξ 2 ) : ξ i ∈ ℝ } ; we define a metric d on ℝ 2 as
d ( x , y ) = ( ξ 1 − η 1 ) 2 + ( ξ 2 − η 2 ) 2 (7)
where x = ( ξ 1 , ξ 2 ) and y = ( η 1 , η 2 ) ∈ ℝ 2 . Moreover, for Euclidean space ℝ n , ℝ n : { ( x = ξ 1 , ξ 2 , ⋯ , ξ i ) } , ξ i are real. Additionally, for C n : z = { ( μ 1 , μ 2 , ⋯ , μ i ) } , z i are complex numbers, and
d ( x , y ) = ( ξ 1 − η 1 ) 2 + ( ξ 2 − η 2 ) 2 + ⋯ + ( ξ n − η n ) n on ℝ n (8)
and
d 1 ( x , y ) = ( ξ 1 − η 1 ) 2 + ( ξ 2 − η 2 ) 2 + ⋯ + ( ξ n − η n ) n on C n (9)
where x = ξ 1 , ξ 2 , ⋯ , ξ n , y = η 1 , η 2 , ⋯ , η n . Then, d 1 ( x , y ) = | ξ 1 − η 1 | + | ξ 2 − η 2 | . To satisfy the triangle inequality, let z = ( μ 1 , μ 2 ) ; then,
d 1 ( x , y ) = | ξ 1 − μ 1 + μ 1 − η 1 | + | ξ 2 − μ 2 + μ 2 − η 2 | ≤ | ξ 1 − μ 1 | + | ξ 2 − μ 2 | + | μ 1 − η 1 | + | μ 2 − η 2 | = d 1 ( x , z ) + d 1 ( z , y ) (10)
Therefore, ( ℝ 2 , d 1 ) is also a metric space. Let X : { x = ( ξ 1 , ξ 2 , ⋯ , ξ n ) : ξ i ∈ ℝ or C } be a set of bounded sequences of real or complex numbers
We define the distance as d ( x , y ) = sup i | ξ i − η i | , where x = ( ξ i ) i = 1 ∞ ∈ L ∞ and y = ( η i ) i = 1 ∞ ∈ L ∞ . Let z = ( μ i ) i = 1 ∞ ∈ L ∞ ,
d ( x , y ) = sup i | ξ i − η i | = sup i | ξ i − μ i + μ i − η i | ≤ sup i | ξ i − μ i | + sup i | μ i − η i | = d ( x , z ) + d ( z , y ) (11)
Thus, ( L ∞ , d ) is a metric space.
Example 1.3. Suppose that S consists of the set of all bounded and unbounded sequences of complex numbers. Let the metric d be defined as d ( x , y ) = ∑ i = 1 ∞ 1 2 i | ξ i − μ i | 1 + | ξ i − μ i | , which is convergent and finite. To prove d ( x , y ) ≤ d ( x , z ) + d ( z , y ) , let z = μ i ∈ S . We consider the function f ( t ) = t 1 + t , where t ∈ ℝ . Since f ′ ( t ) = t ( 1 + t ) 2 > 0 , the function f ( t ) is increasing. Based on the inequality, | a + b | ≤ | a | + | b | , we have f ( | a + b | ) = f ( | a | + | b | ) , which implies
| a + b | 1 + | a + b | ≤ | a | + | b | 1 + | a | + | b | ≤ | a | 1 + | a | + | b | 1 + | b |
Setting a = ξ i − η i and b = μ i − η i , we obtain
| ξ i − η i | 1 + | ξ i − η i | ≤ | ξ i − μ i | 1 + | ξ i − μ i | + | μ i − η i | 1 + | μ i − η i | = ∑ i = 1 ∞ 1 2 i | ξ i − η i | 1 + | ξ i − η i | ≤ ∑ i = 1 ∞ 1 2 i | ξ i − μ i | 1 + | ξ i − μ i | + ∑ i = 1 ∞ 1 2 i | μ i − η i | 1 + | μ i − η i | (12)
Because d ( x , y ) ≤ d ( x , z ) + d ( z , y ) , we conclude that ( S , d ) is a metric space.
Example 2.3. Consider the L b space for p ≥ 1 , where L b : { x = ( ξ 1 , ξ 2 , ⋯ , ξ i , ⋯ ) } such that ∑ i = 1 ∞ | ξ i | b < ∞ and ξ i are scalars.
Result 1. (Holder’s inequality). Let x = ξ j ∈ L b and y = η j ∈ L b . Then, the product of these sequences satisfies
∑ j = 1 ∞ | ξ j η j | ≤ ( ∑ k = 1 ∞ | ξ k | p ) 1 p ( ∑ m = 1 ∞ | η m | q ) 1 q (13)
where p > 1 and 1 p + 1 q = 1 .
Proof. Let ξ ¯ j and η ¯ j be two sequences such that ∑ j = 1 ∞ | ξ ¯ j | p = 1 and ∑ j = 1 ∞ | η ¯ j | q = 1 . Taking α = | ξ ¯ j | and β = | η ¯ j | as real positive numbers, we use the inequality α β = α p p + β q q to obtain
∑ j = 1 ∞ | ξ ¯ η ¯ | ≤ 1 p ∑ j = 1 ∞ | ξ ¯ j | p + 1 q ∑ j = 1 ∞ | η ¯ j | q ≤ 1 p + 1 q = 1 (14)
Let us derive Holder’s inequality. Suppose that x = ξ j ∈ L b and y = η j ∈ L b are non-zero elements with
ξ ¯ j = ξ j ( ∑ k = 1 ∞ | ξ k | p ) 1 p and η ¯ j = η j ( ∑ m = 1 ∞ | η m | q ) 1 q (15)
Clearly, the sequences ξ ¯ j and η ¯ j satisfy (13). Hence, using (13), we obtain ∑ j = 1 ∞ | ξ j η j | ≤ ( ∑ k = 1 ∞ | ξ k | p ) 1 p ( ∑ m = 1 ∞ | η m | q ) 1 q , which is finite.
Result 2. (Minkowski inequality). Let x = ξ j ∈ L b , y = η j ∈ L b , and p ≥ 1 . Then,
( ∑ j = 1 ∞ | ξ j + η j | p ) 1 p ≤ ( ∑ k = 1 ∞ | ξ k | p ) 1 p + ( ∑ m = 1 ∞ | η m | p ) 1 p
Proof. By setting p = 1 and | ξ j + η j | ≤ | ξ j | + | η j | and applying the triangle inequality, we obtain
∑ j = 1 ∞ | ξ j + η j | ≤ ∑ j = 1 ∞ | ξ j | + ∑ j = 1 ∞ | η j | (16)
For simplicity, let ξ j + η j = ω j ; then,
| ω j | = | ξ j + η j | p = | ξ j + η j | p − 1 ≤ | ξ j | | ω j | p − 1 + | η j | | ω j | p − 1 (17)
By choosing j = 1 , 2 , ⋯ , n (any fixed value of n), x = ξ j ∈ L b and | ω j | p − 1 ∈ L q because
( | ω j | p − 1 ) q = | ω j | ( p − 1 ) q (18)
Because ∑ j = 1 ∞ | ω j | ( p − 1 ) q = ∑ j = 1 ∞ | ω j | p < ∞ , we can apply the Holder’s inequality to obtain
∑ j = 1 n | ξ j | | ω j | p − 1 ≤ ( ∑ k = 1 n | ξ k | p ) 1 p ( ( ∑ m = 1 n | ω m | p − 1 ) q ) 1 q = ( ∑ k = 1 n | ξ k | p ) 1 p ( ∑ m = 1 n | ω m | p ) 1 q
∑ j = 1 n | ξ j | | ω j | p − 1 ≤ ( ∑ k = 1 ∞ | ξ k | p ) 1 p ( ∑ m = 1 ∞ | ω m | p ) 1 q (19)
Then, we obtain
∑ j = 1 n | ω j | p ≤ ∑ j = 1 n ( | ξ j | + | η j | ) | ω j | p − 1 ≤ ( ( ∑ k = 1 n | ξ k | p ) 1 p + ( ∑ k = 1 n | η k | p ) 1 p ) ( ∑ m = 1 n | ω m | p ) 1 q (20)
Taking the limit as n → ∞ , we obtain
( ∑ j = 1 ∞ | ω j | p ) 1 − 1 q ≤ ( ( ∑ k = 1 ∞ | ξ k | p ) 1 p + ( ∑ k = 1 ∞ | η k | p ) 1 p ) (21)
Thus,
( ∑ j = 1 ∞ | ω j | p ) 1 p ≤ ( ( ∑ k = 1 ∞ | ξ k | p ) 1 p + ( ∑ k = 1 ∞ | η k | p ) 1 p ) (22)
Finally, we conclude that
( ∑ j = 1 ∞ | ξ j + η j | p ) 1 p ≤ ( ∑ k = 1 ∞ | ξ k | p ) 1 p + ( ∑ k = 1 ∞ | η k | p ) 1 p (23)
Theorem 3.3. A mapping T of a metric space ( X , d ) into a metric space ( X , d ) is continuous if and only if the inverse image of any open subset of Y is an open subset of X.
Definition 3.4. (Complete metric space) A continuous metric space ( X , d ) is said to be complete if every Cauchy sequence in X converges to an element of X.
Example 3.5. Let ( ℝ n , d ) be a complete metric space.
L ∞ d ( x , y ) = ∑ i = 1 n | ξ i − μ i | 2 (24)
where x = ξ 1 , ξ 2 , ⋯ , ξ n , y = η 1 , η 2 , ⋯ , η n ∈ ℝ n . Now, we are ready to state our main result.
Example 3.6. A Banach space under the norm defined by ‖ x ‖ 2 = [ ∑ i = 1 n | ξ i | 2 ] 1 2 ,where x = ( ξ i ) i = 1 n = ( ξ 1 , ξ 2 , ⋯ , ξ n ) ∈ ℝ n , ξ i ∈ ℝ for all i.
Proof. To prove that ‖ x ‖ 2 is a normed linear space, we prove the properties of a norm:
(i) ‖ x ‖ = [ ∑ i = 1 n | ξ i | 2 ] 1 2 ≥ 0 , ξ i ∈ ℝ for all i. ‖ x ‖ = 0 implies that [ ∑ i = 1 n | ξ i | 2 ] 1 2 = 0 ; then, ∑ i = 1 n | ξ i | 2 = 0 , | ξ i | 2 = 0 , | ξ i | = 0 , and ξ i = 0 .
(ii) ‖ x + y ‖ 2 = [ ∑ i = 1 n | ξ i + η i | 2 ] 1 2 = ∑ i = 1 n | ξ i + η i | + | ξ i + η i | . ‖ x + y ‖ 2 ≤ ∑ i = 1 n ( | ξ i | + | η i | ) | ξ i + η i | = ∑ i = 1 n | ξ i | | ξ i + η i | + ∑ i = 1 n | η i | | ξ i + η i | = ∑ i = 1 n | ξ i ( ξ i + η i ) | + ∑ i = 1 n | η i ( ξ i + η i ) | ≤ ‖ x ‖ ‖ x + y ‖ + ‖ y ‖ ‖ x + y ‖ implies that ‖ x + y ‖ 2 ≤ ‖ x + y ‖ ( ‖ x + y ‖ ) . Then, ‖ x + y ‖ ≤ ‖ x ‖ + ‖ y ‖ .
(iii) ‖ α x ‖ = [ ∑ i = 1 n | α ξ i | 2 ] 1 2 = [ ∑ i = 1 n | α | 2 | ξ i | 2 ] 1 2 = | α | [ ∑ i = 1 n | ξ i | 2 ] 1 2 = | α | ‖ x ‖ . Hence, it is a normed space.
Proof of completeness: Let 〈 x m 〉 be a Cauchy sequence in ℝ 2 . Then, for any ε > 0 , ∃ n 0 ∈ N such that ‖ x m − x n ‖ < ε ∃ m , r ≥ n 0 and x m , x r ∈ ℝ 2 .
x m = ( ξ 1 ( m ) , ξ 2 ( m ) , ⋯ , ξ i ( m ) , ⋯ , ξ n ( m ) ) and x n = ( ξ 1 ( r ) , ξ 2 ( r ) , ⋯ , ξ i ( r ) , ⋯ , ξ n ( r ) ) , ξ i ( m ) , ξ i ( r ) ∈ ℝ , ∀ i so [ ∑ i = 1 n | ξ i ( m ) − ξ i ( r ) | 2 ] 1 2 < ε , ∀ m , r ≥ n 0 .
∑ i = 1 n | ξ i ( m ) − ξ i ( r ) | 2 < ε 2 , ∀ m , r ≥ n 0
| ξ i ( m ) − ξ i ( r ) | 2 < ε 2 , ∀ m , r ≥ n 0
| ξ i ( m ) − ξ i ( r ) | < ε , ∀ m , r ≥ n 0
Because 〈 ξ i ( m ) 〉 is a Cauchy sequence in ℝ , ξ i ( m ) → ξ i ∈ ℝ , and ℝ is complete. Let x = ξ 1 , ξ 2 , ⋯ , ξ i , ⋯ , ξ n ∈ ℝ , ξ i ∈ ℝ , ∀ i . Now, ‖ x m − x ‖ = [ ∑ i = 1 n | ξ i ( m ) − ξ i | 2 ] 1 2 , where ξ i ( m ) → ξ i , ∀ i ; then, ξ i ( m ) − ξ i → 0 as m → ∞ , and x m − x → 0 as m → ∞ implies that x m → x ∈ ℝ 2 , so ℝ n is a complete space.
Lemma 3.7. Let ( L ∞ , d ∞ ) be a complete metric space L ∞ : { x = ( ξ i ) , ξ i ∈ ℝ n or C : sup i | ξ i | < ∞ } . d ∞ d ( x , y ) = sup i | ξ i − μ i | , where x = ( ξ i ) i = 1 ∞ and y = ( η i ) i = 1 ∞ ∈ L ∞ .
Claim. We consider L ∞ . Given x m = ( ξ i m ) i = 1 ∞ as a Cauchy sequence in L ∞ , for given ε > 0 , ∃ N ( ε ) such that for n ≥ N , d ∞ ( x m , x n ) < ε , ∃ m , r ≥ N , sup i | x m − μ i ( r ) | < ε . For each fixed | ξ i ( m ) − μ i ( r ) | < ε , we consider ( ξ i ( 1 ) , ξ i ( 2 ) , ⋯ ) . x i behaves as a real or complex Cauchy sequence because x m = ( η 1 , η 2 , ⋯ , η n , ⋯ ) . To show x ∈ L ∞ , we obtain sup i | ξ i ( m ) − μ i ( r ) | < ε for m , r ≥ N , each i, and let r → ∞ . Thus, d ∞ ( x m , x ) = sup i | ξ i ( m ) − μ i ( r ) | < ε , x m → x because
| ξ i | = | ξ i − ξ i ( m ) | + | ξ i ( m ) | < ε + k m (25)
where k m = sup i | ξ i ( m ) | < ∞ , x m ∈ L ∞ . ( L ∞ , d ∞ ) is a complete metric space. Alternative proofs do exist.
Result 3. Every normed space is a metric space, but the converse need not be true in general.
Let S be s set of sequences (bounded or unbounded) of real or complex numbers and define d ( x , y ) = ∑ i = 1 ∞ 1 2 i | ξ i − η i | 1 + | ξ i − η i | , where x = ( ξ i ) i = 1 ∞ ∈ X and y = ( η i ) i = 1 ∞ ∈ X . Clearly, ( S , d ) is a metric space. The question is whether it is a normed space? The answer is no. If it were a normed space, we could define ( x , 0 ) = ‖ x ‖ = ∑ i = 1 ∞ 1 2 i | x i | 1 + | x i | . In that case, we would have the following:
‖ α x ‖ = ∑ i = 1 ∞ 1 2 i | α x i | 1 + | α x i | = | α | ‖ x ‖ (26)
‖ α x ‖ ≠ | α | ‖ x ‖ which fails to satisfy the norm property. So, ( S , ‖ . ‖ ) is not a normed space, but it is a metric space.
Lemma 3.8. Consider L p space and p > 1 : L p : { x = ( ξ i ) i = 1 ∞ , ξ i ∈ ℝ or ℂ : sup i | ξ i | p < ∞ } . Define ‖ x ‖ L p = ( ∑ i = 1 ∞ | ξ i | p ) 1 p . Therefore, ( L p , ‖ ⋅ ‖ p ) is a normed space, so
d ( x , y ) = ( ∑ i = 1 ∞ | ξ i − η i | p ) 1 p = ‖ x − y ‖ L p (27)
where x = ( ξ i ) ∈ L p and y = ( η i ) ∈ L p . Thus, ( L p , d ) is a complete metric space. Hence, ( L p , ‖ ⋅ ‖ p ) is a Banach space.
Theorem 2.9. Let ( X , ≤ ) be a partially ordered set and suppose that there exists a metric d on X such that ( X , d ) is a complete metric space. Let α : X → X and F and G be two self-mappings of ( X , d ) such that for comparable x , y ∈ X ,
ξ d ( α ∘ F ( x ) , α ∘ G ( y ) ) + η d ( α ∘ F ( x ) , α ( x ) ) + μ d ( α ( y ) , α ∘ G ( y ) ) − min { d ( α ∘ F ( x ) , α ( y ) ) , d ( α ∘ G ( y ) , α ( x ) ) } ≤ k max { d ( α ( x ) , α ( y ) ) , d ( α ∘ F ( x ) , α ( x ) ) , d ( α ( y ) , α ∘ G ( y ) , 1 2 d ( α ∘ F ( x ) , α ( y ) ) ) }
For ξ , η , μ > 0 , k > 0 , and ξ > k , we assume the following:
(i) F is G,α-weakly increasing and
(ii) X is regular.
Then, F and G have a unique α-fixed point.
Proof. Let x 0 ∈ X . From the sequence x n with respect to α , we obtain x 2 n + 2 = α ∘ F ( x 2 n + 1 ) = F α ( x 2 n + 1 ) and x 2 n + 1 = α ∘ G ( x 2 n ) = G α ( x 2 n ) for n = 1 , 2 , ⋯ . Let d n = d ( α ∘ ( x n ) ) , ( α ∘ ( x n + 1 ) ) > 0 , n = 1 , 2 , ⋯ . Because G α is F α -weakly increasing, we have
x 1 ≤ α ∘ G ( x 0 ) ≤ α ∘ F ( α ∘ G ( x 0 ) ) = α ∘ F ( x 1 ) = x 2 ≤ ( α ∘ G ) ( α ∘ F ( α ∘ G ( x 0 ) ) ) = α ∘ G ( α ∘ F ( x 1 ) ) = α ∘ G ( x 2 ) = x 3 ≤ α ∘ G ( x 1 ) ≤ α ∘ F ( α ∘ G ( x 2 ) ) = α ∘ F ( x 3 ) = x 4 ≤ ( α ∘ G ) α ∘ F ( α ∘ G ( x 2 ) ) = ( α ∘ G ) ( α ∘ F ( x 3 ) ) = x 5 (28)
By continuing this process, we obtain x 1 ≤ x 2 ≤ x 3 ≤ ⋯ ≤ x n ≤ x n + 1 ≤ ⋯ so x 2 n ≤ x 2 n + 1 , ∀ n = 1 , 2 , ⋯ . Now, with x = x 2 n + 1 and y = x 2 n , we have
[ ξ d ( α ∘ F ( x 2 n + 1 ) ) , α ∘ G ( x 2 n ) + η d ( α ∘ F ( x 2 n + 1 ) , x 2 n + 2 ) ] + μ d ( x 2 n , α ∘ G ( x 2 n ) ) − min { d ( α ∘ F ( x 2 n + 1 ) , x 2 n ) , d ( α ∘ G ( x 2 n ) , x 2 n + 1 ) } ≤ k max { d ( x 2 n + 1 , x 2 n ) , d ( α ∘ F ( x 2 n + 1 ) , x 2 n + 1 ) , d ( x 2 n , α ∘ G ( x 2 n ) , 1 2 d ( α ∘ F ( x 2 n + 1 ) , x 2 n ) ) } (29)
or
[ ξ d ( x 2 n + 2 , x 2 n + 1 ) + η d ( x 2 n + 2 , x 2 n + 1 ) + μ d ( x 2 n , x 2 n + 1 ) − min { d ( x 2 n , x 2 n + 2 ) , d ( x 2 n + 1 , x 2 n + 1 ) } ] ≤ k max { d ( x 2 n + 1 , x 2 n ) , d ( x 2 n + 2 , x 2 n + 1 ) , d ( x 2 n , x 2 n + 1 ) , 1 2 d ( x 2 n + 1 , x 2 n ) } (30)
or
ξ d 2 n + 1 + η d 2 n + 1 + μ d 2 n − min { d 2 n , d 2 n + 1 , 0 } ≤ k max { d 2 n , d 2 n + 1 , 1 2 ( d 2 n , d 2 n + 1 ) } (31)
Letting H = max { d 2 n , d 2 n + 1 } , we have d 2 n ≤ H , and d 2 n + 1 ≤ H implies that 1 2 ( d 2 n , d 2 n + 1 ) ≤ H . Therefore, max { d 2 n , d 2 n + 1 , 1 2 ( d 2 n , d 2 n + 1 ) } ≤ H = max { d 2 n , d 2 n + 1 } . From Equation (30), ( ξ + η ) d 2 n + 1 + μ d 2 n ≤ k max { d 2 n , d 2 n + 1 } if d 2 n ≤ d 2 n + 1 . Then, ( ξ + η ) d 2 n + 1 + μ d 2 n ≤ k d 2 n + 1 , so ( ξ + η − k ) d 2 n + 1 ≤ − μ d 2 n , d 2 n + 1 ≤ g d 2 n , where g = − μ ξ + η − f ≤ 1 , and if d 2 n + 1 ≤ d 2 n , then ( ξ + η ) d 2 n + 1 + μ d 2 n ≤ g d 2 n implies that d 2 n + 1 ≤ g − μ ξ + η d 2 n and d 2 n + 1 ≤ g d 2 n , where g = f − μ ξ + η d 2 n < 1 . Therefore, d 2 n + 1 ≤ g d 2 n ≤ g 2 d 2 n − 1 ≤ ⋯ ≤ g 2 n + 1 d 0 → 0 as n → ∞ . Thus, { x n } is a Cauchy sequence in X, X is complete, and there exists a point z ∈ X such that { x 2 n } converges to z. Hence, lim ( α ∘ F ) ( x 2 n + 1 ) = x 2 n + 1 = z and lim ( α ∘ F ) ( x 2 n + 1 ) = z . Because { x 2 n } is a nondecreasing sequence, if X is regular, it follows that x 2 n ≤ z , ∀ n . Now, if we put x = x 2 n + 1 and y = z , we obtain
[ ξ d ( α ∘ F ) x 2 n + 1 , ( ( α ∘ G ) , z ) + η d ( x 2 n + 1 , ( α ∘ F ) x 2 n + 1 ) + μ d ( z , ( α ∘ G ) z ) − min { d ( ( α ∘ G ) ( x 2 n + 1 ) , z ) , d ( ( α ∘ G ) z , x 2 n + 1 ) } ] ≤ k max { d ( x 2 n + 1 , z ) , d ( ( α ∘ F ) ( x 2 n + 1 ) , ( x 2 n + 1 ) , d ( z , ( α ∘ G ) z ) , 1 2 d ( ( α ∘ F ) ( x 2 n + 1 ) , z ) ) } .
Finally, we arrive at the following conclusion:
ξ d ( z , α ∘ F ) ( z ) + η d ( z , z ) + μ d ( z , α ∘ G ( z ) ) − min { d ( z , z ) , d ( z , α ∘ G ( u ) ) } ≤ k max { d ( z , z ) , d ( z , z ) , d ( z , α ∘ G ( z ) , 1 2 d ( z , z ) ) } .
Alternatively, we can simplify it as ( ξ + μ − g ) d ( z , α ∘ G ( z ) ) ≤ 0 or α ∘ G ( z ) = z , given that ξ > 1 + g . Thus, z is a fixed point of G. Additionally, using similar reasoning with x = z , y = x 2 n , we obtain α ∘ z = α ∘ F ( z ) . Hence, z is a common α-fixed point of F and G.
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
The authors express their gratitude to the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University, which funded their work via Research Group no. RG-21-09-51.
Hassan, E.I. (2023) Generalization of Inequalities in Metric Spaces with Applications. Journal of Applied Mathematics and Physics, 11, 2923-2931. https://doi.org/10.4236/jamp.2023.1110193