Open Access

Non-Archimedean and random HUR-approximation of a Cauchy-Jensen additive mapping

Journal of Inequalities and Applications20142014:209

https://doi.org/10.1186/1029-242X-2014-209

Received: 30 August 2012

Accepted: 26 March 2014

Published: 23 May 2014

Abstract

In this paper, using the fixed point and direct methods, we prove the Hyers-Ulam-Rassias approximation (briefly, HUR-approximation) of a Cauchy-Jensen additive (briefly, CJA) functional equation in various normed spaces.

MSC:39B52, 39B82, 47H10, 46S10.

Keywords

HUR-approximation non-Archimedean normed space random normed spaces direct method fixed point method

1 Introduction

A classical question in the theory of functional equations is the following: When is it true that a function which approximately satisfies a functional equation must be close to an exact solution of the equation? If the problem accepts a solution, we say that the equation is stable. The first stability problem concerning group homomorphisms was raised by Ulam [1] in 1940. In 1941, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. Aoki [3] proved a generalization of Hyers’ theorem for additive mappings and Rassias [4] proved a generalization of Hyers’ theorem for linear mappings.

Theorem 1.1 (ThM Rassias)

Let f : E E be a mapping from a normed vector space E into a Banach space E subject to the inequality f ( x + y ) f ( x ) f ( y ) ε ( x p + y p ) , for all x , y E , where ε and p are constants with ε > 0 and 0 p < 1 . Then the limit L ( x ) = lim n f ( 2 n x ) 2 n exists, for all x E , and L : E E is the unique additive mapping which satisfies
f ( x ) L ( x ) 2 ε 2 2 p x p ,

for all x E . Also, if for each x E the function f ( t x ) is continuous in t R , then L is linear.

This new concept is known as a the Hyers-Ulam stability or the Hyers-Ulam-Rassias stability of functional equations. Furthermore, in 1994, a generalization of Rassias’ theorem was obtained by Gǎvruta [5] by replacing the bound ε ( x p + y p ) by a general control function φ ( x , y ) .

In 1983, a generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [6] for mappings f : X Y , where X is a normed space and Y is a Banach space. In 1984, Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik [8] proved the generalized Hyers-Ulam stability of the quadratic functional equation. The readers are referred to [929] and references therein for detailed information on stability of functional equations.

In 1897, Hensel [30] has introduced a normed space which does not have the Archimedean property. It turned out that non-Archimedean spaces have many nice applications (see [3135]).

Definition 1.1 By a non-Archimedean field we mean a field equipped with a function (valuation) | | : K [ 0 , ) such that, for all r , s K , the following conditions hold: (a)  | r | = 0 if and only if r = 0 ; (b) | r s | = | r | | s | ; (c) | r + s | max { | r | , | s | } .

Clearly, by (b), | 1 | = | 1 | = 1 and so, by induction, it follows from (c) that | n | 1 , for all n 1 .

Definition 1.2 Let X be a vector space over a scalar field with a non-Archimedean non-trivial valuation | | .
  1. (1)

    A function : X R is a non-Archimedean norm (valuation) if it satisfies the following conditions: (a) x = 0 if and only if x = 0 , for all x X ; (b) r x = | r | x , for all r K and x X ; (c) the strong triangle inequality (ultra-metric) holds, that is, x + y max { x , y } , for all x , y X .

     
  2. (2)

    The space ( X , ) is called a non-Archimedean normed space (briefly, NAN-space).

     

Note that x n x m max { x j + 1 x j : m j n 1 } , for all m , n N with n > m .

Definition 1.3 Let ( X , ) be a non-Archimedean normed space.
  1. (a)

    A sequence { x n } is a Cauchy sequence in X if { x n + 1 x n } converges to zero in X.

     
  2. (b)

    The non-Archimedean normed space ( X , ) is said to be complete if every Cauchy sequence in X is convergent.

     

The most important examples of non-Archimedean spaces are p-adic numbers. A key property of p-adic numbers is that they do not satisfy the Archimedean axiom: for all x , y > 0 , there exists a positive integer n such that x < n y .

Example 1.1 Fix a prime number p. For any nonzero rational number x, there exists a unique positive integer n x such that x = a b p n x , where a and b are positive integers not divisible by p. Then | x | p : = p n x defines a non-Archimedean norm on . The completion of with respect to the metric d ( x , y ) = | x y | p is denoted by Q p , which is called the p-adic number field. In fact, Q p is the set of all formal series x = k n x a k p k , where | a k | p 1 . The addition and multiplication between any two elements of Q p are defined naturally. The norm | k n x a k p k | p = p n x is a non-Archimedean norm on Q p and Q p is a locally compact field.

In Section 3, we adopt the usual terminology, notions and conventions of the theory of random normed spaces as in [36]. Throughout this paper, let + denote the set of all probability distribution functions F : R [ , + ] [ 0 , 1 ] such that F is left-continuous and nondecreasing on and F ( 0 ) = 0 , F ( + ) = 1 . It is clear that the set D + = { F + : l F ( ) = 1 } , where l f ( x ) = lim t x f ( t ) , is a subset of +. The set + is partially ordered by the usual point-wise ordering of functions, that is, F G if and only if F ( t ) G ( t ) , for all t R . For any a 0 , the element H a ( t ) of D + is defined by
H a ( t ) = { 0 , if  t a , 1 , if  t > a .

We can easily show that the maximal element in + is the distribution function H 0 ( t ) .

Definition 1.4 A function T : [ 0 , 1 ] 2 [ 0 , 1 ] is a continuous triangular norm (briefly, a t-norm) if T satisfies the following conditions: (a) T is commutative and associative; (b) T is continuous; (c) T ( x , 1 ) = x , for all x [ 0 , 1 ] ; (d) T ( x , y ) T ( z , w ) whenever x z and y w , for all x , y , z , w [ 0 , 1 ] .

Three typical examples of continuous t-norms are as follows: T P ( x , y ) = x y , T max ( x , y ) = max { a + b 1 , 0 } , T M ( x , y ) = min ( a , b ) . Recall that, if T is a t-norm and { x n } is a sequence in [ 0 , 1 ] , then T i = 1 n x i is defined recursively by T i = 1 1 x 1 = x 1 and T i = 1 n x i = T ( T i = 1 n 1 x i , x n ) , for all n 2 . T i = n x i is defined by T i = 1 x n + i .

Definition 1.5 A random normed space (briefly, RN-space) is a triple ( X , μ , T ) , where X is a vector space, T is a continuous t-norm and μ : X D + is a mapping such that the following conditions hold:
  1. (a)

    μ x ( t ) = H 0 ( t ) , for all t > 0 if and only if x = 0 ;

     
  2. (b)

    μ α x ( t ) = μ x ( t | α | ) , for all α R with α 0 , x X and t 0 ;

     
  3. (c)

    μ x + y ( t + s ) T ( μ x ( t ) , μ y ( s ) ) , for all x , y X and t , s 0 .

     

Every normed space ( X , ) defines a random normed space ( X , μ , T M ) , where μ u ( t ) = t t + u , for all t > 0 and T M is the minimum t-norm. This space X is called the induced random normed space.

If the t-norm T is such that sup 0 < a < 1 T ( a , a ) = 1 , then every RN-space ( X , μ , T ) is a metrizable linear topological space with the topology τ (called the μ-topology or the ( ε , δ ) -topology, where ε > 0 and λ ( 0 , 1 ) ) induced by the base { U ( ε , λ ) } of neighborhoods of θ, where
U ( ε , λ ) = { x X : μ x ( ε ) > 1 λ } .
Definition 1.6 Let ( X , μ , T ) be an RN-space.
  1. (a)

    A sequence { x n } in X is said to be convergent to a point x X (write x n x as n ) if lim n μ x n x ( t ) = 1 , for all t > 0 .

     
  2. (b)

    A sequence { x n } in X is called a Cauchy sequence in X if lim n μ x n x m ( t ) = 1 , for all t > 0 .

     
  3. (c)

    The RN-space ( X , μ , T ) is said to be complete if every Cauchy sequence in X is convergent.

     

Theorem 1.2 If ( X , μ , T ) is RN-space and { x n } is a sequence such that x n x , then lim n μ x n ( t ) = μ x ( t ) .

Definition 1.7 Let X be a set. A function d : X × X [ 0 , ] is called a generalized metric on X if d satisfies the following conditions:
  1. (a)

    d ( x , y ) = 0 if and only if x = y , for all x , y X ;

     
  2. (b)

    d ( x , y ) = d ( y , x ) , for all x , y X ;

     
  3. (c)

    d ( x , z ) d ( x , y ) + d ( y , z ) , for all x , y , z X .

     

Theorem 1.3 ([37, 38])

Let ( X , d ) be a complete generalized metric space and J : X X be a strictly contractive mapping with Lipschitz constant L < 1 . Then, for all x X , either d ( J n x , J n + 1 x ) = , for all nonnegative integers n, or there exists a positive integer n 0 such that
  1. (a)

    d ( J n x , J n + 1 x ) < , for all n 0 n 0 ;

     
  2. (b)

    the sequence { J n x } converges to a fixed point y of J;

     
  3. (c)

    y is the unique fixed point of J in the set Y = { y X : d ( J n 0 x , y ) < } ;

     
  4. (d)

    d ( y , y ) d ( y , J y ) 1 L , for all y Y .

     
In this paper, using the fixed point and direct methods, we prove the HUR-approximation of the following CJA functional equation:
2 f ( x + y + z 2 ) = f ( x ) + f ( y ) + f ( z )
(1.1)

in various normed spaces.

2 NAN-stability

In this section, we deal with the stability problem for the Cauchy-Jensen additive functional equation (1.1) in non-Archimedean normed spaces.

Theorem 2.1 Let X be a non-Archimedean normed space and Y is a complete non-Archimedean space. Let φ : X 3 [ 0 , ) be a function such that there exists an α < 1 with
φ ( x 2 , y 2 , z 2 ) α φ ( x , y , z ) | 2 | ,
(2.1)
for all x , y , z X . Let f : X Y be a mapping satisfying
2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) Y φ ( x , y , z ) ,
(2.2)
for all x , y , z X . Then there exists a unique additive mapping : X Y such that
f ( x ) ( x ) Y α φ ( x , 2 x , x ) ( | 2 | | 2 | α ) 1 ,
(2.3)

for all x X .

Proof Putting y = 2 x and z = x in (2.1), we get f ( 2 x ) 2 f ( x ) Y φ ( x , 2 x , x ) , for all x X . So
f ( x ) 2 f ( x 2 ) Y | 2 | 1 α φ ( x , 2 x , x ) ,
(2.4)
for all x X . Consider the set S : = { h : X Y } and introduce the generalized metric on S:
d ( g , h ) = inf μ ( 0 , + ) g ( x ) h ( x ) Y μ φ ( x , 2 x , x ) ,
for all x X , where, as usual, inf ϕ = + . It is easy to show that ( S , d ) is complete (see [39]). Now we consider the linear mapping J : S S such that J g ( x ) : = 2 g ( x 2 ) , for all x X . Let g , h S be given such that d ( g , h ) = ε . Then g ( x ) h ( x ) Y ε φ ( x , 2 x , x ) , for all x X . Hence
J g ( x ) J h ( x ) Y = 2 g ( x 2 ) 2 h ( x 2 ) Y α ε φ ( x , 2 x , x ) ,
for all x X . So d ( g , h ) = ε implies that d ( J g , J h ) α ε . This means that d ( J g , J h ) α d ( g , h ) , for all g , h S . It follows from (2.4) that d ( f , J f ) | 2 | 1 α . By Theorem 1.3, there exists a mapping : X Y satisfying the following:
  1. (1)
    is a fixed point of J, i.e.,
    ( x ) = 2 ( x 2 ) ,
    (2.5)
     
for all x X . The mapping is a unique fixed point of J in the set M = { g S : d ( h , g ) < } . This implies that is a unique mapping satisfying (2.5) such that there exists a μ ( 0 , ) satisfying f ( x ) ( x ) Y μ φ ( x , 2 x , x ) , for all x X ;
  1. (2)
    d ( J n f , ) 0 as n . This implies the equality
    lim n 2 n f ( x 2 n ) = ( x ) ,
    (2.6)
     
for all x X ;
  1. (3)
    d ( f , ) d ( f , J f ) 1 α , which implies the inequality d ( f , ) α ( | 2 | | 2 | α ) 1 . This implies that the inequalities (2.3) holds. It follows from (2.1) and (2.2) that
    2 ( x + y + z 2 ) ( x ) ( y ) ( z ) Y = lim n 2 f ( x + y + z 2 n + 1 ) f ( x 2 n ) f ( y 2 n ) f ( z 2 n ) Y | 2 | n lim n α n φ ( x , y , z ) = 0 ,
     

for all x , y , z X . So 2 ( x + y + z 2 ) = ( x ) + ( y ) + ( z ) , for all x , y , z X . Hence : X Y is an CJA mapping and we get the desired results. □

Corollary 2.1 Let θ be a positive real number and r is a real number with 0 < r < 1 . Let f : X Y be a mapping satisfying
2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) Y θ ( x r + y r + z r ) ,
(2.7)
for all x , y , z X . Then there exists a unique CJA mapping : X Y such that
f ( x ) ( x ) Y | 2 | θ ( 2 + | 2 | r ) ( | 2 | r + 1 | 2 | 2 ) 1 x r ,

for all x X .

Proof The proof follows from Theorem 2.1 by taking φ ( x , y , z ) = ( x r + y r + z r ) , for all x , y , z X . Then we can choose α = | 2 | 1 r and we get the desired result. □

Theorem 2.2 Let X be a non-Archimedean normed space and Y is a complete non-Archimedean space. Let φ : X 3 [ 0 , ) be a function such that there exists an α < 1 with φ ( x , y , z ) | 2 | α φ ( x 2 , y 2 , z 2 ) , for all x , y , z X . Let f : X Y be a mapping satisfying (2.2). Then there exists a unique CJA mapping : X Y such that
f ( x ) ( x ) Y φ ( x , 2 x , x ) ( | 2 | | 2 | α ) 1 ,
(2.8)

for all x X .

Proof Let ( S , d ) be the generalized metric space defined in the proof of Theorem 2.1. Now we consider the linear mapping J : S S such that J g ( x ) : = g ( 2 x ) 2 , for all x X . Let g , h S be given such that d ( g , h ) = ε . Then g ( x ) h ( x ) Y ε φ ( x , 2 x , x ) , for all x X . Hence
J g ( x ) J h ( x ) Y = g ( 2 x ) 2 h ( 2 x ) 2 Y | 2 | α ε φ ( x , 2 x , x ) | 2 | ,
for all x X . So d ( g , h ) = ε implies that d ( J g , J h ) α ε . This means that d ( J g , J h ) α d ( g , h ) , for all g , h S . It follows from (2.4) that d ( f , J f ) | 2 | 1 . By Theorem 1.3, there exists a mapping : X Y satisfying the following:
  1. (1)
    is a fixed point of J, i.e.,
    ( 2 x ) 2 = ( x ) ,
    (2.9)
     
for all x X . The mapping is a unique fixed point of J in the set M = { g S : d ( h , g ) < } . This implies that is a unique mapping satisfying (2.9) such that there exists a μ ( 0 , ) satisfying f ( x ) ( x ) Y μ φ ( x , 2 x , x ) , for all x X ;
  1. (2)

    d ( J n f , ) 0 as n . This implies the equality lim n f ( 2 n x ) 2 n = ( x ) , for all x X ;

     
  2. (3)

    d ( f , ) d ( f , J f ) 1 α , which implies the inequality d ( f , ) ( | 2 | | 2 | α ) 1 . This implies that the inequalities (2.8) holds. The rest of the proof is similar to the proof of Theorem 2.1. □

     
Corollary 2.2 Let θ be a positive real number and r is a real number with r > 1 . Let f : X Y be a mapping satisfying (2.7). Then there exists a unique CJA mapping : X Y such that
f ( x ) ( x ) Y θ ( 2 + | 2 | r ) ( | 2 | | 2 | r ) 1 x r ,

for all x X .

Proof The proof follows from Theorem 2.2 by taking φ ( x , y , z ) = ( x r + y r + z r ) , for all x , y , z X . Then we can choose α = | 2 | r 1 and we get the desired result. □

Theorem 2.3 Let G be an additive semigroup and X is a non-Archimedean Banach space. Assume that λ : G 3 [ 0 , + ) be a function such that
lim n | 2 | n λ ( x 2 n , y 2 n , z 2 n ) = 0 ,
(2.10)
for all x , y , z G . Suppose that, for any x G , the limit
£ ( x ) = lim n max 0 k < n | 2 | k λ ( x 2 k + 1 , x 2 k , x 2 k + 1 )
(2.11)
exists and f : G X be a mapping satisfying
2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) X λ ( x , y , z ) .
(2.12)
Then the limit ( x ) : = lim n 2 n f ( x 2 n ) exists, for all x G , and defines an CJA mapping : G X such that
f ( x ) ( x ) £ ( x ) .
(2.13)

Moreover, if lim j lim n max j k < n + j | 2 | k λ ( x 2 k + 1 , x 2 k , x 2 k + 1 ) = 0 then is the unique CJA mapping satisfying (2.13).

Proof Putting y = 2 x and z = x in (2.12), we get
f ( 2 x ) 2 f ( x ) Y λ ( x , 2 x , x ) ,
(2.14)
for all x G . Replacing x by x 2 n + 1 in (2.14), we obtain
2 n + 1 f ( x 2 n + 1 ) 2 n f ( x 2 n ) | 2 | n λ ( x 2 n + 1 , x 2 n , x 2 n + 1 ) .
(2.15)
Thus, it follows from (2.10) and (2.15) that the sequence { 2 n f ( x 2 n ) } n 1 is a Cauchy sequence. Since X is complete, it follows that { 2 n f ( x 2 n ) } n 1 is convergent. Set ( x ) : = lim n 2 n f ( x 2 n ) . By induction on n, one can show that
2 n f ( x 2 n ) f ( x ) max 0 k < n | 2 | k λ ( x 2 k + 1 , x 2 k , x 2 k + 1 ) ,
(2.16)
for all n 1 and x G . By taking n in (2.16) and using (2.11), one obtains (2.13). By (2.10) and (2.12), we get
2 ( x + y + z 2 ) ( x ) ( y ) ( z ) = lim n | 2 | n 2 f ( x + y + z 2 n + 1 ) f ( x 2 n ) f ( y 2 n ) f ( z 2 n ) lim n | 2 | n λ ( x 2 n , y 2 n , z 2 n ) = 0 ,
for all x , y , z X . So
( x + y + z 2 ) = ( x ) + ( y ) + ( z ) .
(2.17)

Letting x = y = z = 0 in (2.17), we get ( 0 ) = 0 . Letting z = x + y in (2.17), we get ( x + y ) = ( x ) + ( y ) , for all x , y X . Hence the mapping : X Y is Cauchy additive.

To prove the uniqueness property of , let be another mapping satisfying (2.13). Then we have
( x ) ( x ) X = lim n | 2 | n ( x 2 n ) ( x 2 n ) X lim k | 2 | n max { ( x 2 n ) f ( x 2 n ) X , f ( x 2 n ) ( x 2 n ) X } lim j lim n max j k < n + j | 2 | k λ ( x 2 k + 1 , x 2 k , x 2 k + 1 ) = 0 ,

for all x G . Therefore, = . This completes the proof. □

Corollary 2.3 Let ξ : [ 0 , ) [ 0 , ) be a function satisfying ξ ( t | 2 | ) ξ ( 1 | 2 | ) ξ ( t ) , ξ ( 1 | 2 | ) < 1 | 2 | , for all t 0 . Assume that κ > 0 and f : G X be a mapping such that
2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) Y κ ( ξ ( | x | ) + ξ ( | y | ) + ξ ( | z | ) ) ,
for all x , y , z G . Then there exists a unique CJA mapping : G X such that
f ( x ) ( x ) | 2 | 1 ( 2 + | 2 | ) ξ ( | x | ) .
Proof If we define λ : G 3 [ 0 , ) by λ ( x , y , z ) : = κ ( ξ ( | x | ) + ξ ( | y | ) + ξ ( | z | ) ) , then we have lim n | 2 | n λ ( x 2 n , y 2 n , z 2 n ) = 0 , for all x , y , z G . On the other hand, it follows that £ ( x ) = | 2 | 1 ( 2 + | 2 | ) ξ ( | x | ) exists, for all x G . Also, we have
lim j lim n max j k < n + j | 2 | k λ ( x 2 k + 1 , x 2 k , x 2 k + 1 ) = lim j | 2 | j λ ( x 2 j + 1 , x 2 j , x 2 j + 1 ) = 0 .

Thus, applying Theorem 2.3, we have the conclusion. This completes the proof. □

Theorem 2.4 Let G be an additive semigroup and X is a non-Archimedean Banach space. Assume that λ : G 3 [ 0 , + ) be a function such that lim n λ ( 2 n x , 2 n y , 2 n z ) 2 n = 0 , for all x , y , z G . Suppose that, for any x G , the limit
£ ( x ) = lim n max 0 k < n λ ( 2 k x , 2 k + 1 x , 2 k x ) | 2 | k
(2.18)
exists and f : G X be a mapping satisfying (2.12). Then the limit ( x ) : = lim n f ( 2 n x ) 2 n exists, for all x G , and
f ( x ) ( x ) £ ( x ) | 2 | ,
(2.19)

for all x G . Moreover, if lim j lim n max j k < n + j λ ( 2 k x , 2 k + 1 x , 2 k x ) | 2 | k = 0 , then is the unique CJA mapping satisfying (2.19).

Proof It follows from (2.14) that
f ( x ) f ( 2 x ) 2 X λ ( x , 2 x , x ) | 2 | ,
(2.20)
for all x G . Replacing x by 2 n x in (2.20), we obtain
f ( 2 n x ) 2 n f ( 2 n + 1 x ) 2 n + 1 X λ ( 2 n x , 2 n + 1 x , 2 n x ) | 2 | n + 1 .
(2.21)
Thus it follows from (2.21) that the sequence { f ( 2 n x ) 2 n } n 1 is convergent. Set ( x ) : = lim n f ( 2 n x ) 2 n . On the other hand, it follows from (2.21) that
f ( 2 p x ) 2 p f ( 2 q x ) 2 q = k = p q 1 f ( 2 k + 1 x ) 2 k + 1 f ( 2 k x ) 2 k max p k < q { f ( 2 k + 1 x ) 2 k + 1 f ( 2 k x ) 2 k } 1 | 2 | max p k < q λ ( 2 k x , 2 k + 1 x , 2 k x ) | 2 | k ,

for all x G and p , q 0 with q > p 0 . Letting p = 0 , taking q in the last inequality and using (2.18), we obtain (2.19).

The rest of the proof is similar to the proof of Theorem 2.3. This completes the proof. □

Corollary 2.4 Let ξ : [ 0 , ) [ 0 , ) be a function satisfying ξ ( | 2 | t ) ξ ( | 2 | ) ξ ( t ) , ξ ( | 2 | ) < | 2 | , for all t 0 . Let κ > 0 and f : G X be a mapping satisfying
2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) κ ( ξ ( | x | ) ξ ( | y | ) ξ ( | z | ) ) ,
for all x , y , z G . Then there exists a unique CJA mapping : G X such that
f ( x ) ( x ) κ ξ ( | x | ) 3 .

Proof If we define λ : G 3 [ 0 , ) by λ ( x , y , z ) : = κ ( ξ ( | x | ) ξ ( | y | ) ξ ( | z | ) ) and apply Theorem 2.4, then we get the conclusion. □

3 RNS-stability

In this section, using the fixed point and direct methods, we prove the HUR-approximation of the functional equation (1.1) in random normed spaces.

Theorem 3.1 Let X be a real linear space, ( Z , μ , min ) be an RN-space and φ : X 3 Z be a function such that there exists 0 < α < 1 2 such that
μ φ ( x 2 , y 2 , z 2 ) ( t ) μ φ ( x , y , z ) ( t α ) ,
(3.1)
for all x , y , z X and t > 0 and lim n μ φ ( x 2 n , y 2 n , z 2 n ) ( t 2 n ) = 1 , for all x , y , z X and t > 0 . Let ( Y , μ , min ) be a complete RN-space. If f : X Y be a mapping such that
μ 2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) ( t ) μ φ ( x , y , z ) ( t ) ,
(3.2)
for all x , y , z X and t > 0 . Then the limit ( x ) = lim n 2 n f ( x 2 n ) exists, for all x X , and defines a unique CJA mapping : X Y such that
μ f ( x ) A ( x ) ( t ) μ φ ( x , 2 x , x ) ( ( 1 2 α ) t α ) ,
(3.3)

for all x X and t > 0 .

Proof Putting y = 2 x and z = x in (3.2), we see that
μ f ( 2 x ) 2 f ( x ) ( t ) μ φ ( x , 2 x , x ) ( t ) .
(3.4)
Replacing x by x 2 in (3.4), we obtain
μ 2 f ( x 2 ) f ( x ) ( t ) μ φ ( x 2 , x , x 2 ) ( t ) μ φ ( x , 2 x , x ) ( t α ) ,
(3.5)
for all x X . Replacing x by x 2 n in (3.5) and using (3.1), we obtain
μ 2 n + 1 f ( x 2 n + 1 ) 2 n f ( x 2 n ) ( t ) μ φ ( x 2 n + 1 , x 2 n , x 2 n + 1 ) ( t 2 n ) μ φ ( x , 2 x , x ) ( t 2 n α n + 1 )
and so
μ 2 n f ( x 2 n ) f ( x ) ( k = 0 n 1 2 k α k + 1 t ) = μ k = 0 n 1 2 k + 1 f ( x 2 k + 1 ) 2 k f ( x 2 k ) ( k = 0 n 1 2 k α k + 1 t ) T k = 0 n 1 ( μ 2 k + 1 f ( x 2 k + 1 ) 2 k f ( x 2 k ) ( 2 k α k + 1 t ) ) T k = 0 n 1 ( μ φ ( x , 2 x , x ) ( t ) ) = μ φ ( x , 2 x , x ) ( t ) .
This implies that
μ 2 n f ( x 2 n ) f ( x ) ( t ) μ φ ( x , 2 x , x ) ( t k = 0 n 1 2 k α k + 1 ) .
(3.6)
Replacing x by x 2 p in (3.6), we obtain
μ 2 n + p f ( x 2 n + p ) 2 p f ( x 2 p ) ( t ) μ φ ( x , 2 x , x ) ( t k = p n + p 1 2 k α k + 1 ) .
(3.7)
Since lim p , n μ φ ( x , 2 x , x ) ( t k = p n + p 1 2 k α k + 1 ) = 1 , it follows that { 2 n f ( x 2 n ) } n = 1 is a Cauchy sequence in a complete RN-space ( Y , μ , min ) and so there exists a point ( x ) Y such that lim n 2 n f ( x 2 n ) = ( x ) . Fix x X and put p = 0 in (3.7) and so, for any ε > 0 ,
μ ( x ) f ( x ) ( t + ε ) T ( μ ( x ) 2 n f ( x 2 n ) ( ε ) , μ φ ( x , 2 x , x ) ( t k = 0 n 1 2 k α k + 1 ) ) .
(3.8)
Taking n in (3.8), we get μ ( x ) f ( x ) ( t + ε ) μ φ ( x , 2 x , x ) ( ( 1 2 α ) t α ) . Since ε is arbitrary, by taking ε 0 in the previous inequality, we get
μ ( x ) f ( x ) ( t ) μ φ ( x , 2 x , x ) ( ( 1 2 α ) t α ) .
Replacing x, y and z by x 2 n , y 2 n and z 2 n in (3.2), respectively, we get
μ 2 n + 1 f ( x + y + z 2 n + 1 ) 2 n f ( x 2 n ) 2 n f ( y 2 n ) 2 n f ( z 2 n ) ( t ) μ φ ( x 2 n , y 2 n , z 2 n ) ( t 2 n ) ,
for all x , y , z X and t > 0 . Since lim n μ φ ( x 2 n , y 2 n , z 2 n ) ( t 2 n ) = 1 , we conclude that satisfies (1.1). On the other hand
2 ( x 2 ) ( x ) = lim n 2 n + 1 f ( x 2 n + 1 ) lim n 2 n f ( x 2 n ) = 0 .
This implies that : X Y is an CJA mapping. To prove the uniqueness of the CJA mapping , assume that there exists another CJA mapping : X Y which satisfies (3.3). Then we have
μ ( x ) ( x ) ( t ) = lim n μ 2 n ( x 2 n ) 2 n ( x 2 n ) ( t ) lim n min { μ 2 n ( x 2 n ) 2 n f ( x 2 n ) ( t 2 ) , μ 2 n f ( x 2 n ) 2 n ( x 2 n ) ( t 2 ) } lim n μ φ ( x 2 n , 2 x 2 n , x 2 n ) ( ( 1 2 α ) t 2 n ) lim n μ φ ( x , 2 x , x ) ( ( 1 2 α ) t 2 n α n ) .

Since lim n μ φ ( x , 2 x , x ) ( ( 1 2 α ) t 2 n α n ) = 1 . Therefore, we have μ ( x ) ( x ) ( t ) = 1 , for all t > 0 , and so ( x ) = ( x ) . This completes the proof. □

Corollary 3.1 Let X be a real normed linear space, ( Z , μ , min ) be an RN-space and ( Y , μ , min ) be a complete RN-space. Let r be a positive real number with r > 1 , z 0 Z and f : X Y be a mapping satisfying
μ 2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) ( t ) μ ( x r + y r + z r ) z 0 ( t ) ,
(3.9)
for all x , y X and t > 0 . Then the limit ( x ) = lim n 2 n f ( x 2 n ) exists, for all x X , and defines a unique CJA mapping : X Y such that
μ f ( x ) ( x ) ( t ) μ x p z 0 ( ( 2 r 2 ) t 2 r + 2 ) ,

for all x X and t > 0 .

Proof Let α = 2 r and φ : X 3 Z be a mapping defined by φ ( x , y , z ) = ( x r + y r + z r ) z 0 . Then, from Theorem 3.1, the conclusion follows. □

Theorem 3.2 Let X be a real linear space, ( Z , μ , min ) be an RN-space and φ : X 3 Z be a function such that there exists 0 < α < 2 such that μ φ ( 2 x , 2 y , 2 z ) ( t ) μ α φ ( x , y , z ) ( t ) , for all x X and t > 0 , and
lim n μ φ ( 2 n x , 2 n y , 2 n z ) ( 2 n t ) = 1 ,
for all x , y , z X and t > 0 . Let ( Y , μ , min ) be a complete RN-space. If f : X Y be a mapping satisfying (3.2). Then the limit ( x ) = lim n f ( 2 n x ) 2 n exists, for all x X , and defines a unique CJA mapping : X Y such that
μ f ( x ) ( x ) ( t ) μ φ ( x , 2 x , x ) ( ( 2 α ) t ) ,
(3.10)

for all x X and t > 0 .

Proof It follows from (3.4) that
μ f ( 2 x ) 2 f ( x ) ( t ) μ φ ( x , 2 x , x ) ( 2 t ) .
(3.11)
Replacing x by 2 n x in (3.11), we obtain
μ f ( 2 n + 1 x ) 2 n + 1 f ( 2 n x ) 2 n ( t ) μ φ ( 2 n x , 2 n + 1 x , 2 n x ) ( 2 n + 1 t ) μ φ ( x , 2 x , x ) ( 2 n + 1 t α n ) .

The rest of the proof is similar to the proof of Theorem 3.1. □

Corollary 3.2 Let X be a real normed linear space, ( Z , μ , min ) be an RN-space and ( Y , μ , min ) be a complete RN-space. Let r be a positive real number with 0 < r < 1 , z 0 Z and f : X Y be a mapping satisfying (3.9). Then the limit ( x ) = lim n f ( 2 n x ) 2 n exists, for all x X , and defines a unique CJA mapping : X Y such that
μ f ( x ) ( x ) ( t ) μ x p z 0 ( ( 2 2 r ) t 2 r + 2 ) ,

for all x X and t > 0 .

Proof Let α = 2 r and φ : X 3 Z be a mapping defined by φ ( x , y , z ) = ( x r + y r + z r ) z 0 . Then, from Theorem 3.2, the conclusion follows. □

Theorem 3.3 Let X be a linear space, ( Y , μ , T M ) be a complete RN-space and Φ be a mapping from X 3 to D + ( Φ ( x , y , z ) is denoted by Φ x , y . z ) such that there exists 0 < α < 1 2 such that
Φ 2 x , 2 y , 2 z ( t ) Φ x , y , z ( α t ) ,
(3.12)
for all x , y , z X and t > 0 . Let f : X Y be a mapping satisfying
μ 2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) ( t ) Φ x , y , z ( t ) ,
(3.13)
for all x , y , z X and t > 0 . Then, for all x X , ( x ) : = lim n 2 n f ( x 2 n ) exists and : X Y is a unique CJA mapping such that
μ f ( x ) ( x ) ( t ) Φ x , 2 x , x ( ( 1 2 α ) t α ) ,
(3.14)

for all x X and t > 0 .

Proof Putting y = 2 x and z = x in (3.13), we have
μ 2 f ( x 2 ) f ( x ) ( t ) Φ x , 2 x , x ( t α ) ,
(3.15)
for all x X and t > 0 . Consider the set S : = { g : X Y } and the generalized metric d in S defined by
d ( f , g ) = inf u ( 0 , ) { μ g ( x ) h ( x ) ( u t ) Φ x , 2 x , x ( t ) , x X , t > 0 } ,
(3.16)
where inf = + . It is easy to show that ( S , d ) is complete (see [39], Lemma 2.1). Now, we consider a linear mapping J : ( S , d ) ( S , d ) such that
J h ( x ) : = 2 h ( x 2 ) ,
(3.17)
for all x X . First, we prove that J is a strictly contractive mapping with the Lipschitz constant 2α. In fact, let g , h S be such that d ( g , h ) < ε . Then we have μ g ( x ) h ( x ) ( ε t ) Φ x , 2 x , x ( t ) , for all x X and t > 0 , and so
μ J g ( x ) J h ( x ) ( 2 α ε t ) = μ 2 g ( x 2 ) 2 h ( x 2 ) ( 2 α ε t ) = μ g ( x 2 ) h ( x 2 ) ( α ε t ) Φ x 2 , x , x 2 ( α t ) Φ x , 2 x , x ( t ) ,
for all x X and t > 0 . Thus d ( g , h ) < ε implies that d ( J g , J h ) < 2 α ε . This means that d ( J g , J h ) 2 α d ( g , h ) , for all g , h S . It follows from (3.15) that d ( f , J f ) α . By Theorem 1.3, there exists a mapping A : X Y satisfying the following:
  1. (1)
    is a fixed point of J, that is,
    ( x 2 ) = 1 2 ( x ) ,
    (3.18)
     
for all x X . The mapping is a unique fixed point of J in the set Ω = { h S : d ( g , h ) < } . This implies that is a unique mapping satisfying (3.18) such that there exists u ( 0 , ) satisfying μ f ( x ) ( x ) ( u t ) Φ x , 2 x , x ( t ) , for all x X and t > 0 .
  1. (2)

    d ( J n f , ) 0 as n . This implies the equality lim n 2 n f ( x 2 n ) = ( x ) , for all x X .

     
  2. (3)
    d ( f , ) d ( f , J f ) 1 2 α with f Ω , which implies the inequality d ( f , ) α 1 2 α and so
    μ f ( x ) ( x ) ( t ) Φ x , 2 x , x ( ( 1 2 α ) t α ) ,
     
for all x X and t > 0 . This implies that the inequality (3.14) holds. On the other hand
μ 2 n + 1 f ( x + y + z 2 n + 1 ) 2 n f ( x 2 n ) 2 n f ( y 2 n ) 2 n f ( z 2 n ) ( t ) Φ x 2 n , y 2 n , z 2 n ( t 2 n ) ,
for all x , y , z X , t > 0 and n 1 . By (3.12), we know that Φ x 2 n , y 2 n , z 2 n ( t 2 n ) Φ x , y , z ( t ( 2 α ) n ) . Since lim n Φ x , y , z ( t ( 2 α ) n ) = 1 , for all x , y , z X and t > 0 , we have μ 2 ( x + y + z 2 ) ( x ) ( y ) ( z ) ( t ) = 1 , for all x , y , z X and t > 0 . Thus the mapping : X Y satisfying (1.1). Furthermore
( 2 x ) 2 ( x ) = lim n 2 n f ( x 2 n 1 ) 2 lim n 2 n f ( x 2 n ) = 2 [ lim n 2 n 1 f ( x 2 n 1 ) lim n 2 n f ( x 2 n ) ] = 0 .

This completes the proof. □

Corollary 3.3 Let X be a real normed space, θ 0 and r be a real number with r > 1 . Let f : X Y be a mapping satisfying
μ 2 f ( x + y + z 2 ) f ( x ) f ( y ) f ( z ) ( t ) t t + θ ( x r + y r + z r ) ,
(3.19)
for all x , y , z X and t > 0 . Then ( x ) = lim n 2 n f ( x 2 n ) exists, for all x X , and : X Y is a unique CJA mapping such that
μ f ( x ) ( x ) ( t ) ( 2 r 2 ) t ( 2 r 2 ) t + ( 2 r + 2 ) θ x r ,

for all x X and t > 0 .

Proof The proof follows from Theorem 3.3 if we take Φ x , y , z ( t ) = t t + θ ( x r + y r + z r ) , for all x , y , z X and t > 0 . In fact, if we choose α = 2 r , then we get the desired result. □

Theorem 3.4 Let X be a linear space, ( Y , μ , T M ) be a complete RN-space and Φ be a mapping from X 3 to D + ( Φ ( x , y , z ) is denoted by Φ x , y , z ) such that for some 0 < α < 2 , Φ x 2 , y 2 , z 2 ( t ) Φ x , y , z ( α t ) , for all x , y , z X and t > 0 . Let f : X Y be a mapping satisfying (3.13). Then the limit ( x ) : = lim n f ( 2 n x ) 2 n exists, for all x X , and : X Y is a unique CJA mapping such that
μ f ( x ) ( x ) ( t ) Φ x , 2 x , x ( ( 2 α ) t ) ,
(3.20)

for all x X and t > 0 .

Proof Putting y = 2 x and z = x in (3.13), we have
μ f ( 2 x ) 2 f ( x ) ( t ) Φ x , 2 x , x ( 2 t ) ,
(3.21)
for all x X and t > 0 . Let ( S , d ) be the generalized metric space defined in the proof of Theorem 3.1. Now, we consider a linear mapping J : ( S , d ) ( S , d ) such that J h ( x ) : = 1 2 h ( 2 x ) , for all x X . It follows from (3.21) that d ( f , J f ) 1 2 . By Theorem 1.3, there exists a mapping : X Y satisfying the following:
  1. (1)
    is a fixed point of J, that is,
    ( 2 x ) = 2 ( x ) ,
    (3.22)
     
for all x X . The mapping is a unique fixed point of J in the set Ω = { h S : d ( g , h ) < } . This implies that is a unique mapping satisfying (3.22) such that there exists u ( 0 , ) satisfying μ f ( x ) ( x ) ( u t ) Φ x , 2 x , x ( t ) , for all x X and t > 0 .
  1. (2)
    d ( J n f , ) 0 as n . This implies the equality
    lim n f ( 2 n x ) 2 n = ( x ) ,
     
for all x X .
  1. (3)

    d ( f , ) d ( f , J f ) 1 α 2 with f Ω , which implies the inequality μ f ( x ) ( x ) ( t 2 α ) Φ x , 2 x , x ( t ) , for all x X and t > 0 . This implies that the inequality (3.20) holds. The rest of the proof is similar to the proof of Theorem 3.3. □

     
Corollary 3.4 Let X be a real normed space, θ 0 and r be a real number with 0 < r < 1 . Let f : X Y be a mapping satisfying (3.19). Then the limit ( x ) = lim n f ( 2 n x ) 2 n exists, for all x X , and : X Y is a unique CJA mapping such that
μ f ( x ) ( x ) ( t ) ( 2 2 r ) t ( 2 2 r ) t + ( 2 r + 2 ) θ x r ,

for all x X and t > 0 .

Proof The proof follows from Theorem 3.4 if we take Φ x , y , z ( t ) = t t + θ ( x r + y r + z r ) , for all x , y , z X and t > 0 . In fact, if we choose α = 2 r , then we get the desired result. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University
(2)
Department of Mathematics, Beyza Branch, Islamic Azad University

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