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,yE, where ε and p are constants with ε>0 and 0p<1. Then the limit L(x)= lim n f ( 2 n x ) 2 n exists, for all xE, and L:E E is the unique additive mapping which satisfies

f ( x ) L ( x ) 2 ε 2 2 p x p ,

for all xE. Also, if for each xE the function f(tx) is continuous in tR, 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:XY, 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,sK, the following conditions hold: (a) |r|=0 if and only if r=0; (b) |rs|=|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 n1.

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

  1. (1)

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

  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 :mjn1}, for all m,nN 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<ny.

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 ℚ. The completion of ℚ 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 |p1. 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 ℝ 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, FG if and only if F(t)G(t), for all tR. For any a0, 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 xz and yw, for all x,y,z,w[0,1].

Three typical examples of continuous t-norms are as follows: T P (x,y)=xy, T max (x,y)=max{a+b1,0}, 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 n2. 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 αR with α0, xX and t0;

  3. (c)

    μ x + y (t+s)T( μ x (t), μ y (s)), for all x,yX and t,s0.

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 xX (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,yX;

  2. (b)

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

  3. (c)

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

Theorem 1.3 ([37, 38])

Let (X,d) be a complete generalized metric space and J:XX be a strictly contractive mapping with Lipschitz constant L<1. Then, for all xX, 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={yX:d( J n 0 x,y)<};

  4. (d)

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

In this paper, using the fixed point and direct methods, we prove the HUR-approximation of the following CJA functional equation:

2f ( 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,zX. Let f:XY 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,zX. Then there exists a unique additive mapping :XY such that

f ( x ) ( x ) Y αφ(x,2x,x) ( | 2 | | 2 | α ) 1 ,
(2.3)

for all xX.

Proof Putting y=2x and z=x in (2.1), we get f ( 2 x ) 2 f ( x ) Y φ(x,2x,x), for all xX. So

f ( x ) 2 f ( x 2 ) Y | 2 | 1 αφ(x,2x,x),
(2.4)

for all xX. Consider the set S:={h:XY} and introduce the generalized metric on S:

d(g,h)= inf μ ( 0 , + ) g ( x ) h ( x ) Y μφ(x,2x,x),

for all xX, where, as usual, infϕ=+. It is easy to show that (S,d) is complete (see [39]). Now we consider the linear mapping J:SS such that Jg(x):=2g( x 2 ), for all xX. Let g,hS be given such that d(g,h)=ε. Then g ( x ) h ( x ) Y εφ(x,2x,x), for all xX. Hence

J g ( x ) J h ( x ) Y = 2 g ( x 2 ) 2 h ( x 2 ) Y αεφ(x,2x,x),

for all xX. So d(g,h)=ε implies that d(Jg,Jh)αε. This means that d(Jg,Jh)αd(g,h), for all g,hS. It follows from (2.4) that d(f,Jf) | 2 | 1 α. By Theorem 1.3, there exists a mapping :XY satisfying the following:

  1. (1)

    ℑ is a fixed point of J, i.e.,

    (x)=2 ( x 2 ) ,
    (2.5)

for all xX. The mapping ℑ is a unique fixed point of J in the set M={gS:d(h,g)<}. This implies that ℑ is a unique mapping satisfying (2.5) such that there exists a μ(0,) satisfying f ( x ) ( x ) Y μφ(x,2x,x), for all xX;

  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 xX;

  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,zX. So 2( x + y + z 2 )=(x)+(y)+(z), for all x,y,zX. Hence :XY 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:XY 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,zX. Then there exists a unique CJA mapping :XY such that

f ( x ) ( x ) Y |2|θ ( 2 + | 2 | r ) ( | 2 | r + 1 | 2 | 2 ) 1 x r ,

for all xX.

Proof The proof follows from Theorem 2.1 by taking φ(x,y,z)=( x r + y r + z r ), for all x,y,zX. 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|αφ( x 2 , y 2 , z 2 ), for all x,y,zX. Let f:XY be a mapping satisfying (2.2). Then there exists a unique CJA mapping :XY such that

f ( x ) ( x ) Y φ(x,2x,x) ( | 2 | | 2 | α ) 1 ,
(2.8)

for all xX.

Proof Let (S,d) be the generalized metric space defined in the proof of Theorem 2.1. Now we consider the linear mapping J:SS such that Jg(x):= g ( 2 x ) 2 , for all xX. Let g,hS be given such that d(g,h)=ε. Then g ( x ) h ( x ) Y εφ(x,2x,x), for all xX. 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 xX. So d(g,h)=ε implies that d(Jg,Jh)αε. This means that d(Jg,Jh)αd(g,h), for all g,hS. It follows from (2.4) that d(f,Jf) | 2 | 1 . By Theorem 1.3, there exists a mapping :XY satisfying the following:

  1. (1)

    ℑ is a fixed point of J, i.e.,

    ( 2 x ) 2 =(x),
    (2.9)

for all xX. The mapping ℑ is a unique fixed point of J in the set M={gS:d(h,g)<}. This implies that ℑ is a unique mapping satisfying (2.9) such that there exists a μ(0,) satisfying f ( x ) ( x ) Y μφ(x,2x,x), for all xX;

  1. (2)

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

  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:XY be a mapping satisfying (2.7). Then there exists a unique CJA mapping :XY such that

f ( x ) ( x ) Y θ ( 2 + | 2 | r ) ( | 2 | | 2 | r ) 1 x r ,

for all xX.

Proof The proof follows from Theorem 2.2 by taking φ(x,y,z)=( x r + y r + z r ), for all x,y,zX. 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,zG. Suppose that, for any xG, 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:GX 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 xG, and defines an CJA mapping :GX 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 thenis the unique CJA mapping satisfying (2.13).

Proof Putting y=2x and z=x in (2.12), we get

f ( 2 x ) 2 f ( x ) Y λ(x,2x,x),
(2.14)

for all xG. 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 n1 and xG. 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,zX. 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,yX. Hence the mapping :XY is Cauchy additive.

To prove the uniqueness property of ℑ, let ℜ 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 xG. 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 t0. Assume that κ>0 and f:GX 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,zG. Then there exists a unique CJA mapping :GX 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,zG. On the other hand, it follows that £(x)= | 2 | 1 (2+|2|)ξ(|x|) exists, for all xG. 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,zG. Suppose that, for any xG, 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:GX be a mapping satisfying (2.12). Then the limit (x):= lim n f ( 2 n x ) 2 n exists, for all xG, and

f ( x ) ( x ) £ ( x ) | 2 | ,
(2.19)

for all xG. 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, thenis 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 xG. 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 xG and p,q0 with q>p0. 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 t0. Let κ>0 and f:GX 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,zG. Then there exists a unique CJA mapping :GX such that

f ( x ) ( x ) κξ ( | 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,zX 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,zX and t>0. Let (Y,μ,min) be a complete RN-space. If f:XY 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,zX and t>0. Then the limit (x)= lim n 2 n f( x 2 n ) exists, for all xX, and defines a unique CJA mapping :XY such that

μ f ( x ) A ( x ) (t) μ φ ( x , 2 x , x ) ( ( 1 2 α ) t α ) ,
(3.3)

for all xX and t>0.

Proof Putting y=2x 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 xX. 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 xX 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 ε0 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,zX and t>0. Since lim n μ φ ( x 2 n , y 2 n , z 2 n ) ( t 2 n )=1, we conclude that ℑ 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 :XY is an CJA mapping. To prove the uniqueness of the CJA mapping ℑ, assume that there exists another CJA mapping :XY 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:XY 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,yX and t>0. Then the limit (x)= lim n 2 n f( x 2 n ) exists, for all xX, and defines a unique CJA mapping :XY such that

μ f ( x ) ( x ) (t) μ x p z 0 ( ( 2 r 2 ) t 2 r + 2 ) ,

for all xX 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 xX 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,zX and t>0. Let (Y,μ,min) be a complete RN-space. If f:XY be a mapping satisfying (3.2). Then the limit (x)= lim n f ( 2 n x ) 2 n exists, for all xX, and defines a unique CJA mapping :XY such that

μ f ( x ) ( x ) (t) μ φ ( x , 2 x , x ) ( ( 2 α ) t ) ,
(3.10)

for all xX and t>0.

Proof It follows from (3.4) that

μ f ( 2 x ) 2 f ( x ) (t) μ φ ( x , 2 x , x ) (2t).
(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:XY be a mapping satisfying (3.9). Then the limit (x)= lim n f ( 2 n x ) 2 n exists, for all xX, and defines a unique CJA mapping :XY such that

μ f ( x ) ( x ) (t) μ x p z 0 ( ( 2 2 r ) t 2 r + 2 ) ,

for all xX 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 (αt),
(3.12)

for all x,y,zX and t>0. Let f:XY 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,zX and t>0. Then, for all xX, (x):= lim n 2 n f( x 2 n ) exists and :XY 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 xX and t>0.

Proof Putting y=2x 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 xX and t>0. Consider the set S:={g:XY} 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

Jh(x):=2h ( x 2 ) ,
(3.17)

for all xX. First, we prove that J is a strictly contractive mapping with the Lipschitz constant 2α. In fact, let g,hS be such that d(g,h)<ε. Then we have μ g ( x ) h ( x ) (εt) Φ x , 2 x , x (t), for all xX 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 xX and t>0. Thus d(g,h)<ε implies that d(Jg,Jh)<2αε. This means that d(Jg,Jh)2αd(g,h), for all g,hS. It follows from (3.15) that d(f,Jf)α. By Theorem 1.3, there exists a mapping A:XY satisfying the following:

  1. (1)

    ℑ is a fixed point of J, that is,

    ( x 2 ) = 1 2 (x),
    (3.18)

for all xX. The mapping ℑ is a unique fixed point of J in the set Ω={hS:d(g,h)<}. This implies that ℑ is a unique mapping satisfying (3.18) such that there exists u(0,) satisfying μ f ( x ) ( x ) (ut) Φ x , 2 x , x (t), for all xX 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 xX.

  2. (3)

    d(f,) d ( f , J f ) 1 2 α with fΩ, 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 xX 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,zX, t>0 and n1. 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,zX and t>0, we have μ 2 ( x + y + z 2 ) ( x ) ( y ) ( z ) (t)=1, for all x,y,zX and t>0. Thus the mapping :XY 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, θ0 and r be a real number with r>1. Let f:XY 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,zX and t>0. Then (x)= lim n 2 n f( x 2 n ) exists, for all xX, and :XY 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 xX 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,zX 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 (αt), for all x,y,zX and t>0. Let f:XY be a mapping satisfying (3.13). Then the limit (x):= lim n f ( 2 n x ) 2 n exists, for all xX, and :XY is a unique CJA mapping such that

μ f ( x ) ( x ) (t) Φ x , 2 x , x ( ( 2 α ) t ) ,
(3.20)

for all xX and t>0.

Proof Putting y=2x and z=x in (3.13), we have

μ f ( 2 x ) 2 f ( x ) (t) Φ x , 2 x , x (2t),
(3.21)

for all xX 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 Jh(x):= 1 2 h(2x), for all xX. It follows from (3.21) that d(f,Jf) 1 2 . By Theorem 1.3, there exists a mapping :XY satisfying the following:

  1. (1)

    ℑ is a fixed point of J, that is,

    (2x)=2(x),
    (3.22)

for all xX. The mapping ℑ is a unique fixed point of J in the set Ω={hS:d(g,h)<}. This implies that ℑ is a unique mapping satisfying (3.22) such that there exists u(0,) satisfying μ f ( x ) ( x ) (ut) Φ x , 2 x , x (t), for all xX 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 xX.

  1. (3)

    d(f,) d ( f , J f ) 1 α 2 with fΩ, which implies the inequality μ f ( x ) ( x ) ( t 2 α ) Φ x , 2 x , x (t), for all xX 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, θ0 and r be a real number with 0<r<1. Let f:XY be a mapping satisfying (3.19). Then the limit (x)= lim n f ( 2 n x ) 2 n exists, for all xX, and :XY 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 xX 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,zX and t>0. In fact, if we choose α= 2 r , then we get the desired result. □