Open Access

On the Ulam stability of mixed type QA mappings in IFN-spaces

Advances in Difference Equations20132013:203

DOI: 10.1186/1687-1847-2013-203

Received: 13 March 2013

Accepted: 15 June 2013

Published: 8 July 2013

Abstract

We give Ulam-type stability results concerning the quadratic-additive functional equation in intuitionistic fuzzy normed spaces.

Keywords

t-norm t-conorm quadratic-additive functional equation intuitionistic fuzzy normed space Hyers-Ulam stability

1 Introduction

In 1940, Ulam [1] proposed the following stability problem: ‘When is it true that a function which satisfies some functional equation approximately must be close to one satisfying the equation exactly?’. Hyers [2] gave the first affirmative partial answer to the question of Ulam for Banach spaces. Aoki [3] presented a generalization of Hyers results by considering additive mappings, and later on Rassias [4] did for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias has significantly influenced the development of what we now call the Hyers-Ulam-Rassias stability of functional equations. Various extensions, generalizations and applications of the stability problems have been given by several authors so far; see, for example, [524] and references therein.

The notion of intuitionistic fuzzy set introduced by Atanassov [25] has been used extensively in many areas of mathematics and sciences. Using the idea of intuitionistic fuzzy set, Saadati and Park [26] presented the notion of intuitionistic fuzzy normed space which is a generalization of the concept of a fuzzy metric space due to Bag and Samanta [27]. The authors of [2834] defined and studied some summability problems in the setting of an intuitionistic fuzzy normed space.

In the recent past, several Hyers-Ulam stability results concerning the various functional equations were determined in [3546], respectively, in the fuzzy and intuitionistic fuzzy normed spaces. Quite recently, Alotaibi and Mohiuddine [47] established the stability of a cubic functional equation in random 2-normed spaces, while the notion of random 2-normed spaces was introduced by Goleţ [48] and further studied in [4951].

The Hyers-Ulam stability problems of quadratic-additive functional equation
f ( x + y + z ) + f ( x ) + f ( y ) + f ( z ) = f ( x + y ) + f ( y + z ) + f ( x + z )

under the approximately even (or odd) condition were established by Jung [52] and the solution of the above functional equation where the range is a field of characteristic 0 was determined by Kannappan [53]. In this paper we determine the stability results concerning the above functional equation in the setting of intuitionistic fuzzy normed spaces. This work indeed presents a relationship between two various disciplines: the theory of fuzzy spaces and the theory of functional equations.

2 Definitions and preliminaries

We shall assume throughout this paper that the symbol denotes the set of all natural numbers.

A binary operation : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , 1 ] is said to be a continuous t-norm if it satisfies the following conditions:

(a)  is associative and commutative, (b)  is continuous, (c)  a 1 = a for all a [ 0 , 1 ] , (d)  a b c d whenever a c and b d for each a , b , c , d [ 0 , 1 ] .

A binary operation : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , 1 ] is said to be a continuous t-conorm if it satisfies the following conditions:

(a′) is associative and commutative, (b′)  is continuous, (c′) a 0 = a for all a [ 0 , 1 ] , (d′) a b c d whenever a c and b d for each a , b , c , d [ 0 , 1 ] .

The five-tuple ( X , μ , ν , , ) is said to be intuitionistic fuzzy normed spaces (for short, IFN-spaces) [26] if X is a vector space, is a continuous t-norm, is a continuous t-conorm, and μ, ν are fuzzy sets on X × ( 0 , ) satisfying the following conditions. For every x , y X and s , t > 0 ,
  1. (i)

    μ ( x , t ) + ν ( x , t ) 1 ,

     
  2. (ii)

    μ ( x , t ) > 0 ,

     
  3. (iii)

    μ ( x , t ) = 1 if and only if x = 0 ,

     
  4. (iv)

    μ ( α x , t ) = μ ( x , t | α | ) for each α 0 ,

     
  5. (v)

    μ ( x , t ) μ ( y , s ) μ ( x + y , t + s ) ,

     
  6. (vi)

    μ ( x , ) : ( 0 , ) [ 0 , 1 ] is continuous,

     
  7. (vii)

    lim t μ ( x , t ) = 1 and lim t 0 μ ( x , t ) = 0 ,

     
  8. (viii)

    ν ( x , t ) < 1 ,

     
  9. (ix)

    ν ( x , t ) = 0 if and only if x = 0 ,

     
  10. (x)

    ν ( α x , t ) = ν ( x , t | α | ) for each α 0 ,

     
  11. (xi)

    ν ( x , t ) ν ( y , s ) ν ( x + y , t + s ) ,

     
  12. (xii)

    ν ( x , ) : ( 0 , ) [ 0 , 1 ] is continuous,

     
  13. (xiii)

    lim t ν ( x , t ) = 0 and lim t 0 ν ( x , t ) = 1 .

     
In this case ( μ , ν ) is called an intuitionistic fuzzy norm. For simplicity in notation, we denote the intuitionistic fuzzy normed spaces by ( X , μ , ν ) instead of ( X , μ , ν , , ) . For example, let ( X , ) be a normed space, and let a b = a b and a b = min { a + b , 1 } for all a , b [ 0 , 1 ] . For all x X and every t > 0 , consider
μ ( x , t ) : = t t + x and ν ( x , t ) : = x t + x .

Then ( X , μ , ν ) is an intuitionistic fuzzy normed space.

The notions of convergence and Cauchy sequence in the setting of IFN-spaces were introduced by Saadati and Park [26] and further studied by Mursaleen and Mohiuddine [30].

Let ( X , μ , ν ) be an intuitionistic fuzzy normed space. Then the sequence x = ( x k ) is said to be:
  1. (i)

    Convergent to L X with respect to the intuitionistic fuzzy norm ( μ , ν ) if, for every ϵ > 0 and t > 0 , there exists k 0 N such that μ ( x k L , t ) > 1 ϵ and ν ( x k L , t ) < ϵ for all k k 0 . In this case, we write ( μ , ν ) - lim x k = L or x k ( μ , ν ) L as k .

     
  2. (ii)

    Cauchy sequence with respect to the intuitionistic fuzzy norm ( μ , ν ) if, for every ϵ > 0 and t > 0 , there exists k 0 N such that μ ( x k x , t ) > 1 ϵ and ν ( x k x , t ) < ϵ for all k , k 0 . An IFN-space ( X , μ , ν ) is said to be complete if every Cauchy sequence in ( X , μ , ν ) is convergent in the IFN-space. In this case, ( X , μ , ν ) is called an intuitionistic fuzzy Banach space.

     

3 Stability of a quadratic-additive functional equation in the IFN-space

We shall assume the following abbreviation throughout this paper:
D f ( x , y , z ) = f ( x + y + z ) f ( x + y ) f ( y + z ) f ( x + z ) + f ( x ) + f ( y ) + f ( z ) .
Theorem 3.1 Let X be a linear space and ( X , μ , ν ) be an IFN-space. Suppose that f is an intuitionistic fuzzy q-almost quadratic-additive mapping from ( X , μ , ν ) to an intuitionistic fuzzy Banach space ( Y , μ , ν ) such that
μ ( D f ( x , y , z ) , s + t + u ) μ ( x , s q ) μ ( y , t q ) μ ( z , u q ) and ν ( D f ( x , y , z ) , s + t + u ) ν ( x , s q ) ν ( y , t q ) ν ( z , u q ) }
(3.1)
for all x , y , z X and s , t , u > 0 , where q is a positive real number with q 1 2 , 1 . Then there exists a unique quadratic-additive mapping T : X Y such that
μ ( T ( x ) f ( x ) , t ) { sup t < t μ ( x , ( 2 2 p 3 ) q t q ) if q > 1 , sup t < t μ ( x , ( ( 4 2 p ) ( 2 2 p ) 6 ) q t q ) if 1 2 < q < 1 , sup t < t μ ( x , ( 2 p 4 3 ) q t q ) if 0 < q < 1 2 and ν ( T ( x ) f ( x ) , t ) { sup t < t ν ( x , ( 2 2 p 3 ) q t q ) if q > 1 , sup t < t ν ( x , ( ( 4 2 p ) ( 2 2 p ) 6 ) q t q ) if 1 2 < q < 1 , sup t < t ν ( x , ( 2 p 4 3 ) q t q ) if 0 < q < 1 2 , }
(3.2)

for all x X and all t > 0 with t ( 0 , t ) , where p = 1 / q .

Proof Putting x = 0 = y = z in (3.1), it follows that
μ ( f ( 0 ) , t ) μ ( 0 , ( t / 3 ) q ) μ ( 0 , ( t / 3 ) q ) μ ( 0 , ( t / 3 ) q ) = 1
and
ν ( f ( 0 ) , t ) ν ( 0 , ( t / 3 ) q ) ν ( 0 , ( t / 3 ) q ) ν ( 0 , ( t / 3 ) q ) = 0

for all t > 0 . Using the definition of IFN-space, we have f ( 0 ) = 0 . Now we are ready to prove our theorem for three cases. We consider the cases as q > 1 , 1 2 < q < 1 and 0 < q < 1 2 .

Case 1. Let q > 1 . Consider a mapping J n f : X Y to be such that
J n f ( x ) = 1 2 ( 4 n ( f ( 2 n x ) + f ( 2 n x ) ) + 2 n ( f ( 2 n x ) f ( 2 n x ) ) )
for all x X . Notice that J 0 f ( x ) = f ( x ) and
J j f ( x ) J j + 1 f ( x ) = D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 + D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 + D f ( 2 j x , 2 j x , 2 j x ) 2 j + 2 D f ( 2 j x , 2 j x , 2 j x ) 2 j + 2
(3.3)
for all x X and j 0 . Using the definition of IFN-space and (3.1), this equation implies that if n + m > m 0 , then
μ ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 3 2 ( 2 p 2 ) j t p ) = μ ( j = m n + m 1 ( J j f ( x ) J j + 1 f ( x ) ) , j = m n + m 1 3 2 j p 2 j + 1 t p ) j = m n + m 1 μ ( J j ( f ( x ) J j + 1 f ( x ) ) , 3 2 j p 2 j + 1 ) j = m n + m 1 { μ ( ( 2 j + 1 + 1 ) D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 ( 2 j + 1 + 1 ) 2 j p t p 2 4 j + 1 ) μ ( 1 ( 2 j + 1 ) D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 ( 2 j + 1 1 ) 2 j p t p 2 4 j + 1 ) } j = m n + m 1 μ ( 2 j x , 2 j t ) = μ ( x , t )
(3.4)
and
ν ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 3 2 ( 2 p 2 ) j t p ) = ν ( j = m n + m 1 ( J j f ( x ) J j + 1 f ( x ) ) , j = m n + m 1 3 2 j p 2 j + 1 t p ) j = m n + m 1 ν ( J j ( f ( x ) J j + 1 f ( x ) ) , 3 2 j p 2 j + 1 ) j = m n + m 1 { ν ( ( 2 j + 1 + 1 ) D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 ( 2 j + 1 + 1 ) 2 j p t p 2 4 j + 1 ) ν ( 1 ( 2 j + 1 ) D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 ( 2 j + 1 1 ) 2 j p t p 2 4 j + 1 ) } j = m n + m 1 ν ( 2 j x , 2 j t ) = ν ( x , t )
(3.5)
for all x X and t > 0 , where j = 1 n a j = a 1 a 2 a n , j = 1 n a j = a 1 a 2 a n . Let ϵ > 0 and δ > 0 be given. Since lim t μ ( x , t ) = 1 and lim t ν ( x , t ) = 0 , there exists t 0 > 0 such that μ ( x , t 0 ) 1 ϵ and ν ( x , t 0 ) ϵ for all x X . We observe that for some t ˜ > t 0 , the series j = 0 3 2 j p 2 j + 1 t ˜ p converges for p = 1 q < 1 , there exists some n 0 0 such that j = m n + m 1 3 2 j p 2 j + 1 t ˜ p < δ for each m n 0 and n > 0 . Using (3.4) and (3.5), we have
μ ( J m f ( x ) J n + m f ( x ) , δ ) μ ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 3 2 j p 2 j + 1 t ˜ p ) μ ( x , t ˜ ) μ ( x , t 0 ) 1 ϵ
and
ν ( J m f ( x ) J n + m f ( x ) , δ ) ν ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 3 2 j p 2 j + 1 t ˜ p ) ν ( x , t ˜ ) ν ( x , t 0 ) ϵ
for all x X and δ > 0 . Hence { J n f ( x ) } is a Cauchy sequence in the fuzzy Banach space ( Y , μ , ν ) . Thus, we define a mapping T : X Y such that T ( x ) : = ( μ , ν ) lim n J n f ( x ) for all x X . Moreover, if we put m = 0 in (3.4) and (3.5), we get
μ ( f ( x ) J n f ( x ) , t ) μ ( x , t q ( j = 0 n 1 3 2 j p 2 j + 1 ) q ) and ν ( f ( x ) J n f ( x ) , t ) ν ( x , t q ( j = 0 n 1 3 2 j p 2 j + 1 ) q ) }
(3.6)
for all x X and t > 0 . Now we have to show that T is quadratic additive. Let x , y , z X . Then
μ ( D T ( x , y , z ) , t ) μ ( ( T J n f ) ( x + y + z ) , t 28 ) μ ( ( T J n f ) ( x ) , t 28 ) μ ( ( T J n f ) ( y ) , t 28 ) μ ( ( T J n f ) ( z ) , t 28 ) μ ( ( J n f T ) ( x + y ) , t 28 ) μ ( ( J n f T ) ( x + z ) , t 28 ) μ ( ( J n f T ) ( y + z ) , t 28 ) μ ( D J n f ( x , y , z ) , 3 t 4 )
(3.7)
and
ν ( D T ( x , y , z ) , t ) ν ( ( T J n f ) ( x + y + z ) , t 28 ) ν ( ( T J n f ) ( x ) , t 28 ) ν ( ( T J n f ) ( y ) , t 28 ) ν ( ( T J n f ) ( z ) , t 28 ) ν ( ( J n f T ) ( x + y ) , t 28 ) ν ( ( J n f T ) ( x + z ) , t 28 ) ν ( ( J n f T ) ( y + z ) , t 28 ) ν ( D J n f ( x , y , z ) , 3 t 4 )
(3.8)
for all t > 0 and n N . Taking the limit as n in the inequalities (3.7) and (3.8), we can see that first seven terms on the right-hand side of (3.7) and (3.8) tend to 1 and 0, respectively, by using the definition of T. It is left to find the value of the last term on the right-hand side of (3.7) and (3.8). By using the definition of J n f ( x ) , write
μ ( D J n f ( x , y , z ) , 3 t 4 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 2 n , 3 t 16 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 2 n , 3 t 16 )
(3.9)
and, similarly,
ν ( D J n f ( x , y , z ) , 3 t 4 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 2 n , 3 t 16 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 2 n , 3 t 16 )
(3.10)
for all x , y , z X , t > 0 and n N . Also, from (3.1), we have
μ ( D f ( ± 2 n x , ± 2 n y , ± 2 n z ) 2 4 n , 3 t 16 ) = μ ( D f ( ± 2 n x , ± 2 n y , ± 2 n z ) , 3 4 n t 8 ) μ ( 2 n x , ( 4 n t 8 ) q ) μ ( 2 n y , ( 4 n t 8 ) q ) μ ( 2 n z , ( 4 n t 8 ) q ) μ ( x , 2 ( 2 q 1 ) n 3 q t q ) μ ( y , 2 ( 2 q 1 ) n 3 q t q ) μ ( z , 2 ( 2 q 1 ) n 3 q t q )
(3.11)
and
μ ( D f ( ± 2 n x , ± 2 n y , ± 2 n z ) 2 2 n , 3 t 16 ) μ ( x , 2 ( 2 q 1 ) n 3 q t q ) μ ( y , 2 ( 2 q 1 ) n 3 q t q ) μ ( z , 2 ( 2 q 1 ) n 3 q t q )
(3.12)
for all x , y , z X , t > 0 and n N . Since q > 1 , therefore (3.9) tends to 1 as n with the help of (3.11) and (3.12). Similarly, by proceeding along the same lines as in (3.11) and (3.12), we can show that (3.10) tends to 0 as n . Thus, inequalities (3.7) and (3.8) become
μ ( D T ( x , y , z ) , t ) = 1 and ν ( D T ( x , y , z ) , t ) = 0
for all x , y , z X and t > 0 . Accordingly, D T ( x , y , z ) = 0 for all x , y , z X . Now we approximate the difference between f and T in a fuzzy sense. Choose ϵ ( 0 , 1 ) and 0 < t < t . Since T is the intuitionistic fuzzy limit of { J n f ( x ) } such that
μ ( T ( x ) J n f ( x ) , t t ) 1 ϵ and ν ( T ( x ) J n f ( x ) , t t ) ϵ
for all x X , t > 0 and n N . From (3.6), we have
μ ( T ( x ) f ( x ) , t ) μ ( T ( x ) J n f ( x ) , t t ) μ ( J n f ( x ) f ( x ) , t ) ( 1 ϵ ) μ ( x , t q ( j = 0 n 1 3 2 j p 2 j + 1 ) q ) ( 1 ϵ ) μ ( x , ( ( 2 2 p ) t 3 ) q )
and
ν ( T ( x ) f ( x ) , t ) ν ( T ( x ) J n f ( x ) , t t ) ν ( J n f ( x ) f ( x ) , t ) ( 1 ϵ ) ν ( x , ( ( 2 2 p ) t 3 ) q ) .

Since ϵ ( 0 , 1 ) is arbitrary, we get the inequality (3.2) in this case.

To prove the uniqueness of T, assume that T is another quadratic-additive mapping from X into Y, which satisfies the required inequality, i.e., (3.2). Then, by (3.3), for all x X and n N ,
T ( x ) J n T ( x ) = j = 0 n 1 ( J j T ( x ) J j + 1 T ( x ) ) = 0 , T ( x ) J n T ( x ) = j = 0 n 1 ( J j T ( x ) J j + 1 T ( x ) ) = 0 . }
(3.13)
Therefore
μ ( T ( x ) T ( x ) , t ) = μ ( J n T ( x ) J n T ( x ) , t ) μ ( J n T ( x ) J n f ( x ) , t 2 ) μ ( J n f ( x ) J n T ( x ) , t 2 ) μ ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( T f ) ( 2 n x ) 2 2 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 2 n , t 8 ) μ ( ( T f ) ( 2 n x ) 2 2 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 2 n , t 8 ) sup t < t μ ( x , 2 ( q 1 ) n 2 q ( 2 2 p 3 ) q t q )
and
ν ( T ( x ) T ( x ) , t ) = ν ( J n T ( x ) J n T ( x ) , t ) ν ( J n T ( x ) J n f ( x ) , t 2 ) ν ( J n f ( x ) J n T ( x ) , t 2 ) ν ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( T f ) ( 2 n x ) 2 2 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 2 n , t 8 ) ν ( ( T f ) ( 2 n x ) 2 2 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 2 n , t 8 ) sup t < t ν ( x , 2 ( q 1 ) n 2 q ( 2 2 p 3 ) q t q )

for all x X , t > 0 and n N . Since q = 1 / p > 1 and taking limit as n in the last two inequalities, we get μ ( T ( x ) T ( x ) , t ) = 1 and ν ( T ( x ) T ( x ) , t ) = 0 for all x X and t > 0 . Hence T ( x ) = T ( x ) for all x X .

Case 2. Let 1 2 < q < 1 . Consider a mapping J n f : X Y to be such that
J n f ( x ) = 1 2 ( 4 n ( f ( 2 n x ) + f ( 2 n x ) ) + 2 n ( f ( x 2 n ) f ( x 2 n ) ) )
for all x X . Then J 0 f ( x ) = f ( x ) and
J j f ( x ) J j + 1 f ( x ) = D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 + D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 2 j 1 ( D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) )
for all x X and j 0 . Thus, for each n + m > m 0 , we have
μ ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 ( 3 4 ( 2 p 4 ) j + 3 2 p ( 2 2 p ) j ) t p ) j = m n + m 1 { μ ( D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 2 j p t p 2 4 j + 1 ) μ ( D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 2 j p t p 2 4 ( j + 1 ) ) μ ( 2 j 1 D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) , 3 2 j 1 t p 2 ( j + 1 ) p ) μ ( 2 j 1 D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) , 3 2 j 1 t p 2 ( j + 1 ) p ) } j = m n + m 1 { μ ( 2 j x , 2 j t ) μ ( x 2 j + 1 , t 2 j + 1 ) } = μ ( x , t ) and ν ( J m f ( x ) J n + m f ( x ) , j = m n + m 1 ( 3 4 ( 2 p 4 ) j + 3 2 p ( 2 2 p ) j ) t p ) j = m n + m 1 { ν ( D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 2 j p t p 2 4 j + 1 ) ν ( D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 , 3 2 j p t p 2 4 ( j + 1 ) ) ν ( 2 j 1 D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) , 3 2 j 1 t p 2 ( j + 1 ) p ) ν ( 2 j 1 D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) , 3 2 j 1 t p 2 ( j + 1 ) p ) } j = m n + m 1 { ν ( 2 j x , 2 j t ) ν ( x 2 j + 1 , t 2 j + 1 ) } = ν ( x , t ) ,
where ∏ and are the same as in Case 1. Proceeding along a similar argument as in Case 1, we see that { J n f ( x ) } is a Cauchy sequence in ( Y , μ , ν ) . Thus, we define T ( x ) : = ( μ , ν ) - lim n J n f ( x ) for all x X . Putting m = 0 in the last two inequalities, we get
μ ( f ( x ) J n f ( x ) , t ) μ ( x , t p ( j = 0 n 1 ( 3 4 ( 2 p 4 ) j + 3 2 p ( 2 2 p ) j ) ) q ) and ν ( f ( x ) J n f ( x ) , t ) ν ( x , t p ( j = 0 n 1 ( 3 4 ( 2 p 4 ) j + 3 2 p ( 2 2 p ) j ) ) q ) }
(3.14)
for all x X and t > 0 . To prove that t is a quadratic-additive function, it is enough to show that the last term on the right-hand side of (3.7) and (3.8) tends to 1 and 0, respectively, as n . Using the definition of J n f ( x ) and (3.1), we obtain
μ ( D J n f ( x , y , z ) , 3 t 4 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) μ ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) μ ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( x , 2 ( 2 q 1 ) n 3 q t q ) μ ( y , 2 ( 2 q 1 ) n 3 q t q ) μ ( z , 2 ( 2 q 1 ) n 3 q t q ) μ ( x , 2 ( 1 q ) n 3 q t q ) μ ( y , 2 ( 1 q ) n 3 q t q ) μ ( z , 2 ( 1 q ) n 3 q t q )
(3.15)
and
ν ( D J n f ( x , y , z ) , 3 t 4 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) ν ( D f ( 2 n x , 2 n y , 2 n z ) 2 4 n , 3 t 16 ) ν ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( x , 2 ( 2 q 1 ) n 3 q t q ) ν ( y , 2 ( 2 q 1 ) n 3 q t q ) ν ( z , 2 ( 2 q 1 ) n 3 q t q ) ν ( x , 2 ( 1 q ) n 3 q t q ) ν ( y , 2 ( 1 q ) n 3 q t q ) ν ( z , 2 ( 1 q ) n 3 q t q )
(3.16)
for each x , y , z X , t > 0 and n N . Since 1 / 2 < q < 1 and taking the limit as n , we see that (3.15) and (3.16) tend to 1 and 0, respectively. As in Case 1, we have D T ( x , y , z ) = 0 for all x , y , z X . Using the same argument as in Case 1, we see that (3.2) follows from (3.14). To prove the uniqueness of T, assume that T is another quadratic-additive mapping from X into Y satisfying (3.2). Using (3.2) and (3.13), we have
μ ( T ( x ) T ( x ) , t ) = μ ( J n T ( x ) J n T ( x ) , t ) μ ( J n T ( x ) J n f ( x ) , t 2 ) μ ( J n f ( x ) J n T ( x ) , t 2 ) μ ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) μ ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) μ ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) sup t < t μ ( x , 2 ( 2 q 1 ) n 2 q ( ( 4 2 p ) ( 2 p 2 ) 6 ) q t q ) sup t < t μ ( x , 2 2 ( 1 q ) n 2 q ( ( 4 2 p ) ( 2 p 2 ) 6 ) q t q )
(3.17)
and
ν ( T ( x ) T ( x ) , t ) ν ( J n T ( x ) J n f ( x ) , t 2 ) ν ( J n f ( x ) J n T ( x ) , t 2 ) ν ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( T f ) ( 2 n x ) 2 4 n , t 8 ) ν ( ( f T ) ( 2 n x ) 2 4 n , t 8 ) ν ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) sup t < t μ ( x , 2 ( 2 q 1 ) n 2 q ( ( 4 2 p ) ( 2 p 2 ) 6 ) q t q ) sup t < t μ ( x , 2 2 ( 1 q ) n 2 q ( ( 4 2 p ) ( 2 p 2 ) 6 ) q t q )
(3.18)

for all x X , t > 0 and n N . Letting n in (3.17) and (3.18), and using the fact that lim n 2 ( 2 q 1 ) n 2 q = lim n 2 ( 1 q ) n 2 q = together with the definition of IFN-space, we get μ ( T ( x ) T ( x ) , t ) = 1 and ν ( T ( x ) T ( x ) , t ) = 0 for all x X and t > 0 . Hence T ( x ) = T ( x ) for all x X .

Case 3. Let 0 < q < 1 2 . Define a mapping J n f : X Y by
J n f ( x ) = 1 2 ( 4 n ( f ( 2 n x ) + f ( 2 n x ) ) + 2 n ( f ( x 2 n ) f ( x 2 n ) ) )
for all x X . In this case, J 0 f ( x ) = f ( x ) and
J j f ( x ) J j + 1 f ( x ) = 4 j 2 ( D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) + D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) ) 2 j 1 ( D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) )
for all x X and j 0 . Thus, for each n + m > m 0 , we have
μ ( J m f ( x ) J n + m f ( x ) j = m n + m 1 3 2 p ( 4 2 p ) j t p ) j = m n + m 1 { μ ( ( 4 j + 2 j ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) 2 , 3 ( 4 j + 2 j ) t p 2 2 ( j + 1 ) p ) μ ( ( 4 j 2 j ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) 2 , 3 ( 4 j 2 j ) t p 2 2 ( j + 1 ) p ) } j = m n + m 1 μ ( x 2 j + 1 , t 2 j + 1 ) = μ ( x , t ) and ν ( J m f ( x ) J n + m f ( x ) j = m n + m 1 3 2 p ( 4 2 p ) j t p ) j = m n + m 1 { ν ( ( 4 j + 2 j ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) 2 , 3 ( 4 j + 2 j ) t p 2 2 ( j + 1 ) p ) ν ( ( 4 j 2 j ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) 2 , 3 ( 4 j 2 j ) t p 2 2 ( j + 1 ) p ) } j = m n + m 1 ν ( x 2 j + 1 , t 2 j + 1 ) = ν ( x , t )
for all x X and t > 0 . Proceeding along a similar argument as in the previous cases, we see that { J n f ( x ) } is a Cauchy sequence in ( Y , μ , ν ) . Thus, we define T ( x ) : = ( μ , ν ) lim n J n f ( x ) for all x X . Putting m = 0 in the last two inequalities, we get
μ ( f ( x ) J n f ( x ) , t ) μ ( x , t q ( j = 0 n 1 3 2 p ( 4 2 p ) j ) q ) and ν ( f ( x ) J n f ( x ) , t ) ν ( x , t q ( j = 0 n 1 3 2 p ( 4 2 p ) j ) q ) }
(3.19)
for all x X and t > 0 . Write
μ ( D J n f ( x , y , z ) , 3 t 4 ) μ ( 4 n 2 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( 4 n 2 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) μ ( x , 2 ( 1 2 q ) n 3 q t q ) μ ( y , 2 ( 1 2 q ) n 3 q t q ) μ ( z , 2 ( 1 2 q ) n 3 q t q ) μ ( x , 2 ( 1 q ) n 3 q t q ) μ ( y , 2 ( 1 q ) n 3 q t q ) μ ( z , 2 ( 1 q ) n 3 q t q )
(3.20)
and
ν ( D J n f ( x , y , z ) , 3 t 4 ) ν ( 4 n 2 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( 4 n 2 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( 2 n 1 D f ( x 2 n , y 2 n , z 2 n ) , 3 t 16 ) ν ( x , 2 ( 1 2 q ) n 3 q t q ) ν ( y , 2 ( 1 2 q ) n 3 q t q ) ν ( z , 2 ( 1 2 q ) n 3 q t q ) ν ( x , 2 ( 1 q ) n 3 q t q ) ν ( y , 2 ( 1 q ) n 3 q t q ) ν ( z , 2 ( 1 q ) n 3 q t q )
(3.21)
for all x , y , z X , t > 0 and n N . Since 1 / 2 < q < 1 and taking the limit as n , we see that (3.20) and (3.21) tend to 1 and 0, respectively. As in the previous cases, we have that D T ( x , y , z ) = 0 for all x , y , z X . By the same argument as in previous cases, we can see that (3.2) follows from (3.19). To prove the uniqueness of T, assume that T is another quadratic-additive mapping from X into Y satisfying (3.2). From (3.2) and (3.13), for all x X and t > 0 , write
μ ( T ( x ) T ( x ) , t ) = ν ( J n T ( x ) J n T ( x ) , t ) μ ( J n T ( x ) J n f ( x ) , t 2 ) μ ( J n f ( x ) J n T ( x ) , t 2 ) μ ( 4 n 2 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 4 n 2 ( ( f T ) ( x 2 n ) ) , t 8 ) μ ( 4 n 2 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 4 n 2 ( ( f T ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) μ ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) sup t < t μ ( x , 2 ( 1 2 q ) n 2 q ( 2 p 4 3 ) q t q )
and, similarly,
ν ( T ( x ) T ( x ) , t ) ν ( 4 n 2 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 4 n 2 ( ( f T ) ( x 2 n ) ) , t 8 ) ν ( 4 n 2 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 4 n 2 ( ( f T ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( T f ) ( x 2 n ) ) , t 8 ) ν ( 2 n 1 ( ( f T ) ( x 2 n ) ) , t 8 ) sup t < t ν ( x , 2 ( 1 2 q ) n 2 q ( 2 p 4 3 ) q t q )

for n N . Letting n in (3.17) and (3.18), and using the fact that lim n 2 ( 2 q 1 ) n 2 q = lim n 2 ( 1 q ) n 2 q = together with the definition of IFN-space, we get μ ( T ( x ) T ( x ) , t ) = 1 and ν ( T ( x ) T ( x ) , t ) = 0 for all x X and t > 0 . Hence T ( x ) = T ( x ) for all x X . □

Remark 3.2 Let ( X , μ , ν ) be an IFN-space and ( X , μ , ν ) be an intuitionistic fuzzy Banach space ( Y , μ , ν ) . Let f : X Y be a mapping satisfying (3.1) with a real number q < 0 and for all t > 0 . If we choose a real number α with 0 < 3 α < t , then
μ ( D f ( x , y , z ) , t ) μ ( D f ( x , y , z ) , 3 α ) μ ( x , α q ) μ ( y , α q ) μ ( z , α q ) and ν ( D f ( x , y , z ) , t ) ν ( D f ( x , y , z ) , 3 α ) ν ( x , α q ) ν ( y , α q ) ν ( z , α q )
for all x , y , z X , t > 0 and q < 0 . Since q < 0 , we have lim α 0 + α q = . This implies that
lim α 0 + μ ( x , α q ) = 1 = lim α 0 + μ ( y , α q ) = lim α 0 + μ ( z , α q ) and lim α 0 + ν ( x , α q ) = 0 = lim α 0 + ν ( y , α q ) = lim α 0 + ν ( z , α q ) .

Thus, we have μ ( D f ( x , y , z ) , t ) = 1 and ν ( D f ( x , y , z ) , t ) = 0 for all x , y , z X and t > 0 . Hence D f ( x , y , z ) = 0 for all x , y , z X . In other words, if f is an intuitionistic fuzzy q-almost quadratic-additive mapping for the case q < 0 , then f is itself a quadratic-additive mapping.

Corollary 3.3 Suppose that f is an even mapping satisfying the conditions of Theorem 3.1. Then there exists a unique quadratic mapping T : X Y such that
μ ( T ( x ) f ( x ) , t ) sup t < t μ ( x , ( | 4 2 p | t 3 ) q ) and ν ( T ( x ) f ( x ) , t ) sup t < t ν ( x , ( | 4 2 p | t 3 ) q ) }
(3.22)

for all x X and t > 0 , where p = 1 / q .

Proof Since f is an even mapping, we get
J n f ( x ) = { f ( 2 n x ) + f ( 2 n x ) 2 4 n if  q > 1 2 , 1 2 ( 4 n ( f ( 2 n x ) + f ( 2 n x ) ) ) if  0 < q < 1 2 ,
for all x X , where J n f is defined as in Theorem 3.1. In this case, J 0 f ( x ) = f ( x ) . For all x X and j N { 0 } , we have
J j f ( x ) J j + 1 f ( x ) = { D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 + D f ( 2 j x , 2 j x , 2 j x ) 2 4 j + 1 if  q > 1 2 , 4 j 2 ( D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) + D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) ) if  0 < q < 1 2 .
Proceeding along the same lines as in Theorem 3.1, we obtain that T is a quadratic-additive function satisfying (3.22). Notice that T ( x ) : = ( μ , ν ) lim n J n f ( x ) , T is even and D T ( x , y , z ) = 0 for all x , y , z X . Hence, we get
T ( x + y ) + T ( x y ) 2 T ( x ) 2 T ( y ) = D T ( x , y , x ) = 0

for all x , y X . It follows that T is a quadratic mapping. □

Corollary 3.4 Suppose that f is an even mapping satisfying the conditions of Theorem 3.1. Then there exists a unique additive mapping T : X Y such that
μ ( T ( x ) f ( x ) , t ) sup t < t μ ( x , ( | 2 2 p | t 3 ) q ) and ν ( T ( x ) f ( x ) , t ) sup t < t ν ( x , ( | 2 2 p | t 3 ) q ) }
(3.23)

for all x X and t > 0 , where p = 1 / q .

Proof Since f is an odd mapping, we get
J n f ( x ) = { f ( 2 n x ) + f ( 2 n x ) 2 n + 1 if  q > 1 , 2 n 1 ( f ( 2 n x ) + f ( 2 n x ) ) if  0 < q < 1 ,
for all x X , where J n f is defined as in Theorem 3.1. Here J 0 f ( x ) = f ( x ) . For all x X and j N { 0 } , we have
J j f ( x ) J j + 1 f ( x ) = { D f ( 2 j x , 2 j x , 2 j x ) 2 j + 2 D f ( 2 j x , 2 j x , 2 j x ) 2 j + 2 if  q > 1 , 2 j 1 ( D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) D f ( x 2 j + 1 , x 2 j + 1 , x 2 j + 1 ) ) if  0 < q < 1 .
Proceeding along the same lines as in Theorem 3.1, we obtain that T is a quadratic-additive function satisfying (3.23). Here T ( x ) : = ( μ , ν ) lim n J n f ( x ) , T is odd and D T ( x , y , z ) = 0 for all x , y , z X . Hence, we obtain
T ( x + y ) T ( x ) T ( y ) = D f ( x y 2 , x + y 2 , x + y 2 ) = 0

for all x , y X . It follows that T is an additive mapping. □

Declarations

Acknowledgements

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant no. (405/130/1433). The authors, therefore, acknowledge with thanks DSR technical and financial support.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Abdulaziz University

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