Journal of Modern Mathematics Frontier

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

Ulam - Hyers Stability of a 2- Variable AC - Mixed Type Functional Equation: Direct and Fixed Point Methods M. Arunkumar1, Matina J. Rassias2, Yanhui Zhang3 1

2

Department of Mathematics, Government Arts College, Tiruvannamalai - 606 603, TamilNadu, India Department of Statistical Science, University College London, 1-19 Torrington Place, #140, London, WC1E 7HB, UK 3 Department of Mathematics, Beijing Technology and Business University, China 1 annarun2002@yahoo.co.in; 2matina@stats.ucl.ac.uk; 3zhangyanhui@th.btbu.edu.cn was introduced by J.M. Rassias [18]. The stability of generalized mixed type functional equation of the form

Abstract- In this paper, we obtain the general solution and generalized Ulam - Hyers stability of a 2 variable AC mixed type functional equation

f ( x  ky)  f ( x  ky)

f  2 x  y, 2 z  w   f  2 x  y, 2 z  w 

 k 2 [ f ( x  y)  f ( x  y)]  2(1  k 2 ) f ( x) (1.3) for fixed integers k and k  0, 1 in quasi-Banach spaces was introduced by M. Eshaghi Gordji and H. Khodaie [7]. The mixed type (I.3) is having the property additive, quadratic and cubic.

 4  f  x  y, z  w   f  x  y , z  w    6 f  y , w  using direct and fixed point methods. Keywords- Additive Functional Equations; Cubic Functional Equation; Mixed Type AC Functional Equation; Ulam – Hyers Stability

I.

J.H. Bae and W.G. Park proved the general solution of the 2- variable quadratic functional equation

INTRODUCTION

The study of stability problems for functional equations is related to a question of Ulam [30] concerning the stability of group homomorphisms and affirmatively answered for Banach spaces by Hyers [9]. It was further generalized and excellent results obtained by number of authors [2], [6], [16], [20], [23].

f ( x  y, z  w)  f ( x  y, z  w) (1.4)  2 f ( x, z)  2 f ( y, w) and investigated the generalized Hyers-Ulam-Rassias stability of (I.4). The above functional equation has solution

Over the last six or seven deades, the above problem was tackled by numerous authors and its solutions via various forms of functional equations like additive, quadratic, cubic, quartic, mixed type functional equations which involve only these types of functional equations were discussed. We refer the interested readers for more information on such problems to the monographs [1], [5], [8], [10], [12], [13], [15], [17], [19], [21], [24], [25], [26], [27], [28], [29], [31], [32] and [33].

f ( x, y)  ax 2  bxy  cy 2 (1.5) The stability of the functional equation (I.4) in fuzzy normed space was investigated by M. Arunkumar et., al [3]. Using the ideas in [3], the general solution and generalized Hyers-Ulam- Rassias stability of a 3- variable quadratic functional equation

f ( x  y, z  w, u  v)  f ( x  y, z  w, u  v) (1.6)  2 f ( x, z, u)  2 f ( y, w, v) was discussed by K. Ravi and M. Arunkumar [22]. The solution of the (I.6) is of the form

In 2006, K.W. Jun and H.M. Kim [11] introduced the following generalized additive and quadratic type of functional equation

f ( x, y, z )  ax 2  by 2  cz 2  dxy  eyz  fzx (1.7) In this paper, we obtain the general solution and generalized Ulam - Hyers stability of a 2 variable AC mixed type functional equation

n n  n  f   xi    n  2   f  xi    f  xi  x j  i 1 1 i  j  n  i 1 

(1.1) in the class of function between real vector spaces. The general solution and Ulam stability of mixed type additive and cubic functional equation of the form

f  2 x  y,2 z  w  f  2 x  y,2 z  w  4  f  x  y, z  w  f  x  y, z  w  6 f  y, w 

having solutions

3 f ( x  y  z )  f ( x  y  z )  f ( x  y  z )  f ( x  y  z )  4[ f ( x)  f ( y)  f ( z )]  4[ f ( x  y)  f ( x  z)  f ( y  z)]

(1.8)

f ( x, y)  ax  by

(1.9)

and

(1.2)

f ( x, y)  ax3  bx 2 y  cxy 2  dy 3

10

(1.10)

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

In Section 2, we present the general solution of the (I.8). The generalized Ulam-Hyers stability using direct and fixed point method are discussed in Section 3 and Section 4, respectively.

Lemma 2.4: If f : U 2  V is a mapping satisfying (I.8) and let t : U V be a mapping given by

II. GENERAL SOLUTION

t (2x  y)  t (2x  y)  4[t (x  y)  t (x  y)]  6t ( y)

t ( x)  f ( x, x)

(II.13)

In this section, we present the solution of the (I.8). Through out this section let U and V be real vector spaces.

for all x, y U . Proof: From (I.8) and (II.12), we get t (2 x  y )  t (2 x  y )

Lemma 2.1: If f : U 2  V is a mapping satisfying (I.8) and let g : U 2  V be a mapping given by g ( x , x)  f (2x ,2x)  8 f ( x , x)

(II.12)

for all x U , then t satisfies

 f (2 x  y, 2 x  y )  f (2 x  y, 2 x  y )  4[ f ( x  y, x  y )  f ( x  y, x  y )]  6 f ( y, y )

(II.1)

 4[t ( x  y )  t ( x  y )]  6t ( y )

for all x U then

for all x, y U .

g (2x,2x)  2g ( x, x) for all x U such that g is additive.

(II.2) III. STABILITY RESULTS: DIRECT METHOD In this section, we investigate the generalized UlamHyers stability of (I.8) using direct method.

Proof: Letting ( x, y, z, w) by (0,0,0,0) in (I.8), we get f (0,0)  0.

(II.3)

Through out this section let U be a normed space and V be a Banach space. Define a mapping F : U 2  V by

Setting ( x, y, z, w) by (0, y,0, y) in (I.8), we obtain (II.4) f ( y,  y)   f ( y, y) for all y U . Replacing ( x, y, z, w) by ( x, x, x, x) in (I.8), we arrive

F ( x, y, z, w)  f  2 x  y,2 z  w  f  2 x  y,2 z  w  4  f  x  y, z  w  f  x  y, z  w  6 f  y, w

for all x, y, z, w U .

(II.5) f (3x,3x)  4 f (2x,2x)  5 f ( x, x) for all x U . Again replacing ( x, y, z, w) by ( x,2x, x,2x) in

Theorem 3.1: Let j  1 . Let f : U 2  V be a mapping for which there exist a function  : U 4  (0, ] with the condition

(I.8) and using (II.3), (II.4) and (II.5), we have f (4x,4x)  10 f (3x,3x) 16 f ( x, x)

(II.6)

for all x U . From (II.1), we establish lim

g(2x,2x)  2g( x, x)  f (4x,4x) 10 f (2x,2x) 16 f ( x, x) (II.7)

n 

F ( x, y, z, w)   ( x, y, z, w)

(III.2) for all x, y, z, w U . Then there exists a unique 2-variable additive mapping A : U 2  V satisfying (I.8) and

Lemma 2.2: If f : U 2  V be a mapping satisfying (I.8) and let h : U 2  V be a mapping given by (II.8)

f (2 x, 2 x)  8 f ( x, x)  A( x, x) 

for all x U then h(2x,2x)  8h( x, x)

(II.9)

(III.3)

for all x U . The mapping  (2kj x) and A( x, x) are defined by

Proof: It follows from (II.8) that

 (2kj x)  4 (2kj x,2kj x,2kj x,2kj x)   (2 kj x,2( k 1) j x,2 kj x,2( k 1) j x)

h(2x,2x)  8g( x, x)  f (4x,4x) 10 f (2x,2x)  16 f ( x, x)

A( x, x)  lim

(II.10) for all x U . Using (II.6) in (II.10), we desired our result.

n 

1  f (2(n 1) j x,2(n 1) j x)  8 f (2nj x,2nj x)  2nj

(III.4) for all x U .

Remark 2.3: If f : U  V be a mapping satisfying (I.8) 2

Proof: Assume j  1 . Letting ( x, y, z, w) by ( x, x, x, x) in (III.2), we obtain

and let g , h : U 2  V be a mapping defined in (II.1) and (II.8) then 1  h ( x, x )  g ( x, x )  6

1    (2kj x) 2 k  1 j 2

for all x U such that h is cubic.

f ( x, x ) 

(III.1)

such that the functional inequality

for all x U . Using (II.6) in (II.7), we desired our result.

h( x, x)  f (2x,2x)  2 f ( x, x)

1  (2nj x, 2nj y, 2nj z, 2nj w)  0 2nj

f (3x,3x)  4 f (2 x,2 x)  5 f ( x, x)   ( x, x, x, x)

(II.11)

for all x U . Replacing ( x, y, z, w) by (III.2), we get

for all x U .

11

(III.5) ( x,2x, x,2x) in

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

f (4 x,4 x)  4 f (3x,3x)  6 f (2 x,2 x) 4 f ( x, x)

 g (2n x,2n x)   2n  

for all x U . This shows that the sequence 

(III.6)   ( x,2x, x,2x) for all x U . Now, from (III.6) and (III.7), we have

is Cauchy sequence. Since V is complete, there exists a mapping A( x, x) : U 2  V such that

f (4 x,4 x)  10 f (2 x,2 x)  16 f ( x, x)  4 f (3x,3x)  4 f (2 x,2 x)  5 f ( x, x)

g (2n x, 2n x) n  2n

A( x, x)  lim

 f (4 x,4 x)  4 f (3x,3x)  6 f (2 x,2 x)  4 f ( x, x)

 4 ( x, x, x, x)   ( x,2x, x,2x) for all x U . From (III.8), we arrive f (4 x,4 x)  10 f (2 x,2 x)  16 f ( x, x)   ( x)

Letting n   in (III.16) and using (II.1), we see that (III.3) holds for all x U . To show that A satisfies (I.8), replacing ( x, y, z, w) by (2n x,2n y,2n z,2n w) and dividing by 2n in (III.2), we obtain

(III.7) (III.8)

where  ( x)  4 ( x, x, x, x)   ( x,2x, x,2x) for all x U . It is easy to see from (III.9) that

1 1 F (2n x, 2n y, 2n z, 2n w)  n  (2n x, 2n y, 2n z, 2n w) 2n 2 for all x, y, z, w U . Letting n   in the above inequality

(III.9)

f (4 x,4 x)  8 f (2 x,2 x)  2( f (2 x,2 x)  8 f ( x; x))   ( x) (III.10)

and using the definition of A( x, x) , we see that

for all x U . Using (II.1) in (III.11), we obtain g (2 x,2 x)  2 g ( x, x)   ( x)

A  2 x  y, 2 z  w   A  2 x  y, 2 z  w   4  A  x  y , z  w   A  x  y , z  w    6 A  y , w 

(III.11)

for all x U . From (III.12), we arrive

Hence A satisfies (I.8) for all x, y, z, w U . To prove A is unique 2-variable additive function satisfying (I.8), we let B(x,x) be another 2-variable additive mapping satisfying (I.8) and (III.3), then

g (2 x, 2 x )  ( x)  g ( x, x )  (III.12) 2 2 for all x U . Now replacing x by 2x and dividing by 2

in (III.13), we get

A( x, x)  B( x, x)

g (22 x,22 x) g ( x, x)  (2 x)   22 2 22

1 A(2n x, 2n x)  B(2n x, 2n x) 2n 1  n A(2n x, 2n x)  f (2n 1 x, 2n 1 x)  8 f (2n x, 2 n x) 2 

(III.13)

for all x U . From (III.13) and (III.14), we obtain

g (22 x, 2 2 x)  g ( x, x ) 22 

 f (2n 1 x, 2n 1 x)  8 f (2n x, 2 n x)  B(2 n x, 2 n x) 

k 0

1  (2 x)   ( x)  (III.14) 2  2  for all x U . Proceeding further and using induction on a positive integer n , we get 

g (2n x,2n x) 1 n 1  (2k x)  g ( x, x)   n 2 2 k 0 2 

for all n   . Hence A is unique. For j  1 , we can prove a similar stability result. This completes the proof of the theorem. The following Corollary is an immediate consequence of Theorem 3.1 concerning the stability of (I.8).

(III.15)

1   (2k x)  2 2 k 0

Corollary 3.2: Let F : U 2  V be a mapping and there exists real numbers  and s such that

for all x U . In order to prove the convergence of the sequence

F ( x, y, z, w)     x s  y s  z s  w s , s  1 or s  1;   1 1 s s s s   x y z w , s  or s  ; 4 4   x s y s z s w s  x 4 s  y 4 s  z 4 s  w   1 1  s or s  ;  4 4

 g (2 x,2 x)   , 2n   n

n

replacing x by 2m x and dividing by 2m in (III.16), for any m, n > 0 , we deduce g (2n  m x, 2n  m x) g (2m x, 2 m x)  2n  m 2m

1 g (2  2 x, 2  2 x)  m  g (2m x, 2m x) 2 2n n

m

n

 (2

k n

x) 2k  n  0 as n   

g (2 x, 2 x) g (2 2 x, 2 2 x) g (2 x, 2 x)  g ( x, x )   2 22 2

 x U .

m

1   (2k x)  0 as m    2 k  0 2k  m

4s

,

(III.17) for all x, y, z, w U , then there exists a unique 2- variable additive function A : U 2  V such that

12

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

f (2 x,2 x)  8 f ( x, x)  A( x, x) 5  s s 1  18  2   x  2  2s   4s 2s  4  2  x  2  24 s   2s 4s   22  2  2  2  2  24 s  2  22 s 

2n ( y, w),2n ( x  y, z  w),2n ( x  y, z  w), 2n (2 x  y,2 z  w),2n (2 x  y,2 z  w), (1,1).

and  (2n (2 x  y,2 z  w))   (2n (2 x  y,2 z  w)) 4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2n ( y, w))  0

(III.18)   x  

For n  0,1,........k  1 . From the definition of f and (III.21), we obtain that

4s

F ( x, y , z , w) 

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) n 2 n0



for all x U .

4[ (2n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w))

Now we will provide an example to illustrate that (I.8) is not stable for s = 1 in (ii) of Corollary 3.2.

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) n 2 nk



Example 3.3: Let  : R  R be a function defined by

4[ (2n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w))

 x, if | x | 1  , otherwise

 ( x)  

1 2 16   16  k  32  (| x |  | y |  | z |  | w |) n 2 2 nk



  0 is a constant, and define a function f : R  R by

where

Thus f satisfies (III.19) for all x, y, z, w  R with

2

 (2 x)

n0

2n

f ( x, x )  

0 | x |  | y |  | z |  | w |

n

for all x  R.

We claim that the additive (I.8) is not stable for s = 1 in (ii) Corollary 3.2. Suppose on the contrary that there exist a additive mapping A : R2  R and a constant   0 satisfying (III.20). Since f is bounded and continuous for all x  R , A is bounded on any open interval containing the origin and continuous at the origin. In view of Theorem 3.1. A must have the form A( x, x)  cx for any x in R . Thus we obtain that

Then F satisfies the functional inequality F ( x, y, z, w)  32  (| x |  | y |  | z |  | w |)

(III.19)

for all x, y, z, w  R . Then there do not exist a additive mapping A : R2  R and a constant   0 such that f (2 x,2 x)  8 f ( x, x)  A( x, x)   |x |

for all x  R. (III.20)

Proof: Now 

 (2n x)

n 0

2n

f ( x, x )  

n 0

2n



f (2 x,2 x)  8 f ( x, x)  (   | c |) | x |, (III.22) But we can choose a positive integer m with m    | c | .

 2

 If x   0,

Therefore we see that f is bounded. We are going to prove that f satisfies (III.19).

1  n  , then 2 x  (0,1) for all n  0, 1,.....m 1. 2  m 1

For this x, we get

If x = y = z = w = 0 then (III.19) is trivial. If | x |  | y |  | z |  | w |

f (2 x, 2 x)  8 f ( x, x)  

1 , then the left hand side of (III.19) 2

(III.21)

1 4

Example 3.4: Let s be such that 0  s  . Then there is a function F : R2  R and a constant   0 satisfying

1 1 1 1 2k 1 x  , 2k 1 y  , 2k 1 z  , 2k 1 w  2 2 2 2

s

( y, w),2

s

s

1 3 s 4

F ( x, y, z, w)   | x | 4 | y | 4 | z | 4 | w |

and consequently 2

 (2n x)

A counter example to illustrate the non stability in (iii) of Corollary 3.2 is given in the following example.

so that

k 1

m 1

which contradicts (III.22). Therefore (I.8) is not stable in sense of Ulam, Hyers and Rassias if s = 1, assumed in (ii) of (III.17).

integer k such that

k 1

n



2 2n n0  m x  (   | c |) x

Then there exists a positive

1 1 | x |  | y |  | z |  | w | k 1 2k 2

 (2n x)

n0

is less than 32 . Now suppose that 1 0 | x |  | y |  | z |  | w | . 2

1 2

( x  y, z  w),2

k 1

(III.23)

for all x, y, z, w  R and

( x  y, z  w),

2k 1 (2 x  y,2 z  w),2k 1 (2 x  y,2 z  w), (1,1).

sup

Therefore for each n  0,1,........k  1 , we have

x0

13

f (2 x,2 x)  8 f ( x, x)  A( x, x) x

 

(III.24)

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26 f (2 x  y, 2 z  w)  f (2 x  y, 2 z  w)  4 f ( x  y , z  w)

for every additive mapping A( x, x) : R2  R .

4 f ( x  y , z  w)  6 f ( y , w)

Proof: If we take

 ( 2v1  v2 , 2v3  v4 )ln | 2v1  v2 , 2v3  v4 |

if x  0 if x  0

( x, x)ln( x, x), f ( x, x)   0,

 ( 2v1  v2 , 2v3  v4 )ln | 2v1  v2 , 2v3  v4 |  4( v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |

Then from (III.24), it follows that sup

 4( v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |  6( v2 , v4 )ln | v2 , v4 |

f (2 x,2 x)  8 f ( x, x)  A( x, x)

x0

 sup

 4 f (v1  v2 , v3  v4 )

x

 4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 )

f (2n,2n)  8 f ( n, n)  A(n, n)

s

s

n(2,2)ln | 2n,2n | 8n(1,1)ln | n, n | nA(1,1)

s

s

s

1 3 s 4

  | x |4 | y |4 | z |4 | w |

Case (iii): If x, z  0, y, w  0 then 2x  y,

n

n n0

2z  w, x  y, z  w  0,2x  y,2z  w, x  y, z  w  0 in (III.23)

 sup (2,2)ln | 2n,2n | 8(1,1)ln | n, n |  A(1,1)  

then,

n n0

f (2 x  y,2 z  w)  f (2 x  y ,2 z  w)  4 f ( x  y , z  w)

We have to prove (III.23) is true.

4 f ( x  y, z  w)  6 f ( y, w)

Case (i): If x, y, z, w  0 in (III.23) then,

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

f (2 x  y, 2 z  w)  f (2 x  y , 2 z  w)  4 f ( x  y , z  w)

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

4 f ( x  y, z  w)  6 f ( y, w)

 4( x  y, z  w)ln | x  y, z  w |

 (2 x  y, 2 z  w)ln | 2 x  y, 2 z  w |

 4( x  y, z  w)ln | x  y, z  w |  6( y, w)ln | y, w |

 (2 x  y, 2 z  w)ln | 2 x  y , 2 z  w |

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

 4( x  y, z  w)ln | x  y , z  w |

f (2 x  y,2 z  w)  f (2 x  y,2 z  w)  4 f ( x  y, z  w)

 4( x  y, z  w)ln | x  y , z  w |  6( y , w)ln | y, w |

4 f ( x  y, z  w)  6 f ( y, w)

Set x  v1, y  v2 , z  v3 , w  v4 it follows that

 (2v1  v2 ,2v3  v4 )ln | 2v1  v2 ,2v3  v4 |

f (2 x  y,2 z  w)  f (2 x  y,2 z  w)  4 f ( x  y, z  w)

 (2v1  v2 ,2v3  v4 )ln | (2v1  v2 ,2v3  v4 ) |

4 f ( x  y, z  w)  6 f ( y, w)

 4(v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |

 (2v1  v2 ,2v3  v4 )ln | 2v1  v2 ,2v3  v4 |

 4(v1  v2 , v3  v4 )ln | (v1  v2 , v3  v4 ) |  6(v2 , v4 )ln | v2 , v4 |

 (2v1  v2 ,2v3  v4 )ln | 2v1  v2 ,2v3  v4 |

 f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )

 4(v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |

 4 f (v1  v2 , v3  v4 )

 4(v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |  6(v2 , v4 )ln | v2 , v4 |

 4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 )

 f (2v1  v2 ,2v3  v4 )  f (2v1  v2 ,2v3  v4 )

s

s

s 4

s 4

s

s

1 3 s 4

Case (iv): If x, z  0, y, w  0 then 2x  y,

1 3 s 4

s 4

  | v1 | | v2 | | v3 | | v4 | s 4

1 3 s 4

s

  | x |4 | y |4 | z |4 | w |

 4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 ) s 4

s

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 |

 4 f (v1  v2 , v3  v4 )

s 4

1 3 s 4

s

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 |

n

n n0

 sup

 f (2v1  v2 , 2v3  v4 )  f ( 2v1  v2 , 2v3  v4 )

2z  w, x  y, z  w  0,2x  y,2z  w, x  y, z  w  0 in (III.23)

1 3 s 4

then,

  | x| | y | | z | | w|

f (2 x  y,2 z  w)  f (2 x  y ,2 z  w)  4 f ( x  y , z  w)

Case (ii): If x, y, z, w  0 in (III.23) then,

4 f ( x  y, z  w)  6 f ( y, w)

f (2 x  y,2 z  w)  f (2 x  y ,2 z  w)  4 f ( x  y , z  w)

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

4 f ( x  y, z  w)  6 f ( y, w)

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

 4( x  y, z  w)ln | x  y, z  w |

 (2 x  y,2 z  w)ln | 2 x  y,2 z  w |

 4( x  y, z  w)ln | x  y, z  w |  6( y, w)ln | y, w |

 4( x  y, z  w)ln | x  y, z  w |

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

 4( x  y, z  w)ln | x  y, z  w |  6( y, w)ln | y, w |

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

14

Journal of Modern Mathematics Frontier

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

f (2 x  y, 2 z  w)  f (2 x  y, 2 z  w)  4 f ( x  y , z  w)

so that

4 f ( x  y , z  w)  6 f ( y , w)

1 1 1 1 1 1 1 2k 1 | x | 4 | y | 4 | z | 4 | w | 4  , 2 k 1 x  , 2 k 1 y  , 2 2 2 1 k 1 1 k 1 2 z  ,2 w  2 2

 (2v1  v2 , 2v3  v4 )ln | (2v1  v2 , 2v3  v4 ) |  (2v1  v2 , 2v3  v4 )ln | 2v1  v2 , 2v3  v4 |  4(v1  v2 , v3  v4 )ln | (v1  v2 , v3  v4 ) |  4(v1  v2 , v3  v4 )ln | v1  v2 , v3  v4 |  6( v2 , v4 )ln | v2 , v4 |

and consequently

 f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )

2k 1 ( y, w), 2k 1 ( x  y, z  w), 2k 1 ( x  y, z  w),

 4 f (v1  v2 , v3  v4 )

 1 1 2k 1 (2 x  y, 2 z  w), 2k 1 (2 x  y, 2 z  w),   ,  .  4 4 Therefore for each n  0,1,........k  1 , we have

 4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 ) s

s

1 3 s 4

s

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 | s

s

1 3 s 4

s

  | x |4 | y |4 | z |4 | w |

2n ( y, w), 2n ( x  y, z  w), 2 n ( x  y, z  w),

Case (v): If x  y  z  w  0 in (III.23) then it is trivial.

 1 1 2n (2 x  y, 2 z  w), 2 n (2 x  y, 2 z  w)    ,  .  4 4

Now we will provide an example to illustrate that the (I.8) is not stable for s 

and

1 in (iv) of Corollary 3.2. 4

 (2n (2 x  y,2 z  w))   (2n (2 x  y,2 z  w))

Example 3.5: Let  : R  R be a function defined by

4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2n ( y, w))  0 For n  0,1,........k  1 . From the definition of f and (III.27), we obtain that

1    x, if | x | 4  ( x)     , otherwise  4

F ( x, y , z , w) 

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) n 2 n0



  0 is a constant, and define a function 2 f : E  R by

where

 (2 x)

n0

2n

f ( x, x )  

n

4[ (2 n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w)) 

for all x  R.

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) n 2 nk



Then F satisfies the functional inequality

4[ (2 n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w))

F ( x, y, z, w)



  (III.25)  8  | x | | y | | z | | w |  | x |  | y |  | z |  | w |    for all x, y, z, w  R . Then there do not exist a additive 1 4

1 4

1 4

1 4

nk

1  1 1 1   8   | x | 4 | y | 4 | z | 4 | w | 4  | x |  | y |  | z |  | w |   

mapping A : R2  R and a constant   0 such that f (2 x,2 x)  8 f ( x, x)  A( x, x)   |x |

Thus f satisfies (III.25) for all x, y, z, w with

for all x  R. (III.26)

1

f ( x, x )  

 (2n x) 2

n0

n

 n0

1     . 2n 4 2

1

1

We claim that (I.8) is not stable for s 

If x = y = z = w = 0 then (III.25) is trivial. 1

1

If | x | 4 | y | 4 | z | 4 | w | 4  | x |  | y |  | z |  | w | 

1 , then the left 2

hand side of (III.25) is less than 8 . Now suppose that 1

1

1

1

0 | x | 4 | y | 4 | z | 4 | w | 4  | x |  | y |  | z |  | w | 

1 . 2

1   x   0, m 1  , then  2  n  0, 1,.....m 1. For this x, we get

Then there exists a positive integer k such that 1 4

1 4

1 4

1 2

1 in (iii) Corollary 4

3.2. Suppose on the contrary that there exist a additive mapping A : R2  R and a constant   0 satisfying (III.26). Since f is bounded and continuous for all x  R , A is bounded on any open interval containing the origin and continuous at the origin. In view of Theorem 3.1, A must have the form A( x, x)  cx for any x in R . Thus we obtain that f (2 x,2 x)  8 f ( x, x)  (   | c |) | x |, (III.28) But we can choose a positive integer m with m    | c | .

Therefore we see that f is bounded. We are going to prove that f satisfies (III.25). 1

1

0 | x | 4 | y | 4 | z | 4 | w | 4  | x |  | y |  | z |  | w | 

Proof: Now

1

1 16  2  4 k 2n 4 2

If

1 4

1 1 | x | | y | | z | | w |  | x |  | y |  | z |  | w |  k 1 k 2 2

(III.27)

15

2n x  (0,1)

for

all

Journal of Modern Mathematics Frontier 

f (2 x, 2 x)  8 f ( x, x)  

 (2n x) n

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26 m 1



for all x U . Proceeding further and using induction on a positive integer n , we get

 (2n x)

2 2 n0 n0  m x  (   | c |) x

n

h(2n x,2n x) 1 n 1  (2k x)  h( x, x)   n 8 8 k  0 8k

which contradicts (III.28). Therefore (I.8) is not stable in sense of Ulam, Hyers and Rassias if s 

1 , assumed in (iv) 4

of (III.17).

lim

1  (2nj x, 2nj y, 2nj z, 2nj w)  0 8nj

 h(2n x,2n x)   , 8n  

(III.29)

replacing x by 2m x and dividing by 8m in (III.39), for any m, n > 0 , we deduce

such that the functional inequality F ( x, y, z, w)   ( x, y, z, w)

h(2n  m x,2n  m x) h(2m x,2m x)  8n  m 8m

(III.30)

for all x, y, z, w U . Then there exists a unique 2-variable cubic mapping C : U 2  V satisfying (I.8) and f (2 x, 2 x)  2 f ( x, x)  C ( x, x) 

1    (2kj x) 8 k  1 j

 

2

 h(2n x,2n x)   8n  

for all x U . This shows that the sequence 

 (2kj x)  4 (2kj x,2kj x,2kj x,2kj x)   (2 kj x,2( k 1) j x,2 kj x,2( k 1) j x) n 

1 h(2n  2m x,2n  2m x)  h(2m x,2m x) 8m 8n

1   (2k x)  8 k  0 8k  m  0 as m  

(III.31)

for all x U . The mapping  (2kj x) and C( x, x) are defined by

C ( x, x)  lim

is Cauchy sequence. Since V is complete, there exists a mapping C( x, x) : U 2  V such that

1  f (2(n 1) j x,2(n 1) j x)  2 f (2nj x,2nj x)  8nj

(III.32)

h(2n x, 2 n x) n  2n

for all x U .

C ( x, x)  lim

Proof: It is easy to see from (III.9) that (III.34) for all x U . Using (II.8) in (III.34), we obtain h(2 x,2 x)  8h( x, x)   ( x)

The following Corollary is an immediate consequence of Theorem 3.6 concerning the stability of (I.8).

(III.35)

Corollary 3.7: Let F : U 2  V be a mapping and there exists real numbers  and s such that

for all x U . From (III.35), we arrive h(2 x, 2 x)  ( x)  h ( x, x )  8 8

(III.36)

F ( x, y, z, w)   s s s s  x  y  z  w , s  3 or s  3;  3 3 s s s s   x y z w , s  4 or s  4 ;   s s s s  x y z w  4s 4s 4s 4s  x  y  z  w ,   3 3 s or s  ;  4 4 

for all x U . Now replacing x by 2x and dividing by 8 in (III.36), we get h(22 x,22 x) h( x, x)  (2 x)   82 8 82

(III.37)

for all x U . From (III.36) and (III.37), we obtain h(22 x, 22 x)  h ( x, x ) 82 h(2 x, 2 x) h(2 2 x, 2 2 x) h(2 x, 2 x)  h ( x, x )   8 82 8 1  (2 x)     ( x)  8 8 

 x U .

Letting n   in (III.39) and using (II.8), we see that (III.31) holds for all x U . To show that C satisfies (I.8) and it is unique the proof is similar to that of Theorem 3.1

f (4 x,4 x)  2 f (2 x,2 x)  8( f (2 x,2 x)  2 f ( x; x))   ( x)

(III.39)

for all x U . In order to prove the convergence of the sequence

Theorem 3.6: Let j  1 . Let f : U 2  V be a mapping for which there exist a function  : U 4  (0, ] with the condition n 

1   (2k x)  8 k  0 8k



(III.40) for all x, y, z, w U , then there exists a unique 2- variable cubic function C : U 2  V such that

(III.38)

f (2 x,2 x)  2 f ( x, x)  C ( x, x)

16

Journal of Modern Mathematics Frontier 5 / 7  s s 1  18  2   x  7 8  2s   4s 2s  4  2  x  7 8  24 s   2s 2  24 s   22  2   7 8  24 s  7 8  22 s 

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26 2n ( y, w), 2n ( x  y, z  w), 2 n ( x  y, z  w),  1 1 2n (2 x  y, 2 z  w), 2 n (2 x  y, 2 z  w)    ,  .  2 2

and  (2n (2 x  y,2 z  w))   (2n (2 x  y,2 z  w))

(III.41)   x  

4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2n ( y, w))  0 For n  0,1,........k  1 . From the definition of f and (III.44), we obtain that

4s

F ( x, y, z , w)

for all x U .



Now we will provide an example to illustrate that (I.8) is not stable for s = 3 in (ii) of Corollary 3.6.

n0

4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2 n ( y, w))

Example 3.8: Let  : R  R be a function defined by



  x , if | x | 1  ( x)     , otherwise 3

nk

  0 is a constant, and define a function f : R  R by



2

 (2n x)

n0

8n

f ( x, x )  

nk

for all x  R.

f (2 x,2 x)  2 f ( x, x)  C ( x, x)   | x |3 for all x  R. (III.43)

Proof: Now

n 0

8

n

n 0

8n



8 7

f (2 x,2 x)  2 f ( x, x)  (   | c |) | x |3 ,

Therefore we see that f is bounded. We are going to prove that f satisfies (III.42).

1   x   0, m 1  , then  2  n  0, 1,.....m 1. For this x, we get

2n x  (0,1)

If

1 | x |3  | y |3  | z |3  | w |3  , then the left hand side of (III.42) 8 16  8 is less than . Now suppose that 7

f (2 x, 2 x)  2 f ( x, x)   n 0

 (2n x) 8

n

m 1



for

all

 (2n x)3

n 0

 m x  (   | c |) x 3

1 . 8

8n 3

which contradicts (III.45). Therefore (I.8) is not stable in sense of Ulam, Hyers and Rassias if s =3, assumed in the inequality (ii) of (III.41).

Then there exists a positive integer k such that 1 1 | x |3  | y |3  | z |3  | w |3  k 1 8k  2 8

(III.45)

But we can choose a positive integer m with m    | c | .

If x = y = z = w = 0, then (III.42) is trivial. If

0 | x |3  | y |3  | z |3  | w |3 

1 8

We claim that (I.8) is not stable for s =3 in (ii) Corollary 3.7. Suppose on the contrary that there exist a cubic mapping C : R2  R and a constant   0 satisfying (III.43). Since f is bounded and continuous for all x  R , C is bounded on any open interval containing the origin and continuous at the origin. In view of Theorem 3.6, C must have the form C( x, x)  cx3 for any x in R . Thus we obtain that

(III.42) for all x, y, z, w  R . Then there do not exist a cubic mapping C : R2  R and a constant   0 such that

 (2n x)

16  83 (| x |3  | y |3  | z |3  | w |3 ) 7

0 | x |3  | y |3  | z |3  | w |3 

16   83 F ( x, y, z , w)  (| x |3  | y |3  | z |3  | w |3 ) 7

1 16  8 1 16   k 8n 7 8

Thus f satisfies (III.42) for all x, y, z, w with

Then F satisfies the functional inequality

f ( x, x )  

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) 2n

4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2 n ( y, w))

where

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) 2n

(III.44)

A counter example to illustrate the non stability in (iii) of Corollary 3.7 is given in the following example.

so that 1 1 1 1 2k 1 x3  , 2k 1 y 3  , 2k 1 z 3  , 2k 1 w3  8 8 8 8

3 4

Example 3.9: Let s be such that 0  s  . Then there is a

and consequently

function F : R2  R and a constant   0 satisfying

2k 1 ( y, w), 2k 1 ( x  y, z  w), 2 k 1 ( x  y, z  w),  1 1 2k 1 (2 x  y, 2 z  w), 2 k 1 (2 x  y, 2 z  w)    ,  .  2 2 Therefore for each n  0,1,........k  1 , we have

s

s

s

F ( x, y, z, w)   | x | 4 | y | 4 | z | 4 | w |

for all x, y, z, w  R and

17

3 3 s 4

(III.46)

Journal of Modern Mathematics Frontier f (2 x,2 x)  2 f ( x, x)  C ( x, x)

sup x0

x

3

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

 

f (2 x  y, 2 z  w)  f (2 x  y, 2 z  w)  4 f ( x  y , z  w)

(III.47)

4 f ( x  y, z  w)  6 f ( y, w)  (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

for every cubic mapping C( x, x) : R2  R .

 (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

Proof: If we take 3  ( x, x) ln( x, x), f ( x, x)    0,

 4(v1  v2 , v3  v4 )3 ln | v1  v2 , v3  v4 |

if x  0 if x  0

 4(v1  v2 , v3  v4 )3 ln | v1  v2 , v3  v4 |  6(v2 , v4 )3 ln | v2 , v4 |  f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )

Then from (III.47), it follows that

 4 f (v1  v2 , v3  v4 )

f (2 x, 2 x)  2 f ( x, x)  C ( x, x)

sup x0

 sup

s

n

s

s

s

1 3 s 4

1 3 s 4

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 |

f (2n, 2n)  2 f (n, n)  C (n, n)

n n0

 sup

 4 f (v1  v2 , v3  v4 )  6 f ( v2 , v4 )

3

x

s

s

  | x |4 | y |4 | z |4 | w |

3

n3 (2, 2)3 ln | 2n, 2n | 2n3 (1,1)3 ln | n, n | n3C (1,1)

n n0

n

Case (iii): If x, z  0, y, w  0 then 2x  y,

3

2z  w, x  y, z  w  0,2x  y,2z  w, x  y, z  w  0 in (III.46)

 sup (2, 2)3 ln | 2n, 2n | 8(1,1) 3 ln | n, n | C (1,1)  

then,

n n0

f (2 x  y,2 z  w)  f (2 x  y ,2 z  w)  4 f ( x  y , z  w)

We have to prove (III.46) is true. Case (i): If x, y, z, w  0 in (III.46) then,

4 f ( x  y, z  w)  6 f ( y, w)  (2 x  y,2 z  w)3 ln | 2 x  y,2 z  w |

f (2 x  y, 2 z  w)  f (2 x  y , 2 z  w)  4 f ( x  y , z  w) 4 f ( x  y, z  w)  6 f ( y, w)

 (2 x  y,2 z  w)3 ln | 2 x  y,2 z  w |

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |

 4( x  y, z  w)3 ln | x  y, z  w |  4( x  y, z  w)3 ln | x  y, z  w |  6( y, w)3 ln | y, w |

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |  4( x  y, z  w)3 ln | x  y, z  w |

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

 4( x  y, z  w)3 ln | x  y, z  w |  6( y, w)3 ln | y, w |

f (2 x  y, 2 z  w)  f (2 x  y, 2 z  w)  4 f ( x  y, z  w)

Set x  v1, y  v2 , z  v3 , w  v4 it follows that

4 f ( x  y , z  w)  6 f ( y, w)  (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

f (2 x  y , 2 z  w)  f (2 x  y , 2 z  w)  4 f ( x  y , z  w) 4 f ( x  y , z  w)  6 f ( y , w)

 (2v1  v2 , 2v3  v4 )3 ln | (2v1  v2 , 2v3  v4 ) |

 (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

 4(v1  v2 , v3  v4 )3 ln | v1  v2 , v3  v4 |  4(v1  v2 , v3  v4 )3 ln | (v1  v2 , v3  v4 ) |  6( v2 , v4 )3 ln | v2 , v4 |

 (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

 f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )

 4(v1  v2 , v3  v4 )3 ln | v1  v2 , v3  v4 |

 4 f (v1  v2 , v3  v4 )

 4(v1  v2 , v3  v4 ) ln | v1  v2 , v3  v4 |  6(v2 , v4 ) ln | v2 , v4 | 3

3

 4 f (v1  v2 , v3  v4 )  6 f ( v2 , v4 )

 f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )

s

 4 f (v1  v2 , v3  v4 )  4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 ) s 4

s 4

s

  | v1 | | v2 | | v3 | | v4 | s 4

s 4

s 4

Case

1 3 s 4

s

s

1 3 s 4

(iv):

If

x, z  0, y, w  0

then

2 x  y,

2z  w, x  y, z  w  0,2x  y,2z  w, x  y, z  w  0 in (III.46)

  | x| | y | | z | | w|

then,

Case (ii): If x, y, z, w  0 in (III.46) then,

f (2 x  y, 2 z  w)  f (2 x  y , 2 z  w)  4 f ( x  y , z  w)

f (2 x  y, 2 z  w)  f (2 x  y , 2 z  w)  4 f ( x  y , z  w)

4 f ( x  y, z  w)  6 f ( y, w)

4 f ( x  y, z  w)  6 f ( y, w)

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |

 (2 x  y, 2 z  w)3 ln | 2 x  y, 2 z  w |

 4( x  y, z  w)3 ln | x  y, z  w |

 4( x  y, z  w) ln | x  y, z  w | 3

 4( x  y, z  w)3 ln | x  y, z  w |  6( y, w)3 ln | y, w |

 4( x  y, z  w) ln | x  y, z  w |  6( y, w) ln | y, w | 3

1 3 s 4

s

  | x |4 | y |4 | z |4 | w |

1 3 s 4

s 4

s

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 |

3

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

Set x  v1,  y  v2 , z  v3 , w  v4 it follows that

18

Journal of Modern Mathematics Frontier

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

f (2 x  y , 2 z  w)  f (2 x  y, 2 z  w)  4 f ( x  y, z  w) 4 f ( x  y , z  w)  6 f ( y , w)

8

 (2v1  v2 , 2v3  v4 ) ln | (2v1  v2 , 2v3  v4 ) | 3

 f (2v1  v2 , 2v3  v4 )  f (2v1  v2 , 2v3  v4 )  4 f (v1  v2 , v3  v4 )  4 f (v1  v2 , v3  v4 )  6 f (v2 , v4 )

s 4

2k 1 ( y, w), 2k 1 ( x  y, z  w), 2k 1 ( x  y, z  w),

1 3 s 4

s 4

  | x| | y | | z | | w|

 1 1 2k 1 (2 x  y, 2 z  w), 2k 1 (2 x  y, 2 z  w),   ,  .  2 2 Therefore for each n  0,1,........k  1 , we have

Case (v): If x  y  z  w  0 in (III.46) then it is trivial. Now we will provide an example to illustrate that (I.8) is not stable for s 

3 in (iv) of Corollary 3.7. 4

2n ( y, w), 2n ( x  y, z  w), 2 n ( x  y, z  w),  1 1 2n (2 x  y, 2 z  w), 2 n (2 x  y, 2 z  w)    ,  .  2 2

Example 3.10: Let  : R  R be a function defined by and

3  3   x , if | x | 4  ( x)    3 , otherwise  4

 (2n (2 x  y,2 z  w))   (2n (2 x  y,2 z  w)) 4[ (2n ( x  y, z  w))   (2n ( x  y, z  w))]  6 (2n ( y, w))  0 For n  0,1,........k  1 . From the definition of f and (III.50), we obtain that

  0 is a constant, and define a function f : R  R by

where 2

 (2n x)

n0

8n

f ( x, x )  

F ( x, y , z , w) for all x  R.

 n0

Then F satisfies the functional inequality

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) n nk 8



3 96  82  34 34 34 3 3 3 3    | x | | y | | z | | w | 4  | x |  | y |  | z |  | w |   7  

4[ (2 n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w))

(III.48) for all x, y, z, w  R . Then there do not exist a cubic mapping C : R2  R and a constant   0 such that f (2 x,2 x)  2 f ( x, x)  C ( x, x)   |x |

3

1 16   3 12   8 1  n 4 7 8k nk 8 



3  3 3 3   8   | x | 4 | y | 4 | z | 4 | w | 4  | x |3  | y |3  | z |3  | w |3   

for all x  R. (III.49)

Proof: Now

Thus f satisfies (III.48) for all x, y, z, w with 

f ( x, x )  

 (2 x) n

8

n0

n

 n0

1 3 6    . 4 7 8n

3

3

3

left hand side of (III.48) is less than

96  . Now suppose 7

that 3

3

3

3

3

1 8

3 in (iii) Corollary 4

3.7. Suppose on the contrary that there exist a cubic mapping C : R2  R and a constant   0 satisfying (III.49). Since f is bounded and continuous for all x  R , C is bounded on any open interval containing the origin and continuous at the origin. In view of Theorem 3.6, C must have the form C( x, x)  cx3 for any x in R . Thus we obtain that

1 , then the 8

0 | x | 4 | y | 4 | z | 4 | w | 4  | x |3  | y |3  | z |3  | w |3 

3

We claim that (I.8) is not stable for s 

If x = y = z = w = 0, then (III.48) is trivial. If | x | 4 | y | 4 | z | 4 | w | 4  | x |3  | y |3  | z |3  | w |3 

3

0 | x | 4 | y | 4 | z | 4 | w | 4  | x |3  | y |3  | z |3  | w |3 

Therefore we see that f is bounded. We are going to prove that f satisfies (III.48). 3

1  (2n (2 x  y, 2 z  w))   (2 n (2 x  y, 2 z  w)) 8n

4[ (2 n ( x  y, z  w))   (2 n ( x  y, z  w))]  6 (2 n ( y, w))

F ( x, y, z, w)

3

1 8 k 1

and consequently

1 3 s 4

  | v1 | 4 | v2 | 4 | v3 | 4 | v4 | s 4

3

1 1 1 1 1 1 1 2k 1 | x | 4 | y | 4 | z | 4 | w | 4  , 2 k 1 x  , 2 k 1 y  , 2 2 2 1 k 1 1 k 1 2 z  ,2 w  2 2

 4(v1  v2 , v3  v4 )3 ln | v1  v2 , v3  v4 |  6( v2 , v4 )3 ln | v2 , v4 |

s

3

so that

 4(v1  v2 , v3  v4 )3 ln | (v1  v2 , v3  v4 ) |

s

3

| x | 4 | y | 4 | z | 4 | w | 4  | x |3  | y |3  | z |3  | w |3  

(III.50)

 (2v1  v2 , 2v3  v4 )3 ln | 2v1  v2 , 2v3  v4 |

s

3

1 k 2

f (2 x,2 x)  2 f ( x, x)  (   | c |) | x |3 ,

1 . 8

(III.51)

But we can choose a positive integer m with m    | c | .

Then there exists a positive integer k such that

19

Journal of Modern Mathematics Frontier 1   x   0, m 1  , then  2  n  0, 1,.....m 1. For this x, we get

2n x  (0,1)

If

f (2 x, 2 x)  2 f ( x, x)   n 0

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

for

for all x U . Thus we obtain (III.55) by defining

all

1 1 A( x, x) and C1 ( x, x)  C ( x, x) ,  (2kj x) , A( x, x) 6 6 and C( x, x) are defined in (III.4), (III.5) and (III.33) A1 ( x, x) 

 (2n x) 8n

m 1



 (2n x)3

n 0

 m x  (   | c |) x 3

for all x U .

8n 3

The following corollary is the immediate consequence of Theorem 3.11, using Corollaries 3.2 and 3.7 concerning the stability of (I.8).

which contradicts (III.51). Therefore (I.8) is not stable in sense of Ulam, Hyers and Rassias if s 

3 , assumed in (iv) 4

Corollary 3.12: Let F : U 2  V be a mapping and there exists real numbers  and s such that

of (III.41). Now, we are ready to prove our main stability results.

F ( x, y, z, w)

Theorem 3.11: Let j  1 . Let f : U 2  V be a mapping for which there exist a function  : U 4  (0, ] with the condition given in (III.1) and (III.29) respectively, such that the functional inequality

    x s  y s  z s  w s , s  3 or s  3;   3 3 s s s s   x y z w , s  or s  ; 4 4   x s y s z s w s  x 4 s  y 4 s  z 4 s  w   3 3  s or s  ;  4 4

F ( x, y, z, w)   ( x, y, z, w)

(III.52) for all x, y, z, w U . Then there exists a unique 2-variable additive mapping A : U 2  V unique 2-variable cubic mapping C : U 2  V satisfying (I.8) and

f ( x, x)  A( x, x)  C ( x, x)   1 1  1       (2kj x)    (2kj x)  6  2 k  1 j 8 k  1 j  2 2  

(III.53)

  5  1   6 1  7    18  2 s 1   1  1 s    x  s s  22  6  7 8  2      4  22 s   1  1 4s    x   4s 4s   6 2  2 7 8  2     2s  22  2    1  1   x 4 s   2  24 s 7 8  22 s  6     4 s   22  1 1 4s   x   4s 2s   6 2  2 7 8  2   

Proof: By Theorems 3.1 and 3.6, there exists a unique 2variable additive function A1 : U 2  V and a unique 2variable cubic function C1 : U 2  V such that (III.54)

1    (2kj x) 8 k  1 j

(III.55)

2

and f (2 x, 2 x)  2 f ( x, x)  C1 ( x, x) 

2

for all x U . Now from (III.54) and (III.55), one can see that f ( x, x ) 

,

(III.57)

for all x U . Now we will provide an example to illustrate that (I.8) is not stable for s  1 in (ii) of Corollary 3.12.

1 1 A1 ( x, x )  C1 ( x, x ) 6 6

Example 3.13: Let  : R  R be a function defined by

 f (2 x, 2 x) 8 f ( x, x) A1 ( x, x)       6 6 6  

 ( x  x3 ), if | x | 1   , otherwise 

 ( x)  

 f (2 x, 2 x) 2 f ( x, x) C1 ( x, x)      6 6 6   

4s

f ( x, x)  A( x, x)  C ( x, x)

for all x U .

1    (2kj x) 2 k  1 j

(III.56) for all x, y, z, w U , then there exists a unique 2-variable additive mapping A : U 2  V unique 2-variable cubic mapping C : U 2  V such that

for all x U . The mapping  (2kj x) , A( x, x) and C( x, x) are defined in (III.4), (III.5) and (III.33)

f (2 x, 2 x)  8 f ( x, x)  A1 ( x, x) 

  0 is a constant, and define a function f : R  R by

where

1  f (2 x, 2 x)  8 f ( x, x)  A1 ( x, x) 6  f (2 x, 2 x)  8 f ( x, x)  A1 ( x, x) 

2

 (2n x)

n0

2n

f ( x, x )  

  1 1  1       (2kj x)    (2 kj x)  6  2 k  1 j 8 k  1 j  2 2  

for all x  R.

Then F satisfies the functional inequality

20

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

F ( x, y, z, w)  32 (| x |  | y |  | z |  | w |)

given x , either d T n x, T n 1 x    for all n  0 ,or there

(III.58)

for all x, y, z, w  R . Then there do not exist a 2-variable additive mapping A : U 2  V and 2-variable cubic mapping C : U 2  V and a constant   0 such that

exists a natural number n0 such that i) d T n x, T n 1 x    for all n  n0 ; ii) The sequence T n x  is convergent to a fixed to a fixed

f ( x, x)  A( x, x)  C ( x, x)   | x | for all x  R. (III.59)

A counter example to illustrate the non stability in (iii) of Corollary 3.12 is given in the following example.

point y * of T iii) y * is the unique fixed point of T in the set

1 Example 3.14: Let s be such that 0  s  . Then there is a 4

function F : R  R and a constant   0 satisfying s

s

s

1 3 s 4

F ( x, y, z, w)   | x | 4 | y | 4 | z | 4 | w |

x

(III.60)

 

Thorough out this section, let U be a vector space and V Banach space. Define a mapping F : U 2  V by

(III.61)

for every additive mapping A( x, x) : R2  R , and for every cubic mapping C( x, x) : R2  R .

F ( x, y, z, w)  f  2 x  y,2 z  w  f  2 x  y,2 z  w

Now we will provide an example to illustrate that (I.8)

 4  f  x  y, z  w  f  x  y, z  w  6 f  y, w

1 is not stable for s  in (iv) of Corollary 3.12. 4

for all x, y, z, w U .

Example 3.15: Let  : R  R be a function defined by

Theorem 4.2: Let f : U 2  V be a mapping for which there exists a function  : U 4  (0, ] with the condition

1  3   ( x  x ), if | x | 4  ( x)     , otherwise  4

1 n n  i

lim

 (2n x)

n0

2n

F ( x, y, z, w)   ( x, y, z, w)

(IV.1)

(IV.2)

for all x, y, z, w U . If there exists L  L  i   1 such that

for all x  R.

the function

Then F satisfies the functional inequality

x    x   12   x  ,

F ( x, y, z, w)

has the property

1  1 1 1  (III.62)  8  | x | 4 | y | 4 | z | 4 | w | 4  | x |  | y |  | z |  | w |    for all x, y, z, w  R . Then there do not exist a 2-variable

  x   L  i   i x  .

(IV.3)

Then there exists a unique 2-variable additive mapping A : U 2  V satisfying (I.8) and

additive mapping A : U 2  V and a cubic mapping C : R2  R and a constant   0 such that f (2 x,2 x)  2 f ( x, x)  C ( x, x)   |x |3

 ( in x, in y, in z, in w)  0

with i  2 if i  0 and i  12 if i  1 such that the functional inequality

  0 is a constant, and define a function 2 f : R  R by

where

f ( x, x )  

Using the above theorem, we now obtain the generalized Hyers-Ulam-Rassias stability of (I.8).

f ( x, x)  A( x, x)  C ( x, x)

x0

1 iv) d  y, y*   d  y, Ty  for all y  . 1 L

for all x, y, z, w  R and sup

  y  : d T n0 x, y   ;

2

Li 1   x (IV.4) 1 L for all x U . The mapping  (2kj x) and A( x, x) are f (2 x, 2 x )  8 f ( x , x )  A( x , x ) 

for all x  R. (III.63)

defined in (III.10) and (III.5) respectively for all x U .

IV. STABILITY RESULTS:FIXED POINT METHOD In this section, we apply a fixed point method for achieving stability of the 2-variable AC (I.8).

Proof: Consider the set

Now, we present the following theorem due to B. Margolis and J.B. Diaz [14] for fixed point Theory.

and introduce the generalized metric on  ,

   p p : U 2  V , p  0   0

d  g , h   d  g , h 

Theorem 4.1: Suppose that for a complete generalized metric space  ,d  and a strictly contractive mapping

 inf K   0,   p  x   q  x   K   x  , x  U

It is easy to see that  ,d  is complete.

T :   with Lipschitz constant L . Then, for each

21

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26 for all x, y, z, w U , i.e., A satisfies (I.8).

Define T : 2   by Tp  x, x  

1

i

According to the alternative fixed point, since A is the T unique fixed point of in the set    p   : d  f , p   , A is the unique function such that

p  i x, i x 

for all x U . Now for all p, q ,

f (2 x,2 x)  8 f ( x, x)  A( x, x)  K  x 

d  g, h  K  p  x, x   q  x, x   K   x  , x  U ,  

1

i 1

i

p   i x,  i x   p  i x, i x  

1

i

q  i x, i x  

1

i

1

for all x U and K  0 . Again using the fixed point alternative, we obtain

K   i x  , x  U ,

q  i x, i x   LK   x  , x  U ,

 d Tp, Tq   LK .

L1 i d  f , Tf 1 L

f (2 x, 2 x)  8 f ( x, x)  A( x, x) 

strictly contractive mapping of  , with Lipschitz constant L . From (III.12), we arrive g (2 x, 2 x )  ( x)  g ( x, x )  2 2

The following Corollary is an immediate consequence of Theorem 4.2 concerning the stability of (I.8).

(IV.5)

Corollary 4.3: Let F :U 2 V be a mapping and there exists real numbers  and s such that

g (2 x, 2 x)  g ( x, x )  L ( x ) for all x U 2

F ( x, y, z, w)

i.e., d  g ,Tg   L  d  g ,Tg   L  L   x , in (IV.5), we get, 2

(IV.6)

for all x U . Using (IV.3) for the case i  1 it reduces to

 2 s 1 18  2 s 1   x s   2  2s   24 s 1  4  22 s   x 4 s  2  24 s   4 s 1 2s 4s  2  22  2  2  2   x  4s 22 

Now from the fixed point alternative in both cases, it follows that there exists a fixed point A of T in  such that f ( in 1 x, in 1 x)  8 f ( in x, in x)  n 

1

i

(IV.7)

d T n g , A   0 . for all x U , since nlim 

  x, y , z 

F  in x, in y, in z , in w 

 lim

n 

in

   x,  y ,  z ,  w  in

n i

4s

 x s  y s  z s  w s ,  s s s s    x y z w ,  s s s s 4s  x y z w  x  y 

from (IV.1) that

n i

(IV.9)

Proof: Setting

in x, in y, in z, in w in (IV.2) and divide by  in . It follows

n i

,

for all x U .

To prove A : U 2  V is additive. Putting  x, y, z, w by

n i

4s

f (2 x,2 x)  8 f ( x, x)  A( x, x)

i.e., d  g ,Tg   1  d  g ,Tg   1  L0  

n 

(IV.8) for all x, y, z, w U , then there exists a unique 2- variable additive function A : U 2  V such that

 x x g ( x, x)  2 g  ,    ( x) for all x U ,  2 2

A  x, y, z , w   lim

 x s  y s  z s  w s , s  1 or s  1;   1 1 s s s s  x y z w , s  or s  ; 4 4   s s s s 4s 4s 4s  x y z w  x  y  z  w   1 1 s or s  ;  4 4 

1

 x x  x g ( x, x)  2 g  ,      2 2   2

L1 i   x 1 L

this completes the proof of the theorem.

for all x U . Using (IV.3) for the case i  0 it reduces to

n 

d  f , A 

which yields

This gives d Tp,Tq   Ld  p, q  , for all p, q , i.e., T is a

A( x, x)  lim

1 d  f ,Tf  1 L

this implies

 T p  x, x   T q  x, x   LK   x  , x  U ,

Again replacing x 

d  f , A 

0

22

4s

 z

4s

 w

4s

,

Journal of Modern Mathematics Frontier

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

for all x, y, z, w U . Then, for s  2 if i  0 and for s  1 if i  1 we get

Hence the (IV.3) holds either, L  2s 1 for s  1 if i  0 and L

  in x, in y, in z , in w 

in

1 for s  1 if i  1 . 2s 1

Now from (IV.4), we prove the following cases for (i).

s s s  n n n s n   n i x  i y  i z  i w ,  i  n s n s n s n s  n i x i y i z i w ,   i    n x s  n y s  n z s  nw s i i i i n  i 4s 4s 4s  in x  in y  in z  in w   0 as n  ,    0 as n  ,  0 as n  , 

Case (i): L  2s 1 for s  1 if i  0 . f (2 x, 2 x)  8 f ( x, x)  A( x, x)

2 

s 1 1 0

1 2

18  2 s 1  s    || x || 2  

s 1

 2  18  2 s 1

4s



,

1 2

s 1

 

s 1

2

 2   2 18  2 s 1

22

 

s

Thus (IV.1) is holds. But we have x    x   12   x  , has

s 1

2

 s   || x || 

 2  18  2   || x || s 1 1

 s   || x || 

s 1

s

2  2s

1 for s  1 if i  1 . 2s 1 f (2 x, 2 x)  8 f ( x, x)  A( x, x)

the property   x   L  i   i x  , for all x U . Hence

Case (ii): L 

1 2 1  (4 ( x, x, x, x)   ( x, 2 x, x, 2 x)) 2  s s  2 (18 || x || 2 || 2 x || )     (4  2 2 s ) || x ||4 s 2  2s 4s 4s  2 (22  2  2  2 ) || x || 

  x    x

11

 1   s 1  18  2 s 1  2  s    || x || 1  2  1  s 1 2 s 1 18  2 s 1  2 s  s 1    || x || 2 1 2 

 2   2 18  2 s 1

 2s  2 

  (18 || i x ||s 2 || 2 i x ||s )  2   i   1   i x    (4  22 s ) || i x ||4 s i 2  i    (22  22 s  2  24 s ) || i x ||4 s   2i

2

 s   || x || 

 2 18  2   || x || s 1

Now

s 1

s 1

s

2s  2

In similar manner we can prove the following cases 1 for s  1 if i  1 , 2 4 s 1 1 for s  1 if i  0 and L  4 s 1 for s  1 if 2

L  24s 1 for s  1 if i  0 and L 

and L  24s 1

i  1 for (ii) and (iii) respectively. complete.

  s i (18 || x ||s 2 || 2 x ||s )  2   i   4 s  i (4  22 s ) || x ||4 s  2 i   4s i (22  22 s  2  24 s ) || x ||4 s   2i   s 1 s s  2 i (18 || x || 2 || 2 x || )     i4 s 1 (4  22 s ) || x ||4 s 2   4 s 1 2s 4s 4s  2 i (22  2  2  2 ) || x || 

Hence the proof is

Theorem 4.4: Let f : U 2  V be a mapping for which there exist a function  : U 4  (0, ] with the condition lim

n 

1

i3n

 ( in x, in y, in z , in w)  0

with i  2 if i  0 and i 

1 2

(IV.10)

if i  1 such that the

functional inequality F ( x, y, z, w)   ( x, y, z, w)

(IV.11)

for all x, y, z, w U . If there exists L  L  i   1 such that the function

 is 1 ( x)    i4 s 1 ( x)  4 s 1  i  ( x )

x    x   12   x  ,

has the property   x   L  i3   i x  .

for all x U .

23

(IV.12)

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Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

Then there exists a unique 2-variable cubic mapping C : U 2  V satisfying (I.8) and f (2 x, 2 x )  2 f ( x , x )  C ( x , x ) 

Li 1   x 1 L

Now from the fixed point alternative in both cases, it follows that there exists a fixed point C of T in  such that

(IV.4)

C ( x, x )  lim

n 

for all x U . The mapping  (2 x) and C( x, x) are defined in (III.10) and (III.33) respectively for all x U . kj

d  g , h   d  g , h 

 inf K   0,   p  x   q  x   K   x  , x  U

 lim

Define T :    by 1

i3

p  i x, i x  

i3n

1

i3 1

i3

q  i x, i x  

0

According to the alternative fixed point, since C is the T unique fixed point of in the set    p   : d  f , p   , C is the unique function such that

 p  x, x   q  x, x   K   x  , x  U ,

1

n i

for all x, y, z, w U , i.e., C satisfies (I.8).

p  i x, i x 

d  g, h  K

i3

   x,  y, in z, in w  n i

n 

for all x U . Now for all p, q ,

i3n

n 

2

Tp  x, x  

(IV.16)

F  in x, in y, in z , in w 

C  x, y, z , w   lim

It is easy to see that  ,d  is complete.

p   i x,  i x  

x, in 1 x)  2 f ( in x, in x ) 

in x, in y, in z, in w in (IV.11) and divide by i3n . It follows from (IV.1) that

and introduce the generalized metric on  ,

1

n 1 i

To prove C : U 2  V is cubic. Putting  x, y, z, w by

   p p : U 2  V , p  0   0

i3

 f (

d  T n h, C   0 . for all x U , since nlim 

Proof: Consider the set

1

i3n

f (2 x,2 x)  8 f ( x, x)  A( x, x)  K  x 

1

i3

for all x U and K  0 . Again using the fixed point alternative, we obtain

K   i x  , x  U ,

q  i x, i x   LK   x  , x  U ,

 T p  x, x   T q  x, x   LK   x  , x  U ,

d  f ,C  

1 d  f ,Tf  1 L

d  f ,C  

L1 i d  f , Tf 1 L

this implies

 d Tp, Tq   LK .

This gives d Tp,Tq   Ld  p, q  , for all p, q , i.e., T is a strictly contractive mapping of  , with Lipschitz constant L.

which yields f (2 x, 2 x)  2 f ( x, x)  C ( x, x) 

From (III.36), we arrive h(2 x, 2 x)  ( x)  h ( x, x )  8 8

 L1 i   x 1 L

this completes the proof of the theorem. (IV.14)

The following Corollary is an immediate consequence of Theorem 4.4 concerning the stability of (I.8).

for all x U . Using (IV.14) for the case i  0 it reduces to

Corollary 4.5: Let F :U 2 V be a mapping and there exists real numbers  and s such that

h(2 x, 2 x)  h( x, x )  L ( x ) 8

for all x U ,

F ( x, y, z, w)

x Again replacing x  , in (IV.14), we get, 2

 x x  x h( x, x)  8h  ,      2 2 2

 x s  y s  z s  w s , s  1 or s  1;   3 3 s s s s  x y z w , s  or s  ; 4 4   s s s s 4s 4s 4s  x y z w  x  y  z  w   3 3 s or s  ;  4 4 

i.e., d  h,Th   L  d  h,Th   L  L1  

(IV.15)

for all x U . Using (IV.14) for the case i  1 it reduces to

4s

,

(IV.17) for all x, y, z, w U , then there exists a unique 2- variable cubic function C : U 2  V such that

 x x h( x, x)  8h  ,    ( x)  2 2

for all x U ,

f (2 x,2 x)  8 f ( x, x)  A( x, x)

i.e., d  h,Th   1  d  h,Th   1  L0  

24

Journal of Modern Mathematics Frontier  2 s 1 18  2 s 1   x s   8  2s   24 s 1  4  22 s   x 4 s  8  24 s   4 s 1 2s 4s  2  22  2  2  2   x  8  24 s 

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26

 1  Li 1 Li 1     x    x  6 1  L 1 L  for all x U . Thus we obtain (IV.21) by defining 1 1 A1 ( x, x)  A( x, x) and C1 ( x, x)  C ( x, x) ,  (2kj x) , A( x, x) 6 6 and C( x, x) are defined in (III.10), (III.5) and (III.33) for

(IV.18) 4s

for all x U .

all x U .

Proof: The proof of the corollary is similar tracing as that of corollary 4.3.

The following corollary is the immediate consequence of Theorem 4.6, using Corollaries 4.3 and 4.5 concerning the stability of (I.8).

Theorem 4.6: Let f : U 2  V be a mapping for which there exist a function  : U 4  (0, ] with (IV.1) and (IV.10) where i  2 if i  0 and i 

1  f (2 x, 2 x)  8 f ( x, x)  A1( x, x) 6  f (2 x, 2 x)  8 f ( x, x)  A1 ( x, x) 

Corollary 4.7: Let F : U 2  V be a mapping and there exists real numbers  and s such that

1 if i  1 such that 2

the functional inequality F ( x, y, z, w)  ( x, y, z, w)

F ( x, y, z, w)

(IV.19)     x s  y s  z s  w s , s  3 or s  3;   3 3 s s s s   x y z w , s  or s  ; 4 4   x s y s z s w s  x 4 s  y 4 s  z 4 s  w   3 3  s or s  ;  4 4

for all x, y, z, w U . If there exists L  L  i   1 such that the function

x    x    x , 1 2

has the (IV.3) and (IV.12), then there exists a unique 2variable additive mapping A : U 2  V unique 2-variable cubic mapping C : U 2  V satisfying (I.8) and f ( x, x)  A( x, x)  C ( x, x) 

1 Li 1   x 31 L

(IV.20)

 18  2 s 1   1  1 s    x  s s   3  7 82   22  2 s 4  2   1  1 4s    x   4s 4s   3 2  2 7 8  2     2s 4s   22  2  2  2   1 1    4s  3 7 8  22 s   22

variable cubic function C1 : U 2  V such that (IV.21)

and f (2 x, 2 x)  2 f ( x, x)  C1 ( x, x) 

i 1

L   x 1 L

(IV.22)

,

  x  

4s

(III.24) for all x U .

for all x U . Now from (III.21) and (III.22), one can see that f ( x, x ) 

4s

f ( x, x)  A( x, x)  C ( x, x)

Proof: By Theorems 4.1 and 4.4, there exists a unique 2variable additive function A1 : U 2  V and a unique 2Li 1   x 1 L

(IV.23) for all x, y, z, w U , then there exists a unique 2-variable additive mapping A : U 2  V unique 2-variable cubic mapping C : U 2  V such that

for all x U . The mapping  (2kj x) , A( x, x) and C( x, x) are defined in (III.5), (III.10) and (III.33) respectively for all x U .

f (2 x, 2 x)  8 f ( x, x)  A1 ( x, x) 

V.

1 1 A1 ( x, x)  C1 ( x, x) 6 6

CONCLUSION

The (I.8) is stable in Banach spaces via direct and fixed point method in sprit of Hyers, Ulam, Rasssias.

 f (2 x,2 x) 8 f ( x, x) A1 ( x, x)       6 6 6  

REFERENCES

 f (2 x,2 x) 2 f ( x, x) C1 ( x, x)      6 6 6  

[1] J. Aczel and J. Dhombres, Functional Equations in Several Variables, Cambridge Univ, Press, 1989. [2] T. Aoki, On the stability of the linear transformation in Banach spaces, J. Math. Soc. Japan, 2 (1950), 64-66. [3] M. Arunkumar, V. Arasu, N. Balaji, Fuzzy Stability of a 2 Variable Quadratic Functional Equation, International

25

Journal of Modern Mathematics Frontier

Sept. 2012, Vol. 1 Iss. 3, PP. 10-26