1. Introduction

Available online at http://scik.org J. Math. Comput. Sci. 6 (2016), No. 5, 712-729 ISSN: 1927-5307

GENERALIZED STABILITY OF AN AQ-FUNCTIONAL EQUATION IN QUASI-(2;P)-BANACH SPACES MENG LIU∗ , MEIMEI SONG Department of Mathematics, Tianjin University of Technology, Tianjin 300384, P.R. China c 2016 Meng Liu and Meimei Song. This is an open access article distributed under the Creative Commons Attribution License, Copyright which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we introduce and investigate the general solution of a new functional equation f(

x+y z+w x+y z+w + )+ f( − ) = a b a b +

1 [(1 + a) f (x + y) + (1 − a) f (−x − y)] a2 1 [ f (z + w) + f (−z − w)] b2

where a, b ≥ 1 and discuss its Generalized Hyers-Ulam-Rassias stability under the conditions such as even, odd, approximately even and approximately odd in quasi-(2;p)-Banach spaces. Keywords: Generalized Hyers-Ulam-Rassias stability; AQ-functional equation; quasi-(2;p)-normed spaces; quadratic function; quasi-(2;p)-Banach spaces . 2010 AMS Subject Classification: 39B52, 39B72, 39B82.

1. Introduction The stability problem of functional equations originated from a question of Ulam [1] in 1940 concerning the stability of group homomorphisms. Let (G1 , ·) be a group and let (G2 , ∗) be a metric group with the metric d(·, ·). Given ε > 0, does there exist δ > 0 such that if a mapping ∗ Corresponding

author

Received April 18, 2016 712

GENERALIZED STABILITY OF AN AQ-FUNCTIONAL EQUATION IN QUASI-(2;P)-BANACH SPACES

713

h : G1 −→ G2 satisfies the inequality d(h(x · y), h(x) ∗ h(y)) < δ for all x, y ∈ G1 , then a homomorphism H : G1 −→ G2 exists with d(h(x), H(y)) < ε for all x ∈ G1 ? In 1941, Hyers [2] considered the case of approximately additive mappings f : E −→ E

0

0

where E and E are Banach spaces. He proved the following theorem. 0

0

Theorem 1.1 [2] E, E is Banach spaces and let f : E −→ E be a mapping satisfying k f (x + y) − f (x) − f (y)k ≤ ε f (2n x) n n−→∞ 2

for all x ∈ E and ε > 0. Then the limit l(x) = lim

0

exists for all x ∈ E and l : E −→ E is

the unique additive mapping satisfying k f (x) − l(x)k ≤ ε for all x ∈ E. Moreover, if f (tx) is continuous in t(−∞ < t < +∞) for each fixed x ∈ E, then l is linear. From the above property, the additive functional equation f (x + y) = f (x) + f (y) has Hyers0

Ulam stability on (E, E ). The theorem of Hyers was generalized by Aoki [3] for additive mappings. In 1978, Rassias [4] considered an unbounded Cauchy difference for linear mappings. It states as follows: 0

Theorem 1.2 [4] Let E, E be two Banach spaces and let θ ∈ [0, ∞) and p ∈ [0, 1). If a function 0

f : E −→ E satisfies the inequality k f (x + y) − f (x) − f (y)k ≤ θ [kxk p + kyk p ] 0

for all x ∈ E. Then there exists a unique additive mapping T : E −→ E such that k f (x) − T (x)k ≤

2θ kxk p 2 − 2p

for all x ∈ E. Moreover, if f (tx) is continuous in t(−∞ < t < +∞) for each fixed x ∈ E, then l is linear. The work of Rassias [4] has had a lot of influence in the development of a generalization of the Hyers-Ulam stability concept. The terminology Hyers-Ulam-Rassias stability originates

714

MENG LIU, MEIMEI SONG

from these historical backgrounds and this terminology is also applied to the case of other functional equations. In this paper, we introduce and investigate the general solution of a new functional equation f(

x+y z+w x+y z+w 1 + )+ f( − ) = 2 [(1 + a) f (x + y) + (1 − a) f (−x − y)] a b a b a 1 + 2 [ f (z + w) + f (−z − w)] b

(1.1)

where a, b ≥ 1 and discuss its Generalized Hyers-Ulam-Rassias stability in quasi-(2;p)-Banach spaces. It may be noted that f (x) = ax2 + bx + c is a solution of the functional equation.

2. Preliminaries Before giving the main results, we will present some preliminaries results. Definition 2.1 [5] Let X be a linear space over R with dim X > 1. A quasi 2-norm is a realvalued function on X × X satisfying the following conditions: (1) k x, y k= 0 if and only if x and y are linearly dependent, (2) k x, y k=k y, x k, (3) k αx, y k= |α| k x, y k for all α ∈ K, (4) There is a constant K ≥ 1 such thatk x + y, z k≤ K(kx, zk + ky, zk) for all x, y, z ∈ X. The pair (X, k·, ·k) is called a quasi 2-normed space if k·, ·k is a quasi 2-norm on X. A quasi 2-norm k·, ·k is called quasi-(2;p)-norm (0 < p ≤ 1) if ||x + y, z|| p ≤ ||x, z|| p + ||y, z|| p for all x, y, z ∈ X. The pair (X, k·, ·k) is called a quasi-(2; p)-normed space if k·, ·k is a quasi(2; p)-norm on X. Definition 2.2 [10] A sequence {xn } in a quasi-(2; p)-normed space (X, k·, ·k) is called a Cauchy sequence if lim kxn − xm , yk = 0

m,n−→∞

for all y ∈ X.


715

Definition 2.3 [10] A sequence {xn } in a quasi-(2; p)-normed space (X, k·, ·k) is called a convergent sequence if there is an x ∈ X such that lim kxn − x, yk = 0

n−→∞

for all y ∈ X. If {xn } converges to x, write xn −→ x as n −→ ∞ and call x the limit of {xn }.In this case,we also write lim xn = x. n−→∞

Definition 2.4 [10] we say that a quasi-(2;p)-normed spaces (X, k·, ·k) is a quasi-(2;p)-Banach spaces if every Cauchy sequence in X is a convergent sequence. We introduce a basic property of a quasi-(2;p)-normed space as follows. Let (X, k·, ·k) be linear quasi-(2;p)-normed space, x ∈ X and k x, y k= 0 for each y ∈ X. suppose x 6= 0. Since dim X > 1, choose y ∈ X such that {x, y} is linearly independent so we have kx, yk 6= 0, which is a contradiction. Therefore, we have the following lemma. Lemma 2.5 Let (X, k·, ·k) be a linear quasi-(2; p)-normed space. If x ∈ X and kx, yk = 0, for each y ∈ X, then x = 0.

3. odd case In this section, we assume that E1 is a real vector space, E2 is a quasi-(2;p)-Banach space and f (0) = 0. For simplicity, given a mapping f : E1 −→ E2 and D f : E1 × E1 × E1 × E1 −→ E2 by D f (x, y, z, w) = f (

x+y z+w x+y z+w 1 + )+ f( − ) − 2 [(1 + a) f (x + y) + (1 − a) f (−x − y)] a b a b a 1 − 2 [ f (z + w) + f (−z − w)] b

for all x, y, z, w ∈ E1 . Lemma 3.1 [6] Let E1 and E2 denote real vectors spaces, if f : E1 −→ E2 is an even function satisfying (1.1) for all x, y, z, w ∈ E1 , then f is quadratic. Lemma 3.2 [6] Let E1 and E2 denote real vectors spaces, if f : E1 −→ E2 is an odd function satisfying (1.1) for all x, y, z, w ∈ E1 , then f is additive.

716


Theorem 3.3 Let φ : R × R × R × R −→ [0, ∞) be a function such that ∞ y z w x φe(x, y, z, w, v) = ∑ aip φ (k i , vk, k i , vk, k i , vk, k i , vk) p < ∞ a a a a i=0

(3.1)

for all x, y, z, w, v ∈ E1 . If f : E1 −→ E2 is an odd mapping satisfies kD f (x, y, z, w), vk ≤ φ (kx, vk, ky, vk, kz, vk, kw, vk)

(3.2)

for all x, y, z, w, v ∈ E1 . Then there exists a unique additive mapping A : E1 −→ E2 satisfying the equation (1.1) such that 1 a k f (x) − A(x), vk ≤ φe(x, 0, 0, 0, v) p 2

(3.3)

Proof. Using oddness and f (0) = 0 in (3.2) we have kf(

x+y z+w 2 x+y z+w + )+ f( − ) − f (x + y), vk p ≤ φ (kx, vk, ky, vk, kz, vk, kw, vk) p a b a b a (3.4)

for all x, y, z, w, v ∈ E1 . Replace (y, z, w) by (0, 0, 0) in (3.4) we have x 2 k2 f ( ) − f (x), vk p ≤ φ (kx, vk, 0, 0, 0) p a a

(3.5)

Again replacing x by ax in (3.5) and multiply both sides by ( a2 ) p yields a ka f (x) − f (ax), vk p ≤ ( ) p φ (kax, vk, 0, 0, 0) p 2 for all x, v ∈ E1 . Again replacing x by ka f (

x ai+1

(3.6)

in (3.6) we have

x

x a p x p ) − f ( ), vk ≤ ( ) φ (k , vk, 0, 0, 0) p ai+1 ai 2 ai

(3.7)

so, n−1 x x x x n p ka f ( m ) − a f ( n ), vk ≤ ∑ kai f ( i ) − ai+1 f ( i+1 ), vk p a a a a i=m m

n−1

=

x

x

∑ aipka f ( ai+1 ) − f ( ai ), vk p

i=m

≤

n−1 (i+1)p a

∑

i=m

2p

x · φ (k i , vk, 0, 0, 0) p a

(3.8)


717

for all x, v ∈ E1 and for any n > m ≥ 0. Since the right-hand side of inequality (3.8) tend to 0 as m −→ ∞. We conclude that {an f ( axn )} is a Cauchy sequence in E2 and so it converges. Because of the completeness of E2 , we can define a mapping A : E1 −→ E2 by A(x) = lim an f ( n−→∞

x ) an

for all x ∈ E1 . By (3.1) and (3.2), we obtain that x y z w , , , ), vk p n−→∞ an an an an x y z w ≤ lim anp φ (k n , vk, k n , vk, k n , vk, k n , vk) p = 0 n−→∞ a a a a

kDA(x, y, z, w), vk p =

lim anp kD f (

for all x, y, z, w, v ∈ E1 . Hence the mapping A : E1 −→ E2 satisfies (1.1). Note that f is an odd mapping ,we obtain A(x) + A(−x) = lim an f ( n−→∞

x x ) + an f (− n ) = 0 n a a

for all x ∈ E1 . So A(x) = −A(−x). Using lemma 3.2, A is additive. Taking m = 0, n −→ ∞ in (3.8), we get x a(i+1)p ∑ 2 p φ (k ai , vk, 0, 0, 0) p i=0 a = ( ) p φe(x, 0, 0, 0, v) 2

k f (x) − A(x), vk p ≤

∞

so, 1 a k f (x) − A(x), vk ≤ φe(x, 0, 0, 0, v) p 2

We get the inequality (3.3). To prove the uniqueness of the additive mapping A, let us assume 0

that there exists a additive mapping A : E1 −→ E2 satisfies (1.1) and (3.3). Using f (0) = 0 and oddness in (1.1), we get f(

x+y z+w x+y z+w 2 + )+ f( − ) = f (x + y) a b a b a

(3.9)

Replacing (z, w) by (0, 0) in (3.9), we obtain f(

1 x+y ) = f (x + y) a a

(3.10)

718


Replacing x by y in (3.10), we obtain f(

2y 1 ) = f (2y) a a

(3.11)

Replacing 2y by ax in (3.11), we obtain (3.12)

f (ax) = a f (x) 0

0

0

0

Then it follows that A (ax) = aA (x), A (am x) = am A (x). We have 0

0

kA(x) − A (x), vk

p

A(am x) A (am x) p − , vk = k am am 0 1 1 ≤ mp kA(am x) − f (am x), vk p + mp kA (am x) − f (am x), vk p a a p 2 a ≤ mp · p φe(am x, 0, 0, 0, v) −→ 0 as m −→ ∞ a 2

for all x, v ∈ E1 . Therefore A is unique. Corollary 3.4 Let E1 be a quasi-2-normed linear space and E2 be a quasi-(2;p)-Banach space. Let θ , r be real numbers such that θ ≥ 0, r > 1. Suppose that a odd mapping f : E1 −→ E2 satisfies kD f (x, y, z, w), vk ≤ θ (kx, vkr + ky, vkr + kz, vkr + kw, vkr ) for all x, y, z, w, v ∈ E1 . Then there exists a unique additive mapping A : E1 −→ E2 satisfying the equation (1.1) such that k f (x) − A(x), vk ≤

a θ kx, vkr p p 2 1 − a(1−r)p

4. even case Theorem 4.1 Let φ : R × R × R × R −→ [0, ∞) be a function such that ∞

i+1 x, vk, kai+1 y, vk, kai+1 z, vk, kai+1 w, vk) p

φ (ka φe(x, y, z, w, v) = ∑ i=0

a2ip

m ≥ 0. Since the right-hand side of inequality (4.8) tend to 0 as n

x) } is a Cauchy sequence in E2 and so it converges. Because m −→ ∞. We conclude that { f (a a2n

of the completeness of E2 , we can define a mapping A : E1 −→ E2 by A(x) = lim

n−→∞

f (an x) a2n

720


for all x ∈ E1 . By (4.1) and (4.2), we obtain that kD f (x, y, z, w), vk p = ≤

lim

1

n−→∞ a2np

lim

1

n−→∞ a2np

kD f (an x, an y, an z, an w), vk p φ (kan x, vk, kan y, vk, kan z, vk, kan w, vk) p = 0

for all x, y, z, w, v ∈ E1 . Hence the mapping A : E1 −→ E2 satisfies (1.1). Note that f is an even mapping ,we obtain A(x) − A(−x) = lim

n−→∞

f (an x) f (−an x) − =0 a2n a2n

for all x ∈ E1 . So A(x) = A(−x). Using lemma 3.1 A is quadratic. Taking m = 0, n −→ ∞ in (4.8), we get k f (x) − A(x), vk

∞

p

≤

1

1

∑ a2ip · 2 p φ (kai+1x, vk, 0, 0, 0) p

i=o

1 = ( ) p φe(x, 0, 0, 0, v) 2 so, 1 1 k f (x) − A(x), vk ≤ φe(x, 0, 0, 0) p 2

We get the inequality (4.4). To prove the uniqueness of the quadratic mapping A, let us assume 0

that there exists a quadratic mapping A : E1 −→ E2 satisfies (1.1) and (4.3). Using f (0) = 0 and evenness in (1.1), we get f(

x+y z+w x+y z+w 2 2 + )+ f( − ) = 2 f (x + y) + 2 f (z + w) a b a b a b

(4.9)

Replacing (z, w) by (0, 0) in (4.9), we obtain f(

x+y 1 ) = 2 f (x + y) a a

(4.10)

Replacing x by 0 in (4.10), we obtain y 1 f ( ) = 2 f (y) a a

(4.11)

f (ax) = a2 f (x)

(4.12)

Replacing y by ax in (4.11), we obtain

GENERALIZED STABILITY OF AN AQ-FUNCTIONAL EQUATION IN QUASI-(2;P)-BANACH SPACES 0

0

0

721

0

Then it follows that A (ax) = a2 A (x), A (am x) = a2m A (x). We have 0

0

kA(x) − A (x), vk

p

A(am x) A (am x) p = k 2m − , vk a a2m 0 1 1 ≤ 2mp kA(am x) − f (am x), vk p + 2mp kA (am x) − f (am x), vk p a a 2 1 ≤ 2mp · p φe(am x, 0, 0, 0, v) −→ 0 a 2

as m −→ ∞ for all x, v ∈ E1 . Therefore A is unique. Corollary 4.2 Let E1 be a quasi-2-normed linear space and E2 be a quasi-(2;p)-Banach space. Let θ , r be real numbers such that θ ≥ 0, r < 2. Suppose that a even mapping f : E1 −→ E2 satisfies kD f (x, y, z, w), vk ≤ θ (kx, vkr + ky, vkr + kz, vkr + kw, vkr ) for all x, y, z, w, v ∈ E1 . Then there exists a unique quadratic mapping A : E1 −→ E2 satisfying the equation (1.1) such that k f (x) − A(x), vk ≤

ar θ kx, vkr p p 2 1 − a(r−2)p

Corollary 4.3 Let E1 be a quasi-2-normed linear space and E2 be a quasi-(2;p)-Banach space. Let θ be real numbers such that θ ≥ 0. Suppose that a even mapping f : E1 −→ E2 satisfies kD f (x, y, z, w), vk ≤ θ for all x, y, z, w, v ∈ E1 . Then there exists a unique quadratic mapping A : E1 −→ E2 satisfying the equation (1.1) such that k f (x) − A(x), vk ≤

θ 1 √ p 2 1 − a−2p

5. Approximately even case Lemma 5.1 Let φ : R × R × R × R −→ [0, ∞) be a given mapping. Suppose that a mapping f : E1 −→ E2 satisfies kD f (x, y, z, w), vk ≤ φ (kx, vk, ky, vk, kz, vk, kw, vk)

(5.1)

722


for all x, y, z, w, v ∈ E1 . We have 1 + an an − 1 n f (a x) + f (−an x), vk p 2a2n 2a2n n 1 + ak−1 p ak ak ak ak−1 − 1 p ak ≤ ∑ {[( x, vk, k x, vk, k x, vk, k x, vk) p } ) + ( ) ]φ (k 2k−2 2k−2 2 2 2 2 4 · a 4 · a k=1

k f (x) −

(5.2)

for all x, v ∈ E1 and n ∈ N. Proof. We use mathematical induction on n to prove lemma. Putting x = y = z, w = −x in (5.1) yields 2 1+a a−1 k2 f ( x) − 2 f (2x) + 2 f (−2x), vk p ≤ φ (kx, vk, kx, vk, kx, vk, kx, vk) p a a a for all x, v ∈ E1 . Replacing x by k f (x) −

ax 2

(5.3)

in (5.3) and dividing by 2 p gives

1+a ax ax ax a−1 1 ax f (ax) + f (−ax), vk p ≤ p φ (k , vk, k , vk, k , vk, k , vk) p 2 2 2a 2a 2 2 2 2 2

(5.4)

for all x, v ∈ E1 . Note that (5.4) proves the validity of inequality (5.2) for the case n = 1.Assume that inequality (5.2) holds for n ∈ N. Replacing x by an x in (5.4) yields 1+a a−1 n+1 f (a x) + f (−an+1 x), vk p 2 2 2a 2a n+1 n+1 1 a x a x an+1 x an+1 x ≤ p φ (k , vk, k , vk, k , vk, k , vk) p 2 2 2 2 2 We have the following relation: k f (an x) −

k ≤ + + ≤

1 + an+1 an+1 − 1 n+1 f (a x) + f (−an+1 x), vk p 2a2(n+1) 2a2(n+1) 1 + an an − 1 n k f (x) − f (a x) + f (−an x), vk p 2n 2n 2a 2a n 1+a 1+a a−1 ( 2n ) p k f (an x) − f (an+1 x) + f (−an+1 x), vk p 2 2 2a 2a 2a an − 1 p 1 + a a−1 n+1 ( 2n ) k − f (−an x) + f (−a x) − f (an+1 x), vk p 2a 2a2 2a2 n 1 + ak−1 ak−1 − 1 ak ak ak ak ∑ {[( 4 · a2k−2 ) p + ( 4 · a2k−2 ) p]φ (k 2 x, vk, k 2 x, vk, k 2 x, vk, k 2 x, vk) p} k=1 f (x) −

1 + an p 1 an+1 x an+1 x an+1 x an+1 x ) · φ (k , vk, k , vk, k , vk, k , vk) p 2a2n 2p 2 2 2 2 an − 1 1 an+1 x an+1 x an+1 x an+1 x + ( 2n ) p · p φ (k , vk, k , vk, k , vk, k , vk) p 2a 2 2 2 2 2 + (

n+1

1 + ak−1 p ak−1 − 1 p ak ak ak ak ) +( ) ]φ (k x, vk, k x, vk, k x, vk, k x, vk) p } ≤ ∑ {[( 2k−2 2k−2 2 2 2 2 4·a 4·a k=1

(5.5)


723

for all x, v ∈ E1 . This proves the validity of inequality (5.2) for the case n + 1. Theorem 5.2 Let φ : R × R × R × R −→ [0, ∞) be a function such that i

∞

i

i

i

p

φ (ka x, vk, ka y, vk, ka z, vk, ka w, vk) φe(x, y, z, w, v) = ∑ m, we obtain f (am x) f (an x) p − 2n , vk a2m a 1 f (an−m · am x) p = 2mp k f (am x) − , vk a a2(n−m)

k

≤

n−m

1 a2mp

+[

1 + ak−1

ak−1 − 1

∑ [( 4 · a2k−2 ) p + ( 4 · a2k−2 ) p] · φ (k

k=1 n−m a −1

2a2(n−m)

ak+m ak+m ak+m ak+m x, vk, k x, vk, k x, vk, k x, vk) p 2 2 2 2

] p ψ(kan−m x, vk) p (5.12)

for all x, v ∈ E1 and n ∈ N. From (5.6) and (5.7), the right-hand side of inequality (5.12) tends n

x) } is a Cauchy sequence. Completeness of E2 allows us to to 0 as m −→ ∞, the sequence { f (a a2n

assume that there exists a mapping A so that A(x) = lim

n−→∞

f (an x) a2n

for all x ∈ E1 . By (5.9), we obtain that kDA(x, y, z, w), vk p = ≤

lim

1

n−→∞ a2np

kD f (an x, an y, an z, an w), vk p

1 φ (kan x, vk, kan y, vk, kan z, vk, kan w, vk) p −→ 0 n−→∞ anp anp lim

as n −→ ∞ for all x, y, z, w, v ∈ E1 and so the mapping A satisfies (1.1). We have the following results kA(x) − A(−x), vk p = ≤

lim k

n−→∞

1 a2np

f (an x) f (−an x) p − , vk a2n a2n

ψ(kan x, vk) p −→ 0

as n −→ ∞ for all x, v ∈ E1 . So, A(x) = A(−x) and A is quadratic. Taking m = 0, n −→ ∞ in (5.12), we get (5.10). Next, we prove the uniqueness of A. A satisfies (1.1) and putting y = z = w = 0, we have x 1+a a−1 2A( ) − 2 A(x) + 2 A(−x) = 0 a a a for all x ∈ E1 . Using evenness of A and replacing x by ax in (5.13), we have A(ax) = a2 A(x)

(5.13)


725

0

for all x ∈ E1 . So,we assume that A : E1 −→ E2 be another quadratic mapping satisfying (1.1) and (5.10),we calculate 0

kA(x) − A (x), vk p 0

A(an x) A (an x) p = k 2n − , vk a a2n 0 1 1 ≤ 2np kA(an x) − f (an x), vk p + 2np k f (an x) − A (an x), vk p a a ∞ k−1 k−1 2 1+a a −1 p ak+n ak+n ak+n ak+n p ≤ 2np · ∑ [( ) + ( ) ] · φ (k x, vk, k x, vk, k x, vk, k x, vk) p 2k−2 2k−2 a 2 2 2 2 4 · a 4 · a k=1 −→ 0 as n −→ ∞ for all x ∈ E1 .

6. Approximately odd case Lemma 6.1 Let φ : R × R × R × R −→ [0, ∞) be a given mapping. Suppose that a mapping f : E1 −→ E2 satisfies kD f (x, y, z, w), vk ≤ φ (kx, vk, ky, vk, kz, vk, kw, vk)

(6.1)

for all x, y, z, w, v ∈ E1 . We have an + a2n x a2n − an x f( n)− f (− n ), vk p 2 a 2 a (6.2) n 2k k 2k k a +a p a −a p x x x x p ≤ ∑ [( ) +( ) ]φ (k k−1 , vk, k k−1 , vk, k k−1 , vk, k k−1 , vk) 4 4 2a 2a 2a 2a k=1

k f (x) −

for all x, v ∈ E1 and n ∈ N. proof. Replacing x by

x a

in (5.4), we have

a−1 1 x x 1+a x x x kf( )− f (x) + f (−x), vk p ≤ p φ (k , vk, k , vk, k , vk, k , vk) p 2 2 a 2a 2a 2 2 2 2 2

(6.3)

Replacing x by −x in (6.3), we have 1+a a−1 1 x x x x x k f (− ) − f (−x) + f (x), vk p ≤ p φ (k , vk, k , vk, k , vk, k , vk) p 2 2 a 2a 2a 2 2 2 2 2

(6.4)

726


From (6.3) and (6.4), we get k f (x) −

a + a2 x a2 − a x a2 + a p x x x x f( )− f (− ), vk p ≤ ( ) φ (k , vk, k , vk, k , vk, k , vk) p 2 a 2 a 4 2 2 2 2 (6.5) 2 x x x x a −a p p ) φ (k , vk, k , vk, k , vk, k , vk) +( 4 2 2 2 2

for all x, v ∈ E1 . Note that (6.5) proves the validity of inequality (6.2) for the case n = 1.Assume that inequality (6.2) holds for n ∈ N. Replacing x by kf(

x an

in (6.5), we get

a + a2 x a2 − a x a2 + a p x x x x x p ) − f ( ) − f (− ), vk ≤ ( ) φ (k n , vk, k n , vk, k n , vk, k n , vk) p n n+1 n+1 a 2 a 2 a 4 2a 2a 2a 2a 2 a −a p x x x x +( ) φ (k n , vk, k n , vk, k n , vk, k n , vk) p 4 2a 2a 2a 2a

so, x x an+1 + a2(n+1) a2(n+1) − an+1 f ( n+1 ) − f (− n+1 ), vk p 2 a 2 a n 2n 2n n a +a x x a −a ≤ k f (x) − f( n)− f (− n ), vk p 2 a 2 a 2 2n n x x x a+a a2 − a a +a p ) kf( n)− f ( n+1 ) − f (− n+1 ), vk p +( 2 a 2 a 2 a 2n n 2 2 a −a p x x x a+a a −a +( ) k f (− n ) − f (− n+1 ) − f ( n+1 ), vk p 2 a 2 a 2 a n 2k k 2k k a −a p x x x x a +a p ) +( ) ]φ (k k−1 , vk, k k−1 , vk, k k−1 , vk, k k−1 , vk) p ≤ ∑ [( 4 4 2a 2a 2a 2a k=1

k f (x) −

a2n+2 + an+1 p x x x x ) φ (k n , vk, k n , vk, k n , vk, k n , vk) p 4 2a 2a 2a 2a 2n+2 n+1 x x x a −a x +( ) p φ (k n , vk, k n , vk, k n , vk, k n , vk) p 4 2a 2a 2a 2a +(

for all x, v ∈ E1 . This proves the validity of inequality (6.2) for the case n + 1. Theorem 6.2 Let φ : R × R × R × R −→ [0, ∞) be a function such that ∞ x y z w φe(x, y, z, w, v) = ∑ a2ip φ (k i , vk, k i , vk, k i , vk, k i , vk) p < ∞ a a a a i=0

(6.6)

for all x, y, z, w, v ∈ E1 . Let ψ : R −→ [0, ∞) satisfies lim an ψ(k

n−→∞

x , vk) = 0 an

(6.7)


727

for all x ∈ E1 . If f : E1 −→ E2 is a mapping satisfies k f (x) + f (−x), vk ≤ ψ(kx, vk)

(6.8)

kD f (x, y, z, w), vk ≤ φ (kx, vk, ky, vk, kz, vk, kw, vk)

(6.9)

for all x, v ∈ E1 .and

for all x, y, z, w, v ∈ E1 . Then there exists a unique additive mapping A : E1 −→ E2 satisfying the equation (1.1) such that k f (x) − A(x), vk ∞

≤ { ∑ [( k=1

1 a2k − ak p x x x x a2k + ak p ) +( ) ]φ (k k−1 , vk, k k−1 , vk, k k−1 , vk, k k−1 , vk) p } p 4 4 2a 2a 2a 2a (6.10)

proof. It follows from Lemma 6.1 and (6.8) that we have x ), vk p n a 2n a + an x a2n − an x a2n − an p x x ≤ k f (x) − f( n)− f (− n ), vk p + ( ) k f ( n ) + f (− n ), vk p 2 a 2 a 2 a a n 2k k 2k k a −a p x x x x a +a p ) +( ) ]φ (k k−1 , vk, k k−1 , vk, k k−1 , vk, k k−1 , vk) p ≤ ∑ [( 4 4 2a 2a 2a 2a k=1

k f (x) − an f (

+(

a2n − an p x ) ψ(k n , vk) p 2 a

(6.11)

for all x, v ∈ E1 and n ∈ N. By virtue of (6.11), for n, m ∈ N with n > m, we obtain x x ) − an f ( n ), vk p m a a x x = amp k f ( m ) − an−m f ( n−m m ), vk p a a ·a n−m 2k k 2k a +a p a − ak p x x x x ≤ amp ∑ [( ) +( ) ]φ (k k+m−1 , vk, k k+m−1 , vk, k k+m−1 , vk, k k+m−1 , vk) p 4 4 2a 2a 2a 2a k=1

kam f (

+(

a2n−m − an p x ) ψ(k n−m , vk) p 2 a

(6.12)

for all x, v ∈ E1 and n ∈ N. From (6.6) and (6.7), the right-hand side of inequality (6.12) tends to 0 as m −→ ∞, the sequence {an f ( axn )} is a Cauchy sequence. Completeness of E2 allows us

728


to assume that there exists a mapping A so that A(x) = lim an f ( n−→∞

x ) an

for all x ∈ E1 . By (6.9), we obtain that x y z w , n , n , n ), vk p n a a a a y z w x np ≤ lim a φ (k n , vk, k n , vk, k n , vk, k n , vk) p −→ 0 n−→∞ a a a a

kDA(x, y, z, w), vk p =

lim anp kD f (

n−→∞

as n −→ ∞ for all x, y, z, w, v ∈ E1 and so the mapping A satisfies (1.1). We have the following results x x ) + an f (− n ), vk p n a a x np p ≤ lim a ψ(k n , vk) −→ 0 n−→∞ a

kA(x) + A(−x), vk p =

lim kan f (

n−→∞

as n −→ ∞ for all x, v ∈ E1 . So, A(x) = −A(−x) and A is additive. Taking m = 0, n −→ ∞ in (6.12), we get (6.10). Next, we prove the uniqueness of A. A satisfies (1.1) and putting y = z = w = 0, we have 1+a a−1 x 2A( ) − 2 A(x) + 2 A(−x) = 0 a a a

(6.13)

for all x ∈ E1 . Using oddness of A and replacing x by ax in (6.13), we have A(ax) = aA(x) 0

for all x ∈ E1 . So,we assume that A : E1 −→ E2 be another quadratic mapping satisfying (1.1) and (6.5),we calculate 0

kA(x) − A (x), vk p 0

A(an x) A (an x) p =k − , vk an an 0 1 1 ≤ np kA(an x) − f (an x), vk p + np k f (an x) − A (an x), vk p a a ∞ 2k k 2k 2 a +a p a − ak p an x an x an x an x ≤ np ∑ [( ) +( ) ]φ (k k−1 , vk, k k−1 , vk, k k−1 , vk, k k−1 , vk) p −→ 0 a k=1 4 4 2a 2a 2a 2a as n −→ ∞ for all x ∈ E1 . This completes the proof. Conflict of Interests


729

The authors declare that there is no conflict of interests. Acknowledgements The authors also would like to express their appreciation to Professor Meimei Song of Tianjin University of Technology for a careful reading and many very helpful suggestions for the improvement of the original. R EFERENCES [1] Ulam S M. Problems in Modern Mathematics. New York :Wiley, 1964. [2] Hyers D H. On the stability of the linear functional equation. Proc. Natl. Acad. Sci, 27(1941), 222-224. [3] Aoki T. On the stability of the linear transformation in Banach spaces. J. Math. Soc. Jpn. 2(1950), 64-66. [4] Rassias T M. On the stability of the linear mapping in Banach spaces. Proc. Am. Math. Soc. 72(1978), 297300. [5] Mehmet, KJr,Mehmet, Acikgoz. A Study Involving the Completion of a Quasi-2-Normed Space. International Journal of Analysis. 2013(2013), Article ID 512372. [6] Ravi K. Generalized Ulam-Hyers stability of an AQ-functional equation in quasi-beta-normed spaces. Mathematica Aeterna. 1(2011), 217-236. [7] Young-su, L, Yujin, J, Hyemin, H: Stability of a Jensen type quadratic-additive functional equation under the approximately conditions. Advances in Difference Equations. 2015(2015), Article ID 61. [8] Gavruta P. A generalization of the Hyers-Ulam-Rassias stability of approximately additive mappings. J. Math. Anal. Appl. 184(1994), 431-436. [9] Lee Y-S. Stability of a quadratic functional equation in the spaces of generalized functions. J. Inequal. Appl. 2008(2008), Article ID 210615. [10] Park C. Generalized quasi-Banach spaces and normed spaces. J. Chung. Math. Soc. 19(2006),197-206.