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Acta Mathematica Sinica, English Series

Mar., 2012, Vol. 28, No. 3, pp. 529–560

Published online: November 25, 2011

DOI: 10.1007/s10114-011-9663-0

Http://www.ActaMath.com

Acta Mathematica Sinica, English Series© Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 2012

Generalized Hyers–Ulam Stability of a General Mixed Additive-cubic

Functional Equation in Quasi-Banach Spaces

Tian Zhou XUSchool of Mathematics, Beijing Institute of Technology, Beijing 100081, P. R. China

E-mail : [email protected]

John Michael RASSIASPedagogical Department E.E., Section of Mathematics and Informatics,

National and Capodistrian University of Athens, 4, Agamemnonos Str., Aghia Paraskevi,

Athens 15342, Greece

E-mail : [email protected]

Wan Xin XUDepartment of Electrical and Computer Engineering, College of Engineering,

University of Kentucky, Lexington 40506, USA

E-mail : wxbit [email protected]

Abstract In this paper, we establish a general solution and the generalized Hyers–Ulam–Rassias

stability of the following general mixed additive-cubic functional equation

f(kx + y) + f(kx − y) = kf(x + y) + kf(x − y) + 2f(kx) − 2kf(x)

in the quasi-Banach spaces.

Keywords Stability, additive mapping, cubic mapping, quasi-Banach space, p-Banach space

MR(2000) Subject Classification 46S40, 39B52, 39B82, 26E50, 46S50

1 Introduction and Preliminaries

Defining (in some way) the class of approximate solutions of a given functional equation, one canask whether each mapping from the class can be somehow approximated by an exact solutionof the considered equation. Such a problem was formulated by Ulam in 1940 (cf. [1]). In 1941,Hyers [2] has excellently answered the question of Ulam for the Cauchy functional equation.He gave rise to the stability theory for functional equations. In 1950, Aoki [3] generalizedHyers’ theorem for approximately additive mappings. In 1978, Th. M. Rassias [4] provided ageneralized version of Hyers’ result which allows the Cauchy difference to be unbounded. Inaddition, J. M. Rassias [5–10] and Xu et al. [11–14] generalized the Hyers stability result byintroducing two weaker conditions controlled by a product of different powers of norms and

Received November 16, 2009, accepted December 7, 2010

The first author is supported by National Natural Science Foundation of China (Grant Nos. 10671013 and

11171022)

530 Xu T. Z., et al.

a mixed product-sum of powers of norms, respectively. Hyers stability problems for variousfunctional equations has been investigated by a number of authors, we refer the reader to[15–23] and references therein.

Jun and Kim [17] introduced the following cubic functional equation

f(2x + y) + f(2x − y) = 2f(x + y) + 2f(x − y) + 12f(x) (1.1)

and they established the general solution and the generalized Hyers–Ulam–Rassias stability forthe functional equation (1.1). It is easy to see that the function f(x) = cx3 is a solution ofthe functional equation (1.1). Hence, the functional equation (1.1) is called a cubic functionalequation and every solution of the cubic functional equation (1.1) is said to be a cubic mapping.They proved that a function f between real vector spaces X and Y is a solution of (1.1) if andonly if there exits a unique function C : X × X × X → Y such that f(x) = C(x, x, x) for allx ∈ X, and C is symmetric for each fixed one variable and is additive for fixed two variables.The Hyers–Ulam stability for cubic functional equation was proved by Jun and Kim [17]. Junand Kim [18] have obtained the generalized Hyers–Ulam–Rassias stability for a mixed type ofadditive-cubic functional equation.

Najati and Eskandani [19] established the general solution and investigated the Hyers–Ulam–Rassias stability of the following functional equation

f(2x + y) + f(2x − y) = 2f(x + y) + 2f(x − y) + 2f(2x) − 4f(x), (1.2)

with f(0) = 0 in the quasi-Banach spaces. It is easy to see that the mapping f(x) = ax3 + bx isa solution of the functional equation (1.2), which is called a mixed additive and cubic functionalequation and every solution of the mixed additive and cubic functional equation is said to bea mixed additive and cubic mapping. Also, Gordji and Khodaei [24] introduced the followingfunctional equation for fixed integers k with k �= 0,±1:

f(x + ky) + f(x − ky) = k2f(x + y) + k2f(x − y) + 2(1 − k2)f(x), (1.3)

with f(0) = 0. They obtained the general solution of functional equation (1.3) when f is amapping between vector spaces, and established the generalized Hyers–Ulam–Rassias stabilityof the functional equation (1.3) whenever f is a function between two quasi-Banach spaces.

In the present paper we introduce the following functional equation for fixed integers k withk �= 0,±1:

f(kx + y) + f(kx − y) = kf(x + y) + kf(x − y) + 2f(kx) − 2kf(x), (1.4)

with f(0) = 0. It is easy to see that the function f(x) = ax3 + bx is a solution of thefunctional equation (1.4). The main purpose of this paper is to investigate the general solutionof the functional equation (1.4) when f is a mapping between vector spaces, and establishthe generalized Hyers–Ulam–Rassias stability of the functional equation (1.4) whenever f isa function between two quasi-Banach spaces. We generalize the results in paper achieved byNajati and Eskandani [19] and also some other papers.

We recall some basic facts concerning quasi-Banach space and some preliminary results.

Generalized Hyers–Ulam Stability of a General Mixed Additive-cubic Functional Equation 531

Definition 1.1 ([25, 26]) Let X be a real linear space. A quasi-norm is a real-valued functionon X satisfying the following :

(1) ‖x‖ ≥ 0 for all x ∈ X and ‖x‖ = 0 if and only if x = 0.(2) ‖λx‖ = |λ|‖x‖ for all λ ∈ R and all x ∈ X.(3) There is a constant M ≥ 1 such that ‖x + y‖ ≤ M(‖x‖ + ‖y‖) for all x, y ∈ X.

A quasi-normed space is a pair (X, ‖ · ‖), where ‖ · ‖ is a quasi-norm on X. The smallestpossible M is called the modulus of concavity of ‖ · ‖. A quasi-Banach space is a completequasi-normed space.

A quasi-norm ‖ · ‖ is called a p-norm (0 < p ≤ 1) if

‖x + y‖p ≤ ‖x‖p + ‖y‖p

for all x, y ∈ X. In this case, a quasi-Banach space is called a p-Banach space.It follows from the condition (3) that

∥∥∥∥

2n∑

i=1

xi

∥∥∥∥≤ Mn

2n∑

i=1

‖xi‖,∥∥∥∥

2n+1∑

i=1

xi

∥∥∥∥≤ Mn+1

2n∑

i=1

‖xi‖

for all n ≥ 1 and all x1, x2, . . . , x2n+1 ∈ X. By the Aoki–Rolewicz theorem [26] (see also [25]),each quasi-norm is equivalent to some p-norm. Since it is much easier to work with p-normthan quasi-norms, henceforth we restrict our attention mainly to p-norms.

2 General Solution

Throughout this section, X and Y will be real vector spaces. Before proceeding the proof ofTheorem 2.4 which is the main result in this section, we shall need the following three lemmas.

Lemma 2.1 Let f : X → Y be a mapping with f(0) = 0 satisfying (1.4). Then the mappingg(x) = (k − 1)2f(kx) − k(k + 1)f((k − 1)x) − k(k − 1)2f(x) is additive.

Proof Let f satisfy (1.4). Set x = 0 in (1.4) to get f(−y) = −f(y). So the mapping f is odd.Setting y = x in (1.4) and using the oddness of f , we get

f((k + 1)x) = −f((k − 1)x) + kf(2x) + 2f(kx) − 2kf(x) (2.1)

for all x ∈ X. Replacing y by (k − 1)y in (1.4), gives

f(kx + (k − 1)y) + f(kx − (k − 1)y)

= kf(x + (k − 1)y) + kf(x − (k − 1)y) + 2f(kx) − 2kf(x) (2.2)

for all x, y ∈ X. Replacing y by (k + 1)y in (1.4), we see that

f(kx + (k + 1)y) + f(kx − (k + 1)y)

= kf(x + (k + 1)y) + kf(x − (k + 1)y) + 2f(kx) − 2kf(x) (2.3)

for all x, y ∈ X. Replacing y by x + y in (1.4) and using the oddness of f , we have

f((k + 1)x + y) + f((k − 1)x − y) = kf(2x + y) − kf(y) + 2f(kx) − 2kf(x) (2.4)

for all x, y ∈ X. Replacing y by x − y in (1.4), we get

f((k + 1)x − y) + f((k − 1)x + y) = kf(2x − y) + kf(y) + 2f(kx) − 2kf(x) (2.5)


for all x, y ∈ X. It follows from (2.4) and (2.5) that

f((k + 1)x + y) − f((k + 1)x − y) = kf(2x + y) − kf(2x − y) + f((k − 1)x + y)

− f((k − 1)x − y) − 2kf(y) (2.6)

for all x, y ∈ X. Interchange x with y in (2.6) and using the oddness of f , we have

f((k + 1)y + x) + f(x − (k + 1)y) = kf(2y + x) − kf(2y − x) + f((k − 1)y + x)

− f((k − 1)y − x) − 2kf(x) (2.7)


f(kx + (k + 1)y) + f(kx − (k + 1)y) = k2f(2y + x) − k2f(2y − x) + kf((k − 1)y + x)

− kf((k − 1)y − x) + 2f(kx) − 2k(k + 1)f(x) (2.8)

for all x, y ∈ X. We substitute x by x + y in (1.4) and then x by x − y in (1.4), we get

f(kx + (k + 1)y) + f(kx + (k − 1)y)

= kf(x + 2y) + kf(x) + 2f(k(x + y)) − 2kf(x + y) (2.9)

and

f(kx − (k − 1)y) + f(kx − (k + 1)y)

= kf(x) + kf(x − 2y) + 2f(k(x − y)) − 2kf(x − y) (2.10)

for all x, y ∈ X. Adding (2.9) to (2.10) and using (2.2) lead to

f(kx + (k + 1)y) + f(kx − (k + 1)y)

= −[f(kx + (k − 1)y) + f(kx − (k − 1)y)] + kf(x + 2y) + kf(x − 2y)

+ 2f(k(x + y)) + 2f(k(x − y)) − 2kf(x + y) − 2kf(x − y) + 2kf(x)

= −kf(x + (k − 1)y) − kf(x − (k − 1)y) + kf(x + 2y) + kf(x − 2y)

+ 2f(k(x + y)) + 2f(k(x − y)) − 2kf(x + y) − 2kf(x − y) − 2f(kx) + 4kf(x) (2.11)


k2f(2y + x) − k2f(2y − x) + kf((k − 1)y + x) − kf((k − 1)y − x)

+ 2f(kx) − 2k(k + 1)f(x)

= −kf(x + (k − 1)y) − kf(x − (k − 1)y) + kf(x + 2y) + kf(x − 2y)

+ 2f(k(x + y)) + 2f(k(x − y)) − 2kf(x + y) − 2kf(x − y) − 2f(kx) + 4kf(x) (2.12)

for all x, y ∈ X. Thus by the oddness of f , we have

2k[f(x + (k − 1)y) + f(x − (k − 1)y)] + k(k − 1)[f(x + 2y) + f(x − 2y)]

= 2f(k(x + y)) + 2f(k(x − y)) − 2kf(x + y) − 2kf(x − y)

− 4f(kx) + 2k(k + 3)f(x) (2.13)

for all x, y ∈ X.


On the other hand, replacing y by ky in (1.4), we get

f(kx + ky) + f(kx − ky) = kf(x + ky) + kf(x − ky) + 2f(kx) − 2kf(x) (2.14)

for all x, y ∈ X. Interchanging x with y in (2.14), we have

f(kx + ky) + f(ky − kx) = kf(y + kx) + kf(y − kx) + 2f(ky) − 2kf(y) (2.15)

for all x, y ∈ X. Then by the oddness of f ,

kf(kx + y) − kf(kx − y) = f(kx + ky) + f(ky − kx) − 2f(ky) + 2kf(y) (2.16)

for all x, y ∈ X. Therefore, it follows from (1.4) and (2.16) that

2kf(kx + y) = k2f(x + y) + k2f(x − y) + f(kx + ky) + f(ky − kx)

− 2f(ky) + 2kf(y) + 2kf(kx) − 2k2f(x) (2.17)

for all x, y ∈ X. Interchanging x with y in (2.17) and using the oddness of f , we have

2kf(ky + x) = k2f(x + y) − k2f(x − y) + f(kx + ky) − f(ky − kx)

− 2f(kx) + 2kf(x) + 2kf(ky) − 2k2f(y) (2.18)

for all x, y ∈ X. Adding (2.17) to (2.18), we get

k[f(kx + y) + f(ky + x)] = k2f(x + y) + f(kx + ky) + (k − 1)f(kx)

+ (k − 1)f(ky) + k(k − 1)f(x) + k(k − 1)f(y) (2.19)

for all x, y ∈ X. Letting x by x − y in (2.19), we have

k[f(kx − (k − 1)y) + f(x + (k − 1)y)]

= k2f(x) + f(kx) + (k − 1)f(kx − ky)

+ (k − 1)f(ky) + k(k − 1)f(x − y) + k(k − 1)f(y) (2.20)

for all x, y ∈ X. Setting y by −y in (2.20) and using the oddness of f , we obtain

k[f(kx + (k − 1)y) + f(x − (k − 1)y)]

= k2f(x) + f(kx) + (k − 1)f(kx + ky)

− (k − 1)f(ky) + k(k − 1)f(x + y) − k(k − 1)f(y) (2.21)


k[f(kx + (k − 1)y) + f(kx − (k − 1)y)] + k[f(x + (k − 1)y) + f(x − (k − 1)y)]

= 2k2f(x) + 2f(kx) + (k − 1)f(kx + ky) + (k − 1)f(kx − ky)

+ k(k − 1)f(x + y) + k(k − 1)f(x − y) (2.22)


k(k + 1)[f(x + (k − 1)y) + f(x − (k − 1)y)]

= 4k2f(x) − 2(k − 1)f(kx) + (k − 1)f(kx + ky) + (k − 1)f(kx − ky)

+ k(k − 1)f(x + y) + k(k − 1)f(x − y) (2.23)


for all x, y ∈ X. Setting x by x − y in (2.18) and using the oddness of f , we get

2kf(x + (k − 1)y) = k2f(x) + k2f(2y − x) + f(kx) − f(2ky − kx)

− 2f(kx − ky) + 2kf(x − y) + 2kf(ky) − 2k2f(y) (2.24)

for all x, y ∈ X. Setting y by −y in (2.24) and using the oddness of f , we obtain

2kf(x − (k − 1)y) = k2f(x) − k2f(2y + x) + f(kx) + f(2ky + kx)

− 2f(kx + ky) + 2kf(x + y) − 2kf(ky) + 2k2f(y) (2.25)

for all x, y ∈ X. Subtracting (2.25) from (2.24), we get

2k[f(x + (k − 1)y) − f(x − (k − 1)y)]

= −f(2ky − kx) + k2f(2y − x) − 2f(kx − ky) + 4kf(ky) + 2kf(x − y)

− 4k2f(y) − f(kx + 2ky) + k2f(x + 2y) + 2f(kx + ky) − 2kf(x + y) (2.26)

for all x, y ∈ X. Thus by (2.23) and (2.26) (multiply both sides of (2.23) by 2 and (2.26) by(k + 1)), we have

4k(k + 1)f(x + (k − 1)y)

= −(k + 1)f(2ky − kx) + k2(k + 1)f(2y − x) − 4f(kx − ky) + 4k(k + 1)f(ky)

+ 4k2f(x − y) − 4k2(k + 1)f(y) − (k + 1)f(kx + 2ky) + k2(k + 1)f(x + 2y)

+ 4kf(kx + ky) − 4kf(x + y) + 8k2f(x) − 4(k − 1)f(kx) (2.27)

for all x, y ∈ X. Multiplying each side of (2.24) by 2(k + 1) and compare with (2.27), then weconclude that

k2(k + 1)[f(x + 2y) + f(x − 2y)] − (k + 1)[f(kx + 2ky) + f(kx − 2ky)]

= 4kf(x − y) + 4kf(x + y) − 4kf(kx − ky) − 4kf(kx + ky)

+ 2(3k − 1)f(kx) + 2k2(k − 3)f(x) (2.28)

for all x, y ∈ X. Multiplying each side of (2.28) by 2k, we have

2k3(k + 1)[f(x + 2y) + f(x − 2y)] − 2k(k + 1)[f(kx + 2ky) + f(kx − 2ky)]

= 8k2f(x − y) + 8k2f(x + y) − 8k2f(kx − ky) − 8k2f(kx + ky)

+ 4k(3k − 1)f(kx) + 4k2(k − 3)f(x) (2.29)

for all x, y ∈ X.Now, by letting y by 2y − x in (2.17), we get

2kf(2y + (k − 1)x) = k2f(2y) + f(2ky) − f(2kx − 2ky) + k2f(2x − 2y)

− 2f(2ky − kx) + 2kf(kx) + 2kf(2y − x) − 2k2f(x) (2.30)

for all x, y ∈ X. Replacing x by −x in (2.30) and using the oddness of f , we get

2kf(2y − (k − 1)x) = k2f(2y) + f(2ky) + f(2kx + 2ky) − k2f(2x + 2y)

− 2f(2ky + kx) − 2kf(kx) + 2kf(2y + x) + 2k2f(x) (2.31)


for all x, y ∈ X.

On the other hand, replacing x by x + y and y by x − y in (1.4), respectively, we get

f((k + 1)x + (k − 1)y) + f((k + 1)y + (k − 1)x)

= kf(2x) + kf(2y) + 2f(kx + ky) − 2kf(x + y) (2.32)

for all x, y ∈ X. Replacing y by x + (k − 1)y in (1.4) and using the oddness of f , we have

f((k + 1)x + (k − 1)y) + f((k − 1)x − (k − 1)y)

= kf(2x + (k − 1)y) − kf((k − 1)y) + 2f(kx) − 2kf(x) (2.33)

for all x, y ∈ X. Interchanging x with y in (2.33), we obtain

f((k + 1)y + (k − 1)x) + f((k − 1)y − (k − 1)x)

= kf(2y + (k − 1)x) − kf((k − 1)x) + 2f(ky) − 2kf(y) (2.34)

for all x, y ∈ X. Adding (2.33) to (2.34) and using the oddness of f , we get

f((k + 1)x + (k − 1)y) + f((k + 1)y + (k − 1)x)

= kf(2x + (k − 1)y) + kf(2y + (k − 1)x) − kf((k − 1)x) − kf((k − 1)y)

+ 2f(kx) + 2f(ky) − 2kf(x) − 2kf(y) (2.35)

for all x, y ∈ X. Hence it follows from (2.32) and (2.35) that

k[f(2x + (k − 1)y) + f(2y + (k − 1)x)]

= kf(2x) + kf(2y) + 2f(kx + ky) − 2kf(x + y) + kf((k − 1)x)

+ kf((k − 1)y) − 2f(kx) − 2f(ky) + 2kf(x) + 2kf(y) (2.36)

for all x, y ∈ X. Replacing x by x + y and y by x − y in (2.14), respectively, we get

f(2kx) + f(2ky) = kf((k + 1)x − (k − 1)y) + kf((k + 1)y − (k − 1)x)

+ 2f(kx + ky) − 2kf(x + y) (2.37)

for all x, y ∈ X. Replacing y by −y in (2.38) and using the oddness of f , we have

f(2kx) − f(2ky) = k[f((k + 1)x + (k − 1)y) − f((k + 1)y + (k − 1)x)]

+ 2f(kx − ky) − 2kf(x − y) (2.38)

for all x, y ∈ X. Subtracting (2.34) from (2.33) and using the oddness of f , we get

f((k + 1)x + (k − 1)y) − f((k + 1)y + (k − 1)x)

= k[f(2x + (k − 1)y) − f(2y + (k − 1)x)] − 2f((k − 1)x − (k − 1)y)

− kf((k − 1)y) + 2f(kx) − 2kf(x) + kf((k − 1)x) − 2f(ky) + 2kf(y) (2.39)

for all x, y ∈ X. Thus it follows from (2.38) and (2.39) that

k2[f(2x + (k − 1)y) − f(2y + (k − 1)x)]

= f(2kx) − f(2ky) + k2f((k − 1)y) − k2f((k − 1)x)


+ 2k2f(x) − 2k2f(y) − 2kf(kx) + 2kf(ky)

+ 2kf((k − 1)x − (k − 1)y) + 2f(ky − kx) + 2kf(x − y) (2.40)

for all x, y ∈ X. Therefore, by (2.36) and (2.40), we conclude that

2k2f(2x + (k − 1)y)

= 2kf(kx + ky) + 2f(ky − kx) + 2kf((k − 1)x) − (k − 1)y)

+ f(2kx) − f(2ky) − 4kf(kx) + 4k2f(x) + 2k2f((k − 1)y)

+ k2f(2x) + k2f(2y) − 2k2f(x + y) + 2kf(x − y) (2.41)

for all x, y ∈ X. Interchanging x with y in (2.41), then replacing x with −x and using theoddness of f , we get

2k2f(2y − (k − 1)x)

= −2kf(kx − ky) − 2f(kx + ky) + 2kf((k − 1)y) + (k − 1)x)

+ f(2ky) + f(2kx) − 4kf(ky) + 4k2f(y) − 2k2f((k − 1)x)

+ k2f(2y) − k2f(2x) + 2k2f(x − y) + 2kf(x + y) (2.42)

for all x, y ∈ X. Comparing (2.31) with (2.42), we get

2k2f(x + 2y) − 2kf(kx + 2ky)

= −2kf(kx − ky) − 2f(kx + ky) + 2kf((k − 1)x) + (k − 1)y)

+ (1 − k)f(2ky) + f(2kx) − 4kf(ky) + 4k2f(y) − 2k2f((k − 1)x)

+ (k2 − k3)f(2y) − k2f(2x) + 2k2f(x − y) + 2kf(x + y)

− kf(2kx + 2ky) + k3f(2x + 2y) + 2k2f(kx) − 2k3f(x) (2.43)

for all x, y ∈ X. Replacing y with −y in (2.43) and using the oddness of f , we get

2k2f(x − 2y) − 2kf(k(x − 2y))

= −2kf(k(x + y)) − 2f(k(x − y)) + 2kf((k − 1)(x − y))

− (1 − k)f(2ky) + f(2kx) + 4kf(ky) − 4k2f(y) − 2k2f((k − 1)x)

− (k2 − k3)f(2y) − k2f(2x) + 2k2f(x + y) + 2kf(x + y)

− kf(2k(x − y)) + k3f(2(x − y)) + 2k2f(kx) − 2k3f(x) (2.44)


2k2[f(x + 2y) + f(x − 2y)] − 2k[f(k(x + 2y)) + f(k(x − 2y))]

= −2(k + 1)f(k(x + y)) − 2(k + 1)f(k(x − y)) + 2kf((k − 1)(x + y))

+ 2kf((k − 1)(x − y)) + 2f(2kx) − 4k2f((k − 1)x) − 2k2f(2x)

+ 2k(k + 1)f(x + y) + 2k(k + 1)f(x − y) − kf(2k(x + y)) − kf(2k(x − y))

+ k3f(2(x + y)) + k3f(2(x − y)) + 4k2f(kx) − 4k3f(x) (2.45)

for all x, y ∈ X. Multiplying each side of (2.45) by (k + 1), we have

2k2(k + 1)[f(x + 2y) + f(x − 2y)] − 2k(k + 1)[f(k(x + 2y)) + f(k(x − 2y))]


= −2(k + 1)2f(k(x + y)) − 2(k + 1)2f(k(x − y)) + 2k(k + 1)f((k − 1)(x + y))

+ 2k(k + 1)f((k − 1)(x − y)) + 2(k + 1)f(2kx) − 4k2(k + 1)f((k − 1)x)

− 2k2(k + 1)f(2x) + 2k(k + 1)2f(x + y) + 2k(k + 1)2f(x − y)

− k(k + 1)f(2k(x + y)) − k(k + 1)f(2k(x − y)) + k3(k + 1)f(2(x + y))

+ k3(k + 1)f(2(x − y)) + 4k2(k + 1)f(kx) − 4k3(k + 1)f(x) (2.46)

for all x, y ∈ X. Subtracting (2.46) from (2.29) and using the oddness of f , we get

2k2(k2 − 1)[f(x + 2y) + f(x − 2y)]

= −(6k2 − 4k − 2)[f(k(x + y)) + f(k(x − y))] − 2k(k + 1)f((k − 1)(x + y))

− 2k(k + 1)f((k − 1)(x − y)) − 2(k + 1)f(2kx) + 4k2(k + 1)f((k − 1)x)

+ 2k2(k + 1)f(2x) − 2k(k − 1)2f(x + y) − 2k(k − 1)2f(x − y)

+ k(k + 1)f(2k(x + y)) + k(k + 1)f(2k(x − y)) − k3(k + 1)f(2(x + y))

− k3(k + 1)f(2(x − y)) − 4k(k − 1)2f(kx) + 8k3(k − 1)f(x) (2.47)

for all x, y ∈ X. Hence, it follows from (2.13), (2.23) and (2.47) that

− 4k(k − 3)f(kx) + 4k2f((k − 1)x) − 2f(2kx) + 2k2f(2x) + 4k2(k − 3)f(x)

= 2(3k − 1)f(k(x + y)) + 2(3k − 1)f(k(x − y)) + 2kf((k − 1)(x + y))

+ 2kf((k − 1)(x − y)) − kf(2k(x + y)) − kf(2k(x − y)) + k3f(2(x + y))

+ k3f(2(x − y)) − 2k(3k − 1)f(x + y) − 2k(3k − 1)f(x − y) (2.48)

for all x, y ∈ X. Interchanging x with y in (2.48), we obtain

− 4k(k − 3)f(ky) + 4k2f((k − 1)y) − 2f(2ky) + 2k2f(2y) + 4k2(k − 3)f(y)

= 2(3k − 1)f(k(x + y)) + 2(3k − 1)f(k(y − x)) + 2kf((k − 1)(x + y))

+ 2kf((k − 1)(y − x)) − kf(2k(x + y)) − kf(2k(y − x)) + k3f(2(x + y))

+ k3f(2(y − x)) − 2k(3k − 1)f(x + y) − 2k(3k − 1)f(y − x) (2.49)

for all x, y ∈ X. Adding (2.48) to (2.49) and using the oddness of f , we get

2(3k − 1)f(k(x + y)) + 2kf((k − 1)(x + y)) + k3f(2(x + y))

− 2k(3k − 1)f(x + y) − kf(2k(x + y))

= −2k(k − 3)f(kx) + 2k2f((k − 1)x) − f(2kx) + k2f(2x) + 2k2(k − 3)f(x)

− 2k(k − 3)f(ky) + 2k2f((k − 1)y) − f(2ky) + k2f(2y) + 2k2(k − 3)f(y) (2.50)

for all x, y ∈ X. By replacing y with kx in (1.4) and using the oddness of f , we get

f(2kx) = kf((k + 1)x) − kf((k − 1)x) + 2f(kx) − 2kf(x) (2.51)

for all x ∈ X. Then it follows from (2.1) and (2.51) that

f(2kx) = 2(k + 1)f(kx) − 2kf((k − 1)x) + k2f(2x) − 2k(k + 1)f(x) (2.52)


for all x ∈ X. Hence using (2.52), we have

2(3k − 1)f(k(x + y)) + 2kf((k − 1)(x + y)) + k3f(2(x + y))

− 2k(3k − 1)f(x + y) − kf(2k(x + y))

= 2(3k − 1)f(k(x + y)) + 2kf((k − 1)(x + y)) + k3f(2(x + y))

− 2k(3k − 1)f(x + y) − k[2(k + 1)f(k(x + y)) − 2kf((k − 1)(x + y))

+ k2f(2(x + y)) − 2k(k + 1)f(x + y)]

= −2(k − 1)2f(k(x + y)) + 2k(k + 1)f((k − 1)(x + y)) + 2k(k − 1)2f(x + y) (2.53)

and

− 2k(k − 3)f(kx) + 2k2f((k − 1)x) + k2f(2x) + 2k(k − 3)[kf(x) − f(ky)]

+ 2k2f((k − 1)y) − f(2ky) + k2f(2y) + 2k2(k − 3)f(y) − f(2kx)

= −2k(k − 3)f(kx) + 2k2f((k − 1)x) + k2f(2x) + 2k2(k − 3)f(x)

− 2k(k − 3)f(ky) + 2k2f((k − 1)y) − f(2ky) + k2f(2y) + 2k2(k − 3)f(y)

− [2(k + 1)f(kx) − 2kf((k − 1)x) + k2f(2x) − 2k(k + 1)f(x)]

= −2(k − 1)2[f(kx) + f(ky)] + 2k(k + 1)[f((k − 1)x)f((k − 1)y)]

+ 2k(k − 1)2[f(x) + f(y)] (2.54)

for all x, y ∈ X. Then, according to (2.50), (2.53) and (2.54), we get

(k − 1)2f(k(x + y)) − k(k + 1)f((k − 1)(x + y)) − k(k − 1)2f(x + y)

= (k − 1)2[f(kx) + f(ky)] − k(k + 1)[f((k − 1)x) + f((k − 1)y)]

− k(k − 1)2[f(x) + f(y)] (2.55)

for all x, y ∈ X. This means that

g(x + y) = g(x) + g(y) (2.56)

for all x, y ∈ X. Therefore the mapping g : X → Y is additive. �

Lemma 2.2 Let f : X → Y be a mapping with f(0) = 0 satisfying (1.4). Then the mappingF (x) = f(kx) − kf(x) is cubic-additive.

Proof It is enough to prove

F (x + 2y) + F (x − 2y) = 4F (x + y) + 4F (x − y) − 6F (x) (2.57)

for all x, y ∈ X (see [24, Lemma 2.2]). Let g : X → Y be the additive mapping defined inLemma 2.1. Since g(x + y) = g(x) + g(y) and g(x − y) = g(x) + g(−y) for all x, y ∈ X, thenusing the oddness of f , we have

(k − 1)2f(k(x + y)) − k(k + 1)f((k − 1)(x + y)) − k(k − 1)2f(x + y)

= (k − 1)2f(kx) − k(k + 1)f((k − 1)x) − k(k − 1)2f(x)

+ (k − 1)2f(ky) − k(k + 1)f((k − 1)y) − k(k − 1)2f(y) (2.58)


and

(k − 1)2f(k(x − y)) − k(k + 1)f((k − 1)(x − y)) − k(k − 1)2f(x − y)

= (k − 1)2f(kx) − k(k + 1)f((k − 1)x) − k(k − 1)2f(x)

− (k − 1)2f(ky) + k(k + 1)f((k − 1)y) + k(k − 1)2f(y) (2.59)


(k − 1)2[f(k(x + y)) + f(k(x − y))] − k(k − 1)2[f(x + y) + f(x − y)]

− k(k + 1)[f((k − 1)(x + y)) + f((k − 1)(x − y))]

= 2(k − 1)2f(kx) − 2k(k + 1)f((k − 1)x) − 2k(k − 1)2f(x) (2.60)

for all x, y ∈ X. Hence

k(k + 1)[f((k − 1)(x + y)) + f((k − 1)(x − y))]

= (k − 1)2[f(k(x + y)) + f(k(x − y))] − k(k − 1)2[f(x + y) + f(x − y)]

− 2(k − 1)2f(kx) + 2k(k + 1)f((k − 1)x) + 2k(k − 1)2f(x) (2.61)

for all x, y ∈ X.On the other hand, by (2.28) and (2.47), we obtain

2(k2 − 1)[f(k(x + 2y)) + f(k(x − 2y))]

= −2k(k − 1)(k + 3)[f(x + y) + f(x − y)] + 2(k − 1)2[f(k(x + y)) + f(k(x − y))]

+ 4k2(k − 1)(k + 3)f(x) − 4(k3 + k2 − 3k + 1)f(kx) + 4k2(k + 1)f((k − 1)x)

− 2k(k + 1)[f((k − 1)(x + y)) + f((k − 1)(x − y))] − 2(k + 1)f(2kx)

+ 2k2(k + 1)f(2x) + k(k + 1)[f(2k(x + y)) + f(2k(x − y))]

− k3(k + 1)[f(2(x + y)) + f(2(x − y))] (2.62)

for all x, y ∈ X. From (2.47), (2.61) and (2.62), we get

2k(k2 − 1)[F (x + 2y) + F (x − 2y)]

= 2k(k2 − 1)[f(k(x + 2y)) + f(k(x − 2y))] − 2k2(k2 − 1)[f(x + 2y) + f(x − 2y)]

= 2(k − 1)(k + 1)2[f(k(x + y)) + f(k(x − y))] − 2k(k − 1)(k + 1)2[f(x + y) + f(x − y)]

− 4k2(k2 − 1)[f(kx) − kf(x)] − 2(k2 − 1)f(2kx) + 2k2(k2 − 1)f(2x)

+ 4k2(k2 − 1)f((k − 1)x) + k(k2 − 1)[f(2k(x + y)) + f(2k(x − y))]

− k3(k2 − 1)[f(2(x + y)) + f(2(x − y))]

− 2k(k2 − 1)[f((k − 1)(x + y)) + f((k − 1)(x − y))]

= 2(k − 1)(k + 1)2[f(k(x + y)) + f(k(x − y))] − 2k(k − 1)(k + 1)2[f(x + y) − f(x − y)]

− 4k2(k2 − 1)[f(kx) − kf(x)] − 2(k2 − 1)f(2kx) + 2k2(k2 − 1)f(2x)

+ 4k2(k2 − 1)f((k − 1)x) + k(k2 − 1)[f(2k(x + y)) + f(2k(x − y))]

− k3(k2 − 1)[f(2(x + y)) + f(2(x − y))]

− 2(k − 1){(k − 1)2[f(k(x + y)) + f(k(x − y))] − k(k − 1)2[f(x + y) + f(x − y)]


− 2(k − 1)2f(kx) + 2k(k + 1)f((k − 1)x) + 2k(k − 1)2f(x)}= 8k(k − 1)[f(k(x + y)) + f(k(x − y))] − 8k2(k − 1)[f(x + y) + f(x − y)]

− 4(k − 1)(k3 + 2k − 1)[f(kx) − kf(x)] + 4k(k + 1)(k − 1)2f((k − 1)x)

− k3(k2 − 1)[f(2(x + y)) + f(2(x − y))] − 2(k2 − 1)f(2kx) + 2k2(k2 − 1)f(2x)

+ k(k2 − 1)[f(2k(x + y)) + f(2k(x − y))] (2.63)


2k(k + 1)[F (x + 2y) + F (x − 2y)]

= 8k[f(k(x + y)) + f(k(x − y))] − 8k2[f(x + y) + f(x − y)] + 4k(k2 − 1)f((k − 1)x)

− 4(k3 + 2k − 1)[f(kx) − kf(x)] − k3(k + 1)[f(2(x + y)) + f(2(x − y))]

− 2(k + 1)f(2kx) + 2k2(k + 1)f(2x) + k(k + 1){2(k + 1)f(k(x + y))

− 2kf((k − 1)(x + y)) + k2f(2(x + y)) − 2k(k + 1)f(x + y)

+ 2(k + 1)f(k(x − y)) − 2kf((k − 1)(x − y)) + k2f(2(x − y)) − 2k(k + 1)f(x − y)}= 8k(k + 1)[f(k(x + y)) + f(k(x − y))] − 8k2(k + 1)[f(x + y) + f(x − y)]

− 4(2k − 1)(k + 1)[f(kx) − kf(x)] − 4k(k + 1)f((k − 1)x) + 2k2(k + 1)f(2x)

− 2(k + 1){2(k + 1)f(kx) − 2kf((k − 1)x) + k2f(2x) − 2k(k + 1)f(x)}= 8k(k + 1)[f(k(x + y)) + f(k(x − y))] − 8k2(k + 1)[f(x + y) + f(x − y)]

− 12k(k + 1)[f(kx) − kf(x)]

= 8k(k + 1)[f(k(x + y)) + f(k(x − y)) − kf(x + y) − kf(x − y)]

− 12k(k + 1)[f(kx) − kf(x)]

= 8k(k + 1)[F (x + y) + F (x − y)] − 12k(k + 1)F (x) (2.64)

for all x, y ∈ X. Thus

F (x + 2y) + F (x − 2y) = 4F (x + y) + 4F (x − y) − 6F (x) (2.65)

for all x, y ∈ X. Therefore the mapping F : X → Y, F (x) = f(kx) − kf(x) is a cubic-additivemapping [24, Lemma 2.2]. This completes the proof of the lemma. �

Lemma 2.3 Let f : X → Y be a mapping with f(0) = 0 satisfying (1.4). Then the mappingG(x) := f(2x) − 8f(x) is additive and H(x) := f(2x) − 2f(x) is cubic.

Proof Replacing y with (k − 2)x in (1.4), we have

f(2(k − 1)x) = 2f(kx) + kf((k − 1)x) − kf((k − 3)x) − f(2x) − 2kf(x) (2.66)

for all x ∈ X. Hence it follows from (2.1), (2.52) and (2.66) that

f(2(k + 1)x) = −f(2(k − 1)x) + kf(4x) + 2f(2kx) − 2kf(2x)

− [2f(kx) + kf((k − 1)x) − kf((k − 3)x) − f(2x) − 2kf(x)] + kf(4x)

+ 2[2(k + 1)f(kx) − 2kf((k − 1)x) + k2f(2x) − 2k(k + 1)f(x)] − 2kf(2x)

= (4k + 2)f(kx) − 5kf((k − 1)x) + kf((k − 3)x) + kf(4x)


+ (2k2 − 2k + 1)f(2x) − 2k(2k + 1)f(x) (2.67)

for all x ∈ X. Replacing y with 3x in (1.4), we get

f((k + 3)x) = 2f(kx) − f((k − 3)x) + kf(4x) − kf(2x) − 2kf(x) (2.68)

for all x ∈ X. Replacing y with (k + 2)x in (1.4) and using (2.17) and (2.68), we get

f(2(k + 1)x) = 2f(kx) + kf((k − 1)x) − kf((k − 3)x)

+ k2f(4x) + (1 − 2k2)f(2x) − 2kf(x) (2.69)

for all x ∈ X. Thus by (2.67) and (2.69), we conclude that

4f(kx) − 6f((k − 1)x) + 2f((k − 3)x) + (1 − k)f(4x) + 2(2k − 1)f(2x) − 4kf(x) = 0 (2.70)

for all x, y ∈ X.Let g : X → Y be the additive mapping defined in Lemma 2.1. Since g(2x) = 2g(x) for all

x ∈ X, then using (2.52) and (2.66), we have

(2k − 6)f(kx) − (3k − 5)f((k − 1)x) + (k + 1)f((k − 3)x) + (k2 − 3k + 4)f(2x)

− 2k(k − 3)f(x) = 0 (2.71)

for all x ∈ X. Hence by (2.70) and (2.71), we get

f((k − 1)x) = f(kx) − 116

(k2 − 1)f(4x) +18(k − 1)(k + 5)f(2x) − kf(x) (2.72)

for all x ∈ X. So by (2.52) and (2.72), we have

f(2kx) = 2f(kx) + k3f(2x) − 2k3f(x) (2.73)

for all x ∈ X. By Lemma 2.2 and [19, Lemma 2.1, Lemma 2.2], g1(x) = F (2x) − 8F (x) isadditive and h1(x) = F (2x) − 2F (x) is cubic. Hence by (2.73), we get

g1(x) = f(2kx) − kf(2x) − 8(f(kx) − kf(x))

= 2f(kx) + k3f(2x) − 2k3f(x) − kf(2x) − 8(f(kx) − kf(x))

= −6f(kx) + k(k2 − 1)f(2x) − 2k(k2 − 4)f(x) (2.74)

and

h1(x) = f(2kx) − kf(2x) − 2(f(kx) − kf(x))

= 2f(kx) + k3f(2x) − 2k3f(x) − kf(2x) − 2(f(kx) − kf(x))

= (k3 − k)(f(2x) − 2f(x)) (2.75)

for all x ∈ X. This shows that the mapping H(x) = f(2x) − 2f(x) is cubic. Since H(2x) =8H(x) (see [17]) for all x ∈ X, then

f(4x) = 10f(x) − 16f(x) (2.76)

for all x ∈ X. Also by (2.72) and (2.75), we get

f((k − 1)x) = f(kx) − 12k(k − 1)f(2x) + (k2 − k − 1)f(x) (2.77)


for all x ∈ X. Then by Lemma 2.1 and (2.77), we have

g(x) = −(3k − 1)f(kx) +12k2(k2 − 1)f(2x) − k2(k2 + k − 4)f(x) (2.78)

for all x ∈ X. Therefore,

6g(x) − (3k − 1)g1(x) = k(k2 − 1)(f(2x) − 8f(x)) (2.79)

for all x ∈ X. This shows that the mapping G(x) = f(2x) − 8f(x) is additive. �

Theorem 2.4 A mapping f : X → Y with f(0) = 0 satisfies (1.4) for all x, y ∈ X if and onlyif there exist mappings C : X ×X ×X → Y and A : X → Y such that f(x) = C(x, x, x)+A(x)for all x ∈ X, where the mapping C is symmetric for each fixed one variable and is additive forfixed two variables and the mapping A is additive.

Proof Let f : X → Y with f(0) = 0 satisfying (1.4) for all x, y ∈ X, and let G, H : X → Y

be the mappings defined in Lemma 2.3. Thus by [17, Theorem 2.1], there exists a mappingC : X × X × X → Y such that H(x) = 6C(x, x, x) for all x ∈ X, where the mapping C

is symmetric for each fixed one variable and is additive for fixed two variables. Now, letA(x) = −G(x)/6. Then we get f(x) = C(x, x, x) + A(x) for all x ∈ X. The proof of theconverse is trivial. �

3 Stability of Functional Equation

Throughout this section, assume that X is a quasi-normed space with quasi-norm ‖ · ‖X andthat Y is a p-Banach space with p-norm ‖ · ‖Y . Let M be the modulus of concavity of ‖ · ‖Y .

In this section, we prove the stability of Eq. (1.4). We establish the following new stability forgeneralized mixed additive-cubic functional equation in quasi-Banach spaces. For convenience,we use the following abbreviation for a given mapping f : X → Y :

Df(x, y) := f(kx + y) + f(kx − y) − kf(x + y) − kf(x − y) − 2f(kx) + 2kf(x)

for all x, y ∈ X.

Lemma 3.1 ([19]) Let 0 < p ≤ 1 and let x1, x2, . . . , xn be non-negative real numbers. Then( n∑

i=1

xi

)p

≤n∑

i=1

xpi .

Theorem 3.2 Let f : X → Y be a mapping with f(0) = 0 for which there is a functionϕa : X × X → [0,∞) such that

∞∑

i=0

12ip

ϕpa(2ix, 2iy) < ∞ (3.1)

and

‖Df(x, y)‖Y ≤ ϕa(x, y) (3.2)

for all x, y ∈ X. Then the limit

A(x) = limn→∞

12n

[f(2n+1x) − 8f(2nx)]


exists for all x ∈ X and A : X → Y is a unique additive mapping such that

‖f(2x) − 8f(x) − A(x)‖Y ≤ 12[Φa(x)]

1p (3.3)

for all x ∈ X, where

Φa(x) :=M3p

(k3 − k)p

∞∑

i=0

12ip

{

(M4 + M6k)p[ϕpa(2ix, (2k + 1)2ix) + ϕp

a(2ix, (2k − 1)2ix)]

+ M4pϕpa(3 · 2ix, 2ix) + (M4 + 8Mk2)pϕp

a(2ix, 2ix) + M6pϕpa(2ix, 3k2ix)

+ M4pϕpa(2ix, k2ix) + M4pk2pϕp

a(2i+1x, 2i+1x) + Mpϕpa(2i+1x, k2i+1x)

+ 2pM3pϕpa(2ix, (k + 1)2ix) + 2pM3pϕp

a(2ix, (k − 1)2ix) + 2pMpϕpa(2i+1x, 2ix)

+ 2pϕpa(2i+1x, k2ix) + 8pMpϕp

a(2i−1x, k2i−1x) + 8pM3pkpϕpa(2i−1x, (2k − 1)2i−1x)

+ 8pM3pkpϕpa(2i−1x, (2k + 1)2i−1x) + 8pM3pϕp

a(2i−1x, 3k2i−1x)

+(

M4(k + 1)k − 1

)p

ϕpa(0, (k + 1)2ix) +

(M4 + 8Mk2

k − 1

)p

ϕpa(0, (k − 1)2ix)

+(

2M3

k − 1

)p

ϕpa(0, 2ix) +

(M6k

k − 1

)p

ϕpa(0, (3k − 1)2ix) +

(M4k2

k − 1

)p

ϕpa(0, (k − 1)2i+1x)

+(

M4(k2 + k − 1)k − 1

)p

ϕpa(0, k2i+1x) +

(8M3k

k − 1

)p

ϕpa(0, (3k − 1)2i−1x)

+(

8Mk

k − 1

)p

ϕpa(0, (k + 1)2i−1x) +

(2M3k + 8M(k2 − 1)

k − 1

)p

ϕpa(0, k2ix)

}

.

Proof Letting x = 0 in (3.2), we get

‖f(y) + f(−y)‖Y ≤ 1k − 1

ϕa(0, y) (3.4)

for all y ∈ X. Putting y = x in (3.2), we have

‖f((k + 1)x) + f((k − 1)x) − kf(2x) − 2f(kx) + 2kf(x)‖Y ≤ ϕa(x, x) (3.5)

for all x ∈ X. Hence

‖f(2(k + 1)x) + f(2(k − 1)x) − kf(4x) − 2f(2kx) + 2kf(2x)‖Y ≤ ϕa(2x, 2x) (3.6)

for all x ∈ X. Letting y = kx in (3.2), we get

‖f(2kx) − kf((k + 1)x) − kf(−(k − 1)x) − 2f(kx) + 2kf(x)‖Y ≤ ϕa(x, kx) (3.7)

for all x ∈ X. Letting y = (k + 1)x in (3.2), we have

‖f((2k + 1)x) + f(−x) − kf((k + 2)x) − kf(−kx) − 2f(kx) + 2kf(x)‖Y

≤ ϕa(x, (k + 1)x) (3.8)

for all x ∈ X. Letting y = (k − 1)x in (3.2), we have

‖f((2k − 1)x) − (k + 2)f(kx) − kf(−(k − 2)x) + (2k + 1)f(x)‖Y ≤ ϕa(x, (k − 1)x) (3.9)

for all x ∈ X. Replacing x and y by 2x and x in (3.2), respectively, we get

‖f((2k + 1)x) + f((2k − 1)x) − 2f(2kx) − kf(3x) + 2kf(2x) − kf(x)‖Y ≤ ϕa(2x, x) (3.10)


for all x ∈ X. Replacing x and y by 3x and x in (3.2), respectively, we get

‖f((3k + 1)x) + f((3k − 1)x) − 2f(3kx) − kf(4x) − kf(2x) + 2kf(3x)‖Y ≤ ϕa(3x, x) (3.11)

for all x ∈ X. Replacing x and y by 2x and kx in (3.2), respectively, we have

‖f(3kx) + f(kx) − kf((k + 2)x) − kf(−(k − 2)x) − 2f(2kx) + 2kf(2x)‖Y

≤ ϕa(2x, kx) (3.12)

for all x ∈ X. Setting y = (2k + 1)x in (3.2), we have

‖f((3k + 1)x) + f(−(k + 1)x) − kf(2(k + 1)x) − kf(−2kx) − 2f(kx) + 2kf(x)‖Y

≤ ϕa(x, (2k + 1)x) (3.13)

for all x ∈ X. Letting y = (2k − 1)x in (3.2), we have

‖f((3k − 1)x) + f(−(k − 1)x) − kf(−2(k − 1)x) − kf(2kx) − 2f(kx) + 2kf(x)‖Y

≤ ϕa(x, (2k − 1)x) (3.14)

for all x ∈ X. Letting y = 3kx in (3.2), we have

‖f(4kx) + f(−2kx) − kf((3k + 1)x) − kf(−(3k − 1)x) − 2f(kx) + 2kf(x)‖Y

≤ ϕa(x, 3kx) (3.15)

for all x ∈ X. By (3.4), (3.5), (3.11), (3.13) and (3.14), we get

‖kf(2(k + 1)x) + kf(−2(k − 1)x) + 6f(kx) − 2f(3kx) − kf(4x) + 2kf(3x) − 6kf(x)‖Y

≤ M4

[

ϕa(x, (2k + 1)x) + ϕa(x, (2k − 1)x) + ϕa(3x, x) + ϕa(x, x)

+1

k − 1ϕa(0, (k + 1)x) +

1k − 1

ϕa(0, (k − 1)x) +k

k − 1ϕa(0, 2kx)

]

(3.16)

for all x ∈ X. By (3.4), (3.8) and (3.9), we have

‖f((2k + 1)x) + f((2k − 1)x) − kf((k + 2)x) − kf(−(k − 2)x) − 4f(kx) + 4kf(x)‖Y

≤ M2

[

ϕa(x, (k + 1)x) + ϕa(x, (k − 1)x) +1

k − 1ϕa(0, x) +

k

k − 1ϕa(0, kx)

]

(3.17)

for all x ∈ X. It follows from (3.10) and (3.17) that

‖kf((k + 2)x) + kf(−(k − 2)x) − 2f(2kx) + 4f(kx) − kf(3x) + 2kf(2x) − 5kf(x)‖Y

≤ M

[

M2ϕa(x, (k + 1)x) + M2ϕa(x, (k − 1)x) + ϕa(2x, x)

+M2

k − 1ϕa(0, x) +

M2k

k − 1ϕa(0, kx)

]

(3.18)

for all x ∈ X. By (3.12) and (3.18), we have

‖f(3kx) − 4f(2kx) + 5f(kx) − kf(3x) + 4kf(2x) − 5kf(x)‖Y

≤ M

[

M3ϕa(x, (k + 1)x) + M3ϕa(x, (k − 1)x) + Mϕa(2x, x) + ϕa(2x, kx)


+M3

k − 1ϕa(0, x) +

M3k

k − 1ϕa(0, kx)

]

(3.19)

for all x ∈ X. By (3.4), (3.13), (3.14) and (3.15), we have

‖kf(−(k + 1)x) − kf(−(k − 1)x) − k2f(2(k + 1)x) + k2f(−2(k − 1)x)

+ k2f(2kx) − (k2 − 1)f(−2kx) + f(4kx) − 2f(kx) + 2kf(x)‖Y

≤ M2

[

kϕa(x, (2k + 1)x) + kϕa(x, (2k − 1)x) + ϕa(x, 3kx)

+k

k − 1ϕa(0, (3k − 1)x)

]

(3.20)

for all x ∈ X. It follows from (3.4), (3.6), (3.7) and (3.20) that

‖f(4kx) − 2f(2kx) − k3f(4x) + 2k3f(2x)‖Y

≤ M3

[

M2kϕa(x, (2k + 1)x) + M2kϕa(x, (2k − 1)x) + M2ϕa(x, 3kx)

+ ϕa(x, kx) + k2ϕa(2x, 2x) +M2k

k − 1ϕa(0, (3k − 1)x)

+k

k − 1ϕa(0, (k + 1)x) +

k2

k − 1ϕa(0, 2(k − 1)x) + (k + 1)ϕa(0, 2kx)

]

(3.21)

for all x ∈ X. Hence

‖f(2kx) − 2f(kx) − k3f(2x) + 2k3f(x)‖Y

≤ M3

[

M2kϕa

(x

2,(2k + 1)x

2

)

+ M2kϕa

(x

2,(2k − 1)x

2

)

+ M2ϕa

(x

2,3kx

2

)

+ ϕa

(x

2,kx

2

)

+ k2ϕa(x, x) +M2k

k − 1ϕa

(

0,(3k − 1)x

2

)

+k

k − 1ϕa

(

0,(k + 1)x

2

)

+k2

k − 1ϕa(0, (k − 1)x) + (k + 1)ϕa(0, kx)

]

(3.22)

for all x ∈ X. By (3.7), we have

‖f(4kx) − kf(2(k + 1)x) − kf(−2(k − 1)x) − 2f(2kx) + 2kf(2x)‖Y ≤ ϕa(2x, 2kx) (3.23)

for all x ∈ X. From (3.21) and (3.23), we have

‖kf(2(k + 1)x) + kf(−2(k − 1)x) − k3f(4x) + (2k3 − 2k)f(2x)‖Y

≤ M

[

M5kϕa(x, (2k + 1)x) + M5kϕa(x, (2k − 1)x) + M5ϕa(x, 3kx)

+ M3ϕa(x, kx) + M3k2ϕa(2x, 2x) + ϕa(2x, 2kx) +M5k

k − 1ϕa(0, (3k − 1)x)

+M3k

k − 1ϕa(0, (k + 1)x) +

M3k2

k − 1ϕa(0, 2(k − 1)x) + M3(k + 1)ϕa(0, 2kx)

]

(3.24)

for all x ∈ X. Also, from (3.16) and (3.24), we get

‖2f(3kx) − 6f(kx) + (k − k3)f(4x) − 2kf(3x) + (2k3 − 2k)f(2x) + 6kf(x)‖Y

≤ M

{

(M4 + M6k)[ϕa(x, (2k + 1)x) + ϕa(x, (2k − 1)x)] + M4ϕa(3x, x)


+ M4ϕa(x, x) + M6ϕa(x, 3kx) + M4ϕa(x, kx) + M4k2ϕa(2x, 2x)

+ Mϕa(2x, 2kx) +M4(k + 1)

k − 1ϕa(0, (k + 1)x) +

M4

k − 1ϕa(0, (k − 1)x)

+M4(k2 + k − 1)

k − 1ϕa(0, 2kx) +

M6k

k − 1ϕa(0, (3k − 1)x) +

M4k2

k − 1ϕa(0, 2(k − 1)x)

}

(3.25)

for all x ∈ X.On the other hand, it follows from (3.19) and (3.25) that

‖8f(2kx) − 16f(kx) + (k − k3)f(4x) + (2k3 − 10k)f(2x) + 16kf(x)‖Y

≤ M

{

(M5 + M7k)[ϕa(x, (2k + 1)x) + ϕa(x, (2k − 1)x)] + M5ϕa(3x, x)

+ M5ϕa(x, x) + M7ϕa(x, 3kx) + M5ϕa(x, kx) + M5k2ϕa(2x, 2x)

+ M2ϕa(2x, 2kx) + 2M4ϕa(x, (k + 1)x) + 2M4ϕa(x, (k − 1)x)

+ 2M2ϕa(2x, x) + 2Mϕa(2x, kx) +2M4

k − 1ϕa(0, x) +

2M4k

k − 1ϕa(0, kx)

+M5(k + 1)

k − 1ϕa(0, (k + 1)x) +

M5

k − 1ϕa(0, (k − 1)x) +

M5(k2 + k − 1)k − 1

ϕa(0, 2kx)

+M7k

k − 1ϕa(0, (3k − 1)x) +

M5k2

k − 1ϕa(0, 2(k − 1)x)

}

(3.26)

for all x ∈ X. Therefore, by (3.22) and (3.26), we get

‖f(4x) − 10f(2x) + 16f(x)‖Y

≤ M3

k3 − k

{

(M4 + M6k)[ϕa(x, (2k + 1)x) + ϕa(x, (2k − 1)x)]

+ M4ϕa(3x, x) + (M4 + 8Mk2)ϕa(x, x) + M6ϕa(x, 3kx) + M4ϕa(x, kx)

+ M4k2ϕa(2x, 2x) + Mϕa(2x, 2kx) + 2M3ϕa(x, (k + 1)x) + 2M3ϕa(x, (k − 1)x)

+ 2Mϕa(2x, x) + 2ϕa(2x, kx) + 8Mϕa

(x

2,kx

2

)

+ 8M3kϕa

(x

2,(2k − 1)x

2

)

+ 8M3kϕa

(x

2,(2k + 1)x

2

)

+ 8M3ϕa

(x

2,3kx

2

)

+M4(k + 1)

k − 1ϕa(0, (k + 1)x)

+M4 + 8Mk2

k − 1ϕa(0, (k − 1)x) +

2M3

k − 1ϕa(0, x) +

M6k

k − 1ϕa(0, (3k − 1)x)

+M4k2

k − 1ϕa(0, 2(k − 1)x) +

M4(k2 + k − 1)k − 1

ϕa(0, 2kx) +8M3k

k − 1ϕa

(

0,(3k − 1)x

2

)

+8Mk

k − 1ϕa

(

0,(k + 1)x

2

)

+2M3k + 8M(k2 − 1)

k − 1ϕa(0, kx)

}

(3.27)

for all x ∈ X.Now, let g : X → Y be the mapping defined by g(x) := f(2x) − 8f(x) and let

ϕa(x) :=M3

k3 − k

{

(M4 + M6k)[ϕa(x, (2k + 1)x) + ϕa(x, (2k − 1)x)]

+ M4ϕa(3x, x) + (M4 + 8Mk2)ϕa(x, x) + M6ϕa(x, 3kx) + M4ϕa(x, kx)

+ M4k2ϕa(2x, 2x) + Mϕa(2x, 2kx) + 2M3ϕa(x, (k + 1)x) + 2M3ϕa(x, (k − 1)x)


+ 2Mϕa(2x, x) + 2ϕa(2x, kx) + 8Mϕa

(x

2,kx

2

)

+ 8M3kϕa

(x

2,(2k − 1)x

2

)

+ 8M3kϕa

(x

2,(2k + 1)x

2

)

+ 8M3ϕa

(x

2,3kx

2

)

+M4(k + 1)

k − 1ϕa(0, (k + 1)x)

+M4 + 8Mk2

k − 1ϕa(0, (k − 1)x) +

2M3

k − 1ϕa(0, x) +

M6k

k − 1ϕa(0, (3k − 1)x)

+M4k2

k − 1ϕa(0, 2(k − 1)x) +

M4(k2 + k − 1)k − 1

ϕa(0, 2kx) +8M3k

k − 1ϕa

(

0,(3k − 1)x

2

)

+8Mk

k − 1ϕa

(

0,(k + 1)x

2

)

+2M3k + 8M(k2 − 1)

k − 1ϕa(0, kx)

}

(3.28)

for all x ∈ X. Therefore (3.27) means that

‖f(4x) − 10f(2x) + 16f(x)‖Y ≤ ϕa(x) (3.29)

for all x ∈ X. Also, we get

‖g(2x) − 2g(x)‖Y ≤ ϕa(x) (3.30)

for all x ∈ X. Replacing x by 2nx in (3.30) and dividing both sides of (3.30) by 2n+1, we get∥∥∥∥

12n+1

g(2n+1x) − 12n

g(2nx)∥∥∥∥

Y

≤ 12n+1

ϕa(2nx) (3.31)

for all x ∈ X and all non-negative integers n. Since Y is a p-Banach space, hence by (3.31), wehave

∥∥∥∥

12n+1

g(2n+1x) − 12m

g(2mx)∥∥∥∥

p

Y

≤n∑

i=m

∥∥∥∥

12i+1

g(2i+1x) − 12i

g(2ix)∥∥∥∥

p

Y

≤ 12p

n∑

i=m

12ip

ϕpa(2ix) (3.32)

for all x ∈ X and all non-negative integers n and m with m ≤ n. Since 0 < p ≤ 1, then byLemma 3.1 and (3.28), we get

ϕpa(x) ≤ M3p

(k3 − k)p

{

(M4 + M6k)p[ϕpa(x, (2k + 1)x) + ϕp

a(x, (2k − 1)x)]

+ M4pϕpa(3x, x) + (M4 + 8Mk2)pϕp

a(x, x) + M6pϕpa(x, 3kx) + M4pϕp

a(x, kx)

+ M4pk2pϕpa(2x, 2x) + Mpϕp

a(2x, 2kx) + 2pM3pϕpa(x, (k + 1)x)

+ 2pM3pϕpa(x, (k − 1)x) + 2pMpϕp

a(2x, x) + 2pϕpa(2x, kx) + 8pMpϕp

a

(x

2,kx

2

)

+ 8pM3pkpϕpa

(x

2,(2k − 1)x

2

)

+ 8pM3pkpϕpa

(x

2,(2k + 1)x

2

)

+ 8pM3pϕpa

(x

2,3kx

2

)

+(

M4(k + 1)k − 1

)p

ϕpa(0, (k + 1)x) +

(M4 + 8Mk2

k − 1

)p

ϕpa(0, (k − 1)x)

+(

2M3

k − 1

)p

ϕpa(0, x) +

(M6k

k − 1

)p

ϕpa(0, (3k − 1)x) +

(M4k2

k − 1

)p

ϕpa(0, 2(k − 1)x)

+(

M4(k2 + k − 1)k − 1

)p

ϕpa(0, 2kx) +

(8M3k

k − 1

)p

ϕpa

(

0,(3k − 1)x

2

)


+(

8Mk

k − 1

)p

ϕpa

(

0,(k + 1)x

2

)

+(

2M3k + 8M(k2 − 1)k − 1

)p

ϕpa(0, kx)

}

for all x ∈ X. So by (3.1),∞∑

i=0

12ip

ϕpa(2ix) < ∞ (3.33)

for all x ∈ X. This shows that the sequence { 12n g(2nx)} is a Cauchy sequence in Y for all

x ∈ X. Since Y is complete, the sequence { 12n g(2nx)} converges for all x ∈ X. So we can define

a mapping A : X → Y by

A(x) := limn→∞

12n

g(2nx) (3.34)

for all x ∈ X. Putting m = 0 and passing the limit n → ∞ in (3.32), we get

‖g(x) − A(x)‖pY ≤ 1

2p

∞∑

i=0

12ip

ϕpa(2ix) (3.35)

for all x ∈ X. Hence (3.3) follows from (3.1) and (3.35).Now we show that A is additive. By (3.1), (3.31) and (3.34), we have

‖A(2x) − 2A(x)‖Y = limn→∞

∥∥∥∥

12n

g(2n+1x) − 12n−1

g(2nx)∥∥∥∥

Y

= 2 limn→∞

∥∥∥∥

12n+1

g(2n+1x) − 12n

g(2nx)∥∥∥∥

Y

≤ limn→∞

12n

ϕa(2nx) = 0

for all x ∈ X. So

A(2x) = 2A(x) (3.36)

for all x ∈ X. Also, by (3.1), (3.2) and (3.34), we get

‖DA(x, y)‖Y = limn→∞

12n

‖Dg(2nx, 2ny)‖Y

= limn→∞

12n

‖Df(2n+1x, 2n+1y) − 8Df(2nx, 2ny)‖Y

≤ limn→∞

M

2n[‖Df(2n+1x, 2n+1y)‖Y + 8‖Df(2nx, 2ny)‖Y ]

≤ limn→∞

M

2n[ϕa(2n+1x, 2n+1y) + 8ϕa(2nx, 2ny)] = 0

for all x, y ∈ X. Hence the mapping A satisfies (1.4). By Lemma 2.3, the mapping x →A(2x) − 8A(x) is additive. Therefore (3.36) implies that the mapping A is additive.

To prove the uniqueness of A, let T : X → Y be another additive mapping satisfying (3.3).Since

limn→∞

12np

∞∑

i=0

12ip

ϕpa(2i+nx, 2i+ny) = lim

n→∞

∞∑

i=n

12ip

ϕpa(2ix, 2iy) = 0

for all x, y ∈ X, hence

limn→∞

12np

Φa(2nx) = 0


for all x ∈ X. Then we have

‖A(x) − T (x)‖pY = lim

n→∞1

2np‖g(2nx) − T (2nx)‖p

Y ≤ 12p

limn→∞

12np

Φa(2nx) = 0

for all x ∈ X. So A = T . �

Theorem 3.3 Let f : X → Y be a mapping with f(0) = 0 for which there is a functionϕa : X × X → [0,∞) such that

∞∑

i=1

2ipϕpa

(x

2i,

y

2i

)

< ∞ (3.37)

and

‖Df(x, y)‖Y ≤ ϕa(x, y) (3.38)


A(x) = limn→∞ 2n

[

f

(x

2n−1

)

− 8f

(x

2n

)]

(3.39)

exists for all x ∈ X and A : X → Y is a unique additive mapping such that

‖f(2x) − 8f(x) − A(x)‖Y ≤ 12[Φa(x)]

1p (3.40)


Φa(x) :=M3p

(k3 − k)p

∞∑

i=1

2ip

{

(M4 + M6k)p

[

ϕpa

(x

2i,(2k + 1)x

2i

)

+ ϕpa

(x

2i,(2k − 1)x

2i

)]

+ M4pϕpa

(3x

2i,

x

2i

)

+ (M4 + 8Mk2)pϕpa

(x

2i,

x

2i

)

+ M6pϕpa

(x

2i,3kx

2i

)

+ M4pϕpa

(x

2i,kx

2i

)

+ M4pk2pϕpa

(x

2i−1,

x

2i−1

)

+ Mpϕpa

(x

2i−1,

kx

2i−1

)

+ 2pM3pϕpa

(x

2i,(k + 1)x

2i

)

+ 2pM3pϕpa

(x

2i,(k − 1)x

2i

)

+ 2pMpϕpa

(x

2i−1,

x

2i

)

+ 2pϕpa

(x

2i−1,kx

2i

)

+ 8pMpϕpa

(x

2i+1,

kx

2i+1

)

+ 8pM3pkpϕpa

(x

2i+1,(2k − 1)x

2i+1

)

+ 8pM3pkpϕpa

(x

2i+1,(2k + 1)x

2i+1

)

+ 8pM3pϕpa

(x

2i+1,

3kx

2i+1

)

+(

M4(k + 1)k − 1

)p

ϕpa

(

0,(k + 1)x

2i

)

+(

M4 + 8Mk2

k − 1

)p

ϕpa

(

0,(k − 1)x

2i

)

+(

2M3

k − 1

)p

ϕpa

(

0,x

2i

)

+(

M6k

k − 1

)p

ϕpa

(

0,(3k − 1)x

2i

)

+(

M4k2

k − 1

)p

ϕpa

(

0,(k − 1)x

2i−1

)

+(

M4(k2 + k − 1)k − 1

)p

ϕpa

(

0,kx

2i−1

)

+(

8M3k

k − 1

)p

ϕpa

(

0,(3k − 1)x

2i+1

)

+(

8Mk

k − 1

)p

ϕpa

(

0,(k + 1)x

2i+1

)

+(

2M3k + 8M(k2 − 1)k − 1

)p

ϕpa

(

0,kx

2i

)}

.

Proof We can prove the theorem by a similar technique in Theorem 3.2. �


Corollary 3.4 Let θ, r, s be non-negative real numbers such that r, s > 1 or 0 ≤ r, s < 1.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality

‖Df(x, y)‖Y ≤

⎧

⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

θ, r = s = 0;

θ‖x‖rX , r > 0, s = 0;

θ‖y‖sX , r = 0, s > 0;

θ(‖x‖rX + ‖y‖s

X), r, s > 0,

(3.41)

for all x, y ∈ X. Then there exists a unique additive mapping A : X → Y satisfying

‖f(2x) − 8f(x) − A(x)‖Y ≤ M3θ

k3 − k×

⎧

⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

δa, r = s = 0;

αa‖x‖rX , r > 0, s = 0;

βa‖x‖sX , r = 0, s > 0;

γa(x), r, s > 0,

(3.42)


δa ={

12p − 1

[

2(M4 + M6k)p + 2M4p + (M4 + 8Mk2)p + M6p + M4pk2p + Mp

+ 2p+1M3p + 2pMp + 2p + 8pMp + 23p+1M3pkp + 8pM3p

+(

M4(k + 1)k − 1

)p

+(

M4 + 8Mk2

k − 1

)p

+(

2M3

k − 1

)p

+(

M6k

k − 1

)p

+(

M4k2

k − 1

)p

+(

M4(k2 + k − 1)k − 1

)p

+(

8M3k

k − 1

)p

+(

8Mk

k − 1

)p

+(

2M3k + 8M(k2 − 1)k − 1

)p]} 1p

,

αa ={

1|2p − 2pr| [2(M4 + M6k)p + M4p3pr + (M4 + 8Mk2)p + M6p + M4p

+ 2prM4pk2p + Mp2pr + 2p+1M3p + 2p(r+1)Mp + 2p(r+1) + 2p(3−r)Mp

+ 2p(3−r)+1M3pkp + 2p(3−r)M3p]} 1

p

,

βa ={

1|2p − 2ps|

[

(M4 + M6k)p((2k + 1)ps + (2k − 1)ps) + M4p + (M4 + 8Mk2)p

+ M6p(3k)ps + M4pkps + 2psM4pk2p + Mp(2k)ps + 2pM3p(k + 1)ps + 2pMp

+ 2pM3p(k − 1)ps + 2p(3−s)Mpkps + 2p(3−s)M3pkp(2k − 1)ps + 2p(3−s)M3p(3k)ps

+ 2p(3−s)M3pkp(2k + 1)ps + 2pkps(M4 + 8Mk2)p(k − 1)p(s−1) +(

2M3

k − 1

)p

+(

M4(k + 1)k − 1

)p

(k + 1)ps +(

M6k

k − 1

)p

(3k − 1)ps +(

Mk

k − 1

)p

(k + 1)ps2p(3−s)

+(

M4(k2 + k − 1)k − 1

)p

kps2ps +(

M3k

k − 1

)p

(3k − 1)ps2p(3−s)

+ (M4k2)p(k − 1)p(s−1)2ps +(

2M3k + 8M(k2 − 1)k − 1

)p

kps

]} 1p

,

γa(x) = (αpa‖x‖pr

X + βpa‖x‖ps

X )1p .

Proof Define ϕa(x, y) = θ(‖x‖rX +‖y‖s

X) for all x, y ∈ X, and apply Theorems 3.2 and 3.3. �


Corollary 3.5 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r + s �= 1.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality

‖Df(x, y)‖Y ≤ θ‖x‖rX‖y‖s

X (3.43)


‖f(2x) − 8f(x) − A(x)‖Y ≤ M3θ

k3 − kεa‖x‖λ

X


εa ={

1|2p − 2pλ| [(M

4 + M6k)p[(2k + 1)ps + (2k − 1)ps] + 3prM4p + (M4 + 8Mk2)p

+ M6p(3k)ps + M4pkps + M4pk2p2pλ + Mpkps2pλ + 2pM3p(k + 1)ps

+ 2pM3p(k − 1)ps + 2p(r+1)kps + 2p(3−λ)Mpkps + 2p(3−λ)M3pkp(2k − 1)ps

+ 2p(r+1)Mp + 2p(3−λ)M3pkp(2k + 1)ps + 2p(3−λ)M3p(3k)ps]} 1

p

.

Proof Define ϕa(x, y) = θ‖x‖rX‖y‖s

X for all x, y ∈ X, and apply Theorems 3.2 and 3.3. �

Corollary 3.6 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r + s �= 1.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality

‖Df(x, y)‖Y ≤ θ[‖x‖rX‖y‖s

X + (‖x‖r+sX + ‖y‖r+s

X )] (3.44)


‖f(2x) − 8f(x) − A(x)‖Y ≤ M3θ

k3 − k(αp

a + βpa + εp

a)‖x‖λX

for all x ∈ X, where αa, βa and εa are defined as in Corollaries 3.4 and 3.5.

Proof Define ϕa(x, y) = θ[‖x‖rX‖y‖s

X + (‖x‖r+sX + ‖y‖r+s

X )] for all x, y ∈ X, and apply Theo-rems 3.2 and 3.3. �

Theorem 3.7 Let f : X → Y be a mapping with f(0) = 0 for which there is a functionϕc : X × X → [0,∞) such that

∞∑

i=0

18ip

ϕpc(2

ix, 2iy) < ∞ (3.45)

and

‖Df(x, y)‖Y ≤ ϕc(x, y) (3.46)


C(x) = limn→∞

18n

[f(2n+1x) − 2f(2nx)] (3.47)

exists for all x ∈ X and C : X → Y is a unique cubic mapping such that

‖f(2x) − 2f(x) − C(x)‖Y ≤ 18[Φc(x)]

1p (3.48)


Φc(x) :=M3p

(k3 − k)p

∞∑

i=0

18ip

{

(M4 + M6k)p[ϕpc(2

ix, (2k + 1)2ix) + ϕpc(2

ix, (2k − 1)2ix)]


+ M4pϕpc(3 · 2ix, 2ix) + (M4 + 8Mk2)pϕp

c(2ix, 2ix) + M6pϕp

c(2ix, 3k2ix)

+ M4pϕpc(2

ix, k2ix) + M4pk2pϕpc(2

i+1x, 2i+1x) + Mpϕpc(2

i+1x, k2i+1x)

+ 2pM3pϕpc(2

ix, (k + 1)2ix) + 2pM3pϕpc(2

ix, (k − 1)2ix) + 2pMpϕpc(2

i+1x, 2ix)

+ 2pϕpc(2

i+1x, k2ix) + 8pMpϕpc(2

i−1x, k2i−1x) + 8pM3pkpϕpc(2

i−1x, (2k − 1)2i−1x)

+ 8pM3pkpϕpc(2

i−1x, (2k + 1)2i−1x) + 8pM3pϕpc(2

i−1x, 3k2i−1x)

+(

M4(k + 1)k − 1

)p

ϕpc(0, (k + 1)2ix) +

(M4 + 8Mk2

k − 1

)p

ϕpc(0, (k − 1)2ix)

+(

2M3

k − 1

)p

ϕpc(0, 2ix) +

(M6k

k − 1

)p

ϕpc(0, (3k − 1)2ix)

+(

M4k2

k − 1

)p

ϕpc(0, (k − 1)2i+1x) +

(M4(k2 + k − 1)

k − 1

)p

ϕpc(0, k2i+1x)

+(

8M3k

k − 1

)p

ϕpc(0, (3k − 1)2i−1x) +

(8Mk

k − 1

)p

ϕpc(0, (k + 1)2i−1x)

+(

2M3k + 8M(k2 − 1)k − 1

)p

ϕpc(0, k2ix)

}

.

Proof As in the proof of Theorem 3.2, we have

‖f(4x) − 10f(2x) + 16f(x)‖Y ≤ ϕc(x) (3.49)


ϕc(x) :=M3

k3 − k

{

(M4 + M6k)[ϕc(x, (2k + 1)x) + ϕc(x, (2k − 1)x)]

+ M4ϕc(3x, x) + (M4 + 8Mk2)ϕc(x, x) + M6ϕc(x, 3kx) + M4ϕc(x, kx)

+ M4k2ϕc(2x, 2x) + Mϕc(2x, 2kx) + 2M3ϕc(x, (k + 1)x) + 2M3ϕc(x, (k − 1)x)

+ 2Mϕc(2x, x) + 2ϕc(2x, kx) + 8Mϕc

(x

2,kx

2

)

+ 8M3kϕc

(x

2,(2k − 1)x

2

)

+ 8M3kϕc

(x

2,(2k + 1)x

2

)

+ 8M3ϕc

(x

2,3kx

2

)

+M4(k + 1)

k − 1ϕc(0, (k + 1)x)

+M4 + 8Mk2

k − 1ϕc(0, (k − 1)x) +

2M3

k − 1ϕc(0, x) +

M6k

k − 1ϕc(0, (3k − 1)x)

+M4k2

k − 1ϕc(0, 2(k − 1)x) +

M4(k2 + k − 1)k − 1

ϕc(0, 2kx) +8M3k

k − 1ϕc

(

0,(3k − 1)x

2

)

+8Mk

k − 1ϕc

(

0,(k + 1)x

2

)

+2M3k + 8M(k2 − 1)

k − 1ϕc(0, kx)

}

(3.50)

for all x ∈ X.

Now, let h : X → Y be the mapping defined by h(x) := f(2x) − 2f(x). By (3.49), we have

‖h(2x) − 8h(x)‖Y ≤ ϕc(x) (3.51)

for all x ∈ X. Replacing x by 2nx in (3.51) and dividing both sides of (3.51) by 8n+1, we get∥∥∥∥

18n+1

h(2n+1x) − 18n

h(2nx)∥∥∥∥

Y

≤ 18n+1

ϕc(2nx) (3.52)


for all x ∈ X and all non-negative integers n. Since Y is a p-Banach space, hence by (3.52), wehave

∥∥∥∥

18n+1

h(2n+1x) − 18m

h(2mx)∥∥∥∥

p

Y

≤n∑

i=m

∥∥∥∥

18i+1

h(2i+1x) − 18i

h(2ix)∥∥∥∥

p

Y

≤ 18p

n∑

i=m

18ip

ϕpc(2

ix) (3.53)

for all x ∈ X and all non-negative integers n and m with m ≤ n. Since 0 < p ≤ 1, then byLemma 3.1 and (3.50), we get

ϕpc(x) ≤ M3p

(k3 − k)p

{

(M4 + M6k)p[ϕpc(x, (2k + 1)x) + ϕp

c(x, (2k − 1)x)]

+ M4pϕpc(3x, x) + (M4 + 8Mk2)pϕp

c(x, x) + M6pϕpc(x, 3kx) + M4pϕp

c(x, kx)

+ M4pk2pϕpc(2x, 2x) + Mpϕp

c(2x, 2kx) + 2pM3pϕpc(x, (k + 1)x)

+ 2pM3pϕpc(x, (k − 1)x) + 2pMpϕp

c(2x, x) + 2pϕpc(2x, kx) + 8pMpϕp

c

(x

2,kx

2

)

+ 8pM3pkpϕpc

(x

2,(2k − 1)x

2

)

+ 8pM3pkpϕpc

(x

2,(2k + 1)x

2

)

+ 8pM3pϕpc

(x

2,3kx

2

)

+(

M4(k + 1)k − 1

)p

ϕpc(0, (k + 1)x) +

(M4 + 8Mk2

k − 1

)p

ϕpc(0, (k − 1)x)

+(

2M3

k − 1

)p

ϕpc(0, x) +

(M6k

k − 1

)p

ϕpc(0, (3k − 1)x) +

(M4k2

k − 1

)p

ϕpc(0, 2(k − 1)x)

+(

M4(k2 + k − 1)k − 1

)p

ϕpc(0, 2kx) +

(8M3k

k − 1

)p

ϕpc

(

0,(3k − 1)x

2

)

+(

8Mk

k − 1

)p

ϕpc

(

0,(k + 1)x

2

)

+(

2M3k + 8M(k2 − 1)k − 1

)p

ϕpc(0, kx)

}

for all x ∈ X. So by (3.45),∞∑

i=0

18ip

ϕpc(2

ix) < ∞ (3.54)

for all x ∈ X. This shows that the sequence { 18n h(2nx)} is a Cauchy sequence in Y for all

x ∈ X. Since Y is complete, the sequence { 18n h(2nx)} converges for all x ∈ X. So we can

define a mapping C : X → Y by

C(x) := limn→∞

18n

h(2nx) (3.55)

for all x ∈ X. Letting m = 0 and passing the limit n → ∞ in (3.53), we get

‖h(x) − C(x)‖pY ≤ 1

8p

∞∑

i=0

18ip

ϕpc(2

ix) (3.56)

for all x ∈ X. Hence (3.48) follows from (3.45) and (3.56).

Now we show that C is cubic. By (3.45), (3.52) and (3.55), we have

‖C(2x) − 8C(x)‖Y = limn→∞

∥∥∥∥

18n

h(2n+1x) − 18n−1

h(2nx)∥∥∥∥

Y


= 8 limn→∞

∥∥∥∥

18n+1

h(2n+1x) − 18n

h(2nx)∥∥∥∥

Y

≤ limn→∞

18n

ϕc(2nx) = 0

for all x ∈ X. So

C(2x) = 8C(x) (3.57)

for all x ∈ X. Also, by (3.45), (3.46) and (3.47), we get

‖DC(x, y)‖Y = limn→∞

18n

‖Dh(2nx, 2ny)‖Y

= limn→∞

18n

‖Df(2n+1x, 2n+1y) − 2Df(2nx, 2ny)‖Y

≤ limn→∞

M

8n[‖Df(2n+1x, 2n+1y)‖Y + 2‖Df(2nx, 2ny)‖Y ]

≤ limn→∞

M

8n[ϕc(2n+1x, 2n+1y) + 2ϕc(2nx, 2ny)] = 0

for all x, y ∈ X. Hence the mapping C satisfies (1.4). By Lemma 2.3, the mapping x →C(2x) − 2C(x) is cubic. Hence (3.57) implies that the mapping C is cubic.

To prove the uniqueness of C, let S : X → Y be another cubic mapping satisfying (3.48).Since

limn→∞

18np

∞∑

i=0

18ip

ϕpc(2

i+nx, 2i+ny) = limn→∞

∞∑

i=n

18ip

ϕpc(2

ix, 2iy) = 0

for all x, y ∈ X, hence

limn→∞

18np

Φc(2nx) = 0

for all x ∈ X. Then we have

‖C(x) − S(x)‖pY = lim

n→∞1

8np‖h(2nx) − S(2nx)‖p

Y ≤ 18p

limn→∞

18np

Φc(2nx) = 0

for all x ∈ X. So C = S. �

Theorem 3.8 Let f : X → Y be a mapping with f(0) = 0 for which there is a functionϕc : X × X → [0,∞) such that

∞∑

i=1

8ipϕpc

(x

2i,

y

2i

)

< ∞ (3.58)

and

‖Df(x, y)‖Y ≤ ϕc(x, y) (3.59)


C(x) = limn→∞ 8n

[

f

(x

2n−1

)

− 2f

(x

2n

)]

(3.60)

exists for all x ∈ X and C : X → Y is a unique cubic mapping such that

‖f(2x) − 2f(x) − C(x)‖Y ≤ 18[Φc(x)]

1p (3.61)



Φc(x) :=M3p

(k3 − k)p

∞∑

i=1

8ip

{

(M4 + M6k)p

[

ϕpc

(x

2i,(2k + 1)x

2i

)

+ ϕpc

(x

2i,(2k − 1)x

2i

)]

+ M4pϕpc

(3x

2i,

x

2i

)

+ (M4 + 8Mk2)pϕpc

(x

2i,

x

2i

)

+ M6pϕpc

(x

2i,3kx

2i

)

+ M4pϕpc

(x

2i,kx

2i

)

+ M4pk2pϕpc

(x

2i−1,

x

2i−1

)

+ Mpϕpc

(x

2i−1,

kx

2i−1

)

+ 2pM3pϕpc

(x

2i,(k + 1)x

2i

)

+ 2pM3pϕpc

(x

2i,(k − 1)x

2i

)

+ 2pMpϕpc

(x

2i−1,

x

2i

)

+ 2pϕpc

(x

2i−1,kx

2i

)

+ 8pMpϕpc

(x

2i+1,

kx

2i+1

)

+ 8pM3pkpϕpc

(x

2i+1,(2k − 1)x

2i+1

)

+ 8pM3pkpϕpc

(x

2i+1,(2k + 1)x

2i+1

)

+ 8pM3pϕpc

(x

2i+1,

3kx

2i+1

)

+(

M4(k + 1)k − 1

)p

ϕpc

(

0,(k + 1)x

2i

)

+(

M4 + 8Mk2

k − 1

)p

ϕpc

(

0,(k − 1)x

2i

)

+(

2M3

k − 1

)p

ϕpc

(

0,x

2i

)

+(

M6k

k − 1

)p

ϕpc

(

0,(3k − 1)x

2i

)

+(

M4k2

k − 1

)p

ϕpc

(

0,(k − 1)x

2i−1

)

+(

M4(k2 + k − 1)k − 1

)p

ϕpc

(

0,kx

2i−1

)

+(

8M3k

k − 1

)p

ϕpc

(

0,(3k − 1)x

2i+1

)

+(

8Mk

k − 1

)p

ϕpc

(

0,(k + 1)x

2i+1

)

+(

2M3k + 8M(k2 − 1)k − 1

)p

ϕpc

(

0,kx

2i

)}

.

Proof We can prove the theorem by a similar technique in Theorem 3.7. �

Corollary 3.9 Let θ, r, s be non-negative real numbers such that r, s > 3 or 0 ≤ r, s < 3.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality (3.41) for all x, y ∈ X.Then there exists a unique cubic mapping C : X → Y satisfying

‖f(2x) − 2f(x) − C(x)‖Y ≤ M3θ

k3 − k×

⎧

⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

δc, r = s = 0;

αc‖x‖rX , r > 0, s = 0;

βc‖x‖sX , r = 0, s > 0;

γc(x), r, s > 0,

(3.62)


δc ={

18p − 1

[

2(M4 + M6k)p + 2M4p + (M4 + 8Mk2)p + M6p + M4pk2p + Mp

+ 2p+1M3p + 2pMp + 2p + 8pMp + 23p+1M3pkp + 8pM3p

+(

M4(k + 1)k − 1

)p

+(

M4 + 8Mk2

k − 1

)p

+(

2M3

k − 1

)p

+(

M6k

k − 1

)p

+(

M4k2

k − 1

)p

+(

M4(k2 + k − 1)k − 1

)p

+(

8M3k

k − 1

)p

+(

8Mk

k − 1

)p

+(

2M3k + 8M(k2 − 1)k − 1

)p]} 1p

,


αc ={

1|8p − 2pr| [2(M4 + M6k)p + M4p3pr + (M4 + 8Mk2)p + M6p + M4p

+ 2prM4pk2p + Mp2pr + 2p+1M3p + 2p(r+1)Mp + 2p(r+1) + 2p(3−r)Mp

+ 2p(3−r)+1M3pkp + 2p(3−r)M3p]} 1

p

,

βc ={

1|8p − 2ps|

[

(M4 + M6k)p((2k + 1)ps + (2k − 1)ps) + M4p + (M4 + 8Mk2)p

+ M6p(3k)ps + M4pkps + 2psM4pk2p + Mp(2k)ps + 2pM3p(k + 1)ps + 2pMp

+ 2pM3p(k − 1)ps + 2p(3−s)Mpkps + 2p(3−s)M3pkp(2k − 1)ps + 2p(3−s)M3p(3k)ps

+ 2p(3−s)M3pkp(2k + 1)ps + 2pkps(M4 + 8Mk2)p(k − 1)p(s−1) +(

2M3

k − 1

)p

+(

M4(k + 1)k − 1

)p

(k + 1)ps +(

M6k

k − 1

)p

(3k − 1)ps +(

Mk

k − 1

)p

(k + 1)ps2p(3−s)

+(

M4(k2 + k − 1)k − 1

)p

kps2ps +(

M3k

k − 1

)p

(3k − 1)ps2p(3−s)

+ (M4k2)p(k − 1)p(s−1)2ps +(

2M3k + 8M(k2 − 1)k − 1

)p

kps

]} 1p

,

γc(x) = (αpc‖x‖pr

X + βpc ‖x‖ps

X )1p .

Proof Define ϕc(x, y) = θ(‖x‖rX +‖y‖s

X) for all x, y ∈ X, and apply Theorems 3.7 and 3.8. �

Corollary 3.10 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r +s �= 3.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality (3.43) for all x, y ∈ X.Then there exists a unique cubic mapping C : X → Y satisfying

‖f(2x) − 2f(x) − C(x)‖Y ≤ M3θ

k3 − kεc‖x‖λ

X (3.63)


εc ={

1|8p − 2pλ| [(M

4 + M6k)p[(2k + 1)ps + (2k − 1)ps] + 3prM4p + (M4 + 8Mk2)p

+ M6p(3k)ps + M4pkps + M4pk2p2pλ + Mpkps2pλ + 2pM3p(k + 1)ps

+ 2pM3p(k − 1)ps + 2p(r+1)kps + 2p(3−λ)Mpkps + 2p(3−λ)M3pkp(2k − 1)ps

+ 2p(r+1)Mp + 2p(3−λ)M3pkp(2k + 1)ps + 2p(3−λ)M3p(3k)ps]} 1

p

.

Proof Define ϕc(x, y) = θ‖x‖rX‖y‖s

X for all x, y ∈ X, and apply Theorems 3.7 and 3.8. �

Corollary 3.11 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r +s �= 3.Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality (3.44) for all x, y ∈ X.Then there exists a unique cubic mapping C : X → Y satisfying

‖f(2x) − 2f(x) − C(x)‖Y ≤ M3θ

k3 − k(αp

c + βpc + εp

c)‖x‖λX

for all x ∈ X, where αc, βc and εc are defined as in Corollaries 3.9 and 3.10.


Proof Define ϕc(x, y) = θ[‖x‖rX‖y‖s

X + (‖x‖r+sX + ‖y‖r+s

X )] for all x, y ∈ X, and apply Theo-rems 3.7 and 3.8. �

Theorem 3.12 Let f : X → Y be a mapping with f(0) = 0 for which there is a functionϕ : X × X → [0,∞) such that

∞∑

i=0

12ip

ϕp(2ix, 2iy) < ∞ (3.64)

and

‖Df(x, y)‖Y ≤ ϕ(x, y) (3.65)

for all x, y ∈ X. Then there exist a unique additive mapping A : X → Y and a unique cubicmapping C : X → Y such that

‖f(x) − A(x) − C(x)‖Y ≤ M

48{4[Φa(x)]

1p + [Φc(x)]

1p } (3.66)

for all x ∈ X, where Φa(x) and Φc(x) are defined as in Theorems 3.2 and 3.7.

Proof By Theorems 3.2 and 3.7, there exist an additive mapping A0 : X → Y and a cubicmapping C0 : X → Y such that

‖f(2x) − 8f(x) − A0(x)‖Y ≤ 12[Φa(x)]

1p , ‖f(2x) − 2f(x) − C0(x)‖Y ≤ 1

8[Φc(x)]

1p

for all x ∈ X. Hence∥∥∥∥f(x) +

16A0(x) − 1

6C0(x)

∥∥∥∥

Y

≤ M

48{4[Φa(x)]

1p + [Φc(x)]

1p }

for all x ∈ X. So we obtain (3.66) by letting A(x) = −16A0(x) and C(x) = 1

6C0(x) for allx ∈ X.

To prove the uniqueness of A and C, let A1, C1 : X → Y be another additive and cubicmapping satisfying (3.66). Let A′ = A − A1 and C ′ = C − C1. So

‖A′(x) + C ′(x)‖Y ≤ M [‖f(x) − A(x) − C(x)‖Y + ‖f(x) − A1(x) − C1(x)‖Y ]

≤ M2

24{4[Φa(x)]

1p + [Φc(x)]

1p } (3.67)

for all x ∈ X. Since

limn→∞

12np

Φa(2nx) = limn→∞

18np

Φc(2nx) = 0

for all x ∈ X, then (3.67) implies that

limn→∞

18n

‖A′(2nx) + C ′(2nx)‖Y = 0

for all x ∈ X. Therefore C ′ = 0. So it follows from (3.67) that

‖A′(x)‖Y ≤ 5M2

24[Φa(x)]

1p

for all x ∈ X. Therefore A′ = 0. �The next theorem is an alternative result of Theorem 3.12.



∞∑

i=1

8ipϕp

(x

2i,

y

2i

)

< ∞

and

‖Df(x, y)‖Y ≤ ϕ(x, y)


‖f(x) − A(x) − C(x)‖Y ≤ M

48{4[Φa(x)]

1p + [Φc(x)]

1p }



∞∑

i=1

2ipϕp

(x

2i,

y

2i

)

< ∞,

∞∑

i=0

18ip

ϕp(2ix, 2iy) < ∞,

and

‖Df(x, y)‖Y ≤ ϕ(x, y)


‖f(x) − A(x) − C(x)‖Y ≤ M

48{4[Φa(x)]

1p + [Φc(x)]

1p }


Proof The proof is similar to the proof of Theorem 3.12 and the result follows from Theo-rems 3.3 and 3.7. �

Corollary 3.15 Let θ, r, s be non-negative real numbers such that r, s > 3 or 0 ≤ r, s < 1 or1 < r, s < 3. Suppose that a mapping f : X → Y with f(0) = 0 satisfies the inequality (3.41)for all x, y ∈ X. Then there exist a unique additive mapping A : X → Y and a unique cubicmapping C : X → Y such that

‖f(x) − A(x) − C(x)‖Y ≤ M4θ

6(k3 − k)×

⎧

⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(δa + δc), r = s = 0;

(αa + αc)‖x‖rX , r > 0, s = 0;

(βa + βc)‖x‖sX , r = 0, s > 0;

(γa(x) + γc(x)), r, s > 0,

for all x ∈ X, where δa, δc, αa, αc, βa, βc, γa(x) and γc(x) are defined as in Corollaries 3.4and 3.9.

Corollary 3.16 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r +s ∈ (0, 1) ∪ (1, 3) ∪ (3,∞). Suppose that a mapping f : X → Y with f(0) = 0 satisfies the


inequality (3.43) for all x, y ∈ X. Then there exist a unique additive mapping A : X → Y anda unique cubic mapping C : X → Y such that

‖f(x) − A(x) − C(x)‖Y ≤ M4θ

6(k3 − k)(εa + εc)‖x‖λ

X

for all x ∈ X, where εa and εc are defined as in Corollaries 3.5 and 3.10.

Corollary 3.17 Let θ ≥ 0 and r, s > 0 be non-negative real numbers such that λ := r +s ∈ (0, 1) ∪ (1, 3) ∪ (3,∞). Suppose that a mapping f : X → Y with f(0) = 0 satisfies theinequality (3.44) for all x, y ∈ X. Then there exist a unique additive mapping A : X → Y anda unique cubic mapping C : X → Y such that

‖f(x) − A(x) − C(x)‖Y ≤ M4θ

6(k3 − k)(αp

a + βpa + εp

a + αpc + βp

c + εpc)‖x‖λ

X

for all x ∈ X, where αa, βa, εa, αc, βc and εc are defined as in Corollaries 3.4, 3.5, 3.9 and 3.10.

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