Solution and Stability of a Mixed Type Cubic and Quartic Functional Equation in Quasi-Banach Spaces

Abstract

We obtain the general solution and the generalized Ulam-Hyers stability of the mixed type cubic and quartic functional equation 𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)=4(𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦))−24𝑓(𝑦)−6𝑓(𝑥)+3𝑓(2𝑦) in quasi-Banach spaces.

1. Introduction

We recall some basic facts concerning quasiBanach space. A quasinorm is a real-valued function on 𝑋 satisfying the following.

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

The pair (𝑋,‖⋅‖) is called a quasinormed space if ‖⋅‖ is a quasinorm on 𝑋. A quasiBanach space is a complete quasinormed space. A quasinorm ‖⋅‖ is called a 𝑝-norm (0<𝑝≤1) if ‖𝑥+𝑦‖^𝑝≤‖𝑥‖^𝑝+‖𝑦‖^𝑝 (1.1) for all 𝑥,𝑦∈𝑋. In this case, a quasiBanach space is called a 𝑝-Banach space. Given a 𝑝-norm, the formula 𝑑(𝑥,𝑦)∶=‖𝑥−𝑦‖^𝑝 gives us a translation invariant metric on 𝑋. By the Aoki-Rolewicz theorem [1] (see also [2]), each quasinorm is equivalent to some 𝑝-norm. Since it is much easier to work with 𝑝-norms, henceforth we restrict our attention mainly to 𝑝-norms. The stability problem of functional equations originated from a question of Ulam [3] 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 a 𝛿>0, such that if a mapping ℎ∶𝐺₁→𝐺₂ satisfies the inequality 𝑑(ℎ(𝑥⋅𝑦),ℎ(𝑥)∗ℎ(𝑦))<𝛿 for all 𝑥,𝑦∈𝐺₁, then there exists a homomorphism 𝐻∶𝐺₁→𝐺₂ with 𝑑(ℎ(𝑥),𝐻(𝑥))<𝜖 for all 𝑥∈𝐺₁? In the other words, Under what condition does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [4] gave the first affirmative answer to the question of Ulam for Banach spaces. Let 𝑓∶𝐸→𝐸^′ be a mapping between Banach spaces such that ‖𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)‖≤𝛿 (1.2) for all 𝑥,𝑦∈𝐸, and for some 𝛿>0. Then there exists a unique additive mapping 𝑇∶𝐸→𝐸^′ such that ‖𝑓(𝑥)−𝑇(𝑥)‖≤𝛿 (1.3) for all 𝑥∈𝐸. Moreover if 𝑓(𝑡𝑥) is continuous in 𝑡∈ℝ for each fixed 𝑥∈𝐸, then 𝑇 is linear. Rassias [5] succeeded in extending the result of Hyers' Theorem by weakening the condition for the Cauchy difference controlled by (‖𝑥‖^𝑝+‖𝑦‖^𝑝), 𝑝∈[0,1) to be unbounded. This condition has been assumed further till now, through the complete Hyers direct method, in order to prove linearity for generalized Hyers-Ulam stability problem forms. A number of mathematicians were attracted to the pertinent stability results of Rassias [6], and stimulated to investigate the stability problems of functional equations. The stability phenomenon that was introduced and proved by Rassias is called Hyers-Ulam-Rassias stability. And then the stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [4, 5, 7–18]).

The following cubic functional equation, which is the oldest cubic functional equation, was introduced by the third author of this paper, Rassias [6] (in 2001): 𝑓(𝑥+2𝑦)+3𝑓(𝑥)=3𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)+6𝑓(𝑦). (1.4) Jun and Kim [19] introduced the following cubic functional equation: 𝑓(2𝑥+𝑦)+𝑓(2𝑥−𝑦)=2𝑓(𝑥+𝑦)+2𝑓(𝑥−𝑦)+12𝑓(𝑥), (1.5) and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation (1.5). The function 𝑓(𝑥)=𝑥³ satisfies the functional equation (1.5), which is thus called a cubic functional equation. Every solution of the cubic functional equation is said to be a cubic function. Jun and Kim proved that a function 𝑓 between real vector spaces 𝑋 and 𝑌 is a solution of (1.5) if and only if there exists a unique function 𝐶∶𝑋×𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐶(𝑥,𝑥,𝑥) for all 𝑥∈𝑋, and 𝐶 is symmetric for each fixed one variable and is additive for fixed two variables (see also [20]).

The quartic functional equation (1.6) was introduced by Rassias [21] (in 2000) and then (in 2005) was employed by Park and Bae [22] and others, such that: 𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)=4[𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)]+24𝑓(𝑦)−6𝑓(𝑥). (1.6) In fact they proved that a function 𝑓 between real vector spaces 𝑋 and 𝑌 is a solution of (1.6) if and only if there exists a unique symmetric multiadditive function 𝑄∶𝑋×𝑋×𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝑄(𝑥,𝑥,𝑥,𝑥) for all 𝑥 (see also [21–29]). It is easy to show that the function 𝑓(𝑥)=𝑥⁴ satisfies the functional equation (1.6), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function. In this paper we deal with the following functional equation: 𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)=4(𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦))−24𝑓(𝑦)−6𝑓(𝑥)+3𝑓(2𝑦) (1.7) in quasiBanach spaces. It is easy to see that the function 𝑓(𝑥)=𝑎𝑥³+𝑏𝑥⁴ is a solution of the functional equation (1.7). In the present paper we investigate the general solution of functional equation (1.7) when 𝑓 is a mapping between vector spaces, and we establish the generalized Hyers-Ulam-Rassias stability of the functional equation (1.7) whenever 𝑓 is a mapping between two quasiBanach spaces. We only mention here the papers [30, 31] concerning the stability of the mixed type functional equations.

2. General Solution

Throughout this section, 𝑋 and 𝑌 will be real vector spaces. Before proceeding to the proof of Theorem 2.3 which is the main result in this section, we shall need the following two lemmas.

Lemma 2.1. If an even function 𝑓∶𝑋→𝑌 satisfies (1.7), then 𝑓 is quartic.

Proof. Putting 𝑥=𝑦=0 in (1.7), we get 𝑓(0)=0. Setting 𝑥=0 in (1.7), by evenness of 𝑓 we obtain 𝑓(2𝑦)=16𝑓(𝑦) (2.1) for all 𝑦∈𝑋. Hence (1.7) can be written as 𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)=4(𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦))+24𝑓(𝑦)−6𝑓(𝑥). (2.2) This means that 𝑓 is quartic function, which completes the proof of the lemma.

Lemma 2.2. If an odd function 𝑓∶𝑋→𝑌 satisfies (1.7), then f is a cubic function.

Proof. Setting 𝑥=𝑦=0 in (1.7) gives 𝑓(0)=0. Putting 𝑥=0 in (1.7), then by oddness of 𝑓, we have 𝑓(2𝑦)=8𝑓(𝑦). (2.3) Hence (1.7) can be written as 𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥−𝑦)−6𝑓(𝑥). (2.4) Replacing 𝑥 by 𝑥+𝑦 in (2.4), we obtain 𝑓(𝑥+3𝑦)+𝑓(𝑥−𝑦)=4𝑓(𝑥+2𝑦)−6𝑓(𝑥+𝑦)+4𝑓(𝑥). (2.5) Substituting −𝑦 for 𝑦 in (2.5) gives 𝑓(𝑥−3𝑦)+𝑓(𝑥+𝑦)=4𝑓(𝑥−2𝑦)−6𝑓(𝑥−𝑦)+4𝑓(𝑥). (2.6) If we subtract (2.5) from (2.6), we obtain 𝑓(𝑥+3𝑦)−𝑓(𝑥−3𝑦)=4𝑓(𝑥+2𝑦)−4𝑓(𝑥−2𝑦)−5𝑓(𝑥+𝑦)+5𝑓(𝑥−𝑦). (2.7) Let us interchange 𝑥 and 𝑦 in (2.7). Then we see that 𝑓(3𝑥+𝑦)+𝑓(3𝑥−𝑦)=4𝑓(2𝑥+𝑦)+4𝑓(2𝑥−𝑦)−5𝑓(𝑥+𝑦)−5𝑓(𝑥−𝑦). (2.8) With the substitution 𝑦∶=𝑥+𝑦 in (2.4), we have 𝑓(3𝑥+2𝑦)−𝑓(𝑥+2𝑦)=4𝑓(2𝑥+𝑦)−4𝑓(𝑦)−6𝑓(𝑥). (2.9) From the substitution 𝑦∶=−𝑦 in (2.9) it follows that 𝑓(3𝑥−2𝑦)−𝑓(𝑥−2𝑦)=4𝑓(2𝑥−𝑦)+4𝑓(𝑦)−6𝑓(𝑥). (2.10) If we add (2.9) to (2.10), we have 𝑓(3𝑥+2𝑦)+𝑓(3𝑥−2𝑦)=4𝑓(2𝑥+𝑦)+4𝑓(2𝑥−𝑦)+𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)−12𝑓(𝑥). (2.11) Replacing 𝑥 by 2𝑥 in (2.7) and using (2.3), we obtain 𝑓(2𝑥+3𝑦)−𝑓(2𝑥−3𝑦)=32𝑓(𝑥+𝑦)−32𝑓(𝑥−𝑦)−5𝑓(2𝑥+𝑦)+5𝑓(2𝑥−𝑦). (2.12) Interchanging 𝑥 with 𝑦 in (2.12) gives the equation 𝑓(3𝑥+2𝑦)+𝑓(3𝑥−2𝑦)=32𝑓(𝑥+𝑦)+32𝑓(𝑥−𝑦)−5𝑓(𝑥+2𝑦)−5𝑓(𝑥−2𝑦). (2.13) If we compare (2.11) and (2.13) and employ (2.4), we conclude that 𝑓(2𝑥+𝑦)+𝑓(2𝑥−𝑦)=2𝑓(𝑥+𝑦)+2𝑓(𝑥−𝑦)+12𝑓(𝑥). (2.14) This means that 𝑓 is cubic function. This completes the proof of Lemma.

Theorem 2.3. A function 𝑓∶𝑋→𝑌 satisfies (1.7) for all 𝑥,𝑦∈𝑋 if and only if there exists a unique function 𝐶∶𝑋×𝑋×𝑋→𝑌 and a unique symmetric multiadditive function 𝑄∶𝑋×𝑋×𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐶(𝑥,𝑥,𝑥)+𝑄(𝑥,𝑥,𝑥,𝑥) for all 𝑥∈𝑋, and that 𝐶 is symmetric for each fixed one variable and is additive for fixed two variables.

Proof. Let 𝑓 satisfy (1.7). We decompose 𝑓 into the even part and odd part by setting 𝑓_𝑒1(𝑥)=2(𝑓(𝑥)+𝑓(−𝑥)),𝑓_𝑜1(𝑥)=2(𝑓(𝑥)−𝑓(−𝑥)) (2.15) for all 𝑥∈𝑋. By (1.7), we have 𝑓_𝑒(𝑥+2𝑦)+𝑓_𝑒1(𝑥−2𝑦)=2[𝑓(𝑥+2𝑦)+𝑓(−𝑥−2𝑦)+𝑓(𝑥−2𝑦)+𝑓(−𝑥+2𝑦)]=4(𝑓_𝑒(𝑥+𝑦)+𝑓_𝑒(𝑥−𝑦))−24𝑓_𝑒(𝑦)−6𝑓_𝑒(𝑥)+3𝑓_𝑒(2𝑦) (2.16) for all 𝑥,𝑦∈𝑋. This means that 𝑓_𝑒 satisfies in (1.7). Similarly we can show that 𝑓_𝑜 satisfies (1.7). By Lemmas 2.1 and 2.2, 𝑓_𝑒 and 𝑓_𝑜 are quartic and cubic, respectively. Thus there exists a unique function 𝐶∶𝑋×𝑋×𝑋→𝑌 and a unique symmetric multiadditive function 𝑄∶𝑋×𝑋×𝑋×𝑋→𝑌 such that 𝑓_𝑒(𝑥)=𝑄(𝑥,𝑥,𝑥,𝑥) and that 𝑓_𝑜(𝑥)=𝐶(𝑥,𝑥,𝑥) for all 𝑥∈𝑋, and 𝐶 is symmetric for each fixed one variable and is additive for fixed two variables. Thus 𝑓(𝑥)=𝐶(𝑥,𝑥,𝑥)+𝑄(𝑥,𝑥,𝑥,𝑥) for all 𝑥∈𝑋. The proof of the converse is trivial.

3. Stability

Throughout this section, 𝑋 and 𝑌 will be a uniquely two-divisible abelian group and a quasiBanach spaces respectively, and 𝑝 will be a fixed real number in [0,1]. We need the following lemma in the main theorems. Now before taking up the main subject, given 𝑓∶𝑋→𝑌, we define the difference operator 𝐷_𝑓∶𝑋×𝑋→𝑌 by 𝐷_𝑓(𝑥,𝑦)=𝑓(𝑥+2𝑦)+𝑓(𝑥−2𝑦)−4[𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)]−3𝑓(2𝑦)+24𝑓(𝑦)+6𝑓(𝑥) (3.1) for all 𝑥,𝑦∈𝑋. We consider the following functional inequality: ‖𝐷_𝑓(𝑥,𝑦)‖≤𝜙(𝑥,𝑦) (3.2) for an upper bound 𝜙∶𝑋×𝑋→[0,∞).

Lemma 3.1. Let 𝑥₁,𝑥₂,…,𝑥_𝑛 be nonnegative real numbers. Then (^𝑛_𝑖=1𝑥_𝑖)^𝑝≤^𝑛_𝑖=1𝑥_𝑖^𝑝. (3.3)

Theorem 3.2. Let 𝑙∈{1,−1} be fixed and let 𝜑∶𝑋×𝑋→ℝ⁺ be a function such that lim_𝑛→∞16^ln𝑥𝜑(2^ln,𝑦2^ln)=0 (3.4) for all 𝑥,𝑦∈𝑋 and ^∞_𝑖=116^𝑖𝑙𝑝𝜑^𝑝𝑦(0,2^𝑙𝑖)<∞ (3.5) for all 𝑦∈𝑋. Suppose that an even function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜑(𝑥,𝑦), (3.6) for all 𝑥,𝑦∈𝑋. Then the limit 𝑄(𝑥)∶=lim_𝑛→∞16^ln𝑥𝑓(2^ln) (3.7) exists for all 𝑥∈𝑋 and 𝑄∶𝑋→𝑌 is a unique quartic function satisfying ‖𝑓(𝑥)−𝑄(𝑥)‖_𝑌≤𝐾[𝜓16^𝑒(𝑥)]^1/𝑝, (3.8) where 𝜓_𝑒(𝑥)∶=^∞_{𝑖=|𝑙+1|/2}16^𝑖𝑙𝑝𝜑^𝑝𝑥(0,2^𝑙𝑖) (3.9) for all 𝑥∈𝑋.

Proof. Let 𝑙=1. By putting 𝑥=0 in (3.6), we get ‖𝑓(2𝑦)−16𝑓(𝑦)‖_𝑌≤𝜑(0,𝑦) (3.10) for all 𝑦∈𝑋. Replacing 𝑦 by 𝑥 in (3.10) yields ‖𝑓(2𝑥)−16𝑓(𝑥)‖_𝑌≤𝜑(0,𝑥) (3.11) for all 𝑥∈𝑋. Let 𝜓(𝑥)=𝜑(0,𝑥) for all 𝑥∈𝑋, then by (3.11), we get ‖𝑓(2𝑥)−16𝑓(𝑥)‖_𝑌≤𝜓(𝑥) (3.12) for all 𝑥∈𝑋. Interchanging 𝑥 with 𝑥/2^𝑛+1 in (3.12), and multiplying by 16^𝑛 it follows that ‖16^𝑛+1𝑥𝑓(2^𝑛+1)−16^𝑛𝑥𝑓(2^𝑛)‖_𝑌≤𝐾16^𝑛𝑥𝜓(2^𝑛+1) (3.13) for all 𝑥∈𝑋 and all nonnegative integers 𝑛. Since 𝑌 is 𝑝-Banach space, then by (3.13) we have ‖16^𝑛+1𝑥𝑓(2^𝑛+1)−16^𝑚𝑥𝑓(2^𝑚)‖^𝑝_𝑌≤^𝑛_𝑖=𝑚‖16^𝑖+1𝑥𝑓(2^𝑖+1)−16^𝑖𝑥𝑓(2^𝑖)‖^𝑝_𝑌≤𝐾^𝑝𝑛_𝑖=𝑚16^𝑖𝑝𝜓^𝑝(𝑥2^𝑖+1) (3.14) for all nonnegative integers 𝑛 and 𝑚 with 𝑛≥𝑚 and all 𝑥∈𝑋. Since 𝜓^𝑝(𝑥)=𝜑^𝑝(0,𝑥) for all 𝑥∈𝑋. Therefore by (3.5) we have ^∞_𝑖=116^𝑖𝑝𝜓^𝑝(𝑥2^𝑖)<∞ (3.15) for all 𝑥∈𝑋. Therefore we conclude from (3.14) and (3.15) that the sequence {16^𝑛𝑓(𝑥/2^𝑛)} is a Cauchy sequence for all 𝑥∈𝑋. Since 𝑌 is complete, it follows that the sequence {16^𝑛𝑓(𝑥/2^𝑛)} converges for all 𝑥∈𝑋. We define the mapping 𝑄∶𝑋→𝑌 by (3.7) for all 𝑥∈𝑋. Letting 𝑚=0 and passing the limit 𝑛→∞ in (3.14), we get ‖𝑓(𝑥)−𝑄(𝑥)‖^𝑝_𝑌≤𝐾^𝑝∞_𝑖=016^𝑖𝑝𝜓^𝑝(𝑥2^𝑖+1𝐾)=^𝑝16^𝑝∞_𝑖=116^𝑖𝑝𝜓^𝑝(𝑥2^𝑖) (3.16) for all 𝑥∈𝑋. Therefore (3.8) follows from (3.9) and (3.16). Now we show that 𝑄 is quartic. It follows from (3.4), (3.6) and (3.7) ‖𝐷_𝑄(𝑥,𝑦)‖_𝑌=lim_𝑛→∞16^𝑛‖𝐷_𝑓(𝑥2^𝑛,𝑦2^𝑛)‖_𝑌≤lim_𝑛→∞16^𝑛𝑥𝜑(2^𝑛,𝑦2^𝑛)=0 (3.17) for all 𝑥,𝑦∈𝑋. Therefore the mapping 𝑄∶𝑋→𝑌 satisfies (1.7). Since 𝑄(0)=0, then by Lemma 2.1 we get that the mapping 𝑄∶𝑋→𝑌 is quartic. To prove the uniqueness of 𝑄, let 𝑇∶𝑋→𝑌 be another quartic mapping satisfies (3.8). Since lim_𝑛→∞16^∞𝑛𝑝_𝑖=116^𝑖𝑝𝜑^𝑝(𝑥2^𝑖+𝑛,𝑦2^𝑖+𝑛)=lim^∞_{𝑛→∞𝑖=𝑛+1}16^𝑖𝑝𝜑^𝑝(𝑥2^𝑖,𝑦2^𝑖)=0 (3.18) for all 𝑦∈𝑋 and all 𝑥∈{0}, then lim_𝑛→∞16^𝑛𝑝𝜓_𝑒(𝑥2^𝑛)=0 (3.19) for all 𝑥∈𝑋. It follows from (3.8), (3.19) ‖𝑄(𝑥)−𝑇(𝑥)‖^𝑝_𝑌=lim_𝑛→∞16^𝑛𝑝𝑥‖𝑓(2^𝑛𝑥)−𝑇(2^𝑛)‖^𝑝_𝑌≤𝐾^𝑝16^𝑝lim_𝑛→∞16^𝑛𝑝𝜓_𝑒(𝑥2^𝑛)=0 (3.20) for all 𝑥∈𝑋. Hence 𝑄=𝑇. For 𝑙=−1, we obtain ‖𝑓(2^𝑛𝑥)16^𝑛−𝑓(𝑥)‖^𝑝_𝑌≤𝐾^𝑝16^𝑝∞_𝑖=0𝜑^𝑝(0,2^𝑖𝑥)16^𝑖𝑝, (3.21) from which one can prove the result by a similar technique.

Corollary 3.3. Let 𝜃,𝑟,𝑠,𝑢,𝑣 be nonnegative real numbers such that 𝑠≠4≠𝑢+𝑣. Suppose that an even function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) (3.22) for all 𝑥,𝑦∈𝑋. Then there exists a unique quartic function 𝑄∶𝑋→𝑌 satisfying ‖𝑓(𝑥)−𝑄(𝑥)‖_𝑌1≤𝐾𝜃{|16^𝑝−2^𝑠𝑝|}^1/𝑝‖𝑥‖^𝑠_𝑋 (3.23) for all 𝑥∈𝑋.

Proof. It follows from Theorem 3.2that𝜑(𝑥,𝑦)∶=𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) for all 𝑥,𝑦∈𝑋.

Theorem 3.4. Let 𝑙∈{1,−1} be fixed and let 𝜑∶𝑋×𝑋→[0,∞) be a function such that lim_𝑛→∞8^ln𝑥𝜑(2^ln,𝑦2^ln)=0 (3.24) for all 𝑥,𝑦∈𝑋 and ^∞_𝑖=18^𝑖𝑙𝑝𝜑^𝑝𝑦(0,2^𝑖𝑙)<∞ (3.25) for all 𝑦∈𝑋. Suppose that an odd function 𝑓∶𝑋→𝑌 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜑(𝑥,𝑦), (3.26) for all 𝑥,𝑦∈𝑋. Then the limit 𝐶(𝑥)∶=lim_𝑛→∞8^ln𝑥𝑓(2^ln) (3.27) exists for all 𝑥∈𝑋 and 𝐶∶𝑋→𝑌 is a unique cubic function satisfying ‖𝑓(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾[𝜙24^𝑜(𝑥)]^1/𝑝 (3.28) for all 𝑥∈𝑋, where 𝜙_𝑜(𝑥)∶=^∞_{𝑖=|𝑙+1|/2}8^𝑖𝑙𝑝𝜑^𝑝𝑥(0,2^𝑖𝑙). (3.29)

Proof. Let 𝑙=1. Setting 𝑥=0 in (3.26), we get ‖3𝑓(2𝑦)−24𝑓(𝑦)‖_𝑌≤𝜑(0,𝑦) (3.30) for all 𝑦∈𝑋. If we replace 𝑦 in (3.30) by 𝑥 and divide both sides of (3.30) by 3, we get ‖𝑓(2𝑥)−8𝑓(𝑥)‖_𝑌≤13𝜑(0,𝑥) (3.31) for all 𝑥∈𝑋. Let 𝜙(𝑥)=(1/3)𝜑(0,𝑥) for all 𝑥∈𝑋, then by (3.31), we get ‖𝑓(2𝑥)−8𝑓(𝑥)‖_𝑌≤𝜙(𝑥) (3.32) for all 𝑥∈𝑋. Multiply (3.32) by 8^𝑛 and replace 𝑥 by 𝑥/2^𝑛+1, we obtain that ‖8^𝑛+1𝑥𝑓(2^𝑛+1)−8^𝑛𝑥𝑓(2^𝑛)‖_𝑌≤𝐾8^𝑛𝑥𝜙(2^𝑛+1) (3.33) for all 𝑥∈𝑋 and all nonnegative integers 𝑛. Since 𝑌 is a 𝑝-Banach space, (3.33) follows that ‖8^𝑛+1𝑥𝑓(2^𝑛+1)−8^𝑚𝑥𝑓(2^𝑚)‖^𝑝_𝑌≤^𝑛_𝑖=𝑚‖8^𝑖+1𝑥𝑓(2^𝑖+1)−8^𝑖𝑥𝑓(2^𝑖)‖^𝑝_𝑌≤𝐾^𝑝𝑛_𝑖=𝑚8^𝑖𝑝𝜙^𝑝(𝑥2^𝑖+1) (3.34) for all nonnegative integers 𝑛 and 𝑚 with 𝑛≥𝑚 and all 𝑥∈𝑋. Since 𝜙^𝑝(𝑥)=(1/3^𝑝)𝜑^𝑝(0,𝑥) for all 𝑥∈𝑋. Therefore it follows from (3.25) that ^∞_𝑖=18^𝑖𝑝𝜙^𝑝(𝑥2^𝑖)<∞ (3.35) for all 𝑥∈𝑋, therefore we conclude from (3.34) and (3.35) that the sequence {8^𝑛𝑓(𝑥/2^𝑛)} is a Cauchy sequence for all 𝑥∈𝑋. Since 𝑌 is complete, the sequence {8^𝑛𝑓(𝑥/2^𝑛)} converges for all 𝑥∈𝑋. So one can define the mapping 𝐶∶𝑋→𝑌 by (3.27) for all 𝑥∈𝑋. Letting 𝑚=0 and passing the limit 𝑛→∞ in (3.34), we get ‖𝑓(𝑥)−𝐶(𝑥)‖^𝑝_𝑌≤𝐾^𝑝∞_𝑖=08^𝑖𝑝𝜙^𝑝(𝑥2^𝑖+1𝐾)=^𝑝8^𝑝∞_𝑖=18^𝑖𝑝𝜙^𝑝(𝑥2^𝑖) (3.36) for all 𝑥∈𝑋. Therefore (3.28) follows from (3.29) and (3.36). Now we show that 𝐶 is cubic. It follows from (3.24), (3.26) and (3.27) ‖𝐷_𝐶(𝑥,𝑦)‖_𝑌=lim_𝑛→∞8^𝑛‖𝐷_𝑓(𝑥2^𝑛,𝑦2^𝑛)‖_𝑌≤lim_𝑛→∞8^𝑛𝑥𝜑(2^𝑛,𝑦2^𝑛)=0 (3.37) for all 𝑥,𝑦∈𝑋. Therefore the mapping 𝐶∶𝑋→𝑌 satisfies (1.7). Since 𝑓 is an odd function, then (3.27) implies that the mapping odd. Therefore by Lemma 2.2 we get that the mapping 𝐶∶𝑋→𝑌 is cubic. The rest of proof is similar to the proof of Theorem 3.2.

Corollary 3.5. Let 𝜃 be a nonnegative real number and 𝑟,𝑠 be real numbers such that 𝑠≠3≠𝑢+𝑣. Suppose that an odd function 𝑓∶𝑋→𝑌 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) (3.38) for all 𝑥,𝑦∈𝑋. Then there exists a unique cubic function 𝐶∶𝑋→𝑌 satisfying ‖𝑓(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾𝜃3{1|8^𝑝−2^𝑠𝑝|}^1/𝑝‖𝑥‖^𝑠_𝑋 (3.39) for all 𝑥∈𝑋.

Proof. It follows from (3.38) and Theorem 3.4that𝜑(𝑥,𝑦)∶=𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) for all 𝑥,𝑦∈𝑋.

Theorem 3.6. Let 𝑙∈{1,−1} be fixed and let 𝜑∶𝑋×𝑋→[0,∞) be a function which satisfies lim_𝑛→∞{(|𝑙|+𝑙2)16^ln𝑥𝜑(2^ln,𝑦2^ln)+(|𝑙|−𝑙2)8^ln𝑥𝜑(2^ln,𝑦2^ln)}=0 (3.40) for all 𝑥,𝑦∈𝑋 and ^∞_{𝑖=|𝑙+1|/2}{(|𝑙|+𝑙2)16^𝑖𝑙𝑝𝜑^𝑝𝑦(0,2^𝑙𝑖)+(|𝑙|−𝑙2)8^𝑖𝑙𝑝𝜑^𝑝𝑦(0,2^𝑙𝑖)}<∞ (3.41) for all 𝑦∈𝑋. Suppose that a function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜑(𝑥,𝑦) (3.42) for all 𝑥,𝑦∈𝑋. Then there exists a unique quartic function 𝑄∶𝑋→𝑌 and a unique cubic function 𝐶∶𝑋→𝑌 satisfying (1.7) and ‖𝑓(𝑥)−𝑄(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾³[𝜓32^𝑒(𝑥)+𝜓_𝑒(−𝑥)]^1/𝑝+𝐾³[𝜙48^𝑜(𝑥)+𝜙_𝑜(−𝑥)]^1/𝑝 (3.43) for all 𝑥∈𝑋, where 𝜓_𝑒(𝑥) and 𝜙_𝑜(𝑥) that have been defined in (3.9) and (3.29), respectively.

Proof. Let 𝑓_𝑒(𝑥)=(1/2)(𝑓(𝑥)+𝑓(−𝑥)) for all 𝑥∈𝑋. Then 𝑓_𝑒(0)=0,𝑓_𝑒(−𝑥)=𝑓_𝑒(𝑥), and ‖𝐷_𝑓𝑒(𝑥,𝑦)‖≤(𝐾/2)[𝜑(𝑥,𝑦)+𝜑(−𝑥,−𝑦)] for all 𝑥,𝑦∈𝑋. Let 𝐾Φ(𝑥,𝑦)=2[𝜑(𝑥,𝑦)+𝜑(−𝑥,−𝑦)] (3.44) for all 𝑥,𝑦∈𝑋. So lim_𝑛→∞16^ln𝑥Φ(2^ln,𝑦2^ln)=0 (3.45) for all 𝑥,𝑦∈𝑋. Since Φ^𝑝𝐾(𝑥,𝑦)≤^𝑝2^𝑝[𝜑^𝑝(𝑥,𝑦)+𝜑^𝑝(−𝑥,−𝑦)] (3.46) for all 𝑥,𝑦∈𝑋, then ^∞_𝑖=116^𝑖𝑙𝑝Φ^𝑝(𝑥2^𝑖𝑙,𝑦2^𝑖𝑙)<∞ (3.47) for all 𝑦∈𝑋 and all 𝑥∈{0}. Hence, in view of Theorem 3.2, there exists a unique quartic function 𝑄∶𝑋→𝑌 satisfying ‖𝑓_𝑒(𝑥)−𝑄(𝑥)‖_𝑌≤𝐾[Ψ16^𝑒(𝑥)]^1/𝑝 (3.48) for all 𝑥∈𝑋, where Ψ_𝑒(𝑥)∶=^∞_𝑖=116^𝑖𝑙𝑝Φ^𝑝𝑥(0,2^𝑖𝑙). (3.49) We have Ψ_𝑒𝐾(𝑥)≤^𝑝2^𝑝[𝜓_𝑒(𝑥)+𝜓_𝑒(−𝑥)] (3.50) for all 𝑥∈𝑋. Therefore it follows from (3.48) that, ‖𝑓_𝑒(𝑥)−𝑄(𝑥)‖_𝑌≤𝐾²[𝜓32^𝑒(𝑥)+𝜓_𝑒(−𝑥)]^1/𝑝 (3.51) for all 𝑥∈𝑋. Let 𝑓_𝑜(𝑥)=(1/2)(𝑓(𝑥)−𝑓(−𝑥)) for all 𝑥∈𝑋. Then 𝑓_𝑜(0)=0,𝑓_𝑜(−𝑥)=−𝑓_𝑜(𝑥), and ‖𝐷_𝑓𝑜(𝑥,𝑦)‖≤Φ(𝑥,𝑦) for all 𝑥,𝑦∈𝑋. From Theorem 3.4, it follows that there exists a unique cubic function 𝐶∶𝑋→𝑌 satisfying ‖𝑓_𝑜(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾[Φ24^𝑜(𝑥)]^1/𝑝 (3.52) for all 𝑥∈𝑋, where Φ_𝑜(𝑥)∶=^∞_𝑖=18^𝑖𝑝Φ^𝑝𝑥(0,2^𝑖). (3.53) Since Φ_𝑜𝐾(𝑥)≤^𝑝2^𝑝[𝜙_𝑜(𝑥)+𝜙_𝑜(−𝑥)] (3.54) for all 𝑥∈𝑋, it follows from (3.52) that, ‖𝑓_𝑜(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾²[𝜙48^𝑜(𝑥)+𝜙_𝑜(−𝑥)]^1/𝑝 (3.55) for all 𝑥∈𝑋. Hence (3.43) follows from (3.51) and (3.55).

Corollary 3.7. Let 𝜃,𝑟,𝑠 be nonnegative real numbers such that 𝑢+𝑣,𝑠∈(4,∞)∪(−∞,3). Suppose that a function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality ‖𝐷_𝑓(𝑥,𝑦)‖_𝑌≤𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) (3.56) for all 𝑥,𝑦∈𝑋. Then there exists a unique quartic function 𝑄∶𝑋→𝑌 and a unique cubic function 𝐶∶𝑋→𝑌 satisfying (1.7) and ‖𝑓(𝑥)−𝑄(𝑥)−𝐶(𝑥)‖_𝑌≤𝐾³𝜃31{3[|16^𝑝−2^𝑠𝑝|]^1/𝑝1+[|8^𝑝−2^𝑠𝑝|]^1/𝑝}‖𝑥‖^𝑠_𝑋 (3.57) for all 𝑥∈𝑋.

Proof. It follows from Theorem 3.6that 𝜑(𝑥,𝑦)∶=𝜃(‖𝑥‖^𝑢_𝑋‖𝑦‖^𝑣_𝑋+‖𝑥‖^𝑟_𝑋+‖𝑦‖^𝑠_𝑋) (3.58) for all 𝑥,𝑦∈𝑋.

Acknowledgment

The second and fourth authors would like to thank the office of gifted students at Semnan University for its financial support.