An AQCQ-functional equation in paranormed spaces

Abstract

In this article, we prove the Hyers-Ulam stability of an additive-quadratic-cubic-quartic functional equation in paranormed spaces.

Mathematics Subject Classification (2010): Primary 39B82; 39B52; 39B72; 46A99.

1. Introduction and preliminaries

The concept of statistical convergence for sequences of real numbers was introduced by Fast [1] and Steinhaus [2] independently and since then several generalizations and applications of this notion have been investigated by various authors (see [3–7]). This notion was defined in normed spaces by Kolk [8].

We recall some basic facts concerning Fré chet spaces.

Definition 1.1. [9] Let X be a vector space. A paranorm P : X → [0, ∞) is a function on X such that

• (1) P(0) = 0;

• (2) P(-x) = P(x);

• (3) P(x + y) ≤ P(x) + P(y) (triangle inequality)

• (4) If {t _n } is a sequence of scalars with t _n → t and {x _n } ⊂ X with P(x _n - x) → 0, then P(t _n x _n - tx) → 0 (continuity of multiplication).

The pair (X, P) is called a paranormed space if P is a paranorm on X.

The paranorm is called total if, in addition, we have

• (5) P(x) = 0 implies x = 0.

A Fréchet space is a total and complete paranormed space.

The stability problem of functional equations originated from a question of Ulam [10] concerning the stability of group homomorphisms. Hyers [11] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [12] for additive mappings and by Rassias [13] for linear mappings by considering an unbounded Cauchy difference. A generalization of Rassias' theorem was obtained by Găvruta [14] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias' approach.

In 1990, Rassias [15] during the 27th International Symposium on Functional Equations asked the question whether such a theorem can also be proved for p ≥ 1. In 1991, Gajda [16] following the same approach as in Rassias [13], gave an affirmative solution to this question for p > 1. It was shown by Gajda [16], as well as by Rassias and Šemrl [17] that one cannot prove a Rassias-type theorem when p = 1 (cf. the books of Czerwik [18], Hyers et al. [19]).

In 1982, Rassias [20] followed the innovative approach of the Rassias' theorem [13] in which he replaced the factor ∥x∥^p+ ∥y∥^pby ∥x∥^p· ∥y∥^qfor p, q ∈ ℝ with p + q ≠ 1.

The functional equation

$$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$$

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [21] for mappings f : X → Y, where X is a normed space and Y is a Banach space. Cholewa [22] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [23] proved the Hyers-Ulam stability of the quadratic functional equation. The stability problems of several functional equations have extensively been investigated by a number of authors and there are many interesting results concerning this problem (see [24–30]).

Jun and Kim [31] considered the following cubic functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=2f\left(x+y\right)+2f\left(x-y\right)+12f\left(x\right).$$

(1.1)

It is easy to show that the function f(x) = x³ satisfies the functional equation (1.1), which is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic mapping.

Lee et al. [32] considered the following quartic functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=4f\left(x+y\right)+4f\left(x-y\right)+24f\left(x\right)-6f\left(y\right).$$

(1.2)

It is easy to show that the function f(x) = x⁴ satisfies the functional equation (1.2), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic mapping.

Throughout this article, assume that (X, P) is a Fré chet space and that (Y, ∥ · ∥) is a Banach space.

In this article, we prove the Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equation

$$f\left(x+2y\right)+f\left(x-2y\right)=4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right)+f\left(2y\right)+f\left(-2y\right)-4f\left(y\right)-4f\left(-y\right)$$

(1.3)

in paranormed spaces.

One can easily show that an odd mapping f : X → Y satisfies (1.3) if and only if the odd mapping f : X → Y is an additive-cubic mapping, i.e.,

$$f\left(x+2y\right)+f\left(x-2y\right)=4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right).$$

It was shown in [[33], Lemma 2.2] that g(x) := f(2x) - 2f(x) and h(x) := f(2x) - 8f(x) are cubic and additive, respectively, and that $f\left(x\right)=\frac{1}{6}g\left(x\right)-\frac{1}{6}h\left(x\right)$.

One can easily show that an even mapping f : X → Y satisfies (1.3) if and only if the even mapping f : X → Y is a quadratic-quartic mapping, i.e.,

$$f\left(x+2y\right)+f\left(x-2y\right)=4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right)+2f\left(2y\right)-8f\left(y\right).$$

It was shown in [[34], Lemma 2.1] that g(x) := f(2x) - 4f(x) and h(x) := f(2x) - 16f(x) are quartic and quadratic, respectively, and that $f\left(x\right)=\frac{1}{12}g\left(x\right)-\frac{1}{12}h\left(x\right)$.

2. Hyers-Ulam stability of the functional equation (1.3): an odd mapping case

For a given mapping f, we define

$$\begin{array}{c}Df\left(x,y\right):=f\left(x+2y\right)+f\left(x-2y\right)-4f\left(x+y\right)-4f\left(x-y\right)+6f\left(x\right)\\ -f\left(2y\right)-f\left(-2y\right)+4f\left(y\right)+4f\left(-y\right).\end{array}$$

In this section, we prove the Hyers-Ulam stability of the functional equation Df(x, y) = 0 in paranormed spaces: an odd mapping case.

Note that P(2x) ≤ 2P(x) for all x ∈ Y.

Theorem 2.1. Let r, θ be positive real numbers with r > 1, and let f : Y → X be an odd mapping such that

$$P\left(Df\left(x,y\right)\right)\le \theta \left({∥x∥}^r+{∥y∥}^r\right)$$

(2.1)

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

$$P\left(f\left(2x\right)-8f\left(x\right)-A\left(x\right)\right)\le \frac{2^r+9}{2^r-2}\theta {∥x∥}^r$$

(2.2)

for all x ∈ Y.

Proof. Letting x = y in (2.1), we get

$$P\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)\le 2\theta {∥y∥}^r$$

(2.3)

for all y ∈ Y.

Replacing x by 2y in (2.1), we get

$$P\left(f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)\right)\le \left(2^p+1\right)\theta {∥y∥}^r$$

(2.4)

for all y ∈ Y.

By (2.3) and (2.4),

$$\begin{array}{c}P\left(f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)\right)\hfill \\ \le P\left(4\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)\right)\hfill \\ \le 4P\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)\right)\hfill \\ \le 8\theta {∥y∥}^r+\left(2^r+1\right)\theta {∥y∥}^r=\left(2^r+9\right)\theta {∥y∥}^r\hfill \end{array}$$

(2.5)

for all y ∈ Y. Replacing y by $\frac{x}{2}$ and letting g(x) := f(2x) - 8f(x) in (2.5), we get

$$P\left(g\left(x\right)-2g\left(\frac{x}{2}\right)\right)\le \frac{2^r+9}{2^r}\theta {∥x∥}^r$$

for all x ∈ Y. Hence

$$P\left(2^{l}g\left(\frac{x}{2^l}\right)-2^{m}g\left(\frac{x}{2^m}\right)\right)\le \sum _{j=l}^{m-1}\frac{\left(2^r+9\right)2^j}{2^{rj+r}}\theta {∥x∥}^r$$

(2.6)

for all nonnegative integers m and l with m > l and all x ∈ Y. It follows from (2.6) that the sequence $\left\{2^{k}g\left(\frac{x}{2^k}\right)\right\}$ is Cauchy for all x ∈ Y. Since X is complete, the sequence $\left\{2^{k}g\left(\frac{x}{2^k}\right)\right\}$ converges. So one can define the mapping A : Y → X by

$$A\left(x\right):=\underset{k\to \infty }{\text{lim}}{2}^{k}g\left(\frac{x}{2^k}\right)$$

for all x ∈ Y.

By (2.1),

$$P\left(DA\left(x,y\right)\right)=\underset{k\to \infty }{\text{lim}}P\left(2^{k}Dg\left(\frac{x}{2^k},\frac{y}{2^k}\right)\right)\le \frac{2^k\theta }{2^{rk}}\left(2^r+8\right)\left({∥x∥}^r+{∥y∥}^r\right)=0$$

for all x, y ∈ Y. So DA(x, y) = 0. Since g : Y → X is odd, A : Y → X is odd. So the mapping A : Y → X is additive. Moreover, letting l = 0 and passing the limit m → ∞ in (2.6), we get (2.2). So there exists an additive mapping A : Y → X satisfying (2.2).

Now, let T : Y → X be another additive mapping satisfying (2.2). Then we have

$$\begin{array}{c}P\left(A\left(x\right)-T\left(x\right)\right)=P\left(2^{p}A\left(\frac{x}{2^q}\right)-2^{q}T\left(\frac{x}{2^q}\right)\right)\hfill \\ \le P\left(2^q\left(A\left(\frac{x}{2^q}\right)-g\left(\frac{x}{2^q}\right)\right)\right)+P\left(2^q\left(T\left(\frac{x}{2^q}\right)-g\left(\frac{x}{2^q}\right)\right)\right)\hfill \\ \le \frac{2\left(2^r+9\right)2^q}{\left(2^r-2\right)2^{rq}}\theta {∥x∥}^r,\hfill \end{array}$$

which tends to zero as q → ∞ for all x ∈ Y. So we can conclude that A(x) = T(x) for all x ∈ Y. This proves the uniqueness of A. Thus the mapping A : Y → X is a unique additive mapping satisfying (2.2).

Theorem 2.2. Let r be a positive real number with r < 1, and let f : X → Y be an odd mapping such that

$$∥Df\left(x,y\right)∥\le P{\left(x\right)}^r+P{\left(y\right)}^r$$

(2.7)

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

$$∥f\left(2x\right)-8f\left(x\right)-A\left(x\right)∥\le \frac{9+2^r}{2-2^r}P{\left(x\right)}^r$$

(2.8)

for all x ∈ X.

Proof. Letting x = y in (2.7), we get

$$∥f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)∥\le 2P{\left(y\right)}^r$$

(2.9)

for all y ∈ X.

Replacing x by 2y in (2.7), we get

$$∥f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)∥\le \left(2^p+1\right)P{\left(y\right)}^r$$

(2.10)

for all y ∈ X.

By (2.9) and (2.10),

$$\begin{array}{c}∥f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)∥\hfill \\ \le ∥4\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)∥+∥f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)∥\hfill \\ \le 4∥f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)∥+∥f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)∥\hfill \\ \le 8P{\left(y\right)}^r+\left(2^r+1\right)P{\left(y\right)}^r=\left(2^r+9\right)P{\left(y\right)}^r\hfill \end{array}$$

(2.11)

for all y ∈ X. Replacing y by x and letting g(x) := f(2x) - 8f(x) in (2.11), we get

$$∥g\left(x\right)-\frac{1}{2}g\left(2x\right)∥\le \frac{2^r+9}{2}P{\left(x\right)}^r$$

for all x ∈ X. Hence

$$∥\frac{1}{2^l}g\left(2^{l}x\right)-\frac{1}{2^m}g\left(2^{m}x\right)∥\le \sum _{j=l}^{m-1}\frac{\left(2^r+9\right)2^{r}j}{2^{j+1}}P{\left(x\right)}^r$$

(2.12)

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (2.12) that the sequence $\left\{\frac{1}{2^k}g\left(2^{k}x\right)\right\}$ is Cauchy for all x ∈ X. Since Y is complete, the sequence $\left\{\frac{1}{2^k}g\left(2^{k}x\right)\right\}$ converges. So one can define the mapping A : X → Y by

$$A\left(x\right):=\underset{k\to \infty }{\text{lim}}\frac{1}{2^k}g\left(2^{k}x\right)$$

for all x ∈ X.

By (2.7),

$$∥DA\left(x,y\right)∥=\underset{k\to \infty }{\text{lim}}∥\frac{1}{2^k}Dg\left(2^{k}x,2^{k}y\right)∥\le \frac{2^{rk}}{2^k}\left(2^r+8\right)\left(P{\left(x\right)}^r+P{\left(y\right)}^r\right)=0$$

for all x, y ∈ X. So DA(x, y) = 0. Since g : X → Y is odd, A : X → Y is odd. So the mapping A : X → Y is additive. Moreover, letting l = 0 and passing the limit m → ∞ in (2.12), we get (2.8). So there exists an additive mapping A : X → Y satisfying (2.8).

Now, let T : X → Y be another additive mapping satisfying (2.8). Then we have

$$\begin{array}{ll}\hfill ∥A\left(x\right)-T\left(x\right)∥& =∥\frac{1}{2^q}A\left(2^{q}x\right)-\frac{1}{2^q}T\left(2^{q}x\right)∥\\ \le ∥\frac{1}{2^q}\left(A\left(2^{q}x\right)-g\left(2^{q}x\right)\right)∥+∥\frac{1}{2^q}\left(T\left(2^{q}x\right)-g\left(2^{q}x\right)\right)∥\\ \le \frac{2\left(9+2^r\right)2^{rq}}{\left(2-2^r\right)2^q}P{\left(x\right)}^r,\end{array}$$

which tends to zero as q → ∞ for all x ∈ X. So we can conclude that A(x) = T(x) for all x ∈ X. This proves the uniqueness of A. Thus the mapping A : X → Y is a unique additive mapping satisfying (2.8).

Theorem 2.3. Let r, θ be positive real numbers with r > 3, and let f : Y → X be an odd mapping satisfying (2.1). Then there exists a unique cubic mapping C : Y → X such that

$$P\left(f\left(2x\right)-2f\left(x\right)-C\left(x\right)\right)\le \frac{2^r+9}{2^r-8}\theta {∥x∥}^r$$

for all x ∈ Y.

Proof. Replacing y by $\frac{x}{2}$ and letting g(x) := f(2x) - 2f(x) in (2.5), we get

$$P\left(g\left(x\right)-8g\left(\frac{x}{2}\right)\right)\le \frac{2^r+9}{2^r}\theta {∥x∥}^r$$

for all x ∈ Y.

The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 2.4. Let r be a positive real number with r < 3, and let f : X → Y be an odd mapping satisfying (2.7). Then there exists a unique cubic mapping C : X → Y such that

$$∥f\left(2x\right)-2f\left(x\right)-C\left(x\right)∥\le \frac{9+2^r}{8-2^r}P{\left(x\right)}^r$$

for all x ∈ X.

Proof. Replacing y by x and letting g(x) := f(2x) - 2f(x) in (2.11), we get

$$∥g\left(x\right)-\frac{1}{8}g\left(2x\right)∥\le \frac{2^r+9}{8}P{\left(x\right)}^r$$

for all x ∈ X.

The rest of the proof is similar to the proof of Theorem 2.2.

3. Hyers-Ulam stability of the functional equation (1.3): an even mapping case

In this section, we prove the Hyers-Ulam stability of the functional equation Df(x, y) = 0 in paranormed spaces: an even mapping case.

Note that P(2x) ≤ 2P(x) for all x ∈ Y.

Theorem 3.1. Let r, θ be positive real numbers with r > 2, and let f : Y → X be an even mapping satisfying f(0) = 0 and (2.1). Then there exists a unique quadratic mapping Q₂ : Y → X such that

$$P\left(f\left(2x\right)-16f\left(x\right)-Q_2\left(x\right)\right)\le \frac{2^r+9}{2^r-4}\theta {∥x∥}^r$$

for all x ∈ Y.

Proof. Letting x = y in (2.1), we get

$$P\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)\le 2\theta {∥y∥}^r$$

(3.1)

for all y ∈ Y.

Replacing x by 2y in (2.1), we get

$$P\left(f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)\right)\le \left(2^r+1\right)\theta {∥y∥}^r$$

(3.2)

for all y ∈ Y.

By (3.1) and (3.2),

$$\begin{array}{c}P\left(f\left(4y\right)-20f\left(2y\right)+64f\left(y\right)\right)\hfill \\ \le P\left(4\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)\right)\hfill \\ \le 4P\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)\right)\hfill \\ \le 8\theta {∥y∥}^r+\left(2^r+1\right)\theta {∥y∥}^r=\left(2^p+9\right)\theta {∥y∥}^r\hfill \end{array}$$

(3.3)

for all y ∈ Y. Replacing y by $\frac{x}{2}$ and letting g(x) := f(2x) - 16f(x) in (3.3), we get

$$P\left(g\left(x\right)-4g\left(\frac{x}{2}\right)\right)\le \frac{2^r+9}{2^r}\theta {∥x∥}^r$$

for all x ∈ Y.

The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 3.2. Let r be a positive real number with r < 2, and let f : X → Y be an even mapping satisfying f(0) = 0 and (2.7). Then there exists a unique quadratic mapping Q₂ : X → Y such that

$$∥f\left(2x\right)-16f\left(x\right)-Q_2\left(x\right)∥\le \frac{9+2^r}{4-2^r}P{\left(x\right)}^r$$

(3.4)

for all x ∈ X.

Proof. Letting x = y in (2.7), we get

$$∥f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)∥\le 2P{\left(y\right)}^r$$

(3.5)

for all y ∈ X.

Replacing x by 2y in (2.7), we get

$$∥f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)∥\le \left(2^r+1\right)P{\left(y\right)}^r$$

(3.6)

for all y ∈ X.

By (3.5) and (3.6),

$$\begin{array}{c}∥f\left(4y\right)-20f\left(2y\right)+64f\left(y\right)∥\hfill \\ \le ∥4\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)∥+∥f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)∥\hfill \\ \le 4∥f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)∥+∥f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)∥\hfill \\ \le 8P{\left(y\right)}^r+\left(2^r+1\right)P{\left(y\right)}^r=\left(2^p+9\right)P{\left(y\right)}^r\hfill \end{array}$$

(3.7)

for all y ∈ X. Replacing y by x and letting g(x) := f(2x) - 16f(x) in (3.7), we get

$$∥g\left(x\right)-\frac{1}{4}g\left(2x\right)∥\le \frac{2^r+9}{4}P{\left(x\right)}^r$$

for all x ∈ X.

The rest of the proof is similar to the proof of Theorem 2.2.

Theorem 3.3. Let r, θ be positive real numbers with r > 4, and let f : Y → X be an even mapping satisfying f(0) = 0 and (2.1). Then there exists a unique quartic mapping Q₄ : Y → X such that

$$P\left(f\left(2x\right)-4f\left(x\right)-Q_4\left(x\right)\right)\le \frac{2^r+9}{2^r-16}\theta {∥x∥}^r$$

for all x ∈ Y.

Proof. Replacing y by $\frac{x}{2}$ and letting g(x) := f(2x) - 4f(x) in (3.3), we get

$$P\left(g\left(x\right)-16g\left(\frac{x}{2}\right)\right)\le \frac{2^r+9}{2^r}\theta {∥x∥}^r$$

for all x ∈ Y.

The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 3.4. Let r be a positive real number with r < 4, and let f : X → Y be an even mapping satisfying f(0) = 0 and (2.7). Then there exists a unique quartic mapping Q₄ : X → Y such that

$$∥f\left(2x\right)-4f\left(x\right)-Q_4\left(x\right)∥\le \frac{9+2^r}{16-2^r}P{\left(x\right)}^r$$

for all x ∈ X.

Proof. Replacing y by x and letting g(x) := f(2x) - 4f(x) in (3.7), we get

$$∥g\left(x\right)-\frac{1}{16}g\left(2x\right)∥\le \frac{2^r+9}{16}P{\left(x\right)}^r$$

for all x ∈ X.

The rest of the proof is similar to the proof of Theorem 2.2.

Let $f_o\left(x\right):=\frac{f\left(x\right)-f\left(-x\right)}{2}$ and $f_e\left(x\right):=\frac{f\left(x\right)+f\left(-x\right)}{2}$. Then f _o is odd and f _e is even. f _o , f _e satisfy the functional equation (1.3). Let g _o (x) := f _o (2x) - 2f _o (x) and h _o (x) := f _o (2x) - 8f _o (x). Then $f_o\left(x\right)=\frac{1}{6}{g}_o\left(x\right)-\frac{1}{6}{h}_o\left(x\right)$. Let g _e (x) := f _e (2x) - 4f _e (x) and h _e (x) := f _e (2x) - 16f _e (x). Then $f_e\left(x\right)=\frac{1}{12}{g}_e\left(x\right)-\frac{1}{12}{h}_e\left(x\right)$. Thus

$$f\left(x\right)=\frac{1}{6}{g}_o\left(x\right)-\frac{1}{6}{h}_o\left(x\right)+\frac{1}{12}{g}_e\left(x\right)-\frac{1}{12}{h}_e\left(x\right).$$

Theorem 3.5. Let r, θ be positive real numbers with r > 4. Let f : Y → X be a mapping satisfying f(0) = 0 and (2.1). Then there exist an additive mapping A : Y → X, a quadratic mapping Q₂ : Y → X, a cubic mapping C : Y → X and a quartic mapping Q₄ : Y → X such that

$$\begin{array}{c}P\left(24f\left(x\right)-4A\left(x\right)-2Q_2\left(x\right)-4C\left(x\right)-2Q_4\left(x\right)\right)\\ \le \left(\frac{4\left(2^r+9\right)}{2^r-2}+\frac{2\left(2^r+9\right)}{2^r-4}+\frac{4\left(2^r+9\right)}{2^r-8}+\frac{2\left(2^r+9\right)}{2^r-16}\right)\theta {∥x∥}^r\end{array}$$

for all x ∈ Y.

Theorem 3.6. Let r be a positive real number with r < 1. Let f : X → Y be a mapping satisfying f(0) = 0 and (2.7). Then there exist an additive mapping A : X → Y, a quadratic mapping Q₂ : X → Y, a cubic mapping C : X → Y and a quartic mapping Q₄ : X → Y such that

$$\begin{array}{c}∥24f\left(x\right)-4A\left(x\right)-2Q_2\left(x\right)-4C\left(x\right)-2Q_4\left(x\right)∥\\ \le \left(\frac{4\left(2^r+9\right)}{2-2^r}+\frac{2\left(2^r+9\right)}{4-2^r}+\frac{4\left(2^r+9\right)}{8-2^r}+\frac{2\left(2^r+9\right)}{16-2^r}\right)P{\left(x\right)}^r\end{array}$$

for all x ∈ X.