Theory and Modern Applications

# An AQCQ-functional equation in paranormed spaces

## Abstract

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  and Steinhaus  independently and since then several generalizations and applications of this notion have been investigated by various authors (see ). This notion was defined in normed spaces by Kolk .

We recall some basic facts concerning Fré chet spaces.

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

1. (1)

P(0) = 0;

2. (2)

P(-x) = P(x);

3. (3)

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

4. (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

1. (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  concerning the stability of group homomorphisms. Hyers  gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki  for additive mappings and by Rassias  for linear mappings by considering an unbounded Cauchy difference. A generalization of Rassias' theorem was obtained by Găvruta  by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias' approach.

In 1990, Rassias  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  following the same approach as in Rassias , gave an affirmative solution to this question for p > 1. It was shown by Gajda , as well as by Rassias and Šemrl  that one cannot prove a Rassias-type theorem when p = 1 (cf. the books of Czerwik , Hyers et al. ).

In 1982, Rassias  followed the innovative approach of the Rassias' theorem  in which he replaced the factor xp+ ypby xp· yqfor p, q 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  for mappings f : XY, where X is a normed space and Y is a Banach space. Cholewa  noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik  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 ).

Jun and Kim  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.  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.

$f\left(x+2y\right)+f\left(x-2y\right)=4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right)+f\left(2y\right)+f\left(-2y\right)-4f\left(y\right)-4f\left(-y\right)$
(1.3)

in paranormed spaces.

One can easily show that an odd mapping f : XY satisfies (1.3) if and only if the odd mapping f : XY 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 [, 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 : XY satisfies (1.3) if and only if the even mapping f : XY 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 [, 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

$\begin{array}{c}Df\left(x,y\right):=f\left(x+2y\right)+f\left(x-2y\right)-4f\left(x+y\right)-4f\left(x-y\right)+6f\left(x\right)\\ \phantom{\rule{1em}{0ex}}-f\left(2y\right)-f\left(-2y\right)+4f\left(y\right)+4f\left(-y\right).\end{array}$

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 : YX 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 : YX 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),

$\begin{array}{c}P\left(f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le P\left(4\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 4P\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 8\theta {∥y∥}^{r}+\left({2}^{r}+1\right)\theta {∥y∥}^{r}=\left({2}^{r}+9\right)\theta {∥y∥}^{r}\hfill \end{array}$
(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 : YX 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 : YX is odd, A : YX is odd. So the mapping A : YX 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 : YX satisfying (2.2).

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

$\begin{array}{c}P\left(A\left(x\right)-T\left(x\right)\right)=P\left({2}^{p}A\left(\frac{x}{{2}^{q}}\right)-{2}^{q}T\left(\frac{x}{{2}^{q}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le P\left({2}^{q}\left(A\left(\frac{x}{{2}^{q}}\right)-g\left(\frac{x}{{2}^{q}}\right)\right)\right)+P\left({2}^{q}\left(T\left(\frac{x}{{2}^{q}}\right)-g\left(\frac{x}{{2}^{q}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{2\left({2}^{r}+9\right){2}^{q}}{\left({2}^{r}-2\right){2}^{rq}}\theta {∥x∥}^{r},\hfill \end{array}$

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 : YX is a unique additive mapping satisfying (2.2).

Theorem 2.2. Let r be a positive real number with r < 1, and let f : XY 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 : XY 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),

$\begin{array}{c}∥f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le ∥4\left(f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)\right)∥+∥f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 4∥f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)∥+∥f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 8P{\left(y\right)}^{r}+\left({2}^{r}+1\right)P{\left(y\right)}^{r}=\left({2}^{r}+9\right)P{\left(y\right)}^{r}\hfill \end{array}$
(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 : XY 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 : XY is odd, A : XY is odd. So the mapping A : XY 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 : XY satisfying (2.8).

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

$\begin{array}{ll}\hfill ∥A\left(x\right)-T\left(x\right)∥& =∥\frac{1}{{2}^{q}}A\left({2}^{q}x\right)-\frac{1}{{2}^{q}}T\left({2}^{q}x\right)∥\phantom{\rule{2em}{0ex}}\\ \le ∥\frac{1}{{2}^{q}}\left(A\left({2}^{q}x\right)-g\left({2}^{q}x\right)\right)∥+∥\frac{1}{{2}^{q}}\left(T\left({2}^{q}x\right)-g\left({2}^{q}x\right)\right)∥\\ \le \frac{2\left(9+{2}^{r}\right){2}^{rq}}{\left(2-{2}^{r}\right){2}^{q}}P{\left(x\right)}^{r},\end{array}$

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 : XY is a unique additive mapping satisfying (2.8).

Theorem 2.3. Let r, θ be positive real numbers with r > 3, and let f : YX be an odd mapping satisfying (2.1). Then there exists a unique cubic mapping C : YX 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 : XY be an odd mapping satisfying (2.7). Then there exists a unique cubic mapping C : XY 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 : YX be an even mapping satisfying f(0) = 0 and (2.1). Then there exists a unique quadratic mapping Q2 : YX 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),

$\begin{array}{c}P\left(f\left(4y\right)-20f\left(2y\right)+64f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le P\left(4\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 4P\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)+P\left(f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 8\theta {∥y∥}^{r}+\left({2}^{r}+1\right)\theta {∥y∥}^{r}=\left({2}^{p}+9\right)\theta {∥y∥}^{r}\hfill \end{array}$
(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 : XY be an even mapping satisfying f(0) = 0 and (2.7). Then there exists a unique quadratic mapping Q2 : XY 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),

$\begin{array}{c}∥f\left(4y\right)-20f\left(2y\right)+64f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le ∥4\left(f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)\right)∥+∥f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 4∥f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)∥+∥f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)∥\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 8P{\left(y\right)}^{r}+\left({2}^{r}+1\right)P{\left(y\right)}^{r}=\left({2}^{p}+9\right)P{\left(y\right)}^{r}\hfill \end{array}$
(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 : YX be an even mapping satisfying f(0) = 0 and (2.1). Then there exists a unique quartic mapping Q4 : YX 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 : XY be an even mapping satisfying f(0) = 0 and (2.7). Then there exists a unique quartic mapping Q4 : XY 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 : YX be a mapping satisfying f(0) = 0 and (2.1). Then there exist an additive mapping A : YX, a quadratic mapping Q2 : YX, a cubic mapping C : YX and a quartic mapping Q4 : YX such that

$\begin{array}{c}P\left(24f\left(x\right)-4A\left(x\right)-2{Q}_{2}\left(x\right)-4C\left(x\right)-2{Q}_{4}\left(x\right)\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \left(\frac{4\left({2}^{r}+9\right)}{{2}^{r}-2}+\frac{2\left({2}^{r}+9\right)}{{2}^{r}-4}+\frac{4\left({2}^{r}+9\right)}{{2}^{r}-8}+\frac{2\left({2}^{r}+9\right)}{{2}^{r}-16}\right)\theta {∥x∥}^{r}\end{array}$

for all x Y.

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

$\begin{array}{c}∥24f\left(x\right)-4A\left(x\right)-2{Q}_{2}\left(x\right)-4C\left(x\right)-2{Q}_{4}\left(x\right)∥\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \left(\frac{4\left({2}^{r}+9\right)}{2-{2}^{r}}+\frac{2\left({2}^{r}+9\right)}{4-{2}^{r}}+\frac{4\left({2}^{r}+9\right)}{8-{2}^{r}}+\frac{2\left({2}^{r}+9\right)}{16-{2}^{r}}\right)P{\left(x\right)}^{r}\end{array}$

for all x X.

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## Acknowledgements

This study was supported by the Daejin University Research Grants in 2012.

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Correspondence to Jung Rye Lee.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

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Park, C., Lee, J.R. An AQCQ-functional equation in paranormed spaces. Adv Differ Equ 2012, 63 (2012). https://doi.org/10.1186/1687-1847-2012-63 