On the generalized Hyers-Ulam-Rassias stability problem of radical functional equations

Abstract

In this paper, the generalized Hyers-Ulam-Rassias stability problem of radical quadratic and radical quartic functional equations in quasi-β-Banach spaces and then the stability by using subadditive and subquadratic functions for radical functional equations in $(\beta ,p)$-Banach spaces are given.

MSC:39B82, 39B52, 46H25.

1 Introduction

In 1960, the stability problem of functional equations originated from the question of Ulam [1, 2] concerning the stability of group homomorphisms. The famous Ulam stability problem was partially solved by Hyers [3] in Banach spaces. Later, Aoki [4] and Bourgin [5] considered the stability problem with unbounded Cauchy differences. Rassias [6–9] provided a generalization of Hyers’ theorem by proving the existence of unique linear mappings near approximate additive mappings. On the other hand, Rassias [10, 11] considered the Cauchy difference controlled by a product of different powers of norm. The above results have been generalized by Forti [12] and Gǎvruta [13], who permitted the Cauchy difference to become arbitrary unbounded. Gajda and Ger [14] showed that one can get analogous stability results for subadditive multifunctions. Gruber [15] remarked that Ulam’s problem is of particular interest in probability theory and in the case of functional equations of different types. Recently, Baktash et al. [16], Cho et al. [17–20], Gordji et al. [21–24], Lee et al. [25, 26], Najati et al. [27, 28], Park et al. [29], Saadati et al. [30] and Savadkouhi et al. [31] have studied and generalized several stability problems of a large variety of functional equations.

The most famous functional equation is the Cauchy equation

$$f(x+y)=f(x)+f(y),$$

any solution of which is called additive. It is easy to see that the function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=c{x}^2$, where c is an arbitrary constant, is a solution of the functional equation

$$f(x+y)+f(x-y)=2f(x)+2f(y).$$

(1.1)

Thus it is natural that each equation is said to be a quadratic functional equation. In particular, every solution of the quadratic equation (1.1) is said to be a quadratic function.

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

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

(1.2)

The functional equation (1.2) clearly has $f(x)=c{x}^4$ as a solution when f is a real valued function of a real variable. So, it is said to be a quartic functional equation.

Before we present our results, we introduce here some basic facts concerning quasi-β-normed space and preliminary results. Let β be a real number with $0<\beta \le 1$, and $\mathbb{K}$ be either $\mathbb{R}$ or $\mathbb{C}$. Let X be a linear space over $\mathbb{C}$. A quasi-β-norm $\parallel \cdot \parallel$ is a real-valued function on X satisfying the following:

• (1) $\parallel x\parallel \ge 0$ for all $x\in X$ and $\parallel x\parallel =0$ if and only if $x=0$;

• (2) $\parallel \lambda x\parallel ={|\lambda |}^{\beta }\cdot \parallel x\parallel$ for all and $x\in X$;

• (3) There is a constant $K\ge 1$ such that $\parallel x+y\parallel \le K(\parallel x\parallel +\parallel y\parallel )$ for all $x,y\in X$.

Then $(X,\parallel \cdot \parallel )$ is called a quasi-β-normed space if $\parallel \cdot \parallel$ is a quasi-β-norm on X. The smallest possible K is called the module of concavity of $\parallel \cdot \parallel$. A quasi-β-Banach space ia a complete quasi-β-normed space.

A quasi-β-norm $\parallel \cdot \parallel$ is called a $(\beta ,p)$-norm $(0<p\le 1)$ if ${\parallel x+y\parallel }^p\le {\parallel x\parallel }^p+{\parallel y\parallel }^p$ for all $x,y\in X$. In this case, a quasi-β-Banach space is called a $(\beta ,p)$-Banach space.

In [33], Tabor investigated a version of the Hyers-Rassias-Gajda theorem in quasi-Banach spaces. For further details on quasi-β normed spaces and $(\beta ,p)$-Banach space, we refer to the papers [34–37].

In this paper, we consider the following functional equations:

(1.3)

(1.4)

and discuss the generalized Hyers-Ulam-Rassias stability problem in quasi-β-normed spaces and then the stability by using subadditive and subquadratic functions for the functional equations (1.3) and (1.4) in $(\beta ,p)$-Banach spaces.

2 Stability of radical functional equations

Using an idea of Gǎvruta [13], we prove the generalized stability of (1.3) and (1.4) in the spirit of Ulam, Hyers and Rassias.

In [38], Khodaei et al. proved the following result:

Lemma 2.1 Let X be a real linear space and $f:\mathbb{R}\to X$ be a function. Then we have the following:

• (1) If f satisfies the functional equation (1.3), then f is quadratic.

• (2) If f satisfies the functional equation (1.4), then f is quartic.

Proof We will only prove (2). Letting $x=y=0$ in (1.4), we get $f(0)=0$. Setting $x=-x$ in (1.4), we have

$$f\left(\sqrt{x^2+y^2}\right)+f\left(\sqrt{|x^2-y^2|}\right)=2f(-x)+2f(y)$$

(2.1)

for all $x,y\in \mathbb{R}$. If we compare (1.4) with (2.1), we obtain that $f(-x)=f(x)$ for all $x\in \mathbb{R}$. Letting $y=x$ in (1.4) and then using $f(0)=0$ and the evenness of f, we have $f(\sqrt{2}x)=4f(x)$ for all $x\in \mathbb{R}$. Putting $y=\sqrt{2}x$ in (1.4) and using $f(\sqrt{2}x)=4f(x)$, we get $f(\sqrt{3}x)=9f(x)$ for all $x\in \mathbb{R}$. By induction, we lead to $f(\sqrt{n}x)=n^{2}f(x)$ for all $x\in \mathbb{R}$ and $n\in {\mathbb{Z}}^{+}$. We obtain $f(\frac{x}{\sqrt{n}})=\frac{1}{n^2}f(x)$, and so $f(\sqrt{\frac{m}{n}}x)=\frac{m^2}{n^2}f(x)$ for all $x\in \mathbb{R}$ and $m,n\in {\mathbb{Z}}^{+}$. So, we have

$$f(\sqrt{r}x)=r^{2}f(x)$$

(2.2)

for all $x\in \mathbb{R}$ and all $r\in {\mathbb{Q}}^{+}$. Replacing x and y by $x+y$ and $x-y$ in (1.4) respectively, we obtain

$$f\left(\sqrt{2x^2+2y^2}\right)+f\left(\sqrt{|4xy|}\right)=2f(x+y)+2f(x-y)$$

(2.3)

for all $x,y\in \mathbb{R}$. Setting $y=\sqrt{x^2+y^2}$ in (1.4) and using the evenness of f, we get

$$f\left(\sqrt{2x^2+y^2}\right)-2f\left(\sqrt{x^2+y^2}\right)=2f(x)-f(y)$$

(2.4)

for all $x,y\in \mathbb{R}$. Putting $x=2x$ in (2.3) and using (2.2), we get

$$2f\left(\sqrt{4x^2+y^2}\right)+32f\left(\sqrt{|xy|}\right)=f(2x+y)+f(2x-y)$$

(2.5)

for all $x,y\in \mathbb{R}$. Setting $x=\sqrt{2}x$ in (1.4) and using (2.2), we get

$$f\left(\sqrt{4x^2+y^2}\right)-2f\left(\sqrt{2x^2+y^2}\right)=8f(x)-f(y)$$

(2.6)

for all $x,y\in \mathbb{R}$. It follows from (2.2), (2.3), (2.4) and (2.6) that

$$f\left(\sqrt{4x^2+y^2}\right)+16f\left(\sqrt{|xy|}\right)=2f(x+y)+2f(x-y)+12f(x)-3f(y)$$

(2.7)

for all $x,y\in \mathbb{R}$. So it follows from (2.5) and (2.7) that f satisfies (1.2). Therefore, f is quartic. This completes the proof. □

Let $\mathcal{X}$ be a quasi-β-Banach space with the quasi-β-norm ${\parallel \cdot \parallel }_{\mathcal{X}}$ and K be the modulus of concavity of ${\parallel \cdot \parallel }_{\mathcal{X}}$. Let $\varphi ,\psi :{\mathbb{R}}^2\to {\mathbb{R}}^{+}\cup \{0\}$ be functions. A function $f:\mathbb{R}\to \mathcal{X}$ is said to be ϕ-approximatively radical quadratic if

$${\parallel f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y)\parallel }_{\mathcal{X}}\le \varphi (x,y)$$

(2.8)

and a function $f:\mathbb{R}\to \mathcal{X}$ is said to be ψ-approximatively radical quartic if

$${\parallel f\left(\sqrt{x^2+y^2}\right)+f\left(\sqrt{|x^2-y^2|}\right)-2f(x)-2f(y)\parallel }_{\mathcal{X}}\le \psi (x,y)$$

(2.9)

for all $x,y\in \mathbb{R}$.

Theorem 2.2 Let $f:\mathbb{R}\to \mathcal{X}$ be a ϕ-approximatively radical quadratic function with $f(0)=0$. If a function $\varphi :{\mathbb{R}}^2\to {\mathbb{R}}^{+}\cup \{0\}$ satisfies

$$\hat{\varphi }(x):=\sum _{j=0}^{\mathrm{\infty }}{\left(\frac{K}{2^{\beta }}\right)}^j(\varphi (2^{\frac{j}{2}}x,2^{\frac{j}{2}}x)+\varphi (2^{\frac{j+1}{2}}x,0))<\mathrm{\infty }$$

and ${lim}_{n\to \mathrm{\infty }}\frac{1}{2^{\beta n}}\varphi (2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)=0$ for all $x,y\in \mathbb{R}$, then the limit $\mathcal{F}(x):={lim}_{n\to \mathrm{\infty }}\frac{1}{2^n}f(2^{\frac{n}{2}}x)$ exists for all $x\in \mathbb{R}$ and a function $\mathcal{F}:\mathbb{R}\to \mathcal{X}$ is unique quadratic satisfying the functional equation (1.3) and the inequality

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{X}}\le \frac{K}{2^{\beta }}\hat{\varphi }(x)$$

(2.10)

for all $x\in \mathbb{R}$.

Proof Replacing x and y by $\frac{x+y}{\sqrt{2}}$ and $\frac{x-y}{\sqrt{2}}$ in (2.8) respectively, we get

$${\parallel f\left(\sqrt{x^2+y^2}\right)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right)\parallel }_{\mathcal{X}}\le \varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})$$

(2.11)

for all $x,y\in \mathbb{R}$. It follows from (2.8) and (2.11) that

$${\parallel f(x)+f(y)-f\left(\frac{x+y}{\sqrt{2}}\right)-f\left(\frac{x-y}{\sqrt{2}}\right)\parallel }_{\mathcal{X}}\le K(\varphi (x,y)+\varphi (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}))$$

(2.12)

for all $x,y\in \mathbb{R}$. Setting $y=x$ in (2.12), it follows from $f(0)=0$ that

$${\parallel 2f(x)-f(\sqrt{2}x)\parallel }_{\mathcal{X}}\le K(\varphi (x,x)+\varphi (\sqrt{2}x,0))$$

(2.13)

and so

$${\parallel f(x)-\frac{1}{2}f(\sqrt{2}x)\parallel }_{\mathcal{X}}\le \frac{K}{2^{\beta }}(\varphi (x,x)+\varphi (\sqrt{2}x,0))$$

(2.14)

for all $x\in \mathbb{R}$. For any integers m, k with $m>k\ge 0$,

$${\parallel \frac{1}{2^k}f\left(2^{\frac{k}{2}}x\right)-\frac{1}{2^m}f\left(2^{\frac{m}{2}}x\right)\parallel }_{\mathcal{X}}\le \frac{K}{2^{\beta }}\sum _{j=k}^{m-1}{\left(\frac{K}{2^{\beta }}\right)}^j(\varphi (2^{\frac{j}{2}}x,2^{\frac{j}{2}}x)+\varphi (2^{\frac{j+1}{2}}x,0))$$

(2.15)

for all $x\in \mathbb{R}$. Then a sequence $\{\frac{1}{2^n}f(2^{\frac{n}{2}}x)\}$ is a Cauchy sequence in a quasi-β-Banach space $\mathcal{X}$ and so it converges. We can define a function $\mathcal{F}:\mathbb{R}\to \mathcal{X}$ by $\mathcal{F}(x):={lim}_{n\to \mathrm{\infty }}\frac{1}{2^n}f(2^{\frac{n}{2}}x)$ for all $x\in \mathbb{R}$. From (2.8), the following inequality holds:

for all $x,y\in \mathbb{R}$. Then $\mathcal{F}(\sqrt{x^2+y^2})-\mathcal{F}(x)-\mathcal{F}(y)=0$ and, by Lemma 2.1, $\mathcal{F}$ is a quadratic function. Taking the limit $m\to \mathrm{\infty }$ in (2.15) with $k=0$, we find that a function $\mathcal{F}$ satisfies (2.10) near the approximate function f of the functional equation (1.3).

Next, we assume that there exists another quadratic function $\mathcal{G}:\mathbb{R}\to \mathcal{X}$ which satisfies the functional equation (1.3) and (2.10). Since $\mathcal{G}$ satisfies (1.3), it easy to show that $\mathcal{G}(2^{\frac{n}{2}}x)=2^n\mathcal{G}(x)$ for all $x\in \mathbb{R}$ and $n\ge 1$. Then we have

$$\begin{array}{rcl}\parallel \mathcal{F}(x)-\mathcal{G}(x)\parallel & =& \parallel \frac{1}{2^n}\mathcal{F}\left(2^{\frac{n}{2}}x\right)-\frac{1}{2^n}\mathcal{G}\left(2^{\frac{n}{2}}x\right)\parallel \\ \le & \frac{K}{2^{\beta n}}(\parallel \mathcal{F}\left(2^{\frac{n}{2}}x\right)-f\left(2^{\frac{n}{2}}x\right)\parallel +\parallel f\left(2^{\frac{n}{2}}x\right)-\mathcal{G}\left(2^{\frac{n}{2}}x\right)\parallel )\\ \le & \frac{K^{2-n}}{2^{\beta -1}}\sum _{k=n}^{\mathrm{\infty }}{\left(\frac{K}{2^{\beta }}\right)}^k(\varphi (2^{\frac{k}{2}}x,2^{\frac{k}{2}}x)+\varphi (2^{\frac{k+1}{2}}x,0))\end{array}$$

for all $x\in \mathbb{R}$. Letting $n\to \mathrm{\infty }$, we establish $\mathcal{F}(x)=\mathcal{G}(x)$ for all $x\in \mathbb{R}$. This completes the proof. □

Theorem 2.3 Let $f:\mathbb{R}\to \mathcal{X}$ be a ϕ-approximatively radical quadratic function. If a function $\varphi :{\mathbb{R}}^2\to {\mathbb{R}}^{+}\cup \{0\}$ satisfies

$$\hat{\varphi }(x):=\sum _{j=1}^{\mathrm{\infty }}{\left(K{2}^{\beta }\right)}^j(\varphi (2^{-\frac{j}{2}}x,2^{-\frac{j}{2}}x)+\varphi (2^{\frac{1-j}{2}}x,0))<\mathrm{\infty }$$

and ${lim}_{n\to \mathrm{\infty }}{2}^{\beta n}\varphi (2^{-\frac{n}{2}}x,2^{-\frac{n}{2}}y)=0$ for all $x,y\in \mathbb{R}$, then the limit $\mathcal{F}(x):={lim}_{n\to \mathrm{\infty }}{2}^{n}f(2^{-\frac{n}{2}}x)$ exists for all $x\in \mathbb{R}$ and a function $\mathcal{F}:\mathbb{R}\to \mathcal{X}$ is unique quadratic satisfying the functional equation (1.3) and the inequality

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{X}}\le \frac{K}{2^{\beta }}\hat{\varphi }(x)$$

for all $x\in \mathbb{R}$.

Proof If x is replaced by $2^{-\frac{1}{2}}x$ in the inequality (2.13), then the proof follows from the proof of Theorem 2.2. □

Corollary 2.4 Let $r,s\in {\mathbb{R}}^{+}\cup \{0\}$ and $\epsilon \ge 0$. If a function $f:\mathbb{R}\to \mathcal{X}$ satisfies the following inequality:

$${\parallel f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y)\parallel }_{\mathcal{X}}\le \{\begin{array}{c}\epsilon ;\hfill \\ \epsilon {|x|}^r{|y|}^s\hfill \end{array}$$

for all $x,y\in \mathbb{R}$, then there exists a unique quadratic function $\mathcal{F}:\mathbb{R}\to \mathcal{X}$ satisfying the functional equation (1.3) and the following inequality:

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{X}}\le \{\begin{array}{cc}\frac{2\epsilon K}{2^{\beta }-K},\hfill & K<2^{\beta };\hfill \\ \frac{\epsilon K{|x|}^{r+s}}{2^{\beta }-K\sqrt{2^{r+s}}},\hfill & r+s<2\beta -2{log}_{2}K\hfill \end{array}$$

for all $x\in \mathbb{R}$.

Corollary 2.5 Let $r,s\in {\mathbb{R}}^{+}\cup \{0\}$ and $\epsilon \ge 0$. If a function $f:\mathbb{R}\to \mathcal{X}$ satisfies the following inequality:

$${\parallel f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y)\parallel }_{\mathcal{X}}\le \epsilon ({|x|}^r+{|y|}^s)$$

for all $x,y\in \mathbb{R}$, then there exists a unique quadratic function $\mathcal{F}:\mathbb{R}\to \mathcal{X}$ satisfying the functional equation (1.3) and the following inequality:

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{X}}\le \epsilon K(\frac{(1+\sqrt{2^r}){|x|}^r}{2^{\beta }-K\sqrt{2^r}}+\frac{{|x|}^s}{2^{\beta }-K\sqrt{2^s}}),r,s<2\beta -2{log}_{2}K$$

for all $x\in \mathbb{R}$.

Now, we give an example to illustrate that the functional equation (1.3) is not stable for $r=2=s$ in a quasi-1-Banach space with $K=1$.

Example 2.6 Let be defined by

$$\varphi (x):=\{\begin{array}{cc}{x}^2\hfill & \mathit{\text{for}}|x|<1;\hfill \\ 1\hfill & \mathit{\text{for}}|x|\ge 1.\hfill \end{array}$$

Consider the function by the formula

$$f(x):=\sum _{m=0}^{\mathrm{\infty }}\frac{1}{4^m}\varphi \left(2^{m}x\right)$$

for all . It is clear that f is bounded by $\frac{4}{3}$ on . We prove that

$$|f\left(\sqrt{x^2+y^2}\right)-f(x)-f(y)|\le 16({|x|}^2+{|y|}^2)$$

(2.16)

for all . To see this, if ${|x|}^2+{|y|}^2=0$ or ${|x|}^2+{|y|}^2\ge \frac{1}{4}$, then we have (2.16). Now suppose that $0<{|x|}^2+{|y|}^2<\frac{1}{4}$. Then there exists a positive integer k such that

$$\frac{1}{4^{k+1}}<{|x|}^2+{|y|}^2<\frac{1}{4^k},$$

(2.17)

so that $2^m|x|$, $2^m|y|$, $2^m\sqrt{x^2+y^2}\in (-1,1)$ for all $m=0,1,\dots ,k-1$. It follows from the definition of f and (2.17) that

for all with ${|x|}^2+{|y|}^2<\frac{1}{4}$. Thus, the condition (2.16) holds true.

Next, we claim that the quadratic equation (1.3) is not stable for $r=2=s$. Assume that there exist a quadratic function and a positive constant μ such that $|f(x)-\mathcal{F}(x)|\le \mu {|x|}^2$ for all . Then there exists a constant such that $\mathcal{F}(x)=c{x}^2$ for all . So we have

$$|f(x)|\le (\mu +|c|){|x|}^2$$

(2.18)

for all . But, we can choose a $\lambda \in {\mathbb{Z}}^{+}$ with $\lambda >\mu +|c|$. If $x\in (0,2^{-\lambda })$, then $2^{j}x\in (0,1)$ for all $j=0,1,\dots ,\lambda -1$. For this x, we obtain

$$|f(x)|=\left|\sum _{j=0}^{\mathrm{\infty }}\frac{\varphi (2^{j}x)}{4^j}\right|\ge \sum _{j=0}^{\lambda -1}\frac{\varphi (2^{j}x)}{4^j}=\lambda x^2>(\mu +|c|){|x|}^2,$$

which contradicts (2.18). Therefore, the quadratic equation (1.3) is not stable for $r=2=s$ in Corollary 2.5.

Theorem 2.7 Let $f:\mathbb{R}\to \mathcal{X}$ be a ψ-approximatively radical quartic function with $f(0)=0$. If the function $\psi :{\mathbb{R}}^2\to {\mathbb{R}}^{+}\cup \{0\}$ satisfies

$$\hat{\psi }(x):=\sum _{j=0}^{\mathrm{\infty }}{\left(\frac{K}{4^{\beta }}\right)}^j(\psi (2^{\frac{j}{2}}x,2^{\frac{j}{2}}x)+\frac{1}{2}\psi (2^{\frac{j+1}{2}}x,0))<\mathrm{\infty }$$

and ${lim}_{n\to \mathrm{\infty }}\frac{1}{4^{\beta n}}\psi (2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)=0$ for all $x,y\in \mathbb{R}$, then the limit $\mathcal{H}(x):={lim}_{n\to \mathrm{\infty }}\frac{1}{4^n}f(2^{\frac{n}{2}}x)$ for all $x\in \mathbb{R}$ exists and a function $\mathcal{H}:\mathbb{R}\to \mathcal{X}$ is unique quartic satisfying the functional equation (1.4) and the inequality

$${\parallel f(x)-\mathcal{H}(x)\parallel }_{\mathcal{X}}\le \frac{K^2}{2^{3\beta -1}}\hat{\psi }(x)$$

(2.19)

for all $x\in \mathbb{R}$.

Proof Replacing x and y by $\sqrt{2}x$ and $\sqrt{2}y$ in (2.9) respectively, we have

$${\parallel f\left(\sqrt{2x^2+2y^2}\right)+f\left(\sqrt{|2x^2-2y^2|}\right)-2f(\sqrt{2}x)-2f(\sqrt{2}y)\parallel }_{\mathcal{X}}\le \psi (\sqrt{2}x,\sqrt{2}y)$$

(2.20)

for all $x,y\in \mathbb{R}$. Again, replacing x and y by $x+y$ and $x-y$ in (2.9) respectively, we get

$${\parallel f\left(\sqrt{2x^2+2y^2}\right)+f\left(\sqrt{|4xy|}\right)-2f(x+y)-2f(x-y)\parallel }_{\mathcal{X}}\le \psi (x+y,x-y)$$

(2.21)

for all $x,y\in \mathbb{R}$. It follows from (2.20) and (2.21) that

(2.22)

for all $x,y\in \mathbb{R}$. Setting $y=0$ in (2.22) and $f(0)=0$, we have

$${\parallel 2f(\sqrt{2}x)-f\left(\sqrt{2x^2}\right)-4f(x)\parallel }_{\mathcal{X}}\le K(\psi (x,x)+\psi (\sqrt{2}x,0))$$

(2.23)

for all $x\in \mathbb{R}$. Setting $y=x$ in (2.9), we have

$${\parallel f\left(\sqrt{2x^2}\right)-4f(x)\parallel }_{\mathcal{X}}\le \psi (x,x)$$

(2.24)

for all $x\in \mathbb{R}$. It follows from (2.23) and (2.24) that

$${\parallel 2f(\sqrt{2}x)-8f(x)\parallel }_{\mathcal{X}}\le K^2(2\psi (x,x)+\psi (\sqrt{2}x,0))$$

(2.25)

for all $x\in \mathbb{R}$. Since $f(0)=0$, it follows from (2.24) and (2.25) that

$${\parallel f(x)-\frac{1}{4}f(\sqrt{2}x)\parallel }_{\mathcal{X}}\le \frac{K^2}{2^{3\beta -1}}(\psi (x,x)+\frac{1}{2}\psi (\sqrt{2}x,0))$$

(2.26)

for all $x\in \mathbb{R}$. Hence we have

(2.27)

for all $x\in \mathbb{R}$ and $m>k\ge 0$. Then a sequence $\{\frac{1}{4^n}f(2^{\frac{n}{2}}x)\}$ is a Cauchy sequence in $\mathcal{X}$, and so it converges to a point in $\mathcal{X}$. We can define a function $\mathcal{H}:\mathbb{R}\to \mathcal{X}$ by $\mathcal{H}(x):={lim}_{n\to \mathrm{\infty }}\frac{1}{4^n}f(2^{\frac{n}{2}}x)$ for all $x\in \mathbb{R}$. From (2.9), the following inequality holds:

$${\parallel \mathcal{H}\left(\sqrt{x^2+y^2}\right)+\mathcal{H}\left(\sqrt{|x^2-y^2|}\right)-2\mathcal{H}(x)-2\mathcal{H}(y)\parallel }_{\mathcal{X}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{1}{4^{\beta n}}\psi (2^{\frac{n}{2}}x,2^{\frac{n}{2}}y)=0$$

for all $x,y\in \mathbb{R}$. Then, we have $\mathcal{H}(\sqrt{x^2+y^2})+\mathcal{H}(\sqrt{|x^2-y^2|})-2\mathcal{H}(x)-2\mathcal{H}(y)=0$, and by Lemma 2.1, $\mathcal{H}$ is quartic. Taking the limit $m\to \mathrm{\infty }$ in (2.27) with $k=0$, we obtain that a function $\mathcal{H}$ satisfies (2.19) near the approximate function f of the functional equation (1.4). The remaining assertion is similar to the corresponding part of Theorem 2.2. This completes the proof. □

Corollary 2.8 If there exist $r,s,t\in {\mathbb{R}}^{+}\cup \{0\}$ and $\epsilon \ge 0$; if a function $f:\mathbb{R}\to \mathcal{X}$ satisfies the inequality

$${\parallel f\left(\sqrt{x^2+y^2}\right)+f\left(\sqrt{|x^2-y^2|}\right)-2f(x)-2f(y)\parallel }_{\mathcal{X}}\le \{\begin{array}{c}\epsilon ;\hfill \\ \epsilon {|x|}^r{|y|}^s;\hfill \\ \epsilon ({|x|}^r+{|y|}^s)\hfill \end{array}$$

for all $x,y\in \mathbb{R}$, then there exists a unique quartic function $\mathcal{H}:\mathbb{R}\to \mathcal{X}$ which satisfies (1.4) and the inequality

$${\parallel f(x)-\mathcal{H}(x)\parallel }_{\mathcal{X}}\le \{\begin{array}{cc}\frac{3\epsilon K^2}{2^{3\beta }-2^{\beta }K},\hfill & K<4^{\beta };\hfill \\ \frac{\epsilon K^2{|x|}^{r+s}}{2^{2\beta }-\sqrt{2^{r+s}}},\hfill & r+s<4\beta -2{log}_{2}K;\hfill \\ \frac{\epsilon K^2}{2^{\beta -1}}(\frac{2+\sqrt{2^{r-2}}{|x|}^r}{4^{\beta }-K\sqrt{2^r}}+\frac{{|x|}^s}{4^{\beta }-K\sqrt{2^s}}),\hfill & r,s<4\beta -2{log}_{2}K\hfill \end{array}$$

for all $x\in \mathbb{R}$.

We recall that a subadditive function is a function $\varphi :A\to B$ having a domain A and a codomain $(B,\le )$ that are both closed under addition with the following property:

$$\varphi (x+y)\le \varphi (x)+\varphi (y)$$

for all $x,y\in A$. Also, a subquadratic function is a function $\psi :A\to B$ with $\psi (0)=0$ and the following property:

$$\psi (x+y)+\psi (x-y)\le 2\psi (x)+2\psi (y)$$

for all $x,y\in A$.

Let $\ell \in \{-1,1\}$ be fixed. If there exists a constant L with $0<L<1$ such that a function $\varphi :A\to B$ satisfies

$$\ell \varphi (x+y)\le \ell L^{\ell }(\varphi (x)+\varphi (y))$$

for all $x,y\in A$, then we say that ϕ is contractively subadditive if $\ell =1$ and ϕ is expansively superadditive if $\ell =-1$. It follows by the last inequality that ϕ satisfies the following properties:

$$\varphi \left(2^{\ell }x\right)\le 2^{\ell }L\varphi (x),\varphi \left(2^{\ell k}x\right)\le {\left(2^{\ell }L\right)}^k\varphi (x)$$

for all $x\in A$ and $k\ge 1$.

Similarly, if there exists a constant L with $0<L<1$ such that a function $\psi :A\to B$ with $\psi (0)=0$ satisfies

$$\ell \psi (x+y)+\ell \psi (x-y)\le 2\ell L^{\ell }(\psi (x)+\psi (y))$$

for all $x,y\in A$, then we say that ψ is contractively subquadratic if $\ell =1$ and ψ is expansively superquadratic if $\ell =-1$. It follows from the last inequality that ψ satisfies the following properties:

$$\psi \left(2^{\ell }x\right)\le 4^{\ell }L\psi (x),\psi \left(2^{\ell k}x\right)\le {\left(4^{\ell }L\right)}^k\psi (x)$$

for all $x\in A$ and $k\ge 1$.

From now on, we establish the modified Hyers-Ulam-Rassias stability of the equations (1.3) and (1.4) in a $(\beta ,p)$-Banach space $\mathcal{Y}$.

Theorem 2.9 Let $f:\mathbb{R}\to \mathcal{Y}$ be a ϕ-approximatively radical quadratic function with $f(0)=0$. Assume that the function ϕ is contractively subadditive with a constant L satisfying $2^{1-2\beta }L<1$. Then there exists a unique quadratic function $\mathcal{F}:\mathbb{R}\to \mathcal{Y}$ which satisfies (1.3) and the inequality

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Phi }}(x)}{\sqrt[p]{4^{\beta p}-{(2L)}^p}}$$

(2.28)

for all $x\in \mathbb{R}$, where

$$\hat{\mathrm{\Phi }}(x):=K^2(2^{\beta }(\varphi (x,x)+\varphi (\sqrt{2}x,0))+\varphi (\sqrt{2}x,\sqrt{2}x)+\varphi (2x,0)).$$

Proof Using (2.13) in the proof of Theorem 2.2, we get

$${\parallel f(x)-\frac{1}{4}f(2x)\parallel }_{\mathcal{Y}}\le \frac{1}{4^{\beta }}\hat{\mathrm{\Phi }}(x)$$

(2.29)

for all $x\in \mathbb{R}$. Thus it follows from (2.29) that

$$\begin{array}{rcl}{\parallel \frac{1}{4^k}f\left(2^{k}x\right)-\frac{1}{4^m}f\left(2^{m}x\right)\parallel }_{\mathcal{Y}}^p& \le & \frac{1}{4^{\beta p}}\sum _{j=k}^{m-1}{\left(\frac{1}{4^{\beta }}\right)}^{jp}{(2L)}^{jp}\hat{\mathrm{\Phi }}{(x)}^p\\ \le & \frac{\hat{\mathrm{\Phi }}{(x)}^p}{4^{\beta p}}\sum _{j=k}^{m-1}{\left(2^{1-2\beta }L\right)}^{jp}\end{array}$$

(2.30)

for all $x\in \mathbb{R}$ and integers $m>k\ge 0$. Then a sequence $\{\frac{1}{4^n}f(2^{n}x)\}$ is a Cauchy sequence in a $(\beta ,p)$-Banach space ${\mathcal{Y}}^p$, and so we can define a function $\mathcal{F}:\mathbb{R}\to \mathcal{Y}$ by $\mathcal{F}(x):={lim}_{n\to \mathrm{\infty }}\frac{1}{4^n}f(2^{n}x)$ for all $x\in \mathbb{R}$. Then, we get

for all $x,y\in \mathbb{R}$, and so, by Lemma 2.1, $\mathcal{F}$ is a quadratic function. Taking $m\to \mathrm{\infty }$ in (2.30) with $k=0$, we have

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Phi }}(x)}{\sqrt[p]{4^{\beta p}-{(2L)}^p}}.$$

Next, we assume that there exists a quadratic function $\mathcal{G}:\mathbb{R}\to \mathcal{Y}$ which satisfies the functional equation (1.3) and (2.28). Then we have

$${\parallel \mathcal{G}(x)-\frac{1}{4^n}f\left(2^{n}x\right)\parallel }_{\mathcal{Y}}^p\le \frac{1}{4^{\beta np}}{\parallel \mathcal{G}\left(2^{n}x\right)-f\left(2^{n}x\right)\parallel }_{\mathcal{Y}}^p\le \frac{\hat{\mathrm{\Phi }}{(x)}^p}{4^{\beta p}-{(2L)}^p}{\left(2^{1-2\beta }L\right)}^{np}$$

for all $x\in \mathbb{R}$. Letting $n\to \mathrm{\infty }$, we have the uniqueness of $\mathcal{F}(x)$. This completes the proof. □

Theorem 2.10 Let $f:\mathbb{R}\to \mathcal{Y}$ be a ϕ-approximatively radical quadratic function with $f(0)=0$. Assume that the function ϕ is expansively superadditive with a constant L satisfying $2^{2\beta -1}L<1$. Then there exists a unique quadratic function $\mathcal{F}:\mathbb{R}\to \mathcal{Y}$ which satisfies (1.3) and the inequality

$${\parallel f(x)-\mathcal{F}(x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Phi }}(x)}{\sqrt[p]{{(2L^{-1})}^p-4^{\beta p}}}$$

for all $x\in \mathbb{R}$.

Proof From (2.29) in the proof of Theorem 2.9, we get

$${\parallel 4^{k}f\left(2^{-k}x\right)-4^{k+1}f\left(2^{-k-1}x\right)\parallel }_{\mathcal{Y}}\le 4^{\beta k}\hat{\mathrm{\Phi }}\left(2^{-k-1}x\right)$$

for all $x\in \mathbb{R}$. For integers k, m with $m>k\ge 0$, we get

$${\parallel 4^{k}f\left(2^{-k}x\right)-4^{m}f\left(2^{-m}x\right)\parallel }_{\mathcal{Y}}^p\le \frac{\hat{\mathrm{\Phi }}{(x)}^p}{4^{\beta p}}\sum _{j=k+1}^m{\left(2^{2\beta -1}L\right)}^{jp}.$$

The remaining part follows as the proof of Theorem 2.9. This completes the proof. □

Theorem 2.11 Let $f:\mathbb{R}\to \mathcal{Y}$ be a ψ-approximatively radical quadratic function with $f(0)=0$. Assume that the function ψ is contractively subquadratic with a constant L satisfying $2^{2-4\beta }L<1$. Then there exists a unique quartic function $\mathcal{H}:\mathbb{R}\to \mathcal{Y}$ which satisfies (1.4) and the inequality

$${\parallel f(x)-\mathcal{H}(x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Psi }}(x)}{2^{\beta }\sqrt[p]{{16}^{\beta p}-{(4L)}^p}},$$

(2.31)

where

$$\hat{\mathrm{\Psi }}(x)=2K^3(4^{\beta }(\psi (x,x)+\frac{1}{2}\psi (\sqrt{2}x,0))+\psi (\sqrt{2}x,\sqrt{2}x)+\frac{1}{2}\psi (2x,0))$$

for all $x\in \mathbb{R}$.

Proof Using (2.26) in the proof of Theorem 2.7, we get

$${\parallel f(x)-\frac{1}{16}f(2x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Psi }}(x)}{2^{5\beta }}$$

(2.32)

for all $x\in \mathbb{R}$. Then it follows from (2.32) that

(2.33)

for all $x\in \mathbb{R}$ and $m>k\ge 0$. Then $\{\frac{1}{{16}^n}f(2^{n}x)\}$ is a Cauchy sequence in $\mathcal{Y}$, and it converges to a point in $\mathcal{Y}$. We can define a function $\mathcal{H}:\mathbb{R}\to \mathcal{X}$ by

$$\mathcal{H}(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{16}^n}f\left(2^{n}x\right)$$

for all $x\in \mathbb{R}$. From (2.9), the following inequality holds:

for all $x,y\in \mathbb{R}$. Thus the function $\mathcal{H}$ is quartic. Taking the limit $m\to \mathrm{\infty }$ in (2.33) with $k=0$, $\mathcal{H}$ satisfies (2.31) near the approximate function f of the functional equation (1.4). The remaining proof is similar to that of Theorem 2.9. This completes the proof. □

Theorem 2.12 Let $f:\mathbb{R}\to \mathcal{Y}$ be a ψ-approximatively radical quadratic function with $f(0)=0$. Assume that the function ψ is expansively superquadratic with a constant L satisfying $2^{4\beta -2}L<1$. Then there exists a unique quartic function $\mathcal{H}:\mathbb{R}\to \mathcal{Y}$ which satisfies (1.4) and the inequality

$${\parallel f(x)-\mathcal{H}(x)\parallel }_{\mathcal{Y}}\le \frac{\hat{\mathrm{\Psi }}(x)}{2^{\beta }\sqrt[p]{{(4L^{-1})}^p-{16}^{\beta p}}}$$

for all $x\in \mathbb{R}$.