Approximate solution of generalized inhomogeneous radical quadratic functional equations in 2-Banach spaces

Abstract

In this paper, using Brzdȩk and Ciepliński’s fixed point theorems in a 2-Banach space, we investigate approximate solution for the generalized inhomogeneous radical quadratic functional equation of the form

$$f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)=af(x)+bf(y) + D(x,y), $$

where f is a mapping on the set of real numbers, $a, b\in\mathbf {R}_{+}$ and $D(x,y)$ is a given function. Some stability and hyperstability properties are presented.

1 Introduction

In this paper, N and R denote the sets of all positive integers, and real numbers, respectively. We put $\mathbf {N}_{0}:=\mathbf{N}\cup{0}$, $\mathbf{R}_{0}:=\mathbf{R}\setminus{0}$ and $\mathbf{R}_{+}:=[0,\infty)$. Also, $Y^{X}$ denotes the set of all functions from a nonempty set X to a nonempty set Y.

The study of stability problems for functional equations originates from a question of Ulam [28] concerning the stability of group homomorphisms. Popularly speaking, the question was “Under what conditions a mathematical object satisfying a certain property approximately must be close to an object satisfying the property exactly?” In the following year, Hyers [20] first partially answered Ulam’s question, and proved the Ulam stability of Cauchy function in Banach spaces. Aoki [5] and Rassias [27] generalized the Hyers’ results by allowing the Cauchy difference to become unbounded. During the last decades, the Ulam–Hyers–Rassias stability of functional equations has been extensively investigated and generalized by many mathematicians (see [2, 3, 6,7,8,9, 11,12,13,14, 16, 24, 26, 29] and the references therein).

Recently, a lot of papers (see, for instance, [1, 4, 15, 17,18,19, 21,22,23]) on the stability of radical function equations have been published. The functional equation

$$ f \bigl(\sqrt{x^{2}+y^{2}} \bigr)=f(x)+f(y) $$

(1)

is called a radical quadratic functional equation. Kim et al. [23] investigated the generalized Hyers–Ulam–Rassias stability problem of Eq. (1) in quasi-β-Banach spaces using the direct method. Khodaei et al. [22] introduced and solved the generalized radical quadratic functional equation

$$ f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)=af(x)+bf(y). $$

(2)

They established some stability results in 2-normed spaces by using the direct method, and proved new theorems about the generalized Ulam stability by using subadditive and subquadratic functions in p-2-normed spaces. Cho et al. [15] proved the generalized Hyers–Ulam stability results for Eq. (2) in quasi-β-Banach spaces by using subadditive and subquadratic functions. Using Brzdȩk’s fixed point theorem, Aiemsomboom et al. [1] and Kang [21] investigated the stability of Eqs. (1) and (2), respectively, where f is a self-mapping on R.

Let $(Y, \Vert \cdot,\cdot \Vert )$ be a 2-Banach space, $D: \mathbf {R}^{2}\rightarrow Y$ a given function, and let $a, b\in\mathbf{R}_{+}$ be fixed. The purpose of this paper is to prove stability and hyperstability results for the generalized inhomogeneous quadratic radical functional equation

$$ f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)=af(x)+bf(y)+D(x,y),\quad x,y\in\mathbf{R}_{0} $$

(3)

in a 2-Banach space using Brzdȩk and Ciepliński’s fixed point results in [11].

2 Preliminaries

Let us recall some basic definitions and facts concerning 2-Banach spaces (see, for instance, [11, 17, 18, 25]).

Definition 1

Let X be a linear space over R with $\dim X\geq2$ and let $\Vert \cdot,\cdot \Vert :X\times X\rightarrow\mathbf {R}_{+}$ be a function satisfying the following properties:

• (1) $\Vert x,y \Vert = 0$ if and only if x and y are linearly dependent;

• (2) $\Vert x,y \Vert = \Vert y,x \Vert$ for $x, y\in X$;

• (3) $\Vert r x,y \Vert = |r| \Vert x,y \Vert$ for $r\in\mathbf{R}$ and $x, y\in X$;

• (4) $\Vert x + y,z \Vert \leq \Vert x,z \Vert + \Vert y,z \Vert$ for $x, y,z \in X$.

Then the pair $(X, \Vert \cdot,\cdot \Vert )$ is called a 2-normed space.

If $x\in X$ and $\Vert x,y \Vert =0$ for all $y\in X$, then $x=0$. Moreover, the functions $x\rightarrow \Vert x,y \Vert$ are continuous functions of X into $\mathbf{R}_{+}$ for each fixed $y\in X$.

Definition 2

Let $\{x_{n}\}$ be a sequence in a 2-normed space X.

• (1) A sequence $\{x_{n}\}$ in a 2-normed space is called a Cauchy sequence if there are linear independent $y,z\in X$ such that

  $$\lim_{n,m\rightarrow\infty} \Vert x_{n}-x_{m},y \Vert =0= \lim_{n,m\rightarrow\infty} \Vert x_{n}-x_{m},z \Vert ; $$

• (2) A sequence $\{x_{n}\}$ is said to be convergent if there exists an $x\in X$ such that $\lim_{n\rightarrow\infty} \Vert x_{n}-x,y \Vert =0$ for all $y\in X$. Then, the point x is called the limit of the sequence $\{x_{n}\}$, which is denoted by $\lim_{n\rightarrow\infty}x_{n}=x$;

• (3) If every Cauchy sequence in X converges, then the 2-normed space X is called a 2-Banach space.

It is easily seen that $(\mathbf{R}^{2}, \Vert \cdot,\cdot \Vert )$ is a 2-Banach space, where the Euclidean 2-norm $\Vert \cdot,\cdot \Vert$ is defined by

$$\bigl\Vert (x_{1},x_{2}), (y_{1},y_{2}) \bigr\Vert := \vert x_{1} y_{2}- x_{2} y_{1} \vert , \quad (x_{1},x_{2}), (y_{1},y_{2})\in\mathbf{R}^{2}. $$

The next example following from [11, Proposition 2.3].

Example 1

If $(X, \langle\cdot, \cdot\rangle)$ is a real Hilbert space, then $(X, \Vert \cdot,\cdot \Vert )$ is a 2-Banach space, where $\Vert \cdot,\cdot \Vert$ is given by

$$\Vert x,y \Vert := \sqrt{ \Vert x \Vert ^{2} \Vert y \Vert ^{2}-\langle x, y\rangle^{2}},\quad x,y\in X. $$

3 Fixed point theorems

Recently, Brzdȩk and Ciepliński [11] proved a new fixed point theorem in 2-Banach spaces and showed its applications to the Ulam stability of some single-variable equations and the most important functional equation in several variables, namely, the Cauchy equation. And they extended the fixed point result to the n-normed spaces in [10].

Let us introduce the following hypotheses:

• (H1) X is a nonempty set, $(Y, \Vert \cdot, \cdot \Vert )$ is a 2-Banach space, $Y_{0}$ is a subset of Y containing two linearly independent vectors;

• (H2) $j\in\mathbf{N}, f_{1},\dots,f_{j}: X\rightarrow X$, $g_{1},\dots,g_{j}: Y_{0}\rightarrow Y_{0}$, and $L_{1},\dots,L_{j}:X\times Y_{0}\rightarrow\mathbf{R}_{+}$ are given maps;

• (H3) $\mathcal{T}: Y^{X}\rightarrow Y^{X}$ is an operator satisfying the inequality

  $$ \bigl\Vert (\mathcal{T} \xi) (x)-(\mathcal{T} \eta) (x),z \bigr\Vert \leq\sum_{i=1}^{j} L_{i}(x, y) \bigl\Vert \xi \bigl(f_{i}(x) \bigr)-\eta \bigl(f_{i}(x) \bigr),g_{i}(z) \bigr\Vert , $$

  (4)

  where $\xi, \eta\in Y^{X}, x\in X, z\in Y_{0}$;

• (H4) $\varLambda:\mathbf{R}_{+}^{X\times Y_{0}}\rightarrow\mathbf {R}_{+}^{X\times Y_{0}}$ is an operator defined by

  $$ (\varLambda\delta) (x, z):=\sum_{i=1}^{j} L_{i}(x,z)\delta \bigl(f_{i}(x), g_{i}(z) \bigr), \delta\in\mathbf{R}_{+}^{X\times Y_{0}},\quad x\in X, z\in Y_{0}. $$

  (5)

Now, we are in a position to present the above mentioned fixed point result. We use it to assert the existence of a unique fixed point of operator $\mathcal{T}: Y^{X}\rightarrow Y^{X}$.

Theorem 1

Let hypotheses (H1)–(H4) hold and functions $\epsilon: X\times Y_{0}\rightarrow\mathbf{R}_{+}$ and $\varphi: X\rightarrow Y$ fulfill the following two conditions:

$$ \bigl\Vert (\mathcal{T} \varphi) (x)-\varphi(x),z \bigr\Vert \leq \epsilon(x,z), \quad x\in X, z\in Y_{0} $$

(6)

and

$$ \epsilon^{*}(x,z):=\sum_{l=0}^{\infty} \bigl(\varLambda^{l}\epsilon \bigr) (x,z)< \infty ,\quad x\in X, z\in Y_{0}. $$

(7)

Then, there exists a unique fixed point ψ of $\mathcal{T}$ with

$$ \bigl\Vert \varphi(x)-\psi(x),z \bigr\Vert \leq \epsilon^{*}(x,z),\quad x\in X, z\in Y_{0}. $$

(8)

Moreover,

$$\psi(x)=\lim_{l\rightarrow\infty} \bigl(\mathcal{T}^{l}\varphi \bigr) (x),\quad x\in X. $$

4 The main results

In this section, we investigate the stability and hyperstability of the generalized inhomogeneous radical quadratic functional equation (3) in 2-Banach spaces by using Theorem 1. In what follows, we assume that $a, b\in\mathbf{N}$ are fixed, $(Y, \Vert \cdot ,\cdot \Vert )$ is a 2-Banach space, and $Y_{0}$ is a subset of Y containing two linearly independent vectors.

Theorem 2

Let $h_{1},h_{2}:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be two functions such that

$$\begin{aligned} M_{0} &:= \biggl\{ n\in\mathbf{N}: k_{n}:= \frac{1}{a} \lambda _{1} \bigl(a+bn^{2} \bigr) \lambda_{2} \bigl(a+bn^{2} \bigr)+ \frac{b}{a} \lambda_{1} \bigl(n^{2} \bigr)\lambda _{2} \bigl(n^{2} \bigr)< 1 \biggr\} \\ &\neq \emptyset, \end{aligned}$$

(9)

where

$$\lambda_{i}(n):=\inf \bigl\{ t\in\mathbf{R}_{+}: h_{i} \bigl(nx^{2},z \bigr)\leq th_{i} \bigl(x^{2},z \bigr) \bigr\} , $$

where $x\in\mathbf{R}_{0}, z\in Y_{0}, i=1,2, n\in\mathbf{N}$. Suppose that $f:\mathbf{R}\rightarrow Y$ satisfies the inequality

$$ \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y),z \bigr\Vert \leq h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr), $$

(10)

where $x,y\in\mathbf{R}_{0},z\in Y_{0}$. Then there exists a unique solution $Q: \mathbf{R}\rightarrow Y$ of (2) such that

$$ \bigl\Vert f(x)-Q(x),z \bigr\Vert \leq\lambda_{0}(x,z), \quad x\in\mathbf {R}_{0}, z\in Y_{0}, $$

(11)

where

$$\lambda_{0}(x,z):=\inf_{m\in M_{0}} \biggl\{ \frac{\lambda _{2}(m^{2})h_{1}(x^{2},z)h_{2}(x^{2},z)}{a(1-k_{m})}, x\in\mathbf{R}_{0}, z\in Y_{0} \biggr\} . $$

Proof

Putting $y=mx$ in (10), we obtain that

$$ \bigl\Vert f \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-af(x)-bf(mx),z \bigr\Vert \leq h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(m^{2}x^{2},z \bigr), $$

(12)

where $m\in\mathbf{N}, x\in\mathbf{R}_{0}, z\in Y_{0}$, and so

$$ \biggl\Vert \frac{1}{a}f \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-\frac {b}{a}f(mx)-f(x),z \biggr\Vert \leq \frac{1}{a}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(m^{2}x^{2},z \bigr), $$

(13)

where $m\in\mathbf{N}, x\in\mathbf{R}_{0}, z\in Y_{0}$.

For each $m\in\mathbf{N}$, we define the operators $\mathcal{T}_{m}: Y^{\mathbf{R}_{0}} \rightarrow Y^{\mathbf{R}_{0}}$ and $\varLambda_{m}: \mathbf {R}_{+}^{\mathbf{R}_{0}\times Y_{0} } \rightarrow\mathbf{R}_{+}^{\mathbf {R}_{0}\times Y_{0} }$ by

$$\begin{aligned} &(\mathcal{T}_{m} \xi ) (x):=\frac{1}{a}\xi \bigl(\sqrt { \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-\frac{b}{a}\xi(mx), \\ &(\varLambda_{m} \delta ) (x, z):=\frac{1}{a}\delta \bigl(\sqrt { \bigl(a+bm^{2} \bigr)x^{2}}, z \bigr)+ \frac{b}{a} \delta(mx, z), \end{aligned}$$

where $x\in\mathbf{R}_{0}, \xi\in Y^{\mathbf{R}_{0}}, \delta\in\mathbf {R}_{+}^{\mathbf{R}_{0}\times Y_{0} }, z\in Y_{0}$. Then the operator $\varLambda _{m}$ has the form (5) with $X:=\mathbf{R}_{0}$, $j=2$, $f_{1}(x):=\sqrt{(a+bm^{2})x^{2}},f_{2}(x):=mx$, $g_{1}(z)=g_{2}(z):=z, L_{1}(x,z):=\frac{1}{a}$ and $L_{2}(x,z):= \frac{b}{a}$ for all $x\in \mathbf{R}_{0}$ and $z\in Y_{0}$. Next, put

$$\epsilon_{m}(x,z):=\frac{1}{a}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(m^{2}x^{2},z \bigr),\quad m\in\mathbf {N},x\in\mathbf{R}_{0},z\in Y_{0} $$

and observe that

$$\epsilon_{m}(x,z)=\frac{1}{a}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(m^{2}x^{2},z \bigr)\leq\frac {1}{a}\lambda_{2} \bigl(m^{2} \bigr)h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr), $$

where $m\in\mathbf{N}, x\in\mathbf{R}_{0},z\in Y_{0}$. Then, inequality (13) can be rewritten as

$$\bigl\Vert (\mathcal{T}_{m} f ) (x)-f(x),z \bigr\Vert \leq \epsilon _{m}(x,z),\quad m\in\mathbf{N}, x\in\mathbf{R}_{0}, z\in Y_{0}, $$

and we have

$$\begin{aligned} & \bigl\Vert (\mathcal{T}_{m} \xi) (x)-( \mathcal{T}_{m} \eta) (x)),z \bigr\Vert \\ &\quad= \biggl\Vert \frac{1}{a}\xi \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-\frac{b}{a}\xi (mx) \\ &\qquad{} -\frac{1}{a}\eta \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)+\frac{b}{a}\eta (mx),z \biggr\Vert \\ &\quad\leq \biggl\Vert \frac{1}{a}\xi \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-\frac {1}{a}\eta \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr),z \biggr\Vert \\ &\qquad{}+ \biggl\Vert \frac{b}{a}\xi(mx)-\frac{b}{a}\eta(mx),z \biggr\Vert \\ &\quad=\frac{1}{a} \bigl\Vert \xi \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)-\eta \bigl(\sqrt { \bigl(a+bm^{2} \bigr)x^{2}} \bigr),z \bigr\Vert \\ &\qquad{}+ \frac{b}{a} \bigl\Vert \xi(mx)-\eta(mx),z \bigr\Vert \\ &\quad=L_{1}(x,z) \bigl\Vert \xi \bigl(f_{1}(x) \bigr)- \eta \bigl(f_{1}(x) \bigr),z \bigr\Vert \\ &\qquad{}+L_{2}(x,z) \bigl\Vert \xi \bigl(f_{2}(x) \bigr)- \eta \bigl(f_{2}(x) \bigr),z \bigr\Vert \end{aligned}$$

(14)

for any $x\in\mathbf{R}_{0}, \xi, \eta\in Y^{\mathbf{R}_{0}}, z\in Y_{0}$. Therefore,

$$ \bigl\Vert (\mathcal{T}_{m} \xi) (x)-( \mathcal{T}_{m} \eta) (x)),z \bigr\Vert \leq\sum _{i=1}^{2}L_{i}(x,z) \bigl\Vert \xi \bigl(f_{i}(x) \bigr)-\eta \bigl(f_{i}(x) \bigr),z \bigr\Vert , $$

(15)

so (4) holds for $\mathcal{T}:= \mathcal{T}_{m}$ with $m\in \mathbf{N}$. By the definition of $\lambda_{i}(n)$, we have

$$ h_{i} \bigl(nx^{2},z \bigr)\leq \lambda_{i}(n) h_{i} \bigl(x^{2},z \bigr),\quad x\in \mathbf{R}_{0}, z\in Y_{0}, i=1,2, n\in\mathbf{N}, $$

(16)

whence, using induction, we get

$$ \bigl(\varLambda^{n}_{m} \epsilon_{m} \bigr) (x,z)\leq\frac{1}{a}\lambda _{2} \bigl(m^{2} \bigr)k_{m}^{n}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr),\quad n\in \mathbf{N}_{0}, x\in\mathbf {R}_{0}, z\in Y_{0}. $$

(17)

Indeed, for $n=0$, (17) is obviously true. Next, we will assume that (17) holds for $n=j$, where $j\in \mathbf{N}$. Then, we have

$$\begin{aligned} &\bigl(\varLambda_{m} ^{j+1}\epsilon_{m} \bigr) (x,z) \\ &\quad= \bigl(\varLambda_{m} \bigl(\varLambda_{m}^{j} \epsilon_{m}(x,z) \bigr) \bigr) \\ &\quad=\frac{1}{a} \bigl(\varLambda_{m}^{j} \epsilon_{m} \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr)x^{2}}, z \bigr)+ \frac{b}{a} \bigl(\varLambda_{m}^{j} \epsilon _{m} \bigr) (mx, z) \\ &\quad\leq \frac{1}{a^{2}}\lambda _{2} \bigl(m^{2} \bigr)k_{m}^{j}h_{1} \bigl( \bigl(a+bm^{2} \bigr)x^{2},z \bigr)h_{2} \bigl( \bigl(a+bm^{2} \bigr)x^{2},z \bigr) \\ &\qquad{}+\frac{b}{a}\frac{1}{a}\lambda _{2} \bigl(m^{2} \bigr)k_{m}^{j}h_{1} \bigl(m^{2}x^{2},z \bigr)h_{2} \bigl(m^{2}x^{2},z \bigr) \\ &\quad\leq\frac{1}{a}\lambda_{2} \bigl(m^{2} \bigr)k_{m}^{j}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr)\biggl[\frac{1}{a} \lambda_{1} \bigl(a+bm^{2} \bigr) \lambda_{2} \bigl(a+bm^{2} \bigr) \\ &\qquad {} + \frac{b}{a} \lambda_{1} \bigl(m^{2} \bigr)\lambda_{2} \bigl(m^{2} \bigr) \biggr] \\ &\quad=\frac{1}{a}\lambda_{2} \bigl(m^{2} \bigr)k_{m}^{j+1}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr),\quad x\in\mathbf {R}_{0}, z\in Y_{0}, m\in M_{0}. \end{aligned}$$

This shows that (17) holds for $n=j+1$. Now we can conclude that inequality (17) holds for all $n\in \mathbf{N}_{0}$. Therefore, by (17), we obtain that

$$\begin{aligned} \epsilon_{m}^{*}(x,z)&=\sum_{n=0}^{\infty} \bigl(\varLambda_{m}^{n}\epsilon _{m} \bigr) (x,z) \\ &\leq \frac{1}{a}\lambda_{2} \bigl(m^{2} \bigr)h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr)\sum_{n=0}^{\infty }k_{m}^{n} \\ &= \frac{\lambda_{2}(m^{2})h_{1}(x^{2},z)h_{2}(x^{2},z)}{a(1-k_{m})} \end{aligned}$$

for all $x\in\mathbf{R}_{0},z\in Y_{0}$ and $m\in M_{0}$. Thus, according to Theorem 1, for any $m\in M_{0}$, there exists a unique fixed point $Q_{m}': \mathbf{R}_{0}\rightarrow Y$ of $\mathcal{T}_{m}$, which satisfies the estimate

$$ \bigl\Vert f(x)-Q_{m}'(x),z \bigr\Vert \leq\epsilon_{m}^{*}(x,z)\leq\frac {\lambda_{2}(m^{2})h_{1}(x^{2},z)h_{2}(x^{2},z)}{a(1-k_{m})}, $$

(18)

where $x\in\mathbf{R}_{0}, z\in Y_{0}, m\in M_{0}$. Moreover,

$$Q_{m}'(x):=\lim_{n\rightarrow\infty} \bigl( \mathcal{T}_{m}^{n}f \bigr) (x),\quad x\in\mathbf{R}_{0}, m\in M_{0}. $$

and for any $m\in M_{0}$, the function $Q_{m}: \mathbf{R}\rightarrow Y$, given by the formula

$$Q_{m}(0)=0,\qquad Q_{m}(x):= Q_{m}'(x),\quad x\in\mathbf{R}_{0}, $$

is a solution of the equation

$$ Q(x)=\frac{1}{a}Q \bigl(\sqrt{ \bigl(a+bm^{2} \bigr)x^{2}} \bigr)- \frac{b}{a} Q(mx),\quad x\in\mathbf{R}, m\in M_{0}. $$

(19)

Now, we show that

$$\begin{aligned} & \bigl\Vert \bigl(\mathcal{T}_{m}^{n}f \bigr) \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-a \bigl( \mathcal{T}_{m}^{n}f \bigr) (x)-b \bigl(\mathcal{T}_{m}^{n}f \bigr) (y),z \bigr\Vert \\ &\quad\leq k_{m}^{n}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr) \end{aligned}$$

(20)

for any $x,y\in\mathbf{R}_{0}, z\in Y_{0}, m\in M_{0}$ and $n\in\mathbf{N}_{0}$.

Since the case $n=0$ follows immediately from (10), take $j\in\mathbf{N}_{0}$ and assume that (20) holds for $n=j$, $x,y\in\mathbf{R}_{0}$, $m\in M_{0}$ and $z\in Y$. Then, by (16), we get

$$\begin{aligned} &\bigl\Vert \bigl(\mathcal{T}_{m}^{j+1}f \bigr) \bigl( \sqrt {ax^{2}+by^{2}} \bigr)-a \bigl(\mathcal{T}_{m}^{j+1}f \bigr) (x)-b \bigl(\mathcal{T}_{m}^{j+1}f \bigr) (y),z \bigr\Vert \\ &\quad= \biggl\Vert \frac{1}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr) \bigl(ax^{2}+by^{2} \bigr)} \bigr)-\frac{b}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(m\sqrt{ax^{2}+by^{2}} \bigr) \\ &\qquad{}- \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr)x^{2}} \bigr)+b \bigl( \mathcal{T}_{m}^{j}f \bigr) (mx) \\ &\qquad{}-\frac{b}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr)y^{2}} \bigr)+ \frac{b^{2}}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) (my),z \biggr\Vert \\ &\quad\leq \biggl\Vert \frac{1}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr) \bigl(ax^{2}+by^{2} \bigr)} \bigr)- \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl( \sqrt { \bigl(a+bm^{2} \bigr)x^{2}} \bigr) \\ &\qquad{}-\frac{b}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(\sqrt { \bigl(a+bm^{2} \bigr)y^{2}} \bigr),z \biggr\Vert \\ &\qquad{}+ \biggl\Vert \frac{b}{a} \bigl(\mathcal{T}_{m}^{j}f \bigr) \bigl(m\sqrt {ax^{2}+by^{2}} \bigr)-b \bigl( \mathcal{T}_{m}^{j}f \bigr) (mx)-\frac {b^{2}}{a} \bigl( \mathcal{T}_{m}^{j}f \bigr) (my),z \biggr\Vert \\ &\quad\leq \frac{1}{a} k_{m}^{j}h_{1} \bigl( \bigl(a+bm^{2} \bigr)x^{2},z \bigr)h_{2} \bigl( \bigl(a+bm^{2} \bigr)y^{2},z \bigr) \\ &\qquad{}+ \frac{b}{a} k_{m}^{j}h_{1} \bigl(m^{2}x^{2},z \bigr)h_{2} \bigl(m^{2}y^{2},z \bigr) \\ &\quad \leq k_{m}^{j}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr) \biggl[\frac{1}{a} \lambda _{1} \bigl(a+bm^{2} \bigr)\lambda_{2} \bigl(a+bm^{2} \bigr)+ \frac{b}{a} \lambda_{1} \bigl(m^{2} \bigr)\lambda _{2} \bigl(m^{2} \bigr) \biggr] \\ &\quad=k_{m}^{j+1}h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr),\quad x,y\in\mathbf{R}_{0}, z\in Y_{0}, m\in M_{0}. \end{aligned}$$

Thus, by induction, we have shown that (20) holds for any $n\in\mathbf{N}_{0}$, $x,y\in\mathbf{R}_{0}, z\in Y_{0}$ and $m\in M_{0}$. Letting $n\rightarrow\infty$ in (20) and using Lemmas 2.1 and 2.2 in [11], we obtain that

$$ Q_{m} \bigl(\sqrt{ax^{2}+by^{2}} \bigr)=aQ_{m}(x)+bQ_{m}(y),\quad x,y\in\mathbf{R}_{0}, m\in M_{0}. $$

(21)

This way, for each $m\in M_{0}$, we obtain a function $Q_{m}$ such that (21) holds for $x,y\in\mathbf{R}$ and

$$ \bigl\Vert f(x)-Q_{m}(x),z \bigr\Vert \leq \frac{\lambda _{2}(m^{2})h_{1}(x^{2},z)h_{2}(x^{2},z)}{a(1-k_{m})},\quad x\in\mathbf{R}, z\in Y_{0}, m\in M_{0}. $$

(22)

Let $L>0$ be a constant. Next, we will see that every generalized radical quadratic mapping $Q:\mathbf{R}\rightarrow Y$ satisfying the inequality

$$ \bigl\Vert f(x)-Q(x),z \bigr\Vert \leq L h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr),\quad x\in \mathbf{R}_{0}, z\in Y_{0} $$

(23)

is equal to $Q_{m}$ for any $m\in M_{0}$. To do this, fix $s \in M_{0}$ and let $Q:\mathbf{R}\rightarrow Y$ be a generalized radical quadratic mapping satisfying (23). By (18), we have

$$\begin{aligned} \bigl\Vert Q_{s}(x)-Q(x),z \bigr\Vert &\leq \bigl\Vert Q_{s}(x)-f(x),z \bigr\Vert + \bigl\Vert f(x)-Q(x),z \bigr\Vert \\ & \leq \biggl(\frac{\lambda_{2}(s^{2})}{a(1-k_{s})}+L \biggr) h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr) \\ &= L_{0} h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr) \sum _{n=0}^{\infty}k_{s}^{n},\quad x\in \mathbf {R}_{0}, z\in Y_{0}, \end{aligned}$$

(24)

where $L_{0}=aL(1-k_{s})+ \lambda_{2}(s^{2})$. Observe also that Q and $Q_{s}$ are solutions of equation (19) for any $m\in M_{0}$.

Now, we will see that, for any $j\in\mathbf{N}_{0}$,

$$ \bigl\Vert Q_{s}(x)-Q(x),z \bigr\Vert \leq L_{0} h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr) \sum_{n=j}^{\infty}k_{s}^{n},\quad x\in\mathbf{R}_{0}, z\in Y_{0}. $$

(25)

The case $j=0$ follows from the previous inequality. Fix a $j\in\mathbf {N}_{0}$ and assume that (25) holds. Then, by (16), we get

$$\begin{aligned} &\bigl\Vert Q_{s}(x)-Q(x),z \bigr\Vert \\ & \quad = \biggl\Vert \frac{1}{a}Q_{s} \bigl(\sqrt{ \bigl(a+bs^{2} \bigr)x^{2}} \bigr)- \frac {b}{a} Q_{s}(sx) -\frac{1}{a}Q \bigl(\sqrt{ \bigl(a+bs^{2} \bigr)x^{2}} \bigr)+ \frac {b}{a} Q(sx),z \biggr\Vert \\ &\quad\leq \biggl\Vert \frac{1}{a}Q_{s} \bigl(\sqrt{ \bigl(a+bs^{2} \bigr)x^{2}} \bigr)-\frac {1}{a}Q \bigl( \sqrt{ \bigl(a+bs^{2} \bigr)x^{2}} \bigr), z \biggr\Vert \\ &\qquad{}+ \biggl\Vert \frac{b}{a} Q_{s}(sx)- \frac{b}{a} Q(sx),z \biggr\Vert \\ &\quad\leq \frac{1}{a} L_{0} h_{1} \bigl( \bigl(a+bs^{2} \bigr)x^{2},z \bigr)h_{2} \bigl( \bigl(a+bs^{2} \bigr)x^{2},z \bigr) \sum _{n=j}^{\infty}k_{s}^{n} \\ &\qquad{}+ \frac{b}{a}L_{0} h_{1} \bigl(s^{2}x^{2},z \bigr)h_{2} \bigl(s^{2}x^{2},z \bigr) \sum _{n=j}^{\infty}k_{s}^{n} \\ &\quad \leq L_{0} h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr) \biggl( \frac{1}{a} \lambda _{1} \bigl(a+bs^{2} \bigr)\lambda_{2} \bigl(a+bs^{2} \bigr)+ \frac{b}{a} \lambda_{1} \bigl(s^{2} \bigr)\lambda _{2} \bigl(s^{2} \bigr) \biggr)\sum_{n=j}^{\infty}k_{s}^{n} \\ &\quad= L_{0} h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(x^{2},z \bigr) \sum _{n=j+1}^{\infty}k_{s}^{n},\quad x\in \mathbf {R}_{0}, z\in Y_{0}. \end{aligned}$$

(26)

Thus (25) is valid for any $j\in\mathbf{N}_{0}$. Letting $j\rightarrow\infty$ in (25) and using Lemma 2.1 in [11], we get

$$ Q(x)=Q_{s}(x),\quad x\in\mathbf{R}_{0}, $$

(27)

which, together with $Q(0)=Q_{s}(0)=0$, gives $Q=Q_{s}$. This means that $Q_{m}=Q_{s}$ for any $m\in M_{0}$. Therefore, by (18), we have

$$ \bigl\Vert f(x)-Q_{s}(x),z \bigr\Vert \leq \frac{\lambda _{2}(m^{2})h_{1}(x^{2},z)h_{2}(x^{2},z)}{a(1-k_{m})},\quad x\in\mathbf{R}_{0}, z\in Y_{0}, m\in M_{0}. $$

(28)

Hence, we get inequality (11) with $Q:= Q_{s}$. □

In a similar way, one can prove the following.

Theorem 3

Let $H:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be a function such that

$$ \mathcal{M}:= \biggl\{ n\in\mathbf{N}: \frac{1}{a} \rho \bigl(a+bn^{2} \bigr) + \frac{b}{a} \rho \bigl(n^{2} \bigr)< 1 \biggr\} \neq\emptyset, $$

(29)

where

$$\rho(n): =\inf \bigl\{ t\in\mathbf{R}_{+}: H \bigl(nx^{2},z \bigr)\leq t H \bigl(x^{2},z \bigr), x\in \mathbf{R}_{0}, z\in Y_{0}, n\in\mathbf{N} \bigr\} . $$

Suppose that $f:\mathbf{R}\rightarrow Y$ satisfies the inequality

$$ \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y),z \bigr\Vert \leq H \bigl(x^{2},z \bigr)+ H \bigl(y^{2},z \bigr), $$

(30)

where $x,y\in\mathbf{R}_{0},z\in Y_{0}$. Then there exists a unique solution $Q: \mathbf{R}\rightarrow Y$ of (2) such that

$$ \bigl\Vert f(x)-Q(x),z \bigr\Vert \leq\rho_{0}(x,z),\quad x\in\mathbf{R}_{0}, z\in Y_{0}, $$

(31)

where

$$\rho_{0}(x,z):=\inf_{m\in\mathcal{M}} \biggl\{ \frac{ (1+\rho (m^{2}) ) H(x^{2},z)}{a-\rho(a+bm^{2})-b\rho(m^{2})}, x\in\mathbf{R}_{0}, z\in Y_{0} \biggr\} . $$

From the above theorems we can obtain results analogous to Theorems 2 and 3 for the inhomogeneous radical quadratic functional equation.

Corollary 1

Let $h_{1},h_{2}:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$, $f:\mathbf{R}\rightarrow Y$ and $D: \mathbf{R}^{2} \rightarrow Y$ such that (9) holds, and

$$ \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)- D(x,y),z \bigr\Vert \leq h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr), $$

(32)

where $x,y\in\mathbf{R}_{0},z\in Y_{0}$. Assume that (3) has a solution $f_{0}:\mathbf{R}\rightarrow Y$. Then there exists a unique solution $F: \mathbf{R}\rightarrow Y$ of (3) such that

$$ \bigl\Vert f(x)-F(x),z \bigr\Vert \leq\lambda_{0}(x,z),\quad x\in\mathbf {R}_{0}, z\in Y_{0}, $$

(33)

where $\lambda_{0}(x,z)$ is defined as in Theorem 2.

Proof

Write $f_{1}:=f-f_{0}$. Then, we have

$$\begin{aligned} &\bigl\Vert f_{1} \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af_{1}(x)-bf_{1}(y),z \bigr\Vert \\ &\quad= \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)- D(x,y) \\ &\qquad{}- \bigl( f_{0} \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af_{0}(x)-bf_{0}(y)-D(x,y) \bigr),z \bigr\Vert \\ &\quad= \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)- D(x,y),z \bigr\Vert \\ &\quad \leq h_{1} \bigl(x^{2},z \bigr)h_{2} \bigl(y^{2},z \bigr),\quad x,y\in\mathbf{R}_{0},z\in Y_{0}, \end{aligned}$$

and, according to Theorem 2, there is a unique solution $Q: \mathbf{R}\rightarrow Y$ of (2) such that

$$\bigl\Vert f_{1}(x)-Q(x),z \bigr\Vert \leq\lambda_{0}(x,z),\quad x\in\mathbf {R}_{0}, z\in Y_{0}, $$

where $\lambda_{0}(x,z)$ is defined as in Theorem 2. Let $F=f_{0}+Q$. Then F is a solution to (3) and (33) holds. The uniqueness of F follows from the uniqueness of Q (see [6, Corollary 4]). □

Corollary 2

Let $H:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$, $f:\mathbf {R}\rightarrow Y$ and $D: \mathbf{R}^{2} \rightarrow Y$ such that (29) holds, and

$$ \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)- D(x,y),z \bigr\Vert \leq H \bigl(x^{2},z \bigr)+ H \bigl(y^{2},z \bigr), $$

(34)

where $x,y\in\mathbf{R}_{0},z\in Y_{0}$. Assume that (3) admits a solution $f_{0}:\mathbf{R}\rightarrow Y$. Then there exists a unique solution $F: \mathbf{R}\rightarrow Y$ of (3) such that

$$ \bigl\Vert f(x)-F(x),z \bigr\Vert \leq\rho_{0}(x,z),\quad x\in\mathbf{R}_{0}, z\in Y_{0}, $$

(35)

where $\rho_{0}(x,z)$ is defined as in Theorem 3.

Corollaries 1 and 2 yield at once the following hyperstability results.

Corollary 3

Let $h_{1},h_{2}:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be functions such that

$$\begin{aligned} \begin{aligned}& \sup_{n\in\mathbf{N}} \bigl\{ \lambda_{1} \bigl(a+bn^{2} \bigr)\lambda_{2} \bigl(a+bn^{2} \bigr)+ b \lambda_{1} \bigl(n^{2} \bigr)\lambda_{2} \bigl(n^{2} \bigr) \bigr\} < a, \\ &\inf_{n\in\mathbf{N}} \bigl\{ \lambda_{2} \bigl(n^{2} \bigr) \bigr\} =0, \end{aligned} \end{aligned}$$

(36)

where $\lambda_{i}(\cdot)\ (i=1,2)$ are defined as in Theorem 2. Assume that Eq. (3) has a solution $f_{0}$. Then any function $f:\mathbf{R}\rightarrow Y$, which satisfies $f(0)=f_{0}(0)$ and inequality (32), is a solution of (3).

Corollary 4

Let $H:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be function such that

$$ \sup_{n\in\mathbf{N}} \bigl\{ \rho \bigl(a+bn^{2} \bigr)+ b\rho \bigl(n^{2} \bigr) \bigr\} < a,\qquad \inf_{n\in\mathbf{N}} \bigl\{ \rho \bigl(n^{2} \bigr) \bigr\} = -1, $$

(37)

where $\rho(\cdot)$ is defined as in Theorem 3. Assume that (3) has a solution $f_{0}$. Then any function $f:\mathbf{R}\rightarrow Y$, which satisfies $f(0)=f_{0}(0)$ and inequality (34), is a solution of (3).

According to Corollaries 3 and 4, we derive the following particular cases.

Corollary 5

Let $h_{1},h_{2}:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be functions such that

$$ \lim_{n\rightarrow\infty} \bigl(\lambda_{1} \bigl(a+bn^{2} \bigr)\lambda_{2} \bigl(a+bn^{2} \bigr)+ b \lambda_{1} \bigl(m^{2} \bigr)\lambda_{2} \bigl(n^{2} \bigr) \bigr)=0,\qquad \lim_{n\rightarrow\infty} \lambda_{2} \bigl(n^{2} \bigr)=0, $$

(38)

where $\lambda_{i}(\cdot)\ (i=1,2)$ are defined as in Theorem 2. Assume that (3) has a solution $f_{0}$. Then any function $f:\mathbf{R}\rightarrow Y$, which satisfies $f(0)=f_{0}(0)$ and inequality (32), is a solution of (3).

Corollary 6

Let $H:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ be function such that

$$\lim_{n\rightarrow\infty} \bigl(\rho \bigl(a+bn^{2} \bigr)+ b\rho \bigl(n^{2} \bigr) \bigr)=0,\qquad \lim_{n\rightarrow\infty} \rho \bigl(n^{2} \bigr)= -1, $$

where $\rho(\cdot)$ is defined as in Theorem 3. Assume that (3) has a solution $f_{0}$. Then any function $f:\mathbf{R}\rightarrow Y$, which satisfies $f(0)=f_{0}(0)$ and inequality (34), is a solution of (3).

Next, we derive some hyperstability results for particular forms of $h_{1},h_{2},H$ and (3).

Corollary 7

Let $\theta\in\mathbf{R}_{+}$ and let $p,q\in\mathbf{R}$ be such that $p+q<0$. Assume that Eq. (3) has a solution $f_{0}$. If $f:\mathbf{R}\rightarrow Y$ satisfies $f(0)=f_{0}(0)$ and the inequality

$$ \bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)-D(x,y), z \bigr\Vert \leq\theta \vert x \vert ^{p} \vert y \vert ^{q},\quad x,y\in\mathbf{R}_{0}, z\in Y, $$

(39)

then f is a solution of (3).

Proof

Define $h_{1},h_{2}:\mathbf{R}_{0}\times Y_{0}\rightarrow\mathbf{R}_{+}$ by $h_{1}(x^{2},z)=\theta_{1}|x|^{p}$ and $h_{2}(y^{2},z)=\theta_{2} |y|^{p}$, where $\theta_{1},\theta_{2}\in\mathbf{R}_{+}$ with $\theta=\theta_{1} \theta_{2}$. Then, we have

$$\begin{aligned} \lambda_{1}(n) &= \inf \bigl\{ t\in\mathbf{R}_{+}: h_{1} \bigl(nx^{2}, z \bigr)\leq t h_{1} \bigl(x^{2},z \bigr), x\in\mathbf{R}_{0},z\in Y_{0} \bigr\} \\ &= \inf \bigl\{ t\in\mathbf{R}_{+}: \theta_{1} \bigl\vert n^{1/2}x^{2} \bigr\vert ^{p} \leq t \theta_{1} \vert x \vert ^{p}, x\in\mathbf{R}_{0},z \in Y_{0} \bigr\} \\ &= n^{p/2},\quad n\in\mathbf{N}. \end{aligned}$$

Similarly, we get $\lambda_{2}(n)=n^{q/2}$ for any $n\in\mathbf{N}$. Thus,

$$\begin{aligned} &\lim_{n\rightarrow\infty} \bigl( \lambda_{1} \bigl(a+bn^{2} \bigr)\lambda _{2} \bigl(a+bn^{2} \bigr)+ b \lambda_{1} \bigl(n^{2} \bigr)\lambda_{2} \bigl(n^{2} \bigr) \bigr) \\ &\quad=\lim_{n\rightarrow\infty} \bigl( \bigl(a+bn^{2} \bigr)^{(p+q)/2}+ b n^{p+q} \bigr) \\ &\quad=0. \end{aligned}$$

As $p,q\in\mathbf{R}$ with $p+q<0$, either $p<0$ or $q<0$. Hence, by inequality (39), one can see that it is sufficient to consider only the case $q<0$, and thus

$$\lim_{n\rightarrow\infty} \lambda_{2} \bigl(n^{2} \bigr) = \lim_{n\rightarrow\infty} n^{q}=0. $$

So, we can apply Corollary 5. □

Corollary 8

Let $\theta\in\mathbf{R}_{+}$ and let $p,q\in\mathbf{R}$ be such that $p+q<0$. If $f:\mathbf{R}\rightarrow Y$ satisfies $f(0)=0$ and the inequality

$$\bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y), z \bigr\Vert \leq \theta \vert x \vert ^{p} \vert y \vert ^{q},\quad x,y\in\mathbf{R}_{0}, z\in Y, $$

then f is a solution of (2).

Similarly, we can prove the following.

Corollary 9

Let $\theta\in\mathbf{R}_{+}$ and consider $p\in\mathbf{R}$ with $p<0$. Assume that (3) has a solution $f_{0}$. If $f:\mathbf{R}\rightarrow Y$ satisfies $f(0)=f_{0}(0)$ and the inequality

$$\bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y)-D(x,y), z \bigr\Vert \leq\theta \bigl( \vert x \vert ^{p} + \vert y \vert ^{p} \bigr), $$

where $x,y\in\mathbf{R}_{0}, z\in Y$, then f is a solution of (3).

Corollary 10

Let $\theta\in\mathbf{R}_{+}$ and consider $p\in\mathbf{R}$ with $p<0$. If $f:\mathbf{R}\rightarrow Y$ satisfies $f(0)=0$ and the inequality

$$\bigl\Vert f \bigl(\sqrt{ax^{2}+by^{2}} \bigr)-af(x)-bf(y), z \bigr\Vert \leq \theta \bigl( \vert x \vert ^{p} + \vert y \vert ^{p} \bigr),\quad x,y\in\mathbf {R}_{0}, z\in Y, $$

then f is a solution of (2).