On superstability of exponential functional equations

Abstract

The aim of this paper is to prove the superstability of the following functional equations:

$$\begin{aligned}& f \bigl(P(x,y) \bigr)= g(x)h(y), \\& f(x+y)=g(x)h(y). \end{aligned}$$

1 Introduction and preliminaries

The stability problem of functional equations was raised by Ulam from a question concerning the stability of group homomorphisms [1]. Hyers [2] obtained the first important result in this field. See [3–10] for more information on functional equations and applications. In 1979, Baker, Lawrence, and Zorzitto [11] proved the superstability of the exponential functional equation: Let X be a real vector space and $f:X\to \mathbb{R}$ be an approximately exponential function, i.e., there exists a nonnegative number ε such that

$$ \bigl\Vert f(x+y)-f(x)f(y) \bigr\Vert \leqslant \varepsilon ,\quad x,y\in X. $$

Then f is either bounded or exponential. The same result is also true for approximately exponential mappings f from a semigroup $(G, +)$ with values in a normed algebra with the property that the norm is multiplicative [12]. Gǎvruta [13] proved the superstability of the Lobacevski functional equation

$$ f \biggl( {\frac{x+y}{2}} \biggr)^{2}= f(x)f(y) $$

under the condition bounded by a constant. Kim [14] investigated the solution and the superstability of the Pexiderized Lobacevski functional equation

$$ f \biggl( {\frac{x+y}{2}} \biggr)^{2}= g(x)h(y). $$

Kim and Park [15] considered the superstability of the generalized Pexider exponential functional equation

$$ f \biggl( {\frac{x+y}{m}} \biggr)^{m}= g(x)h(y) $$

in unital normed algebras, where m is a positive integer. For more information on the superstability of functional equations and applications, see [16–18].

The aim of this paper is to prove the superstability of the following generalized Pexider exponential functional equation:

$$ f \bigl(P(x,y) \bigr) = g(x)h(y) $$

(GPE)

in unital normed algebras.

2 Superstability of the generalized Pexider exponential functional equation (GPE)

In this section, assume that $(G,*)$ is a semigroup with identity e, A is a commutative unital normed algebra with unit I, and $P:G\times G\to G$ is a function such that

$$ P(x,y*z)=P(x*y,z),\quad x,y,z\in G. $$

It is clear that if $x*y=y*x$, then $P(x,y)=P(x,y*e)=P(x*y,e)=P(y,x)$.

Example 2.1

• (1) Let G be an operator algebra and $P : G \times G\to G$ be given by $P(a,b)= ab$ for all $a,b \in G$. Then $P(x,yz) = x(yz) = (xy)z = P(xy, z)$ for all $x,y,z \in G$.

• (2) Let $G=GL_{2} ({\mathbb{C}})$ be the set of invertible $2\times 2$ complex matrices and $P : G \times G\to G$ be given by $P(a, b) = b^{-1} a^{-1}$ for all $a,b \in G$. Then

  $$ P(x,yz) = (yz)^{-1} x^{-1} = \bigl(z^{-1} y^{-1} \bigr) x^{-1} = z^{-1} \bigl(y^{-1} x^{-1}\bigr) = z^{-1} (xy)^{-1} = P(xy, z) $$

  for all $x,y,z \in G$.

• (3) Let $G=U(A)$ be the unitary group of a unital $C^{*}$-algebra A and $P : G \times G\to G$ be given by $P(a,b)= b^{*} a^{*}$ for all $a,b \in G$. Then

  $$ P(x,yz) = (yz)^{*} x^{*} = \bigl(z^{*} y^{*} \bigr) x^{*} = z^{*} \bigl(y^{*} x^{*}\bigr) = z^{*} (xy)^{*} = P(xy, z) $$

  for all $x,y,z \in G$ (see [19]).

We prove the superstability of the generalized Pexider exponential equation (GPE).

Theorem 2.2

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{y\in G}\varphi (x, y)<\infty$ for each $x\in G$, and that $f, g, h : G \rightarrow A$ satisfy the inequality

$$ \bigl\Vert {f \bigl(P(x,y) \bigr)}- g(x)h(y) \bigr\Vert \leqslant \varphi (x, y) $$

(2.1)

for all $x,y\in G$. If there exists a sequence $\{y_{n}\}_{n}$ in G such that $\|h(y_{n})^{-1}\| \to 0$ as $n \to \infty$, then g satisfies $g(x*y)g(e)=g(x)g(y)$ for all $x,y\in G$.

Proof

It follows from (2.1) that

$$ \begin{aligned} \bigl\Vert f \bigl(P(x,y_{n}) \bigr) h(y_{n})^{-1} - g(x) \bigr\Vert & = \bigl\Vert \bigl[f \bigl(P(x,y_{n}) \bigr) - g(x) h(y_{n}) \bigr]h(y_{n})^{-1} \bigr\Vert \\ &\leqslant \bigl\Vert f \bigl(P(x,y_{n}) \bigr) - g(x) h(y_{n}) \bigr\Vert \cdot \bigl\Vert h(y_{n})^{-1} \bigr\Vert \\ &\leqslant \bigl\Vert h(y_{n})^{-1} \bigr\Vert \varphi (x, y_{n}) \end{aligned} $$

(2.2)

for all $x\in G$. So we have

$$ g(x)=\lim_{n \to \infty } f (P(x,y_{n} ) h(y_{n})^{-1},\quad x\in G. $$

(2.3)

For $x,y\in G$, let $\Delta (x,y)=f ( P(x,y) )- g(x)h(y)$. Then (2.1) implies that

$$\begin{aligned} \bigl\Vert \Delta (x,y*y_{n})g(z) \bigr\Vert \leqslant \varphi (x, y*y_{n}) \bigl\Vert g(z) \bigr\Vert \end{aligned}$$

for all $x,y,z\in G$. Since $\lim_{n\to \infty }\|h(y_{n})^{-1}\|=0$, we get

$$ \lim_{n\to \infty }\Delta (x,y*y_{n})g(z)h(y_{n})^{-1}=0,\quad x,y,z \in G. $$

Therefore it follows from (2.5) that

$$\begin{aligned} g(x*y)g(z)={}&\lim_{n \to \infty } f (P(x*y,y_{n} ) h(y_{n})^{-1} g(z) \\ ={}&\lim_{n \to \infty } f (P(x,y*y_{n} ) h(y_{n})^{-1} g(z) \\ ={}&\lim_{n \to \infty } \bigl[\Delta (x,y*y_{n})h(y_{n})^{-1}g(z)+g(x)h(y*y_{n})h(y_{n})^{-1}g(z) \bigr] \\ ={}&\lim_{n \to \infty } \bigl[\Delta (x,y*y_{n})g(z)h(y_{n})^{-1}+f \bigl(P(z,y*y_{n}) \bigr)g(x)h(y_{n})^{-1} \\ &{} -\Delta (z,y*y_{n})g(x)h(y_{n})^{-1} \bigr] \\ ={}&\lim_{n \to \infty } \bigl[\Delta (x,y*y_{n})g(z)h(y_{n})^{-1}+f \bigl(P(z*y,y_{n}) \bigr)g(x)h(y_{n})^{-1} \\ &{} -\Delta (z,y*y_{n})g(x)h(y_{n})^{-1} \bigr] \\ ={}&g(x)g(z*y). \end{aligned}$$

Hence

$$ g(x*y)g(z)=g(x)g(z*y),\quad x,y,z\in G. $$

(2.4)

Putting $z=e$ in (2.4), we obtain $g(x*y)g(e)=g(x)g(y)$ for all $x,y\in G$. □

Using the proof of Theorem 2.2, we get the following result.

Theorem 2.3

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $f, g, h : G \rightarrow A$ satisfy inequality (2.1) for all $x,y\in G$. If there exists a sequence $\{y_{n}\}_{n}$ in G such that $\varphi (x,y*y_{n})\|h(y_{n})^{-1}\| \to 0$ for all $x,y\in G$, as $n \to \infty$, then g satisfies $g(x*y)g(e)=g(x)g(y)$ for all $x,y\in G$.

Proof

According to the proof of Theorem 2.2, we get (2.2). By the assumption (with $y=e$), we have $\varphi (x,y_{n})\|h(y_{n})^{-1}\| \to 0$ for all $x\in G$. Then (2.2) implies (2.5). Let $\Delta (x,y):=f ( P(x,y) )- g(x)h(y)$. By (2.1), we have

$$\begin{aligned} \bigl\Vert \Delta (x,y*y_{n})g(z) \bigr\Vert \leqslant \varphi (x, y*y_{n}) \bigl\Vert g(z) \bigr\Vert ,\quad x,y,z\in G. \end{aligned}$$

Since $\varphi (x,y*y_{n})\|h(y_{n})^{-1}\| \to 0$, we get

$$ \lim_{n\to \infty }\Delta (x,y*y_{n})g(z)h(y_{n})^{-1}=0,\quad x,y,z \in G. $$

The rest of the proof is the same as the proof of Theorem 2.2. □

The proof of the following theorem is similar to the proof of Theorem 2.2.

Theorem 2.4

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{x\in G}\varphi (x, y)<\infty$ for each $y\in G$, and that $f, g, h : G \rightarrow A$ satisfy inequality (2.1) for all $x,y\in G$. If there exists a sequence $\{y_{n}\}_{n}$ in G such that $\|g(y_{n})^{-1}\| \to 0$ as $n \to \infty$, then h satisfies $h(x*y)h(e)=h(x)h(y)$ for all $x,y\in G$.

Proof

It follows from (2.1) that

$$ \begin{aligned} &\bigl\Vert f \bigl(P(y_{n},x) \bigr) g(y_{n})^{-1} - h(x) \bigr\Vert \\ &\quad = \bigl\Vert \bigl[f \bigl(P(y_{n},x) \bigr) - g(y_{n}) h(x) \bigr]g(y_{n})^{-1} \bigr\Vert \\ &\quad \leqslant \bigl\Vert f \bigl(P(y_{n},x) \bigr) - g(y_{n}) h(x) \bigr\Vert \cdot \bigl\Vert g(y_{n})^{-1} \bigr\Vert \\ &\quad \leqslant \bigl\Vert g(y_{n})^{-1} \bigr\Vert \varphi (y_{n},x) \end{aligned} $$

for all $x\in G$. So we have

$$ h(x)=\lim_{n \to \infty } f (P(y_{n},x ) g(y_{n})^{-1},\quad x\in G. $$

(2.5)

Let $\Delta (x,y)=f ( P(x,y) )- g(x)h(y)$. Then (2.1) implies that

$$ \bigl\Vert \Delta (y_{n}*y,z)h(x) \bigr\Vert \leqslant \varphi ( y_{n}*y,z) \bigl\Vert h(x) \bigr\Vert ,\quad x,y,z\in G. $$

Since $\lim_{n\to \infty }\|g(y_{n})^{-1}\|=0$, we get

$$ \lim_{n\to \infty }\Delta (y_{n}*y,z)h(x)g(y_{n})^{-1}=0,\quad x,y,z \in G. $$

Therefore it follows from (2.5) that

$$\begin{aligned} h(x*y)h(z)&=\lim_{n \to \infty } f (P(y_{n}, x*y ) g(y_{n})^{-1} h(z) \\ &=\lim_{n \to \infty } f (P(y_{n}*x,y ) g(y_{n})^{-1} h(z) \\ &=\lim_{n \to \infty } \bigl[\Delta (y_{n}*x,y)h(z)g(y_{n})^{-1}+g(y_{n}*x)h(y)g(y_{n})^{-1}h(z) \bigr] \\ &=\lim_{n \to \infty } \bigl[\Delta (y_{n}*x,y)h(z)g(y_{n})^{-1}+f \bigl(P(y_{n}*x,z) \bigr)h(y)g(y_{n})^{-1} \\ & -\Delta (y_{n}*x,z)h(y)g(y_{n})^{-1} \bigr] \\ &=\lim_{n \to \infty } f \bigl(P(y_{n}, x*z) \bigr)g(y_{n})^{-1}h(y) \\ &=h(x*z)h(y). \end{aligned}$$

Hence

$$ h(x*y)h(z)=h(x*z)h(y),\quad x,y,z\in G. $$

(2.6)

Letting $z=e$ in (2.6), we obtain $h(x*y)h(e)=h(x)h(y)$ for all $x,y\in G$. □

Remark 2.5

It is clear that the results in Theorems 2.2 and 2.4 are valid if $\varphi (x,y)=\varepsilon$ for all $x,y\in G$, where $\varepsilon \geqslant 0$ is a constant.

Corollary 2.6

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{y\in G}\varphi (x, y)<\infty$ for each $x\in G (\sup_{x\in G}\varphi (x, y)<\infty \textit{ for each }y\in G)$, and that $f, g : G \rightarrow A$ satisfy the inequality

$$ \bigl\Vert f \bigl( P(x,y) \bigr)- g(x)g(y) \bigr\Vert \leqslant \varphi (x, y) $$

for all $x,y\in G$ and $g(e)=I$. If there exists a sequence $\{y_{n}\}_{n}$ in G such that $\|g(y_{n})^{-1}\| \to 0$ as $n \to \infty$, then g satisfies $g(x*y)=g(x)g(y)$ for all $x,y\in G$.

Proof

Letting $h=g$ in Theorems 2.2 and 2.4 and using $g(e)=I$, we get the desired result. □

Corollary 2.7

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{y\in G}\varphi (x, y)<\infty$ for each $x\in G (\sup_{x\in G}\varphi (x, y)<\infty \textit{ for each }y\in G)$, and that $f : G \rightarrow A$ satisfies the inequality

$$ \bigl\Vert f \bigl( P(x,y) \bigr)- f(x)f(y) \bigr\Vert \leqslant \varphi (x,y) $$

for all $x,y\in G$. If there exists a sequence $\{y_{n}\}_{n}$ in G such that $\|f(y_{n})^{-1}\| \to 0$ as $n \to \infty$, then f satisfies $f(x*y)f(e)=f(x)f(y)$ for all $x,y\in G$.

Proof

Letting $h=g=f$ in Theorems 2.2 and 2.4, we get the desired result. □

Corollary 2.8

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{y\in G}\varphi (x, y)<\infty$ for each $x\in G$, and that $f, g, h : G \rightarrow \mathbb{C}$ satisfy the inequality

$$ \bigl\vert f \bigl( P(x,y) \bigr)- g(x)h(y) \bigr\vert \leqslant \varphi (x,y) $$

for all $x,y\in G$. If h is not bounded, then g satisfies $g(x*y)g(e)=g(x)g(y)$ for all $x,y\in G$.

Proof

Since h is not bounded, one can choose $\{y_{n}\}_{n}$ such that $|h(y_{n})^{-1}| = \frac{1}{|h(y_{n})|} \rightarrow 0$ as $n \rightarrow \infty$. Hence one can get the desired result by Theorem 2.2. □

Corollary 2.9

Let $\varphi : G\times G \rightarrow [0,+\infty )$ be a function. Assume that $\sup_{x\in G}\varphi (x, y)<\infty$ for each $y\in G$, and that $f, g, h : G \rightarrow \mathbb{C}$ satisfy the inequality

$$ \bigl\vert f \bigl( P(x,y) \bigr)- g(x)h(y) \bigr\vert \leqslant \varphi (x,y) $$

for all $x,y\in G$. If g is not bounded, then h satisfies $h(x*y)h(e)=h(x)h(y)$ for all $x,y\in G$.

Proof

Since g is not bounded, one can choose $\{y_{n}\}_{n}$ such that $|g(y_{n})^{-1}| = \frac{1}{|g(y_{n})|} \rightarrow 0$ as $n \rightarrow \infty$. Hence one can get the desired result by Theorem 2.4. □

Corollary 2.10

Suppose that $f: G \rightarrow \mathbb{C}$ satisfies the inequality

$$ \bigl\vert f \bigl( P(x,y) \bigr)- f(x)f(y) \bigr\vert \leqslant \varepsilon $$

for all $x,y\in G$. If f is not bounded, then f satisfies $f(x*y)f(e)=f(x)f(y)$ for all $x,y\in G$.

Remark 2.11

When the semigroup G and the function $\varphi (x,y)$ in the above results are replaced with an algebra B and $\varphi (x)$ or $\varphi (y)$, respectively, we have similar results.

Given a semigroup $(G,*)$ and a commutative field $\mathbb{F}$, let W be a vector space of functions f from G into $\mathbb{F}$. The vector space W is called right invariant if, for each $f\in W$, the mapping $\psi _{y}:G\to \mathbb{F}$, for each $y\in G$, defined by $\psi _{y}(x)=f(x*y)$ belongs to W. Left invariant spaces are defined similarly. The vector space W is called invariant if it is both right and left invariant. Following Székelyhidi [20], we obtain the following results.

Theorem 2.12

Given a semigroup $(G,*)$ with identity e, a commutative field $\mathbb{F}$, and a right invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,g,h:G\to \mathbb{F}$ be such that the function $\psi _{y}:G\to \mathbb{F}$ defined by $\psi _{y}(x)=f (P(x,y) )-g(x)h(y)$ belongs to W for each $y\in G$. If $h(e)=1$, then either $g\in W$ or $h(x*y)=h(x)h(y)$ for all $x,y\in G$.

Proof

Suppose that there are $y_{0},z_{0}\in G$ such that $h(y_{0}*z_{0})\neq h(y_{0})h(z_{0})$. Hence

$$\begin{aligned} \bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]g(x)={}&\bigl[f \bigl(P(x*y_{0},z_{0}) \bigr)-g(x*y_{0})h(z_{0})\bigr] \\ & {}-\bigl[f \bigl(P(x,y_{0}*z_{0}) \bigr)-g(x)h(y_{0}*z_{0})\bigr] \\ & {}+h(z_{0})\bigl[f \bigl(P(x,y_{0}) \bigr)-g(x)h(y_{0})\bigr] \\ & {}-h(z_{0})\bigl[f \bigl(P(x*y_{0},e) \bigr)-g(x*y_{0})h(e)\bigr] \end{aligned}$$

for all $x\in G$. Since $\psi _{y}\in W$ for each $y\in G$, we get that the function $\phi _{y,z}:G\to \mathbb{F}$ defined by $\phi _{y,z}(x)=\psi _{y}(x*z)$ for all $z\in G$ belongs to W. Therefore

$$ g=\bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]^{-1}\bigl[\phi _{z_{0},y_{0}}-\phi _{y_{0}*z_{0},e}+h(z_{0}) \phi _{y_{0},e}-h(z_{0})\phi _{e,y_{0}}\bigr]. $$

So we conclude $g\in W$. □

Let $(G,*)$ be a semigroup and $B(G,\mathbb{C})$ be the linear space of bounded functions with complex values on G. It is clear that $B(G,\mathbb{C})$ is an invariant vector space. Hence Corollary 2.9 is a consequence of Theorem 2.12.

Theorem 2.13

Given a semigroup $(G,*)$ with identity e, a commutative field $\mathbb{F}$, and an invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,g,h:G\to \mathbb{F}$ be such that the functions $\phi _{x},\psi _{y}:G\to \mathbb{F}$ defined by $\phi _{x}(y)=f (P(x,y) )-g(x)h(y)$ and $\psi _{y}(x)=f (P(x,y) )-g(x)h(y)$ belong to W for each $x,y\in G$. If $h(e)=1$, then either $g\in W$ or $h(x*y)=h(x)h(y)$ and $g(x)=g(e)h(x)$ for all $x,y\in G$.

Proof

Suppose that $g\notin W$. By Theorem 2.12, h satisfies $h(x*y)=h(x)h(y)$ for all $x,y\in G$. Since $\phi _{e}\in W$ and W is left invariant, the function $\xi _{x}:G\to \mathbb{F}$ for each $x\in G$, defined by $\xi _{x}(y)=\phi _{e}(x*y)$, belongs to W. For all $x,y\in G$, we have

$$\begin{aligned} f \bigl(P(e,x*y) \bigr)-g(e)h(x*y) &=\bigl[f \bigl(P(x,y) \bigr)-g(x)h(y)\bigr]+ \bigl[g(x)h(y)-g(e)h(x*y)\bigr] \\ &=\bigl[f \bigl(P(x,y) \bigr)-g(x)h(y)\bigr]+\bigl[g(x)-g(e)h(x)\bigr]h(y). \end{aligned}$$

Therefore

$$ \bigl[g(x)-g(e)h(x)\bigr]h=\xi _{x}-\phi _{x},\quad x\in G. $$

(2.7)

We claim that $g(x)=g(e)h(x)$ for all $x\in G$. If there is $x_{0}\in G$ such that $g(x_{0})-g(e)h(x_{0})\neq 0$, then (2.7) implies that $h\in W$. Since $h, \phi _{e}\in W$, we get that the function $\vartheta :G\to \mathbb{F}$ defined by $\vartheta (y)=f (P(e,y) )=f (P(y,e) )$ belongs to W. Therefore $g\in W$, since $\vartheta =\psi _{e}+g$. This contradiction implies that $g(x)=g(e)h(x)$ for all $x\in G$. □

Corollary 2.14

Given a semigroup $(G,*)$ with identity e and a function $\varphi :G\times G\to [0,+\infty )$ with $\sup_{x\in G}\varphi (x,y)<\infty$ and $\sup_{y\in G}\varphi (x,y)<\infty$ for all $x,y\in G$. Let $f,g,h:G\to \mathbb{F}$ be such that

$$ \bigl\vert f \bigl( P(x,y) \bigr)- g(x)h(y) \bigr\vert \leqslant \varphi (x,y),\quad x,y\in G. $$

If $h(e)=1$, then either g is bounded or $h(x*y)=h(x)h(y)$ and $g(x)=g(e)h(x)$ for all $x,y\in G$.

Proof

Let W be the vector space of all bounded functions from G to $\mathbb{F}$. Then W is an invariant vector space. Hence the desired result follows from Theorem 2.13. □

Corollary 2.15

Given a semigroup $(G,*)$ with identity e, a commutative field $\mathbb{F}$, and an invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,g,h:G\to \mathbb{F}$ be such that the functions $\phi _{x},\psi _{y}:G\to \mathbb{F}$ defined by $\phi _{x}(y)=f(x*y)-g(x)h(y)$ and $\psi _{y}(x)=f(x*y)-g(x)h(y)$ belong to W for each $x,y\in G$. If $h(e)=1$, then either $g\in W$ or $h(x*y)=h(x)h(y)$ and $g(x)=g(e)h(x)$ for all $x,y\in G$. Moreover, if $g\in W$, then $f\in W$.

Proof

The desired result follows from Theorem 2.13 by letting $P(x,y)=x*y$ for all $x,y\in G$. □

Specially, we have the following result without any assumption on h.

Corollary 2.16

Given a semigroup $(G,*)$ with identity e, a commutative field $\mathbb{F}$, and an invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,h:G\to \mathbb{F}$ be such that the functions $\phi _{x},\psi _{y}:G\to \mathbb{F}$ defined by $\phi _{x}(y)=f(x*y)-f(x)h(y)$ and $\psi _{y}(x)=f(x*y)-f(x)h(y)$ belong to W for each $x,y\in G$. Then either $f\in W$ or $h(x*y)=h(x)h(y)$ and $f(x)=f(e)h(x)$ for all $x,y\in G$.

Proof

Suppose that $f\notin W$. We claim that $h(x*y)=h(x)h(y)$ for all $x,y\in G$. Let $y_{0},z_{0}\in G$ such that $h(y_{0}*z_{0})\neq h(y_{0})h(z_{0})$. Hence

$$\begin{aligned} \bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]f(x)={}&\bigl[f(x*y_{0}*z_{0})-f(x*y_{0})h(z_{0}) \bigr] \\ &{} -\bigl[f(x*y_{0}*z_{0})-f(x)h(y_{0}*z_{0}) \bigr] \\ & {}+h(z_{0})\bigl[f(x*y_{0})-f(x)h(y_{0}) \bigr] \end{aligned}$$

for all $x\in G$. Since $\psi _{y}\in W$ for each $y\in G$, we get that the function $\theta _{y,z}:G\to \mathbb{F}$ defined by $\theta _{y,z}(x)=\psi _{y}(x*z)$ for all $z\in G$ belongs to W. Therefore

$$ f=\bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]^{-1}[\theta _{z_{0},y_{0}}- \theta _{y_{0}*z_{0},e}+h(z_{0}) \theta _{y_{0},e}. $$

So we conclude $f\in W$, which is a contradiction. Now, we prove $f(x)=f(e)h(x)$ for all $x\in G$. Suppose that there is $x_{0}\in G$ such that $f(x_{0})-f(e)h(x_{0})\neq 0$. For each $y\in G$, we have

$$\begin{aligned} f(x_{0}*y))-f(e)h(x_{0}*y) &=\bigl[f(x_{0}*y)-f(x_{0})h(y) \bigr]+\bigl[f(x_{0})h(y)-f(e)h(x_{0}*y)\bigr] \\ &=\bigl[f(x_{0}*y)-f(x_{0})h(y)\bigr]+ \bigl[f(x_{0})-f(e)h(x_{0})\bigr]h(y). \end{aligned}$$

Therefore, $[g(x_{0})-f(e)h(x_{0})]h\in W$, and so $h\in W$. Since $h, \phi _{e}\in W$ and $\phi _{e}=f-f(e)h$, we get $f\in W$, which is again a contradiction. □

Theorem 2.17

Given a semigroup $(G,*)$, a commutative field $\mathbb{F}$, and a right invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,g,h:G\to \mathbb{F}$ be such that the function $\psi _{y}:G\to \mathbb{F}$ defined by $\psi _{y}(x)=f(x*y)-g(x)h(y)$ belongs to W for each $y\in G$. If $f-g\in W$ or $h(e)=1$ (when G has the identity e), then either $f,g\in W$ or $h(x*y)=h(x)h(y)$ for all $x,y\in G$.

Proof

Let G have not an identity. Suppose that $f-g\in W$ and there are $y_{0},z_{0}\in G$ such that $h(y_{0}*z_{0})\neq h(y_{0})h(z_{0})$. Then

$$\begin{aligned} \bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]g(x)=&\bigl[f(x*y_{0}*z_{0})-g(x*y_{0})h(z_{0}) \bigr] \\ & {}-\bigl[f(x*y_{0}*z_{0})-g(x)h(y_{0}*z_{0}) \bigr] \\ &{} +h(z_{0})\bigl[f(x*y_{0})-g(x)h(y_{0}) \bigr] \\ &{} -h(z_{0})\bigl[f(x*y_{0})-g(x*y_{0}) \bigr] \end{aligned}$$

for all $x\in G$. Since $\psi _{y}, f-g\in W$ for each $y\in G$, we get that the functions $\phi _{y}, \varphi :G\to \mathbb{F}$ defined by $\phi _{y}(x)=\psi _{y}(x*y_{0})$ and $\varphi (x)=(f-g)(x*y_{0})$ belong to W. Therefore

$$ g=\bigl[h(y_{0}*z_{0})-h(y_{0})h(z_{0}) \bigr]^{-1}\bigl[\phi _{z_{0}}-\psi _{y_{0}*z_{0}}+h(z_{0}) \psi _{y_{0}}-h(z_{0})\varphi \bigr]. $$

So $g\in W$, and we conclude $f\in W$. For $h(e)=1$, if G has the identity e, we get $f-g\in W$ by the hypothesis. Hence the result follows from the previous case. □

Specially, we get Székelyhidi’s result.

Corollary 2.18

([20])

Given a semigroup $(G,*)$, a commutative field $\mathbb{F}$, and a right invariant vector space W of $\mathbb{F}$-valued functions on G. Let $f,g:G\to \mathbb{F}$ be such that the function $\psi _{y}:G\to \mathbb{F}$ defined by $\psi _{y}(x)=f(x*y)-f(x)g(y)$ belongs to W for each $y\in G$. Then either $f\in W$ or $g(x*y)=g(x)g(y)$ for all $x,y\in G$.

Proof

It follows from Theorem 2.17 by replacing g with f, and h with g. □

3 Superstability of the Pexider exponential equation

Using an idea from [21], we establish the superstability of the Pexider exponential equation $f(x+y)=g(x)h(y)$.

Theorem 3.1

Let X and E be a real normed space and a normed algebra with multiplicative norm, respectively. Let $a\in (E\setminus \{0\})\cup (\mathbb{R}\setminus \{0\})$ and $f:X\to E$ be a function such that $af(z)=f(z)a$ for all $z\in X$. If f satisfies the inequality

$$ \bigl\Vert af(x+y)-f(x)f(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p},\quad x,y\in X $$

(3.1)

for some $\varepsilon , \theta , p\geqslant 0$, then either $\sup_{\|x\|\geqslant 1}\frac{\|f(x)\|}{\|x\|^{p}}<\infty$ or

$$ af(x+y)=f(x)f(y),\quad x,y\in X. $$

Proof

We assume that $\varepsilon +\theta >0$ and continue to employ the notation $\sharp a \sharp$ to denote $\|a\|$ (if $a\in E$) and $|a|$ (if $a\in \mathbb{R}$), respectively. Let $\{ \frac{\|f(x)\|}{\|x\|^{p}}: \|x\|\geqslant 1 \}$ be not bounded. Then there exists a sequence $\{x_{n}\}_{n=1}^{\infty }\subseteq X$ such that

$$ \Vert x_{n} \Vert \geqslant 1,\qquad \frac{ \Vert f(x_{n}) \Vert }{ \Vert x_{n} \Vert ^{p}} \geqslant n,\quad n\in \mathbb{N}. $$

Therefore

$$ \lim_{n\to \infty } \frac{ \Vert x_{n} \Vert ^{p}}{ \Vert f(x_{n}) \Vert }=0, \qquad \lim _{n \to \infty } \bigl\Vert f(x_{n}) \bigr\Vert =+ \infty . $$

(3.2)

Choose $x,y,z\in X$ with $f(x)\neq 0$. It then follows from (3.1) that

$$\begin{aligned}& \bigl\Vert af(x+y+z)-f(z)f(x+y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x+y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr)+ \theta \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}, \\& \bigl\Vert af(x+y+z)-f(y+z)f(x) \bigr\Vert \leqslant \varepsilon \bigl( \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}\bigr)+ \theta \Vert y+z \Vert ^{p} \Vert x \Vert ^{p}. \end{aligned}$$

Hence we get

$$ \begin{aligned} \bigl\Vert f(z)f(x+y)-f(y+z)f(x) \bigr\Vert \leqslant {}&\varepsilon \bigl( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\theta \bigl( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert y+z \Vert ^{p} \Vert x \Vert ^{p}\bigr). \end{aligned} $$

(3.3)

In view of (3.1), we have

$$ \begin{aligned} \bigl\Vert af(z)f(x+y)-f(z)f(y)f(x) \bigr\Vert \leqslant \bigl\Vert f(z) \bigr\Vert \bigl[\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \bigr]. \end{aligned} $$

(3.4)

Inequalities (3.3) and (3.4) yield

$$ \begin{aligned} \bigl\Vert af(y+z)f(x)-f(z)f(y)f(x) \bigr\Vert \leqslant{}& \varepsilon \sharp a \sharp \bigl( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\theta \sharp a \sharp \bigl( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert y+z \Vert ^{p} \Vert x \Vert ^{p}\bigr) \\ & {}+ \bigl\Vert f(z) \bigr\Vert \bigl[\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \bigr]. \end{aligned} $$

(3.5)

Since E is a normed algebra with multiplicative norm, it follows from (3.5) that

$$ \begin{aligned} & \bigl\Vert af(y+z)-f(z)f(y) \bigr\Vert \\ &\quad \leqslant \frac{\varepsilon \sharp a \sharp ( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p})}{ \Vert f(x) \Vert } \\ &\qquad {} + \frac{\theta \sharp a \sharp ( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert y+z \Vert ^{p} \Vert x \Vert ^{p})}{ \Vert f(x) \Vert } \\ &\qquad {} + \frac{ \Vert f(z) \Vert [\varepsilon ( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p})+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} ]}{ \Vert f(x) \Vert }. \end{aligned} $$

(3.6)

If we put $x=x_{n}$ in (3.6) and take the limit as $n\to +\infty$, then it follows from (3.2) that

$$ af(z+y)=f(z)f(y), $$

as desired. □

Theorem 3.2

Let X and E be a real normed space and a normed algebra with multiplicative norm, respectively. Let $f, g:X\to E$ be functions such that $f(0)f(z)=f(z)f(0)$ for all $z\in X$ and satisfy

$$ \bigl\Vert g(x+y)-f(x)f(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p} \bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p},\quad x,y\in E $$

(3.7)

for some $\varepsilon , \theta , p\geqslant 0$. Then either $\sup_{\|x\|\geqslant 1}\frac{\|g(x)\|}{\|x\|^{p}}<\infty$ or

$$ f(0)f(x+y)=f(x)f(y),\quad x,y\in X. $$

Proof

Let $\{ \frac{\|g(x)\|}{\|x\|^{p}}: \|x\|\geqslant 1 \}$ be not bounded. Then (3.7) implies that $f(0)\neq 0$ and $\{ \frac{\|f(x)\|}{\|x\|^{p}}: \|x\|\geqslant 1 \}$ is not bounded. In view of (3.7), we have

$$ \bigl\Vert g(x+y)-f(0)f(x+y) \bigr\Vert \leqslant \varepsilon \Vert x+y \Vert ^{p}\leqslant 2^{p} \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr), \quad x,y\in X. $$

Therefore

$$\begin{aligned} \bigl\Vert f(0)f(x+y)-f(x)f(y) \bigr\Vert &\leqslant \bigl\Vert f(0)f(x+y)-g(x+y) \bigr\Vert + \bigl\Vert g(x+y)-f(x)f(y) \bigr\Vert \\ &\leqslant \bigl(2^{p} +1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \quad x,y\in X. \end{aligned}$$

By Theorem 3.1, we conclude that $f(0)f(x+y)=f(x)f(y)$ for all $x,y\in X$. □

Theorem 3.3

Let X and E be a real normed space and a normed algebra with multiplicative norm, respectively. Let $a\in (E\setminus \{0\})\cup (\mathbb{R}\setminus \{0\})$ and $f,g:X\to E$ be functions satisfying one of the following conditions:

• (i) $af(z)=f(z)a$, $\|af(x+y)-f(x)g(y)\|\leqslant \varepsilon (\|x\|^{p}+ \|y\|^{p})+\theta \|x\|^{p}\|y\|^{p}$, $x,y,z\in X$;

• (ii) $ag(z)=g(z)a$, $\|af(x+y)-g(x)f(y)\|\leqslant \varepsilon (\|x\|^{p}+ \|y\|^{p})+\theta \|x\|^{p}\|y\|^{p}$, $x,y,z\in X$,

for some $\varepsilon , \theta , p\geqslant 0$. Then either $\sup_{\|x\|\geqslant 1}\frac{\|f(x)\|}{\|x\|^{p}}<\infty$ or

$$ ag(x+y)=g(x)g(y),\quad x,y\in X. $$

Proof

We use the notation $\sharp a \sharp$ to denote $\|a\|$ (if $a\in E$) and $|a|$ (if $a\in \mathbb{R}$), respectively. Let f, g satisfy (i) and $\{ \frac{\|f(x)\|}{\|x\|^{p}}: \|x\|\geqslant 1 \}$ be unbounded. Then there exists a sequence $\{x_{n}\}_{n=1}^{\infty }\subseteq X$ such that (3.2) holds true. In view of (i), we have

$$\begin{aligned} &\bigl\Vert af(x+y+z)-f(x+y)g(z) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x+y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr)+ \theta \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}, \\ &\bigl\Vert af(x+y+z)-f(x)g(y+z) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y+z \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y+z \Vert ^{p},\quad x,y,z\in X. \end{aligned}$$

Therefore

$$ \begin{aligned} \bigl\Vert f(x+y)g(z)-f(x)g(y+z) \bigr\Vert \leqslant {}&\varepsilon \bigl( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\theta \bigl( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert x \Vert ^{p} \Vert y+z \Vert ^{p}\bigr) \end{aligned} $$

(3.8)

for all $x,y,z\in X$. On the other hand, (i) implies

$$ \bigl\Vert af(x+y)g(z)-f(x)g(y)g(z) \bigr\Vert \leqslant \bigl\Vert g(z) \bigr\Vert \bigl[\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \bigr] $$

(3.9)

for all $x,y,z\in X$. It follows from (3.8) and (3.9) that

$$ \begin{aligned} \bigl\Vert af(x)g(y+z)-f(x)g(y)g(z) \bigr\Vert \leqslant{}& \varepsilon \sharp a \sharp \bigl( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ & {}+\theta \sharp a \sharp \bigl( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert x \Vert ^{p} \Vert y+z \Vert ^{p}\bigr) \\ &{} + \bigl\Vert g(z) \bigr\Vert \bigl[\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \bigr] \end{aligned} $$

(3.10)

for all $x,y,z\in X$. Since $af(x)=f(x)a$ and E is a normed algebra with multiplicative norm, it follows from (3.10) that

$$ \begin{aligned} \bigl\Vert ag(y+z)-g(y)g(z) \bigr\Vert &\leqslant \frac{\varepsilon \sharp a \sharp ( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p})}{ \Vert f(x) \Vert } \\ & + \frac{\theta \sharp a \sharp ( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert x \Vert ^{p} \Vert y+z \Vert ^{p})}{ \Vert f(x) \Vert } \\ & + \frac{ \Vert g(z) \Vert [\varepsilon ( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p})+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} ]}{ \Vert f(x) \Vert } \end{aligned} $$

(3.11)

for all $x,y,z\in X$. If we put $x=x_{n}$ in (3.11) and take the limit as $n\to +\infty$, then it follows from (3.2) that

$$ ag(z+y)=g(z)g(y),\quad y,z\in X. $$

Similarly, we get the result if f, g satisfy condition (ii). □

Theorem 3.4

Let X and E be a real normed space and a normed algebra with multiplicative norm, respectively. Let $f,g, h:X\to E$ be functions satisfying the inequality

$$ \bigl\Vert f(x+y)-g(x)h(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p} \bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \quad x,y\in X $$

(3.12)

for some $\varepsilon , \theta , p\geqslant 0$.

• (i) If $g(0)h(x)=h(x)g(0)$ for all $x\in X$, then either $\sup_{\|x\|\geqslant 1}\frac{\|h(x)\|}{\|x\|^{p}}<\infty$ or

  $$ g(0)g(x+y)=g(x)g(y),\quad x,y\in X. $$

• (ii) If $h(0)g(x)=g(x)h(0)$ for all $x\in X$, then either $\sup_{\|x\|\geqslant 1}\frac{\|g(x)\|}{\|x\|^{p}}<\infty$ or

  $$ h(0)h(x+y)=h(x)h(y),\quad x,y\in X. $$

Proof

In view of (3.12), we have

$$\begin{aligned} &\bigl\Vert f(x+y)-g(0)h(x+y) \bigr\Vert \leqslant \varepsilon \Vert x+y \Vert ^{p}\leqslant 2^{p} \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr), \\ &\bigl\Vert f(x+y)-g(x+y)h(0) \bigr\Vert \leqslant \varepsilon \Vert x+y \Vert ^{p}\leqslant 2^{p} \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr),\quad x,y,z\in X. \end{aligned}$$

Therefore

$$ \begin{aligned} & \bigl\Vert g(0)h(x+y)-g(x)h(y) \bigr\Vert \\ &\quad \leqslant \bigl\Vert f(x+y)-g(0)h(x+y) \bigr\Vert + \bigl\Vert f(x+y)-g(x)h(y) \bigr\Vert \\ &\quad \leqslant \bigl(2^{p}+1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \end{aligned} $$

(3.13)

and

$$ \begin{aligned} & \bigl\Vert g(x+y)h(0)-g(x)h(y) \bigr\Vert \\ &\quad \leqslant \bigl\Vert f(x+y)-g(x+y)h(0) \bigr\Vert + \bigl\Vert f(x+y)-g(x)h(y) \bigr\Vert \\ &\quad \leqslant \bigl(2^{p}+1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \end{aligned} $$

(3.14)

for all $x,y,z\in X$. To prove (i), if $g(0)\neq 0$, the result follows by Theorem 3.3. For the case $g(0)=0$, inequality (3.13) yields

$$ \bigl\Vert g(x)h(y) \bigr\Vert \leqslant \bigl(2^{p}+1\bigr) \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \quad x,y\in X. $$

Hence, if $\{ \frac{\|h(x)\|}{\|x\|^{p}}: \|x\|\geqslant 1 \}$ is unbounded, then the last inequality implies that $g(x)=0$ for all $x\in X$. This completes the proof of (i).

Similarly, one can prove (ii). □

We now show some counterparts of Shtern’s theorem (see [7]).

Theorem 3.5

Let E be a normed linear space and $\mathcal{A}$ be a complex Banach algebra. Assume that $a\in \mathcal{A}\cup \mathbb{R}$ and the mapping $f:E\to \mathcal{A}$ is such that $af(z)=f(z)a$ for all $z\in E$, and

$$ \bigl\Vert af(x+y)-f(x)f(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \quad x,y\in E $$

for some $\varepsilon , \theta , p\geqslant 0$. If, for each nonzero element $b\in \mathcal{A}$, the E-orbit of b

$$ O_{RE}(f, b):= \biggl\{ \frac{f(x)b}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} , \quad \textit{or}\quad O_{LE}(b, f):= \biggl\{ \frac{bf(x)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

is unbounded, then f satisfies $af(x+y)=f(x)f(y)$ for all $x,y\in E$. Moreover, if $a=0$, then $f\equiv 0$.

Proof

We assume that $\varepsilon +\theta >0$ and $O_{RE}(f, b)$ is unbounded for each $b\neq 0$. We continue to employ the notation $\sharp a \sharp$, to denote $\|a\|$ (if $a\in \mathcal{A}$) and $|a|$ (if $a\in \mathbb{R}$), respectively. For every $x, y, z\in E$, we have

$$\begin{aligned} \bigl\Vert f(x)\bigl[f(y)f(z)-af(y+z)\bigr] \bigr\Vert \leqslant{}& \bigl\Vert a^{2}f(x+y+z)-af(x)f(y+z) \bigr\Vert \\ &{} + \bigl\Vert af(x+y)f(z)-a^{2}f(x+y+z) \bigr\Vert \\ & {}+ \bigl\Vert f(x)f(y)f(z)-af(x+y)f(z) \bigr\Vert \\ \leqslant {}&\varepsilon \sharp a\sharp \bigl( \Vert x \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x+y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\theta \sharp a\sharp \bigl( \Vert x \Vert ^{p} \Vert y+z \Vert ^{p}+ \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}\bigr) \\ &{} +\varepsilon \bigl\Vert f(z) \bigr\Vert \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \bigl\Vert f(z) \bigr\Vert \Vert x \Vert ^{p} \Vert y \Vert ^{p} \\ \leqslant {}&\bigl( \bigl\Vert f(z) \bigr\Vert +2^{p+1}\sharp a \sharp \bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\bigl( \bigl\Vert f(z) \bigr\Vert +2^{p}\sharp a\sharp +2^{p}\bigr) \\ &{}\times\theta \bigl( \Vert x \Vert ^{p} \Vert y \Vert ^{p}+ \Vert x \Vert ^{p} \Vert z \Vert ^{p}+ \Vert y \Vert ^{p} \Vert z \Vert ^{p} \bigr). \end{aligned}$$

Then, for $\|x\|\geqslant 1$, we have

$$\begin{aligned} \bigl\Vert f(x)\bigl[f(y)f(z)-af(y+z)\bigr] \bigr\Vert \leqslant{}& \bigl( \bigl\Vert f(z) \bigr\Vert +2^{p+1}\sharp a\sharp \bigr) \varepsilon \Vert x \Vert ^{p}\bigl(1+ \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\bigl( \bigl\Vert f(z) \bigr\Vert +2^{p}\sharp a\sharp +2^{p}\bigr)\theta \Vert x \Vert ^{p} \bigl( \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}+ \Vert y \Vert ^{p} \Vert z \Vert ^{p} \bigr) \\ ={}&M \Vert x \Vert ^{p}, \end{aligned}$$

where

$$\begin{aligned} M :=&\varepsilon \bigl( \bigl\Vert f(z) \bigr\Vert +2^{p+1}\sharp a \sharp \bigr) \bigl(1+ \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{}+\theta \bigl( \bigl\Vert f(z) \bigr\Vert +2^{p}\sharp a\sharp +2^{p} \bigr) \bigl( \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}+ \Vert y \Vert ^{p} \Vert z \Vert ^{p} \bigr). \end{aligned}$$

Therefore,

$$ \biggl\Vert f(x)\frac{f(y)f(z)-af(y+z)}{M} \biggr\Vert \leqslant \Vert x \Vert ^{p}. $$

Letting $b:=\frac{f(y)f(z)-af(y+z)}{M}$, we get $O_{RE}(f, b)$ is bounded. By assumption, this implies $b=0$. Hence $af(y+z)=f(y)f(z)$. Moreover, if $a=0$, then we get $f(x)f(y)=0$ for all $x,y\in E$. Let $y\in E$ be an arbitrary element. Then $O_{RE}(f, f(y))=\{0\}$ is bounded, and by assumption we conclude that $f(y)=0$. Hence $f\equiv 0$. If we assume that $O_{LE}(b)$ is unbounded for each nonzero $b\in \mathcal{A}$, the proof proceeds in a similar way. □

Corollary 3.6

Let E be a normed linear space and $\mathcal{A}$ be a commutative semisimple complex Banach algebra. Assume that a mapping $f:E\to A$ satisfies

$$ \bigl\Vert f(x+y)-f(x)f(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p},\quad x,y\in E $$

for some $\varepsilon , \theta , p\geqslant 0$. If, for every nonzero linear multiplicative functional φ on $\mathcal{A}$, the set

$$ G_{\varphi }:= \biggl\{ \frac{(\varphi o f)(x)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

is unbounded, then f is exponential.

Proof

Let $b\neq 0$ be an element in $\mathcal{A}$. Since $\mathcal{A}$ is semisimple, there is a linear multiplicative functional φ on $\mathcal{A}$ such that $\varphi (b)\neq 0$. By assumption, $G_{\varphi }$ is unbounded. Then the set

$$ G_{\varphi }.\varphi (b)= \biggl\{ \frac{\varphi (f(x)b)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} =\varphi \bigl(O_{RE}(b)\bigr) $$

is unbounded, and we conclude that $O_{RE}(b)$ is unbounded. By Theorem 3.5, f is exponential. □

Corollary 3.7

Let E be a normed linear space and $\mathcal{A}$ be a complex Banach algebra. Assume that mappings $f, g:E\to \mathcal{A}$ satisfy $f(0)f(z)=f(z)f(0)$ for all $z\in E$, and (3.7). If, for each nonzero element $b\in \mathcal{A}$, the E-orbit of b

$$ O_{RE}(f,b):= \biggl\{ \frac{f(x)b}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} \quad \textit{or}\quad O_{LE}(b,f):= \biggl\{ \frac{bf(x)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

is unbounded, then f satisfies $f(0)f(x+y)=f(x)f(y)$ for all $x,y\in E$. Moreover, if $f(0)=0$, then $f\equiv 0$.

Proof

As in the proof of Theorem 3.2, we obtain

$$\begin{aligned} \bigl\Vert f(0)f(x+y)-f(x)f(y) \bigr\Vert &\leqslant \bigl\Vert f(0)f(x+y)-g(x+y) \bigr\Vert + \bigl\Vert g(x+y)-f(x)f(y) \bigr\Vert \\ &\leqslant \bigl(2^{p} +1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p},\quad x,y\in E. \end{aligned}$$

By Theorem 3.5, we get the desired result. □

Theorem 3.8

Let E be a normed linear space and $\mathcal{A}$ be a complex Banach algebra. Assume that $\varepsilon , \theta , p\geqslant 0$, $a\in \mathcal{A}\cup \mathbb{R}$ and $f,g:E\to \mathcal{A}$ satisfy one of the following conditions:

• (i) $af(z)=f(z)a$, $\|af(x+y)-f(x)g(y)\|\leqslant \varepsilon (\|x\|^{p}+ \|y\|^{p})+\theta \|x\|^{p}\|y\|^{p}$, $x,y,z\in E$; and for each nonzero element $b\in \mathcal{A}$, the E-orbit of b

  $$ O_{RE}(f,b):= \biggl\{ \frac{f(x)b}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

  is unbounded.

• (ii) $ag(z)=g(z)a$, $\|af(x+y)-g(x)f(y)\|\leqslant \varepsilon (\|x\|^{p}+ \|y\|^{p})+\theta \|x\|^{p}\|y\|^{p}$, $x,y,z\in E$; and for each nonzero element $b\in \mathcal{A}$, the E-orbit of b

  $$ O_{LE}(b,f):= \biggl\{ \frac{bf(x)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

  is unbounded.

Then g satisfies $ag(x+y)=g(x)g(y)$ for all $x,y\in E$. Moreover, if $a=0$, then $g\equiv 0$.

Proof

Let f, g satisfy (i) and $O_{RE}(f,b)$ be unbounded for each nonzero element $b\in \mathcal{A}$. Using the same argument as in the proof of Theorem 3.3, we obtain

$$ \begin{aligned} \bigl\Vert af(x)g(y+z)-f(x)g(y)g(z) \bigr\Vert \leqslant{}& \varepsilon \sharp a \sharp \bigl( \Vert x+y \Vert ^{p}+ \Vert y+z \Vert ^{p}+ \Vert x \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} +\theta \sharp a \sharp \bigl( \Vert x+y \Vert ^{p} \Vert z \Vert ^{p}+ \Vert x \Vert ^{p} \Vert y+z \Vert ^{p}\bigr) \\ &{} + \bigl\Vert g(z) \bigr\Vert \bigl[\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \bigr] \end{aligned} $$

for all $x,y,z\in E$. Since $af(x)=f(x)a$, for $\|x\|\geqslant 1$, we obtain

$$\begin{aligned} \bigl\Vert f(x)\bigl[ag(y+z)-g(y)g(z)\bigr] \bigr\Vert \leqslant{}& \bigl( \bigl\Vert g(z) \bigr\Vert +2^{p+1}\sharp a \sharp \bigr)\varepsilon \Vert x \Vert ^{p}\bigl(1+ \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{} + \bigl( \bigl\Vert g(z) \bigr\Vert +2^{p}\sharp a\sharp +2^{p} \bigr)\theta \Vert x \Vert ^{p} \bigl( \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}+ \Vert y \Vert ^{p} \Vert z \Vert ^{p} \bigr) \\ ={}&M \Vert x \Vert ^{p}, \end{aligned}$$

where

$$\begin{aligned} M :=&\varepsilon \bigl( \bigl\Vert g(z) \bigr\Vert +2^{p+1}\sharp a \sharp \bigr) \bigl(1+ \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}\bigr) \\ &{}+\theta \bigl( \bigl\Vert g(z) \bigr\Vert +2^{p}\sharp a\sharp +2^{p} \bigr) \bigl( \Vert y \Vert ^{p}+ \Vert z \Vert ^{p}+ \Vert y \Vert ^{p} \Vert z \Vert ^{p} \bigr). \end{aligned}$$

Therefore

$$ \biggl\Vert f(x)\frac{ag(y+z)-g(y)g(z)}{M} \biggr\Vert \leqslant \Vert x \Vert ^{p}. $$

Letting $b:=\frac{ag(y+z)-g(y)g(z)}{M}$, by assumption, we get that $b=0$. Therefore $ag(y+z)=g(y)g(z)$ for all $y,z\in E$. Moreover, if $a=0$, then (i) implies that $g\equiv 0$.

Similarly, we get the result if f, g satisfy condition (ii). □

Theorem 3.9

Let E be a normed linear space and $\mathcal{A}$ be a complex Banach algebra. Let $f,g, h:E\to \mathcal{A}$ satisfy the inequality

$$ \bigl\Vert f(x+y)-g(x)h(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p} \bigr)+ \theta \Vert x \Vert ^{p} \Vert y \Vert ^{p}, \quad x,y\in E $$

(3.15)

for some $\varepsilon , \theta , p\geqslant 0$.

• (i) If $g(0)g(x)=g(x)g(0)$ for all $x\in E$, and for each nonzero element $b\in \mathcal{A}$ the E-orbit of b

  $$ O_{LE}(b,h):= \biggl\{ \frac{bh(x)}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

  is unbounded, then

  $$ g(0)g(x+y)=g(x)g(y),\quad x,y\in E. $$

  Moreover, if $g(0)=0$, then $g\equiv 0$.

• (ii) If $h(0)g(x)=g(x)h(0)$ for all $x\in E$, and for each nonzero element $b\in \mathcal{A}$ the E-orbit of b

  $$ O_{RE}(g,b):= \biggl\{ \frac{g(x)b}{ \Vert x \Vert ^{p}}: x \in E, \Vert x \Vert \geqslant 1 \biggr\} $$

  is unbounded, then

  $$ h(0)h(x+y)=h(x)h(y),\quad x,y\in E. $$

  Moreover, if $h(0)=0$, then $h\equiv 0$.

Proof

Using the same argument as in the proof of Theorem 3.4, we obtain

$$ \begin{aligned} \bigl\Vert g(0)h(x+y)-g(x)h(y) \bigr\Vert \leqslant \bigl(2^{p}+1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \end{aligned} $$

and

$$ \begin{aligned} \bigl\Vert g(x+y)h(0)-g(x)h(y) \bigr\Vert \leqslant \bigl(2^{p}+1\bigr)\varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr)+\theta \Vert x \Vert ^{p} \Vert y \Vert ^{p} \end{aligned} $$

for all $x,y,z\in E$. Therefore the result follows from Theorem 3.8. □

Remark 3.10

We can replace $\varepsilon (\|x\|^{p} +\|y\|^{p} ) + \theta \|x\|^{p} \|y\|^{p}$ given in the main results of this section by more general control functions $\varphi (x, y)$ given by Gǎvruta [22]. The proofs are similar to the proofs given in this section.

4 Conclusion

We have proved the superstability of the following functional equations:

$$\begin{aligned}& f \bigl(P(x,y) \bigr)= g(x)h(y), \\& f(x+y)=g(x)h(y). \end{aligned}$$