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On superstability of exponential functional equations

Abstract

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

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

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 [310] 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 [1618].

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.

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. (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. (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. (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. □

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:

  1. (i)

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

  2. (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 \(\varepsilon , \theta , p\geqslant 0\).

  1. (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

    $$ g(0)g(x+y)=g(x)g(y),\quad x,y\in X. $$
  2. (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:

  1. (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.

  2. (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 \(\varepsilon , \theta , p\geqslant 0\).

  1. (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 \(g\equiv 0\).

  2. (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.

Conclusion

We have proved the superstability of the following functional equations:

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

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Acknowledgements

We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments that will help to improve the quality of the manuscript.

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

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Noori, B., Moghimi, M.B., Najati, A. et al. On superstability of exponential functional equations. J Inequal Appl 2021, 76 (2021). https://doi.org/10.1186/s13660-021-02615-w

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MSC

  • 39B72
  • 39B82
  • 39B52

Keywords

  • Banach algebra
  • Pexider exponential equation
  • Exponential function
  • Superstability