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On superstability of exponential functional equations
Journal of Inequalities and Applications volume 2021, Article number: 76 (2021)
Abstract
The aim of this paper is to prove the superstability of the following functional equations:
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
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
under the condition bounded by a constant. Kim [14] investigated the solution and the superstability of the Pexiderized Lobacevski functional equation
Kim and Park [15] considered the superstability of the generalized Pexider exponential functional equation
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:
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
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
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
for all \(x\in G\). So we have
For \(x,y\in G\), let \(\Delta (x,y)=f ( P(x,y) )- g(x)h(y)\). Then (2.1) implies that
for all \(x,y,z\in G\). Since \(\lim_{n\to \infty }\|h(y_{n})^{-1}\|=0\), we get
Therefore it follows from (2.5) that
Hence
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
Since \(\varphi (x,y*y_{n})\|h(y_{n})^{-1}\| \to 0\), we get
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
for all \(x\in G\). So we have
Let \(\Delta (x,y)=f ( P(x,y) )- g(x)h(y)\). Then (2.1) implies that
Since \(\lim_{n\to \infty }\|g(y_{n})^{-1}\|=0\), we get
Therefore it follows from (2.5) that
Hence
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
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
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
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
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
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
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
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
Therefore
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
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
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
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
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
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
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
for some \(\varepsilon , \theta , p\geqslant 0\), then either \(\sup_{\|x\|\geqslant 1}\frac{\|f(x)\|}{\|x\|^{p}}<\infty \) or
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
Therefore
Choose \(x,y,z\in X\) with \(f(x)\neq 0\). It then follows from (3.1) that
Hence we get
In view of (3.1), we have
Inequalities (3.3) and (3.4) yield
Since E is a normed algebra with multiplicative norm, it follows from (3.5) that
If we put \(x=x_{n}\) in (3.6) and take the limit as \(n\to +\infty \), then it follows from (3.2) that
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
for some \(\varepsilon , \theta , p\geqslant 0\). Then either \(\sup_{\|x\|\geqslant 1}\frac{\|g(x)\|}{\|x\|^{p}}<\infty \) or
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
Therefore
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
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
Therefore
for all \(x,y,z\in X\). On the other hand, (i) implies
for all \(x,y,z\in X\). It follows from (3.8) and (3.9) that
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
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
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
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
Therefore
and
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
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
for some \(\varepsilon , \theta , p\geqslant 0\). If, for each nonzero element \(b\in \mathcal{A}\), the E-orbit of b
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
Then, for \(\|x\|\geqslant 1\), we have
where
Therefore,
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
for some \(\varepsilon , \theta , p\geqslant 0\). If, for every nonzero linear multiplicative functional φ on \(\mathcal{A}\), the set
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
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
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
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
for all \(x,y,z\in E\). Since \(af(x)=f(x)a\), for \(\|x\|\geqslant 1\), we obtain
where
Therefore
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
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
and
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:
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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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DOI: https://doi.org/10.1186/s13660-021-02615-w
MSC
- 39B72
- 39B82
- 39B52
Keywords
- Banach algebra
- Pexider exponential equation
- Exponential function
- Superstability