Bihomomorphisms and biderivations in Lie Banach algebras

1.   Introduction and preliminaries

The stability problem of functional equations originated from a question of Ulam ^[1] concerning the stability of group homomorphisms. The functional equation $f(x+y) = f(x) + f(y)$ is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping. Hyers ^[2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' Theorem was generalized by Aoki ^[3] for additive mappings and by Rassias ^[4] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by Găvruta ^[5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias' approach.

Gilányi ^[6] showed that if $f$ satisfies the functional inequality

then $f$ satisfies the Jordan-von Neumann functional equation

See also ^[7]. Fechner ^[8] and Gilányi ^[9] proved the Hyers-Ulam stability of the functional inequality $\rm(1.1)$. Park ^[10,11] defined additive $\rho$-functional inequalities and proved the Hyers-Ulam stability of the additive $\rho$-functional inequalities in Banach spaces and non-Archimedean Banach spaces. The stability problems of various functional equations and functional inequalities have been extensively investigated by a number of authors (see ^{[12,13,14,15,16,17]}).

Bae and Park ^[18] proved the Hyers-Ulam stability of bihomomorphisms and biderivations in $C^*$-ternary algebras, Shokri, Park and Shin ^[19] proved the Hyers-Ulam stability of bihomomorphisms and biderivations in intuitionistic fuzzy ternary normed algebras, and Park ^[20] proved the Hyers-Ulam stability of biderivations and bihomomorphisms in Banach algebras.

Definition 1.1. Let $A, B$ be Lie Banach algebras. A bi-additive mapping $H: A\times A \to B$ is called a bihomomorphism if $H$ satisfies

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

Definition 1.2. Let $A$ be a Lie Banach algebra. A bi-additive mapping $\delta : A\times A \to A$ is called a biderivation if $\delta$ satisfies

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

This paper is organized as follows: In Section 2, we solve the bi-additive $s$-functional inequality $\rm(0.1)$ and prove the Hyers-Ulam stability of the bi-additive $s$-functional inequality $\rm(0.1)$ in complex Banach spaces. In Section 3, we prove the Hyers-Ulam stability of bihomomorphisms and biderivations in Lie Banach algebras, associated with the bi-additive $s$-functional inequality $\rm(0.1)$.

Assume that $s$ is a fixed nonzero complex number with $|s| < 1$.

2.   Bi-additive $s$-functional inequality $\rm(0.1)$

Throughout this section, let $X$ be a complex normed space and $Y$ a complex Banach space.

We solve and investigate the bi-additive $s$-functional inequality $\rm(0.1)$ in complex normed spaces.

Theorem 2.1. If a mapping $f:{X}^{2}\to{Y}$ satisfies $\rm(0.1)$ for all $x, y, z \in X$, then $f:{X}^{2}\to{Y}$ is bi-additive.

Proof. Assume that $f$ satisfies $\rm(0.1)$. Replacing $x$ by $y$, $y$ by $z$ and $z$ by $x$ in $\rm(0.1)$, we get

for all $x, y, z\in X$. Replacing $x$ by $z$, $y$ by $x$, and $z$ by $y$ in $\rm(0.1)$, we get

for all $x, y, z\in X$. By $\rm(0.1)$, $\rm(2.1)$, $\rm(2.2)$, we obtain

for all $x, y, z\in X$. From $\rm(2.3)$, we get the equality

for all $x, y, z\in X$. By putting $x = y = z = 0$ in $\rm(2.4)$, we get

Then let's put in $\rm(2.4)$ $x = z = 0.$ We have

Next we take in $\rm(2.4)$ $y = z = 0.$ Then

But we already know that $f(x, 0) = 0.$ Therefore,

Replacing $x$ and $y$ by $\frac{x-y}{2}$ and $z$ by $\frac{x+y}{2}$ in $\rm(2.4)$, we get

for all $x, y\in X$. Replacing $x$ by $\frac{x}{2}$, $y$ by $-\frac{x}{2}$, $z$ by $\frac{x}{2}+y$ in $\rm(2.4)$, we get

for all $x, y\in X$. Replacing $x$ by $x+y$ in $\rm(2.5)$, we get

for all $x, y\in X$. It follows from $\rm(2.6)$ and $\rm(2.7)$ that

for all $x, y\in X$. Replacing $x$ by $\frac{x+y}{2}$, $y$ by $\frac{y-x}{2}$, and $z$ by $-\frac{x+y}{2}$ in $\rm(2.4)$, we get

for all $x, y\in X$. Replacing $y$ by $x$ in $\rm(2.5)$, we get $f(x, x) = -f(x, -x)$ for all $x\in X$, and together with $\rm(2.9)$, we obtain

for all $x, y\in X$. Replacing $x$ by $\frac{x}{2}+y$, $y$ by $-\frac{x}{2}+y$, and z by $\frac{x}{2}$ in $\rm(2.4)$, we get

for all $x, y\in X$. Adding $\rm(2.8)$ and $\rm(2.11)$, we obtain

where the first equality comes from replacing $y$ by $x$ in $\rm(2.5)$. Replacing $x$ by $x-y$ and $y$ by $y+z$ in $\rm(2.8)$, we get

for all $x, y, z\in X$. Replacing $x$ by $-y-z$ and $y$ by $x+y$ in $\rm(2.10)$, we get

for all $x, y, z\in X$. Replacing $x$ by $y+z$ and letting $y = 0$ in $\rm(2.10)$, we get

for all $x, y, z\in X$. Adding $\rm(2.4)$, $\rm(2.13)$, $\rm(2.14)$ and $\rm(2.15)$, we obtain

for all $x, y, z\in X$. Replacing $x$ by $\frac{x+y}{2}$, $y$ by $\frac{y+z-x}{2}$, $z$ by $\frac{x-y}{2}$ in $\rm(2.16)$, we get

for all $x, y, z\in X$. Replacing $x$ by $\frac{x-y-z}{2}$, $y$ by $-\frac{x+y}{2}$, $z$ by $\frac{x+y+z}{2}$ in $\rm(2.16)$, we get

for all $x, y, z\in X$. Adding $\rm(2.17)$ and $\rm(2.18)$, and from $\rm(2.5)$ and $\rm(2.12)$, we obtain

and so

for all $x, y, z\in X$. Thus $f:{X}^{2}\to Y$ is additive in the second variable.

Replacing $y$ by $y-x$ and $z$ by $x$ in $\rm(2.19)$, we get

for all $x, y\in X$ and replacing $y$ by $y-x$ in $\rm(2.10)$, we get

for all $x, y\in X$. Adding (2.20) and (2.21), we get

for all $x, y\in X$. Replacing $y$ by $z$ and $z$ by $y$ in $\rm(2.14)$, we get

for all $x, y, z\in X$. Using $\rm(2.5)$, $\rm(2.22)$ and $\rm(2.23)$, we obtain

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

Define a mapping $g:{X}^{2}\to{Y}$ by $g(x, y) = f(y, x)$ for all $x, y \in X$. Then, from $\rm(2.24)$, $g$ also satisfies $\rm(2.4)$. Thus, in a similar way, we can prove that $g$ is additive in the second variable, which yields that $f:{X}^{2}\to{Y}$ is additive in the first variable.

Therefore, $f:{X}^{2}\to{Y}$ is a bi-additive mapping.

Now, we prove the Hyers-Ulam stability of the bi-additive $s$-functional inequality $\rm(0.1)$.

Theorem 2.2. Let $0 < r < 2$ and $\theta$ be nonnegative real number. If a mapping $f:{X}^{2}\to{Y}$ satisfies $f(0, x) = f(x, 0) = 0$ and

for all $x, y, z \in X$, then there exists a unique bi-additive mapping $B:{X}^{2}\to{Y}$ such that

for all $x, y \in X$, where the function $E:{X}^{2}\to \mathbb{R}$ is defined as

for all $x, y\in X$.

Proof. Replacing $x$ by $y$, $y$ by $z$, $z$ by $x$ in $\rm(2.25)$, we get

for all $x, y, z\in X$. Replacing $x$ by $z$, $y$ by $x$, $z$ by $y$ in $\rm(2.25)$, we get

for all $x, y, z\in X$. By $\rm(2.25)$, $\rm(2.28)$ and $\rm(2.29)$, we obtain

and so

for all $x, y, z\in X$. Replacing $x$, $y$ by $\frac{x-y}{2}$ and $z$ by $\frac{x+y}{2}$ in $\rm(2.25)$, we get

for all $x, y\in X$. Replacing $x$ by $\frac{x}{2}$, $y$ by $-\frac{x}{2}$, $z$ by $\frac{x}{2} +y$ in $\rm(2.25)$, we get

for all $x, y\in X$. Replacing $x$ by $-\frac{x}{2}$, $y$ by $\frac{x}{2} +y$, $z$ by $\frac{x}{2}$ in $\rm(2.25)$, we get

for all $x, y\in X$. Replacing $x$ by $x+y$ in $\rm(2.31)$, we get

for all $x, y\in X$. Adding $\rm(2.32)$, $\rm(2.33)$ and $\rm(2.34)$, we obtain

for all $x, y\in X$. Replacing $x$ by $\frac{x+y}{2}$, $y$ by $\frac{-x+y}{2}$, $z$ by $-\frac{x+y}{2}$ in $\rm(2.25)$, we get

for all $x, y\in X$. Setting $x = y = 0$ and replacing $z$ by $x$ in $\rm(2.25)$, we get

for all $x\in X$. Adding $\rm(2.36)$ and $\rm(2.37)$, we get

for all $x, y\in X$. Replacing $x$ by $\frac{x}{2}+y$ and $y$ by $-\frac{x}{2}+y$ and $z$ by $\frac{x}{2}$ in $\rm(2.30)$, we get

for all $x, y\in X$. Adding $\rm(2.35)$, $\rm(2.39)$ and $\rm(2.37)$ (here, we replace $x$ by $y$), we obtain

for all $x, y, z\in X$, where the function ${E}_{1}:{X}^{2} \to \mathbb{R}$ is defined as

for all $x, y\in X$. Replacing $x$ by $x-y$ and $y$ by $y+z$ in $\rm(2.35)$, we get

for all $x, y, z\in X$. Replacing $x$ by $-y-z$ and $y$ by $x+y$ in $\rm(2.38)$ gives

for all $x, y, z\in X$. Replacing $x$ by $y+z$ and setting $y = 0$ in $\rm(2.38)$, we get

for all $x, y, z\in X$. Adding $\rm(2.30)$, $\rm(2.41)$, $\rm(2.42)$ and $\rm(2.43)$, we obtain

for all $x, y, z\in X$, where the function ${E}_{2}:{X}^{3} \to \mathbb{R}$ is defined as

for all $x, y, z\in X$. Replacing $x$ by $\frac{y}{2}$, $y$ by $\frac{-x+y}{2}$, $z$ by $x-\frac{y}{2}$ in $\rm(2.44)$, we get

for all $x, y\in X$. Replacing $x$ by $\frac{x-y}{2}$, $y$ by $-\frac{y}{2}$, $z$ by $\frac{x+y}{2}$ in $\rm(2.44)$, we get

for all $x, y\in X$. Replacing $x$ by $-\frac{x}{2}$ and $y$ by $-\frac{x}{2} +y$ in $\rm(2.31)$, we get

for all $x, y\in X$. Replacing $x$ by $\frac{x}{2}$ and $y$ by $\frac{x}{2} -y$ in $\rm(2.31)$, we get

for all $x, y\in X$. Replacing $y$ by $\frac{x}{2}$ in $\rm(2.40)$, we get

for all $x\in X$. Replacing $y$ by $y-x$ in $\rm(2.38)$, we get

for all $x, y\in X$. Adding $\rm(2.31)$, $\rm(2.46)$, $\rm(2.47)$, $\rm(2.48)$, $\rm(2.49)$, $\rm(2.50)$ and $\rm(2.51)$, we obtain

for all $x, y\in X$, where the function ${E}_{3}:{X}^{2} \to \mathbb{R}$ is defined as

for all $x, y\in X$. Replacing $y$ by $z$ and $z$ by $y$ in $\rm(2.30)$, we get

for all $x, y, z\in X$. Replacing $x$ by $-x+z$ and $y$ by $y+z$ in $\rm(2.52)$, we get

for all $x, y, z\in X$. Replacing $x$ by $x-z$ and $y$ by $y+z$ in $\rm(2.31)$, we get

for all $x, y, z\in X$. Replacing $x$ by $y+z$ and $y$ by $x-y$ in $\rm(2.31)$, we get

for all $x, y, z\in X$. Adding $\rm(2.53)$, $\rm(2.54)$, $\rm(2.55)$ and $\rm(2.56)$, we obtain

for all $x, y, z\in X$, where the function ${E}_{4}:{X}^{3} \to \mathbb{R}$ is defined as

for all $x, y, z\in X$. Replacing $x$ by $x+y$, $y$ by $x-y$, and $z$ by $y$ in $\rm(2.57)$, we get

for all $x, y\in X$. Replacing $x$ by $y$ and $z$ by $x+y$ $\rm(2.57)$, we get

for all $x, y\in X$. Replacing $x$ by $y$, $y$ by $-y$, and $z$ by $x+y$ $\rm(2.57)$, we get

for all $x, y\in X$. Setting $x = y = 0$ and replacing $z$ by $x$ in $\rm(2.57)$, we get

for all $x\in X$. Adding $\rm(2.58)$, $\rm(2.59)$, $\rm(2.60)$ and $\rm(2.61)$ and adding $\rm(2.40)$ twice, we obtain

for all $x, y\in X$, where the function ${E}_{5}:{X}^{2} \to \mathbb{R}$ is defined as

for all $x, y\in X$.

For any positive integer $n$, replacing $x$ by ${2}^{n-1}x$ and $y$ by ${2}^{n-1}y$ in $\rm(2.62)$, and dividing both sides by ${4}^{n}$, we obtain

for all $x, y\in X$. For any nonnegative integers $u, v$ satisfying $u < v$, by $\rm(2.63)$, we obtain

for all $x, y\in X$. It follows from $\rm(2.64)$ that the sequence $\{ \frac{1}{{4}^{n}}f({2}^{n}x, {2}^{n}y)\}$ is Cauchy for all $x, y \in X$. Since $Y$ is a Banach space, then this sequence converges. So we can define the mapping $B:{X}^{2} \to Y$ by

for all $x, y\in X$. Setting $v = 0$ and passing the limit $u \to \infty$ in $\rm(2.64)$, we get $\rm(2.26)$. Also, from $\rm(2.25)$,

for all $x, y, z \in X$. So, by Theorem 2.1, the mapping $B:{X}^{2} \to Y$ is bi-additive.

Now, let $A:{X}^{2} \to Y$ be another bi-additive specified what the conditions are. Then, for any positive integer $n$, we have

for all $x, y\in X$. When $n$ tends to infinity in $\rm(2.65)$, we have $B(x, y) = A(x, y)$ for all $x, y \in X$. This proves the uniqueness of the bi-additive mapping $B$, as desired.

Theorem 2.3. Let $r > 2$ and $\theta$ be nonnegative real number. If a mapping $f:{X}^{2}\to{Y}$ satisfies $f(0, x) = f(x, 0) = 0$ and $\rm(2.25)$ for all $x, y, z \in X$, then there exists a unique bi-additive mapping $B:{X}^{2}\to Y$ such that

for all $x, y \in X$, where the function $E:{X}^{2}\to \mathbb{R}$ is defined in $\rm(2.27)$.

Proof. Assume that $f$ satisfies $\rm(0.1)$. By the same inappropriate as in the proof of Theorem 2.2, we obatin $\rm(2.62)$. For any positive integer $n$, replacing $x$ by $\frac{x}{2^{n+1}}$ and $y$ by $\frac{y}{2^{n+1}}$ in $\rm(2.62)$, and multiplying both sides by $4^n$, we obtain

for all $x, y\in X$. For any nonnegative integer $u, v$ satisfying $u < v$, by $\rm(2.67)$, we obtain

for all $x, y\in X$. It follows from $\rm(2.68)$ that the sequence $\{ 4^n f(\frac{x}{2^{n}}, \frac{y}{2^{n}}) \}$ is Cauchy for all $x, y \in {X}$. Since $Y$ is a Banach space, this sequence converges. So we can define the mapping $B:{X}^{2} \to Y$ by

for all $x, y\in X$. Setting $v = 0$ and passing the limit $u \to \infty$ in $\rm(2.68)$, we obtain $\rm(2.66)$. Also, from $\rm(2.25)$,

for all $x, y, z \in X$. So, by Theorem 2.1, the mapping $B:{X}^{2} \to Y$ is bi-additive.

Now, let $A:{X}^{2} \to Y$ be another bi-additive specified what the conditions are. Then, for any positive integer $n$, we have

for all $x, y\in X$. When $n$ tends to infinity in $\rm(2.69)$, we have $B(x, y) = A(x, y)$ for all $x, y \in X$. This proves the uniqueness of the bi-additive mapping $B$, as desired.

3.   Hyers-Ulam stability of bihomomorphisms and biderivations in Lie Banach algebras

Throughout this section, let $X$ and $Y$ be complex Lie Banach algebras.

We prove the Hyers-Ulam stability of bihomomorphisms associated with the bi-additive $s$-functional inequality $\rm(0.1)$.

Theorem 3.1. Let $r\neq2$ and $\theta$ be nonnegative real number. If a mapping $f:{X}^{2}\to{Y}$ satisfies $f(0, x) = f(x, 0) = 0$ and $\rm(2.25)$, and

for all $x, y, z \in X$, then there exists a unique bihomomorphism $H:{X}^{2}\to Y$ such that

for all $x, y \in X$, where the function $E:{X}^{2}\to \mathbb{R}$ is defined as $\rm(2.27)$

Proof. First, we deal with the case $r < 2$.

By Theorem 2.2, $H(x, y) : = \lim_{n \to \infty}{\frac{1}{4^{n}} f({2}^{n}x, {2}^{n}y)}$ is a unique bi-additive mapping which satisfies $\rm(3.3)$.

Replacing $x$ by $2^{n} x$, $y$ by $2^{n}y$, and $z$ by $2^{n}z$ in $\rm(3.1)$, we obtain

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

Adding $\rm(3.4)$, we get

for all $x, y, z \in X$. Thus we have $H([x, y], [z, z]) = [H(x, z), H(y, z)]$ for all $x, y, z \in X$. By a similar method, we can also prove that $H([x, x], [y, z]) = [H(x, y), H(x, z)]$, and thus $H:X^2 \to Y$ is a bihomomorphism.

Now, assume that $r > 2$.

By Theorem 2.3, $H(x, y) : = \lim_{n \to \infty}{4^{n} f(\frac{x}{{2}^{n}}, \frac{y}{{2}^{n}})}$ is a unique bi-additive mapping which satisfies $\rm(3.3)$.

Replacing $x$ by $\frac{x}{2^{n}}$, $y$ by $\frac{y}{2^{n}}$, and $z$ by $\frac{z}{2^{n}}$ in $\rm(3.1)$, we obtain

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

Adding $\rm(3.5)$, we get

for all $x, y, z \in X$. Thus we have $H([x, y], [z, z]) = [H(x, z), H(y, z)]$ for all $x, y, z \in X$. By a similar method, we can also prove that $H([x, x], [y, z]) = [H(x, y), H(x, z)]$ and thus $H:X^2 \to Y$ is a bihomomorphism.

Remark 3.2. We have defined the new useful bi-additive functional inequality (0.1), which was not appeared in any papers or any books, and solved the bi-additive functional inequality (0.1). Furthermore, we have proved the Hyers-Ulam-Rassais stability of the bi-additive functional inequality (0.1) by the direct method.

Many authors have only tried to investigate bihomomorphisms and biderivations in Banach algebras, $C^*$-ternary algebras and $C^*$-algebras. But in this paper, we have proved the Hyers-Ulam-Rassias stability of bihomomorphisms and biderivations in Lie Banach algebras associated with the bi-additive functional inequality (0.1).

4.   Conclusions

In this paper, we have introduced and solved the bi-additive $s$-functional inequality (0.1) and we have proved the Hyers-Ulam stability of bihomomorphisms and biderivations in Lie Banach algebras associated with the bi-additive $s$-functional inequality $\rm(0.1)$.

Acknowledgments

This work was supported by the Seoul Science High School R & E program in 2018. C. Park was supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF -2017R1D1A1B04032937).

Conflict of interest

The authors declare that they have no competing interests.