Nonlinear mixed type product $[\mathscr{K}, \mathscr{F}]_\ast \odot \mathscr{D}$ on $\ast$-algebras

1.   Introduction

Consider an algebra $\mathcal{A}$ defined over the complex field $\mathbb{C}$. A map $\ast: \mathcal{A} \to \mathcal{A}$ is called an involution if the following conditions hold for all $\mathscr{K}, \mathscr{F} \in \mathcal{A}$ and $\alpha \in \mathbb{C}$. (i) $(\mathscr{K} +\mathscr{F})^\ast = \mathscr{K}^\ast + \mathscr{F}^\ast$; (ii) $(\alpha \mathscr{K})^\ast = \bar{\alpha} \mathscr{K}^\ast$; (iii) $(\mathscr{K}\mathscr{F})^\ast = (\mathscr{F})^\ast (\mathscr{K})^\ast$ and $(\mathscr{K}^\ast)^\ast = \mathscr{K}$. An algebra $\mathcal{A}$ with the involution $\ast$ is called the $\ast$-algebra. For $\mathscr{K}, \mathscr{F}\in \mathcal{A}$, we call $[\mathscr{K}, \mathscr{F}]_\ast = \mathscr{K}\mathscr{F}- \mathscr{F}\mathscr{K}^\ast$ the skew Lie product, $[\mathscr{K}, \mathscr{F}]_\bullet = \mathscr{K}\mathscr{F}^\ast - \mathscr{F}\mathscr{K}^\ast$ denotes the bi-skew Lie product and $\mathscr{K} \odot \mathscr{F} = \mathscr{K}^\ast \mathscr{F}+ \mathscr{F}\mathscr{K}^\ast$ denotes the bi-skew Jordan product. The skew Lie product, the Jordan product, and the bi-skew Jordan product have become increasingly relevant in various research fields, and numerous authors have shown a keen interest in their exploration. This is evident from the numerous studies by authors (see ^{[1,2,3,5,7,8,9,10,13,15,16]}). Recall that an additive map $\Omega: \mathcal{A} \to \mathcal{A}$ is called an additive derivation if $\Omega(\mathscr{K}\mathscr{F}) = \Omega(\mathscr{K}) \mathscr{F}+ \mathscr{K} \Omega(\mathscr{F})$ for all $\mathscr{K}, \mathscr{F}\in \mathcal{A}$. If $\Omega(\mathscr{K}^\ast) = \Omega (\mathscr{K})^\ast$ for all $\mathscr{K} \in \mathcal{A}$, then $\Omega$ is an additive $\ast$-derivation. Let $\Omega: \mathcal{A} \to \mathcal{A}$ be a map (without the additivity assumption). We say $\Omega$ is a nonlinear skew Lie derivation or nonlinear skew Lie triple derivation if

or

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}$. Similarly, a map $\Omega : \mathcal{A} \to \mathcal{A}$ is said to be a nonlinear bi-skew Lie derivation or nonlinear bi-skew Lie triple derivation if

or

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}.$ In 2021, A. Khan ^[4] established a proof demonstrating that any multiplicative or nonadditive bi-skew Lie triple derivation acting on a factor Von Neumann algebra can be characterized as an additive $\ast$-derivation.

Numerous authors have recently explored the derivations and isomorphisms corresponding to the novel products created by combining Lie and skew Lie products, skew Lie and skew Jordan products see ^[6,11,12,14]. As an illustration, Li and Zhang ^[6] delved into an investigation focused on understanding the arrangement and properties of the nonlinear mixed Jordan triple $\ast$-derivation within the domain of $\ast$-algebras. In 2022, Rehman et. al. ^[12] mixed the concepts of Jordan and Jordan $\ast$-product and gave the complete characterization of nonlinear mixed Jordan $\ast$-triple derivation on $\ast$-algebras. Inspired by the above results, in the present paper, we combined the skew Lie product and bi-skew Jordan product and defined nonlinear mixed bi-skew Jordan triple derivation on $\ast$-algebras. A map $\Omega$: $\mathcal{A}$ $\to$ $\mathcal{A}$ is called nonlinear mixed bi-skew Jordan triple derivations if

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}.$ Our proof establishes that when $\Omega$ represents a nonlinear mixed bi-skew Lie triple derivation acting on $\ast$-algebras, it necessarily possesses an additive $\ast$-derivation. In simpler terms, the study demonstrates that specific properties, such as additivity and self-adjointness, can be attributed to the nature of nonlinear mixed bi-skew Jordan triple derivations on $\ast$-algebras.

2.   Main result

Theorem 2.1. Let $\mathcal{A}$ be a unital $\ast$-algebra with unity $\mathscr{I}$ containing a non-trivial projection $P$. Suppose that $\mathcal{A}$ satisfies

and

Define a map $\Omega : \mathcal{A} \to \mathcal{A}$ such that

then $\Omega$ is an additive $\ast$-derivation.

Let $P = \mathscr{P}_1$ be a non-trivial projection in $\mathcal{A}$, and $\mathscr{P}_2 = \mathfrak{\mathscr{I}} - \mathscr{P}_1,$ where $\mathfrak{\mathscr{I}}$ is the unity of this algebra. Then by Peirce decomposition of $\mathcal{A}$, we have $\mathcal{A} = \mathscr{P}_1 \mathcal{A} \mathscr{P}_1 \oplus \mathscr{P}_1 \mathcal{A} \mathscr{P}_2 \oplus \mathscr{P}_2 \mathcal{A} \mathscr{P}_1 \oplus \mathscr{P}_2 \mathcal{A} \mathscr{P}_2$ and, denote $\mathcal{A}_{11} = \mathscr{P}_1 \mathcal{A} \mathscr{P}_1, \mathcal{A}_{12} = \mathscr{P}_1 \mathcal{A} \mathscr{P}_2, \mathcal{A}_{21} = \mathscr{P}_2 \mathcal{A} \mathscr{P}_1$ and $\mathcal{A}_{22} = \mathscr{P}_2 \mathcal{A} \mathscr{P}_2$. Note that any $\mathscr{K} \in \mathcal{A}$ can be written as $\mathscr{K} = \mathscr{K}_{11} + \mathscr{K}_{12} + \mathscr{K}_{21} + \mathscr{K}_{22},$ where $\mathscr{K}_{ij} \in \mathcal{A}_{ij}$ and $\mathscr{K}_{ij}^\ast \in \mathcal{A}_{ji}$ for $i, j = 1, 2.$

Several lemmas are used to prove Theorem 2.1.

Lemma 2.1. $\Omega(0) = 0$ and $\Omega(\mathscr{I}) = \Omega(\mathscr{I})^\ast$.

Proof. It is trivial that

We can easily see that

From the other side, we yield

From the equations above, we can deduce

The proof is now concluded. □

Lemma 2.2. For any $\mathscr{K}_{12} \in \mathcal{A}_{12}, \mathscr{K}_{21} \in \mathcal{A}_{21}$, we have

Proof. Let $M = \Omega(\mathscr{K}_{12}) + \Omega (\mathscr{K}_{21}) - \Omega(\mathscr{K}_{12}) - \Omega(\mathscr{K}_{21})$. We have

Alternatively, it follows from $[\mathscr{K}_{12}, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$ that

From the last two expressions, we conclude $[M, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$. That means $\mathscr{P}_1 M^\ast \mathscr{P}_2 - \mathscr{P}_2 M\mathscr{P}_1 = 0$. By multiplying $\mathscr{P}_2$ from the left, we find $\mathscr{P}_2 M\mathscr{P}_1 = 0$. In similar way, we can easily show that $\mathscr{P}_1 M \mathscr{P}_2 = 0$.

Also, $[i(\mathscr{P}_1 - \mathscr{P}_2), \mathscr{I}]_\ast \odot \mathscr{K}_{12} = 0$. Thus,

On the other side, we have

From the last two expressions, we obtain $[i (\mathscr{P}_1 - \mathscr{P}_2), \mathscr{I}]_\ast \odot M = 0$. That means $-2i \mathscr{P}_1 M+ 2i \mathscr{P}_2 M-2i M \mathscr{P}_1 + 2i M \mathscr{P}_2 = 0$. By pre and post multiplying by $\mathscr{P}_1$ from both sides, we get $\mathscr{P}_1 M \mathscr{P}_1 = 0$. In the similar way, we can show that $\mathscr{P}_2 M \mathscr{P}_2 = 0$. Hence, $M = 0$, i.e.,

The proof is now concluded. □

Lemma 2.3. For any $\mathscr{K}_{ii} \in \mathcal{A}_{ii}, \mathscr{K}_{ij} \in \mathcal{A}_{ij}, 1 \leq i, j \leq 2$, we have

Proof. First, we will demonstrate the case when \(i = 1\) and \(j = 2\). Let $M = \Omega(\mathscr{K}_{11} + \mathscr{K}_{12} + \mathscr{K}_{21}) - \Omega (\mathscr{K}_{11}) - \Omega(\mathscr{K}_{12}) - \Omega (\mathscr{K}_{21})$. Since $[\mathscr{K}_{11}, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$ and using Lemma 2.2, we have

On the other side, we have

From the above two expressions, we find $[M, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$, and so, $\mathscr{P}_2 M\mathscr{P}_1 = 0$. Similarly, $\mathscr{P}_1 M \mathscr{P}_2 = 0$. Now, for all $\mathscr{X}_{12} \in \mathcal{A}_{12}$, we have

Also, $[\mathscr{X}_{12}, \mathscr{K}_{11}]_\ast \odot \mathscr{P}_2 = 0$ and using Lemma 2.2, we get

From the above two relations, we get $[\mathscr{X}_{12}, M]_\ast \odot \mathscr{P}_2 = 0$. That means $-\mathscr{X}_{12} M^\ast \mathscr{P}_2 + \mathscr{P}_2 M^\ast \mathscr{X}_{12}^\ast = 0$. By post-multiplying by $\mathscr{P}_2$ on both sides, we get $-\mathscr{X}_{12}M^\ast \mathscr{P}_2 = 0$. Therefore, by using (▲) and (▼), we get $\mathscr{P}_2 M \mathscr{P}_2 = 0$. Similarly, $\mathscr{P}_1 M \mathscr{P}_1 = 0$. Hence, $M = 0$. i.e.,

By using the same technique, we can also show for $i = 2, j = 1$. The proof is now concluded. □

Lemma 2.4. For any $\mathscr{K}_{ij} \in \mathcal{A}_{ij}, 1 \leq i, j \leq 2,$ we have

Proof. Let $M = \Omega (\mathscr{K}_{11} + \mathscr{K}_{12} + \mathscr{K}_{21} + \mathscr{K}_{22}) - \Omega(\mathscr{K}_{11}) - \Omega(\mathscr{K}_{12}) - \Omega(\mathscr{K}_{21}) - \Omega(\mathscr{K}_{22}).$ Since, $[\mathscr{K}_{11}, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$ and using Lemma 2.3 that

Alternatively, we have

From the last two relations, we get $[M, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$. Thus, $\mathscr{P}_1 M^\ast \mathscr{P}_2 - \mathscr{P}_2 M \mathscr{P}_1 = 0$. Hence, $\mathscr{P}_2 M \mathscr{P}_1 = 0$. Similarly, $\mathscr{P}_1 M \mathscr{P}_2 = 0$.

Now, for any $\mathscr{X}_{12} \in \mathcal{A}_{12}$, we have

Also, $[\mathscr{X}_{12}, \mathscr{K}_{11}]_\ast \odot \mathscr{P}_2 = 0$, and using Lemma 2.3, we find

Upon comparing the aforementioned two equations, we observe that $[\mathscr{X}_{12}, M]_\ast \odot \mathscr{P}_2 = 0$. On solving, we get $\mathscr{P}_2 M \mathscr{P}_2 = 0$. Similarly, we can show that $\mathscr{P}_1 M \mathscr{P}_1 = 0$. Hence, $M = 0$, i.e.,

This ends the proof. □

Lemma 2.5. For each $\mathscr{K}_{12}, \mathscr{F}_{12} \in \mathcal{A}_{12}$ and $\mathscr{K}_{21}, \mathscr{F}_{21} \in \mathcal{A}_{21}$, we have

(1) $\Omega(\mathscr{K}_{12} + \mathscr{F}_{12}) = \Omega(\mathscr{K}_{12}) + \Omega(\mathscr{F}_{12}).$

(2) $\Omega(\mathscr{K}_{21} + \mathscr{F}_{21}) = \Omega(\mathscr{K}_{21}) + \Omega(\mathscr{F}_{21}).$

Proof. (1) Let $M = \Omega(\mathscr{K}_{12} + \mathscr{F}_{12}) - \Omega(\mathscr{K}_{12}) - \Omega(\mathscr{F}_{12}).$ We have,

On the other hand, it follows from $[\mathscr{K}_{12}, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$ that

From the last two relations, we get $[M, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0.$ This means that $\mathscr{P}_1 M^\ast \mathscr{P}_2 - \mathscr{P}_2 M \mathscr{P}_1 = 0$. By pre-multiplying $\mathscr{P}_2$ on both sides, we get $\mathscr{P}_2 M \mathscr{P}_1 = 0$. Similarly, we can show that $\mathscr{P}_1 M \mathscr{P}_2 = 0$. Now, for any $\mathscr{X}_{12} \in \mathcal{A}_{12}$, we have

On the other hand, it follows from $[\mathscr{X}_{12}, \mathscr{K}_{12}]_\ast \odot \mathscr{P}_2 = 0$ that

On comparing the above two relations, we get $[\mathscr{X}_{12}, M]_\ast \odot \mathscr{P}_2 = 0$. On solving, we get $\mathscr{P}_2 M \mathscr{P}_2 = 0$. Similarly, we can show that $\mathscr{P}_1 M \mathscr{P}_1 = 0$. Hence, $M = 0$, i.e.,

\text{(2)} By using the same technique, we can show that

The proof is now concluded. □

Lemma 2.6. For each $\mathscr{K}_{ii}, \mathscr{F}_{ii} \in \mathcal{A}_{ii}$ such that $1 \leq i \leq 2$, we have

Proof. First, it is prove for $i = 1$. Let $M = \Omega(\mathscr{K}_{11} + \mathscr{F}_{11}) - \Omega (\mathscr{K}_{11}) - \Omega(\mathscr{F}_{11}).$ Since, $[\mathscr{K}_{11}, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$, we have

On the other hand, we have

Upon comparing the aforementioned two equations, we observe that $[M, \mathscr{P}_1]_\ast \odot \mathscr{P}_2 = 0$. On solving, we get $\mathscr{P}_1 M^\ast \mathscr{P}_2 - \mathscr{P}_2 M \mathscr{P}_1 = 0$. By pre-multiplying by $\mathscr{P}_2$ on both sides, we get $\mathscr{P}_2 M \mathscr{P}_1 = 0$. Similarly, $\mathscr{P}_1 M \mathscr{P}_2 = 0$. Now, for any $\mathscr{X}_{12} \in \mathcal{A}_{12}$, we have

It follows from $[\mathscr{X}_{12}, \mathscr{K}_{11}]_\ast \odot \mathscr{P}_1 = 0$ that

By comparing, we get $[\mathscr{X}_{12}, M]_\ast \odot \mathscr{P}_2 = 0$. That means $-\mathscr{X}_{12} M^\ast P_2 + P_2 M^\ast \mathscr{X}_{12}^\ast = 0$. By pre-multiplying $P_2$ on both sides, we get $P_2 M^\ast \mathscr{X}_{12}^\ast = 0$. Thus, by using (▲) and (▼), we get $P_2 M P_2 = 0$. Similarly, $P_1 M P_1 = 0$. Hence, $M = 0$. This completes the proof. □

Lemma 2.7. $\Omega$ is an additive map.

Proof. For any $\mathscr{K}, \mathscr{F}\in \mathcal{A}$, we write $\mathscr{K} = \mathscr{K}_{11} + \mathscr{K}_{12} + \mathscr{K}_{21} + \mathscr{K}_{22}$ and $\mathscr{F} = \mathscr{F}_{11} + \mathscr{F}_{12} + \mathscr{F}_{21} + \mathscr{F}_{22}$. By using Lemmas 2.4–2.6, we get

Hence, $\Omega$ is additive. □

Lemma 2.8. The following conditions holds:

(i) $\Omega(i\mathscr{I})^\ast = \Omega(i\mathscr{I}) = 0$.

(ii) $\Omega(\mathscr{I}) = 0$.

Proof. (i) It follows from Lemma 2.7 that

and

From the last two expressions, we get

Also, we can evaluate

Alternatively, we can write

By comparing above two equations, and also using Lemma 2.1, we find

By using Eqs (2.1) and (2.2), we have

(ii) In the similar way, we can show that $\Omega(\mathscr{I}) = 0$. □

Lemma 2.9. $\Omega$ preserves star, i.e., $\Omega(\mathscr{K}^{*}) = \Omega(\mathscr{K})^{*}$ for all $\mathscr{K} \in \mathcal{A}$.

Proof. From Lemma 2.7, we have

Alternatively, it follows from Lemma 2.8 that

From the above two equations, we obtain

for all $\mathscr{K} \in \mathcal{A}$. This completes the proof. □

Lemma 2.10. We prove that $\Omega(i \mathscr{K}) = i\Omega(\mathscr{K})$ for all $\mathscr{K} \in\mathcal{A}$.

Proof. For any $\mathscr{K} \in \mathcal{A}$, we have

Alternatively, it follows from Lemma 2.8 that

From the above two expressions, we obtain

□

Proof of Theorem 2.1. For any $\mathscr{K}, \mathscr{F}\in \mathcal{A}$, it follows from Lemmas 2.7 that

Also, by using Lemma 2.9 that

for all $\mathscr{K} \in \mathcal{A}$. Now, we only have to show that $\Omega$ is an derivation.

Now, for any $\mathscr{K}, \mathscr{F}\in \mathcal{A}$, and using Lemma 2.7, we have

Also, using Lemma 2.8 that

By comparing the two equations above, we obtain

On the other hand, according to Lemma 2.7, we can infer that

Alternatively, by using Lemma 2.8, we find

From the above two expressions, we find

Subtracting Eq (2.5) to Eq (2.6), we get

By using Lemma 2.10 and the above equation, we find

Adding Eqs (2.7) and (2.8), we get

From Eqs (2.3), (2.4) and (2.9), we get $\Omega$ is an additive $\ast$-derivation. This completes the proof.

Now, we provide an example to demonstrate the necessity of the conditions (▲) and (▼) in Theorem 2.1.

Example 2.1. Consider $\mathcal{A} = \{\left(\begin{array}{cc} a & 0 \\ c & d \\ \end{array} \right)\}$, the algebra of all lower triangular matrix of order $2$ over the field of complex numbers $\mathbb{C}$ and $\mathscr{I} = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ be the unity of $\mathcal{A}$. The map $\ast: \mathcal{A} \to \mathcal{A}$ given by $\ast(\mathscr{K}) = \mathscr{K}^\theta$, where $\mathscr{K}^\theta$ denotes the conjugate transpose of matrix $A$, is an involution. Hence, $\mathcal{A}$ is a unital $\ast$-algebra with unity $\mathscr{I}$. Now, define a map $\Omega:\mathcal{A} \to \mathcal{A}$ such that $\Omega\left(\begin{array}{cc} a & 0 \\ c & d \\ \end{array} \right) = \left(\begin{array}{cc} 0 & 0 \\ -ic & 0 \\ \end{array} \right)$. Note that $\Omega$ is a derivation on $\mathcal{A}$. So, it also satisfies

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}.$ Let $P = \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)$ is a non-trivial projection, so $P^2 = P$ and $P^* = P$. For $W = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right) \neq 0 \in \mathcal{A}$ and hence $W\mathcal{A}P = (0)$ but $0\neq W\in \mathcal{A}.$ However, $\Omega$ is not an additive $*$-derivation because $\Omega(\mathscr{K}^*)\neq(\Omega(\mathscr{K}))^*$ for some $\mathscr{K}\in \mathcal{A}.$

3.   Corollaries

The following corollaries arise directly from Theorem 2.1: The algebra of all bounded linear operators on $\mathcal{H}$ is denoted by $\mathcal{B}(\mathcal{H})$. Let $\mathcal{H}$ be a Hilbert space over the field $\mathbb{F}$ of real or complex numbers. The dimension of an operator's range is known as its rank. An operator with a finite dimensional range is therefore said to have a finite rank. $\mathcal{F}(\mathcal{H})$ is the subalgebra of all bounded linear operators of finite rank on $\mathcal{H}$.

Let $\mathcal{H}$ be a Banach space over the real or complex number field $\mathbb{F}$. In the case of $\mathcal{F}(\mathcal{H})\subseteq\mathrm{\mathscr{K}}(\mathcal{H})$, a subalgebra $\mathcal{\mathscr{K}}(\mathcal{H})$ of $\mathcal{B}(\mathcal{H})$ is referred to as a standard operator algebra.

Corollary 3.1. Let $\mathcal{A}$ be a standard operator algebra on an infinite dimensional complex Hilbert space $\mathcal{H}$ containing an identity operator $\mathfrak{\mathscr{I}}$. Suppose that $\mathcal{A}$ is closed under adjoint operation. Define $\Omega : \mathcal{A} \to \mathcal{A}$ such that

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}$, then $\Omega$ is an additive $\ast$-derivation.

Proof. Every standard operator algebra $\mathcal{A}$ being a prime algebra is a direct consequence of the Hahn-Banach theorem. As a prime algebra, $\mathcal{A}$ naturally fulfills the conditions specified in (▲) and (▼). Consequently, according to Theorem 2.1, it follows that the map $\Omega$ described earlier is an additive $\ast$-derivation. □

A von Neumann algebra is defined as a weakly closed self-adjoint subalgebra of $\mathcal{B}(\mathcal{H})$ that includes the identity operator, where $\mathcal{B}(\mathcal{H})$ is the space of all bounded linear operators on a complex Hilbert space $\mathcal{H}$. In other words, a self-adjoint subalgebra of $\mathcal{B}(\mathcal{H})$ that satisfies the double commutant property, that is, $\mathcal{M}'' = \mathcal{M}$, is considered a von Neumann algebra. In this context, a factor von Neumann algebra is one with a trivial center, which is equal to the intersection of $\mathcal{M}$ and its double commutant, $\mathcal{M} \cap \mathcal{M}' = \mathbb{C} \mathscr{I}$. Additionally, an abelian von Neumann algebra is one where the center is equal to the algebra itself, that is, $\mathcal{Z}(\mathcal{M}) = \mathcal{M}$.

Corollary 3.2. Let $\mathcal{M}$ ba a factor von Neumann algebra with $dim \mathcal{M} \geq 2.$ Define $\Omega : \mathcal{M} \to \mathcal{M}$ such that

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}$, then $\Omega$ is an $\ast$-derivation.

Proof. By using [16,Lemma 2.2], it is established that every factor von Neumann algebra $\mathcal{M}$ satisfies the conditions outlined in (▲) and (▼). Therefore, applying Theorem 2.1, we conclude that the map $\Omega$ described earlier is an additive $\ast$-derivation within the context of factor von Neumann algebras. □

An algebra $\mathcal{A}$ is called prime algebra, if $\mathscr{K} \mathcal{A} \mathscr{K} = \{0\}$ for $\mathscr{K}, \mathscr{F} \in \mathcal{A}$ implies either $\mathscr{K} = 0$ or $\mathscr{F} = 0$.

Corollary 3.3. Let $\mathcal{A}$ be a prime $\ast$-algebra with unit $\mathfrak{\mathscr{I}}$ containing non-trivial projection $P$. A map $\Omega : \mathcal{A} \to \mathcal{A}$ satisfies

for all $\mathscr{K}, \mathscr{F}, \mathscr{D} \in \mathcal{A}$, then $\Omega$ is an additive $\ast$-derivation.

Proof. By the definition of primeness of $\mathcal{A}$, it is straightforward to observe that $\mathcal{A}$ also satisfies (▲) and (▼). Therefore, by Theorem 2.1, we conclude that $\Omega$ is an additive $\ast$-derivation. □

Author contributions

All authors are contributed equally.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This study was carried out with financial support from Researchers Supporting Project Number (RSPD2024R934), King Saud University, Riyadh, Saudi Arabia.

Conflict of interest

The authors declare that they have no conflicts of interest.