Some zero product preserving additive mappings of operator algebras

1.   Introduction

Let $\mathcal{A}$ be a $^{\ast}$-ring, meaning a ring with involution $^{\ast},$ and let $\mathcal{B}$ be a subring of $\mathcal{A}$. An additive mapping $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is called a Jordan $^{\ast}$-derivation (though in some literature, this term may carry a different meaning) if

for all $T\in\mathcal{B}.$ It can be easily verified that if $\mathcal{A}$ is 2-torsion-free, meaning $2A = 0$ implies $A = 0$ for every $A$ in $\mathcal{A},$ then a Jordan $^{\ast}$-derivation can be equivalently defined as

for all $A, B\in\mathcal{B},$ where $A\circ B = AB+BA$. For each $A\in\mathcal{A},$ one can define a Jordan $^{\ast}$-derivation $\delta_{A}$ by $\delta_{A}(T) = TA-AT^{\ast}$, for all $T\in\mathcal{B}.$ Such Jordan $^{\ast}$-derivations are referred to as inner Jordan $^{\ast}$-derivations.

The significance of Jordan $^{\ast}$-derivations lies in their structural importance in problems concerning the representability of quadratic functionals by sesquilinear forms on modules (see ^[10,11,12]).

Brešar and Vukman ^[3] established that if a unital $^{\ast}$-ring $\mathcal{A}$ contains $\frac{1}{2}$ and a central invertible element $A$ such that $A^{\ast} = -A,$ then every Jordan $^{\ast}$-derivation from $\mathcal{A}$ to itself is inner. Consequently, every Jordan $^{\ast}$-derivation on a unital complex $^{\ast}$-algebra is inner. To adapt the approach employed in the proof of [3, Theorem 1], the following lemma can be derived:

Lemma 1.1. Let $\mathcal{A}$ be a complex $^{\ast}$-algebra with the unity $\mathbf{1}$, and let $\mathcal{B}$ be an arbitrary subalgebra of $\mathcal{A}$. Then every Jordan $^{\ast}$-derivation from $\mathcal{B}$ into $\mathcal{A}$ is inner.

Let $\mathcal{B}(\mathcal{H})$ denote the algebra of all bounded linear operators on a real or complex Hilbert space $\mathcal{H}$ with dim $\mathcal{H} > 1$, and let $\mathcal{A}$ be a standard operator algebra on $\mathcal{H}$. Šemrl ^[11] proved that every Jordan $^{\ast}$-derivation from $\mathcal{A}$ to $\mathcal{B}(\mathcal{H})$ is inner.

Let $\mathcal{R}$ be a 2-trision-free, noncommutative prime $^{\ast}$-ring with a nontrivial projection. Qi and Zhang ^[8] demonstrated that if $\delta:\mathcal{R}\rightarrow \mathcal{R}$ satisfies the condition

then $\delta$ is a Jordan $^{\ast}$-derivation.

Consider a real Hilbert space $\mathcal{H}$ with dim $\mathcal{H} = \infty$, and let $\delta:\mathcal{B}(\mathcal{H})\rightarrow \mathcal{B}(\mathcal{H})$ be a real linear mapping. Qi and Wang in ^[9] established that if $\delta$ satisfies

then $\delta$ is inner. In the same paper, the authors constructed an example of an additive mapping that satisfies condition $(\mathbb{P}_{2})$ but is not a Jordan $^{\ast}$-derivation on the algebra of all $2\times2$ real matrices. This implies that, on a large class of $^{\ast}$-rings, an additive mapping $\delta$ that only satisfies condition $(\mathbb{P}_{2})$ is not sufficient to ensure it is a Jordan $^{\ast}$-derivation.

Motivated by these results, in this paper, we aim to characterize an additive mapping $\delta:\mathcal{B}\rightarrow \mathcal{A}$ satisfying the following condition:

Clearly, condition $(\mathbb{P})$ is weaker than condition $(\mathbb{P}_{1}).$

In the paper, our main focus is on investigating the aforementioned preservation problem of operator algebras, specifically in von Neumann algebras and $C^{\ast}$-algebras. For a von Neumann algebra $\mathcal{M}$, we approach the study within a broader context by considering $\mathcal{M}$ as a subalgebra of the $^{\ast}$-algebra of all locally measurable operators with respect to $\mathcal{M}$. Regarding $C^{\ast}$-algebras, achieving results for the preservation problem discussed above is challenging in general $C^{\ast}$-algebras. Hence, we primarily focus on properly infinite, primitive, and AF (approximately finite) $C^{\ast}$-algebras. In the paper, we present the following main results:

${\rm(1)}$Let $\mathcal{M}$ be a von Neumann algebra without direct commutative summands, and let $\mathcal{A}$ be an arbitrary subalgebra of $LS(\mathcal{M})$ containing $\mathcal{M}$. An additive mapping $\delta:\mathcal{A}\rightarrow LS(\mathcal{M})$ is an inner Jordan $^{\ast}$-derivation if and only if it satisfies condition $(\mathbb{P}).$

${\rm(2)}$ Let $\mathcal{A}$ be a properly infinite $C^{\ast}$-algebra. An additive mapping $\delta:\mathcal{A}\rightarrow \mathcal{A}$ is inner if and only if it satisfies condition $(\mathbb{P}).$

${\rm(3)}$ Let $\mathcal{A}$ be a unital noncommutative primitive $C^{\ast}$-algebra with a nonzero $soc(\mathcal{A}),$ and let $\mathcal{B}$ be an arbitrary subalgebra of $\mathcal{A}$ containing $soc(\mathcal{A})$. An additive mapping $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is inner if and only if it satisfies condition $(\mathbb{P}).$

${\rm(4)}$ Suppose $\mathcal{A} = \overline {\bigcup\mathcal{A}_{n}}$ is an AF algebra such that $\mathcal{A}_{1}$ has no direct commutative summands. Let $\mathcal{B}$ be chosen from $\mathcal{A}$ or $\bigcup_{n = 1}^{N}\mathcal{A}_{n},$ where $N$ is a finite integer or infinite. Then an additive mapping $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is inner if and only if it satisfies condition $(\mathbb{P}).$

2.   Main results

An element $P$ in a $^{\ast}$-ring is called a projection if $P^* = P = P^2$. Let $\mathcal{G}$ be a $^{\ast}$-ring with unity $\mathbf{1}$ and a nontrivial projection $P_{1}$. By $P_{2},$ we shall always mean $\mathbf{1} - P_{1}$ unless otherwise specified. To obtain the main results of the paper, we first present the following theorem, which generalizes [8, Theorem 2.2]. Additionally, we place the proof of the theorem at the end of the paper to keep the focus on its main points.

Theorem 2.1. Let $\mathcal{G}$ be a 2-torsion-free $^{\ast}$-ring with unity $\mathbf{1}$ and a nontrivial projection $P_{1}$. Suppose $\mathcal{U}$ is a subring of $\mathcal{G}$ that satisfies the following conditions:

${\rm(1)} \; P_{1}, P_{2}\in\mathcal{U}$;

${\rm(2)} \; \mathcal{U}P_{2}$ left separates $P_{1}\mathcal{G}P_{1},$ i.e. for each $A$ in $P_{1}\mathcal{G}P_{1},$ $A\mathcal{U}P_{2} = \{0\}$ implies that $A = 0;$

${\rm(3)} \; P_{1}\mathcal{U}$ right separates $P_{2}\mathcal{G}P_{2},$ i.e. for each $A$ in $P_{2}\mathcal{G}P_{2},$ $P_{1}\mathcal{U}A = \{0\}$ implies that $A = 0.$

If $\delta:\mathcal{U}\rightarrow \mathcal{G}$ is an additive mapping satisfying condition $(\mathbb{P}),$ then it is a Jordan $^{\ast}$-derivation.

Recall that a ring $\mathcal{R}$ is prime if $A\mathcal{R}B = 0\; (A, B\in\mathcal{R})$, which implies that $A = 0$ or $B = 0.$

Applying Theorem 2.1, we can get the following corollary immediately.

Corollary 2.1. Let $\mathcal{A}$ be a unital, 2-torsion-free, noncommutative prime $^{\ast}$-ring with a nontrivial projection. If $\delta:\mathcal{A}\rightarrow \mathcal{A}$ is an additive mapping satisfying condition $(\mathbb{P})$, then $\delta$ is a Jordan $^{\ast}$-derivation.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{M}$ be a von Neumann algebra in $\mathcal{B}(\mathcal{H})$. Let $\mathcal{P}(\mathcal{M})$ be the set of all projections in $\mathcal{M}$, and $\mathcal{P}_{\mathrm{fin}}(\mathcal{M})$ be the subset of all finite projections of $\mathcal{P}(\mathcal{M})$.

A linear subspace $\mathcal{D}$ in $\mathcal{H}$ is affiliated with $\mathcal{M}$ (denoted as $\mathcal{D}\; \eta\; \mathcal{M}$) if $u(\mathcal{D})\subseteq \mathcal{D}$ for every unitary operator $u$ in $\mathcal{M}^{\prime}$, the commutant of $\mathcal{M}$. $\mathcal{D}$ is strongly dense in $\mathcal{H}$ with respect to $\mathcal{M}$, if $\mathcal{D}\; \eta\; \mathcal{M}$ and there is a sequence of projections $\{P_{n}\}_{n = 1}^{\infty}\subseteq \mathcal{P}(\mathcal{M}),$ such that $P_{n}\uparrow \mathbf{1}, \; p_{n}(\mathcal{H})\subseteq \mathcal{D}$, and $\mathbf{1}-P_{n} \in\mathcal{P}_{\mathrm{fin}}(\mathcal{M})$, for every $n\in \mathbb{N}.$ A linear operator $x$ on $\mathcal{H}$ with a dense domain $\mathcal{D}(x)$ is said to be affiliated with $\mathcal{M}$ (denoted as $x\; \eta\; \mathcal{M}$) if $\mathcal{D}(x)\; \eta\; \mathcal{M}$ and $ux(\xi) = xu(\xi)$ for all $\xi \in\mathcal{D}(x)$ and for every unitary operator in $\mathcal{M}^{\prime}$. A closed linear operator $x$ acting in $\mathcal{H}$ is measurable with respect to $\mathcal{M}$ if $x\; \eta\; \mathcal{M}$ and $\mathcal{D}(x)$ strongly dense in $\mathcal{H}$. Let $S(\mathcal{M})$ denote the set of all measurable operators.

A closed linear operator $x$ acting in $\mathcal{H}$ is called locally measurable with respect to $\mathcal{M}$ if $x\; \eta\; \mathcal{M}$ and there is a sequence $\{P_{n}\}_{n = 1}^{\infty}$ of central projections in $\mathcal{M}$ such that $P_{n}\uparrow \mathbf{1}$ and $xP_{n}\in S(\mathcal{M})$ for every $n\in \mathbb{N}.$

The set $LS(\mathcal{M})$ of all locally measurable operators with respect to $\mathcal{M}$ forms a unital $\ast$-algebra with respect to algebraic operators of strong addition and multiplication and taking the adjoint of an operator. Both $\mathcal{M}$ and $S(\mathcal{M})$ are subalgebras of $LS(\mathcal{M})$. Refer to ^[1,5] and related literature for further details.

Theorem 2.2. Let $\mathcal{M}$ be a von Neumann algebra without direct commutative summands, and let $\mathcal{A}$ be an arbitrary subalgebra of $LS(\mathcal{M})$ containing $\mathcal{M}$. If $\delta:\mathcal{A}\rightarrow LS(\mathcal{M})$ is an additive mapping satisfying condition $(\mathbb{P}),$ then it is an inner Jordan $^{\ast}$-derivation.

Proof. By assumption, $\mathcal{M}$ has no direct commutative summands; there exists a projection $P_{1}$ in $\mathcal{M}$ such that $C_{P_{1}} = C_{P_{2}} = \mathbf{1},$ where $C_{P_{i}}$ denotes the central carrier of $P_{i}$ for $i = 1, 2$.

To prove that $\delta$ is an inner Jordan $^{\ast}$-derivation, according to Lemma 1.1 and Theorem 2.1, it is sufficient to show that $\mathcal{M}P_{2}$ left separates $P_{1}LS(\mathcal{M})P_{1}$ and $P_{1}\mathcal{M}$ right separates $P_{2}LS(\mathcal{M})P_{2}.$

Assume that $A\in P_{1}LS(\mathcal{M})P_{1}$ and $AX = 0$ for each $X\in \mathcal{M}P_{2}$. It follows from [4, Proposition 6.1.8] that there are projections $Q_{1}$ and $T_{1}$ such that $Q_{1}\leq P_{1}, \; T_{1}\leq P_{2}$, and $Q_{1}\sim T_{1}.$ Then there exists a partial isometry $V\in\mathcal{M}$ such that $V^{\ast}V = Q_{1}$ and $VV^{\ast} = T_{1}.$ Thus,

If $Q_{1} = P_{1}$, then the proof is complete. If $P_{1}-Q_{1}\neq 0,$ it implies that $C_{{P}_{1}-Q_{1}}C_{{P}_{2}}\neq 0.$ By [4, Proposition 6.1.8], there exist $Q_{2}\leq P_{1}-Q_{1}$ and $T_{2}\leq P_{2}$ with $Q_{2}\sim T_{2}.$ Let ${Q_{\alpha}}$ be an orthogonal family of projections in $\mathcal{M}$ maximal with respect to the property that $Q_{\alpha}\leq P_{1},$ and $P_{1}AQ_{\alpha} = 0$ for each $\alpha.$ By maximality of ${Q_{\alpha}}$, we have $P_{1} = \sum Q_{\alpha}.$ Therefore,

Using a similar technique, we can show that $P_{1}\mathcal{M}$ right separates $P_{2}LS(\mathcal{M})P_{2},$ and we omit it here. The proof is complete. □

In a $C^{\ast}$-algebra $\mathcal{A}$, projections $P$ and $Q$ are considered (Murray-von Neumann) equivalent, denoted by $P\sim Q,$ if there exists a partial isometry $V\in \mathcal{A}$ such that $V^{\ast}V = P, \; VV^{\ast} = Q,$ and $P\precsim Q$ if $P$ is equivalent to a subprojection of $Q$. Note that $P\precsim Q$ and $Q\precsim P$ do not necessarily imply $P\sim Q$ in general $C^{\ast}$-algebras. In other words, there is no Schröder–Bernstein theorem for the equivalence of projections in general $C^{\ast}$-algebras.

A nonzero projection $P$ in $\mathcal{A}$ is termed properly infinite if there exist mutually orthogonal subprojections $Q_{1}$ and $Q_{2}$ of $P$ such that $Q_{1}\sim P\sim Q_{2}$. A unital $C^{\ast}$-algebra is properly infinite if its unity $\mathbf{1}$ is properly infinite. For example, the Calkin algebra and Cuntz algebras are properly infinite.

Theorem 2.3. Let $\mathcal{A}$ be a properly infinite $C^{\ast}$-algebra. If $\delta:\mathcal{A}\rightarrow \mathcal{A}$ is an additive mapping satisfying condition $(\mathbb{P})$, then $\delta$ is an inner Jordan $^{\ast}$-derivation.

Proof. The first step is to establish the following claim:

Claim 1. If $P$ and $Q$ are two projections in $\mathcal{A}$ such that $P\precsim Q,$ then $\mathcal{A}Q \; (Q\mathcal{A})$ left (right) separates $P\mathcal{A}P.$

Since $P\precsim Q,$ there exists a partial isometry $V\in \mathcal{A}$ with $V^{\ast}V = P$ and $VV^{\ast} = Q_{1}\leqslant Q.$ Now, suppose $A \in P\mathcal{A}P$ such that $A\mathcal{A}Q = 0.$ This implies

which leads to $A = 0.$ Similarly, we can demonstrate that $Q\mathcal{A}$ right separates $P\mathcal{A}P.$ Thus, Claim 1 is validated.

Next, consider mutually orthogonal projections $P_{1}, \; Q_{1}$ in $\mathcal{A}$ such that $P_{1}\sim \mathbf{1} \sim Q_{1}.$

Claim 2. $P_{2}\precsim P_{1}.$

Let $U\in \mathcal{A}$ be a partial isometry such that $U^{\ast}U = \mathbf{1}$ and $UU^{\ast} = P_{1}.$ Define $T = UP_{2}U^{\ast},$ which is a projection satisfying $T\leq P_{1}.$ Let $S = UP_{2}$. Then $S$ is a partial isometry operator. Clearly, $SS^{\ast} = T$ and $S^{\ast}S = P_{2}$. Therefore,

Moreover, it is evident that $P_{1}\precsim P_{2}.$ From Claims 1 and 2, we conclude that $\mathcal{A}P_{2}$ left separates $P_{1}\mathcal{A}P_{1}$, and $P_{1}\mathcal{A}$ right separates $P_{2}\mathcal{A}P_{2}.$ By Lemma 1.1 and Theorem 2.1, we conclude that the statement holds. The proof is complete. □

A complex unital Banach $^{\ast}$-algebra $\mathcal{A}$ is called proper if $A^{\ast}A = 0$ implies $A = 0$ for each $A\in\mathcal{A}$. Suppose a proper $^{\ast}$-algebra $\mathcal{A}$ has a minimal left ideal $\mathcal{J}$, or equivalently, there exists a minimal projection $P\in \mathcal{A}$ such that $\mathcal{J} = \mathcal{A}P$. The sum of all minimal left ideals is referred to as the socle of $\mathcal{A}$, denoted by soc$(\mathcal{A})$. If $\mathcal{A}$ does not have minimal left ideal, we define soc($\mathcal{A}) = 0$. It is well known that the socle of $\mathcal{B}(\mathcal{H})$ is identical to $\mathcal{F}(\mathcal{H})$, the ideal of all finite rank operators in $\mathcal{B}(\mathcal{H})$ (cf. [6, p.1142 and 1143]).

Theorem 2.4. Let $\mathcal{A}$ be a complex proper Banach $^{\ast}$-algebra with unity $\mathbf{1}$, and $\mathcal{B}$ be a subalgebra of $\mathcal{A}$ containing $soc(\mathcal{A})$. Suppose there is a minimal projection $P_{1}$ in $\mathcal{A}$ such that $P_{1}soc(\mathcal{A})$ right separates $P_{2}\mathcal{A}P_{2}.$ If $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is an additive mapping satisfying condition $(\mathbb{P})$, then $\delta$ is an inner Jordan $^{\ast}$-derivation.

Proof. Without loss of generality, we assume that $\mathbf{1}$ in $\mathcal{B}$. If $\mathbf{1}\notin\mathcal{B}$, let $\mathcal{B}_{1} = \mathcal{B}+\mathbb{C}\mathbf{1}.$ In this case, consider the mapping $\widetilde{\delta}:\mathcal{B}_{1}\rightarrow \mathcal{A}$ by $\widetilde{\delta}(B+\lambda\mathbf{1}) = \delta(B)$ for each $B\in\mathcal{B}.$ Clearly, $\widetilde{\delta}|_{\mathcal{B}} = \delta.$

To prove that $\delta$ is an inner Jordan $^{\ast}$-derivation, according to Lemma 1.1 and Theorem 2.1, it is sufficient to show that $soc(\mathcal{A})P_{2}$ left separates $P_{1}\mathcal{A}P_{1}.$

For any $A\in\mathcal{A},$ it follows from [6, Theorem 10.6.2, p.1143] that there exists a continuous linear functional $f$ on $\mathcal{A}$ such that $P_{1}AP_{1} = f(A)P_{1}.$ Given the assumption that $P_{1}soc(\mathcal{A})$ right separates $P_{2}\mathcal{A}P_{2}$, it implies $P_{1}soc(\mathcal{A})P_{2}\neq 0.$ If $P_{1}AP_{1}soc(\mathcal{A})P_{2} = 0,$ it follows that $f(A) = 0.$ Consequently, $P_{1}AP_{1} = 0.$ □

Let $\mathcal{A}$ be a $C^{\ast}$-algebra. A representation $\pi:\mathcal{A}\to \mathcal{B}(\mathcal{H})$ is said to be irreducible if $\pi(\mathcal{A})$ has no nontrivial invariant subspace. $\mathcal{A}$ is called primitive if it has a faithful irreducible representation. It is easy to verify that every primitive $C^{\ast}$-algebra is prime, and for separable algebras, the converse is also true (cf. [2, p. 112]).

Corollary 2.2. Let $\mathcal{A}$ be a unital noncommutative primitive $C^{\ast}$-algebra with a nonzero $soc(\mathcal{A}),$ and let $\mathcal{B}$ be an arbitrary subalgebra of $\mathcal{A}$ containing $soc(\mathcal{A})$. If $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is an additive mapping satisfying condition $(\mathbb{P})$, then it is an inner Jordan $^{\ast}$-derivation.

Proof. Consider $\pi:\mathcal{A}\to \mathcal{B}(\mathcal{H})$, a faithful irreducible representation of $\mathcal{A}.$ If $soc(\mathcal{A})\neq 0$, it implies $soc(\pi(\mathcal{A}))\neq 0$. According to [7, Theorem 6.1.5], we have $soc(\pi(\mathcal{A}))\supseteq\mathcal{F}(\mathcal{H}).$ This implies that $P_{1}soc(\mathcal{A})$ right separates $P_{2}\mathcal{A}P_{2}$ for every minimal projection $P_{1}$ in $\mathcal{A}$. The conclusion then follows from Theorem 2.4. □

The following theorem improves the main result of ^[11].

Theorem 2.5. Let $\mathcal{H}$ be a real or complex Hilbert space, dim $\mathcal{H} > 1,$ and let $\mathcal{A}$ be a standard operator algebra on $\mathcal{H}$. Suppose that $\delta:\mathcal{A}\rightarrow \mathcal{B}(\mathcal{H})$ is an additive mapping satisfying condition $(\mathbb{P})$. Then there exists a unique linear operator $A\in\mathcal{B}(\mathcal{H})$ such that $\delta(X) = XA-AX^{\ast}$ for all $X\in\mathcal{A}.$

Proof. In the real space setting of $\mathcal{H},$ Theorem 2.1 establishes $\delta$ as a Jordan $^{\ast}$-derivation. When $\mathcal{H}$ is a complex space, an immediate application of Corollary 2.2 confirms $\delta$ as an inner Jordan $^{\ast}$-derivation. Therefore, the conclusion holds true in both cases, as supported by [11, Theorem]. □

Let $\mathcal{B}$ be a $C^{\ast}$-subalgebra of a $C^{\ast}$-algebra $\mathcal{A}$. A conditional expectation from $\mathcal{A}$ to $\mathcal{B}$ is a completely positive contraction $\phi: \mathcal{A}\rightarrow \mathcal{B}$ such that $\phi(B) = B, \; \phi(BA) = B\phi(A)$, and $\phi(AB) = \phi(A)B$ for all $A\in \mathcal{A}, \; B\in \mathcal{B}.$ If $\mathcal{B}$ is injective, then there exists a conditional expectation from $\mathcal{A}$ to $\mathcal{B}$ (cf. [2, IV.2.1]).

Recall that an approximately finite (AF) algebra is a unital $C^{\ast}$-algebra $\mathcal{A},$ which is an inductive limit of an increasing sequence of finite-dimensional $C^{\ast}$-algebras $\mathcal{A}_{n}$, $1\leq n < \infty,$ with unital embeddings $\jmath_{n}:\mathcal{A}_{n} \hookrightarrow\mathcal{A}_{n+1}.$ Equivalently, $\mathcal{A}$ is an AF algebra if it can be represented as the closed union of an ascending sequence of finite-dimensional $C^{\ast}$-algebras. Clearly, every finite-dimensional $C^{\ast}$-algebra is injective; thus, there exists a sequence $\phi_{n}:\mathcal{A}\rightarrow \mathcal{A}_{n}$ of conditional expectations such that

Theorem 2.6. Suppose $\mathcal{A} = \overline{\bigcup_{n = 1}^{\infty}\mathcal{A}_{n}}$ is an AF algebra such that $\mathcal{A}_{1}$ has no direct commutative summands. Let $\mathcal{B}$ be either $\mathcal{A}$ or $\bigcup_{n = 1}^{N}\mathcal{A}_{n},$ where $N$ is a finite integer or infinite. If $\delta:\mathcal{B}\rightarrow \mathcal{A}$ is an additive mapping satisfying condition $(\mathbb{P})$, then $\delta$ is an inner Jordan $^{\ast}$-derivation.

Proof. We divide the proof into two cases.

Case 1. Let $\mathcal{B} = \bigcup_{n = 1}^{N}\mathcal{A}_{n}$. For any positive integer $k$ ($k < N+1$), we consider the mapping $\phi_{k}\circ\delta:\mathcal{B}\rightarrow \mathcal{A}_{k}.$ Let $A, B\in\mathcal{A}_{k}$ such that $AB = BA = 0$, then

Thus, $\phi_{k}\circ\delta\mid_{\mathcal{A}_{k}}:\mathcal{A}_{k}\rightarrow \mathcal{A}_{k}$ satisfies condition $(\mathbb{P})$. Since $\mathcal{A}_{1}$ has no direct commutative summands, it implies that $\mathcal{A}_{k}$ has no direct commutative summands. Thus, $\mathcal{A}_{k} = \mathbf{M}_{k_{1}}\oplus\cdot\cdot\cdot\oplus\mathbf{M}_{k_{l}},$ where $k_{i}\geq 2$ for each $i$. Through routine calculation, we can show that $\phi_{k}\circ\delta\mid_{\mathcal{A}_{k}}(\mathbf{M}_{k_{i}})\subseteq \mathbf{M}_{k_{i}}.$ By Theorem 2.1, we have $\phi_{k}\circ\delta\mid_{\mathcal{A}_{k}}$ is a Jordan $^{\ast}$-derivation.

Fix $n$ and let $n\leq k < N+1.$ For each $A_{n} \in \mathcal{A}_{n}\subseteq\mathcal{A}_{k},$ we have

It follows from Eq (2.1) that

on $\bigcup _{n = 1}^{N}\mathcal{A}_{n}.$ Therefore, $\delta$ is a Jordan $^{\ast}$-derivation. By Lemma 1.1, $\delta$ is inner. Hence, we finish the proof of the first statement.

Case 2. Assume $\delta$ is defined from $\mathcal{A}$ to itself. Fix $n$ and choose a nontrivial projection $P_{1}$ in $\mathcal{A}_{n}.$ Now, consider an element $A\in\mathcal{A}$ such that $P_{1}AP_{1}\neq0.$ This implies the existence of a subsequence $A_{k_{m}}$ converging to $A$, where $P_{1}A_{k_{m}}P_{1}\neq0$ and $A_{k_{m}}\in\mathcal{A}_{k_{m}}$ for each $k_{m}.$ Since $\mathcal{A}_{k_{m}}$ is finite dimensional, it is also prime. Therefore,

Hence, $P_{1}A_{k_{m}}P_{1}\mathcal{A}P_{2}\neq0,$ implying $P_{1}AP_{1}\mathcal{A}P_{2}\neq0.$

Similarly, we can show that $P_{1}\mathcal{A}$ right separates $P_{2}\mathcal{A}P_{2}$. By applying Lemma 1.1 and Theorem 2.1, we conclude that $\delta$ is an inner Jordan $^{\ast}$-derivation. □

Next, we prove Theorem 2.1. Before providing its proof, we introduce the following lemmas, established under the assumptions of Theorem 2.1. For convenience, we denote $P_{i}\mathcal{G}P_{j}$ and $P_{i}\mathcal{U}P_{j}$ as $\mathcal{G}_{ij}$ and $\mathcal{U}_{ij}$, respectively. Then, the Peirce decomposition of $\mathcal{G}$ and $\mathcal{U}$ is as follows:

Lemma 2.1. The following statements hold:

${\rm(1)} \; P_{1}\delta(P_{2})P_{1} = P_{2}\delta(P_{1})P_{2} = 0;$

${\rm(2)} \; P_{1}\delta(\mathbf{1})P_{1} = P_{1}\delta(P_{1})P_{1};$

${\rm(3)} \; P_{2}\delta(\mathbf{1})P_{2} = P_{2}\delta(P_{2})P_{2}.$

Proof. Since $P_{1}P_{2} = P_{2}P_{1} = 0$, it follows from the assumption that

Multiplying the above equation from both sides by $P_{1}$, we have $2P_{1}\delta(P_{2})P_{1} = 0.$ Given that $\mathcal{G}_{11}$ is 2-torsion-free by assumption, we have $P_{1}\delta(P_{2})P_{1} = 0.$ Similarly, we have $P_{2}\delta(P_{1})P_{2} = 0.$ Thus, $(1)$ holds. Statements $(2)$ and $(3)$ are easily verified from $(1)$, and we omit the details. Hence, the proof is complete. □

Lemma 2.2. If $E$ is an idempotent in $\mathcal{U}$, then $E\delta(\mathbf{1}) = \delta(\mathbf{1})E^{\ast}.$

Proof. Since $E(\mathbf{1}-E) = (\mathbf{1}-E)E = 0$, it follows from the assumption that

Hence,

Multiplying by $E^{\ast}$ from the right side in Eq (2.2), we have

Multiplying by $E$ from the left side in Eq (2.2), we have

Combining the above two equations, we obtain $E\delta(\mathbf{1}) = \delta(\mathbf{1})E^{\ast}.$ □

Applying the above result, we can get the following lemma immediately.

Lemma 2.3. $\delta(\mathbf{1}) = P_{1}\delta(\mathbf{1})P_{1}+P_{2}\delta(\mathbf{1})P_{2}.$

Lemma 2.4. $\delta(\mathbf{1}) = 0.$

Proof. For any $G_{12}\in\mathcal{U}_{12},$ then $P_{1}+G_{12}$ is an idempotent in $\mathcal{U}$. By Lemma 2.2, we have

It follows from Lemma 2.2 that

By Lemma 2.3, we have $\delta(\mathbf{1})G_{12}\in\mathcal{G}_{12}$ and $G_{12}^{\ast}\delta(\mathbf{1})\in\mathcal{G}_{21}.$ This means that $\delta(\mathbf{1})G_{12} = 0$ for any $G_{12}\in\mathcal{U}_{12}.$ By assumption, $\mathcal{U}P_{2}$ left separates $\mathcal{G}_{11},$ it follows that $P_{1}\delta(\mathbf{1})P_{1} = 0.$ Similarly, $P_{2}\delta(\mathbf{1})P_{2} = 0.$ Using Lemma 2.3, then $\delta(\mathbf{1}) = 0.$ The proof is complete. □

For every $A, B\in\mathcal{G}$, let $[A, B]_{\ast} = AB-BA^{\ast}.$ Define an additive mapping $\sigma:\mathcal{U}\rightarrow \mathcal{G}$ by the formula:

It is evident that $\sigma$ satisfies condition $(\mathbb{P})$. Additionally, it is straightforward to verify that $\sigma(P_{1}) = \sigma(P_{2}) = 0$.

Lemma 2.5. For each $G\in\mathcal{U},$ the following statements hold:

${\rm(1)} \; \sigma(G_{11}) \in\mathcal{G}_{11};$

${\rm(2)} \; \sigma(G_{22})\in\mathcal{G}_{22};$

${\rm(3)} \; P_{1}\sigma(G_{12})P_{1} = P_{2}\sigma(G_{12})P_{2} = 0;$

${\rm(4)} \; P_{1}\sigma(G_{21})P_{1} = P_{2}\sigma(G_{21})P_{2} = 0.$

Proof. For any $G_{11}\in\mathcal{U}_{11},$ we have $G_{11}P_{2} = P_{2}G_{11} = 0.$ Therefore,

Simplifying further, we obtain

Multiplying both sides of Eq (2.3) by $P_{1}$ from the left side, we have

Similarly, multiplying both sides of Eq (2.3) by $P_{1}$ from the right side, we obtain

Furthermore, multiplying both sides of Eq (2.3) by $P_{2}$ and using the assumption that $\mathcal{G}$ is 2-torsion-free, we have $P_{2}\sigma(G_{11})P_{2} = 0.$ This implies that

which proves statement $(1)$. The proof for statement $(2)$ follows a similar pattern as statement $(1).$ Therefore, we omit it here.

implies that

Simplifying further, we obtain

Consequently, we have

Thus, $(3)$ holds. The proof of statement $(4)$ follows a similar approach to that of $(3)$, so we omit it. The proof is complete. □

Lemma 2.6. For each $G_{ij}$ in $\mathcal{U}_{ij} \; (i, j = 1, 2),$ the following statements hold:

${\rm(1)} \; \sigma(G_{11}\circ G_{12}) = \sigma(G_{11})G_{12}^{\ast}+G_{11}\sigma(G_{12})+\sigma(G_{12})G_{11}^{\ast}+G_{12}\sigma(G_{11});$

${\rm(2)} \; \sigma(G_{22}\circ G_{21}) = \sigma(G_{22})G_{21}^{\ast}+G_{22}\sigma(G_{21})+\sigma(G_{21})G_{22}^{\ast}+G_{21}\sigma(G_{22});$

${\rm(3)} \; \sigma(G_{11}\circ G_{21}) = \sigma(G_{11})G_{21}^{\ast}+G_{11}\sigma(G_{21})+\sigma(G_{21})G_{11}^{\ast}+G_{21}\sigma(G_{11});$

${\rm(4)} \; \sigma(G_{22}\circ G_{12}) = \sigma(G_{22})G_{12}^{\ast}+G_{22}\sigma(G_{12})+\sigma(G_{12})G_{22}^{\ast}+G_{12}\sigma(G_{22});$

${\rm(5)} \; \sigma(G_{11}^{2}) = \sigma(G_{11})G_{11}^{\ast}+G_{11}\sigma(G_{11});$

${\rm(6)} \; \sigma(G_{22}^{2}) = \sigma(G_{22})G_{22}^{\ast}+G_{22}\sigma(G_{22});$

${\rm(7)} \; \sigma(G_{12}\circ G_{21}) = \sigma(G_{12})G_{21}^{\ast}+G_{12}\sigma(G_{21})+\sigma(G_{21})G_{12}^{\ast}+G_{21}\sigma(G_{12}).$

Proof.

implies

Simplifying further, we have

Lemma 2.5 $(3)$ implies that

Combining the above two equations, we obtain

It follows from Lemma 2.5 $(1)$ that

Thus, statement $(1)$ holds. The proof of statement $(2)$ follows a similar approach to that of $(1)$, so we omit it.

implies

In Lemma 2.5, we have

which means that

Thus, $(3)$ holds. Similarly, statement $(4)$ is true.

For each $G_{11}\in\mathcal{U}_{11}$, by Eq (2.4), we have

where $\sigma(G_{11}^{2})G_{12}+G_{11}^{2}\sigma(G_{12}^{\ast})\in\mathcal{G}_{12}.$ On the other hand,

where $\sigma(G_{11})(G_{12}^{\ast}G_{11})^{\ast}+G_{11}\sigma(G_{11})G_{12}+G_{11}^{2}\sigma(G_{12}^{\ast})\in\mathcal{G}_{12}.$ Thus

Therefore,

By assumption, $\mathcal{U}_{12}$ left separates $\mathcal{G}_{11};$ it follows that

Thus, $(5)$ holds. Similarly, statement $(6)$ is also true.

Since

It follows that

The above equation implies that

By assumption, $\mathcal{G}$ is 2-torsion-free; it follows that

Thus, $(7)$ holds. The proof is complete. □

Using Lemma 2.6, we obtain

Lemma 2.7. $\sigma$ is a Jordan $^{\ast}$-derivation.

Proof of Theorem 2.1. Based on the definition of $\sigma$, we have

Since the mapping $G\longmapsto [G, P_{1}\delta(P_{1})P_{2}+P_{2}\delta(P_{2})P_{1}]_{\ast}$ is a Jordan $^{\ast}$-derivation, by Lemma 2.7, it follows that $\delta$ is a Jordan $^{\ast}$-derivation. □

Author contributions

Wenbo Huang: validation, resources, writing original draft preparation, writing review and editing, funding acquisition; Jiankui Li: methodology, validation, resources, writing original draft preparation, writing review and editing, funding acquisition; Shaoze Pan: resources, writing original draft preparation, writing review and editing.

Use of AI tools declaration

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

Acknowledgments

The first author was partially supported by National Natural Science Foundation of China (Grant No. 12026252 and 12026250). The second author was partially supported by National Natural Science Foundation of China (Grant No. 11871021).

Conflict of interest

The authors declare no conflicts of interest.