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}, $ where $ LS(\mathcal{M}) $ is the $ ^{\ast} $-algebra of all locally measurable operators with respect to $ \mathcal{M} $. Suppose $ \delta $ is an additive mapping from $ \mathcal{A} $ to $ LS(\mathcal{M}) $ that satisfies the condition $ \delta(A)B^{\ast}+A\delta(B)+\delta(B)A^{\ast}+B\delta(A) = 0 $ whenever $ AB = BA = 0. $ In this paper, we prove that there exists an element $ Y $ in $ LS(\mathcal{M}) $ such that $ \delta(X) = XY-YX^{\ast}, $ for every $ X $ in $ \mathcal{A}. $
Citation: Wenbo Huang, Jiankui Li, Shaoze Pan. Some zero product preserving additive mappings of operator algebras[J]. AIMS Mathematics, 2024, 9(8): 22213-22224. doi: 10.3934/math.20241080
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}, $ where $ LS(\mathcal{M}) $ is the $ ^{\ast} $-algebra of all locally measurable operators with respect to $ \mathcal{M} $. Suppose $ \delta $ is an additive mapping from $ \mathcal{A} $ to $ LS(\mathcal{M}) $ that satisfies the condition $ \delta(A)B^{\ast}+A\delta(B)+\delta(B)A^{\ast}+B\delta(A) = 0 $ whenever $ AB = BA = 0. $ In this paper, we prove that there exists an element $ Y $ in $ LS(\mathcal{M}) $ such that $ \delta(X) = XY-YX^{\ast}, $ for every $ X $ in $ \mathcal{A}. $
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Wenbo Huang, Jiankui Li, Shaoze Pan. Some zero product preserving additive mappings of operator algebras[J]. AIMS Mathematics, 2024, 9(8): 22213-22224. doi: 10.3934/math.20241080
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