Jordan derivations of reflexive algebras
Let \(\mathcal{A}\) be an algebra and \(\mathcal{M}\) be an \(\mathcal{A}\)-bimodule. A linear map \(\delta : \mathcal{A}\rightarrow \mathcal{M}\) is called a Jordan derivation (resp., a derivation) if \(\delta(A^2) = \delta(A)A + A\delta(A)\) for all \(A\in \mathcal{A}\) (resp., \(\delta(AB) = \delta(A)B + A\delta(B)\) for all \(A,B\in \mathcal{A}\)). Let \(X\) be a Banach space, \(B(X)\) be the space of all bounded linear operators on \(X\), \(\mathcal{L}\) be a complete lattice of closed subspaces of \(X\) that contains \(\{0\}\) and \(X\), and \(\text{Alg}\mathcal{L} = \{T\in B(X) : TL\subseteq L, L \in \mathcal{L}\}\). For \(E\in \mathcal{L}\), let \(E_- = \vee\{F\in \mathcal{L} : E\not\leq F\}\) and \(\mathcal{J}(\mathcal{L}) = \{L\in \mathcal{L} : L \neq 0 \;\text{\text{and }} \;L_- \neq X\}\). In the paper under review, the author shows that, if \(\vee\mathcal{J}(\mathcal{L}) = X\) or \(\wedge\{L_- : L\in \mathcal{J}(\mathcal{L})\} = 0\), then every Jordan derivation \(\delta : \text{Alg}\mathcal{L}\rightarrow B(X)\) is a derivation. The proof is based on a careful analysis of the images of the rank one operators in \(\text{Alg}\mathcal{L}\) under \(\delta\).
scientific article; zbMATH DE number 5774261