Mappings on some reflexive algebras characterized by action on zero products or Jordan zero products (1106.4371v1)
Abstract: Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$ and let $\delta:\mathrm{Alg}\mathcal{L}\rightarrow B(X)$ be a linear mapping. If $\vee{L\in \mathcal{L}: L_-\nsupseteq L}=X$ or $\wedge{L_-:L\in \mathcal{L}, L_-\nsupseteq L}=(0)$, we show that the following three conditions are equivalent: (1) $\delta(AB)=\delta(A)B+A\delta(B)$ whenever $AB=0$; (2) $\delta(AB+BA)=\delta(A)B+A\delta(B)+\delta(B)A+B\delta(A)$ whenever $AB+BA=0$; (3) $\delta$ is a generalized derivation and $\delta(I)\in (\mathrm{Alg}\mathcal{L})\prime$. If $\vee{L\in \mathcal{L}: L_-\nsupseteq L}=X$ or $\wedge{L_-:L\in \mathcal{L}, L_-\nsupseteq L}=(0)$ and $\delta$ satisfies $\delta(AB+BA)=\delta(A)B+A\delta(B)+\delta(B)A+B\delta(A)$ whenever $AB=0$, we obtain that $\delta$ is a generalized derivation and $\delta(I)A\in(\mathrm{Alg}\mathcal{L})\prime$ for every $A\in \mathrm{Alg}\mathcal{L}$. We also prove that if $\vee{L\in \mathcal{L}: L_-\nsupseteq L}=X$ and $\wedge{L_-:L\in \mathcal{L}, L_-\nsupseteq L}=(0)$, then $\delta$ is a local generalized derivation if and only if $\delta$ is a generalized derivation.