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.
Categories: math.FA