Characterizations of *-antiderivable mappings on operator algebras

Let A{\mathcal{A}} be a ∗\ast -algebra, ℳ{\mathcal{ {\mathcal M} }} be a ∗\ast -A{\mathcal{A}}-bimodule, and δ\delta be a linear mapping from A{\mathcal{A}} into ℳ{\mathcal{ {\mathcal M} }}. δ\delta is called a ∗\ast -derivation if δ(AB)=Aδ(B)+δ(A)B\delta \left(AB)=A\delta \left(B)+\delta \left(A)...

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Bibliographic Details
Main Authors: An Guangyu, Zhang Xueli, He Jun
Format: Article
Language:English
Published: De Gruyter 2022-06-01
Series:Open Mathematics
Subjects:
Online Access:https://doi.org/10.1515/math-2022-0047
Description
Summary:Let A{\mathcal{A}} be a ∗\ast -algebra, ℳ{\mathcal{ {\mathcal M} }} be a ∗\ast -A{\mathcal{A}}-bimodule, and δ\delta be a linear mapping from A{\mathcal{A}} into ℳ{\mathcal{ {\mathcal M} }}. δ\delta is called a ∗\ast -derivation if δ(AB)=Aδ(B)+δ(A)B\delta \left(AB)=A\delta \left(B)+\delta \left(A)B and δ(A∗)=δ(A)∗\delta \left({A}^{\ast })=\delta {\left(A)}^{\ast } for each A,BA,B in A{\mathcal{A}}. Let GG be an element in A{\mathcal{A}}, δ\delta is called a ∗\ast -antiderivable mapping at GG if AB∗=G⇒δ(G)=B∗δ(A)+δ(B)∗AA{B}^{\ast }=G\Rightarrow \delta \left(G)={B}^{\ast }\delta \left(A)+\delta {\left(B)}^{\ast }A for each A,BA,B in A{\mathcal{A}}. We prove that if A{\mathcal{A}} is a C∗{C}^{\ast }-algebra, ℳ{\mathcal{ {\mathcal M} }} is a Banach ∗\ast -A{\mathcal{A}}-bimodule and GG in A{\mathcal{A}} is a separating point of ℳ{\mathcal{ {\mathcal M} }} with AG=GAAG=GA for every AA in A{\mathcal{A}}, then every ∗\ast -antiderivable mapping at GG from A{\mathcal{A}} into ℳ{\mathcal{ {\mathcal M} }} is a ∗\ast -derivation. We also prove that if A{\mathcal{A}} is a zero product determined Banach ∗\ast -algebra with a bounded approximate identity, ℳ{\mathcal{ {\mathcal M} }} is an essential Banach ∗\ast -A{\mathcal{A}}-bimodule and δ\delta is a continuous ∗\ast -antiderivable mapping at the point zero from A{\mathcal{A}} into ℳ{\mathcal{ {\mathcal M} }}, then there exists a ∗\ast -Jordan derivation Δ\Delta from A{\mathcal{A}} into ℳ♯♯{{\mathcal{ {\mathcal M} }}}^{\sharp \sharp } and an element ξ\xi in ℳ♯♯{{\mathcal{ {\mathcal M} }}}^{\sharp \sharp } such that δ(A)=Δ(A)+Aξ\delta \left(A)=\Delta \left(A)+A\xi for every AA in A{\mathcal{A}}. Finally, we show that if A{\mathcal{A}} is a von Neumann algebra and δ\delta is a ∗\ast -antiderivable mapping (not necessary continuous) at the point zero from A{\mathcal{A}} into itself, then there exists a ∗\ast -derivation Δ\Delta from A{\mathcal{A}} into itself such that δ(A)=Δ(A)+Aδ(I)\delta \left(A)=\Delta \left(A)+A\delta \left(I) for every AA in A{\mathcal{A}}.
ISSN:2391-5455