Articles
Publicationes Mathematicae Debrecen (00333883)105(1-2)pp. 91-105
Let A and B be commutative and semisimple Banach algebras. Suppose that ∥ · ∥γ is an algebra cross-norm on A B such that ∥ · ∥γ ≥ ∥ · ∥e, and AbγB is a semisimple Banach algebra. In this paper, we verify the BED property for AbγB. In fact, we show that if AbγB is of BED, then both A and B are so, whenever either A or B is unital. We also show that if B (resp., A) is unital and Ab ⊆ CBSE0 (∆(A)) (resp., Bb ⊆ CBSE0 (∆(B))), then A\bγB ⊆ CBSE0 (∆(AbγB)). We also establish that if B (resp., A) is finite dimensional, then AbγB is of BED if and only if A (resp., B) is of BED. © 2024 Institute of Mathematics, University of Debrecen. All rights reserved.
Banach Journal of Mathematical Analysis (26622033)18(3)
Let (X, d) be a compact metric space and A be a commutative and semisimple Banach algebra. Some of our recent works are related to the several BSE concepts of the vector-valued Lipschitz algebra Lip(X,A). In this paper as the main purpose, we verify the BED property for Lip(X,A), which is actually different from the BSE feature. We first prove as an elementary result that Lip(X,A) is regular if and only if A is so. Then we prove that A is a BED algebra, whenever Lip(X,A) is so. Afterwards, we verify the converse of this statement. Indeed, we prove that if A is a BED algebra then CBSE0(Δ(Lip(X,A)))⊆Lip(X,A)^ and LipX⊗A^⊆CBSE0(Δ(Lip(X,A))). It follows that if LipX⊗A is dense in Lip(X,A) then Lip(X,A) is a BED algebra, provided that A is so. Moreover, we conclude that the necessary and sufficient condition for the unital and in particular finite dimensional Banach algebra A, to be a BED algebra is that Lip(X,A) is a BED algebra. Finally, regarding to some known results which disapproves the BSE property for lipα(X,A)(0<α<1), we show that for any commutative and semisimple Banach algebra A with A0≠∅, lipα(X,A) fails to be a BED algebra, as well. © Tusi Mathematical Research Group (TMRG) 2024.
