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Zbl 0980.47062
Tao, Changli; Lu, Shijie
Separating vectors, separating functionals and reflexivity.
(Chinese)
[J] Chin. Ann. Math., Ser. A 21, No.4, 513-516 (2000). ISSN 1000-8314

Let $X$ and $Y$ be Banach spaces and $B(X, Y)$ denote the set of all bounded linear operators from $X$ and $Y$. For a subspace $\bbfS$ of $B(X,Y)$, let $\text{Ref}(\bbfS)= \{T\in B(X,Y); Tx\in[\bbfS x],\forall x\in X\}$ where $[\bbfS x]= \bigvee\{\bbfS x,S\in\bbfS\}$. $\bbfS$ is said to be reflexive if $\bbfS= \text{Ref}(\bbfS)$. A vector $x\in X$ is said to be separating $\bbfS$ if $Tx=0$ implies $T= 0$ for arbitrary $T\in\bbfS$. For a natural number $n$ write \align X^{(n)} & = X\oplus X\oplus\cdots\oplus X,\\ Y^{(n)} & = Y\oplus Y\oplus\cdots\oplus Y.\endalign In the paper it is shown that if $\bbfS$ is a closed subspace of $B(X,Y)$ with a separating vector, $n\ge 2$, $F_i$, $i= 1,\dots, n$, are bounded operators on $\bbfS$ with closed ranges and at least two of them are bounded below, then $$\bbfS_n= \left\{\left(\matrix F_1(T)\\ & F_2(T)\\ &&\ddots\\ &&&F_n(T)\endmatrix\right), T\in\bbfS\right\}\subset B(X^{(n)},\ Y^{(n)})$$ is reflexive.
[Shanli Sun (Changchun)]
MSC 2000:
*47L35 Nest algebras, CSL algebras
47L10 Algebras of operators on Banach spaces, etc.
47L30 Abstract operator algebras on Hilbert spaces

Keywords: reflexive; separating vector

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