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arXiv · 1803.09343

$\xi$-completely continuous operators and $\xi$-Schur Banach spaces

Abstract

For each ordinal $0\leqslant \xi\leqslant \omega_1$, we introduce the notion of a $\xi$-completely continuous operator and prove that for each ordinal $0< \xi< \omega_1$, the class $\mathfrak{V}_\xi$ of $\xi$-completely continuous operators is a closed, injective operator ideal which is not surjective, symmetric, or idempotent. We prove that for distinct $0\leqslant \xi, \zeta\leqslant \omega_1$, the classes of $\xi$-completely continuous operators and $\zeta$-completely continuous operators are distinct. We also introduce an ordinal rank $\textsf{v}$ for operators such that $\textsf{v}(A)=\omega_1$ if and only if $A$ is completely continuous, and otherwise $\textsf{v}(A)$ is the minimum countable ordinal such that $A$ fails to be $\xi$-completely continuous. We show that there exists an operator $A$ such that $\textsf{v}(A)=\xi$ if and only if $1\leqslant \xi\leqslant \omega_1$, and there exists a Banach space $X$ such that $\textsf{v}(I_X)=\xi$ if and only if there exists an ordinal $\gamma\leqslant \omega_1$ such that $\xi=\omega^\gamma$. Finally, prove that for every $0<\xi<\omega_1$, the class $\{A\in \mathcal{L}: \textsf{v}(A) \geqslant \xi\}$ is $\Pi_1^1$-complete in $\mathcal{L}$, the coding of all operators between separable Banach spaces. This is in contrast to the class $\mathfrak{V}\cap \mathcal{L}$, which is $\Pi_2^1$-complete in $\mathcal{L}$.

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BibTeXRIS

R. M. Causey, K. Navoyan. 2018-03-25. $\xi$-completely continuous operators and $\xi$-Schur Banach spaces. https://arxiv.org/abs/1803.09343

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