Infinite dimensional $L$-spaces do not have preduals of all orders
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- by Thomas E. Armstrong
- Proc. Amer. Math. Soc. 74 (1979), 285-290
- DOI: https://doi.org/10.1090/S0002-9939-1979-0524301-0
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Abstract:
It is shown that if E is an infinite dimensional Banach space with first dual E’, second dual E", and nth dual ${E^{[n]}}$ and if ${E^{[n]}}$ is either an L- or M-space all duals are either L- or M-spaces except possibly E which could be a Lindenstrauss space. If E is an L- or M-space there is an integer $n(E)$ so that if $m > n(E)$ there is no Banach space F with $E = {F^{[m]}}$. The linear isomorphic analogues to these isometric results are also established. In particular if E is an ${\mathcal {L}_1}$ or ${\mathcal {L}_\infty }$ space there is an integer $\bar n(E)$ so that E is not linearly isomorphic to ${F^{[m]}}$ for any Banach space F when $m > \bar n(E)$.References
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Bibliographic Information
- © Copyright 1979 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 74 (1979), 285-290
- MSC: Primary 46B10
- DOI: https://doi.org/10.1090/S0002-9939-1979-0524301-0
- MathSciNet review: 524301