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arxiv: 1708.01570 · v1 · pith:2XBS3SUMnew · submitted 2017-08-04 · 🧮 math.FA · math.MG

There is no finitely isometric Krivine's theorem

classification 🧮 math.FA math.MG
keywords isometrickrivinetheoremfinitesubsettherebanachevery
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We prove that for every $p\in(1,\infty)$, $p\ne 2$, there exist a Banach space $X$ isomorphic to $\ell_p$ and a finite subset $U$ in $\ell_p$, such that $U$ is not isometric to a subset of $X$. This result shows that the finite isometric version of the Krivine theorem (which would be a strengthening of the Krivine theorem (1976)) does not hold.

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