nLab
isomorphism classes of Banach spaces

This page is inspired by the following question, which appeared on MathOverflow.

Let p,q(1,)p,q \in (1,\infty) with pqp\neq q. Are the Banach spaces L p()L^p(\mathbb{R}), L q()L^q(\mathbb{R}) isomorphic?

More generally, one can ask:

Given two Banach spaces, XX and YY, when are they isomorphic?

The following started out as an adapted version of Bill Johnson’s answer to the MathOverflow question.

One way to prove that a Banach space XX is not isomorphic to a Banach space YY is to exhibit a property which is preserved under isomorphisms that XX has but YY does not. For example, among the spaces L p()L_p(\mathbb{R}) for p[1,]p \in [1,\infty], L L_\infty is the only nonseparable space, and L 1L_1 is the only separable space with a nonseparable dual. Thus L 1L_1 and L L_\infty are not isomorphic to each other or to any L pL_p with p(1,)p \in (1,\infty).

To distinguish among the L pL_p with p(1,)p \in (1,\infty) finer properties are needed. Type and cotype are examples of such properties. The (best) type and cotype of L pL_p are standard calculations: if p[1,2]p \in [1,2] then L pL_p has type pp and cotype 22 (and no better), and if p[2,)p \in [2,\infty) then L pL_p has type 22 and cotype pp (and no better). See for example in Theorem 6.2.14 of AK06. From that, one can see that if pqp \ne q, then L pL_p and L qL_q either have different (best) type or different (best) cotype.

Type and cotype depend only on the collection of finite dimensional subspaces of a space (we call such a property a local property?). So neither can be used to prove, e.g., that for p2p \ne 2, L pL_p is not isomorphic to p\ell_p. One way of proving this is to show that for p2p \ne 2, 2\ell_2 embeds isomorphically into L pL_p but not into p\ell_p (see also AK).

References

  • AK06 Albiac, Fernando and Kalton, Nigel. Topics in Banach space theory. Graduate Texts in Mathematics, 233. Springer, New York, 2006. xii+373 pp. ISBN: 978-0387-28141-4; MR2192298
Revised on November 8, 2011 17:49:19 by Mark Meckes? (129.22.117.158)