Every 𝑙-space is isomorphic to a strictly convex space

Abstract

In the preceding paper' a normed linear space B is called sc (sm) if B is isomorphic to a space for which the unit sphere is strictly convex (smooth). Many special spaces were investigated to determine which, if either, of these properties they possessed, but at that time the behavior of the spaces of summable functions over arbitrary measure spaces was not determined in full. Let ,u be a measure defined over the class of measurable sets in some space X. It was proved in SCM, Theorem 11, that L(,), the space of /A-summable functions on X, is sm if and only if X is a-finite under A; that is, if and only if X is the union of countably many sets of finite MA-measure. It was also proved in SCM, (v) that every L(,) is sc if and only if every L(+) is sc, where ? is a measure for which the measure of the whole space is finite. In this paper we use some results of Maharam2 on the structure of measure algebras to prove that every L(c4) is sc. This gives the necessary information to determine the position of the L(,u) spaces in the table of the introduction of SCM, and gives the general result stated in the title of this note. The theorems of Maharam which we need for our proof are stated in terms of measure algebras. Maharam, Theorem 2, asserts that a measure algebra which is nonatomic is a union of countably many homogeneous measure algebras. Maharam, Theorem 1, characterizes homogeneous measures as simple product measures. Since we are interested in the L-space over the measure algebra rather than in the measure algebra itself, we shall state here the results we want in terms of L-spaces. If I is an index set, let 2' be the product of I copies of the set containing only the real numbers -1 and 1. In 2 define the product measure ir of I copies of the elementary measure which assigns to each of the points -1 and I the measure 1/2. Then if X is a space with finite, nonatomic measure 4, Maharam's results assert that L(+) can be written as a countable direct stum L(+) = 1n>O Ln so that