On a converse of the Hölder inequality

Abstract

Let X be a measure space with measure pt. We consider the associated dual metric spaces LP and L , where 1 M.(Fn) for all n. The stronger condition will be implied if we show that 1u(F) = + oo. If not, then A(E F) = +00, since M (E) = +00. By the assumed weaker condition on X, E-F contains a subset F' of finite but positive measure. Then, if 1u(F) < + oo, we have M< (FU F') < + 0, contrary to the choice of M. To prove necessity of the condition, we note that if X contains a subset E, of measure + oo, whose subsets have measure 0 or + 0o, then the characteristic function of E is evidently not LP. But if g is any Lq function, the set {xIg(x) #0 } is a countable union of sets of finite measure, and so intersects E in a set of measure 0. So the product of g with the characteristic function of E is 0 almost everywhere and thus is integrable, giving a denial of the converse of the Holder inequality. Before proving sufficiency, we shall show that if f is any measurable function whose product with every Lq function is integrable, and if {xjf(x) } #0 is a countable union of sets of finite measure, then fCLP, independent of any condition on X. In this case there is a sequence, {f.,}, of functions each of which is bounded and vanishes except on a set of finite measure, and such that the sequence { Iffl P} converges monotonely to Ifj P. Each of the functions fn, determines a