A Second Note on Weak Differentiability of Pettis Integrals

In a recent paper the author proved that if Q is any compact metric space containing non-denumerably many points and C(Q) is the Banach space of all continuous functional over ft, then there is a Pettis integrable function from the unit interval to C(0) whose integral fails to be weakly differentiable on a set of positive measure. The purpose of this note is to obtain the same result, assuming only that ft contains infinitely many points. This leads to a certain necessary and sufficient condition for weak differentiability of Pettis integrals. The author's previous paper (cited above) will be referred to hereafter as Note I. I t will be assumed that the reader is familiar with the notation, terminology, and results of that paper. Let B be the non-dense perfect set described in Note I, and let B be its complement. Let the intervals of 2? be arranged in a sequence h, hy Izy • • • in such a way that if the order of /,• is greater than the order of I*, then j >k. For each positive integer k, let n(k) be the order of IkAs in Note I, we now define a function {xy t) over the unit square so that for each x, (x, t) is continuous in t. This will serve to define a function from the unit interval to the space C. For xÇzB, let (x, / ) = 0 ; for # £ I i , let {x, 0—0; for #£ /& and /=l/jfe (£ = 2, 3, 4, • • • ), let 0(a, 0=2<*7n(Jfe); for x^Ik and t = l/(k±l/2) (fe = 2, 3, 4, • • • ), let 4>(x, *)=0; for * = 0 or 1, let (x, t) = 0. Now for each xy let 4>{x, t) be extended linearly as a function of / between successive points already determined. We now denote by (x) the function whose values are the elements of C determined by <£(#, /) .