= E(flMd+flMd) = E(J1Md)+E(flMd) =
m (M) + m (M). The computation for z is completely analogous.
x1,z x2,Z
Also, we have Im z(M)n = jE(f R21Md**)l E(If 21MdI) <
+ +
E(f 21MdJI) a E(f 21MiXllzldlBl ) = llxl-zJE(flMdJBIv)
= lx z|l Bm(M), so Imz(M)I xS IxlzlpBl (M). Then for given
M c M, the map (x,z) + m (M) is continuous, bilinear. Then there
XZ
is a continuous linear map m(M) E L(E,Z') satisfying
m ,z(M) = E(f 21Md).
(More precisely, any continuous, bilinear function f(x,z): ExZ R
is continuous and linear in each component. Then the map x f(x,.)
is a continuous linear map from E into Z'. In our situation, we
have m(M)x = mx,.(M), so = mx ,(M), i.e., for M e M, we
have m(M) E L(E,Z').) We also have Im(M)I | unBl(M), since
|m(M)I = sup Im(M)xl = sup Im (M) = sup sup up mm (M)) 5
jxl:i1 xls1 x' xls 1 1zl1 X'Z
sup ( sup xlz1I BI (M)) = PI (M). This gives us a map
IxS1 Iz 1 B B
m: M L(E,Z'). We now verify that m is o-additive and has finite
variation Iml (in particular Iml S VIBI):
First of all, m is additive. Let M,N e M be disjoint; we show
m(M UN) = m(M) + m(N), i.e., that m(M UN)x = m(M)x + m(N)x for all
**