ABSTRACT
As was done in the proof of Theorem 3.6a, we can assume that G is univalent
182 FIRST-ORDER DIFFERENTIAL SUBORDINATIONS
We now need to show that F -< G, and we will do this by using a
subordination chain argument involving Lemma 1.2a. The function
is analytic in I z I < r < 1, for sufficiently small r, and for all t ;;:: 0. It is also continuously differentiable on [0, oo ), for each I z I < r < 1. According to (3.6-9) we have L(z, 0) = g(z). From (3.6-15) we obtain
(3.6-16) z(dLidz)
this leads to
(3.6-17) g(z) = L(z, 0) -< L(z, t) for all t ;;:: 0.
