ABSTRACT
Since h is univalent, we have z0 = w(z0 ). By applying the Schwarz lemma
128 FIRST-ORDER DIFFERENTIAL SUBORDINATIONS
we deduce that w(z) = z, and this leads to q(z) = h(z). However, since h is univalent we have h'(Zo) :f:. 0, which contradicts q'(z0 ) = 0 . Hence q' (z) :f:. 0 for z E U, and q is locally univalent.
