... ·n!f(n)(0),(3:10)thenf(z)∈K.Thus,theresultistrueforj = k. Using the mathematical induction,we complete the proof the theorem.Example 3.1 We consider the functionf(z)= z ... =2)j−1n=2|f(n)(0)| (j 3).(2:2)Proof For j = 2, the inequality (2.1) becomes (1.2) of Lemma 1.2. Thus, the theoremis hold true for j = 2. We need to prove the inequality for j ≧ 3. Note thatf(z)=z0f(t ... |f(k−1)(0)|2√5−k−2n=2|f(n)(0)|,(2:8)thenf(z)∈ S∗. This is equivalent to|f(k)(z)| Nk2√5−k−1n=2|f(n)(0)|.(2:9)Therefore, the theorem holds true for j = k. Thus, applying the mathem atical induc-tion,...