Uniformly convex functions II

Lemma 1. If p1(z) = 1 + c1z + c2z2 + c3z3 + . . . ∈P, then |c2n −1 2c2 n| ≤2 −1 2|cn|2 , (5) |µc2 nc2n −c4 n| ≤8(µ −2) (µ ≥4) (6) and (7) |µcnc2n −c3 n| ≤4(µ −2)

Theorem 1. Let f(z) = z + a2z2 + a3z3 + . . . ∈UCV. Then |an| ≤An (n = 4, 5, 6). Equality holds if and only if f(z) is k(z) or one of its rotations. P r o o f. When n = 4, we see that the coefficients in the expression of a4 in (4.3) are all positive. So we get |a4| ≤A4 from |ck| ≤2 (k = 1, 2, 3). When n = 5, from (4.4) we have 20a5 = 4 π2 c4 + 4 3π2 16 π2 −1  c3c1 + 2 π2  4

Lemma 1 for n = 1 and µ = 6, |c3 −2c2c1 + c3 1| ≤2 [LZ] and |ck| ≤2 (k = 1, . . . , 5) to derive that the upper bound of 30|a6| is achieved when we replace all ck by 2. Thus we have |a6| ≤A6. In each case, we have used the inequality |c1| ≤2 in our proof. Hence equality holds only if p1(z) = (1 + z)/(1 −z) or one of its rotations, which implies that f = k or one of its rotations. On the other hand, it is clear that inequalities become equalities for k and its rotations. This completes the proof

Theorem 2. Let f(z) = z + a2z2 + a3z3 + . . . ∈UCV. Then |µa2 2 −a3| ≤            

Theorem 2. To discuss coefficient bounds for the inverses of functions in UCV, we first observe that for the inverse function Kn(w) of kn(z), Kn(w) = w − 8 (n −1)nπ2 wn + . . . . Hence for f ∈UCV with F(w) = f −1(w) = w + d2w2 + d3w3 + . . . , max{|dn| : f ∈UCV} ≥ 8 (n −1)nπ2 . For n = 2, 3, 4, we can prove that equality holds. Note that the series expan- sion for f −1(w) converges when |w| < ϱf, where 0 < −k(−1) ≤ϱf [MM] depends on f.

Theorem 3. Let f ∈UCV and F(w) = f −1(w) = w+d2w2+d3w3+. . . . Then |dn| ≤ 8 (n −1)nπ2 (n = 2, 3, 4) . Equality holds if and only if f is equal to kn or one of its rotations. P r o o f. As F(f(z)) = z, we have d2 = −a2 , d3 = 2a2 2 −a3 , d4 = −a4 + 5a3a2 −5a3 2 . By using (4.1)–(4.3), we can express dn in terms of cn as follows: d2 = −2

Theorem 4. Let f ∈UCV and |z| = r < 1. Then |f ′′(z)| ≤k′′(r) . Equality holds for any z ∈D if and only if f is k or one of its rotations. P r o o f. Let p(z)=1+zf ′′(z)/f ′(z). Then p ≺q implies that p−1 ≺q−1. As all coefficients of q −1 are positive, the subordination principle yields that for |z| = r, |p(z) −1| ≤q(r) −1 . This is the same as |f ′′(z)/f ′(z)| ≤k′′(r)/k′(r) . From |f ′(z)| ≤k′(r) [MM], we see that |f ′′(z)| ≤k′(r)|f ′′(z)/f ′(z)| ≤k′′(r) . We also know that equality holds in |f ′