A note on successive coefficients of convex functions

Theorem 1.1. Let D+ n and D− n be given in (1.2). Then the following hold.

Theorem 1.2. 1/n < D− n < 2/(n + 1) for each n ≥4. It is an open problem to find the value of D− n for n ≥4. As the triangle inequality implies ||an| −|an+1|| ≤|an+1 −an|, one may think that the study of the functional |an+1 −an| would be helpful to our problem. However, we immediately see that the sharp bound of |an+1 −an| for convex functions is 2 as the function f(z) = z/(1 + z) serves as an extremal one. On the other hand, it is indeed helpful to consider the functional |an+1 −an| for refined

Theorem 1.3. Let 0 ≤p ≤2. Suppose that f(z) = z + a2z2 + a3z3 + · · · is a function in K(p). Then the following sharp inequalities hold: |a3 −a2| ≤(2p + 1)(2 −p) 6 , and (1.5) |a4 −a3| ≤       

Lemma 2.1. For a function P ∈P, the sharp inequality |pn| ≤2 holds for each n. The sharpness can be observed through the example P0(z) = (1+z)/(1−z) = 1+2z + 2z2 + · · · . We will use also the following result due to Carath´eodory and Toeplitz (see [3] or [11]).

Lemma 2.2 (Carath´eodory-Toeplitz theorem). Let P(z) = 1 + P∞ n=1 pnzn be a formal power series with complex coefficients. Then P represents a Carath´eodory function if and only if (2.1) Dn =

Lemma 2.3. Let P(z) = 1 + p1z + p2z2 + · · · be a Carath´eodory function with p1 ∈R and p2 = p2 1 −2. Then P must be of the form P(z) = 1 −z2 1 −p1z + z2. Moreover, the functions f ∈S∗and g ∈K determined by zf ′(z)/f(z) = 1+zg′′(z)/g′(z) = P(z) have the forms f = Kφ and g = Lφ, where φ = arccos[p1/2].

Lemma 2.5. Let −2 ≤p ≤2 and p2, p3 ∈C. There exists a function P ∈P with P(z) = 1 + pz + p2z2 + p3z3 + · · · if and only if (2.5) 2p2 = p2 + x(4 −p2). and (2.6) 4p3 = p3 + 2(4 −p2)px −p(4 −p2)x2 + 2(4 −p2)(1 −|x|2)y

Lemma 2.7. Let a, b, c ∈R with a ≥0 and c ≥0. Then Y (a, b, c) =    a + |b| + c if |b| ≥2(1 −c), 1 + a + b2 4(1 −c) if |b| ≤2(1 −c). The maximum in the definition of Y (a, b, c) is attained at z = ±1 in the first case according as b = ±|b|. When |b| = 2(1 −c) = 0, we set b2/4(1 −c) to be 0 in the above. The lemma can easily be verified by the fact that

Lemma 2.8. Let P(t) and Q(t) be (possibly degenerated) real quadratic polynomials. Suppose that P > 0 and Q > 0 on an interval I ⊂R and that ∆P > 0. If there is a positive constant T such that (i) ∆Q ≥T 3/2∆P, and (ii) TP(t) ≥Q(t) for t ∈I, then the function G(t) = p P(t) − p Q(t) is convex on I.

Lemma 2.9. Let u = 6p2/(4 −p2), v = 2, a = 3p3/(4 −p2), b = 5p/2 and c = p/2 for 4/3 ≤p ≤ √ 2 and consider the function F(z) = |u + vz| − a + bz −cz2. Then F(z) ≤F(−|z|) for z ∈D.