A GENERAL APPROACH TO THE FEKETE-SZEGÖ PROBLEM

JAE HO CHOI, YONG CHAN KIM, AND TOSHIYUKI SUGAWA

Abstract. In this article, we provide a new method solving the Fekete-Szegö problem for classes of close-to-convex functions defined in terms of subordination. As an example, we apply it to the class of strongly close-to-convex functions.

1. Introduction

Fekete and Szegö [5] proved the striking result that the inequality

|a_{3} − μa_{2}^{2}| ≤ 1 + 2 exp (\frac{−2μ}{1 − μ})

holds for any normalized univalent function f(z) = z + a_{2}z^{2} + a_{3}z^{3} + ⋯ in the unit disk 𝔻 = {z ∈ ℂ : |z| < 1} and for 0 ≤ μ ≤ 1 and that this inequality is sharp for each μ (see also [3]). The coefficient functional

(1.1)  Λ_{μ}(f) = a_{3} − μa_{2}^{2} = \frac{1}{6} (f′′′(0) − \frac{3μ}{2}\{f′′(0)\}^{2})

on normalized analytic functions f(z) = z +a_{2}z^{2} +a_{3}z^{3} +⋯ in the unit disk is important in the sense that this can represent various geometric quantities as well as in the sense that this behaves well with respect to the rotation, namely, Λ_{μ}(R_{θ}f) = e^{2iθ}Λ_{μ}(f) for θ ∈ ℝ. Here R_{θ}f denotes the rotation of f by angle θ, more precisely, R_{θ}f(z) = e^{−iθ}f(e^{iθ}z).

In fact, other than the simplest case when Λ_{0}(f) = a_{3}, we have several important ones. For example, Λ_{1}(f) = a_{3} − a_{2}^{2} represents S_{f}(0)/6, where S_{f} denotes the Schwarzian derivative (f′′/f′)′ − (f′′/f′)^{2}/2 of f. Moreover, the first two non-trivial coefficients of the n-th root transform {f(z^{n})}^{1/n} = z + c_{n+1}z^{n+1} + c_{2n+1}z^{2n+1} + ⋯ of f(z) = z + a_{2}z^{2} + ⋯ are written by c_{n+1} = a_{2}/n and c_{2n+1} = (Λ_{(n−1)/2n}f)/n = a_{3}/n − (n − 1)a_{2}^{2}/2n^{2}.

Thus it is quite natural to ask about inequalities for Λ_{μ} corresponding to subclasses of normalized univalent functions in the unit disk. This is sometimes called the Fekete-Szegö problem. Actually, many authors have considered this problem for typical classes of univalent functions (see, for instance, [1], [2], [7], [9], [10], [11], [12], [13]).

We denote by A the set of all normalized analytic functions on the unit disk 𝔻 and denote by S the subclass of A consisting of all univalent functions as usual. Let M be the class of analytic functions φ on the unit disk with φ(0) = 1 and let N be the subclass of M consisting of functions with positive real part. For φ ∈ M, we will denote by M(φ) the subset defined by {h ∈ M : h ≺ φ}. Here and hereafter, we use the notation f ≺ g, or, f(z) ≺ g(z) in 𝔻 for analytic functions f and g on 𝔻 to mean the subordination, namely, that there exists a holomorphic map ω of the unit disk 𝔻 into itself with ω(0) = 0 such that f = g ∘ ω. Note that if g is univalent, f ≺ g is equivalent to the condition that f(0) = g(0) and f(𝔻) ⊂ g(𝔻).

Date: August 25, 2006.

1991 Mathematics Subject Classification. 30C45, 30C50.

Key words and phrases. univalent function, subordination, coefficient bound.

The second named author was supported by Korea Basic Science Research Foundation (Grant No. DP0022). A part of the present work was carried out during the third author’s visit to the University of Helsinki under the bilateral exchange program of scientists between the Academy of Finland and the JSPS. He was partially supported also by the Ministry of Education, Grant-in-Aid for Encouragement of Young Scientists, 11740088.

Ma and Minda [12] gave a complete answer to the Fekete-Szegö problem for the classes S^{∗}(φ) and K(φ) for φ ∈ M with some mild conditions. (Actually, it is enough to assume that φ is univalent and satisfies φ(z̄) = \overline{φ(z)} on 𝔻 and φ′(0) > 0.) These classes are defined for φ ∈ M by

S^{∗}(φ) = \{f ∈ A : \frac{zf′(z)}{f(z)} ≺ φ(z) in 𝔻\} , and

K(φ) = \{f ∈ A : 1 + \frac{zf′′(z)}{f′(z)} ≺ φ(z) in 𝔻\} .

Note that, with the special choice of φ(z) = (1 + z)/(1 − z) the above classes consist of starlike and convex functions in the standard sense.

In our previous paper [8], we gave a partial answer to the problem for the classes

C(φ, ψ) = \{f ∈ A : there exists a g ∈ K(φ) such that \frac{f′}{g′} ≺ ψ in 𝔻\} .

These classes cover close-to-convex functions in some sense as we shall see later. The authors would like to take this opportunity to note that in Lemma 2.2 and in Theorem 3.1 of [8] some conditions for φ such as above was dropped, for instance, the conditions that φ is univalent and satisfies φ(z̄) = \overline{φ(z)} on 𝔻 and φ′(0) > 0 should be assumed.

It is worth noting that R_{θ}f ∈ S^{∗}(φ) if and only if f ∈ S^{∗}(φ) for a fixed real number θ. The same thing can be said for K(φ) and for C(φ, ψ) as well. In particular, these classes are rotation invariant.

The main aim of this paper is to provide a general method to compute the quantity sup {|Λ_{μ}(f)| : f ∈ C(φ, ψ)} in terms of φ, ψ and μ. Actually, we will see that this quantity is determined only by the first two non-trivial coefficients of φ and ψ and the parameter μ provided that φ′(0) ≠ 0 and ψ′(0) ≠ 0. Our method also enables us to find all the extremal functions for the functional |Λ_{μ}|.

Our method developed below can easily be modified to apply to the classes S^{∗}(φ) and K(φ) as well. We, however, will not state it separately because we have satisfactory results due to Ma and Minda [12] already.

As a typical example, we treat the class

SCC(α, β) = ⋃_{−πα/2<γ<πα/2} SCC_{γ}(α, β)

of strongly close-to-convex functions of order (α, β) for 0 < α ≤ 1 and 0 < β ≤ 1. Here, f ∈ SCC_{γ}(α, β) if and only if f belongs to A and there exists a function g ∈ A with | arg(1 + zg′′(z)/g′(z))| < πβ/2 such that | arg(f′(z)/g′(z)) − γ| < πα/2. Though the class SCC(α, β) seems not to be written in the form C(φ, ψ), each SCC_{γ}(α, β) can be described in such a form, see Section 4. Note that SCC(α, 1) is the class of strongly close-to-convex functions of order α (see [6, II, Definition 11.4]), and that SCC(1, 1) is the class of closeto-convex functions in the standard sense. Note also that SCC_{0}(α, α) contains the class of strongly starlike functions of order α.

These classes have been considered by many authors, however, no complete answer to the Fekete-Szegö problem has been given in the literature so far. Actually, many authors treated only “normalized” classes contained in SCC_{0}(α, β) for some α, β. Only the case when α = β = 1 was completely settled by Eenigenburg and Silvia [4] (see also [9]). For other cases, only partial results are known beyond the normalized classes ([10], [1]).

In order to exhibit effectiveness of our method, we will give another proof for a recent result of Darus and Thomas [2] on Fekete-Szegö inequalities for the class SCC_{0}(α, β) for all 0 < α, β ≤ 1 and, moreover, we will prove that those inequalities hold still true for the class SCC(α, β) when 2/3 ≤ μ ≤ 1.

2. General approach to the Fekete-Szegö problem

Let φ(z) = 1 + A_{1}z + A_{2}z^{2} + ⋯, ψ(z) = 1 + B_{1}z + B_{2}z^{2} + ⋯ be analytic in the unit disk with A_{1}B_{1} ≠ 0. For an arbitrary complex number μ, consider the functional Λ_{μ} on A given by (1.1). Since the class C(φ, ψ) is rotation invariant, the range set Δ(φ, ψ, μ) = {Λ_{μ}(f) : f ∈ C(φ, ψ)} of Λ_{μ} is rotation invariant, too. Therefore, the outer boundary of the set is a circle centered at the origin. (We will describe this set in a detailed way in Proposition 2.5 below.) Then the radius, denoted by ρ(φ, ψ, μ), of the circle is nothing but the quantity which we want to compute. Namely,

(2.1)  ρ(φ, ψ, μ) = sup {|Λ_{μ}(f)| : f ∈ C(φ, ψ)}

= sup {Re Λ_{μ}(f) : f ∈ C(φ, ψ)} . (2.2)

First of all, we remark that the quantity ρ(φ, ψ, μ) has an obvious convexity property in μ.

Lemma 2.1. Let s and t be non-negative numbers with s + t = 1. Then, the inequality

ρ(φ, ψ, sμ_{0} + tμ_{1}) ≤ sρ(φ, ψ, μ_{0}) + tρ(φ, ψ, μ_{1})

holds for μ_{0}, μ_{1} ∈ ℂ.

Proof. By definition,

Λ_{sμ_{0}+tμ_{1}}(f) = sΛ_{μ_{0}}(f) + tΛ_{μ_{1}}(f)

holds for each f ∈ C(φ, ψ). Therefore, the triangle inequality

|Λ_{sμ_{0}+tμ_{1}}(f)| ≤ s|Λ_{μ_{0}}(f)| + t|Λ_{μ_{1}}(f)|

yields the required inequality. □

For each function f(z) = z + a_{2}z^{2} + a_{3}z^{3} + ⋯ in C(φ, ψ), by definition, we can take a function g(z) = z + b_{2}z^{2} + b_{3}z^{3} + ⋯ from K(φ) such that h := f′/g′ ≺ ψ. If we write h(z) = 1 + c_{1}z + c_{2}z^{2} + ⋯, by a simple calculation, we have the relation

Λ_{μ}(f) = a_{3} − μa_{2}^{2} = (b_{3} − μb_{2}^{2}) + \frac{1}{3} (c_{2} − \frac{3μ}{4}c_{1}^{2}) + (\frac{2}{3} − μ) b_{2}c_{1}
= Π_{μ}(b_{2}, b_{3}, c_{1}, c_{2}),

where Π_{μ} is a polynomial given by the above formula. Note also that this quantity can be written in the form Λ_{μ}(g) + \frac{1}{3}Λ_{3μ/4}(zh) + (\frac{2}{3} − μ) b_{2}c_{1}. By the triangle inequality, we could get an estimate for |Λ_{μ}(f)| by using the knowledge of |Λ_{μ}| for K(φ), |Λ_{3μ/4}| for zh(z) where h ∈ M(ψ) and |b_{2}c_{1}|. This was the basic idea in our previous paper [8]. Unfortunately, this estimate is not always sharp. Therefore, in order to get a sharp result, we may not divide the terms in this way generally.

To obtain information about ρ(φ, ψ, μ), we now consider the coefficient regions

U_{φ} = {(u_{2}, u_{3}) ∈ ℂ^{2} : ∃g ∈ K(φ) s.t. g(z) = z + u_{2}z^{2} + u_{3}z^{3} + O(z^{4})}, and
W_{ψ} = {(w_{1}, w_{2}) ∈ ℂ^{2} : ∃h ∈ M(ψ) s.t. h(z) = 1 + w_{1}z + w_{2}z^{2} + O(z^{3})}.

Then, by definition, a ∈ Δ(φ, ψ, μ) if and only if a = Π_{μ}(u_{2}, u_{3}, w_{1}, w_{2}) for some (u_{2}, u_{3}) ∈ U_{φ} and (w_{1}, w_{2}) ∈ W_{ψ}. In other words,

Δ(φ, ψ, μ) = Π_{μ}(U_{φ} × W_{ψ}).

To describe these sets, we introduce the universal set V defined by

V = {(v_{1}, v_{2}) ∈ ℂ^{2} : ∃ω : 𝔻 → 𝔻 holomorphic and satisfying ω(z) = v_{1}z + v_{2}z^{2} + O(z^{3})}.

Then the following result can immediately be obtained by applying the Schwarz-Pick lemma to the function ω(z)/z (see, for instance, [14, p. 108]).

Lemma 2.2.

V = {(v_{1}, v_{2}) ∈ ℂ^{2} : |v_{1}|^{2} + |v_{2}| ≤ 1}.

Furthermore, (v_{1}, v_{2}) ∈ ∂V = {|v_{1}|^{2} + |v_{2}| = 1} for an analytic function ω(z) = v_{1}z + v_{2}z^{2}+⋯ on the unit disk with |ω| < 1 if and only if either ω(z) = v_{1}z (in case of |v_{1}| = 1) or ω(z) = az(z + āv_{1})/(1 + a\overline{v_{1}}z) (in case of |v_{1}| < 1), where a = v_{2}/(1 − |v_{1}|^{2}) ∈ ∂𝔻.

In particular, V is a compact connected set and its interior is a Reinhardt domain in ℂ^{2}.

Let g ∈ K(φ). We now take a holomorphich map ω : 𝔻 → 𝔻 with ω(0) = 0 so that 1 + zg′′/g′ = φ ∘ ω. Writing ω(z) = v_{1}z + v_{2}z^{2} + ⋯, we have the relation (u_{2}, u_{3}) = F_{φ}(v_{1}, v_{2}), where F_{φ} : ℂ^{2} → ℂ^{2} is the map determined by

F_{φ}(v_{1}, v_{2}) = (A_{1}v_{1}/2, (A_{1}v_{2} + (A_{2} + A_{1}^{2})v_{1}^{2})/6).

Since F_{φ} is an analytic automorphism, namely, a biholomorphic map of ℂ^{2}, we have the following.

Lemma 2.3. U_{φ} = F_{φ}(V). In particular, U_{φ} is a compact connected set and ∂U_{φ} = F_{φ}(∂V).

Next, suppose that h(z) = 1 + w_{1}z + w_{2}z^{2} + ⋯ = ψ(ω(z)) for some analytic function ω(z) = v_{1}z+v_{2}z^{2}+⋯ on 𝔻 with |ω| < 1. Then, we have the relation (w_{1}, w_{2}) = G_{ψ}(v_{1}, v_{2}), where

G_{ψ}(v_{1}, v_{2}) = (B_{1}v_{1}, B_{2}v_{1}^{2} + B_{1}v_{2}).

Since G_{ψ} : ℂ^{2} → ℂ^{2} is also biholomorphic, we have

Lemma 2.4. W_{ψ} = G_{ψ}(V). In particular, W_{ψ} is a compact connected set and ∂W_{ψ} = G_{ψ}(∂V).

Let (u_{1}, u_{2}) and (v_{1}, v_{2}) be points in ∂V. Then, by Lemma 2.2, functions ω_{1}, ω_{2} are uniquely determined by the conditions |ω_{j}| < 1, ω_{1}(z) = u_{1}z + u_{2}z^{2} + … and ω_{2}(z) = v_{1}z + v_{2}z^{2} + …. Therefore, a function f ∈ C(φ, ψ) is determined by f′/g′ = ψ ∘ ω_{2}, where 1+zg′′/g′ = φ∘ω_{1}. We denote by E[u_{1}, u_{2}, v_{1}, v_{2}] or, more specifically, by E_{φ,ψ}[u_{1}, u_{2}, v_{1}, v_{2}] the function f determined above.

By Lemmas 2.3 and 2.4, we have Δ(φ, ψ, μ) = Π_{μ} ∘(F_{φ} ×G_{ψ})(V ×V). In particular, the quantity ρ = ρ(φ, ψ, μ) depends only on A_{1}, A_{2}, B_{1}, B_{2} and μ. Therefore, we can express ρ also as a function H in these variables, namely, we can write

(2.3)  ρ(φ, ψ, μ) = H(A_{1}, A_{2}, B_{1}, B_{2}, μ),

where φ(z) = 1 + A_{1}z + A_{2}z^{2} + ⋯ and ψ(z) = 1 + B_{1}z + B_{2}z^{2} + · · · . Also, we can now show the following.

Proposition 2.5. Let φ, ψ ∈ M and suppose that φ′(0) ≠ 0 and ψ′(0) ≠ 0. Then the range set Δ(φ, ψ, μ) of the functional Λ_{μ} on C(φ, ψ) is a closed disk centered at the origin.

Proof. The relation Π_{μ}(0, 0, 0, 0) = 0 implies that 0 ∈ Δ = Δ(φ, ψ, μ). Since Δ = Π_{μ}(U_{φ}, V_{ψ}) is closed and connected, there is a point z ∈ Δ with |z| = r for each r ∈ [0, ρ(φ, ψ, μ)]. As we have seen, the set Δ is rotation invariant, and therefore, Δ must be a closed disk {|z| ≤ ρ(φ, ψ, μ)}. □

A direct calculation shows

Π_{μ}(F_{φ}(u_{1}, u_{2}), G_{ψ}(v_{1}, v_{2})) = ((\frac{1}{6} − \frac{μ}{4}) A_{1}^{2} + \frac{A_{2}}{6} ) u_{1}^{2} + \frac{A_{1}}{6} u_{2}

+ (\frac{B_{2}}{3} − \frac{B_{1}^{2}μ}{4}) v_{1}^{2} + \frac{B_{1}}{3} v_{2} + (\frac{1}{3} − \frac{μ}{2}) A_{1}B_{1}u_{1}v_{1}

=Au_{2} + Bv_{2} + Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2}.

Here, for later convenience as well, we have set

A = \frac{A_{1}}{6} ,
B = \frac{B_{1}}{3} ,
(2.4)  K = \frac{A_{2}}{6} + \frac{1}{4} (\frac{2}{3} − μ) A_{1}^{2},
L = \frac{B_{2}}{3} − \frac{μ}{4}B_{1}^{2},
M = \frac{1}{4} (\frac{2}{3} − μ) A_{1}B_{1}.

Based on the above facts, we can reduce our problem to an algebraic one.

Lemma 2.6. The quantity ρ(φ, ψ, μ) is given by the function Ω defined by

(2.5)  Ω(A, B, K, L, M) = max_{{|u_{1}|^{2}+|u_{2}|≤1 \atop |v_{1}|^{2}+|v_{2}|≤1}} |Au_{2} + Bv_{2} + Ku_{1}^{2} + Lv_{1}^{2} + 2Mu_{1}v_{1}|

= max_{{|u_{1}|^{2}+|u_{2}|≤1 \atop |v_{1}|^{2}+|v_{2}|≤1}} Re [Au_{2} + Bv_{2} + Ku_{1}^{2} + Lv_{1}^{2} + 2Mu_{1}v_{1}] , (2.6)

where A, B, K, L, M are related to φ and ψ by (2.4). Furthermore, one can replace the range in the above maxima by |u_{1}|^{2} + |u_{2}| = 1 and |v_{1}|^{2} + |v_{2}| = 1.

Actually, expressions (2.5) and (2.6) follows from (2.1) and (2.2), respectively, together with the above lemmas. The last assertion is an immediate consequence of the maximum principle. (This also follows from the fact that ∂V × ∂V is the Sirov boundary of the domain Int V × Int V.)

The function H given by (2.3) is now described by the relation

H(A_{1}, A_{2}, B_{1}, B_{2}, μ) = Ω [\frac{A_{1}}{6} , \frac{B_{1}}{3} , \frac{A_{2}}{6} + \frac{1}{4} (\frac{2}{3} − μ) A_{1}^{2}, \frac{B_{2}}{3} − \frac{μ}{4}B_{1}^{2}, \frac{1}{4} (\frac{2}{3} − μ) A_{1}B_{1}] .

In the rest of the section, we give several simplifications of the expression of Ω and, equivalently, of H. The expressions (2.5) and (2.6) have their own advantage. We begin with examination of the first expression. Note that each variable of (v_{1}, v_{2}) ∈ V can have arbitrary argument independently in view of the shape of V (Lemma 2.2). This means that, for (u_{1}, u_{2}), (v_{1}, v_{2}) ∈ V, in the chain of trivial inequalities

|Au_{2} + Bv_{2} + Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2}|
(2.7)  ≤|A||u_{2}| + |B||v_{2}| + |Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2}|
(2.8)  ≤|A|(1 − |u_{1}|^{2}) + |B|(1 − |v_{1}|^{2}) + |Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2}|

all equalities can hold at once. Therefore, we can deduce the following expression of Ω from (2.5):

Ω(A, B, K, L, M) = max_{|u|≤1,|v|≤1} [|A|(1 − |u|^{2}) + |B|(1 − |v|^{2}) + |Ku^{2} + 2Muv + Lv^{2}|].

Now we introduce the auxiliary functions P and Q on [0, +∞) × [0, +∞) defined by

P (s, t) = Q(s, t) − |A|s^{2} − |B|t^{2},
Q(s, t) = max_{|u|=s,|v|=t} |Ku^{2} + 2Muv + Lv^{2}|

= max_{θ,τ∈ℝ} |Ks^{2}e^{2iθ} + 2Mste^{i(θ+τ)} + Lt^{2}e^{2iτ} |

= max_{θ∈ℝ} |Ks^{2} + 2Mste^{iθ} + Lt^{2}e^{2iθ}| .

Then, we have the simple-looking expression

(2.9)  Ω = |A| + |B| + max_{0≤s,t≤1} P (s, t).

By definition, these functions are homogeneous of degree 2, precisely, P (rs, rt) = r^{2}P (s, t) and Q(rs, rt) = r^{2}Q(s, t) hold for each r ≥ 0. Therefore, setting F (r) = P (rs, rt), we observe that F (r) is non-decreasing if F (1) ≥ 0 and non-increasing if F (1) ≤ 0. In particular, we can see the relation

max_{0≤s,t≤1} P (s, t) = max{Φ_{1}, Φ_{2}, 0},

where Φ_{1} and Φ_{2} are the functions in A, B, K, L, M defined by

(2.10)  Φ_{1} = max_{0≤s≤1} P (s, 1) and Φ_{2} = max_{0≤t≤1} P (1, t).

Thus, we have reached the following result.

Theorem 2.7. The quantity Ω = Ω(A, B, K, L, M) can be represented by

Ω = |A| + |B| + max{Φ_{1}, Φ_{2}, 0},

where Φ_{j} = Φ_{j}(A, B, K, L, M) is defined by (2.10) for j = 1, 2.

Unfortunately, it is not always easy to calculate Q(s, t) and thus P (s, t). If K, L and M are all real, however, we can do that as we shall see in the next section.

Now we make a brief discussion on extremal points. Let f(z) = z + a_{2}z^{2} + a_{3}z^{3} + ⋯ be an extremal function in C(φ, ψ) for the functional |Λ_{μ}|. Note here that the rotation R_{θ}f is an extremal one because Λ_{μ}(R_{θ}f) = e^{2iθ}Λ_{μ}(f). Thus, we may regard R_{θ}f, θ ∈ ℝ, as a trivial one-parameter family of extremal functions.

By definition, there is a function g ∈ K(φ) such that f′/g′ ≺ ψ. We take holomorphic maps ω_{1}, ω_{2} : 𝔻 → 𝔻 with ω_{1}(0) = ω_{2}(0) = 0 so that 1 + zg′′/g′ = φ ∘ ω_{1} and that f′/g′ = ψ ∘ ω_{2}. Let (u_{1}, u_{2}, v_{1}, v_{2}) be a corresponding point in V × V, namely, ω_{1}(z) = u_{1}z + u_{2}z^{2} + … and ω_{2}(z) = v_{1}z + v_{2}z^{2} + …. For this point, equalities must hold in (2.7) and (2.8) simultaneously. In particular, (u_{1}, u_{2}, v_{1}, v_{2}) ∈ ∂V ×∂V. Now, by Lemma 2.2, the forms of ω_{1}, ω_{2} are exactly determined by u_{1}, u_{2}, v_{1}, v_{2}. In this way, the extremal function f can be expressed in terms of φ, ψ, u_{1}, u_{2}, v_{1}, v_{2} and will be denoted by E_{φ,ψ}[u_{1}, u_{2}, v_{1}, v_{2}].

We further analyze several possible cases and explain how to determine u_{1}, u_{2}, v_{1}, v_{2}. Recall that we have set s = |u_{1}| and t = |v_{1}|.

Case 1: max{Φ_{1}, Φ_{2}} > 0. Assume, for instance, Φ_{1} ≥ Φ_{2}. In this case, we have Ω = |A| + |B| + Φ_{1}. Then there exists an s_{0} ∈ [0, 1] such that Φ_{1} = P (s_{0}, 1). As we shall see by example in the next section, such s_{0} is not unique in general, however, Φ_{1} > P (s, t) holds for any 0 ≤ s ≤ 1 and for any 0 ≤ t < 1 by the homogeneity of P. Therefore, in this case, |v_{1}| = 1 and v_{2} = 0 hold. Choose u_{1} and v_{1} with |u_{1}| = s_{0}, |v_{1}| = 1 so that Q(s_{0}, 1) = |Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2}|. Then choose u_{2} with |u_{2}| = 1 − s_{0}^{2} so that inequality holds in (2.7), in other words, Au_{2} and Ku_{1}^{2} + 2Mu_{1}v_{1} + Lv_{1}^{2} have the same argument. The case when Φ_{2} ≥ Φ_{1} can be treated smilarly.

Case 2: max{Φ_{1}, Φ_{2}} = 0. This is a degenerate case. Assume, for instance, Φ_{1} = 0. Then, there exsits an s_{0} ∈ [0, 1] such that P (s_{0}, 1) = 0. Let (u_{1}, u_{2}, v_{1}, v_{2}) be the point in ∂V × ∂V determined in the same way as in Case 1.

In the present case, P (s_{0}t, t) = 0 holds for any t ∈ [0, 1] by the homogeneity of P (s, t). We choose a complex number b with |b| = 1 so that Bb and Ku_{1}^{2} +2Mu_{1}v_{1} +Lv_{1}^{2} have the same argument. For simplicity, we consider only the generic case when s_{0} < 1. Choose a non-negative number k = k(t) so that |tu_{1}|^{2} + |ku_{2}| = 1, namely, k = (1 − s_{0}^{2}t^{2})/(1 − s_{0}^{2}). We then have a non-trivial one-parameter family of extremal functions f_{t} = E_{φ,ψ}[tu_{1}, (1− s_{0}^{2}t^{2})u_{2}/(1 − s_{0}^{2}), tv_{1}, (1 − t^{2})b)] (0 ≤ t ≤ 1).

Case 3: max{Φ_{1}, Φ_{2}} < 0. This is one of the simplest cases. The maximum in (2.5) is attained only in the case when u_{1} = v_{1} = 0. Therefore, by Lemma 2.2, ω_{1} and ω_{2} above must take the forms ω_{1}(z) = az^{2} and ω_{2}(z) = bz^{2}, where a and b are unimodular constants chosen so that Aa and Bb have the same argument.

We end this section with examination of expression (2.6). Actually, in some cases, this method is more appropriate. As above, we easily have the expression

(2.11) Ω(A, B, K, L, M) = |A|+|B|+ max_{|u|≤1,|v|≤1} Re [Ku^{2}+2Muv+Lv^{2}−|A||u|^{2}−|B||v|^{2}].

Introducing real coordinates u = x_{1} + ix_{2} and v = x_{3} + ix_{4}, we obtain the expression of Ω in terms of a real quadratic form in x_{1}, x_{2}, x_{3}, x_{4}.

Theorem 2.8. The quantity Ω can be described in the form

Ω(A, B, K, L, M) = |A| + |B| + max_{{x_{1}^{2}+x_{2}^{2}≤1 \atop x_{3}^{2}+x_{4}^{2}≤1}} ∑_{j,k=1}^{4} c_{jk}x_{j}x_{k},

where c_{jk} are the entries of the real symmetric matrix

C = (\begin{matrix}Re K − |A| & −Im K & Re M & −Im M \\ −Im K & −Re K − |A| & Im M & −Re M \\ Re M & Im M & Re L − |B| & −Im L \\ −Im M & −Re M & −Im L & −Re L − |B| \end{matrix}) .

It seems quite difficult to treat the above matrix, however, when some of entries vanish, the situation becomes easier to analyze. For example, if the matrix C is non-positive, we know that Ω = |A| + |B|. Sometimes it is easier to check this condition than to calculate Φ_{j} directly in the above if truly Ω = |A| + |B| holds.

3. The case when K, L, M are all real

When K, L and M are all real, we can actually calculate the quantity Ω(A, B, K, L, M). For convenience, we define Sign[x] for x ∈ ℝ by

Sign[x] = \begin{cases}1 & if x > 0 \\ −1 & if x < 0 \\ ±1 & if x = 0. \end{cases}

For example, the assertion K = Sign[x]L means that K = L or K = −L when x = 0. We are now ready to state our main result in this section.

Theorem 3.1. Let A, B ∈ ℂ and K, L, M ∈ ℝ. If KL ≥ 0, the quantity Ω(A, B, K, L, M) in (2.5) is given by

Ω(A, B, K, L, M)

= \begin{cases}|A| + |B| & if |A| + |B| ≥ |K| + |L| and D ≥ 0 \\ |A| + |L| − \frac{M^{2}}{|K| − |A|} & if |A| > |M| + |K| and D < 0 \\ |B| + |K| − \frac{M^{2}}{|L| − |B|} & if |B| > |M| + |L| and D < 0 \\ |K| + 2|M| + |L| & otherwise, \end{cases}

where D = (|K|−|A|)(|L|−|B|)−M^{2}. If KL < 0, the quantity Ω(A, B, K, L, M) is given by

Ω(A, B, K, L, M) = |A| + |B| + max{0, R},

where R is defined by the table

(a)  (b)  (c)
(1)  0  0  |L| − |B| + \frac{M^{2}}{|A| + |K|}
(2)  0  (no possibility)  −|L| − |B| + \frac{M^{2}}{|A| − |K|}
(3) |K| − |A| + \frac{M^{2}}{|B| + |L|} −|K| − |A| + \frac{M^{2}}{|B| − |L|}  S − |A| − |B|

according to the combination of the cases

(1) |A| ≥ max \{|K|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |K|\},

(2) |K| + |M| ≤ |A| < |K|\sqrt{1 − \frac{M^{2}}{KL}},

(3) otherwise, and the cases

(a) |B| ≥ max \{|L|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |L|\},

(b) |L| + |M| ≤ |B| < |L|\sqrt{1 − \frac{M^{2}}{KL}},

(c) otherwise. Here, S is defined by

S = \begin{cases}−|K| + 2|M| + |L| & if |MK| − |ML| ≤ −2|KL| \\ |K − L|\sqrt{1 − \frac{M^{2}}{KL}} & if − 2|KL| ≤ |MK| − |ML| ≤ 2|KL| \\ |K| + 2|M| − |L| & if 2|KL| ≤ |MK| − |ML|. \end{cases}

In the rest of the section, we prove the above theorem. The information on extremal functions can be obtained from Proposition 3.4 when KL ≥ 0 and Proposition 3.7 when KL < 0. We start with the investigation of the preparatory quantity

q(a, b, c, θ) = |a + 2be^{iθ} + ce^{2iθ}|

for a, b, c, θ ∈ ℝ. First we need the following lemma.

Lemma 3.2. When ac ≥ 0, one has

q(a, b, c, θ) ≤ |a| + 2|b| + |c|,

where equality holds when e^{iθ} = Sign(b(a + c)).

Note that a and c have the same signature if ac ≥ 0. In the case when ac < 0 the situation becomes a bit complicated.

Lemma 3.3. When ac < 0, letting ξ = −b(a + c)/2ac, one has

q(a, b, c, θ) ≤ \begin{cases}| − a + 2b − c| = Sign[b](−a + 2b − c) & if ξ ≤ −1 \\ \sqrt{1 − \frac{b^{2}}{ac}} |a − c| & if − 1 ≤ ξ ≤ 1 \\ |a + 2b + c| = Sign[b](a + 2b + c) & if 1 ≤ ξ, \end{cases}

where equality holds when cos θ = −1, cos θ = ξ and cos θ = 1, respectively.

Proof. We set x = cos θ. Then

q(a, b, c, θ)^{2} = 4acx^{2} + 4b(a + c)x + a^{2} + 4b^{2} + c^{2} − 2ac
= 4ac(x − ξ)^{2} + (1 − b^{2}/ac)(a − c)^{2}.

By analyzing the behaviour of this quadratic polynomial in x, we obtain the desired result. □

We now return to our problem. Noting the relation Q(s, t) = max_{θ∈ℝ} q(Ks^{2}, Mst, Lt^{2}, θ), we can explicitly calculate Q(s, t) by Lemmas 3.2 and 3.3.

First, we treat the easier case when KL ≥ 0. In this case, P (s, t) = (|K| − |A|)s^{2} + 2|M|st + (|L| − |B|)t^{2}. We denote by D the descriminant of this quadratic form, precisely, D = (|K| − |A|)(|L| − |B|) − M^{2}. Then the next statement is easily verified.

Proposition 3.4. Assume that KL ≥ 0. Then

Ω(A, B, K, L, M) − |A| − |B| = max_{0≤s,t≤1} P (s, t)

= \begin{cases}P (0, 0) = 0 & if |K| + |L| ≤ |A| + |B| and D ≥ 0 \\ P (\frac{|M|}{|A|−|K|}, 1) = |L| − |B| − \frac{M^{2}}{|K| − |A|} & if |M| < |A| − |K| and D < 0 \\ P (1, \frac{|M|}{|B|−|L|}) = |K| − |A| − \frac{M^{2}}{|L| − |B|} & if |M| < |B| − |L| and D < 0 \\ P (1, 1) = |K| + 2|M| + |L| − |A| − |B| & otherwise. \end{cases}

In each case, the maximum point of P (s, t) is unique in (s, t) ∈ [0, 1] × [0, 1] except for the following cases:

(a) When D = 0 and |K| + |L| < |A| + |B|, one has P (s, t) = 0 if and only if s(|K| − |A|) + t|M| = 0 and s|M| + t(|L| − |B|) = 0.

(b) When M = 0, |K| = |A| and |L| = |B|, one has P (s, t) = 0 for all s, t.

Next, we consider the case when KL < 0 and M ≠ 0. In what follows, the terms “increasing” and “decreasing” will be used in the strict sense.

Let r_{0} and r_{1} be the positive numbers determined by the relations

|\frac{M}{2KL}| (|K|r_{0} − \frac{|L|}{r_{0}}) = −1 and |\frac{M}{2KL}| (|K|r_{1} − \frac{|L|}{r_{1}}) = 1.

Since the function |K|r − |L|/r is increasing in r > 0, the inequality r_{0} < r_{1} holds. We set I_{1} = [0, r_{0}], I_{2} = [r_{0}, r_{1}], I_{3} = [r_{1}, +∞]. Then we can show the following.

Lemma 3.5. Let K, L, M ∈ ℝ with KL < 0 and M ≠ 0. Then

Q(s, t) = \begin{cases}−|K|s^{2} + 2|M|st + |L|t^{2} & if s/t ∈ I_{1} = [0, r_{0}] \\ \sqrt{1 − \frac{M^{2}}{KL}} |Ks^{2} − Lt^{2}| & if s/t ∈ I_{2} = [r_{0}, r_{1}] \\ |K|s^{2} + 2|M|st − |L|t^{2} & if s/t ∈ I_{3} = [r_{1}, ∞]. \end{cases}

Proof. Let ε = Sign[KM]. We now apply Lemma 3.3 to the case when a = Ks^{2}, b = Mst, c = Lt^{2}. Then we have

ξ = \frac{−b(a + c)}{2ac} = −\frac{M}{2KL} (K \frac{s}{t} + L\frac{t}{s}) = ε |\frac{M}{2KL}| (|K|\frac{s}{t} − |L|\frac{t}{s}) .

When ε = +1, the condition ξ ≤ −1 holds precisely if s/t ≤ r_{0}, i.e., s/t ∈ I_{1}. In this case, by Lemma 3.3, we have

Q(s, t) = Sign[M](−Ks^{2} + 2Mst − Lt^{2}) = −|K|s^{2} + 2|M|st + |L|t^{2}.

When ε = −1, the condition ξ ≥ 1 holds precisely if s/t ≤ r_{0}, i.e., s/t ∈ I_{1}. In this case, again by Lemma 3.3, we have

Q(s, t) = Sign[M](Ks^{2} + 2Mst + Lt^{2}) = −|K|s^{2} + 2|M|st + |L|t^{2}.

Therefore, we show the assertion in the case when s/t ∈ I_{1}. The other two cases can be dealt with similarly. □

We need later the following properties of r_{0} and r_{1}.

Lemma 3.6. Let K, L, M ∈ ℝ with KL < 0 and M ≠ 0 and consider the following conditions:

(a) r_{0} ≤ |M|/(|A| + |K|),

(b) r_{1} ≤ |M|/(|A| − |K|), and

(c) |A| ≤ |K|\sqrt{1 − \frac{M^{2}}{KL}}.

Then (a) and (c) are equivalent. Furthermore, when |A| > |K|, (b) and (c) are equivalent. Also,

r_{0} ≤ 1 ⇔ −2|KL| ≤ |MK| − |ML| and r_{1} ≤ 1 ⇔ 2|KL| ≤ |MK| − |ML|.

Proof. We observe

r_{0} ≤ \frac{|M|}{|A| + |K|}

⇔ − 1 ≤ |\frac{M}{2KL}| (|K|\frac{|M|}{|A| + |K|} − |L|\frac{|A| + |K|}{|M|})
⇔ − 2|KL| ≤ \frac{|K|M^{2}}{|A| + |K|} − |L|(|A| + |K|)
⇔ |L|(|A| − |K|) ≤ \frac{|K|M^{2}}{|A| + |K|}
⇔ A^{2} − K^{2} ≤ \frac{|K|M^{2}}{|L|} = −\frac{KM^{2}}{L}
⇔ A^{2} ≤ K^{2} (1 − \frac{M^{2}}{KL}) .

Next, r_{0} ≤ 1 if and only if −1 ≤ |\frac{M}{2KL} | (|K| − |L|) , which is equivalent to −2|KL| ≤ |MK| − |ML|. The other cases can be treated in the similar way. □

Now we are ready to show the following result.

Proposition 3.7. Let A, B ∈ ℂ and K, L, M ∈ ℝ with AB ≠ 0 and KL < 0. Then,

Φ_{1} = \begin{cases}P (\frac{|M|}{|A|+|K|}, 1) = \frac{M^{2}}{|A| + |K|} + |L| − |B| & if |A| ≥ max \{|K|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |K|\} \\ P (\frac{|M|}{|A|−|K|}, 1) = \frac{M^{2}}{|A| − |K|} − |L| − |B| & if |K| + |M| ≤ |A| < |K|\sqrt{1 − \frac{M^{2}}{KL}} \\ P (1, 1) & otherwise. \end{cases}

In each case, the maximum point of P (s, 1) is unique in s ∈ [0, 1] except for the following two cases:

(a) When M = 0 and |A| = |K|, the function P (s, 1) is constant in 0 ≤ s ≤ 1.

(b) When M ≠ 0, |A| = |K|\sqrt{1 − \frac{M^{2}}{KL}} and −2|KL| < |MK| − |ML|, the maximum of P (s, 1) in s ∈ [0, 1] is attained at each point in the interval I_{2} ∩ [0, 1] which has positive length.

Proof. We first consider the case when KM > 0. When s ∈ I_{1} ∩ [0, 1], by Lemma 3.5, we have P (s, 1) = (−|K|s^{2} + 2|M|s + |L|) − |A|s^{2} − |B| = −(|A| + |K|)s^{2} + 2|M|s + |L| − |B|. The axis of symmetry of the graph of this quadratic polynomial in s is s = |M|/(|A| + |K|). By Lemma 3.6, we see that |M|/(|A| + |K|) ∈ I_{1} ∩ [0, 1] if and only if

|A| ≥ max\{|K|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |K|\}. Hence, we can see the following.

Claim 1.

(1) If |A| ≥ max\{|K|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |K|\}, then s = M/(|A| + |K|) is the unique maximum point of P (s, 1) in I_{1} ∩ [0, 1], thus P (s, 1) ≤ P (M/(|A| + |K|), 1) there.

(2) Otherwise, P (s, 1) is increasing in I_{1} ∩ [0, 1].

When s ∈ I_{2} ∩ [0, 1], we have P (s, 1) = \sqrt{1 − \frac{M^{2}}{KL}}(|K|s^{2} + |L|) − |A|s^{2} − |B|. Then it is easy to see the following

Claim 2.

(1) If |K|\sqrt{1 − \frac{M^{2}}{KL}} > |A|, the function P (s, 1) is increasing in I_{2} ∩ [0, 1].

(2) If |K|\sqrt{1 − \frac{M^{2}}{KL}} < |A|, the function P (s, 1) is decreasing in I_{2} ∩ [0, 1].

(3) If |K|\sqrt{1 − \frac{M^{2}}{KL}} = |A|, the function P (s, 1) is constant in I_{2} ∩ [0, 1].

Here, we note that the interval I_{2}∩[0, 1] has positive length if and only if r_{0} < 1. In view of Lemma 3.6, the last condition is equivalent to the inequality −2|KL| < |MK| − |ML|.

When s ∈ I_{3} ∩ [0, 1], we have P (s, 1) = (|K|s^{2} + 2|M|s − |L|) − |A|s^{2} − |B| = −(|A| − |K|)s^{2} + 2|M|s − |L| − |B|. If |A| = |K|, then P (s, 1) is increasing in s. We now assume that |A| ≠ |K|. Then the axis of symmetry is s = |M|/(|A| − |K|) and, by Lemma 3.6, this value is contained in I_{3} ∩ [0, 1] if and only if |A| − |K| > 0, |M| ≤ |A| − |K| and if

|A| ≥ |K|\sqrt{1 − \frac{M^{2}}{KL}}(> |K|). Thus, we obtain

Claim 3.

(1) If |A| ≤ |K| + |M|, the function P (s, 1) is increasing in I_{3} ∩ [0, 1].

(2) If |A| ≥ |K|\sqrt{1 − \frac{M^{2}}{KL}}, then the function P (s, 1) is decreasing in I_{3} ∩ [0, 1].

(3) If |K|+|M| < |A| < |K|\sqrt{1 − \frac{M^{2}}{KL}}, then s = M/(|A|−|K|) is the unique maximum point of P (s, 1) in I_{3} ∩ [0, 1], thus P (s, 1) ≤ P (M/(|A| − |K|), 1) there.

Summarizing the above three claims, we obtain the statement in our proposition including the discussion on the uniqueness of maximum points.

When M = 0, we have P (s, 1) = (K − |A|)s^{2}+const., so the desired conclusion can be directly deduced. □

By interchanging the roles of K, L and A, B, respectively, we can deduce the corresponding result for Φ_{2} from the above proposition. Concretely, we have

Φ_{2} = \begin{cases}P (1, \frac{|M|}{|B|+|L|}) = \frac{M^{2}}{|B| + |L|} + |K| − |A| & if |B| ≥ max \{|L|\sqrt{1 − \frac{M^{2}}{KL}}, |M| − |L|\} \\ P (1, \frac{|M|}{|B|−|L|}) = \frac{M^{2}}{|B| − |L|} − |K| − |A| & if |L| + |M| ≤ |B| < |L|\sqrt{1 − \frac{M^{2}}{KL}} \\ P (1, 1) & otherwise. \end{cases}

Proof of Theorem 3.1. The first part is an immediate consequence of Proposition 3.4. We show now the second part of the theorem. Assume that KL < 0. By Theorem 2.8, we see that the quantity R̃ = max{Φ_{1}, Φ_{2}} satisfies Ω = |A| + |B| + max{0, R̃}. Therefore, it is enough to see that max{0, R̃} = max{0, R}, where R is the quantity given in the theorem. Below, for instance, by case (1a), we mean the case when conditions (1) and (a) in the theorem both hold.

Case (1a): We show that R̃ ≤ 0, and hence, max{0, R̃} = max{0, R}. By condition (a), we have |B| ≥ |L|\sqrt{1 − M^{2}/KL}. By Proposition 3.7,

Φ_{1} = \frac{M^{2}}{|A| + |K|} + |L| − |B|

≤ \frac{M^{2}}{|A| + |K|} + |L| − |L|\sqrt{1 − \frac{M^{2}}{KL}}

= \frac{M^{2}}{|A| + |K|} − \frac{M^{2}}{|K|(1 + \sqrt{1 − M^{2}/KL})}

= \frac{M^{2}(|K|\sqrt{1 − M^{2}/KL} − |A|)}{(|A| + |K|)|K|(1 + \sqrt{1 − M^{2}/KL})}.

Now condition (1) implies the inequality Φ_{1} ≤ 0. In the same way, one can show that Φ_{2} ≤ 0.

Cases (2a) and (1b): First consider case (2a). Note that (|A| − |K|)(|L| + |B|) ≥ M^{2} holds. Then, by Proposition 3.7, we see that

Φ_{1} = \frac{M^{2}}{|A| − |K|} − |L| − |B| = \frac{M^{2} − (|A| − |K|)(|L| + |B|)}{|A| − |K|} ≤ 0

and

Φ_{2} = \frac{M^{2}}{|B| + |L|} + |K| − |A| = \frac{M^{2} − (|A| − |K|)(|L| + |B|)}{|B| + |L|} ≤ 0.

Therefore, we conclude that R̃ ≤ 0. Case (1b) can be treated similarly.

Case (2b): This case never occurs. To show this, suppose that conditions (2) and (b) both hold. Set a = |A|, b = |B|, k = |K|, l = |L| and m = |M|. Then k + m ≤ a < k\sqrt{1 + m^{2}/kl} and l + m ≤ b < l\sqrt{1 + m^{2}/kl}. In particular, m ≠ 0. By squaring k +m < k\sqrt{1 + m^{2}/kl}, we have 2km+m^{2} < km^{2}/l, which is equivalent to 2kl < (k −l)m. By using the other inequality, we also have 2kl < (l − k)m. This is a contradiction.

The other cases: By Proposition 3.7, we obtain the required expression for R up to the relation S = Q(1, 1) in case (3c). When M = 0, this relation is verified in a straightforward way. We now suppose that M ≠ 0. In view of Lemma 3.5, the quantity Q(1, 1) can be computed according to the interval I_{j} to which 1 belongs. For instance, 1 ∈ I_{1} = (0, r_{0}] if and only if 1 ≤ r_{0}, which is equivalent to the condition |MK| − |ML| ≤ −2|KL| by Lemma 3.6. In this way, the relation S = Q(1, 1) is deduced. □

4. Strongly close-to-convex functions

In Section 1, we introduced the class SCC(α, β) of strongly close-to-convex functions of order (α, β). This class itself is out of our category of the classes C(φ, ψ), however, each subclass SCC_{γ}(α, β) can be described in this form. Therefore, our method is applicable.

Indeed, we define the univalent function φ_{α,γ} on 𝔻 for 0 < α ≤ 1 and −πα/2 < γ < πα/2 by

φ_{α,γ}(z) = (\frac{1 + e^{iγ/α}z}{1 − e^{−iγ/α}z} )^{α}

= 1 + (2α cos \frac{γ}{α}) z + 2α cos \frac{γ}{α} (α cos \frac{γ}{α} − i sin \frac{γ}{α}) z^{2} + ⋯

and set φ_{α} = φ_{α,0} for simplicity. Then the function φ_{α,γ} maps the unit disk conformally onto the sector {z ∈ ℂ^{∗} : | arg z −γ| < πα/2}. Note also that φ_{α,γ}(0) = 1 and φ′_{α,γ}(0) > 0. Therefore, by definition, we now see SCC_{γ}(α, β) = C(φ_{β}, φ_{α,γ}).

In the following, we concentrate on the case when φ = φ_{β} and ψ = φ_{α,γ}. According to (2.4), set

A = \frac{β}{3}

B_{γ} = \frac{2α}{3} cos \frac{γ}{α}

K = β^{2}(1 − μ), (4.1)

L_{γ} = \frac{2α}{3} cos \frac{γ}{α} ((1 − \frac{3μ}{2}) α cos \frac{γ}{α} − i sin \frac{γ}{α}) ,

M_{γ} = (\frac{2}{3} − μ) αβ cos \frac{γ}{α}.

Especially, if γ = 0, we have

A = \frac{β}{3} , B_{0} = \frac{2α}{3} , K = β^{2}(1 − μ), L_{0} = α^{2} (\frac{2}{3} − μ) , M_{0} = αβ (\frac{2}{3} − μ) .

Therefore, Theorem 3.1 can be used to deduce the following result due to Darus and Thomas [2].

Theorem 4.1. Let 0 < α, β ≤ 1 and μ ∈ ℝ. Then, every function f(z) = z + a_{2}z^{2} + a_{3}z^{3} + ⋯ in the class SCC_{0}(α, β) satisfies

|a_{3} − μa_{2}^{2}| ≤ \begin{cases}(α + β)^{2} (\frac{2}{3} − μ) + \frac{β^{2}}{3} & if μ ≤ μ_{1}, \\ \frac{1}{3α}[2α^{2} + αβ^{2} − 2β^{2} + \frac{4β^{2}}{2 − 2α + 3αμ}] & if μ_{1} ≤ μ ≤ μ_{2}, \\ \frac{2α + β}{3} & if μ_{2} ≤ μ ≤ μ′_{2}, \\ \frac{1}{3α}[2α^{2} − αβ^{2} − 2β^{2} + \frac{4β^{2}}{2 + 2α − 3αμ}] & if μ′_{2} ≤ μ ≤ μ′_{1}, \\ (α + β)^{2} (μ − \frac{2}{3}) − \frac{β^{2}}{3} & if μ′_{1} ≤ μ, \end{cases}

where μ_{1} = \frac{2}{3} (1 − \frac{1}{α + β} ) , μ_{2} = \frac{2}{3} (1 − \frac{1 − β}{α + 2β − αβ} ) , μ′_{2} = \frac{2}{3} (1 + \frac{1 + β}{α + 2β + αβ} ) ,

μ′_{1} = \frac{2}{3} (1 + \frac{1}{α + β} ) . Furthermore, the above inequality is sharp in the sense that for

each μ there is a function in SCC_{0}(α, β) for which equality holds.

Note that the inequalities μ_{1} < μ_{2} ≤ \frac{2}{3} < 1 ≤ μ′_{2} ≤ μ′_{1} hold always, and that μ_{2} = \frac{2}{3} and μ′_{2} = μ′_{1} if β = 1.

Proof. For simplicity, we write B = B_{0}, L = L_{0} and M = M_{0} in the proof. Set D = (|K| − A)(|L| − B) − M^{2} and set ρ = ρ(φ_{β}, φ_{α}, μ). First noting that KL ≥ 0 if and only if either μ ≤ 2/3 or μ ≥ 1, we divide the proof into three cases according to the value of μ.

Case 1: μ ≤ 2/3. In this case, K ≥ 0, L ≥ 0 and M ≥ 0. A simple computation shows that D < 0 if and only if μ < μ_{2}. For a while, we consider this case. Also, |M| + |K| < |A| is equivalent to the condition μ > ν_{1} := (2 − (1 − β)/(α + β))/3. However, since ν_{1} − μ_{2} = α(1 − β^{2})/(α + β)(α + 2β − αβ) ≥ 0, this case never occurs. On the other hand, |M| + |L| < |B| is equivalent to the condition μ > μ_{1}. In this case, |M|/(|B| − |L|) = β(2 − 3μ)/(2 − α(2 − 3μ)) ∈ (0, 1), and Theorem 3.1 yields ρ = K + B − M^{2}/(L − B) = (2α^{2} + αβ^{2} − 2β^{2} + 4β^{2}/(2 − α(2 − 3μ)))/3α.

The other case, namely, when μ ≤ μ_{1}, by Theorem 3.1, we see that ρ = K + 2M + L = (α + β)^{2}(3/2 − μ) + β^{2}/3.

Next, we consider the case D ≥ 0, in other words, μ_{2} ≤ μ ≤ 2/3. Then, A+B ≥ |K|+|L| if and only if μ ≥ ν_{2} := (2 − (2α + β(1 − β))/(α^{2} + β^{2}))/3. Since ν_{2} < μ_{2} is true, the above condition holds. Hence, by Theorem 3.1, we obtain ρ = A + B = (2α + β)/3.

Case 2: μ ≥ 1. Then, K ≤ 0, L ≤ 0 and M ≤ 0. Note first that D < 0 if and only if μ > μ′_{2}. Then, |M|+|K| < |A| is equivalent to the condition μ < (2+(1+β)/(α+β))/3(< μ′_{2}). Therefore, this case never happens. On the other hand, |M|+|L| < |B| is equivalent to the condition μ < μ′_{1} and in this case −M/(B + L) = β(3μ − 2)/(2 − α(3μ − 2)) ∈ (0, 1). Now Theorem 3.1 implies ρ = B−K +M^{2}/(B+L) = (2α−αβ^{2}−2β^{2}+4β^{2}/(2−α(3μ−2)))/3α when μ′_{2} < μ < μ′_{1}. When μ ≥ μ′_{1}, we have ρ = −K − 2M − L = (α + β)^{2}(μ − 3/2) − β^{2}/3.

Next, consider the case when 1 ≤ μ ≤ μ′_{2}. Then, |K| + |L| ≤ |A| + |B| if and only if μ ≤ (2 + (2α + β + β^{2})/(α^{2} + β^{2}))/3(> μ′_{2}). Therefore, by Theorem 3.1, we have ρ = A + B = (2α + β)/3.

Case 3: 2/3 < μ < 1. Though the present case will be covered by Theorem 4.2, we give a proof as an application of Theorem 3.1. In this case, K > 0, L < 0 and M < 0. Since |K|\sqrt{1 − M^{2}/KL} = β^{2}\sqrt{(1 − μ)/3} ≤ β^{2}/3 ≤ A and |M|−|K| = β[(α+β)μ−2α/3−β] ≤ αβ/3 ≤ A, condition (1) is fulfilled. On the other hand, since

|B| − |L| − |M| = α [\frac{2}{3} − (α + β) (μ − \frac{2}{3})] ≥ 0,

one of the conditions (a) and (b) is satisfied. Therefore, by Theorem 3.1, we obtain ρ = A + B = (2α + β)/3. □

As an application of our results, we finally treat the case when K, L and M are not necessarily real numbers.

Theorem 4.2. Let 2/3 ≤ μ ≤ 1 and α, β ∈ (0, 1]. For each function f(z) = z + a_{2}z^{2} + a_{3}z^{3} + … in SCC(α, β), then the inequality

|a_{3} − μa_{2}^{2}| ≤ \frac{2α + β}{3}

holds. Furthermore, each inequality is sharp in the sense that there is a function f ∈ SCC_{0}(α, β) for which equality holds.

This theorem means that the inequality in Theorem 4.1 is still valid for the full class SCC(α, β) in the case when 2/3 ≤ μ ≤ 1. It is naturally expected that the same thing can be said to an arbitrary μ ∈ ℝ. Indeed, the case when α = β = 1 was confirmed by Eenigenburg and Silvia [4].

Proof. We use the notation given by (4.1) and set ρ_{γ}(μ) = ρ(φ_{β}, φ_{α,γ}, μ).

Case 1: μ = 2/3. In this case, K = β^{2}/3 > 0 and M_{γ} = 0. Therefore,

Q(s, t) = max_{θ∈ℝ}|Ks^{2} + L_{γ}t^{2}e^{2iθ}| = Ks^{2} + |L_{γ}|t^{2},

and thus, P (s, t) = (K − A)s^{2} + (|L_{γ}| − B_{γ})t^{2}. Since K − A = −β(1 − β)/3 ≤ 0 and |L_{γ}| − B_{γ} = −(2α/3) cos(γ/α)(1 − sin(γ/α)) ≤ 0, we see that P (s, t) ≤ P (0, 0) = 0. By (2.9), we obtain

ρ_{γ}(2/3) = A + B_{γ} = \frac{2α cos(γ/α) + β}{3}

≤ \frac{2α + β}{3} = ρ_{0}(2/3).

Case 2: μ = 1. In this case, K = 0 and M_{γ} < 0. Therefore,

Q(s, t) = max_{θ∈ℝ}|2M_{γ}ste^{iθ} + L_{γ}t^{2}e^{2iθ}| = −2M_{γ}st + |L_{γ}|t^{2}

and thus, P (s, t) = −As^{2} − 2M_{γ}st + (|L_{γ}| − B_{γ})t^{2}. We write x = cos(γ/α) and see that |L_{γ}| = (αx/3)\sqrt{4 − (4 − α^{2})x^{2}} < 2αx/3 = B_{γ}. We now investigate the descriminant

D = A(B_{γ} − |L_{γ}|) − M_{γ}^{2} = \frac{αβx}{9} (2 − \sqrt{4 − (4 − α^{2})x^{2}} − αβx) .

Since

D ≥ 0 ⇔ (2 − αβx)^{2} ≥ 4 − (4 − α^{2})x^{2}

⇔ x ≥ x_{0} := \frac{αβ}{1 − α^{2}(1 − β^{2})/4}.

Note that 0 < x_{0} ≤ 1. When x ≥ x_{0}, we obtain P (s, t) ≤ P (0, 0) = 0. When 0 < x < x_{0}, since the inequality |M_{γ}| = αβx/3 < β/3 = A − |K| holds, Proposition 3.4 implies that P (s, t) ≤ P (|M_{γ}|/A, 1) = P (αx, 1) = |L_{γ}| − B_{γ} + M_{γ}^{2}/A. Summarizing the above, we obtain

(4.2)  ρ_{γ}(1) = \begin{cases}\frac{2αx + β}{3} & if x ≥ x_{0} \\ \frac{2αx\sqrt{1 − (1 − α^{2}/4)x^{2}} + α^{2}βx^{2} + β}{3} & if x < x_{0}, \end{cases}

where x = cos(γ/α). We now claim that ρ_{γ}(1) ≤ ρ_{0}(1) = (2α + β)/3 holds for any γ with x = cos(γ/α) < x_{0}. The claim is equivalent to 2αx\sqrt{1 − (1 − α^{2}/4)x^{2}} ≤ 2α − α^{2}βx^{2}. We compute

(2 − αβx^{2})^{2} − (2x\sqrt{1 − (1 − α^{2}/4)x^{2}})^{2}
= (4 − α^{2} + α^{2}β^{2})x^{4} − 4(1 + αβ)x^{2} + 4
= (4 − α^{2} + α^{2}β^{2}) (x^{2} − \frac{2(1 + αβ)}{4 − α^{2} + α^{2}β^{2}} )^{2} + \frac{4(3 − α^{2} − 2αβ)}{4 − α^{2} + α^{2}β^{2}}.

The last term is obviously non-negative, and thus, the claim has been confirmed.

Case 3: 2/3 < μ < 1. Choose positive numbers s and t so that s + t = 1 and μ = 2s/3 + t. By Lemma 2.1 and the previous two cases, we conclude that

ρ_{γ}(μ) = ρ_{γ}(2s/3 + t) ≤ sρ_{γ}(2/3) + tρ_{γ}(1) ≤ \frac{2α + β}{3} .

Thus, we have shown that ρ_{γ}(μ) ≤ (2α + β)/3 for 2/3 ≤ μ ≤ 1 and for γ ∈ (−πα/2, πα/2). The last assertion in the theorem is a direct consequence of the relation ρ_{0}(μ) = (2α + β)/3. □

We end the article with the remark that the quantity ρ_{γ}(1) is not necessarily a monotone function of x = cos(γ/α) even when β = 1 as one can check it by (4.2). The lack of monotonicity in γ seems to cause difficulty in verification of the inequality ρ_{γ}(μ) ≤ ρ_{0}(μ) for an arbitrary μ.

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Department of Mathematics Education, Daegu National University of Education, 1797-6 Daemyong 2 dong, Namgu, Daegu 705-715, Korea

E-mail address: choijh@dnue.ac.kr

Department of Mathematics, College of Education, Yeungnam University, 214-1 Daedong Gyongsan 712-749, Korea

E-mail address: kimyc@yu.ac.kr

Department of Mathematics, Graduate School of Science, Hiroshima University, Higashi-Hiroshima, 739-8526 Japan

E-mail address: sugawa@math.sci.hiroshima-u.ac.jp