STRONG STARLIKENESS FOR A CLASS OF CONVEX FUNCTIONS

STANISŁAWA KANAS AND TOSHIYUKI SUGAWA

Abstract. By means of the Briot-Bouquet differential subordination, we estimate the order of strong starlikeness of strongly convex functions of a prescribed order. We also make numerical experiments to examine our estimates.

1. Introduction

We denote by 𝒜 the class of functions f analytic in the unit disk 𝔻 = {z ∈ ℂ : |z| < 1} and normalized by f(0) = 0 and f′(0) = 1. Let 𝒮 denote the class of normalized univalent analytic functions and, for each 0 ≤ k < 1, let 𝒮 (k) denote the subclass of 𝒮 consisting of those functions which extend to k-quasiconformal mappings of the extended plane Let g and h be meromorphic functions in 𝔻. We say that g is subordinate to h and express it by g ≺ h or conventionally by g(z) ≺ h(z) if g = h ∘ ω for some analytic map ω : 𝔻 → 𝔻 with ω(0) = 0. When h is univalent, the condition g ≺ h is equivalent to g(𝔻) ⊂ h(𝔻) and g(0) = h(0).

An analytic function f in the unit disk 𝔻 is called starlike if f is univalent and f(𝔻) is starlike with respect to f(0). Also, f is called convex if f is univalent and f(𝔻) is convex.

It is well known that f ∈ 𝒜 is starlike if and only if Re (zf′(z)/f(z)) > 0 in 𝔻 and f ∈ 𝒜 is convex if and only if Re (1 + zf′′(z)/f′(z)) > 0 in 𝔻 (see, for instance, [3]). The sets of starlike functions and convex functions in 𝒜 are denoted by 𝒮^{∗} and 𝒦 , respectively. Let α be a positive real number. A function f in 𝒜 is said to be strongly starlike of order α if | arg(zf′(z)/f(z))| < πα/2 for z ∈ 𝔻. Similarly, f ∈ 𝒜 is said to be strongly convex of order α if | arg(1+zf′′(z)/f′(z))| < πα/2 for z ∈ 𝔻. The sets of strongly starlike functions of order α and strongly convex functions of order α are denoted by 𝒮_{α}^{∗} and 𝒦_{α}, respectively. Many geometric characterizations of the class 𝒮_{α}^{∗}, 0 < α < 1, are known (for a short survey, see [13]).

Throughout the paper, we will use the symbol T to stand for the mapping of the unit disk onto the right-half plane which is defined by

T (z) = \frac{1 + z}{1 − z} .

1991 Mathematics Subject Classification. Primary 30C45; Secondary 30C80.

Key words and phrases. strongly starlike, strongly convex, differential subordination.

The second author was partially supported by the Ministry of Education, Grant-in-Aid for Encouragement of Young Scientists (B), 14740100 and Grant-in-Aid for Scientific Research (B), 17340039.

2 S. KANAS AND T. SUGAWA

For 0 < κ ≤ 1, we consider also the subclass 𝒮^{∗}[κ] of 𝒮^{∗} consisting of functions f with zf′(z)/f(z) ≺ T (κz) = (1+κz)/(1−κz). Hence, for 0 < κ < 1, a function f ∈ 𝒜 belongs to the class 𝒮^{∗}[κ] if and only if the inequality

|\frac{zf′(z)}{f(z)} − \frac{1 + κ^{2}}{1 − κ^{2}} | < \frac{2κ}{1 − κ^{2}}

holds in 𝔻. Here are useful criteria for quasiconformal extensions.

Theorem A.

(i) 𝒮_{α}^{∗} ⊂ 𝒮 (sin(πα/2)) for 0 < α < 1. (ii) 𝒮^{∗}[κ] ⊂ 𝒮 (κ) for 0 < κ < 1.

Relation (i) is due to Fait, Krzyż and Zygmunt [4], and (ii) is due to Brown [1] (see also [12]). It is easy to see that 𝒮^{∗}[κ] ⊂ 𝒮_{α}^{∗} for α = (2/π) arcsin(2κ/(1 + κ^{2})) because of T (κz) ≺ T^{α}(z).

Obviously, a convex function is starlike, in other words, 𝒦 ⊂ 𝒮^{∗}. Moreover, Mocanu showed the relation 𝒦_{α} ⊂ 𝒮_{α}^{∗} for 0 < α ≤ 2 in [6]. Therefore, it is natural to consider the problem of finding the number

β^{∗}(α) = inf{β : 𝒦_{α} ⊂ 𝒮_{β}^{∗}}

for each α > 0. By the maximum principle, we have 𝒦_{α} ⊂ 𝒮_{β^{∗}(α)}^{∗}. Hence, β^{∗}(α) is the minimal number β so that 𝒦_{α} ⊂ 𝒮_{β}^{∗}.

Later, Mocanu proved the following in [7]. For 0 < β < 1, set

(1.1)  γ(β) = \frac{2}{π} arctan [tan \frac{πβ}{2} + \frac{β}{(1 + β)^{\frac{1+β}{2}} (1 − β)^{\frac{1−β}{2}} cos(πβ/2)}]

= β + \frac{2}{π} arctan [\frac{β cos(πβ/2)}{(1 + β)^{\frac{1+β}{2}} (1 − β)^{\frac{1−β}{2}} + β sin(βπ/2)}] .

Theorem B (Mocanu). A strongly convex function of order γ(β) is strongly starlike of order β for 0 < β < 1.

The function γ(β) is continuous and strictly increases from 0 to 1 when β moves from 0 to 1. We denote by γ^{−1} : (0, 1) → (0, 1) the inverse function of γ. The theorem then implies the relation 𝒦_{α} ⊂ 𝒮_{γ^{−1}(α)}^{∗}, namely, β^{∗}(α) ≤ γ^{−1}(α) for 0 < α < 1. The same result was re-proved later by Nunokawa [8] and by Nunokawa and Thomas [9]. It is further claimed in [9] that the result is best possible, namely, β^{∗}(α) = γ^{−1}(α) for 0 < α < 1. This is, however, wrong as we see in the following result.

Theorem 1.1. The function β^{∗}(α) is continuous and strictly increasing in 0 < α < 1. Moreover, β^{∗}(α) < γ^{−1}(α) holds for each 0 < α < 1.

STRONG STARLIKENESS FOR A CLASS OF CONVEX FUNCTIONS 3

In the same way as above, we denote by κ^{∗}(α) the minimal number κ so that 𝒦_{α} ⊂ 𝒮^{∗}[κ]. It is clear that κ^{∗}(α) ≤ 1. It seems, however, that no bounds of κ^{∗}(α) were given in the literature. The next theorem implies that for each 0 < α < 1, there exists a κ ∈ (0, 1) such that 𝒦_{α} ⊂ 𝒮^{∗}[κ].

Theorem 1.2. The function κ^{∗}(α) is continuous and strictly increasing in 0 < α < 1. Moreover, κ^{∗}(α) < 1 holds for each 0 < α < 1.

Explicit expressions of β^{∗}(α) and κ^{∗}(α) will be given in Section 3 in terms of a solution to a Briot-Bouquet differential equation (Proposition 3.1). The proof of our theorems depends on geometric properties of the solution. Section 3 will also be devoted to investigation of the solution.

The above two theorems are, however, not quantitative. In order to obtain better and concrete upper bounds for β^{∗}(α) and κ^{∗}(α), we need more efforts. We propose a method of giving a better estimate for them. Due to some technicality, the presentation of the method will be postponed to Section 4. The next section is used to the preparation of necessary materials for the proof of the theorems and for development of our methods.

We end this introduction with the remark that, using Theorem A, we obtain quasiconformal extension criteria for the class 𝒦_{α}, though we do not state them separately.

2. Preliminaries

Our arguments will be largely based on results proved by Miller and Mocanu. We state it in convenient forms for the present aim. The first result is the following.

Theorem C (Miller and Mocanu [5, Theorems 3.2a and 3.2j]). Let h be a convex function in the unit disk with h(0) = 1 and Re h(z) > 0 in |z| < 1 and let q be the analytic solution to the differential equation

q(z) + \frac{zq′(z)}{q(z)} = h(z), z ∈ 𝔻, and q(0) = 1.

Then q is univalent and subordinate to h. Moreover, if an analytic function p in the unit disk with p(0) = 1 satisfies the subordination

p(z) + \frac{zp′(z)}{p(z)} ≺ h(z),

then p(z) ≺ q(z).

The second result that we will need is contained in [5, Theorem 3.4h] with the choice of θ(w) = w and ϕ(w) = 1/w.

Theorem D (Miller and Mocanu). Let q be a non-vanishing univalent function in 𝔻 with q(0) = 1 and set Q(z) = zq′(z)/q(z) and h = q + Q. Suppose that Q is starlike and that Re (q(z) + zQ′(z)/Q(z)) > 0 in 𝔻. Then h is univalent. Furthermore, if an analytic function p in 𝔻 with p(0) = 1 satisfies the relation

p(z) + \frac{zp′(z)}{p(z)} ≺ q(z) + \frac{zq′(z)}{q(z)} = h(z),

then p(z) ≺ q(z).

Note that, when Re q(z) > 0, the condition Re (q(z) + zQ′(z)/Q(z)) > 0 is fulfilled automatically because Re (zQ′(z)/Q(z)) > 0.

We also use the following version of the Julia-Wolff lemma, which is a combination of known facts.

Lemma 2.1. Let z_{0} ∈ ∂𝔻. Suppose that an analytic function ω in 𝔻 ∪ {z_{0}} satisfies |ω(z)| < 1 in |z| < 1, ω(0) = 0 and |ω(z_{0})| = 1. Then

m = \frac{z_{0}ω′(z_{0})}{ω(z_{0})}

is a positive real number with m ≥ 1. Furthermore, m = 1 only if ω(z) ≡ ω(z_{0})z/z_{0}.

Proof. First, by the Julia-Wolff lemma [11, Proposition 4.13], we see that m = z_{0}ω′(z_{0})/ω(z_{0}) is a positive real number. Here we recall the boundary Schwarz lemma due to Osserman [10, Lemma 1]: |ω′(z_{0})| ≥ 2/(1 + |ω′(0)|). The remaining assertion now follows from the Schwarz lemma: |ω′(0)| ≤ 1 and equality holds only if ω is a rotation about the origin.

The next strange-looking result generates a new family of starlike functions from a single starlike function.

Lemma 2.2. Let a and b be complex numbers with a ≠ 0 and |a| + |b| ≤ 1. For a starlike function f : 𝔻 → ℂ with f(0) = 0, the function g : 𝔻 → ℂ defined by

g(z) = \frac{f(\frac{az}{1−bz})}{1 − bz}

is also starlike.

Note that az/(1 − bz) ∈ 𝔻 for z ∈ 𝔻 whenever |a| + |b| ≤ 1. In this lemma, f and g satisfy only the condition f(0) = g(0) = 0, thus, f and g might not be normalized so that f′(0) = g′(0) = 1. In order to obtain a transformation of 𝒮^{∗} into itself, we may consider the operator I_{a,b} defined by I_{a,b}[f](z) = f(az/(1 − bz))/(a(1 − bz)). It might be interesting to observe that I_{a,b}[f](z) = z/(1 − bz)^{2} → z/(1 − z)^{2} (the Koebe function) as (a, b) → (0, 1) for each f ∈ 𝒮^{∗}.

Proof. Let φ(z) = zf′(z)/f(z). Then the starlikeness implies Re φ > 0 in 𝔻. We need to see that

\frac{zg′(z)}{g(z)} = \frac{φ(\frac{az}{1−bz}) + bz}{1 − bz}

has positive real part. First we consider the special case when φ = φ_{ζ} for some ζ ∈ ∂𝔻, where φ_{ζ}(z) = (1 + ζz)/(1 − ζz). Then, a straightforward computation gives

\frac{zg′(z)}{g(z)} = \frac{1 + (aζ + b)z}{1 − (aζ + b)z} .

By assumption, we have |aζ + b| ≤ |a| + |b| ≤ 1 and thus Re (zg′(z)/g(z)) > 0.

To show the general case, we use the Herglotz representation of a function with positive real part (cf. [3]). For the general φ, there exists a Borel probability measure μ on the unit circle ∂𝔻 such that

φ(z) = ∫_{∂𝔻} \frac{1 + ζz}{1 − ζz} dμ(ζ) = ∫_{∂𝔻} φ_{ζ}(z)dμ(ζ).

Therefore,

\frac{φ(\frac{az}{1−bz}) + bz}{1 − bz} = ∫_{∂𝔻} \frac{φ_{ζ}(\frac{az}{1−bz}) + bz}{1 − bz} dμ(ζ)

= ∫_{∂𝔻} \frac{1 + (aζ + b)z}{1 − (aζ + b)z} dμ(ζ).

This shows that zg′(z)/g(z) has positive real part.

We recall also the following simple fact (cf. [2]).

Lemma 2.3. Let f : 𝔻 → ℂ be a convex univalent function and Δ be an open disk contained in 𝔻. Then f(Δ) is also convex.

3. Proof of Theorems 1.1 and 1.2

In this section, we prove Theorems 1.1 and 1.2. To this end, we introduce a mapping associated with the function T^{α}.

Let q_{α} be the analytic function in the unit disk determined by

q_{α}(z) + \frac{zq′_{α}(z)}{q_{α}(z)} = (\frac{1 + z}{1 − z} )^{α} = T^{α}(z), and q_{α}(0) = 1. (3.1)

Since T^{α} is analytic in \overline{𝔻} \ \{±1\}, q_{α} is also analytically continued in a neighborhood of \overline{𝔻} \ \{±1\}. Hence, we can argue the value of q_{α}(e^{iθ}) for 0 < |θ| < π.

As an immediate consequence of Theorem C, we obtain explicit expressions of the quantities β^{∗}(α) and κ^{∗}(α) in terms of the function q_{α} for 0 < α < 1.

Proposition 3.1. Let 0 < α < 1 and let β^{∗}(α) and κ^{∗}(α) be the minimal numbers β and κ, respectively, such that 𝒦_{α} ⊂ 𝒮_{β}^{∗}, 𝒦_{α} ⊂ 𝒮^{∗}[κ]. Then they are expressed by

β^{∗}(α) = sup_{0<θ<π} arg q_{α}(e^{iθ}) and κ^{∗}(α) = sup_{0<θ<π} |\frac{q_{α}(e^{iθ}) − 1}{q_{α}(e^{iθ}) + 1}| .

6 S. KANAS AND T. SUGAWA

Proof. Let f ∈ 𝒦_{α} and set p(z) = zf′(z)/f(z). By the relation p(z) + zp′(z)/p(z) = 1 + zf′′(z)/f′(z) ≺ T^{α}(z), we conclude that p ≺ q_{α} by Theorem C. In particular,

sup_{z∈𝔻}| arg p(z)| ≤ sup_{z∈𝔻}| arg q_{α}(z)| = sup_{0<θ<π} arg q_{α}(e^{iθ}).

Here, equality holds when we take f so that 1+zf′′(z)/f(z) = T^{α}(z), namely, zf′(z)/f(z) = q_{α}(z). Thus, we have shown the first relation. The second one can be deduced in the same way.

Remark. A direct computation gives q_{1}(z) = 1/(1 − z). In the case when α = 1, the above argument thus yields that for f ∈ 𝒦_{1} = 𝒦 , p(z) = zf′(z)/f(z) is subordinate to q_{1}(z) = 1/(1 − z), that is, p(𝔻) ⊂ q_{1}(𝔻) = {Re w > 1/2}. In other words, a convex function is starlike of order 1/2. This is known as Strohhäcker’s theorem, see [3, p. 251].

In order to get information about β^{∗}(α) and κ^{∗}(α), it is thus important to know about the function q_{α}. We summarize geometric properties of q_{α} in the following proposition.

Proposition 3.2. For 0 < α < 1, the function q_{α} maps the unit disk univalently onto a symmetric bounded Jordan domain contained in {w ∈ ℂ : | arg w| < πα/2, Re w > 1/2} in such a way that q_{α}(x) ∈ ℝ for x ∈ (−1, 1) and q′_{α}(0) = α.

The proof is divided into several steps as follows.

Since T^{α} is convex and has positive real part, the functions h = T^{α} and q = q_{α} satisfy the assumptions of Theorem C. Therefore, q_{α} is univalent and | arg q_{α}(z)| < πα/2 in 𝔻. By the symmetry of the equation (3.1), the solution q_{α} is symmetric, namely, \overline{q_{α}(z)} = q_{α}(z̄). Therefore, q_{α}(𝔻) is symmetric in the real axis and q_{α} maps real numbers to real numbers. Since T^{α} ≺ T, the theorem yields also the relation q_{α} ≺ q_{1}. Therefore, q_{α}(𝔻) lies in the domain q_{1}(𝔻) = {Re w > 1/2}. The relation q′_{α}(0) = α > 0 can be verified directly. In particular, we see that Im q_{α}(z) > 0 for Im z > 0.

The following lemma gives an upper bound for |q_{α}(z)|.

Lemma 3.3. The function q_{α} is bounded in the unit disk for 0 < α < 1.

Proof. For a fixed θ ∈ ℝ, we consider the function u(r) = q_{α}(re^{iθ}) in 0 ≤ r ≤ 1. We express u also in the form u(r) = R(r)e^{iΘ(r)}, where R(r) > 0 and Θ(r) ∈ ℝ with Θ(0) = 0. Then

R(r)e^{iΘ(r)} + \frac{rR′(r)}{R(r)} + irΘ′(r) = u(r) + \frac{ru′(r)}{u(r)} = T^{α}(re^{iθ}).

As observed above, |Θ(r)| ≤ πα/2 holds. Therefore, taking the real part of the above relation, we obtain

cR(r) + \frac{rR′(r)}{R(r)} ≤ Re T^{α}(re^{iθ}) ≤ (\frac{1 + r}{1 − r})^{α} ,

where c = cos(πα/2). We now define the positive function g(r), 0 < r ≤ 1, by the relation

log g(r) = ∫_{1/2}^{r} \frac{cR(x)}{x}dx.

Then, rg′(r)/g(r) = cR(r) and thus

\frac{g′′(r)}{g′(r)} = \frac{1}{r} [cR(r) + \frac{rR′(r)}{R(r)} − 1] ≤ \frac{1}{r} [(\frac{1 + r}{1 − r})^{α} − 1] .

An integration yields

log \frac{g′(r)}{g′(1/2)} ≤ Φ(r) = ∫_{1/2}^{r} \frac{1}{x} [(\frac{1 + x}{1 − x})^{α} − 1] dx

for 1/2 < r < 1. Since

Φ(r) < 2^{1+α} ∫_{1/2}^{1} \frac{dx}{(1 − x)^{α}} = \frac{4^{α}}{1 − α},

and g′(1/2) = 2cR(1/2), the inequality

g′(r) < 2cR(\frac{1}{2}) exp \frac{4^{α}}{1 − α}

follows for 1/2 < r < 1. Since g′ > 0, we have g(r) > g(1/2) = 1 for 1/2 < r < 1. Therefore,

R(r) = \frac{rg′(r)}{cg(r)} < 2R(\frac{1}{2}) exp \frac{4^{α}}{1 − α}

for 1/2 < r < 1. We now recall the growth theorem for functions f in 𝒮 (cf. [3, p. 33]):

\frac{r}{(1 + r)^{2}} ≤ |f(z)| ≤ \frac{r}{(1 − r)^{2}} , |z| = r < 1.

Since (q_{α} −1)/α belongs to 𝒮 , we have |q_{α}(z)| ≤ 1+αr/(1−r)^{2}, |z| = r. Letting r = 1/2, we obtain the estimate R(1/2) ≤ 1 + 2α. Therefore,

|q_{α}(z)| = R(r) ≤ 2(1 + 2α) exp \frac{4^{α}}{1 − α}

for z = re^{iθ}, 1/2 < r < 1. The last inequality is valid for all z ∈ 𝔻 by the maximum modulus principle.

Let 0 < α < β < 1. Since T^{α} ≺ T^{β}, Theorem C implies that q_{α} ≺ q_{β}. Note also that ω = q_{β}^{−1} ∘ q_{α} is analytically continued across the border ∂𝔻 \ {±1}. We now show the following.

Lemma 3.4. Let ω = q_{β}^{−1} ∘ q_{α} : 𝔻 → 𝔻 for 0 < α < β < 1. Then |ω(e^{iθ})| < 1 for each θ ∈ (0, π).

8 S. KANAS AND T. SUGAWA

Proof. Suppose, to the contrary, that |ω(z_{0})| = 1 for some z_{0} = e^{iθ_{0}}, θ_{0} ∈ (0, π). Lemma 2.1 implies that m = z_{0}ω′(z_{0})/w_{0} ≥ 1, where we set w_{0} = ω(z_{0}). Note that |w_{0}| = 1 and Im w_{0} > 0. On the other hand, the relation q_{α} = q_{β} ∘ ω yields

T^{α}(z_{0}) = q_{α}(z_{0}) + \frac{z_{0}q′_{α}(z_{0})}{q_{α}(z_{0})}

= q_{β}(w_{0}) + \frac{z_{0}q′_{β}(w_{0})ω′(z_{0})}{q_{β}(w_{0})}

= q_{β}(w_{0}) + \frac{mw_{0}q′_{β}(w_{0})}{q_{β}(w_{0})}

= −(m − 1)q_{β}(w_{0}) + mT^{β}(w_{0}).

By Theorem B, 0 ≤ arg q_{β}(w_{0}) ≤ πγ^{−1}(β)/2 < πβ/2 = arg T^{β}(w_{0}). Therefore, an elementary geometry tells us that the argument of −(m − 1)q_{β}(w_{0}) + mT^{β}(w_{0}) is in between πβ/2 and πβ/2 + π. This contradicts the fact that arg T^{α}(z_{0}) = πα/2 < πβ/2. Thus the inequality in question has been shown.

Lemma 3.5. Let 0 < α < 1. The curve γ_{α} : (0, π) → ℂ defined by γ_{α}(θ) = q_{α}(e^{iθ}) is a Jordan arc of finite length.

Proof. If γ_{α} is not injective, then γ_{α} bounds a domain D. If β > α is close enough to α, the boundary of q_{β}(𝔻) must go through D, which implies that the curve γ_{β} has a common point with the curve γ_{α}. This is, however, impossible by Lemma 3.4. In order to see finiteness of the length of γ_{α}, we use the Hardy spaces. Since T ∈ H^{p} for all p < 1, we see that T^{α} ∈ H^{p} for all p < 1/α. In particular, T^{α} ∈ H^{1}. Here, by (3.1), zq′_{α}(z) = q_{α}(z)(T^{α}(z) − q_{α}(z)). Since q_{α} is bounded by Lemma 3.3, we have q′_{α} ∈ H^{1}. The length of γ_{α} is now estimated by

∫_{0}^{π} |γ′_{α}(θ)|dθ = \frac{1}{2} ∫_{0}^{2π} |q′_{α}(e^{iθ})|dθ = π∥q′_{α}∥_{H^{1}} < ∞.

Completion of the proof of Proposition 3.2. The remaining part is to show that q_{α}(𝔻) is a Jordan domain.

Since the curve γ_{α} has finite length, it extends to a continuous map on [0, π], which will be denoted by the same symbol γ_{α}. By the symmetry of q_{α}, it is now obvious that the image q_{α}(𝔻) is bounded by γ_{α}([0, π]) and its reflection in ℝ. Thus the assertion has been proved.

We are now ready to prove Theorems 1.1 and 1.2. It is a standard fact in the theory of ordinary differential equations that the solutions of initial value problems are continuous

STRONG STARLIKENESS FOR A CLASS OF CONVEX FUNCTIONS 9

with respect to parameters. The continuity of β^{∗}(α) and κ^{∗}(α) immediately follows from this fact. Proposition 3.2 yields that q_{α}(𝔻) is bounded and contained in the half-plane Re w > 1/2, which implies κ^{∗}(α) < 1. (A crude estimate of κ^{∗}(α) can be given by using the estimate in the proof of Lemma 3.3.)

It follows from the next lemma that β^{∗}(α) and κ^{∗}(α) are strictly increasing.

Lemma 3.6. Let 0 < α_{1} < α_{2} < 1. Then q_{α_{1}}(𝔻) is a relatively compact subdomain of q_{α_{2}}(𝔻).

Proof. Since q_{α}(𝔻) is a Jordan domain, Carathéodory’s theorem ensures that q_{α} extends to a homeomorphism of \overline{𝔻} onto \overline{q_{α}(𝔻)}, which will still be denoted by q_{α}. We show now that q_{α_{1}}(1) < q_{α_{2}}(1). Let f_{α} be the function in 𝒜 determined by the relation 1+zf′′_{α}(z)/f′_{α}(z) = T^{α}(z). Then q_{α} is expressed by q_{α}(z) = zf′_{α}(z)/f_{α}(z). Letting S_{α}(z) = (T^{α}(z) − 1)/z, we obtain f′′_{α}/f′_{α} = S_{α}. Integrating both sides of the last relation, we have log f′_{α}(z) = ∫_{0}^{z} S_{α}(ζ)dζ =: U_{α}(z). Therefore,

\frac{1}{q_{α}(1)} = \frac{f_{α}(1)}{f′_{α}(1)} = ∫_{0}^{1} e^{U_{α}(x)−U_{α}(1)}dx.

Since U_{α}(x)−U_{α}(1) = − ∫_{x}^{1} S_{α}(t)dt is strictly decreasing in 0 < α < 1 for a fixed x ∈ (0, 1), it is concluded that q_{α}(1) is strictly increasing. Thus the claim follows. In the same way, we can show that q_{α_{2}}(−1) < q_{α_{1}}(−1).

Let ω = q_{α_{2}}^{−1} ∘ q_{α_{1}} : \overline{𝔻} → \overline{𝔻}. Note that ω is continuous in \overline{𝔻}. What we have seen above means that |ω(ζ)| < 1 for ζ = ±1. On the other hand, Lemma 3.4 asserts that this is valid for ζ ∈ ∂𝔻 \ {1, −1}. Therefore, we conclude that max_{|ζ|=1} |ω(ζ)| < 1, which implies the required assertion.

Finally, we show that β^{∗}(α) < γ^{−1}(α) for 0 < α < 1. To this end, we recall the proof of Theorem B. Set p_{β} = T^{β} and

h_{β}(z) = p_{β}(z) + \frac{zp′_{β}(z)}{p_{β}(z)} = (\frac{1 + z}{1 − z} )^{β} + \frac{2βz}{1 − z^{2}}

for 0 < β < 1. Mocanu showed that h_{β} is univalent in 𝔻 and γ(β) is obtained as the minimum of (2/π) arg h_{β}(e^{iθ}) over 0 < θ < π.

Suppose now that β^{∗}(α) and γ^{−1}(α) are the same number, say β, for some 0 < α < 1. Then, Proposition 3.1 implies that a point in the boundary of q_{α}(𝔻) has argument πβ/2. Therefore, if we set ω = p_{β}^{−1} ∘ q_{α}, |ω(z_{0})| = 1 holds for some z_{0} = e^{iθ}, θ ∈ (0, π). Since ω(0) = 0 and ω(z)/z is not constant, Lemma 2.1 implies that m = z_{0}ω′(z_{0})/w_{0} > 1, where w_{0} = ω(z_{0}). The relation (3.1) now turns to

T^{α}(z_{0}) = p_{β}(w_{0}) + \frac{z_{0}p′_{β}(w_{0})ω′(z_{0})}{p_{β}(w_{0})}

= p_{β}(w_{0}) + \frac{mw_{0}p′_{β}(w_{0})}{p_{β}(w_{0})}

= −(m − 1)p_{β}(w_{0}) + mh_{β}(w_{0}).

Since π > arg h_{β}(w_{0}) ≥ πγ(β)/2 = πα/2 and arg p_{β}(w_{0}) = πβ/2, we have

\frac{πα}{2} < arg \{ − (m − 1)p_{β}(w_{0}) + mh_{β}(w_{0})\} < \frac{πβ}{2} + π.

This is, however, impossible because arg T^{α}(z_{0}) = πα/2. Thus we have shown that β^{∗}(α) < γ^{−1}(α).

The proof of Theorems 1.1 and 1.2 is now complete.

4. Concrete bounds for the order of strong starlikeness

In the present section, we propose elementary bounds for the quantities β^{∗}(α) and κ^{∗}(α) for certain α. For α ∈ (0, 1), u ∈ (0, 1), v ∈ (0, +∞), c ∈ (0, 1], we consider the function

q_{α,u,v,c}(z) = \frac{(1 + v)u(1 + cz)^{α} + (1 − u)v(1 − z)^{α}}{u(1 + cz)^{α} + v(1 − z)^{α}} .

We further set

h_{α,u,v,c}(z) = q_{α,u,v,c}(z) + \frac{zq′_{α,u,v,c}(z)}{q_{α,u,v,c}(z)}.

Then our theorem is now stated as follows.

Theorem 4.1. Let α ∈ (0, 1), u ∈ (0, 1), v ∈ (0, +∞), and c ∈ (0, 1]. The function q = q_{α,u,v,c} is univalent in 𝔻 and the image q(𝔻) is a convex subdomain of the right halfplane. Moreover, h = h_{α,u,v,c} is univalent, and if an analytic function p in 𝔻 with p(0) = 1 satisfies p(z) + zp′(z)/p(z) ≺ h(z), then p(z) ≺ q(z).

The following lemma will be needed to prove the theorem and it may be of independent interest.

Lemma 4.2. Let α be a real number with 0 < α < 1 and let a, b, c, d be non-negative numbers with ad − bc ≠ 0. If q = (aT^{α} + b)/(cT^{α} + d), the function zq′(z)/q(z) is starlike and, in particular, univalent in 𝔻. Here, T^{α}(z) = ((1 + z)/(1 − z))^{α}.

Proof. Set φ(z) = zq′(z)/q(z), ψ(z) = zφ′(z)/φ(z) and p = T^{α}. We have to show that u := Re ψ > 0 on 𝔻. First of all, we have expressions

φ(z) = \frac{zp′(z)}{p(z)} · \frac{(ad − bc)p(z)}{(ap(z) + b)(cp(z) + d)} = \frac{2αz}{1 − z^{2}} · \frac{(ad − bc)p(z)}{(ap(z) + b)(cp(z) + d)}, (4.1)

and

ψ(z) = \frac{1 + z^{2}}{1 − z^{2}} + \frac{zp′(z)}{p(z)} − \frac{azp′(z)}{ap(z) + b} − \frac{czp′(z)}{cp(z) + d}

= \frac{1 + z^{2}}{1 − z^{2}} + \frac{2αz}{1 − z^{2}} · \frac{bd − acp(z)^{2}}{(ap(z) + b)(cp(z) + d)} (4.2)

At this stage, we can see that φ and ψ are both analytic in 𝔻. Since ψ is analytically continued up to the boundary of 𝔻 except for the points z = ±1, the harmonic function u has a harmonic continuation across the curve ∂𝔻 \ {±1}. Therefore, in order to prove positivity of u, by the minimum principle, it is enough to see the following three properties:

(i) u is symmetric in the real axis, namely, u(z̄) = u(z) for z ∈ 𝔻, (ii) u(e^{iθ}) ≥ 0 for θ ∈ (0, π), and (iii) lim inf_{z→±1} u(z) ≥ 0.

Property (i) is straightforward to see. We proceed to property (ii). For θ ∈ (0, π), setting ω = p(e^{iθ}), we have

ψ(e^{iθ}) = i cot θ + \frac{iα}{sin θ} · \frac{bd − acω^{2}}{(aω + b)(cω + d)}

and, in particular,

u(e^{iθ}) = −\frac{α}{sin θ} · Im \frac{bd − acω^{2}}{(aω + b)(cω + d)}.

We now have the following sequence of equivalent conditions:

u(e^{iθ}) ≥ 0
⇔ Im \frac{bd − acω^{2}}{(aω + b)(cω + d)} ≤ 0
⇔ Im \frac{(aω + b)(cω + d)}{bd − acω^{2}} = Im \frac{(ad + bc)ω + 2bd}{bd − acω^{2}} ≥ 0
⇔ Im [((ad + bc)ω̄ + 2bd)(acω^{2} − bd)] ≥ 0
⇔ Im [ω((ad + bc)|ω|^{2} + bd(ad + bc) + 2abcdω)] ≥ 0.

The last condition can be verified by arg ω = πα/2 and

arg((ad + bc)|ω|^{2} + bd(ad + bc) + 2abcdω) ≤ arg ω = πα/2.

We finally show property (iii). Observe first

ψ(z) = \frac{1 + z^{2}}{1 − z^{2}} − \frac{2αz}{1 − z^{2}} · (1 − \frac{b}{ap(z) + b} − \frac{d}{cp(z) + d})

= α · \frac{1 − z}{1 + z} + (1 − α) · \frac{1 + z^{2}}{1 − z^{2}} + \frac{2bαz}{(1 − z^{2})(ap(z) + b)} + \frac{2dαz}{(1 − z^{2})(cp(z) + d)}.

Since

Re \frac{z}{(1 − z^{2})(ap(z) + b)} = Re \frac{(z − z̄|z|^{2})(a\overline{p(z)} + b)}{|1 − z^{2}|^{2}|ap(z) + b|^{2}}

= \frac{1 − |z|^{2}}{|1 − z^{2}|^{2}} [\frac{bRe z}{|ap(z) + b|^{2}} + \frac{aRe z · Re p(z)}{|ap(z) + b|^{2}} + a · \frac{1 + |z|^{2}}{1 − |z|^{2}} · \frac{Im z · Im p(z)}{|ap(z) + b|^{2}}]

and Im z · Im p(z) ≥ 0, we can easily show the inequality

lim inf_{z→1} Re \frac{z}{(1 − z^{2})(ap(z) + b)} ≥ 0.

Similarly, we have

lim inf_{z→1} Re \frac{z}{(1 − z^{2})(cp(z) + d)} ≥ 0.

Thus we obtain the inequality

lim inf_{z→1} u(z) = lim inf_{z→1} Re ψ(z) ≥ 0.

In order to show lim inf_{z→−1} u(z) ≥ 0, letting ψ = ψ_{a,b,c,d}, we observe the relation ψ_{a,b,c,d}(−z) = ψ_{b,a,d,c}(z) by the second expression of ψ in (4.2). Thus, the case can be reduced to the above by interchanging a, c and b, d.

The proof is now complete.

Proof of Theorem 4.1. Firstly, we note that q = q_{α,u,v,c} can be written as p ∘ ω. Here, p = q_{α,u,v,1} and the map ω : 𝔻 → 𝔻 is given by

ω(z) = ω_{c}(z) = \frac{(1 + c)z}{2 − (1 − c)z} = \frac{az}{1 − bz} , (4.3)

where a = (1 + c)/2 and b = (1 − c)/2. Note here that a + b = 1. The function p can be written in the form L ∘ T^{α}, where L is the Möbius transformation given by

L(z) = \frac{(1 + v)uz + (1 − u)v}{uz + v} ,

which maps the right half-plane ℍ onto the disk with diameter (1 − u, 1 + v) in such a way that L(0) = 1 − u, L(1) = 1 and L(∞) = 1 + v. In particular, p is convex. Since Δ = ω(𝔻) is a disk contained in 𝔻, by Lemma 2.3, the image p(Δ) = q(𝔻) is a convex subdomain of the right half-plane.

We next show that Q(z) = zq′(z)/q(z) is starlike. Lemma 4.2 implies that P (z) = zp′(z)/p(z) is starlike. The relation q = p ∘ ω now yields

Q(z) = P (ω(z))\frac{zω′(z)}{ω(z)} = \frac{P (\frac{az}{1−bz})}{1 − bz},

By Lemma 2.2, we conclude that Q is starlike.

Since q has positive real part, all the assumptions in Theorem D are fulfilled. Hence, the assertions in the theorem now follow.

We now set

β(α, u, v, c) = sup_{z∈𝔻} \frac{2}{π} | arg q_{α,u,v,c}(z)|,
κ(α, u, v, c) = sup_{z∈𝔻} |\frac{q_{α,u,v,c}(z) − 1}{q_{α,u,v,c}(z) + 1}| and
Γ(α, u, v, c) = inf_{0<θ<π} \frac{2}{π} arg h_{α,u,v,c}(e^{iθ})

for α, u ∈ (0, 1), v ∈ (0, ∞) and c ∈ (0, 1]. Here, the argument is taken to be the principal value. As a corollary of Theorem 4.1, we have

Corollary 4.3. Let β = β(α, u, v, c), κ = κ(α, u, v, c) and γ = Γ(α, u, v, c) for α, u ∈ (0, 1), v ∈ (0, ∞) and c ∈ (0, 1]. If γ > 0, then 𝒦_{γ} ⊂ 𝒮_{β}^{∗} ∩𝒮^{∗}[κ]. In particular, β^{∗}(γ) ≤ β and κ^{∗}(γ) ≤ κ.

Example. We try to estimate β^{∗}(1/2) with the aid of Mathematica. By numerical experiments, we found that the choice α = 0.4731, u = 0.9285, v = 4.2506, c = 0.9285 yields Γ(α, u, v, c) ≈ 1/2 and β(α, u, v, c) ≈ 0.32104. Therefore, we obtain numerically, β^{∗}(1/2) < 0.3211.

Mocanu’s theorem, in turn, gives the estimate β^{∗}(1/2) ≤ γ^{−1}(1/2) ≈ 0.35046. On the other hand, by numerically solving the differential equation (3.1), we obtain an experimental value β^{∗}(1/2) ≈ 0.309, though we do not know how reliable it is.

We next try to estimate κ^{∗}(1/2). For α = 1/2, u = 0.95, v = 3.47, c = 0.49, we obtain Γ(α, u, v, c) ≈ 1/2 and κ(α, u, v, c) ≈ 0.634. Therefore, κ^{∗}(1/2) < 0.635. By a numerical computation, we have an experimental value κ^{∗}(1/2) ≈ 0.613.

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Department of Mathematics, Rzeszów University of Technology, W. Pola 2, PL-35-959 Rzeszów, Poland

  E-mail address: skanas@prz.rzeszow.pl

Department of Mathematics, Graduate School of Science, Hiroshima University, Higashi-Hiroshima, 839-8526 Japan

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