NORM ESTIMATES AND UNIVALENCE CRITERIA FOR MEROMORPHIC FUNCTIONS

S. PONNUSAMY AND TOSHIYUKI SUGAWA

Abstract. Norm estimates of the pre-Schwarzian derivatives are given for meromorphic functions in the outside of the unit circle and used to deduce several univalence criteria.

1. Introduction

Let 𝒜 denote the set of analytic functions f in the unit disk 𝔻 = {z ∈ ℂ : |z| < 1} normalized so that f(0) = 0 and f′(0) = 1. The set 𝒮 of univalent functions in 𝒜 has been intensively studied by many authors. It is well recognized that the set Σ of univalent meromorphic functions F in the domain Δ = {ζ : |ζ| > 1} of the form

(1.1)  F (ζ) = ζ + ∑_{n=0}^{∞} b_{n}ζ^{−n}

plays an indispensable role in the study of 𝒮 .

In paralell with the analytic case, we consider the set ℳ of meromorphic functions in Δ with the expansion (1.1) around ζ = ∞. For some technical reason, we also consider the set ℳ_{n} of functions F in Σ of the form

F (ζ) = ζ + \frac{b_{n}}{ζ^{n}} + \frac{b_{n+1}}{ζ^{n+1}} + ⋯

for each nonnegative integer n. Note that ℳ_{0} = ℳ .

Practically, it is an important problem to determine univalence of a given function in 𝒜 or in ℳ . The best known conditions for univalence are probably those involving pre-Schwarzian or Schwarzian derivatives, which are defined by

T_{f} = \frac{f′′}{f′} and S_{f} = (\frac{f′′}{f′} )′ − \frac{1}{2} (\frac{f′′}{f′} )^{2} .

Date: 23 May, 2006, File:  ps200605.tex.

1991 Mathematics Subject Classification. Primary 30C55; Secondary 33C45.

Key words and phrases. univalence criterion, pre-Schwarzian derivative.

The second author was partially supported by NBHM (DAE, India) grant and JSPS Grant-in-Aid for Scientific Research (B), 17340039.

We define quantities for functions f ∈ 𝒜 and F ∈ ℳ by

B(f) = sup_{|z|<1}(1 − |z|^{2}) |\frac{zf′′(z)}{f′(z)}| ,

B^{∗}(F) = sup_{|ζ|>1}(|ζ|^{2} − 1) |\frac{ζF′′(ζ)}{F′(ζ)}| ,
N(f) = sup_{|z|<1}(1 − |z|^{2})^{2} |S_{f}(z)| ,
N^{∗}(F) = sup_{|ζ|>1}(|ζ|^{2} − 1)^{2} |S_{F} (ζ)| .

Note that these quantities may take ∞ as their values. For example, if F has a pole at a finite point, then B^{∗}(F) = ∞.

If f ∈ 𝒜 and F ∈ ℳ have the relation f(z) = 1/F (1/z), then we can easily see that the relation

(1 − |z|^{2})^{2}S_{f}(z) = (|ζ|^{2} − 1)^{2}S_{F} (ζ)

holds for z = 1/ζ. In particular, we have N(f) = N^{∗}(F).

Nehari [16] proved the following univalence criteria except for the quasiconformal extension property, which is due to Ahlfors and Weill [1].

Theorem A. Every f ∈ 𝒮 satisfies N(f) ≤ 6. Conversely, if f ∈ 𝒜 satisfies N(f) ≤ 2 then f must be univalent. Moreover, if N(f) ≤ 2k < 2, then f extends to a kquasiconformal mapping of the extended plane. The constants 6 and 2 are best possible. The same is true for meromorphic F.

Here and hereafter, a quasiconformal mapping g is called k-quasiconformal if its Beltrami coefficient μ = gz̄/g_{z} satisfies ∥μ∥_{∞} ≤ k.

Though zf′(z)/f(z) = ζF′(ζ)/F (ζ), there is no such a simple relation between zf′′(z)/f′(z) and ζF′′(ζ)/F′(ζ), and thus, between B(f) and B^{∗}(F) for f(z) = 1/F (ζ), ζ = 1/z. Indeed, we have the formula

(1.2)  F′(ζ) = (\frac{z}{f(z)})^{2} f′(z),

which leads to

−\frac{ζF′′(ζ)}{F′(ζ)} = 2 (1 − \frac{zf′(z)}{f(z)}) + \frac{zf′′(z)}{f′(z)}.

Nevertheless, it is rather surprising that formally the same conclusion can be deduced for f and F. Compare Theorem B with Theorem C.

Theorem B. Every f ∈ 𝒮 satisfies B(f) ≤ 6. Conversely, if f ∈ 𝒜 satisfies B(f) ≤ 1 then f ∈ 𝒮 . Moreover, if B(f) ≤ k < 1, then f extends to a k-quasiconformal mapping of the extended plane. The constants 6 and 1 are best possible.

The sufficiency of univalence and quasiconformal extendibility is due to Becker [7]. The sharpness of the constant 1 is due to Becker and Pommerenke [9]. The sharp inequality B(f) ≤ 6 follows from a standard inequality appearing in coefficient estimation (see, e.g., [10, Theorem 2.4]).

Theorem C. Every F ∈ Σ satisfies B^{∗}(F) ≤ 6. Conversely, if F ∈ ℳ satisfies B^{∗}(F) ≤ 1 then F ∈ Σ. Moreover, if B^{∗}(F) ≤ k < 1, then F extends to a k-quasiconformal mapping of the extended plane. The constants 6 and 1 are best possible.

The sufficiency of univalence and quasiconformal extendibility is due to Becker [8]. The sharpness of the constant 1 is again due to Becker and Pommerenke [9]. On the other hand, the estimate B^{∗}(F) ≤ 6 lies deeper. Avhadiev [4] first showed the sharp inequality B^{∗}(F) ≤ 6 by appealing to Goluzin’s inequality (see [11, p. 139]).

Note that many authors use a different norm for the pre-Schwarzian derivative of f ∈ 𝒜 , namely, ∥T_{f}∥ = sup_{|z|<1}(1 − |z|^{2})|T_{f}(z)|, see [14], [13], [12] and [17]. By definition, we observe B(f) ≤ ∥T_{f}∥.

Recall that a plane domain Ω ⊂ ℂ is called hyperbolic if ∂Ω contains at least two points. The uniformization theorem ensures existence of the (complete) hyperbolic metric ρ_{Ω}(w)|dw| on Ω with constant Gaussian curvature −4. Let Ω be a hyperbolic plane domain such that 1 ∈ Ω but 0 ∉ Ω and set

Π(Ω) = {F ∈ ℳ : F′(ζ) ∈ Ω for all ζ ∈ Δ}.

Set also Π_{n}(Ω) = Π(Ω) ∩ ℳ_{n} for n = 0, 1, 2, ….

In [13], the quantity

W (Ω) = sup_{w∈Ω} \frac{1}{|w|ρ_{Ω}(w)},

is studied and called the circular width of Ω. Note that the circular width can also be expressed by W (Ω) = sup_{z∈𝔻}(1 − |z|^{2})|p′(z)/p(z)|, where p : 𝔻 → Ω is any analytic universal covering projection of 𝔻 onto Ω. (We do not demand the condition p(0) = 1.)

One of our main results in the present paper is an estimate of B^{∗}(F) for F ∈ Π_{n}(Ω). The proof of the following theorem will be given in Section 2.

Theorem 1. Let Ω be a hyperbolic domain such that 1 ∈ Ω but 0 ∉ Ω. For every F ∈ Π_{n}(Ω), n ≥ 0, the inequality

B^{∗}(F) ≤ C_{n}W (Ω)

holds, where C_{n} is the constant given by C_{0} = 2 and

(1.3)  C_{n} = sup_{0<r<1} \frac{(n + 1)(1 − r^{2})r^{n−1}}{1 − r^{2n+2}} , n ≥ 1.

As we shall show later (see Proposition 5), we have C_{1} = 2 and 1 < C_{n} < (n + 1)/n for n ≥ 2. We note that an analytic counterpart of this theorem is known and it is much simpler to prove (see [14, Theorem 4.1]);

B(f) ≤ ∥T_{f}∥ ≤ W (Ω)

holds for f ∈ 𝒜 with f′(𝔻) ⊂ Ω.

The univalence criterion in the following is due to Aksent’ev [2] (see also [6, p. 11]). Later, Krzyż [15] gave quasiconformal extensions.

Theorem D (Aksent’ev, Krzyż). Let 0 ≤ k ≤ 1. If F ∈ ℳ satisfies the inequality

(1.4)  |F′(ζ) − 1| ≤ k, |ζ| > 1,

then F is univalent. Furtheremore, if k < 1, then F extends to a k-quasiconformal mapping of the extended plane. The radii 1 and k are best possible.

The above criterion implies univalence of F ∈ ℳ when the range of F′ is contained in the disk |w − 1| < 1. We remind the reader of the fact that the Noshiro-Warschawski theorem asserts that the condition Re f′ > 0 is sufficient for f ∈ 𝒜 to be univalent (cf. [10, Theorem 2.16]). However, the meromorphic counterpart does not hold. Moreover, the range of F′ cannot be enlarged to any disk of the form |w − r| < r, r > 1, to ensure univalence of F (Aksent’ev and Avhadiev [3], see also §4).

With the aid of Theorem 1, we have several results similar to Theorerm D. The following are a couple of examples. Note that the univalence criteria in Theorems 2 and 3 for the case n = 0 were first given by Avhadiev and Aksent’ev [5].

Let x_{m} be the unique solution to the equation

2F_{1}(1, −\frac{1}{m}; 1 − \frac{1}{m}; x) = \frac{1}{2}

in the interval 0 < x < 1 for each integer m ≥ 2 (see Section 4 for details). Put also x_{1} = x_{2}.

Theorem 2. Let n ≥ 0 and 0 ≤ k ≤ 1. Suppose that a function F ∈ ℳ_{n} satisfies the condition

| arg F′(ζ)| ≤ \frac{kπ}{4C_{n}} , |ζ| > 1,

then F must be univalent. Furtheremore, if k < 1, then F extends to a k-quasiconformal mapping of the extended plane. As for univalence, the constant π/(4C_{n}) cannot be replaced by any larger number than (4/π) arctan x_{n+1}.

Note that x_{1} = x_{2} ≈ 0.4198 and that (4/π) arctan x_{1} ≈ 0.506057 ≈ 1.28866(π/8).

In the following univalence criterion, F′ is even allowed to take values with negative real part. Let β_{m} be the unique solution to the equation

(1.5)  2β ∫_{0}^{π/4}(cot x)^{1/m}e^{2β(x−π/4)}dx = 1

in 0 < β < ∞ for each integer m ≥ 2 (see Example 11 in Section 4). Set β_{1} = β_{2}.

Theorem 3. Let n ≥ 0 and 0 ≤ k ≤ 1. Suppose that a function F ∈ ℳ_{n} satisfies the condition

| log |F′(ζ)|| ≤ \frac{kπ}{4C_{n}} , |ζ| > 1,

then F must be univalent. Furtheremore, if k < 1, then F extends to a k-quasiconformal mapping of the extended plane. As for univalence, the constant π/(4C_{n}) cannot be replaced by any larger number than πβ_{n+1}/2.

A numerical computation gives πβ_{1}/2 ≈ 0.719122 ≈ 1.83123(π/8). These results can also be translated into those for the functions f ∈ 𝒜 by using the relation (1.2). The proofs of the above theorems and slightly more refined results will be presented in Section

2. Proof of Theorem 1

Let Ω be a plane domain with 1 ∈ Ω and 0, ∞ ∈ ℂ̂\Ω and let p be an analytic universal covering map of 𝔻 onto Ω with p(0) = 1. Let F ∈ Π_{n}(Ω) be given. When n = 0, the function F can be expressed in the form F = F_{0} +b_{0}, where F_{0} ∈ ℳ_{1} and b_{0} is a constant, thus F_{0} ∈ Π_{1}(Ω). Recall that C_{0} = C_{1} = 2. Therefore, we may further assume that n ≥ 1.

Let ω : 𝔻 → 𝔻, ω(0) = 0, be the lift of the mapping z ↦ F′(1/z) of 𝔻 into Ω via the covering map p : 𝔻 → Ω, namely,

(2.1)  F′(\frac{1}{z}) = p(ω(z)), |z| < 1.

Since F ∈ ℳ_{n}, it can be expressed in the form

F (ζ) = ζ + ∑_{k=n}^{∞} b_{k}ζ^{−k}, |ζ| > 1,

we have

F′(1/z) = 1 − ∑_{k=n}^{∞} kb_{k}z^{k+1} = 1 − ∑_{k=n+1}^{∞}(k − 1)b_{k−1}z^{k}, |z| < 1.

In particular, ω has a zero of at least order n + 1 at the origin. This implies that the function φ(z) = ω(z)/z^{n+1} is analytic and satisfies |φ(z)| ≤ 1 by the maximum modulus principle. We now apply the Schwarz-Pick lemma to the function φ to get

|φ′(z)| ≤ \frac{1 − |φ(z)|^{2}}{1 − |z|^{2}} , |z| < 1,

and equivalently,

(2.2)  |zω′(z) − (n + 2)ω(z)| ≤ \frac{|z|^{2n+2} − |ω(z)|^{2}}{|z|^{n}(1 − |z|^{2})} , |z| < 1.

In particular, we obtain

(2.3)  |zω′(z)| ≤ (n + 2)|ω(z)| + \frac{|z|^{2n+2} − |ω(z)|^{2}}{|z|^{n}(1 − |z|^{2})} , |z| < 1.

The last inequality can be expressed by

(2.4)  (1 − |z|^{2})|z|^{−1}|ω′(z)| ≤ (1 − |ω(z)|^{2})F (|z|, |ω(z)|), |z| < 1,

where the function F (r, s) is defined by

F (r, s) = \frac{(n + 1)(1 − r^{2})r^{n}s + r^{2n+2} − s^{2}}{r^{n+2}(1 − s^{2})} .

Since |φ(z)| ≤ 1, we see that |ω(z)| ≤ |z|^{n+1} holds. We now show the following elementary result.

Lemma 4.

F (r, s) ≤ F (r, r^{n+1}) = \frac{(n + 1)(1 − r^{2})r^{n−1}}{1 − r^{2n+2}} , 0 ≤ s ≤ r^{n+1}.

Proof. We first see the inequality

\frac{∂F}{∂s} (r, s) = \frac{1 + s^{2}}{r^{n+2}(1 − s^{2})^{2}} [(n + 1)r^{n}(1 − r^{2}) − 2(1 − r^{2n+2})\frac{s}{1 + s^{2}} ]

≥ \frac{1 + s^{2}}{r^{n+2}(1 − s^{2})^{2}} [(n + 1)r^{n}(1 − r^{2}) − 2(1 − r^{2n+2})\frac{r^{n+1}}{1 + r^{2n+2}} ]

= \frac{(1 + s^{2})}{r^{2}(1 − s^{2})^{2}(1 + r^{2n+2})}G(r), 0 ≤ s ≤ r^{n+1},

because the function s/(1 + s^{2}) is increasing in 0 < s < 1 and s ≤ r^{n+1} is assumed. Here,

G(r) = (n + 1)(1 − r^{2})(1 + r^{2n+2}) − 2r(1 − r^{2n+2})

= (1 − r^{2}) [(n + 1)(1 + r^{2n+2}) − 2r ∑_{j=0}^{n} r^{2j}]

= (1 − r^{2}) [(n + 1)(1 + r^{2n+2}) − r ∑_{j=0}^{n}(r^{2j} + r^{2n−2j})]

= (1 − r^{2}) ∑_{j=0}^{n} [(1 + r^{2n+2}) − r(r^{2j} + r^{2n−2j})]

= (1 − r^{2}) ∑_{j=0}^{n}(1 − r^{2j+1})(1 − r^{2n+1−2j}) > 0.

Therefore, we conclude that (∂F/∂s)(r, s) > 0 in 0 < s < r^{n+1}, which implies the monotonicity of the function F (r, s) in s. Thus the inequality F (r, s) ≤ F (r, r^{n+1}) holds in 0 ≤ s ≤ r^{n+1}. □

We now complete the proof of Theorem 1. By taking the logarithmic derivative of the both sides of (2.1), we have the relation

\frac{−F′′(1/z)}{z^{2}F′(1/z)} = \frac{p′(ω(z))}{p(ω(z))}ω′(z), |z| < 1.

Letting ζ = 1/z, we thus obtain

(|ζ|^{2} − 1) |\frac{ζF′′(ζ)}{F′(ζ)}| = (1 − |z|^{2})|z|^{−1} |\frac{p′(ω(z))}{p(ω(z))}| |ω′(z)|.

Recall here that C_{n} is nothing but the supremum of F (r, r^{n+1}) over 0 < r < 1. We then make use of (2.4) and Lemma 4 to deduce the inequality

(|ζ|^{2} − 1) |\frac{ζF′′(ζ)}{F′(ζ)}| ≤ (1 − |ω(z)|^{2}) |\frac{p′(ω(z))}{p(ω(z))}| F (|z|, |z|^{n+1})

≤ C_{n}(1 − |ω(z)|^{2}) |\frac{p′(ω(z))}{p(ω(z))}|

≤ C_{n}W (Ω).

The assertion of the theorem now follows.

Remark. The theorem is sharp if the relation ρ_{0} = r_{0}^{n+1} is satisfied by chance, where r = r_{0} is the point where the maximum is attained in the definition of C_{n} and r = ρ_{0} is the radius where the maximum is attained for (1 − |z|^{2})|p′(z)/p(z)|. Let w_{0} be the maximum point of (1 − |z|^{2})|p′(z)/p(z)| with |w_{0}| = ρ_{0}, and set z_{0} = r_{0}. Then we choose ω_{0} so that ω_{0}(z_{0}) = w_{0} and equalities hold in (2.2) and (2.3) at z = z_{0} simultaneously (see the proof of Dieudonné’s lemma in [10, p. 198]). Then, we actually have B^{∗}(F) = C_{n}W (Ω) in this case, where F is determined by F′(1/z) = p(ω_{0}(z)) in |z| < 1.

As we promised in Introduction, we give some information about the constants C_{n}.

Proposition 5. The constants C_{n} given by (1.3) satisfy the following:

(2.5)  C_{0} = C_{1} = 2, C_{2} = \frac{3\sqrt{6(\sqrt{13} − 1)}}{5 + \sqrt{13}} ≈ 1.37838,
(2.6)  1 < C_{n} < \frac{n + 1}{n}, n = 2, 3, ….

Proof. The relations in (2.5) can be checked in a straightforward way. We omit the details. We show only (2.6). Let n ≥ 2 and set

g_{n}(x) = \frac{1 − x^{n+1}}{x^{(n−1)/2}(1 − x)}, x ∈ (0, 1).

Then clearly, C_{n} = (n + 1)/ inf_{0<x<1} g_{n}(x). First note that

lim_{x→1} g_{n}(x) = n + 1.

Therefore, we have C_{n} ≥ 1. In order to show strictness, we set x = 1 − ε, ε > 0. Then

g_{n}(1 − ε) = (n + 1) − \frac{n + 1}{2}ε + O(ε^{2}), ε → 0,

which implies that g_{n}(x) is smaller than n + 1 when x < 1 is close enough to 1. Therefore, C_{n} > 1.

We next show the reverse inequality. Since g_{n}(x) → +∞ as x → 0+, the function g_{n} takes its minimum at a point in (0, 1). We now estimate g_{n}(x) from below;

g_{n}(x) = x^{(1−n)/2} ∑_{j=0}^{n} x^{j}

> x^{(1−n)/2} ∑_{j=0}^{n−1} x^{j}

= x^{(1−n)/2} ∑_{j=0}^{n−1} \frac{x^{j} + x^{n−1−j}}{2}

= ∑_{j=0}^{n−1} \frac{x^{j−(n−1)/2} + x^{(n−1)/2−j}}{2}

≥ ∑_{j=0}^{n−1} 1 = n.

Thus we get the inequality min_{0<x≤1} g_{n}(x) > n, which in turn implies C_{n} < (n+1)/n. □

3. A variant of Theorem 1

We give a variant of Theorem 1 in the present section. In the following theorem, the condition p(0) = 1 for the analytic universal covering map p of 𝔻 onto Ω is required and the involved constant might not be computed easily, but the estimate is independent of n and better than Theorem 1 at least when n = 0.

Theorem 6. Let Ω be a plane domain with 1 ∈ Ω but 0, ∞ ∉ Ω and let p be an analytic universal covering map of the unit disk 𝔻 onto Ω with p(0) = 1. Then, for every F ∈ Π(Ω) the inequality

B^{∗}(F) ≤ 2 sup_{|z|<1}(1 − |z|) |\frac{p′(z)}{p(z)}|

holds.

Proof. The proof proceeds basically in the same line as in the previous section. In order to show that the constant is really independent of n for which F ∈ Π_{n}(Ω) holds, we prove the assertion under the additional assumption that F ∈ Π_{n}(Ω) for a fixed n ≥ 1. We replace the inequality (2.4) by

(3.1)  (1 − |z|^{2})|z|^{−1}|ω′(z)| ≤ (1 − |ω(z)|)H(|z|, |ω(z)|), |z| < 1,

where

H(r, s) = \frac{(n + 1)(1 − r^{2})r^{n}s + r^{2n+2} − s^{2}}{r^{n+2}(1 − s)} .

Recall here that |ω(z)| ≤ |z|^{n+1} holds. Since the function s^{2} − 2s is decreasing in 0 < s < r^{n+1}, we have

\frac{∂H}{∂s} (r, s) = \frac{s^{2} − 2s + (n + 1)(1 − r^{2})r^{n} + r^{2n+2}}{r^{n+2}(1 − s)^{2}}

≥ \frac{r^{2n+2} − 2r^{n+1} + (n + 1)(1 − r^{2})r^{n} + r^{2n+2}}{r^{n+2}(1 − s)^{2}} .

The numerator of the last term can be written in the form

r^{n} [(n + 1)(1 − r^{2}) − 2r(1 − r^{n+1})]
= r^{n}(1 − r) [(n + 1)(1 + r) − 2r(1 + r + r^{2} + ⋯ + r^{n})]
= r^{n}(1 − r) ∑_{j=0}^{n} (1 + r − 2r^{j+1}) .

It is now clear that (∂H/∂s)(r, s) > 0 in 0 < s ≤ r^{n+1}. Thus H(r, s) is increasing in s and therefore

H(r, s) ≤ H(r, r^{n+1}) = \frac{(n + 1)(1 − r^{2})r^{n−1}}{1 − r^{n+1}} = g(r).

Since

g′(r) = \frac{(n + 1)r^{n−2}((n − 1)(1 − r^{2}) − 2r^{2}(1 − r^{n−1}))}{(1 − r^{n+1})^{2}}

= \frac{(n + 1)r^{n−2}(1 − r)}{(1 − r^{n+1})^{2}} ∑_{j=0}^{n−2} [1 − r^{j+2} + r(1 − r^{j+1})] > 0,

the function g(r) is increasing and thus g(r) < g(1−) = 2 for 0 ≤ r < 1. Therefore, we obtain

sup_{0<s≤r^{n+1}<1} H(r, s) = sup_{0<r<1} g(r) = 2,

which is, indeed, independent of n.

The rest is same as in the previous section. We omit the details. □

Since 1 − r ≤ 1 − r^{2} = (1 + r)(1 − r) ≤ 2(1 − r), the inequalities

sup_{|z|<1}(1 − |z|) |\frac{p′(z)}{p(z)}| ≤ sup_{|z|<1}(1 − |z|^{2}) |\frac{p′(z)}{p(z)}| ≤ 2 sup_{|z|<1}(1 − |z|) |\frac{p′(z)}{p(z)}|

hold. Thus, when n = 0, the estimate in Theorem 6 is better than that in Theorem 1.

4. Examples of non-univalent functions

In this section, we present non-univalent meromorphic functions in the class ℳ to examine our univalence criteria given in Introduction. First, we introduce the example given by Aksent’ev and Avhadiev [3].

Example 7. Let r > 1 be given and set Ω = {w ∈ ℂ : |w − r| < r}. For a number c ∈ (0, 1/2], we set Φ = G ∘ F, where F (ζ) = ζ + c/ζ and G(ζ) = ζ + (1 + c)^{2}/ζ. Then

Φ′(ζ) = 1 − ζ^{−2} + cψ(ζ^{−2}), where ψ(z) = ψ_{c}(z) = −\frac{(c + 3) − (c^{2} + 3)z + (c^{2} − c)z^{2}}{(1 + cz)^{2}} .

Note that the functions 1 − ζ^{−2} and ψ(ζ^{−2}) take the value 0 at ζ = ±1. Since ψ_{c} is uniformly bounded in 𝔻 and ψ′(1) > 0, in order to see that F′(𝔻) ⊂ Ω for sufficiently small c, it is enough to check that the (signed) curvature of the curve θ ↦ ψ(e^{iθ}) is positive at θ = 0, in other words, Re (1 + zψ′′(z)/ψ′(z))/|ψ′(z)| is positive at z = 1. A direct computation gives

1 + \frac{zψ′′(z)}{ψ′(z)} = \frac{3 − 10c + 2(c^{2} + c)z − c^{2}z^{2}}{(3 − cz)(1 + cz)} ,

which shows Re (1 + ψ′′(1)/ψ′(1))/|ψ′(1)| > 0 for a small enough c > 0 as required.

We see now that Φ is not univalent in Δ by observing that the two points ±i(1 + c + \sqrt{1 + 6c + c^{2}})/2 in Δ are zeros of Φ.

The above example is qualitatively very nice but somewhat implicit because it is not simple to give a right value of c for a given r > 1. The next two examples are more concrete.

Example 8. We consider the function F_{m} ∈ ℳ given by

F_{m}(ζ) = ζ − 2 ∑_{j=1}^{∞} \frac{ζ^{1−mj}}{mj − 1}
= ζ(2_{2}F_{1}(1, −\frac{1}{m}; 1 − \frac{1}{m}; ζ^{−m}) − 1), |ζ| > 1,

for each integer m ≥ 2, where _{2}F_{1}(a, b; c; x) stands for the hypergeometric function. Note that F_{m} has the m-fold symmetry

F_{m}(e^{2πi/m}ζ) = e^{2πi/m}F_{m}(ζ)

and belongs to the class ℳ_{m−1}. Since the function h_{m} defined by

h_{m}(x) = 2_{2}F_{1}(1, −\frac{1}{m}; 1 − \frac{1}{m}; x) − 1 (x ∈ (0, 1))

has the properties that h_{m} is monotone decreasing, h_{m}(0) = 1 and lim_{x→1^{−}} h_{m}(x) = −∞, there is the unique point x_{m} such that h(x_{m}) = 0 in the interval 0 < x < 1. Hence, the function F_{m} has the m zeros e^{2πij/m}x_{m}^{−1/m}, j = 0, 1, …, m − 1, in Δ and, in particular, is not univalent in Δ. On the other hand, we have

F′_{m}(ζ) = 1 + 2 ∑_{j=1}^{∞} ζ^{−mj} = p(ζ^{−m}),

where p(z) = (1 + z)/(1 − z). It is a standard fact that p maps the unit disk onto the right half-plane ℍ = {w ∈ ℂ : Re w > 0}. Therefore, F′_{m} maps Δ onto ℍ in an m-to-1 way and Re F′_{m} > 0 holds.

In particular, we have shown the following.

Proposition 9. For each integer n ≥ 0, there is a non-univalent function F in the class ℳ_{n} such that Re F′(ζ) > 0 in |ζ| > 1.

Note that the function F_{2} in the above example can be expressed also by

F_{2}(ζ) = ζ − log \frac{ζ + 1}{ζ − 1}, |ζ| > 1.

A numerical computation yields, for instance,

x_{2} ≈ 0.419798,
x_{3} ≈ 0.667508,
x_{4} ≈ 0.808289.

The above functions can be used to examine univalence criteria. Note that, for a function F ∈ ℳ , the new function

F^{t}(ζ) = tF (\frac{ζ}{t} ), |ζ| > 1,

for t ∈ (0, 1) satisfies the relation (F^{t})′(ζ) = F′(ζ/t). For instance, for m ≥ 2, the function F_{m}^{t}(ζ) = tF_{m}(ζ/t) is not univalent as far as t^{m} > x_{m}, because (ζ/t)^{−m} = x_{m} has m roots in |ζ| > 1 in this case. On the other hand, (F_{m}^{t})′ has the range {w ∈ ℂ : w = (1+t^{m}z)/(1−t^{m}z) for some z ∈ 𝔻} = {w ∈ ℂ : |w−(1+t^{2m})/(1−t^{2m})| < 2t^{m}/(1−t^{2m})}. In this way, we have shown the following.

Proposition 10. Let Ω_{s} = {w ∈ ℂ : |w − (1 + s^{2})/(1 − s^{2})| < 2s/(1 − s^{2})} and n ≥ 1. If s > x_{n+1}, then the class Π_{n}(Ω_{s}) contains non-univalent functions.

Example 11. The construction is similar to that of Example 8. First note that the analytic function ((1 + z)/(1 − z))^{iβ} gives a universal covering projection of the unit disk onto the annulus A = {w ∈ ℂ : e^{−πβ/2} < |w| < e^{πβ/2}} for a positive constant β. Let G ∈ ℳ_{m−1} be the function detemined by the relation G′(ζ) = ((ζ^{m} +1)/(ζ^{m} −1))^{iβ} for an integer m ≥ 2. Then G also has the m-fold symmetry. Let h_{β}(z) = 1/z −∫_{0}^{z} t^{m−2}q_{β}(t^{m})dt, where ((1 + z)/(1 − z))^{iβ} = 1 + zq_{β}(z), so that G(ζ) = h_{β}(1/ζ). Now take any root ω of the polynomial z^{m} + i and set φ(β) = ωh_{β}(ω). Since 1 + ixq_{β}(ix) = ((1 + ix)/(1 − ix))^{iβ} = exp(2iβarctanh(ix)) = exp(−2β arctan x), we have for 0 < r ≤ 1

ωh_{β}(ωr) = \frac{1}{r} + ∫_{0}^{r} it^{n−2}q_{β}(−it^{m})dt
= \frac{1}{r} − ∫_{0}^{r} (exp(2β arctan(t^{m})) − 1) t^{−2}dt.

Thus,

φ(β) = 1 − ∫_{0}^{1} (exp(2β arctan(t^{m})) − 1) t^{−2}dt.

Since φ(0) = 1, φ(+∞) = −∞ and

φ′(β) = − ∫_{0}^{1} t^{−2} arctan(t^{m}) exp(2β arctan(t^{m}))dt < 0,

there exists a unique β_{m} such that φ(β_{m}) = 0. We now simplify the equation φ(β) = 0. Performing integration by parts and then setting x = arctan(t^{m}), we have

φ(β) = e^{πβ/2} − 2β ∫_{0}^{π/4} e^{2βx}(tan x)^{−1/m}dx

= e^{πβ/2} (1 − 2β ∫_{0}^{π/4} e^{2β(x−π/4)}(cot x)^{1/m}dx) .

Thus we have arrived at the form in (1.5).

We now fix any β > β_{m}. Then ωh_{β}(ωr) > 0 for a small enough r > 0 whereas φ(β) = ωh_{β}(ω) < 0. Therefore, there exists an ρ ∈ (0, 1) such that G(1/(ωρ)) = h_{β}(ωρ) = 0. In particular, G has at least m zeros in Δ and thus is not univalent. By the above observations, we have the following proposition. Proposition 12. Let n be an integer with n ≥ 1 and let β > β_{n+1}. Then there exists a non-univalent function G ∈ ℳ_{n} such that e^{−πβ/2} < |G′(ζ)| < e^{πβ/2} for |ζ| > 1.

By a numerical computation, one has

β_{2} ≈ 0.457807,
β_{3} ≈ 0.786518,
β_{4} ≈ 1.03144.

5. Applications to univalence criteria

We combine Theorem 1 or Theorem 6 with Theorem C to deduce several univalence criteria for functions in ℳ . The same method can be applied also to ℳ_{n} for n ≥ 1, but we do not go into details here. In order to make statements concise, we introduce the notation Σ(k) to designate the set of those functions in Σ which can be extended to k-quasiconformal mappings of the extended plane. For k = 1, simply we define Σ(1) = Σ for convenience.

To examine Theorems 1 and 6, we assume Ω to be a disk containing 1 but not containing 0. Then we can express Ω as 𝔻(a, ρ) = {w : |w−a| < ρ}, where 0 < ρ ≤ |a| and |1−a| < ρ. If we put p(z) = a + ρz, then we compute

W (𝔻(a, ρ)) = sup_{z∈𝔻}(1 − |z|^{2})\frac{ρ}{|a + ρz|}

= sup_{0<r<1}(1 − r^{2})\frac{ρ}{|a| − ρr}

= \frac{ρ}{|a|} sup_{0<r<1} \frac{1 − r^{2}}{1 − (ρ/|a|)r}

= \frac{2ρ/|a|}{1 + \sqrt{1 − (ρ/|a|)^{2}}} ,

where we have made a standard but tedious computation at the final step (see, for instance, [13, Lemma 4.2]). Therefore, by Theorem 1, we conclude that

(5.1)  B^{∗}(F) ≤ \frac{2C_{n}ρ/|a|}{1 + \sqrt{1 − (ρ/|a|)^{2}}}

for F ∈ Π_{n}(𝔻(a, ρ)). It is easy to see that the right-hand side of the last inequality is less than or equal to k if and only if ρ/|a| ≤ 4C_{n}k/(4C_{n}^{2} +k^{2}). Thus we can show the following by appealing to Theorem C.

Theorem 13. Let n be an integer with n ≥ 0 and a ∈ ℂ, ρ > 0 satisfy ρ ≤ |a| and |a − 1| < ρ. Suppose that

\frac{ρ}{|a|} ≤ \frac{4C_{n}k}{4C_{n}^{2} + k^{2}}

for a constant k with 0 ≤ k ≤ 1. Then Π_{n}(𝔻(a, ρ)) ⊂ Σ(k).

We recall that Theorem D gives the stronger assertion Π(𝔻(1, k)) ⊂ Σ(k) when a = 1 and ρ = k.

We next consider to apply Theorem 6. It is not simple to treat the case when a is not real. Therefore, we further assume that a > 0 for simplicity. Then the conformal map p of 𝔻 onto 𝔻(a, ρ) with p(0) = 1 can be taken in the form p(z) = (1 + Az)/(1 + Bz), where −1 < B < A ≤ 1. A simple computation gives us the relations A = (ρ^{2} − a^{2} + a)/ρ and B = (1 − a)/ρ.

First observe the expression (see [13, Lemma 4.1])

W = sup_{z∈𝔻}(1 − |z|) |\frac{p′(z)}{p(z)}| = \begin{cases}(A − B) sup_{0<r<1} \frac{1 − r}{(1 − Ar)(1 − Br)} if A + B ≤ 0, \\ (A − B) sup_{0<r<1} \frac{1 − r}{(1 + Ar)(1 + Br)} if A + B ≥ 0. \end{cases}

At any event, we can easily see that W = A − B. Therefore, by Theorem 6, we obtain the estimate

(5.2)  B^{∗}(F) ≤ 2(A − B) = \frac{2(ρ^{2} − (a − 1)^{2})}{ρ}

for F ∈ Π(𝔻(a, ρ)). In the same way as above, we have the following.

Theorem 14. Let a > 0, ρ > 0 satisfy ρ ≤ a and |a − 1| < ρ. Suppose that

ρ^{2} − (a − 1)^{2} ≤ \frac{kρ}{2}

for a constant k with 0 ≤ k ≤ 1. Then Π(𝔻(a, ρ)) ⊂ Σ(k).

As an example, let us consider the disk Ω_{s} = {w ∈ ℂ : |w − (1 + s^{2})/(1 − s^{2})| < 2s/(1 − s^{2})}. In this case, A = s, B = −s, and therefore,

\frac{4ρ/|a|}{1 + \sqrt{1 − (ρ/|a|)^{2}}} = 4s = 2(A − B),

which means that the esimates (5.1) with n = 0 and (5.2) are identical in this case. In particular, Theorems 13 and 14 both imply that Π(Ω_{s}) ⊂ Σ if s ≤ 1/4. This is, however, weaker than Theorem D because Ω_{s} ⊂ 𝔻(1, 1) for s ≤ 1/3. On the other hand, Proposition 10 implies that Π(Ω_{s}) is not contained in Σ for s > x_{2} ≈ 0.4198.

However, Theorems 13 and 14 may imply the inclusion Π(𝔻(a, ρ)) ⊂ Σ for a disk 𝔻(a, ρ) which is not contained in 𝔻(1, 1). For instance, by Theorem 14, we see that Π(𝔻(3/2, 4/5)) ⊂ Σ but 𝔻(3/2, 4/5) is not a subset of 𝔻(1, 1). By the way, this is not implied by Theorem 13.

We next recall basic results for the values of W (Ω) for special domains Ω. We set

S(α, γ) = {w ∈ ℂ : | arg w − γ| < πα/2}
A(r_{1}, r_{2}) = {w ∈ ℂ : r_{1} < |w| < r_{2}},

where α > 0, γ ∈ ℝ and 0 < r_{1} < r_{2} < ∞. The domain S(α, γ) is called a sector with opening πα and vertex at 0. The domain A = A(r_{1}, r_{2}) is called a round annulus centered at 0 with modulus m = log(r_{2}/r_{1}). We write m = mod A. Then we have the following.

Lemma 15 ([13]).

W (S(α, γ)) = 2α, 0 < α < 2,
W (A(r_{1}, r_{2})) = \frac{2}{π} log \frac{r_{2}}{r_{1}} = \frac{2}{π} mod A(r_{1}, r_{2}), 0 < r_{1} < r_{2} < ∞.

Combining this lemma with Theorems 1 and C, we can prove the following two results. Theorems 2 and 3 are just special cases of them up to non-univalent examples, which were supplied in the previous section.

Theorem 16. Let 0 ≤ k ≤ 1. If Ω is a sector with opening kπ/4 and vertex at 0 such that 1 ∈ Ω. Then, Π(Ω) ⊂ Σ(k).

Theorem 17. Let 0 ≤ k ≤ 1. If Ω is a round annulus centered at 0 with modulus kπ/4 such that 1 ∈ Ω. Then, Π(Ω) ⊂ Σ(k).

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Department of Mathematics, Indian Institute of Technology, IIT-Madras, Chennai-600 036, India

  E-mail address: samy@iitm.ac.in

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

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