Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 47–56

ON THE DERIVATIVE OF HYPERBOLICALLY CONVEX FUNCTIONS

Diego Mejía and Christian Pommerenke

Universidad Nacional de Colombia, Departamento de Matemáticas Apartado Aéreo 3840, Medellín, Colombia; dmejia@persesus.unalmed.edu.co Technische Universität, FB Mathematik, MA 8-2 D-10623 Berlin, Germany; pommeren@math.tu-berlin.de

Abstract. A conformal mapping f of the unit disk D into itself is called hyperbolically convex if the non-euclidean segment between any two points of f(D) also belongs to f(D). In this paper we obtain the exact order of growth for the derivative of these functions and investigate their boundary behaviour.

1. The growth of the derivative

Let D be the unit disk and T = ∂D. The analytic univalent function f: D → D is called hyperbolically convex (or simply h-convex) if the non-euclidean segment between any two points of f(D) also belongs to f(D). An h-convex function is continuous in \overline{D}.

Hyperbolically convex functions were first systematically studied by William Ma and David Minda [MM1]. Among many other results they obtained the characterization

(1.1)  Re[1 + z \frac{f′′(z)}{f′(z)} + \frac{2zf′(z)\overline{f(z)}}{1 − |f(z)|^{2}}] > 0  (z ∈ D);

see also [MM2]. The present authors [MP1], [MP2] and Alexandre Vasil′ev [MPV] derived a number of estimates for h-convex functions. The upper bound for the derivative remained an open problem and it was conjectured [MP2] that

(1.2)  f′(z) = O(\frac{1}{1 − |z|}(log \frac{1}{1 − |z|})^{−2})  (|z| → 1);

this was proved with the exponent −1 instead of −2.

The property of being h-convex is invariant under Möbius transformations of D onto itself. This fact can be used to achieve the normalization

(1.3)  f(z) = αz + a_{2}z^{2} + ⋯,  0 < α ≤ 1.

2000 Mathematics Subject Classification: Primary Primary 30C45.

Research supported by COLCIENCIAS and Deutsche Forschungsgemeinschaft (DFG).

48 Diego Mejía and Christian Pommerenke The h-convex function

(1.4)  k_{α}(z) = 2αz/(1 − z + \sqrt{(1 − z)^{2} + 4α^{2}z} ) = αz + ⋯

often plays the role of the extremal function and it was shown in [MPV] that

(1.5) |f′(z)| ≤ k′_{α}(|z|) for |z| ≤ \sqrt{2} − 1,

however not for |z| close to 1.

Let λ(z, ς) denote the non-euclidean distance in D normalized such that

(1.6) λ(z, 0) = \frac{1}{2} log \frac{1 + |z|}{1 − |z|} (z ∈ D).

Theorem 1. Let f be hyperbolically convex. Then

(1.7) \frac{(1 − |z|^{2})|f′(z)|(1 − |ς|^{2})|f′(ς)|}{|f(z) − f(ς)|^{2}}λ(z, ς)^{2} ≤ c_{1} for z, ς ∈ D,

where c_{1} is an absolute constant. In particular, if f(z) = αz + ⋯, then

(1.8) (1 − |z|^{2})|f′(z)|λ(z, 0)^{2} ≤ \frac{c_{1}}{α} |f(z)|^{2} for z ∈ D.

The proof will be very geometric. It follows from (1.8) that the conjecture (1.2) is true. The stronger conjecture [MP1]

a_{n} = O(n^{−1}(log n)^{−2}) (n → ∞)

however remains open.

Theorem 2. Let f be h-convex. Then

(1.9) b(ς) = lim_{r→1}(1 − r^{2})|f′(rς)|λ(r, 0)^{2} < ∞ (ς ∈ T)

exists. If b(ς) > 0 then f(ς) ∈ T and

(1.10)  (f(ς) − f(z))L(ς̄z) → f(ς)b(ς)  as z → ς, z ∈ Δ,
(1.11)  (1 − ς̄^{2}z^{2})f′(z)L(ς̄z)^{2} → ς̄f(ς)b(ς)  as z → ς, z ∈ Δ,

where L(z) = \frac{1}{2} log[(1 + z)/(1 − z)]. Here Δ is any Stolz angle at ς.

Martin Chuaqui and the second author [CP] have proved that a meromorphic function f: D → Ĉ satisfies

(1.12)  \frac{(1 − |z|^{2})|f′(z)|(1 − |ς|^{2})|f′(ς)|}{|f(z) − f(ς)|^{2}}λ(z, ς)^{2} ≤ 1  (z, ς ∈ D),

if and only if its Schwarzian derivative S_{f} satisfies

(1.13)  (1 − |z|^{2})^{2}|S_{f}(z)| ≤ 2  (z ∈ D).

It was shown in [MP2, Example 5.3] that (1.13) does not hold for all h-convex functions. Hence (1.12) does not hold for all h-convex functions and it follows that the constant in Theorem 1 satisfies c_{1} > 1.

See [MP2] and [MM2] for bounds for the Schwarzian derivative of h-convex functions. In [MP2] it is conjectured that

sup_{f} sup_{z}(1 − |z|^{2})^{2}|S_{f}(z)| ≈ 2.384.

2. Boundary points on the unit circle

We now study the closed set T ∩ f(T) where f(D) meets the unit circle. It follows from the McMillan twist theorem [Mc], [P, p. 142] that f has a finite angular derivative for almost all ς ∈ T with f(ς) ∈ T, which implies that f(T) is tangential to T at f(ς).

Let c_{2}, c_{3}, … denote suitable positive absolute constants.

Theorem 3. Let f(z) = αz + ⋯ be h-convex and let ς ∈ T, f(ς) ∈ T. If there exists r_{0} < 1 such that

(2.1)  (1 − r^{2})|f′(rς)| ≤ c_{2}|f(ς) − f(rς)| ≤ c_{3}α  (r_{0} ≤ r < 1),

then, for ω ∈ T ∩ f(T), ω ≠ f(ς),

(2.2)  |ω − f(ς)| ≥ c_{4}|f(ς) − f(rς)|  (r_{0} ≤ r < 1).

The assumption (2.1) roughly says that, near f(ς), the image domain lies in a narrow sector. The assertion (2.2) implies that f(ς) is an isolated point of T ∩ f(T), of which there can be only countably many.

Theorem 4. Let f(z) = αz+⋯ be h-convex and let b(ς) be defined by (1.9). Then b(ς) = 0 except possibly for countably many ς and moreover

(2.3)  ∑_{b(ς)>0} b(ς) ≤ \frac{c_{5}}{α^{2}} .

3. Proof of Theorem 1

(a) First we prove inequality (1.8) for h-convex functions of the form f(z) = a_{1}z + ⋯ with |a_{1}| = α. Let z ∈ D be given. By rotational invariance, we may assume that

u = f(z) > 0.

We write

(3.1)  d(s) = dist (s, ∂f(D))  for 0 ≤ s ≤ u.

Let C^{±} be the circles orthogonal to T that are tangential to {w : |w| < α/2} and {|w −u| < d(u)} and let A^{±} be the arcs of C^{±} between the points of contact. The centers ω and \overline{ω} and the radius ϱ of C^{±} satisfy

(3.2)  |ω| = (1 + α^{2}/4)/α,  ϱ = (1 − α^{2}/4)/α,

in particular |ω|^{2} − ϱ^{2} = 1. We write ξ = Re ω.

Let G be the domain between A^{+} and A^{−} including the disks {|w| < α/2} and {|w − u| < d(u)} touched by A^{±}. We have {|w − u| < d(u)} ⊂ f(D) by (3.1) and {|w| < α/2} ⊂ f(D) by [MM1, Theorem 2]. Hence G ⊂ f(D) because f is h-convex.

Now let 0 < s < u. Geometric considerations show that the points of ∂G nearest to s lie on the arcs A^{±} of C^{±}. Hence

(3.3) d(s) ≥ |ω − s| − ϱ = \frac{|ω − s|^{2} − ϱ^{2}}{|ω − s| + ϱ} ≥ \frac{1 − 2ξs + s^{2}}{2 + 2ϱ} > \frac{α}{4}(1 − 2ξs + s^{2}).

The construction of C^{±} shows that |ω − u| = d(u) + ϱ, which implies

(3.4)  d(u) < \frac{d(u)^{2} + 2ϱd(u)}{2ϱ} = \frac{1 − 2ξu + u^{2}}{2ϱ} < α(1 − 2ξu + u^{2})

by (3.2).

We will show in part (b) that

(3.5)  y ≡ \sqrt{1 − 2ξu + u^{2}} ∫_{0}^{u} \frac{ds}{1 − 2ξs + s^{2}} < c_{5}u.

It follows from (3.3), (3.4) and (3.5) that

\sqrt{d(u)} ∫_{0}^{u} \frac{ds}{d(s)} < \frac{4c_{5}}{\sqrt{α}} u.

Since u = f(z) and d(u) ≤ (1 − |z|^{2})|f′(z)| ≤ 4d(u) by (3.1), we conclude that

\begin{aligned}\sqrt{(1 − |z|^{2})|f′(z)|} λ(z, 0) & ≤ 2\sqrt{d(u)} ∫_{f^{−1}([0,u])} \frac{|f′(ς)|}{(1 − |ς|^{2})|f′(ς)|}| dς| \\ & ≤ 2\sqrt{d(u)} ∫_{0}^{u} \frac{ds}{d(s)} ≤ \frac{8c_{5}}{\sqrt{α}} |f(z)| \end{aligned}

which implies (1.8).

On the derivative of hyperbolically convex functions 51 (b) Now we prove (3.5). We distinguish three cases. Case I. Let −∞ < ξ ≤ −1. We write x = −ξu and obtain

y ≤ \sqrt{1 + 2x + u^{2}} ∫_{0}^{u} \frac{ds}{1 + 2xs/u} ≤ \sqrt{2 + 2x} \frac{u}{2x} log(1 + 2x) ≤ c_{6}u.

Case II. Let −1 < ξ < 1. We can write

(3.6)  1 − 2ξs + s^{2} = (1 − s)^{2} + 2(1 − ξ)s ≥ \frac{1}{2}(1 − s)^{2} + \frac{1}{2}(1 − ξ).

First let (1 − u)^{2} < 1 − ξ. Then, by (3.6),

(3.7)  1 − 2ξu + u^{2} = (1 − u)^{2} + 2(1 − ξ)u < 3(1 − ξ).

If u ≥ \frac{1}{2} then, with y defined in (3.5), we see from (3.6) that

y < ∫_{0}^{u} \frac{2\sqrt{3(1 − ξ)} ds}{(1 − s)^{2} + (1 − ξ)} < ∫_{0}^{∞} \frac{2\sqrt{3}}{t^{2} + 1} dt ≤ 2π\sqrt{3} u.

If u < \frac{1}{2} then \sqrt{1 − ξ} > 1 − u > \frac{1}{2} and thus, by (3.6),

y < ∫_{0}^{u} \frac{2\sqrt{3(1 − ξ)} ds}{1 − ξ} < 4\sqrt{3} u.

Now let (1 − u)^{2} ≥ 1 − ξ. Then 1 − 2ξu + u^{2} < 3(1 − u)^{2} as in (3.7) and thus, by (3.6),

y < ∫_{0}^{u} \frac{2\sqrt{3} (1 − u)}{(1 − s)^{2}} ds = 2\sqrt{3} u.

Case III. Let 1 ≤ ξ < +∞. We define

(3.8) η = \sqrt{ξ^{2} − 1}, v = ξ − η = 1/(ξ + η).

Then 1 − 2ξv + v^{2} = 0 and therefore u < v by (3.4). This time we can write

(3.9)  1 − 2ξs + s^{2} = (v − s)^{2} + 2η(v − s).

First let v − u ≤ η. Then, by (3.9),

(3.10)  1 − 2ξu + u^{2} = (v − u)^{2} + 2η(v − u) ≤ 3η(v − u).

If u ≥ \frac{1}{2} then, by (3.9),

∫_{0}^{u} \frac{ds}{1 − 2ξs + s^{2}} ≤ ∫_{0}^{v−η} \frac{ds}{(v − s)^{2}} + ∫_{v−η}^{u} \frac{ds}{2η(v − s)} ≤ \frac{1}{η} + \frac{1}{2η} log \frac{η}{v − u};

if v−η < 0 then the first integral is omitted. We deduce from (3.10) and v−u ≤ η that

y ≤ \sqrt{\frac{v − u}{η}} (\sqrt{3} + \frac{\sqrt{3}}{2} log \frac{η}{v − u}) ≤ c_{7} ≤ 2c_{7}u.

On the other hand, if u < \frac{1}{2} then, by (3.8),

2\sqrt{ξ^{2} − 1} = 2η = ξ − (v − η) ≥ ξ − u > ξ − \frac{1}{2} ≥ \frac{1}{2}

which implies ξ ≥ ξ_{1} = \sqrt{17}/4 > 1. Hence (3.9) and (3.10) show that

y ≤ \frac{\sqrt{3η(v − u)}}{2η} ∫_{0}^{u} \frac{ds}{v − s} ≤ \frac{\sqrt{3}}{2\sqrt{η}} ∫_{0}^{u} \frac{ds}{\sqrt{v − s}} = \sqrt{\frac{3}{η}} \frac{u}{\sqrt{v} + \sqrt{v − u}} < \sqrt{\frac{3}{vη}} u

and (3.5) follows because, by (3.8),

\frac{1}{vη} = \frac{ξ}{\sqrt{ξ^{2} − 1}} + 1 ≤ \frac{ξ_{1}}{\sqrt{ξ_{1}^{2} − 1}} + 1.

Now let v − u > η. Then (3.9) shows that 1 − 2ξu + u^{2} < 3(v − u)^{2} and thus, by (3.9),

y < \sqrt{3} (v − u) ∫_{0}^{u} \frac{ds}{(v − s)^{2}} = \sqrt{3} \frac{u}{v}.

We obtain from (3.8) that ξ − η = v > η + u ≥ η and therefore ξ > 2η which implies ξ ≤ \sqrt{4/3} . Hence we have

y ≤ \sqrt{3} \frac{u}{v} = \sqrt{3} (ξ + \sqrt{ξ^{2} − 1} )u < 4u.

This completes the proof of (3.5) and thus of assertion (1.8) of Theorem 1. (c) Now we prove assertion (1.7). Let z ∈ D be fixed and let |b| = 1,

(3.11)  φ(t) = \frac{t + z}{1 + z̄t},  g(t) = b\frac{f(φ(t)) − f(z)}{1 − \overline{f(z)}f(φ(t))}  (t ∈ D).

Then g is again h-convex and moreover

g(0) = 0, g′(0) = (1 − |z|^{2})|f′(z)|/(1 − |f(z)|^{2}),

if b ∈ T is suitably chosen. Applying (1.8) to g, we obtain from (3.11) that

\frac{(1 − |z|^{2})|f′(z)|(1 − |φ(t)|^{2})|f′(φ(t))|}{|f(z) − f(φ(t))|^{2}}λ(t, 0)^{2} = g′(0)\frac{(1 − |t|^{2})|g′(t)|}{|g(t)|^{2}}λ(t, 0)^{2} ≤ c_{1}

which implies (1.7) because λ(t, 0) = λ(φ(t), φ(0)) = λ(φ(t), z).

4. Proof of Theorem 2

First let f(ς) ∈ D. We prove that the limit b(ς) exists and is = 0. We may assume that f has the form (1.3). It follows from [MP1, (2.6)] that

(1 − r^{2})|f′(rς)| ≤ 4|f(ς) − f(rς)| ≤ 8(1 − r)^{α^{2}δ/4}

for r close to 1, where δ = 1 − |f(ς)|^{2} > 0.

Now let f(ς) ∈ T. We may assume that ς = 1 and f(ς) = 1; otherwise we consider the h-convex function \overline{f(ς)}f(ςz). For the proof of (1.9) we may furthermore assume that there exist r_{n} → 1 such that

(4.1)  (1 − r_{n}^{2})|f′(r_{n})|λ(r_{n}, 0)^{2} → b ≠ 0  as n → ∞;

see (1.8). By Theorem 1, we have,

\frac{(1 − |z|^{2})|f′(z)|(1 − r_{n}^{2})|f′(r_{n})|}{|f(r_{n}) − f(z)|^{2}}λ(r_{n}, z)^{2} ≤ c_{1}.

Since λ(r_{n}, z) ∼ λ(r_{n}, 0) as n → ∞, we obtain from (4.1) that

(4.2)  \frac{(1 − |z|^{2})|f′(z)|}{|1 − f(z)|^{2}} ≤ \frac{c_{1}}{b}  for z ∈ D.

Ma and Minda [MM1, Theorem 4] have shown that the function

(4.3)  p(z) = \frac{(1 − f(z))^{2}}{(1 − z)^{2}f′(z)}  (z ∈ D)

satisfies Re p > 0. The Julia–Wolff lemma shows that

(4.4)  \frac{(1 − f(z))^{2}}{(1 − z^{2})f′(z)} = \frac{1 − z}{1 + z} p(z) → a  as z → 1, z ∈ Δ

for every Stolz angle Δ, where 0 ≤ a < +∞. It follows from (4.2) that a ≥ b/c_{1} > 0. We obtain from (4.4) that

\frac{f′(z)}{(1 − f(z))^{2}} = (\frac{1}{a} + o(1))\frac{1}{1 − z^{2}} as z → 1, z ∈ Δ,

and by integration we deduce that

(4.5)  \frac{1}{1 − f(z)} = (\frac{1}{a} + o(1))L(z)  as z → 1, z ∈ Δ,

where

L(z) = \frac{1}{2} log \frac{1 + z}{1 − z} .

Together with (4.4) this implies

(1 − z^{2})f′(z)L(z)^{2} → a as z → 1, z ∈ Δ.

Hence the limit (1.9) exists and b(1) = a. The assertions (1.10) and (1.11) now follow from (4.5) and (4.4).

5. Proofs of Theorems 3 and 4

Let c_{8}, c_{9}, … denote suitable positive absolute constants.

Proposition 5. Let f be h-convex and f(0) = 0, |f′(0)| = α. Let ς ∈ T, f(ς) ∈ T, and ω ∈ T ∩ f(T) with ω ≠ f(ς). For given r ∈ (0, 1), there are only two possibilities:

(i) |f(ς) − f(rς)| < c_{8}|f(ς) − ω|,

(ii) (1 − r^{2})|f′(rς)| > c_{9} min(α, |f(ς) − ω|).

Proof. We may assume that ς = 1 and f(ς) = 1. Let G = f(D). We write

(5.1)  q = |f(ς) − ω| = |1 − ω|.

There exists a smooth crosscut Q = Q(r) of G with f(r) ∈ Q that separates 0 from 1 and satisfies

(5.2)  length Q < c_{10}(1 − r^{2})|f′(r)|.

Let C be the circle orthogonal to T through 1 and ω. Since 0 ∈ G and 1, ω ∈ ∂G and since G is h-convex, the non-euclidean triangle T bounded by [0, 1], [0, ω] and D ∩ C lies in G.

The crosscut Q has to meet [0, 1] because Q separates 0 and 1. Let A^{±} be the arcs of Q from the last points of intersection with [0, 1] to ∂G. Then A^{+} and A^{−} go to different sides of [0, 1]. Hence one of these arcs, say A^{+} , has to enter the triangle T. Since T ⊂ G the endpoint of A^{+} on ∂G cannot lie in T. Hence A^{+} has to meet C or [0, ω].

First we consider the case that there exists a ∈ A^{+} ∩ C ⊂ Q ∩ C. Since f(r) ∈ Q we obtain from (5.1) and (5.2) that

|1 − f(r)| ≤ |1 − a| + |a − f(r)| ≤ q + c_{10}(1 − r^{2})|f′(r)|.

If (1 − r^{2})|f′(r)| < q then (i) holds; if (1 − r^{2})|f′(r)| ≥ q then (ii) holds trivially by (5.1).

Now we consider the case that there exists a ∈ A^{+} ∩ [0, ω]. If 0 ≤ r < \frac{1}{2} then

(1 − r^{2})|f′(r)| ≥ c_{11}|f′(0)| = c_{11}α

by the Koebe distortion theorem. Hence (ii) holds.

Hence we may assume that \frac{1}{2} ≤ r < 1. Then |f(r)| ≥ c_{12}α. Now A^{+} ⊂ Q. Thus Q intersects [0, 1] and [0, ω] and furthermore f(r) ∈ Q. It follows that

length Q ≥ max[dist (f(r), [0, 1]), dist (f(r), [0, ω])] ≥ c_{13} min(α, |1 − ω|).

Hence (ii) holds by (5.2).

Proof of Theorem 3. Let c_{8} and c_{9} be the constants of the proposition and put c_{2} = c_{9}/(2c_{8}), c_{3} = c_{2}c_{8} . It follows from (i), (ii) and (2.1) that, for every fixed r ∈ [r_{0}, 1) there are only the two cases

(i′) |f(ς) − f(rς)| < c_{8} min(α, |f(ς) − ω|),

(ii′) |f(ς) − f(rς)| > 2c_{8} min(α, |f(ς) − ω|).

Since |f(ς) − f(rς)| is continuous in [r_{0}, 1), we conclude that either (i′) or (ii′) holds for all r ∈ [r_{0}, 1). But (ii′) is impossible for r close to 1. Hence (i′) holds, which implies our assertion (2.2) with c_{4} = 1/c_{8} .

Proof of Theorem 4. Let ς, f(ς) ∈ T and b(ς) > 0. It follows from (4.2) (where f(ς) = 1) that

(5.3)  (1 − r^{2})|f′(rς)| ≤ \frac{c_{1}}{b(ς)}|f(ς) − f(rς)|^{2}.

Let c_{2} , c_{3} , c_{4} be the constants of Theorem 3 and let

(5.4)  s = s(ς) min(\frac{c_{2}}{c_{1}} b(ς), \frac{c_{3}}{c_{2}} α, α).

Since |f(ς) − f(rς)| = 1 ≥ α for r = 0 and = 0 for r = 1, there exists r_{0} ∈ [0, 1) such that

(5.5)  |f(ς) − f(rς)| ≤ s  for r_{0} ≤ r < 1,  = s  for r = r_{0}.

We obtain from (5.3), (5.4) and (5.5) that, for r_{0} ≤ r < 1,

(1 − r^{2})|f′(rς)| ≤ \frac{c_{1}s}{b(ς)}|f(ς) − f(rς)| ≤ c_{2}|f(ς) − f(rς)|.

Furthermore we see from (5.4) and (5.5) that

c_{2}|f(ς) − f(rς)| ≤ c_{2}s ≤ c_{3}α.

Hence it follows from Theorem 3 and (5.5) that

|ω − f(ς)| ≥ c_{4}|f(ς) − f(r_{0}ς)| = c_{4}s

for ω ∈ T ∩ f(T), ω ≠ f(ς).

The open set T\f(T) is the union of disjoint open arcs I_{n}. We have just shown that, for every ς with b(ς) > 0, there exists n = n(ς) such that f(ς) is the right-hand endpoint of I_{n} and

(5.6)  diam I_{n} ≥ c_{4}s(ς).

56 Diego Mejía and Christian Pommerenke

Let X be the set of ς for which s(ς) = c_{2}b(ς)/c_{1} and let Y be the set for which s(ς) < c_{2}b(ς)/c_{1} ; see (5.4). Then, by (5.6),

∑_{ς∈X} b(ς) ≤ \frac{c_{1}}{c_{2}c_{4}} ∑_{n} diam I_{n} < \frac{2πc_{1}}{c_{2}c_{4}} = c_{14},

∑_{ς∈Y} c_{4} min(\frac{c_{3}}{c_{2}} α, α) ≤ ∑_{n} diam I_{n} < 2π

and thus card Y < c_{15}/α, furthermore b(ς) ≤ c_{1}/α by (1.8) and (1.9). Hence

∑_{b(ς)>0} b(ς) < c_{14} + \frac{c_{1}c_{15}}{α^{2}} .

This proves (2.3). In particular we can have b(ς) > 0 only for countably many ς.

References

[CP] Chuaqui, M., and Ch. Pommerenke: Characteristic properties of Nehari functions. - Pacific J. Math. 188, 1999, 83–94.

[MM1] Ma, W., and D. Minda: Hyperbolically convex functions. - Ann. Polon. Math. 60, 1994, 81–100.

[MM2] Ma, W., and D. Minda: Hyperbolically convex functions II. - Ann. Polon. Math. 71, 1999, 273–285.

[Mc] McMillan, J.E.: Boundary behaviour of a conformal map. - Acta Math. 123, 1969, 43–67.

[MP1] Mejía, D., and Ch. Pommerenke: Sobre aplicaciones conformes hiperbólicamente convexas. - Rev. Colombiana Mat. 32, 1998, 29–43.

[MP2] Mejía, D., and Ch. Pommerenke: On hyperbolically convex functions. - J. Geom. Anal. 10, 2000, 361–374.

[MPV] Mejía, D., Ch. Pommerenke and A. Vasil′ev: Distortion theorems for hyperbolically convex functions. - Complex Variables 44, 2001, 117–130.

[P] Pommerenke, Ch.: Boundary Behaviour of Conformal Maps. - Springer-Verlag, Berlin, 1992.

Received 16 May 2000