Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 18, 1993, 249–271

UNIVALENT FUNCTIONS OF GIVEN TRANSFINITE DIAMETER: A MAXIMUM MODULUS PROBLEM

Robin Cunningham

Trenton State College, Department of Mathematics and Statistics Hillwood Lakes, CN 4700, Trenton, NJ 08650-4700, U.S.A.

Abstract. A variational method is used to find the maximum of |f(ζ)| for fixed ζ ∈ D as f ranges over the class of normalized univalent functions which map the unit disk D onto regions of prescribed transfinite diameter (logarithmic capacity).

0. Introduction

Let S be the usual class of functions f(z) = z + a_{2}z^{2} + ⋯ that are analytic and univalent in the unit disk D. We say f ∈ S_{R} if f ∈ S and the image f(D) of the unit disk under f has transfinite diameter R. We consider the following problem: given a point ζ ∈ D and a transfinite diameter R (1 < R < ∞), find the maximum of |f(ζ)| among all functions f ∈ S_{R}. In the solution to this problem, it turns out that the extremal function has as its range one of five geometrically distinct types. In each case, the boundary of the range lies in a trajectory of a determined quadratic differential. Depending on R and ζ, the extremal values of |f(ζ)| are given either explicitly or parametrically in terms of complete elliptic integrals. For each choice of R and ζ, the extremal function is unique. The proof uses a special variational method devised by Duren and Schiffer [6] in their recent paper on a related problem.

For basic facts about transfinite diameter, see Goluzin [7, Chapter VII], Hille [8, Chapter 16], Pólya–Szegő [10, Part IV, Chapter 2] or Tsuji [14]. Here we note only that if R > 0 is the transfinite diameter of a compact set E in the complex plane and if

g(w) = g(w, ∞) = log |w| + γ + O(\frac{1}{w} )

is Green’s function (with pole at infinity) for the unbounded component of the complement of E, then R = e^{−γ}, the logarithmic capacity of E. A set E with R < ∞ is clearly bounded.

1991 Mathematics Subject Classification: Primary 30C45, 30C70, 30C85, 33C25.

This paper constitutes a portion of the author’s Ph.D. thesis at the University of Michigan under the supervision of Professor Peter Duren.

1. Existence of solution

Given R (1 < R < ∞) and ζ ∈ D, we wish to determine the maximum of |f(ζ)| for f ∈ S_{R}. Without loss of generality we may assume that 0 < ζ < 1, since other values ζ may then be treated via rotations of D. In order to employ a variational method, we must first prove the existence of an extremal function. It is easily seen that S_{R} is a normal family. Although various pinching constructions suggest that S_{R} is not compact, Duren and Schiffer [6] have shown that the larger family Ŝ_{R} = ∪_{Q≤R}S_{Q} is a compact normal family. This ensures that the continuous functional |f(ζ)| attains a maximum over the class Ŝ_{R}. Since the extremal values for |f(ζ)| will be shown to be increasing with R, the maximum of |f(ζ)| over the class Ŝ_{R} is actually attained within the subclass S_{R}. Thus we are assured of the existence of an extremal function within the class S_{R} itself.

2. The Duren–Schiffer variation

A special variation devised by Duren and Schiffer to preserve the class S_{R} will be used to find the extremal functions. First an interior variation will be used to determine (p′(w))^{2} , where p(w) is the analytic completion of Green’s function for the complement of \overline{f(D)}. Consider the perturbation

(1)  w^{∗} = w + ∑_{j=1}^{3} ε_{j} \frac{w}{w − w_{j}} ,

where w_{1}, w_{2}, w_{3} ∉ \overline{f(D)} are all distinct and the ε_{j} are small complex parameters. For ε ≡ max_{j} |ε_{j}| sufficiently small, it is easily seen that

(2)  f^{∗}(z) = f(z) + ∑_{j=1}^{3} ε_{j} \frac{f(z)}{f(z) − w_{j}} = ∑_{n=1}^{∞} a_{n}^{∗}z^{n}

is univalent in D. Notice that

a_{1}^{∗} = 1 − ∑_{j=1}^{3} ε_{j}w_{j}^{−1};

therefore, in order to ensure that f^{∗} ∈ S, we set

(3)  ∑_{j=1}^{3} ε_{j}w_{j}^{−1} = 0.

Now, using the method of interior variation, one may calculate an expression for Green’s function g^{∗}(w, ∞) of the complement of \overline{f^{∗}(D)}. Hence one determines the following relation between R and the transfinite diameter R^{∗} of the region f^{∗}(D):

R^{∗} = R(1 − Re { ∑_{j=1}^{3} ε_{j}w_{j}p′(w_{j})^{2}}) + O(ε^{2}).

(See [13] for a derivation of this formula.) For the above expression, p(w) = p(w, ∞) is the multiple-valued analytic completion of g(w, ∞), Green’s function of the complement of \overline{f(D)}. We wish to ensure that R^{∗} = R, so we set

(4)  ∑_{j=1}^{3} ε_{j}w_{j}p′(w_{j})^{2} = 0

and modify (1) by a term of order ε^{2} (cf. Duren and Schiffer [6]).

Without affecting (3) and (4) we may replace ε_{j} in (1) by ϱe^{iθ}ε_{j}, where ϱ > 0 and θ is an arbitrary real constant. Modifying the variation (2) accordingly, we obtain f ∈ S_{R} and

f^{∗}(ζ) = f(ζ)(1 + ϱe^{iθ} ∑_{j=1}^{3} \frac{ε_{j}}{f(ζ) − w_{j}} ) + O(ϱ^{2}).

It follows that

log f^{∗}(ζ) = log f(ζ) + ϱe^{iθ} ∑_{j=1}^{3} \frac{ε_{j}}{f(ζ) − w_{j}} + O(ϱ^{2}).

For any extremal function f, we know log |f^{∗}(ζ)| ≤ log |f(ζ)|, which implies

Re \{ϱe^{iθ} ∑_{j=1}^{3} \frac{ε_{j}}{f(ζ) − w_{j}} + O(ϱ^{2})\} ≤ 0.

But this is true for arbitrary θ, so

(5)  ∑_{j=1}^{3} \frac{ε_{j}}{f(ζ) − w_{j}} = 0.

(Equation (5) may also be obtained merely by comparing |f^{∗}(ζ)| with |f(ζ)| and making a judicious choice of θ.)

Equations (3), (4) and (5) provide three linear homogeneous equations in ε_{1} , ε_{2} and ε_{3} with a nontrivial solution. This implies the vanishing of a determinant:

(6)  | \begin{matrix}w_{1}^{−1} & w_{2}^{−1} & w_{3}^{−1} \\ (f(ζ) − w_{1})^{−1} & (f(ζ) − w_{2})^{−1} & (f(ζ) − w_{3})^{−1} \\ w_{1}p′(w_{1})^{2} & w_{2}p′(w_{2})^{2} & w_{3}p′(w_{3})^{2} \end{matrix} | = 0.

Every extremal function f satisfies this condition, which remains valid for any choice of distinct points w_{j} ∉ \overline{f(D)}. Setting w_{1} = w and expanding the determinant (6) along the first column, we get

λ_{1}w^{−1} + λ_{2}(f(ζ) − w)^{−1} + λ_{3}wp′(w)^{2} = 0.

Since w_{2} ≠ w_{3} , we see that λ_{3} ≠ 0, so we have

(7)  p′(w)^{2} = aw^{−2} + \frac{b}{(f(ζ) − w)w} .

Now, since p(w) is the analytic completion of Green’s function, it has the form

p(w) = log w + c_{0} + c_{1}w^{−1} + c_{2}w^{−2} + ⋯

near infinity; therefore,

(8)  p′(w)^{2} = w^{−2} − 2c_{1}w^{−3} + · · · .

After a short calculation comparing (7) and (8), we obtain a = b + 1 and

(9)  p′(w)^{2} = \frac{(b + 1)B − w}{w^{2}(B − w)} ,

where B = f(ζ).

From the fact that p(w) is the analytic completion of Green’s function, we know that Re \{p(w)\} = 0 for w in the boundary of \overline{f(D)}. As we shall see later, the boundary of \overline{f(D)} consists of analytic arcs so that p′(w) has an extension to the boundary. This implies that [p′(w)^{2}w′(t)^{2}] < 0 on the boundary of \overline{f(D)}, which suggests that the “outer boundary” of f(D) is composed of arcs w = w(t) lying on trajectories of the quadratic differential

(10)  \frac{(b + 1)B − w}{(w − B)w^{2}} dw^{2} > 0.

As we shall see, the entire boundary of f(D) consists of arcs w = w(t) which satisfy (10). To prove this we shall apply a special boundary variation developed by Duren and Schiffer [6] which preserves the class S_{R}. This additional variation is necessary partly because p(w) will not detect any internal boundary components of f(D) (e.g., internal spires) since Green’s function for the complement of \overline{f(D)} is independent of such boundary components.

Let Γ be the boundary of f(D). Choose an arbitrary w_{0} ∈ Γ and two distinct arbitrary points w_{1}, w_{2} ∉ \overline{f(D)}. We modify the standard boundary variation (see [12] or [15, Chapter 10])

w^{∗} = w + \frac{aϱ^{2}w}{w_{0}(w − w_{0})} + O(ϱ^{3})

by including two additional terms to produce the variation

(11)  w^{∗} = V_{ϱ}(w) = w + \frac{aϱ^{2}w}{w_{0}(w − w_{0})} + \frac{ε_{1}w}{w − w_{1}} + \frac{ε_{2}w}{w − w_{2}} + O(ϱ^{3}),

where ε_{1} , ε_{2} are of order ϱ^{2} and will be specified later. We make the further assumption that a(ϱ) is bounded away from 0; the reason for this will be made evident later. The variation (11) is analytic and univalent off the union of a small subcontinuum of Γ and two small disks about w_{1} and w_{2} respectively. If we let f^{∗} ≡ V_{ϱ} ∘ f, then f^{∗} is analytic and univalent on D and f^{∗}(0) = 0. By requiring that V′_{ϱ}(0) = 1, we ensure that f^{∗} ∈ S. Thus we require that

(12)  aϱ^{2}w_{0}^{−2} + ε_{1}w_{1}^{−1} + ε_{2}w_{2}^{−1} = 0.

Using the method of interior variation to determine the change in Green’s function [12], we calculate the transfinite diameter of R^{∗} in terms of R:

R^{∗} = R [1 − Re {ε_{1}w_{1}p′(w_{1})^{2} + ε_{2}w_{2}p′(w_{2})^{2}}] + O(ϱ^{3}).

We also require that R^{∗} = R, so we set

(13)  ε_{1}w_{1}p′(w_{1})^{2} + ε_{2}w_{2}p′(w_{2})^{2} = 0

and modify (11) by a term of order ϱ^{3} (see [6]) to produce a variation with the properties R^{∗} = R and f^{∗} ∈ S. Thus we have f^{∗} ∈ S_{R}.

Combining equations (9), (12) and (13), we arrive at the equation

(14)  aϱ^{2}w_{0}^{−2} = b[\frac{ε_{1}B}{w_{1}(B − w_{1})} + \frac{ε_{2}B}{w_{2}(B − w_{2})}].

Since w_{0} ≠ ∞ and since we have assumed a ≠ 0, we see from (14) that b ≠ 0. Also, we see from (11) that

(15)  log f^{∗}(ζ) = log B + \frac{aϱ^{2}}{w_{0}(B − w_{0})} + \frac{ε_{1}}{B − w_{1}} + \frac{ε_{2}}{B − w_{2}} + O(ϱ^{3}).

After some calculation involving (12), (14) and the fact that b ≠ 0, we infer from (15) that

log f^{∗}(ζ) = log B + \frac{aϱ^{2}}{b} [\frac{(b + 1)B − w_{0}}{w_{0}^{2}(B − w_{0})} ] + O(ϱ^{3}).

If f is extremal, then Re { log f^{∗}(ζ)} ≤ Re{log B}, which implies

Re \{\frac{aϱ^{2}}{b} [\frac{(b + 1)B − w_{0}}{w_{0}^{2}(B − w_{0})} ] + O(ϱ^{3})\} ≤ 0.

We now invoke Schiffer’s theorem (see [5, p. 297]) to conclude that Γ consists of analytic arcs lying on trajectories of the quadratic differential

(16)  \frac{1}{b}[\frac{(b + 1)B − w}{w^{2}(B − w)} ] dw^{2} > 0.

We note here that our assumption that a is bounded away from 0 goes beyond the hypothesis of Schiffer’s theorem. However, Schiffer’s result is actually stronger than the statement of his theorem indicates, because the proof goes through without modification if we assume a ≠ 0 (see [5, p. 297–302]. Since Γ consists of analytic arcs, the quadratic differentials (10) and (16) must both be positive on the outer boundary of f(D). Thus b < 0 for every choice of ζ and R and we have established that the entire boundary of f(D) consists of arcs lying on trajectories of (10).

3. Analysis of the quadratic differential

The quadratic differential (10) has a simple zero at w = (b + 1)B, a double pole at the origin and infinity, and a simple pole at w = B. (Note that b ≠ 0 implies (b + 1)B ≠ B.)

It is not hard to see that the trajectories of (10) are symmetric in the real axis. First, symmetry in the line w = Bt can be seen by making the substitution w = Bω and recognizing the resulting symmetry in the real axis of the ω -plane. Next, note that f′(0) = 1 and consider the image under f of the curve z = ζt, (0 ≤ t ≤ 1). This image curve has positive direction at the origin and contains B. A symmetry argument enables us to conclude that the image curve is in fact the line segment joining the origin with B in the ω -plane. Therefore the line w = Bt, whose tangent direction is of course a constant, must lie in the real axis. We now see that B is real and positive and that all trajectories are symmetric in the real axis. We are now able to show the trajectories for (10) in Figures 1, 2, and 3 below. (See Chapter 8 in [11] for the local trajectory structure of quadratic differentials.)

Note that of the trajectories pictured, only those surrounding B and the origin may form boundaries of the images of extremal functions f. All the candidate

Figure 1. Sketch of trajectories of (10) for −1 < b < 0.

Figure 2. Sketch of trajectories of (10) for b = −1.

solution regions pictured do in fact occur as boundaries of extremal functions for suitable values of R and ζ.

We now parametrize Γ by w(t) = f(e^{it}) and use (10) to obtain

(18)  F (z) = \frac{z^{2}f′(z)^{2}}{f(z)^{2}} \frac{[(b + 1)B − f(z)]}{B − f(z)} ≥ 0,  |z| = 1.

Since f ∈ S, it is clear that F is analytic in D except for a removable singularity at z = 0 and a simple pole at z = ζ with residue

a_{−1} = \frac{−bζ^{2}f′(ζ)}{f(ζ)}.

Figure 3. Sketch of trajectories of (10) for b < −1.

In order to arrive at a convenient expression for F (z), one shows that F (z) is analytic and non-vanishing on |z| = 1 except for at most one double zero when Γ has an internal spire or a corner. (The proof for this follows just as in [6, Section 3]). Then by the Schwarz reflection principle, F (z) has a meromorphic extension to the Riemann sphere satisfying F (1/z̄) = \overline{F (z)}. Therefore F (z) has the form

(19)  F (z) = \frac{A(z − z_{0})(1 − zz_{0})}{(z − ζ)(1 − ζz)}

where f(z_{0}) = (b + 1)B when w = (b + 1)B is inside Γ. When w = (b + 1)B lies on Γ, we have z_{0} = 1. A symmetry argument shows that the point z_{0} in (19) is real and negative. Therefore we may conclude from the Schwarz reflection principle that A is real. The point (b + 1)B is in \overline{f(D)} for each of the trajectories above which represent possible solutions.

As suggested by the illustrations above, our work breaks into three cases which we handle in the order (i) b = −1; (ii) −1 < b < 0; (iii) b < −1.

4. The case b = −1: ellipses

In this case, (b + 1)B = 0 and z_{0} = 0 so that (19) becomes

(20)  F (z) = \frac{Az}{(z − ζ)(1 − ζz)}

and (10) becomes

(21)  \frac{dw^{2}}{w(B − w)} ≥ 0.

The trajectories of (21) are ellipses with foci w = 0 and w = B [4, p. 309–310]. A comparison of the equations (18), (20) and (21) establishes the relation

(22)  ∫_{0}^{B} \frac{dw}{\sqrt{w(B − w)}} = \sqrt{A} ∫_{0}^{ζ} \frac{dz}{\sqrt{z(ζ − z)(1 − ζz)}}.

An elementary calculation shows that the left-hand integral is equal to π, while the substitution u = \sqrt{z/ζ} shows that the right-hand side is equal to 2\sqrt{A}K(ζ) where

K(k) = ∫_{0}^{1} \frac{du}{\sqrt{(1 − u^{2})(1 − k^{2}u^{2})}}, 0 < k < 1,

is the normal complete elliptic integral of the first kind. Thus (22) allows us to evaluate A = \frac{1}{4}π^{2} · K(ζ)^{−2} . On the other hand, a comparison of (18) and (20) gives

lim_{z→0} \frac{F (z)}{z} = −\frac{A}{ζ} = −\frac{1}{B} .

Consequently, for the case b = −1, we conclude that

(23)  B = \frac{4ζK(ζ)^{2}}{π^{2}} .

This expression is independent of R because, as we shall see, the ellipse case (b = −1) occurs for only one value of R for each choice of ζ.

By calculating G(w), the analytic completion of Green’s function for the complement of the solution ellipses, we shall now obtain a relation between R and B which determines for which values of R the ellipse solution may occur.

Let w_{0} = f(−1) (w_{0} < 0 by symmetry considerations), and define

(24) G_{0}(w) ≡ ∫_{w_{0}}^{w} \frac{dw}{\sqrt{w(w − B)}} = ∫_{0}^{w} \frac{dw}{\sqrt{w(w − B)}}−∫_{0}^{w_{0}} \frac{dw}{\sqrt{w(w − B)}} ≡ I_{1}−I_{2}.

It is easily seen that

I_{1} = log (w\sqrt{w(w − B)} + 2w − B) − log(−B)

where the chosen branch of the logarithm is unimportant since we will be concerned only with Re {G_{0}(w)}. Note that by (18) and (20)

I_{2} = \sqrt{A} ∫_{0}^{−1} \frac{dz}{\sqrt{z(z − ζ)(1 − ζz)}}

and substitute z = (t + 1)/(t − 1), τ = (1 + ζ)/(1 − ζ) to conclude

I_{2} = \frac{−2\sqrt{A}}{1 + ζ}K(\frac{1}{τ} ).

Thus (24) takes the form

G_{0}(w) = log 4w − log(−B) + \frac{2\sqrt{A}}{1 + ζ} K(\frac{1}{τ} ) + O(\frac{1}{w} ).

Since the boundary of f(D) lies on a trajectory of the quadratic differential (21),

(25)  g(w) = Re \{G_{0}(w)\} = log |w| − log \frac{B}{4} + \frac{2\sqrt{A}}{1 + ζ} K(\frac{1}{τ} ) + O(\frac{1}{w} )

is harmonic in the complement of \overline{f(D)} except for a logarithmic singularity at infinity and is constant on the boundary. But G_{0}(w_{0}) = 0, so g(w) = 0 on the boundary. This shows that g is Green’s function of the complement of \overline{f(D)}. The equivalence of transfinite diameter and logarithmic capacity (see Section 0) now shows by (25) that

(26)  R = \frac{B}{4} exp [\frac{2\sqrt{A}}{1 + ζ} K(\frac{1}{τ} )] = \frac{ζK(ζ)^{2}}{π^{2}} exp [\frac{π}{(1 + ζ)K(ζ)}K(\frac{1}{τ} )].

Thus the ellipse solution may occur only for the single value of R satisfying (26). This solves the case b = −1. For information on the explicit form of functions f(z) which map the unit disk conformally onto the interior of an ellipse, see [9, Chapter VII, Section 3].

5. The case −1 < b < 0

In the case −1 < b < 0, Γ is an analytic Jordan curve surrounding w = 0, w = B and w = (b + 1)B. In this case we return to (19) in its original form

(19)  F (z) = \frac{A(z − z_{0})(1 − zz_{0})}{(z − ζ)(1 − ζz)}

where f(z_{0}) = (b + 1)B and 0 < z_{0} < ζ. Now upon considering (18) along with (19) we have four parameters; B, A, b, z_{0} . To express them in terms of R and ζ we require four relations among the variables.

Relation 1. Equate the expressions (18) and (19) for F (z) and let z approach 0 to get

(27)  (b + 1) = \frac{Az_{0}}{ζ} ,  (A > 0).

Relation 2. We find the analytic completion H(w) of Green’s function for the complement of \overline{f(D)} to establish a relation among R, b, B and ζ. If w_{1} = f(1), then

H(w) = G(w) − G(w_{1})

where

G(w) = ∫_{B}^{w} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w} .

A calculation now reveals that

(28)  G(w) = log w + log \frac{−4}{bB} + \sqrt{b + 1} log (\frac{1 − \sqrt{b + 1}}{1 + \sqrt{b + 1}}) + O(\frac{1}{w} )

and

(29)  G(w_{1}) = ∫_{B}^{w_{1}} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w} = ∫_{ζ}^{1} \sqrt{\frac{A(z − z_{0})(1 − z_{0}z)}{(z − ζ)(1 − ζz)}} \frac{dz}{z} .

We may substitute t = (z−1)/(z+1), τ = (1+ζ)/(1−ζ) and t_{0} = (1+z_{0})/(1−z_{0}) in (29) to obtain

G(w_{1}) = \sqrt{A} \frac{2(1 + z_{0})}{1 + ζ} ∫_{−1/τ}^{0} \frac{(t^{2} − t_{0}^{−2}) dt}{(t^{2} − 1)\sqrt{(t^{2} − t_{0}^{−2})(t^{2} − τ^{−2})}}.

Now let ω = −τt. This gives

(30) G(w_{1}) = \sqrt{A}[\frac{2(1 − z_{0})t_{0}^{2}}{1 + ζ} K(\frac{t_{0}}{τ} ) − \frac{8z_{0}}{(1 − z_{0})(1 + ζ)}Π(\frac{1}{τ^{2}} , \frac{t_{0}}{τ} )],

where

Π(α^{2}, k) = ∫_{0}^{1} \frac{dt}{(1 − α^{2})\sqrt{(1 − t^{2})(1 − k^{2}t^{2})}},

the normal complete elliptic integral of the third kind. Therefore,

H(w) = log w + log \frac{−4}{bB} + \sqrt{b + 1} log (\frac{1 − \sqrt{b + 1}}{1 + \sqrt{b + 1}}) − G(w_{1}) + O(\frac{1}{w} ).

An argument similar to that used for the ellipse case shows that Re {H(w)} is Green’s function for the complement of \overline{f(D)}. Thus if G(w_{1}) is given by (30), then

(31)  log R = log \frac{−bB}{4} + \sqrt{b + 1} log (\frac{1 + \sqrt{b + 1}}{1 − \sqrt{b + 1}}) + G(w_{1}).

Relation 3. We use the equality

(32)  ∫_{(b+1)B}^{B} \sqrt{\frac{(b + 1)B − w}{w − B}} \frac{dw}{w} = ∫_{z_{0}}^{ζ} \sqrt{\frac{A(z − z_{0})(1 − z_{0}z)}{(ζ − z)(1 − ζz)}} \frac{dz}{z} .

The left side of (32) is easily seen to be π(1 − \sqrt{b + 1} ) while the right side is an elliptic integral. We refer to formulas 253.00, 253.11, 253.12 and 340.01 in [3] to conclude that the right side of (32) is equal to

(33) \frac{2\sqrt{A}}{z_{0}^{−1} − ζ} [(\frac{1}{ζ} −ζ)Π(α^{2}, k)+(ζ − \frac{1}{ζ})Π(\frac{α^{2}}{ζ^{2}} , k)+(\frac{1}{z_{0}} +z_{0} − \frac{1}{ζ} −ζ)K(k)],

where

α^{2} = \frac{ζ − z_{0}}{ζ^{−1} − z_{0}} , k^{2} = α^{2} · \frac{1 − ζ^{−1}z_{0}}{1 − ζz_{0}} .

A special addition formula for elliptic integrals of the third kind ([3, 117.01]) allows us to simplify (33) to

\frac{2\sqrt{A}}{z_{0}^{−1} − ζ} [2(\frac{1}{ζ} − ζ)Π(α^{2}, k) + (\frac{1}{z_{0}} + z_{0} − \frac{2}{ζ})K(k)] − \frac{π}{2} \sqrt{\frac{Az_{0}}{ζ}} .

Now equality in (32) tells us that

(34) \sqrt{b + 1} = \frac{π}{2} \sqrt{\frac{z_{0}}{ζ}} (\frac{1}{z_{0}} − ζ)[2(\frac{1}{ζ} − ζ)Π(α^{2}, k) + (\frac{1}{z_{0}} + z_{0} − \frac{2}{ζ})K(k)]^{−1},

where (27) has been used. Equation (34) enables us to express b (and thus A) in terms of z_{0} and ζ.

Relation 4. We now have three equations (27), (31) and (34) relating the parameters B, A, b, z_{0} to R and ζ. For a final fourth relation, we use the fact that f′(0) = 1 along with the equality of the integrals,

∫_{f(ε)}^{(b+1)B} \sqrt{\frac{(b + 1)B − w}{B − w}} \frac{dw}{w} = \sqrt{\frac{Az_{0}}{ζ}} ∫_{ε}^{z_{0}} \sqrt{\frac{(z_{0} − z)(z_{0}^{−1} − z)}{(ζ − z)(ζ^{−1} − z)}} \frac{dz}{z} .

For small ε > 0 the left side of this equation is calculated to be

\sqrt{b + 1} log \frac{−4(b + 1)B}{bf(ε)} + log \frac{1 − \sqrt{b + 1}}{1 + \sqrt{b + 1}} + O(ε).

The right integral is equal to

\sqrt{\frac{Az_{0}}{ζ}} (J_{ε} + log \frac{z_{0}}{ε} ) + O(ε),

where

J_{ε} = ∫_{ε}^{z_{0}} (\sqrt{\frac{(z_{0} − z)(z_{0}^{−1} − z)}{(ζ − z)(ζ^{−1} − z)}} − 1) \frac{dz}{z} .

Let ε → 0 in both sides and recall that

lim_{ε→0} \frac{f(ε)}{ε} = 1,

to obtain the final fourth relation,

(35)  log B = −\frac{1}{\sqrt{b + 1}} log \frac{1 − \sqrt{b + 1}}{1 + \sqrt{b + 1}} + log \frac{−bz_{0}}{4(b + 1)} + J_{0}.

The elliptic integral J_{0} will be left in open form for simplicity.

We now prove that the three relations (31), (34) and (35) give a well-defined relation between R, ζ and B. To do this we show that for fixed ζ, R and B both depend monotonically on z_{0} for values of z_{0} between 0 and ζ.

First, the relationship between R and B must be monotone as proven in the following lemma.

Lemma. Suppose f ∈ S_{R} and |f(ζ)| ≥ |g(ζ)| for all g ∈ S_{R}. Then for each ε > 0 there exist δ satisfying 0 < δ < ε, and f^{∗} ∈ S_{R+δ} such that |f^{∗}(ζ)| > |f(ζ)|.

Since ζ is positive we have B = f(ζ) > 0. We now choose a particular variation of the type found in (7) to preserve S while increasing R slightly. First select the points w_{0} , w_{1} and w_{2} in (7) so that w_{1} > B > 0 and w_{0} < 0. After choosing ε_{1} = −ϱ^{2} and ε_{2} = 0 in (7), it is easily verified that for arbitrarily small ϱ > 0, we may choose a(ϱ) such that:

(a) equation (8) is satisfied, thus S is preserved,

(b) the transfinite diameter R^{∗} of the function f^{∗} = V_{ϱ} ∘ f satisfies R < R^{∗} < R + ε;

(c) Re { log f^{∗}(ζ)} > Re { log f(ζ)}. Note that (c) is satisfied using (11). This proves the lemma.

Next we divide both sides of (32) by \sqrt{b + 1} = \sqrt{Az_{0}/ζ} and differentiate with respect to z_{0} to obtain

(36)  \frac{∂b}{∂z_{0}} > 0.

Equation (35) may now be used along with (36) to show that log B and thus B has positive derivative with respect to z_{0} . Therefore both R and B are monotone with respect to z_{0} .

This monotonicity indicates that the case −1 < b < 0 may occur only for values of R and ζ corresponding to 0 < z_{0} < ζ. Since z_{0} = 0 corresponds to R = 1 and z_{0} = ζ corresponds to the ellipse case (b = −1), we see that the case −1 < b < 0 may occur only in the sub-region of the R, ζ half-strip ({(R, ζ) : R > 1, 0 < ζ < 1}) lying below the curve (26). Further, the monotonicity of R with respect to z_{0} shows that when the extremal function falls only into this case, the extremal function is unique.

6. The case b < −1

Here there are two subcases:

Subcase α: Γ consists of a Jordan curve passing through w = (b + 1)B along with a slit [(b + 1)B, η] on the real axis (the slit may have length zero).

Subcase β: Γ is an analytic Jordan curve surrounding w = 0, w = B and w = (b + 1)B.

We handle the former subcase first. In Subcase α, F as given by (18) and (19) has a double zero on the unit circle so that (19) becomes

(37)  F (z) = \frac{A(z + 1)^{2}}{(z − ζ)(1 − ζz)}.

Upon considering (18) along with (37) we have three parameters; A, B, b. To express them in terms of R and ζ we require three relations among the variables. Evaluating lim_{z→0} F (z) in each of the expressions (18) and (37) gives us the first relation:

(38)  b + 1 = \frac{−A}{ζ}.

For the second relation we calculate G(w):

G(w) = ∫_{(b+1)B}^{w} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w}

= log w + log \frac{4}{bB} + 2\sqrt{\frac{A}{ζ}} arg (\sqrt{\frac{A}{ζ}} + i) + O(\frac{1}{w} ),

where the fundamental function element is chosen and (38) has been used. Since Re \{G(w)\} is Green’s function for the complement of \overline{f(D)}, we see that

(39)  log R = log \frac{−bB}{4} + 2\sqrt{\frac{A}{ζ}} arg (\sqrt{\frac{A}{ζ}} + i).

To establish a third relation we use the equality of the integrals,

∫_{f(ε)}^{B} \sqrt{\frac{w − (b + 1)B}{B − w}} \frac{dw}{w} = \sqrt{A} ∫_{ε}^{ζ} \frac{z + 1}{\sqrt{(ζ − z)(1 − ζz)}} \frac{dz}{z} .

Upon calculation it is clear that the first integral is

−i log \frac{b}{2i\sqrt{−(b + 1)} + b + 2} − \sqrt{−(b + 1)} log \frac{bf(ε)}{4(b + 1)B} + O(ε),

while the second integral is

−\sqrt{\frac{A}{ζ}} log \frac{(1 − ζ)^{2}}{4ζ} ε + O(ε).

Since the two integrals are equal for all small ε > 0, we let ε approach zero and use the fact that f′(0) = 1 to conclude

(40)  \sqrt{\frac{ζ}{A}} arg \frac{b}{2i\sqrt{−(b + 1)} + b + 2} = log [\frac{b}{(b + 1)B} · \frac{ζ}{(1 − ζ)^{2}} ].

Employing (39) to eliminate B and (38) to eliminate b we find (41)

log R = log \frac{ζ}{(1 − ζ)^{2}} + log (\frac{A}{ζ} + 2 + \frac{ζ}{A}) + 2(\sqrt{\frac{A}{ζ}} − \sqrt{\frac{ζ}{A}} ) arg (\sqrt{\frac{A}{ζ}} + i) − log 4.

A calculation reveals that this expression for log R is monotone with respect to A. Thus (41) determines A for given values of R. Therefore, equations (39) and (41) determine B parametrically in terms of R and ζ for Subcase α.

To determine for which values of R and ζ Subcase α may occur, we will determine for what values of R the “no-slit” domain described above has inner radius r ≥ 1. If r = 1, then the “no-slit” domain possibly corrresponds to an extremal function. If r > 1 the “no-slit” domain can be modified by the addition of a horizontal radial slit of suitable length to reduce the inner radius of the domain to exactly 1. For r < 1 this subcase may not occur.

Our work will be simplified by obtaining an explicit expression for A in terms of ζ alone for the “no-slit” region. To do this we make use of a special contour integral. In the w-plane, let Γ_{ε} be the union of the two line segments [(b + 1)B, f(−ε)], [f(ε), B] and the lower semicircle joining w = f(−ε) to w = f(ε). Note that f(−ε) = −f(ε) + O(ε^{2}) so the semicircle construction will remain valid when we allow ε to approach zero later. In the z -plane let γ_{ε} be the union of the segments [−1, −ε], [ε, ζ] and the lower semicircle joining z = −ε to z = ε. Then the semicircle in the w-plane is path-homotopic to the image under f of the prescribed semicircle in the z -plane. Since f(−1) = (b + 1)B for the “no-slit” region, it is clear that

∫_{Γ_{ε}} \sqrt{\frac{(b + 1)B − w}{w − B}} \frac{dw}{w} = \sqrt{A} ∫_{γ_{ε}} \frac{z + 1}{\sqrt{(ζ − z)(1 − ζz)}} \frac{dz}{z} .

A calculation reveals that the first and second integrals are

π + \sqrt{−(b + 1)}πi + O(ε)

and

2\sqrt{\frac{A}{ζ}} log \frac{1 − \sqrt{ζ}}{1 + \sqrt{ζ}} + \sqrt{\frac{A}{ζ}} πi + O(ε),

respectively. Now by letting ε approach zero and comparing imaginary parts, we again obtain (38), while a comparison of real parts tells us that

(42)  A = \frac{π^{2}ζ}{4[ log (\frac{1 − \sqrt{ζ}}{1 + \sqrt{ζ}} )]^{2}} .

We define

(43)  Ψ(ζ) = \frac{π}{2 log (\frac{1 + \sqrt{ζ}}{1 − \sqrt{ζ}} )}

and combine (39) and (42) to establish the relation

(44)  B = \frac{4R}{Ψ(ζ)^{2} + 1}e^{−2Ψ(ζ) arg(Ψ(ζ)+i)}.

The same calculation as that used to obtain (40) tells us for the “no-slit”domain that

\sqrt{\frac{ζ}{A}} arg \frac{b}{2i\sqrt{−(b + 1)} + b + 2} + log \frac{(b + 1)B}{b} · \frac{(1 − ζ)^{2}}{ζ} = log r.

Using (44) we see that r ≥ 1 exactly when

(45) log R ≥ log \frac{ζ}{(1 − ζ)^{2}} + log \frac{(Ψ(ζ)^{2} + 1)^{2}}{4Ψ(ζ)^{2}} + 2\frac{Ψ(ζ)^{2} − 1}{Ψ(ζ)} arg (Ψ(ζ) + i).

Thus the extremal function can have the form of Subcase α only if R satisfies (45). If R does not satisfy (45), then the extremal function takes the form of Subcase β or one of the cases b = −1, b > −1. Note that when, in Subcase α, the solution region has no slit, then

B = \frac{ζ}{(1 − ζ)^{2}} [\frac{Ψ^{2} + 1}{Ψ^{2}}e^{−(2/Ψ) arg(Ψ+i)}],

which compares favorably with the Koebe distortion theorem since the term in square brackets takes values between 0 and 1 for all ζ ∈ (0, 1). This is easily proven by noting that the expression in brackets has limits 1 and 0 at ζ = 1, respectively; while it is monotone with respect to ζ between those two values. Note that letting ζ approach zero in (45) leads to the inequality R ≥ ε^{2}π^{2}/64, which agrees with previous work in [6]. This completes Subcase α.

In Subcase β, as in the case b > −1, both R and B will be expressed in terms of a third parameter z_{0} ; further, the solutions will involve the use of elliptic integrals. Note that work in the previous cases and Subcase α above implies that Subcase β certainly occurs for values of R and ζ lying between the curve (26) and the curve obtained by taking equality in (45).

Again F (z) is given by (18) and (19) so

(46)  b + 1 = \frac{Az_{0}}{ζ} .

Now let w_{−1} = f(−1), the point where Γ meets the negative real axis, and consider

H_{0}(w) = ∫_{w_{−1}}^{w} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w} = G̃(w) − G̃(w_{−1}),

where

G̃(w) = ∫_{(b+1)B}^{w} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w} .

Since Γ satisfies (10) we see that H_{0}(w) is the analytic completion of Green’s function for the complement of \overline{f(D)}. Considering (45) and the analysis leading up to it, we have

log R = log \frac{−bB}{4} + 2\sqrt{\frac{−Az_{0}}{ζ}} arg (\sqrt{\frac{−Az_{0}}{ζ}} + i) + Re \{G̃(w_{−1})\},

where (46) has been used. Also, by (19),

G̃(w_{−1}) = ∫_{(b+1)B}^{w_{−1}} \sqrt{\frac{w − (b + 1)B}{w − B}} \frac{dw}{w} = \sqrt{A} ∫_{z_{0}}^{−1} \sqrt{\frac{(z − z_{0})(1 − z_{0}z)}{(z − ζ)(1 − ζz)}} \frac{dz}{z} .

Now a series of substitutions similar to those used to obtain (33) shows that

G̃(w_{−1}) = \sqrt{A} \frac{2(1 − z_{0})}{1 + ζ}K(\frac{t_{0}}{τ} ) − \sqrt{A} \frac{8z_{0}}{(1 + ζ)(1 − z_{0})}Π(t_{0}^{2}, \frac{t_{0}}{τ} ),

where as before, K and Π are normal complete elliptic integrals of the first and third kind, respectively. Hence

(47)  log R = log \frac{−bB}{4} + 2\sqrt{\frac{−Az_{0}}{ζ}} arg (\sqrt{\frac{−Az_{0}}{ζ}} + i) + G̃(w_{−1}).

We now establish a third relation that will determine A (and thus b by (46)) in terms of z_{0} and ζ. Let T be the unit circle and as usual, let Γ be the boundary of f(D). Then the equivalence of (18) and (19) implies that

(48)  ∫_{T} \sqrt{\frac{A(z − z_{0})(1 − z_{0}z)}{(ζ − z)(1 − ζz)}} \frac{dz}{z} = ∫_{T} \sqrt{\frac{(b + 1)B − w}{w − B}} \frac{dw}{w} .

By calculating the residue at infinity of the integrand in the second integral, we see that its value is 2π. The first integral, which is elliptic, will be left in open form after a simplifying substitution. With z = e^{iθ}, the left side of (48) becomes

2\sqrt{\frac{−Az_{0}}{ζ}} J(z_{0}, ζ),

where

(49)  J(z_{0}, ζ) = ∫_{0}^{π} \sqrt{\frac{cos θ − \frac{1}{2}(z_{0} + z_{0}^{−1})}{\frac{1}{2}(ζ + ζ^{−1}) − cos θ}} dθ.

Note that J(z_{0}, ζ) is positive for −1 < z_{0} < 0 and 0 < ζ < 1 and

(50)  J(−1, ζ) = 2 log \frac{1 + \sqrt{ζ}}{1 − \sqrt{ζ}} .

Equation (48) implies that

(51)  b + 1 = \frac{Az_{0}}{ζ} = \frac{−π^{2}}{J(z_{0}, ζ)^{2}} .

We may now view (47) as a relation expressing B in terms of R, ζ and z_{0} .

To express B parametrically in terms of R and ζ alone we need one further relation. It is obtained by observing that since (18) and (37) are equivalent,

(52)  ∫_{f(ε)}^{B} \sqrt{\frac{(b + 1)B − w}{w − B}} \frac{dw}{w} = ∫_{ε}^{ζ} \sqrt{\frac{A(z − z_{0})(1 − z_{0}z)}{(ζ − z)(1 − ζz)}} \frac{dz}{z} .

After dividing both sides by

\sqrt{−(b + 1)} = \sqrt{\frac{−Az_{0}}{ζ}} ,

the left side of this expression may be rewritten as

∫_{f(ε)/B}^{1} (\sqrt{\frac{1 − w(b + 1)^{−1}}{1 − w}} − 1) \frac{dw}{w} − log f(ε) + log B,

while the right side is equal to

log ζ − log ε + ∫_{ε}^{ζ} (\sqrt{\frac{(z − z_{0})(z − z_{0}^{−1})}{(ζ − z)(ζ^{−1} − z)}} − 1) \frac{dz}{z} .

We now let ε approach 0 and use the fact that f′(0) = 1 to see that

(53)  \begin{aligned}log B & = ∫_{0}^{ζ} (\sqrt{\frac{(z − z_{0})(z − z_{0}^{−1})}{(ζ − z)(ζ^{−1} − z)}} − 1) \frac{dz}{z} + log ζ \\ & − ∫_{0}^{1} (\sqrt{\frac{1 − w(b + 1)^{−1}}{1 − w}} − 1) \frac{dw}{w} . \end{aligned}

Together, equations (53) and (47) indicate that R and B are related parametrically in terms of z_{0} for a fixed value of ζ.

The solution to the problem will be complete once it is shown that R varies monotonically with z_{0} . This monotonicity will imply that the extremal function is unique when it falls into Subcase β. Further it will imply that all the cases we have described actually occur and the solution is unique in each case.

To prove monotonicity, it is sufficient in view of the lemma in Section 5 to show that ∂B/∂z_{0} < 0. Toward this end we differentiate with respect to z_{0} in (53) to obtain

(54)  \begin{aligned}\frac{∂ log B}{∂z_{0}} & = \frac{1}{2}(z_{0}^{−2} − 1) ∫_{0}^{ζ} \frac{dz}{\sqrt{q̃(z)}} \\ & − \frac{1}{2} \frac{∂b}{∂z_{0}} ∫_{0}^{1} \frac{dw}{[ − (b + 1)]^{3/2}\sqrt{(1 − w)(w − b + 1)}}, \end{aligned}

where

q̃(z) = (ζ − z)(\frac{1}{ζ} − z)(z − z_{0})(z − \frac{1}{z_{0}} ).

Although we could now obtain an expression for ∂b/∂z_{0} directly from (51), our calculations will be facilitated if we instead use the following relation,

∫_{(b+1)B}^{f(−ε)} \sqrt{\frac{w − (b + 1)B}{B − w}} \frac{dw}{w} = ∫_{z_{0}}^{−ε} \sqrt{\frac{A(z − z_{0})(1 − z_{0}z)}{(ζ − z)(1 − ζz)}} \frac{dz}{z} .

We proceed just as with (52) above to obtain

(55)  \begin{aligned}log B & = − ∫_{z_{0}}^{0} (\sqrt{\frac{(z − z_{0})(z − z_{0}^{−1})}{(ζ − z)(ζ^{−1} − z)}} − 1) \frac{dz}{z} + log \frac{z_{0}}{b + 1} \\ & + ∫_{b+1}^{0} (\sqrt{\frac{1 − w(b + 1)^{−1}}{1 − w}} − 1) \frac{dw}{w} . \end{aligned}

After differentiating (55) and comparing the result with (54) we conclude that

\frac{∂b}{∂z_{0}} = \frac{1}{π} (z_{0}^{−2} − 1)[ − (b + 1)]^{3/2} ∫_{z_{0}}^{ζ} \frac{dz}{\sqrt{q̃(z)}}.

This in turn allows us to conclude that

(56)  \frac{∂ log B}{∂z_{0}} = \frac{1}{2}(z_{0}^{−2} − 1)[\frac{2}{π} arctan \sqrt{−(b + 1)} ∫_{z_{0}}^{ζ} \frac{dz}{\sqrt{q̃(z)}} − ∫_{z_{0}}^{0} \frac{dz}{\sqrt{q̃(z)}}].

Though an analytic proof is yet to be found, computer calculations have shown numerically that the expression in (56) is negative over a wide range of values of z_{0} and ζ. The method of numerical integration used relies upon a modified Newton– Cotes formula known as Weddle’s rule ([2, Chapter 3]) after a change of variable. A few sample values are shown in Table 1 below. These data suggest that B and thus R varies monotonically with respect to z_{0} in Subcase β. Thus the data also indicate that the extremal function for all values of R and ζ is unique.

ζ = .9  ζ = .5  ζ = .1
z_{0} = −.99 −1.7422 −1.1367 −0.3092
z_{0} = −.9 −0.8747 −0.5450 −0.1408
z_{0} = −.5 −0.2206 −0.1204 −0.0271
z_{0} = −.1 −0.0108 −0.0051 −0.0010
z_{0} = −.01 −0.00012 −0.00005 −0.00001

Table 1. Table entries are numerical values for the bracketed terms in (56) above.

7. Summary of results

Before stating our results in the form of a theorem, we summarize the five types of extremal functions which may occur.

I. The extremal function f maps D onto a region bounded by a Jordan curve which is analytic except for a corner at the point w = (b + 1) · |f(ζ)| along with a radial slit from (b + 1) · |f(ζ)| toward the origin. The length of this slit is determined by the fact that f′(0) = 1. Extremal functions of this type are unique when they occur. |f(ζ)| is given implicitly by (39) and (41). This solution type may occur only for values of R and ζ lying above the curve (45).

II. The extremal function f maps D onto a region similar to that described for Type I solutions except that the slit has length zero. Again |f(ζ)| is given by (39) and (41). This type of solution may occur only for values of R and ζ lying on (45).

III. The extremal function f maps D onto a region bounded by an analytic Jordan curve surrounding the point w = (b + 1) · |f(ζ)|, where b > −1. Extremal functions of this type are unique when they occur and they may occur only for values of R and ζ lying below the curve (26). |f(ζ)| is given parametrically in terms of complete elliptic integrals by equations (31) and (35).

IV. The extremal function f maps D onto the interior of an ellipse with foci at w = 0 and w = |f(ζ)|. This type of solution occurs only for R and ζ lying on the curve (26). |f(ζ)| is given directly in terms of complete elliptic integrals by (30).

V. The extremal function f maps D onto a region bounded by an analytic Jordan curve surrounding the point w = (b + 1) · |f(ζ)|, where b < −1. |f(ζ)| is given parametrically in terms of complete elliptic integrals by (47) and (53).

We have established the following results.

Theorem. Given ζ ∈ D and R ∈ (1, ∞), each function f maximizing |g(ζ)| over the class S_{R} maps D onto a region bounded by arcs lying on trajectories of the quadratic differential (10). In (10) the negative parameter b is determined by R and ζ. For all values of R and ζ, arg f(ζ) = arg ζ. For values of R and ζ lying below the curve (26) each extremal function is either of Type III or Type V. On the curve (26) itself each extremal function is of Type IV or Type V. In the region between the curves (26) and (45) the solution is of Type V. On the curve (45) the solution is of Type II or Type V. Finally, above the curve (45) the solution is of Type I or Type V.

As mentioned in Section 6, numerical evidence indicates that extremal functions of Type V may occur only in the region between the curves (26) and (45). This numerical evidence also suggests that the extremal function is unique for each value of R and ζ. For each type of solution, the value of the derivative of the extremal function at ζ may be obtained using the calculations in Section 3.

It should be mentioned that one can show for values of R and ζ lying on the curve (26) that the extremal function is of Type IV and is unique. This is done by defining R_{f} to be the transfinite diameter of f(D) and then finding the maximum of

\frac{|f(ζ)|}{R_{f}}

over the class S. The extremal functions for this problem all map D onto the interior of an ellipse.

Figure 4. A graph of the curves (26) and (45) as drawn using Matlab. Roman numerals are used to indicate which solution types may occur in each region of the (R, ζ) half-strip.

Note added in proof. During the publication process it was brought to the author’s attention that certain of the results (particularly those established in Section 2) were previously established by A. Solynin and M. Gavrilyuk in their article “Moduli estimates for certain classes of univalent functions” in the Russian language publication Dynamic Problems in Mechanics of a Continuous Medium 63, 1983. Their approach employs the method of extremal lengths. Our approach is new as are the results involving the distribution of solutions for various values of R and ζ and the quantitative estimates of solutions.

Univalent functions of given transfinite diameter 271

References

[1] Bateman Manuscript Project: Higher transcendental functions, Volume II. - A. Erdélyi, Editor, McGraw-Hill, New York, 1953.

[2] Buckingham, R.A.: Numerical methods. - Pitman and Sons, London, 1957.

[3] Byrd, P.F., and M.D. Friedman: Handbook of elliptic integrals for engineers and physicists. - Springer-Verlag, Berlin–Göttingen–Heidelberg, 1954.

[4] Courant, R.: Dirichlet’s principle, conformal mapping and minimal surfaces. - Interscience Publishers Inc., New York, 1950; Reprint: Springer-Verlag, Heidelberg–New York, 1977.

[5] Duren, P.L.: Univalent functions. - Springer-Verlag, Heidelberg–New York, 1983.

[6] Duren, P.L., and M. Schiffer: Univalent functions which map onto regions of given transfinite diameter. - Trans. Amer. Math. Soc. 323, 1991, 413–428.

[7] Goluzin, G.M.: Geometric theory of functions of a complex variable. - Gosudarstv. Izdat. Tehn.-Teor. Lit., Moscow-Leningrad, 1952 (Russian). English translation: Amer. Math. Soc., 1969.

[8] Hille, E.: Analytic function theory, Volume II. - Ginn and Co., Boston, 1962; Second edition, Chelsea Publ. Co., New York, 1987.

[9] Nehari, Z.: Conformal mapping. - McGraw-Hill, New York, 1952; Dover reprint, New York, 1975.

[10] Pólya, G., and G. Szegő: Problems and theorems in analysis, Volume II. - Springer-Verlag, New York, 1976.

[11] Pommerenke, Ch.: Univalent functions. - Vandenhoeck and Ruprecht, Göttingen, 1975.

[12] Schiffer, M.: A method of variation within a family of simple functions. - Proc. London Math. Soc. 44, 1938, 432–449.

[13] Schiffer, M.: Hadamard’s formula and variation of domain-functions. - Amer. J. Math. 68, 1946, 417–448.

[14] Tsuji, M.: Potential theory in modern function theory. - Maruzen, Tokyo, 1959; Chelsea reprint, New York, 1975.

Received 28 June 1991