Planar Harmonic and Quasiregular Mappings

CMFT, RMS-Lecture Notes Series No. 19, 2013, pp. 267–333.

Planar Harmonic and Quasiregular Mappings Saminathan Ponnusamy and Antti Rasila

Abstract. Planar harmonic functions are complex-valued functions whose real and imaginary parts are not necessarily conjugate, i.e., do not need to satisfy the Cauchy-Riemann equations. Univalent harmonic functions can be regarded as generalizations of conformal mappings, although they seem to have attracted the attention of the geometric function theorist only recently, after the appearance of the paper of Mocanu for C 1 -functions, and later the pioneering work of Clunie and Sheil-Small in 1984. In this article our emphasize will be on introducing basic theory of harmonic univalent functions and related concepts from the well-established theory of quasiconformal mappings of the plane. Our presentation will be based on a selection of results relevant to our precise objective of introducing certain topics at the basic level, see also [55, 56, 58].

Contents 1.

2.

3.

Planar Harmonic Mappings

268

1.1.

Differential Operators ∂/∂z and ∂/∂z

269

1.2.

Conformal Mappings

271

1.3.

Canonical Representation

273

Lewy’s Theorem and Schwarz’s Lemma for Harmonic Mappings

274

2.1.

Jacobian and Local Univalence

275

2.2.

Solutions of Elliptic Partial Differential Equation

278

2.3.

Composition Rule for Harmonic Functions

279

2.4.

Schwarz’s Lemma

280

2.5.

Landau-Bloch Theorems

283

2.6.

Harmonic Mappings of the Plane C

285

Quasiconformal Mappings in the Plane

286

3.1.

Quasiconformal Diffeomorphisms

287

3.2.

General Quasiconformal Mappings

290

3.3.

Conformal Modulus

292

Version : February 12, 2012. .

268

4.

5.

6.

S. Ponnusamy and A. Rasila

3.4.

Basic Properties of Quasiconformal Mappings

294

3.5.

Estimates with Modulus

294

3.6.

Ring Domains

296

3.7.

Canonical Ring Domains

297

3.8.

Schwarz’s Lemma for Quasiconformal Mappings

299

3.9.

Heinz’s Inequality

300

Univalent Harmonic Mappings

303

4.1.

0 Normality of the families SH and SH

304

4.2.

0 Area Minimizing Property of the Class S H

308

4.3.

Harmonic Univalent Polynomials

309

Method of Shearing

310

5.1.

Shearing Theorem of Clunie and Sheil-Small

312

5.2.

Harmonic Koebe Function

315

5.3.

Coefficient Conjectures for Univalent Harmonic Mappings

317

Criteria for Harmonic Close-to-Convex and Univalent Mappings 6.1.

Some Basic Examples

320 323

Questions and Exercises

327

References

331

1. Planar Harmonic Mappings Let Ω be an open subset of the complex plane C. A real-valued function u : Ω → R is harmonic if u ∈ C 2 (Ω) (continuous first and second partial derivatives in Ω) and satisfies the Laplace equation Δu = 0 on Ω. Here Δ=

∂2 ∂2 + ∂x2 ∂y 2

is the 2-dimensional Laplacian. Thus, a C 2 -solution of Δu = 0 is called a (real) harmonic function. A complex-valued function f = u + iv is called harmonic on Ω if the coordinate functions u and v are (real) harmonic on Ω. Often we study harmonic functions on a domain (open and connected) even when connectivity is not needed. Later we shall be mainly interested in the case in which Ω is the unit disk D = {z : |z| < 1}.

Planar Harmonic and Quasiregular Mappings

269

1.1. Differential Operators ∂/∂z and ∂/∂z. Through this paper, we use the standard notations. For example, a convenient notation is to treat the pair of conjugate complex variables z := x+iy and z := x−iy as two independent variables by writing z+z z−z x= and y = . 2 2i This leads to the following differential operators: � � ∂ ∂ ∂x ∂ ∂y 1 ∂ ∂ := + = −i ∂z ∂x ∂z ∂y ∂z 2 ∂x ∂y and

� � ∂ ∂x ∂ ∂y 1 ∂ ∂ ∂ := + = +i . ∂z ∂x ∂z ∂y ∂z 2 ∂x ∂y In view of this observation, we may treat f (x + iy) as a function of z and z, and so � � � � z+z z−z z+z z−z , , f (x + iy) = u + iv := U (z, z) + iV (z, z). 2 2i 2 2i Consequently, for a complex-valued function f = u + iv with continuous partial derivatives, we may use the formal notations fx = ux + ivx and fy = uy + ivy . Then and

2fz = fx − ify = (ux + vy ) − i(uy − vx ),

2fz = fx + ify = (ux − vy ) + i(uy + vx ), where subscripts denote partial derivatives. Also, we see that |fz |2 − |fz |2 = ux vy − uy vx .

The motivation for these notations is two-fold. We start by observing that ∂(z) ∂(z) =1= ; ∂z ∂z

∂(z) ∂z =0= , ∂z ∂z

and thus, fz = 0

⇐⇒ fx = −ify

ux = vy and uy = −vx .

⇐⇒

Thus, the Cauchy-Riemann equations in the Cartesian form can be equivalently written in a single concise equation: f z = 0. Often this is referred to as the complex form of the Cauchy-Riemann equations (see [54] for details). Moreover, for the Laplacian of f , we have � � ∂ ∂f 4 = 2 (fx − ify )z ∂z ∂z = (fx − ify )x + i(fx − ify )y = fxx − ifyx + ifxy + fyy = fxx + fyy = Δf, and so, the Laplacian takes the form Δ=4

∂2 , ∂z∂z

270


which is referred to as the complex form of Laplacian operator or simply (2-dimensional) Laplacian operator. Thus, we have an equivalent formulation of the definition of the harmonic function. Definition 1. We say that a complex-valued function f is harmonic if and only if f ∈ C 2 (Ω) with Δf = 4fzz = 0. The following properties are easy to verify. • The operators ∂/∂z and ∂/∂z are linear and have the usual properties of differential operators. For example, the product and quotient rules hold: �f � gfz − f gz = , (f g)z = f gz + gfz and g z g2 and similarly for ∂/∂z. • The two derivatives are connected by the property � � (fz ) = f z . As

∂2 Δ(Δf ) = Δ f = 4 ∂z∂z 2

� 2 � ∂ f ∂4f 4 = 16 2 2 , ∂z∂z ∂ z∂ z

we have: Definition 2. We say that a complex-valued function f is biharmonic if and only if f ∈ C 4 (Ω), and f satisfies the bi-harmonic equation ∂4f = 0. ∂2z ∂2z

One of the fundamental theorems in the complex function theory concerned with necessary and sufficient conditions for analyticity. We have an equivalent formulation of this result. Definition 3. A (planar) harmonic function f = u + iv is analytic on a domain Ω if and only if u and v are harmonic conjugates; i.e. u, v ∈ C 2 (Ω) satisfy the Cauchy–Riemann equations: ux = vy , uy = −vx on Ω. The most important examples of harmonic functions arise naturally from the Cauchy-Riemann equations and there is a close interrelation between analytic functions and harmonic functions. Theorem 1. The real and imaginary parts of an analytic function in an open set are harmonic. In particular, every analytic function f and its conjugate f are harmonic. Proof. Suppose that f is analytic. Then f z = f � is also analytic and fz is independent of z so that fzz = 0. That is, f is harmonic. Alternately, if f = u + iv is analytic then it satisfies the Cauchy-Riemann equations which implies that

ux = vy and uy = −vx , uxx = vyx and uyy = −vxy .


271

Adding these two equations it follows that u = Re f is harmonic. Similarly, we can prove that v = Im f is also harmonic. Consequently, f and f are harmonic. Using the complex form of Cauchy-Riemann equations, and the formula Δf = 4f zz , we conclude the following: Proposition 1. Analytic functions f are necessarily independent of z, whereas fz is independent of z for planar harmonic functions f . From this proposition, we observe that the function f is harmonic in a planar domain Ω if and only if fz is analytic in Ω. Also, if f is analytic in Ω then f z (z) = f � (z) in Ω. Clearly, real and imaginary parts of a harmonic function are not necessarily conjugates. Moreover, the most natural way of passing from harmonic to an analytic function is remembered in the following (see [54] for a proof): Theorem 2. Let u(z) be a real-valued harmonic function in a simply connected domain Ω. Then there exists an analytic function f (z) such that Re f (z) = u(z) on Ω. For basic results concerning the theory of analytic functions and examples, exercises and related applications, we refer to the standard texts such as Ahlfors [1], advanced texts such as [12], and the recent books of Ponnusamy [54], and Ponnusamy and Silverman [60]. We use the following notations: for a ∈ C and δ > 0, D(a; δ)

= {z ∈ C : |z − a| < δ},

D(a; δ) = {z ∈ C : |z − a| ≤ δ}, ∂D(a; δ) = {z ∈ C : |z − a| = δ}

denote the open disk (about a), the closed disk, and the circle, respectively. Further, we let D(0; δ) = Dδ and D1 = D, the open unit disk. Notations such as ∂D and D are defined in the obvious way. 1.2. Conformal Mappings. If f is an analytic function defined on a domain Ω, then by the open mapping theorem, f (Ω) is a domain, and if Ω is a simply connected domain then so is f (Ω) (for a proof, we refer to [54]). The function f is said to be conformal at a point z0 ∈ Ω if f preserves the angle at z 0 between any pair of smooth curves γ1 and γ2 passing through z0 . That is the angle between the image curves Γ1 and Γ2 at the image point w0 = f (z0 ) is the same as that between the curves γ1 and γ2 at z0 . If the analytic function f is conformal at every point of Ω, then we say that f is conformal in Ω. Thus, a conformal mapping is simply an angle-preserving (i.e. both sense and magnitude) homeomorphism of some domain onto another. Then, we have the following celebrated theorem due to Riemann. Theorem 3 (The Riemann Mapping Theorem). There exists a unique conformal map f of D onto a simply connected domain (except the whole complex plane C) such that f (z0 ) and arg f � (z0 ) take given values.

272


f Ω

Ω ´= f (Ω) Φ

φ=Φ f

0

Real line

Figure 1. Description for invariance of Laplacian under a conformal map For a proof of the theorem, we refer to standard texts (see also [1, 12, 54, 60]). It is important that solutions of the Laplace equation remain invariant if the original domain is subjected to a conformal mapping. Consequently, complicated domains can be transformed into more convenient ones without having any change in the Laplace equation. Thus, one aims at developing a relationship between a harmonic function φ(x, y) on Ω (called a physical plane) and the corresponding harmonic function Φ(u, v) in Ω� (called a model plane) such that φ at (x, y) ∈ Ω has the relation φ(x, y) = Φ(u(x, y), v(x, y)), where f (z) = u(x, y) + iv(x, y) is conformal on Ω with f (Ω) = Ω � . But then we ask, how do conformal maps help us to solve boundary value problems? We state the following result which actually shows that the Laplace equation remains invariant under conformal maps (see Figure 1). Theorem 4. Assume that φ ∈ C 2 (Ω) and f = u + iv is a conformal mapping of Ω onto Ω� = f (Ω). Then, for φ(x, y) = Φ(u, v) with u = u(x, y) and v = v(x, y), we have Δφ = |f � (z)|2 ΔΦ. In particular, Δφ = 0 on Ω if and only if ΔΦ = 0 on Ω� . Proof. The proof follows by applying the rule of change of variables. Indeed, we have φx

= Φu ux + Φv vx ,

φy

= Φu uy + Φv vy ,

so that φxx φyy

= =

(Φuu ux + Φuv vx )ux + (Φvu ux + Φvv vx )vx + Φu uxx + Φv vxx , (Φuu uy + Φuv vy )uy + (Φvu uy + Φvv vy )vy + Φu uyy + Φv vyy .

Then, by addition, we easily obtain the desired conclusion. An analytic function f defined on a domain Ω is said to be univalent (or one-to-one or schlicht) on Ω if f (z) assumes different values for different values of z so that


273

the equation w = f (z) has at most one root in Ω for every complex w. For an analytic function f (z) to be univalent in a small neighborhood of a point z 0 ∈ Ω, it is necessary and sufficient that f � (z0 ) �= 0. We remark that the local univalence at all points of a domain is however insufficient for the univalence in the whole domain. For instance, f (z) = ez is locally univalent in C but is not univalent in |z| < R if R > π. At this place it is important to observe that one does not require the analyticity for defining univalence. For instance, f (z) = z is univalent in C although it is nowhere analytic. Another important and basic result concerning conformal mapping is the following (see [54] for a proof). Theorem 5. Suppose that f is analytic in a domain Ω. If f � (z0 ) �= 0 at z0 ∈ Ω, then f is conformal at z = z0 . The converse is also true. For instance, • • • •

1 + ez is conformal everywhere on C, z 2 is conformal everywhere except at the origin, cos(πz) is conformal everywhere except at integer points, z + 1/z is conformal at all points of C except at 0, ±1.

Conformal mappings have been successfully used to solve the two-dimensional Poisson equations which appear in problems related to electric fields with space charges, and thermal fields with heat generation, to mention but a few applications. 1.3. Canonical Representation. We now present our first basic result for harmonic functions. Lemma 1 (Canonical Representation). Let Ω be a simply connected domain in C. Then f : Ω → C is a (planar) harmonic function if and only if the function f has the representation f = h + g, where h and g are analytic in Ω. The representation is unique up to an additive constant. We call the functions h and g the analytic and the co-analytic parts of f , respectively. Proof. Set f = u + iv, where u and v both are harmonic in a simply connected plane domain Ω. Then there exist analytic functions F and G on Ω such that G−G F +F and v = Im G = . 2 2i This observation gives the representation � � � � F +F G−G F +G F −G f= + = + := h + g, 2 2 2 2 u = Re F =

where both h and g are clearly analytic in Ω, and g denotes the function z �→ g(z).

Alternately, suppose that f is harmonic. Then, f z is analytic in the simply connected domain Ω, and so we may define h by h � = fz , where h is analytic in Ω and h is determined up to an additive constant. To obtain the desired representation for known f and h, we define g by g =f −h=f −h

274


so that f = h + g. It suffices to show that g is analytic in Ω, i.e., g z = 0. Now ∂ (f − h) = fz − hz = fz − h� = 0 in Ω, gz = ∂z and thus, g is analytic in Ω. The converse part is obvious. The uniqueness follows from the fact that a function which is both analytic and anti-analytic1 is constant. Indeed if f has two representations f = h 1 + g1 and f = h2 + g2 , then h1 − h2 = −(g1 − g2 ), showing that h1 − h2 = −(g1 − g2 ) = a, i.e., h1 = h2 + a and g1 = g2 − a,

for some complex constant a.

Remark 1. If the function f in Lemma 1 is real-valued, then f may be expressed as f = h + h = 2Re h so that 2h represents the analytic completion of f and is • unique up to an additive imaginary constant.

2. Lewy’s Theorem and Schwarz’s Lemma for Harmonic Mappings A complex-valued function f : Ω → C is said to be a harmonic univalent mapping if it is harmonic and one-to-one in Ω, i.e., f (z 1 ) �= f (z2 ) for all z1 , z2 ∈ Ω with z1 �= z2 . In particular, the class of (planar) harmonic univalent mappings on the unit disk D includes the subclass of univalent functions that are also analytic in D, a popular topic in geometric function theory (see [21, 26, 53]). Example 1. Consider f : C → C by f (z) = 4x + i4xy. Then, f is a harmonic function. Also, we easily have the decomposition f = h + g, where f = h + g and h(z) = 2z + z 2 , g(z) = 2z − z 2 .

Is this f a univalent harmonic mapping in C? If not, how about the same function when the domain of the function f is restricted to the right half-plane Ω = {z : Re z > 0}? Does the inverse of f exist on Ω? Must the inverse be univalent harmonic on Ω? •

Example 2. For n ≥ 2, consider the harmonic function 1 f (z) = z − z n , z ∈ D. n It is easy to verify the univalency of f in D. Suppose that f (z 1 ) = f (z2 ) for z1 , z2 ∈ D. Then, we have n(z1 − z2 ) = z1 n − z2 n = (z1 − z2 )(z1 n−1 + z1 n−2 z2 + · · · + z2 n−1 ).

By taking absolute values on both sides, we see that this is impossible unless z 1 = z2 because |z1 n−1 + z1 n−2 z2 + · · · + z2 n−1 | < n. Thus, f is a harmonic mapping of the open unit disk D onto the domain bounded by a hypocycloid of (n + 1) cusps, inscribed in the circle |w| = (n + 1)/n. Note also that F (z) = z + n1 z n is analytic and univalent in D because |F � (z) − 1| < 1 for all z ∈ D. Thus, F is a normalized analytic univalent function in D. The images of f and F for n = 2 are illustrated in Figures 2 and 3 whereas the images of f and F for n = 4 are illustrated in Figures 1 Conjugate

of an analytic function is called an anti-analytic function.


1

1

0.5

0.5

0

0

−0.5

−0.5

−1

−1

−2

−1.5

−1

−0.5

0

0.5

1

−1

Figure 2. Image of D under f (z) = z − 12 z 2 1

0.5

0.5

0

0

−0.5

−0.5

−1

−1 −1

−0.5

0

0.5

Figure 4. Image of D under f (z) = z − 14 z 4

0

0.5

1

1.5

2

Figure 3. Image of D under F (z) = z + 12 z 2

1

−1.5

−0.5

275

1

−1

−0.5

0

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1

1.5

Figure 5. Image of D under F (z) = z + 14 z 4

4 and 5. Later, we shall easily obtain the univalence of f as a simple consequence of some other results (see Example 7). • 2.1. Jacobian and Local Univalence. The Jacobian of a complex-valued function f = u + iv at a point z is defined to be � � � u (z) vx (z) � � = ux (z)vy (z) − uy (z)vx (z), Jf (z) = �� x uy (z) vy (z) � provided that all the partial derivatives exist at z. Because |f z |2 − |fz |2 = ux vy − uy vx , the Jacobian may be expressed equivalently and conveniently in terms of zand z-derivatives Jf (z) = |fz (z)|2 − |fz (z)|2 . If f is analytic on Ω, then fz (z) = 0 on Ω and so fz (z) = f � (z). Thus, the Jacobian takes the form Jf (z) = (ux (z))2 + (vx (z))2 = |f � (z)|2 .

276


Let f be a univalent function defined on a domain Ω and belong to C 1 (Ω) such that Jf (z) �= 0 in Ω. Then f is said to be a diffeomorphism or more accurately, a C 1 -diffeomorphism of Ω onto its range. We remark that, if f : Ω → C is a diffeomorphism, then either Jf (z) > 0 everywhere in Ω or Jf (z) < 0 throughout the domain Ω. This follows from the fact that Ω is connected and J f : Ω → R is a continuous and zero-free function, i.e., the set {J f (z) : z ∈ Ω} is a connected set of real numbers that does not contain zero and hence, J f (Ω) is either a subset of (−∞, 0) or a subset of (0, ∞). When Jf is positive in Ω, then the diffeomorphism f is called orientation-preserving mapping or sense-preserving mapping. A diffeomorphism with a negative Jacobian is said to be orientation-reversing mapping or sensereversing mapping. We see that the conjugate f of a diffeomorphism f : Ω → C is also a diffeomorphism, i.e., the one for which Jf (z) = −Jf (z).

Therefore, f is orientation-reversing when f is orientation-preserving, and vice versa. For example, in the unit disk D, 1. f (z) = z is sense-preserving, as J f (z) = 1 in D, 2. f (z) = (1 + z)2 is sense-preserving, as Jf (z) = |2(1 + z)|2 > 0 in D, 3. f (z) = z is sense-reversing, as Jz (z) = −1 < 0 in D.

Let f : Ω → f (Ω) be a sense-preserving C 1 -diffeomorphism, z0 ∈ Ω, and w = f (z) = u(z) + iv(z). Then, by using the properties of the differential operators from advanced calculus, we obtain du = ux dx + uy dy,

dv = vx dx + vy dy,

and the differential

df = fx dx + fy dy, which can be written as (because f x = fz + fz and ify = fz − fz ) df = fz dz + fz dz = h� (z) dz + g � (z) dz.

Next, we see that if f = h + g and Jf (z) > 0 in Ω, then we have � � 2 Jf (z) = |h� (z)|2 − |g � (z)|2 := |h� (z)|2 1 − |ω(z)| ,

where ω = g � /h� is analytic in Ω such that |ω(z)| < 1 for z ∈ Ω. Note that ω satisfies g � = ωh� and h� (z) �= 0 whenever Jf (z) > 0.

Definition 4. The equation fz = ωfz , where |ω(z)| ≤ k < 1 for z ∈ Ω, is called Beltrami equation. A complex-valued function f satisfying the Beltrami equation is called a quasiregular mapping. A well-known classical result for analytic functions states that an analytic function f is locally univalent at z0 if and only if Jf (z0 ) �= 0 (see for example, [54] and [60, Theorems 11.2 and 11.3]). In 1936, Hans Lewy [42] showed that this remains true for harmonic functions. Definition 5 (Critical points). If u ∈ C 1 (Ω) and Ω ⊂ R2 is open, then a point (x0 , y0 ) is said to be a critical point of u if ux (x0 , y0 ) = 0 and uy (x0 , y0 ) = 0. The points that are not critical are often referred to as regular points. Critical points of a non-constant harmonic function are isolated (See Exercise 30).


277

Theorem 6 (Lewy’s Theorem). A harmonic function f is locally univalent in a neighborhood of z0 if and only if Jf (z0 ) �= 0. That is, f is locally univalent in a domain Ω if and only if Jf (z) �= 0 throughout Ω. Proof. Suppose that Jf (z0 ) �= 0 at some point z0 , then f is locally univalent by the inverse mapping theorem (see for example [54, 60]). Conversely, let f = u + iv be a complex-valued function which is locally univalent in Ω. Assume on the contrary that there exists z 0 ∈ Ω such that Jf (z0 ) = 0. Thus, the determinant of the Jacobian matrix � � ux vx uy vy at z0 must be zero and therefore, the homogeneous system of linear equations in a and b, namely, � aux + bvx = 0, auy + bvy = 0, has a non-trivial solution (a, b) �= (0, 0) at z0 . But then the function ψ = au + bv is a real-valued harmonic function and has a critical point z 0 : ψx (z0 ) = ψy (z0 ) = 0. Without loss of generality, we may now assume that f (z 0 ) = 0. Let φ be its harmonic conjugate and consider the analytic function F = ψ + iφ. As ψ x (z0 ) = ψy (z0 ) = 0, we see from the Cauchy-Riemann equations that φx (z0 ) = φy (z0 ) = 0, and therefore, F (z0 ) = 0. �

Now, we consider the level set L ψ = {z : ψ(z) = 0} near the point z0 . Since Re F (z) = ψ(z) = 0 for z ∈ Lψ ,

the level set is the pre-image of the imaginary axis under F . Again, as F is analytic with F (z0 ) = 0 and F � (z0 ) = 0, for z near z0 ∞ ∞ � � F (z) = ak (z − z0 )k = (z − z0 )n G(z), G(z) = ak (z − z0 )k−n , k=n

k=n

where n > 1 with an �= 0 and G(z) �= 0 for z near z0 . Thus, F is n-valent function near z0 . Indeed we can choose a branch of n-th root and obtain the function H(z) = (z − z0 )G1/n (z) := h1 (z − z0 ) + h2 (z − z0 )2 + · · ·

(h1 = a1/n n )

which is analytic and univalent near z 0 . It follows that F (z) = H n (z) and F (z) is purely imaginary exactly when H(z) lies on a system of n-lines passing through the point z0 and meeting there with angles π/n. As H is locally univalent at H(z 0 ) = 0, the pre-image of that system of lines is locally a system of n distinct analytic arcs intersecting at z0 with equal angles π/n. On the other hand, f maps the levels set L ψ into the line au + bv = 0 which means that f maps n distinct arcs from the level sets into the line au + bv = 0. Because f is assumed to be locally univalent at z 0 , f cannot carry a set composed of several intersecting arcs onto a line segment. This is a contradiction to the assumption that Jf (z0 ) = 0. An immediate consequence of Lemma 1 and Theorem 6 is the following.

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Corollary 1. Let f be a complex-valued harmonic function on a simply connected domain Ω with the decomposition f = h + g. Then f is locally univalent and sense-preserving in Ω if and only if |h� (z)| > |g � (z)| in Ω. Equivalently, f is a sense-preserving local homeomorphism if and only if Jf (z) > 0. The set of all critical points of a C 1 -function consists of those points where the Jacobian vanishes. Thus, for a harmonic function f , the set of critical points consists of those points for which f is not locally univalent. Remark 2. Lewy’s theorem does not hold for harmonic mappings in higher dimensions (n ≥ 3). The following example is due to Wood [72]. Consider f : R 3 → R3 given by f (x, y, z) = (x3 − 3xz 2 + yz, y − 3xz, z). The three coordinate functions u = x3 −3xz 2 +yz, v = y −3xz, w = z are harmonic as they satisfy the 3-dimensional Laplace equation: Δu = 0 = Δv = Δw, where Δ=

∂2 ∂2 ∂2 + + . ∂x2 ∂y 2 ∂z 2

Thus, the function f is harmonic in R 3 . The Jacobian of the given function is � � � 3x2 − 3z 2 −3z 0 � � � z 1 0 �� = 3x2 . Jf (x, y, z) = �� −6xz + y −3x 1 �

To find the inverse function, we need to solve x and y in terms of u, v, w. Substituting z = w, the expressions for u and v become u = x3 + w(y − 3xw) and v = y − 3xw.

Using the second equation, the first one may be rewritten as √ u = x3 + vw, or x = 3 u − vw, and so v = y − 3xw gives

√ y = v + 3w 3 u − vw. Thus, the inverse function f −1 : R3 → R3 is given by √ √ f −1 (u, v, w) = ( 3 u − vw, v + 3w 3 u − vw, w).

Thus, the given function is a homeomorphism of R 3 but the Jacobian vanishes on the plane x = 0. •

2.2. Solutions of Elliptic Partial Differential Equation. Suppose that f is a univalent harmonic function defined on a simply connected domain Ω. Then, because f has the form f = h + g and f is univalent in Ω, we have Jf = |fz |2 − |fz |2 = |h� |2 − |g � |2 �= 0.

Thus, f is either sense-preserving or sense-reversing. In the case of sense-preserving harmonic mappings, we have the following Theorem 7. Let f ∈ C 2 (Ω) with Jf (z) > 0 on Ω. Then the function f is harmonic on Ω if and only if f is the solution of the elliptic partial differential equation (1)

fz (z) = ω(z)fz (z),

z ∈ Ω,

for some analytic function ω on Ω with |ω(z)| < 1 on Ω.


279

Proof. ⇒: Suppose that f = h + g is a harmonic function with Jf (z) > 0 on Ω. Then, as fz (z) = h� (z) �≡ 0 and fz (z) = g � (z), we can define a function ω(z) by ω(z) := νf (z) =

fz (z) g � (z) = � h (z) fz (z)

which is analytic on Ω and |ω(z)| < 1 on Ω, because J f = |h� |2 − |g � |2 > 0. The desired form (1) follows from the last relation. ⇐: Conversely, suppose that f is a C 2 -solution of (1) with Jf (z) > 0 on Ω. Differentiating the equation (1) with respect to z, one finds fzz (z) = ω(z)fzz (z) + ωz (z)fz (z) so that fzz = ωfzz as ωz = 0. Further, as |ω(z)| < 1, we have fzz (z) = 0 and therefore, f is harmonic on Ω. We remark that on each compact subset of the unit disk, ω(z) is away from one, and so f is locally quasiconformal. However, a harmonic mapping need not be quasiconformal since its distortion may be unbounded at the boundary. 2.3. Composition Rule for Harmonic Functions. In the linear space H(Ω) of analytic functions in Ω, analytic functions are preserved under product and composition rules, but harmonic functions are not. For example, the functions x and x 2 show that the product of two harmonic functions is not necessarily harmonic. Also, it is easy to see that the composition of two harmonic functions is not necessarily harmonic. Moreover, inverse or square of a harmonic function need not be harmonic. At this point, it is important to emphasize that the class of harmonic functions is not conformally invariant. Proposition 2. 1. If f : Ω → C is analytic and g : f (Ω) → C is harmonic, then g ◦ f is harmonic. 2. If f : Ω → C is harmonic and g : f (Ω) → C is analytic, then g ◦ f is not necessarily harmonic. 3. If f : Ω → C is a harmonic, then f n is not necessarily harmonic. 4. If f : Ω → C is a harmonic, then f −1 is not necessarily harmonic. Proof. 1. Set h = g ◦ f , where f is analytic with w = f (z), and g is harmonic. Now hz (w) = gw (w)f � (z) and hzz (w) = (gw f � )z = (gw )z f � + gw (f � )z dgw dw � f + 0 (since (f � )z = 0) dw dz = gww f � f � = 0 (since g is harmonic).

=

2. Consider f (z) = z + αz (0 < |α| < 1), g(z) = z 2 and h = g ◦ f . Then, f is a sense-preserving harmonic univalent function and g is analytic. The composition gives hzz = 2α �= 0 showing that h is not harmonic. 3. For the harmonic function f (z) = z + αz (α �= 0), we have g(z) = f 2 (z) = (z + αz)2 , gz = 2(z + αz) and gzz = 2α �= 0

so that g = f 2 is not harmonic.

280


4. For n ≥ 2, consider the harmonic function f (z) = z −

1 n z , z ∈ D. n

It follows that (see Example 2), f is univalent in D and therefore, f is invertible. We see that the inverse function is not harmonic. We leave it as an exercise to verify this for n ≥ 2.

2.4. Schwarz’s Lemma. In this subsection we develop an appropriate analogue of the classical Schwarz lemma for complex-valued harmonic function in the unit disk. The classical Schwarz’s lemma is one of the cornerstones of the geometric function theory because of its relation to many important metrics both in planar and higher dimensional theory. This lemma gives information about the behaviour of an analytic function on the disk about the origin, under mild conditions. Later generalizations by Pick allow one replace the origin by other points of the disk. Schwarz’s lemma is instrumental in obtaining Landau’s theorem which provides the largest schlicht disk for the properly normalized class of bounded analytic functions in the unit disk D. Here by a schlicht disk, we mean a disk which is univalent image of some subregion of D. However, there are many generalizations of the classical Schwarz lemma including the counterpart for quasiconformal maps [2, 41, 66, 69]. Both for analytic functions and for quasiconformal mappings it has a form that is conformally invariant under conformal automorphisms of the unit disk D. The invariance property is no longer valid for harmonic functions. Indeed if φ : D → D is a conformal automorphism of the unit disk D and f : D → D is a harmonic mapping, then φ ◦ f is harmonic in D only in rare exceptional cases! In its most basic form, the familiar Schwarz lemma is the following:

Lemma 2. Let f be analytic in the unit disk D with f (0) = 0 and |f (z)| < 1 for z ∈ D. Then |f (z)| ≤ |z| for all z ∈ D, and |f � (0)| ≤ 1.

Moreover, if |f � (0)| = 1, or |f (z)| = |z| for some z �= 0, then f is a rotation of the disk, i.e., there is a constant α ∈ C with |α| = 1 such that f (z) = αz for all z in D.

Proof. Consider g given by g(z) = f (z)/z for z �= 0 and g(0) = f � (0) and apply the maximum modulus principle to the function g for |z| = r, 0 < r < 1, and then let r to approach 1. Definition 6. A function f analytic in D is said to be subordinate to an analytic univalent function g (written as f ≺ g or f (z) ≺ g(z)) if f (0) = g(0) and f (D) ⊂ g(D). If f ≺ g, then the Schwarz lemma applied to (g −1 ◦ f )(z) shows that there exists an analytic function ω on D with ω(0) = 0 and |ω(z)| < 1 for z ∈ D, and f (z) = g(ω(z)). This definition is extended to general analytic functions g univalent or not, by requiring that f have this form. There are many interesting results concerning subordinations, see for example [21].


281

Lemma 3. Let Ω = {w ∈ C : |Re w| < 1} and f : D → Ω be analytic with f (0) = 0. Then the inequalities 2 1 + |z| 4 |Re f (z)| ≤ arctan |z| and | Im f (z)| ≤ log π π 1 − |z| hold and are sharp for all z ∈ D. Proof. Define g on the unit disk D by

2i 1+z log . π 1−z Then it is a simple exercise to see that g maps D univalently onto the strip Ω. Then f (D) ⊂ g(D), and f (0) = g(0) = 0. Hence, there exists an analytic function ω : D → D with ω(0) = 0 such that f = g ◦ ω. We obtain � � 1 + ω(z) 2 Re f (z) = − arg . π 1 − ω(z) We note that the M¨ obius transformation defined by 1+z w(z) = 1−z is univalent in D and it maps the circle |z| = r (0 < r < 1) onto the circle |w − w0 | = δ, where 1 + r2 2r and δ = . w0 = 1 − r2 1 − r2 Thus, � � � �� 2r �arg 1 + z � ≤ arctan = 2 arctan r, � 1−z � 1 − r2 and � � �1 + z� � ≤ log 1 + r . � log � 1 − z� 1−r By Schwarz’s lemma, we have |ω(z)| ≤ |z| for z ∈ D so that � � �� 2 1 + ω(z) �� 4 4 � ≤ arctan |ω(z)| ≤ arctan |z|, |Re f (z)| = �− arg π 1 − ω(z) � π π g(z) =

since x �→ arctan x is increasing on (0, 1). The equalities hold in the last expression if and only if ω(z) = αz, i.e. f (z) = g(αz), for some constant α with |α| = 1. Similarly, 2 1 + |z| | Im f (z)| ≤ log π 1 − |z| and the result is sharp. A well-known harmonic version of the classical Schwarz lemma due to Heinz [30] (see also [22]) is the following: Lemma 4. If f : D → D is harmonic such that f (0) = 0, then 4 4 (2) |f (z)| ≤ arctan |z| ≤ |z|, z ∈ D. π π This inequality is sharp for each z ∈ D. Furthermore, the bound is sharp everywhere (but is attained only at the origin) for univalent harmonic mappings f of D onto itself with f (0) = 0.

282


Proof. For θ ∈ [0, 2π), let uθ (z) = Re (eiθ f (z)) and observe that |uθ (z)| = |Re (eiθ f (z))| ≤ |eiθ f (z)| < 1.

By Lemma 3,

4 arctan |z|. π As a consequence of the arbitrariness of θ in [0, 2π), it follows that |Re (eiθ f (z))| ≤ |f (z)| ≤ The equality holds when eiθ f (z) = g(αz) =

2 log π

�

4 arctan |z|. π

1 + αz 1 − αz

� and Im(eiθ f (z)) = 0,

so we conclude that the harmonic function has the form � � 1 + αz 2β ϕ(z) = arg , |α| = |β| = 1. π 1 − αz

Up to rotation, ϕ is the Poisson integral of e iθ(t) , where � 0 for 0 ≤ t ≤ π, θ(t) = π for π < t < 2π. Indeed, if 1 f (z) = 2π

� 0

2π

eit + z iθ(t) 1 e Re it dt = e −z 2π

�

π 0

1 eit + z dt − Re it e −z 2π

�

2π

π

Re

eit + z dt, eit − z

then (with the change of variable t = s + π for the second integral), we have � π � π 1 1 eit + z eit − z f (z) = dt − dt Re it Re it 2π 0 e −z 2π 0 e +z � � π �� ∞ ∞ � 1 −itn n −itn n n = Re e z − e (−1) z dt π 0 n=1 n=1 �∞ � � e−iπn − 1 1 n n = Re (1 − (−1) )z π −in n=1 ∞ � 1 1 (1 − (−1)n )2 z n = Re π in n=1

∞ ∞ � � 1 1 1 4 Im (1 − (−1)n )2 z n = − Im z 2k−1 π n π 2k − 1 n=1 k=1 � � 1+z 1+z 2 2 . = − Im log = − arg π 1−z π 1−z

=−

Approximating this step function by a continuous increasing function θ(t) with � 2π θ(2π) = 2π, θ(0) = 0 and 0 eiθ(t) = 0, by Rad´ o-Kneser-Choquet theorem below, we can produce univalent harmonic mapping f of D onto D with f (0) = 0 and |f (z)| arbitrary close to π4 arctan |z|, for z a prescribed point of the disk. But for this case, |f (z)| = π4 arctan |z| only when z = 0. Thus, the bound is sharp even for harmonic mappings of the disk onto itself.


283

We recall the following without proof: Theorem 8 (Rad´ o-Kneser-Choquet). A function f : D → D is a sense preserving harmonic mapping from the unit disk D onto itself if and only if there exists a mapping t �→ θ(t) which is continuous and non-decreasing for t ∈ [0, 2π] and such that θ(2π) − θ(0) = 2π with harmonic extension � 2π 1 − |z|2 iθ(t) 1 f (z) = e dt. 2π 0 |eit − z|2 2.5. Landau-Bloch Theorems. Heinz’s lemma played an important role in determining bounds for Bloch and Landau constants (see [13, 14, 15, 16, 43]). We begin the discussion by recalling: Theorem 9 (Landau). Suppose that f is analytic in D, |f (z)| ≤ M in D, f (0) = f � (0) − 1 = 0. Then (i) f is univalent for |z| < ρ; i.e. ρ−1 f (ρz) ∈ S, (ii) f (Dρ ) ⊇ DR ,

where

1 √ and R = M ρ2 . M + M2 − 1 The numbers ρ and R are the best possible. Here S denotes the class of all analytic functions f that are univalent in the unit disk D, with the normalization f (0) = f � (0) − 1 = 0. ρ=

Remark 3. (i) If M = 1, then the Schwarz lemma gives that f (z) = z. (ii) For the class H of normalized analytic functions in D without the boundedness condition on f , there is still the Bloch theorem which asserts the existence of a positive constant b such that for any f ∈ H the image f (D) contains a schlicht disk of radius b. The Bloch constant is then defined to be the “largest” such constant, i.e., the supremum of all such b. One of the outstanding open problems in the classical complex analysis is perhaps that of determining the precise value of the Bloch constant for H. We refer to the book of Ponnusamy • [54] for general discussion on Bloch constant. Although the Bloch theorem does not hold for normalized harmonic functions, the authors in [13] obtained Bloch’s theorem for normalized quasiregular harmonic mappings and open harmonic mappings. Analogous problem of estimating the Bloch constant for certain class of harmonic mappings has been done by a number of authors. It turns out that one requires suitable additional assumption other than the usual normalization in order to obtain Landau and Bloch theorems and the best known lower estimate is in [16]. We refer to the paper of Bochner [8] for the existence of the Bloch constant for K-quasiregular harmonic mappings (even in higher dimensional case). In [19, Theorem 1.3], Colonna considered harmonic Bloch mappings and obtained the following result which seems to be important for further discussion in this line. Theorem 10 ([19, Theorem 1]). Let f be a harmonic mapping in D. Then (1) Bf = sup(1 − |z|2 )(|h� (z)| + |g � (z)|). z∈D

284


(2) Bf ≤ 4/π, if in addition f (D) ⊂ D.

Here Bf denotes the Bloch constant for f . The following result is crucial in improving the estimate for the Landau-Bloch constants, see [15, 16]: Lemma 5. Suppose that f = h + g is harmonic in D and |f (z)| ≤ 1 for z ∈ D, where ∞ ∞ � � an z n and g(z) = bn z n . h(z) = n=0

n=1

Then |a0 | ≤ 1 and for each n ≥ 1,

4 . π The estimate (3) is sharp for any n ≥ 1. For each n ≥ 1, the extremal function is � � 1 + βz n 2α fn (z) = arg , |α| = |β| = 1 π 1 − βz n

(3)

|an | + |bn | ≤

or f (z) ≡ M.

Proof. Without loss of generality, we assume f (z) = h(z) + g(z) and |f (z)| < 1. For θ ∈ [0, 2π), let vθ (z) = Im(eiθ f (z)) and observe that vθ (z) = Im(eiθ h(z) + e−iθ g(z)) = Im(eiθ h(z) − e−iθ g(z)).

Because |vθ (z)| < 1, it follows that iθ

e h(z) − e

−iθ

2 g(z) ≺ K(z) = λ + log π

�

1 + zξ 1−z

� ,

where ξ = e−iπ Im(λ) and λ = eiθ h(0) − e−iθ g(0). It is a simple exercise to see that the superordinate function K(z) maps D onto a convex domain with 2 (1 + ξ), π and therefore, by a theorem of Rogosinski [63, Theorem 2.3] (see also [21, Theorem 6.4]), it follows that K(0) = λ and K � (0) =

4 2 |1 + ξ| ≤ for n = 1, 2, . . . π π and the desired inequality (3) is a consequence of the arbitrariness of θ in [0, 2π). For the proof of sharpness part, consider the functions � � 2α 1 + βz n fn (z) = , |α| = |β| = 1, Im log π 1 − βz n |an − e−2iθ bn | ≤

whose values are confined to a diametral segment of the unit disk D. Also, �∞ � ∞ � 1 2α � 1 n 2k−1 n 2k−1 fn (z) = (βz ) (βz ) − , iπ 2k − 1 2k − 1 k=1

k=1


which gives |an | + |bn | =

The proof of the lemma is complete.

285

4 . π

2.6. Harmonic Mappings of the Plane C. For entire functions (analytic in C), it is well-known that the only univalent analytic self mappings of C are the linear mappings of the form f (z) = a0 + a1 z, where a0 , a1 are constants with a1 �= 0 (see for example [54] and [60]). It is natural to ask for the harmonic analog of this result. Theorem 11. [18] The only univalent harmonic mappings of C onto C are the affine mappings f (z) = αz + γ + βz, where α, β and γ are complex constants and |α| �= |β|. Proof. Let f map C harmonically onto C. Then f has the form f = h + g, where h and g are entire functions, and we may assume without loss of generality that f is sense-preserving. As f is sense-preserving, we have |g � (z)| < |h� (z)| in C and so g � /h� , being a bounded entire function, reduces to a constant (by Liouville’s theorem). Consequently, g � (z) = bh� (z) so that integration gives g(z) = bh(z) + c for some complex constants b and c with |b| < 1. Thus, f reduces to the form f (z) = h(z) + b h(z) + c.

Setting w = h(z), we may write f (z) = F (h(z)) = (F ◦ h)(z) with F (w) = w + bw + c.

Note that F is an (invertible) affine mapping. It follows that h = F −1 ◦ f is analytic and maps C univalently onto C, and so h(z) = a0 + a1 z, where a0 , a1 are complex constants with a1 �= 0. This shows that f has the form f (z) = a0 + a1 z + b(a0 + a1 z) + c,

which is the affine mapping in the desired form. We remark that the simplest example of sense-preserving harmonic mapping on the plane that is not necessarily conformal is an affine mapping f (z) = αz + βz,

|α| > |β| > 0.

According to Theorem 11, there exists no univalent harmonic mapping of C onto proper sub-domain of C. In particular, there exists no univalent harmonic mapping of C onto the unit disk D. This fact can be also obtained as a simple consequence of the Liouville theorem. Also, there can be no analytic function which can map D univalently onto C, because otherwise, inverse function exists which is analytic mapping of C into D and by Liouville’s theorem reduces to constant. This argument cannot be used for harmonic functions as inverse of a harmonic function need not be harmonic. However, the analog of this result for the harmonic functions does hold. This is indeed due to an important result due to Rad´ o (see Theorem 15).

286


1 0.5 0 −0.5 −1

−1.5

−1

−0.5

0

0.5

1

1.5

Figure 6. Image of unit disk under f n (z) = z + (n/(n + 1))z for n = 1, 2, 3, 4. Example 3. For n ≥ 1, consider (compare with Lemma 1) n z. fn (z) = z + n+1 Then, each fn is a univalent harmonic mapping in C (and in particular, in the unit disk D). Note that {fn } converges uniformly to f (z) = z + z which is clearly a harmonic function but not univalent, see Figure 6. What does this mean? •

Example 4. Consider

1 + 2 ln |z|. z Then it is easy to see that f is a harmonic mapping on the exterior Ω = C \ D of the unit disk D onto the punctured complex plane C \ {0}. Note that f (∂D) is simply the origin. We write 1 f = h + g, h(z) = z + log z and g(z) = − + log z z and note that h and g are not (globally) analytic on C \ D which is not simply connected in C. So, we require an analog of Lemma 1 for multiply connected • domains. f (z) = z −

3. Quasiconformal Mappings in the Plane We shall outline basic definitions and properties of quasiconformal mappings in the plane. Quasiconformal mappings are a natural generalization of conformal mappings. Standard references in this topic are [2], [4] and [41]. Quasiconformal


287

mappings are useful tools in the theory of planar harmonic mappings. This arises from the fact that planar harmonic mappings are locally quasiconformal, as remarked in the beginning. Some familiarity with real analysis and measure theory is required for this study. There are many areas in which quasiconformal mappings are used. See [2, 4, 40, 41]. For the higher dimensional theory, we refer to the books of V¨ aisälä [66] and Vuorinen [69] although at this stage it is too early for us to discuss higher dimensional results. First we recall the problem of Gr¨ otzsch: Problem 1. Let Q be a square and R be a rectangle which is not a square. In 1928, H. Gr¨ otzsch, posed the following problem: Does there exist a conformal mapping of Q onto R which maps vertices onto vertices? Because of the extra condition that the vertices mapping onto vertices, the Riemann mapping theorem does not guarantee existence of such a mapping and so the problem of Gr¨ otzsch is interesting in itself. It is this extra condition that led us to the development of the theory of quasiconformal mappings. Actually, there does not exist a conformal mapping of Q onto R taking vertices onto vertices and hence, the question is to find most nearly conformal mapping of this kind, and this needs a measure of approximate conformality. What is important about quasiconformal mappings? We recall the following facts: 1. Quasiconformal maps are considered to be a natural generalization of conformal maps. 2. In many of the results on conformal mappings, one requires just the quasiconformality and hence, it is of interest to know when conformality is necessary and when it is not. 3. Quasiconformal maps behave less rigidly than conformal maps and thus can be used as a tool in complex analysis. 4. Later development shows that the class of quasiconformal mappings plays an important role in the study of elliptic partial differential equations. 5. Extremal problems in quasiconformal mappings lead to analytic functions connected with regions or Riemann surfaces. 6. The family of conformal mappings degenerates to M¨ obius transformations when generalized to several variables, but the family of quasiconformal mappings is interesting in the higher dimensions also (see [66]). 3.1. Quasiconformal Diffeomorphisms. Let f : Ω → f (Ω) be a sense-preserving C 1 -diffeomorphism, z0 ∈ Ω, and w = f (z) = u(z) + iv(z). Then, dw = df = fz dz + fz dz

and, since f is a diffeomorphism, it is locally linear (in this case at z 0 ). Indeed, the affine map L defined by L(z) := fz (z0 )dz + fz (z0 )dz sends a circle with center 0 in the dz-plane onto an ellipse in the dw-plane, with major axis of length L = |fz (z0 )| + |fz (z0 )| and minor axis of length l = |fz (z0 )| − |fz (z0 )|. It follows that (|fz | − |fz |)|dz| ≤ |dw| ≤ (|fz | + |fz |)|dz|,

288


dy

dz−plane

df r

Λfr

λfr

0

dw−plane

dv

dx

0

du

Figure 7. Image of the circle |dz| = r under the differential map dw. where both the limits are attained. The differential dw maps the circle |dz| = r onto the ellipse, see Figure 7. The ratio D f (z) between the major and the minor axes is |fz | + |fz | ≥ 1. Df := |fz | − |fz | This quantity Df (z) is called dilatation of f at the point z ∈ Ω. Definition 7. A sense-preserving diffeomorphism f is said to be K-quasiconformal, if Df (z) ≤ K throughout the given region Ω, where K ∈ [1, ∞) is a constant.

We define the quantity df (z) as follows: df :=

|fz | < 1, |fz |

because, for sense-preserving maps, |f z | > |fz |. Thus, we have (4)

Df (z) =

1 + df (z) Df (z) − 1 and df (z) = . 1 − df (z) Df (z) + 1

We may use Df as a measure of the local distortion of the mapping f at z. Now we define complex dilatation μf by fz (5) μf = with |μf | = df . fz Thus, by (4) and (5), the two dilatations μf and Df are related by |μf (z)| =

Df (z) − 1 . Df (z) + 1

If ∂α f (z) denotes the directional derivative of a C 1 -mapping f in a direction making an angle α with the positive x-direction, then f (z + reiα ) − f (z) iα r→0 � re u(x + r cos α, y + r sin α) − u(x, y) = e−iα lim r→0 r � v(x + r cos α, y + r sin α) − v(x, y) +i . r

∂α f (z) =

lim


289

By adding and subtracting the terms u(x, y + r sin α) and v(x, y + r sin α) in the numerators of the first and the second terms on the right, respectively, we obtain ∂α f (z) = e−iα [(ux cos α + uy sin α) + i(vx cos α + vy sin α)] = e−iα [fx cos α + fy sin α] = e−iα [(fz + fz ) cos α + i(fz − fz ) sin α] = e−iα [fz eiα + fz e−iα ]

= fz + fz e−2iα � � fz −2iα = fz 1 + e fz

for fz �= 0.

Hence, for a sense-preserving C 1 -map f between planar domains, we have Λf (z) := max |∂α f (z)| = |fz (z)| + |fz (z)| α

and λf (z) := min |∂α f (z)| = |fz (z)| − |fz (z)|. α

It is an easy exercise to see that if z 1 �= 0, then and similarly,

|z1 + z2 | = |z1 | + |z2 | ⇐⇒ arg(z2 /z1 ) = 0,

|z1 + z2 | = |z1 | − |z2 | ⇐⇒ arg(z2 /z1 ) = ±π.

Therefore the maximum is attained provided that � � � � fz fz −2iα 1 arg e = 0, i.e., α = arg . fz 2 fz Thus, the maximum corresponds to the direction 1 arg μf , 2 and the minimum to the orthogonal direction π arg dz = β = α ± . 2 arg dz = α =

We define the quantity ω := νf of f by ω(z) := νf (z) =

fz (z) . fz (z)

The function νf (z) is called the second complex dilatation which turns out to be more natural for studying harmonic mappings than the first complex dilatation μ f . Because |νf | = |μf |, f is quasiconformal if and only if |νf (z)| ≤ k < 1. Finally, we remark that K −1 . Df ≤ K ⇔ df ≤ k := K +1 The mapping f is conformal if and only if μ f (z) = 0 on Ω, i.e., fz vanishes identically on Ω. From our notation, ∂α f (z) is then independent of α so that ∂α f (z) = fz (z) = f � (z).

290


This is equivalent to the dilatation quotient being identically equal to 1. Thus, f is conformal if and only if Df = 1 and μf = 0. If K, defined by K = sup Df = sup z∈Ω

z∈Ω

1 + |μf (z)| , 1 − |μf (z)|

is bounded, then we call f a K-quasiconformal mapping of Ω to f (Ω). We call K the maximal dilatation of f . The maximal dilatation is a finite number larger or equal to 1. Sometimes, f is called a K-quasiconformal mapping with the Beltrami coefficient μf . The simplest example of quasiconformal mapping, which is not conformal, is the affine mapping f (z) = az + bz (a, b ∈ C, 0 < |b| < |a|);

for fz = a, fz = b, |μf | = |b|/|a| and f maps the unit circle to an ellipse, and the ratio of the major and minor axes of this ellipse is K=

1 + |μf | |a| + |b| = . |a| − |b| 1 − |μf |

As another example, we consider � z f (z) = x + iky

if z ∈ U = {w : Re w > 0}, if z ∈ C \ U, k > 1.

Then fz = 1 and fz = 0 if z ∈ U. If z ∈ C \ U, then we may rewrite f as � � z−z z+z + ik f (z) = 2 2i so that fz = Thus,

1+k 1−k and fz = . 2 2

(1 + k)/2 + (k − 1)/2 = k, (1 + k)/2 − (k − 1)/2 and hence, f is a K-quasiconformal mapping of C with K = k. Df =

3.2. General Quasiconformal Mappings. We now recall a general definition of quasiconformal mappings, where we do not need to assume that the function is a diffeomorphism. Definition 8. Let Ω1 , Ω2 be domains in the extended complex plane C∞ . Suppose that f : Ω1 → Ω2 is a sense-preserving homeomorphism. For each z ∈ Ω1 \ {∞, f −1 (∞)}, we define the linear dilatation of f at z by Hf (z) = lim sup r→0

Lf (z, r) , �f (z, r)

where Lf (z, r) = max |f (z) − f (w)|, and

|z−w|

�f (z, r) = min |f (z) − f (w)|. |z−w|


291

We say that f is K-quasiconformal, if Hf is bounded in Ω \ {∞, f −1 (∞)} and Hf (z) ≤ K a.e. in Ω,

where 1 ≤ K < ∞ is a uniform constant.

f z f(z)

r

L

Figure 8. Description for quasiconformality. Examples 1.

(1) A homeomorphism f : Ω → f Ω satisfying |z − w|/L ≤ |f (z) − f (w)| ≤ L|z − w|

for all z, w ∈ Ω is called L-bilipschitz. It is easy to see that L-bilipschitz mappings are L2 -quasiconformal. (2) Not all quasiconformal mappings are bilipschitz. The standard counterexample is the quasiconformal radial stretching z �→ |z|α−1 z,

z ∈ D,

where α ∈ (0, 1). (3) Not all quasiconformal mappings are diffeomorphisms. For example, one may • take a piecewise affine mapping.

Definition 9 (Class ACL(Ω)). Let f be a complex-valued function defined on subinterval I of R. Suppose that for all ε > 0, there is a δ > 0 such that n � k=1

|f (ak ) − f (bk )| < ε

for every finite, pairwise disjoint family of open subintervals [a k , bk ], k = 1, 2, . . . , n of I such that n � |ak − bk | < δ. k=1

Then f is said to be absolutely continuous in I. Obviously, an absolutely continuous function is continuous: simply take n = 1. Let Ω be a domain in C. We say that a function u : Ω → R is absolutely continuous on lines (ACL) in Ω if, for each rectangle R = [a, b] × [c, d] ⊂ Ω,

the function u(x + iy) is absolutely continuous with respect to the variable x a.e. in [a, b] and with respect to y a.e. in [c, d].

292


The next important result [41, IV.2.3] gives an analytic characterization of quasiconformality. A sense preserving homeomorphism f : Ω 1 → Ω2 is K-quasiconformal if and only if f is in ACL(Ω1 ) and (6)

max |∂α f (z)|2 ≤ K Jf (z) a.e. in Ω1 , α

where ∂α denotes the derivative of f to the direction α and J f (z) is the Jacobian of f at z. This characterization is said to be the analytic definition of quasiconformality. Remark 4. If the inequality (6) holds for an ACL(Ω1 ) map f a.e. in Ω1 , but f is not necessarily a homeomorphism, then f is called K-quasiregular. By a result of S. Stoilow a quasiregular mapping f of D onto a domain Ω can be represented as f = g ◦ h, where h is a quasiconformal mapping of D onto itself and g is an analytic • function [41]. 3.3. Conformal Modulus. Suppose that Γ is a path family in C ∞ . We will assign a conformally invariant quantity, modulus, to Γ, which measures the size of Γ. Let ρ : C → [0, ∞] be a Borel measurable function. We call ρ admissible for Γ (denoted ρ ∈ F(Γ)) if � ρ(z) |dz| ≥ 1, γ

for each locally rectifiable γ ∈ Γ. We define the modulus of Γ by �� (7) M(Γ) = inf ρ2 (z) dx dy, ρ

C

where the infimum is taken over all ρ ∈ F(Γ).

The following basic properties of the conformal modulus are well-known: Lemma 6. (1) The modulus is an outer measure in the space of all path families in C, i.e., (i) M(∅) = 0, (ii) If �Γ� 1 ⊂ Γ�2 then � 1 )�≤ M(Γ2 ), and � M(Γ (iii) M j Γj ≤ j M Γj . (2) We say that Γ2 is minorized by Γ1 and write Γ1 < Γ2 if every γ ∈ Γ2 has a subpath in Γ1 . If Γ1 < Γ2 then M(Γ1 ) ≥ M(Γ2 ). (3) If Γ is the family of paths in Ω such that �(γ) ≥ r, then M(Γ) ≤ m(Ω)r −2 , where m(Ω) is the Lebesgue measure of Ω. Proof. (1) (i) Since the zero function is admissible for ∅, M(∅) = 0. (ii) If Γ1 ⊂ Γ2 then F(Γ2 ) ⊂ F(Γ1 ) and hence M(Γ1 ) ≤ M(Γ2 ).

(iii) We may assume that M(Γj ) < ∞ for all j. Let ε > 0. Then we may choose for each j a function ρj admissible for Γj such that �� ρ2j dx dy ≤ M(Γj ) + 2−j ε. Now let

C

ρ = sup ρj j

and Γ =

� j

Γj .


293

Then ρ : C → [0, ∞] is a Borel function. Moreover, if γ ∈ Γ is locally rectifiable, then γ ∈ Γj for some j, � � ρ |dz| ≥ ρj |dz| ≥ 1, γ

γ

and hence, ρ is admissible for Γ. Now M(Γ)

≤ ≤ ≤

�� C

ρ2 dx dy

�� C

�

ρ2j dx dy

j

M(Γj ) + ε,

j

and the claim follows by letting ε → 0. (2) If Γ1 < Γ2 then obviously F(Γ1 ) ⊂ F(Γ2 ). Hence M(Γ1 ) ≥ M(Γ2 ). (3) The claim follows immediately from (7) and the fact that the function ρ = χ Ω /r is admissible for Γ. Let Ω1 , Ω2 ⊂ C and f : Ω1 → Ω2 be a continuous function. Suppose that Γ is a family of paths in Ω1 . Then Γ� = {f ◦ γ : γ ∈ Γ}, the image of Γ under f , is a family of paths in Ω2 . Theorem 12. If f : Ω1 → Ω2 is conformal, then M(f (Γ)) = M(Γ) for all path families Γ in Ω1 . Proof. Let ρ1 ∈ F(f (Γ)), and define

� � ρ(z) = ρ1 f (z) |f � (z)|

for z ∈ Ω1 and ρ(z) = 0 otherwise. Because f is conformal, � � � � ρ |dz| = ρ1 f (z) |f � (z)| |dz| γ γ � ρ1 |dz| ≥ 1 = f ◦γ

for every locally rectifiable γ ∈ Γ. Hence, ρ ∈ F(Γ), and �� M(Γ) ≤ ρ2 dx dy C �� = ρ21 f (z) |Jf (x)| dx dy �� Ω1 ρ21 dx dy = ρ21 dx dy = Ω2

C

for all ρ1 ∈ F(f (Γ)), and thus M(Γ) ≤ M(f (Γ)). The inverse inequality follows from the fact that f −1 is conformal. We now recall the geometric definition of quasiconformality. A sense preserving homeomorphism f : Ω1 → Ω2 is K-quasiconformal if and only if it satisfies the inequalities (8)

M(Γ)/K ≤ M(f Γ) ≤ KM(Γ)

for every family of paths Γ in Ω1 (see [41, IV.3.3]).

294


3.4. Basic Properties of Quasiconformal Mappings. The following properties of quasiconformal mappings are well-known. (1) If f : Ω1 → Ω2 is K1 -quasiconformal and g : Ω2 → Ω3 is K2 -quasiconformal, then g ◦ f : Ω1 → Ω3 is K1 K2 -quasiconformal. The inverse mapping f −1 : Ω2 → Ω1 of a K-quasiconformal mapping is K-quasiconformal. These properties follow immediately from the geometric definition of quasiconformality. (2) A mapping f of Ω is 1-quasiconformal if and only if it is a conformal mapping in Ω \ {∞, f −1 (∞)} [41, I.5.1].

(3) If f : Ω1 → Ω2 is K-quasiconformal and Ω1 , Ω2 are Jordan domains, then f can be extended to a homeomorphism from Ω1 onto Ω2 [41, I.8.2]. (4) Let E ⊂ Ω1 be a closed set, and suppose that E can be expressed as an enumerable union of rectifiable curves. If f : Ω 1 → Ω2 is a homeomorphism, and f is K-quasiconformal in each component of Ω 1 \ E, then f is K-quasiconformal in Ω1 [41, V.3.4]. 3.5. Estimates with Modulus. In general, it is difficult to calculate the modulus of a given path family. However, it is possible to obtain effective upper and lower bounds for the modulus in many situations. Estimates of this type can be used in the geometric theory of quasiconformal mappings. In particular, this technique is used for obtaining distortion results for quasiconformal mappings. At first we introduce the concept of conformal modulus or simply modulus of a quadrilateral. A Jordan domain Ω in C with marked (positively ordered) points z1 , z2 , z3 , z4 ∈ ∂Ω is a quadrilateral and denoted by (Ω; z 1 , z2 , z3 , z4 ) . We use the canonical conformal mapping of quadrilateral onto a rectangle (Ω � ; 1 + ih, ih, 0, 1), with the vertices corresponding, to define the modulus mod(Q) = h of a quadrilateral Q = (Ω; z1 , z2 , z3 , z4 ) . The modulus of (Ω; z2 , z3 , z4 , z1 ) is 1/h . By this we mean that for h = mod(Q) and for this value only there exists a unique conformal mapping from Q onto the rectangle Ω� , which takes the four points z1 , z2 , z3 , z4 ∈ ∂Ω onto the four points 1 + ih, ih, 0, 1 of the rectangle Ω � . Moduli of quadilaterals and path families are connected to each other as follows. Lemma 7. Let Q = (Ω; z1 , z2 , z3 , z4 ) be a quadrilateral and denote by γj the boundary arc of Ω connecting zj and zj+1 for j = 1, 2, 3 and z4 , z1 for j = 4. Then mod(Q) = 1/M(Γ), where Γ is the family of paths connecting γ2 and γ4 in Ω. Proof. By conformal invariance, we may assume that Q is the rectangle with boundary points 1 + ih, ih, 0 and 1. We show that M(Γ) = 1/h. Choose ρ ∈ F(Γ), and let γy be the vertical segment [y, y + ih], where y ∈ (0, 1). older’s inequality Then γy ∈ Γ. We note that by H¨ �� 2 1 ≤ ρ |dz| γy

��

≤ =

��

γy

|dz|

γy

ρ2 |dz|.

� h

γy

�

ρ2 |dz|


295

This holds for all y ∈ (0, 1), and hence by Fubini’s theorem ��

�

2

C

ρ dx dy ≥

1

� 1 ρ |dz| dt ≥ . h γy

��

0

2

Since the above holds for any ρ ∈ F(Γ), M(Γ) ≥

1 . h

Next we choose ρ = 1/h inside Ω, and ρ = 0 otherwise. Then ρ is admissible for Γ and �� 1 ρ2 dx dy = . M(Γ) ≤ h C Remark 5. Suppose that Ω is a region in the complex plane whose boundary ∂Ω consists of a finite number of regular Jordan curves, so that at every point, except possibly at finitely many points, of the boundary a normal ∂n is defined. Then the following problem is known as the Dirichlet-Neumann problem. Let ψ be a real-valued continuous function defined on ∂Ω. Let ∂Ω = A ∪ B where A, B both are unions of Jordan arcs. Find a function u satisfying the following conditions: 1. u is continuous and differentiable in Ω. 2. u(t) = ψ(t), t ∈ A. 3. If ∂/∂n denotes differentiation in the direction of the exterior normal, then ∂ u(t) = ψ(t), ∂n

t ∈ B.

It is possible to express the modulus of a quadrilateral (Ω; z 1 , z2 , z3 , z4 ) in terms of the solution of the Dirichlet-Neumann problem as follows. Let γj , j = 1, 2, 3, 4, be the arcs of ∂Ω between (z 1 , z2 ) , (z2 , z3 ) , (z3 , z4 ) , (z4 , z1 ), respectively. If u is the (unique) harmonic solution of the Dirichlet-Neumann problem with boundary values of u equal to 0 on γ2 , equal to 1 on γ4 and with ∂u/∂n = 0 on γ1 ∪ γ3 , then by [3, p. 65/Thm 4.5]: �� (9) mod(Ω; z1 , z2 , z3 , z4 ) = |�u|2 dx dy. Ω

The solutions of the Dirichlet-Neumann problems can be approximated by the method of finite elements, see [33, pp. 305–314], [49]. Hence, this method can be used to obtain numerical approximations of the conformal modulus, see e.g. [6, 24]. Other numerical methods for approximating conformal modulus are presented in [23, 50] (see also [49]). It is also possible to construct the canonical conformal mapping from the solution • of the boundary value problem, see [28].

296


3.6. Ring Domains. A domain Ω in C∞ is called a ring, if C \ Ω has exactly two components. If the boundary components are E and F , we denote the ring by R(E, F ). For E, F, Ω ⊂ C∞ we denote by Δ(E, F ; Ω) the family of all non-constant paths joining E and F in Ω. The conformal modulus of a ring domain R(E, F ) is defined by modR(E, F ) =

2π � �, M Δ(E, F ; Ω)

provided that M(Δ(E, F ; Ω)) �= 0. Otherwise, we define modR(E, F ) = ∞. Note that this happens only if either E or F is a singleton. Lemma 8. Let 0 < a < b < ∞, A = D(b) \ D(a) and � � ΓA = Δ S(a), S(b); A , where D(r) and S(r) denote the disk {z : |z| < r} and the circle {z : |z| = r}, respectively. Then 2π . M(ΓA ) = log ab / A. For each w with Proof. Let ρ ∈ F(ΓA ). We may assume that ρ(z) = 0 for z ∈ |w| = 1, let γw : [a, b] → C be the radial line segment defined by γ w (z) = |z|w. By H¨ older’s inequality we obtain �� 2 1 ≤ ρ |dz| γw b

��

=

��

b

ρ(sw)2 s ds

≤

a

log

b a

�

a

� 1 ds s

b

ρ(sw)2 s ds. a

By integrating over the unit circle, we have �� b ρ2 dx dy. 2π ≤ log a C Taking the infimum over all admissible ρ yields � b� M(ΓA ). 2π ≤ log a � � Next we define ρ(z) = 1/ |z| log(b/a) for z ∈ A, and ρ(z) = 0 otherwise. Clearly ρ is admissible for ΓA , and hence �� ρ2 dx dy M(ΓA ) ≤ C

�

b �−2 = 2π log a 2π . = log ab

�

a

b

1 ds s


297

Every ring domain R can be mapped conformally onto the annulus {z : 1 < |z| < eM }, where M = modR is the conformal modulus of the ring domain R. Hence the conformal modulus and Lemma 8 can be used to characterize conformally equivalent ring domains. This result may now be stated without proof (see [56]). Theorem 13. Let 0 < ri < Ri < ∞ (i = 1, 2). The annular domains A(r1 , R1 ) and A(r2 , R2 ) are conformally equivalent if and only if they are similar, i.e., Rr11 = Rr22 ; Equivalently, if and only if the annuli have the same moduli: � � � � R1 R2 1 1 log = log . 2 r1 2 r2 Remark 6. The connection between ring and quadrilateral moduli is given in [39, p. 102] or [41, p. 36]. If we map the annulus Ar = {z : 1 < |z| < r} ,

with the segment [1, r] on the real axis removed, by z �→ log z = log |z| + i arg z,

0 < arg z < 2π,

the image is the rectangle R with vertices (log r + 2πi, 2πi, 0, log r) and we see that mod(R; log r + 2πi, 2πi, 0, log r) =

2π . log r

Thus 2π/ log r has this interpretation as the modulus of a quadrilateral. Lemma 9. Let Q be either a quadrilateral or ring domain Q such that Q ⊆ Ω. Then, for a K-quasiconformal mapping f of a domain Ω onto f (Ω), mod(Q) ≤ mod(f (Q)) ≤ Kmod(Q). K Proof. Since every quadrilateral is conformally equivalent to a rectangle, we may assume that f maps a rectangle onto a rectangle. If f is K-quasiconformal, then the claim follows from (8) and Lemma 7. Similarly, the result for ring domains follows from (8) and Lemma 8. Remark 7. In fact, a sense-preserving homeomorphism f : Ω 1 → Ω2 is quasiconformal if and only if the moduli of quadrilaterals are K-quasi-invariant (see [2, Chapter II] or [41, I.5.3]). 3.7. Canonical Ring Domains. The complementary components of the Gr¨ otzsch ring BG (s) in C∞ are the unit circle S = {z : |z| = 1} and [s, ∞], s > 1, and those of the Teichm¨ uller ring BT (s) are [−1, 0] and [s, ∞], s > 0. We define two special functions γ(s), s > 1 and τ (s), s > 0 by � � � γ(s) = M�Δ(S, [s, ∞]; C) , � τ (s) = M Δ([−1, 0], [s, ∞]; C) ,

respectively. We shall refer to these functions as the Gr¨ otzsch modulus and the Teichm¨ uller modulus, respectively. We also use the bounded version of the Gr¨ otzsch ring B(s) = D \ [0, s], where s ∈ (0, 1). By conformal invariance, the modulus of the path family connecting boundary components of B(s) is γ(1/s).

298


Γ ∞

0 s D

Figure 9. Gr¨ otzsch ring domain BG (s)

Γ ∞ −1

0

s

Figure 10. Teichm¨ uller ring domain BT (s) Gr¨ otzsch and Teichm¨ uller moduli functions are strictly decreasing and continuous with range (0, ∞). They are connected by the identity [5, Theorem 8.37]: γ(s) = 2τ (s2 − 1),

(10)

s > 1.

For s > 1, we also have the identity [5, 8.57]: 4 �s − 1� , γ(s) = μ π s+1 where, for r ∈ (0, 1), μ(r) is the quantity μ(r) = √

π K(r� ) , 2 K(r)

K(r) =

� 0

1

dx � , (1 − x2 )(1 − r2 x2 )

and r = 1 − r2 . The function μ(r) is the conformal modulus of the bounded Gr¨ otzsch ring domain B(r) = D \ [0, r] and hence for r > 1, �

(11)

γ(r) =

2π . μ(1/r)

These relations allow us to obtain useful estimates for the functions γ and τ (see e.g. [5, Chapters 5 and 8]). The following functional identities for μ(r) are well known [5, 5.2]: μ(r)μ(r� ) = �1 − r� μ(r)μ = 1+r (12)

μ(r)

=

π2 , 4 π2 , 2 � 2√r � 2μ . 1+r


299

Canonical ring domains are useful in the theory of quasiconformal mappings because they can be used for obtaining a lower bound for the conformal modulus of any given ring domain. Lemma 10. [5, Theorem 8.44] Let R = R(E, F ) be a ring in C∞ and let a, b ∈ E and c, ∞ ∈ F be distinct points. Then 2π ≥ τ (s), modR

s=

|a − c| . |a − b|

Obviously, equality holds for the Teichm¨ uller ring BT (s). 3.8. Schwarz’s Lemma for Quasiconformal Mappings. From the estimates above, it is easy to obtain a version of Schwarz’s lemma for quasiconformal mappings. This result is also useful in studying harmonic mappings. Lemma 11. Let f : D → f (D) ⊂ D be a K-quasiconformal mapping with f (0) = 0. Then |f (z)| ≤ ϕK (|z|), where ϕK is the distortion function � ϕK (r) = μ−1

� μ(r) . K

For a fixed K > 1, the bound ϕK (r) increases with r from 0 to 1. Proof. Fix r ∈ (0, 1), and let s = maxz∈S(r) |f (z)|. We may assume that f (r) = s. Then by Lemma 6 (2) and Lemma 9, � � μ(r) = modB(r) ≤ KmodR f ([0, r]), S(1) . Let

� g1 (z) =

1+z 1−z

�2 and g2 (z) =

(1 − z)(1 − s)2 . 4s

We note that g = g2 ◦ g1 maps D conformally onto the domain C \ {x ∈ R : x ≥ (1 − s)2 /(4s)} with g(0) = 0 and g(s) = −1. Then by Lemma 10, (10), (11) and (12) we have � modR f ([0, r]), S(1)) ≤ = =

2π � � τ 4s/(1 − s)2 4π � √ � γ (1 + s)/(2 s) � √ � 2 s 2μ = μ(s). 1+s

Because μ(s) is a strictly decreasing function, the claim follows.

300


3.9. Heinz’s Inequality. In this section we will study the function � � Df (z) = |fz (z)|2 + |fz (z)|2 = (1/2) u2x (z) + u2y (z) + vx2 (z) + vy2 (z) ,

where f = u + iv. This function is the square of the norm of the first differential and it obviously coincides with the square of the modulus of the complex derivative if the function f is analytic. According to Lewy’s theorem we obtain that the Jacobian of every sense preserving harmonic mapping f of the unit disk D onto itself satisfying the condition f (0) = 0, is positive at the origin but it does not give a positive lower bound. Consequently, distortion theorem in its natural form is false. In contrast, Heinz’s lemma [30] below asserts that for all such mappings Df (0) = |fz (0)|2 + |fz (0)|2 is bounded below by a positive constant and so we get useful information for such class of mappings. In the discussion, we use the notation fz (eiθ ) = lim fz (reiθ ) and fz (eiθ ) = lim fz (reiθ ) r→1−

r→1−

provided the limits exist. Theorem 14 (Heinz’s lemma). Let f be a univalent harmonic mapping of the unit disk D onto D satisfying the condition f (0) = 0. Then Df (0) = |fz (0)|2 + |fz (0)|2 ≥ c

for some absolute constant c > 0. More generally, we have 1 (13) |fz (z)|2 + |fz (z)|2 ≥ 2 π which is valid throughout the unit disk D. Proof. Case (a): Let us first assume that f and its first partial derivatives are continuous in the closed disk D = {z : |z| ≤ 1}.

We may assume without of loss of generality that J f (z) > 0 in D. Consequently, fz (z) �= 0 in D. Furthermore, because f z z (z) = 0, fz is analytic in D and continuous in the closure D and so, by the minimum principle for analytic functions, (14)

min |fz (ζ)| ≤ |fz (z)| for all |z| ≤ 1.

|ζ|=1

By assumption |f (z)| = 1 for |z| = 1 and so, for r ∈ (0, 1), Schwarz’s lemma for harmonic function gives � � � f (eiθ ) − f (reiθ ) � 1 − |f (reiθ )| 1 − (4/π) arctan r � �≥ ≥ . � � 1−r 1−r 1−r Letting r → 1− yields � �� ∂α f (reiθ )�� (15)

2 1 − (4/π) arctan r = 1−r π (see Subsection 3 for the definition ∂ α f (z) of the directional derivative of f in the radial direction which makes an angle α with the positive x-direction). As J f (z) > 0 in D, for r ∈ (0, 1), we have � � � � �∂α f (reiθ )� = �fz (reiθ ) + e−2iα fz (reiθ )� ≤ |fz (z)| + |fz (z)| ≤ 2|fz (z)| r=1

≥ lim

r→1

which, by (15), gives

2|fz (z)| ≥

2 1 , i.e. |fz (z)| ≥ for all |z| = 1. π π


Now, by (14), we observe that for |z| ≤ 1,

1 1 1 ≤ |fz (z)| = |fx (z) − ify (z)| ≤ π 2 2

301

� � √ � 2 |fx (z)|2 + |fy (z)|2

so that

2 ≤ |fx (z)|2 + |fy (z)|2 = 2(|fz (z)|2 + |fz (z)|2 ), π2 which gives the desired inequality. Case (b): Let us now prove the theorem under the hypotheses of our theorem. Let {rn }n≥2 be a sequence of real numbers in the interval (0, 1) such that r n → 1 and let Ωn be the sub-domain of the unit disk such that f (Ω n ) = Drn and satisfies the inclusion (16)

D1−1/n ⊂ Ωn ⊂ D.

Then Ωn is a simply connected domain. By the Riemann mapping theorem, there exists a unique function z = ϕn (ζ) which maps the unit disk conformally onto Ωn (we may suppose that 0 ∈ Ωn for large enough n) such that ϕn (0) = 0 and ϕ�n (0) > 0. Furthermore, ϕn (ζ) is analytic for |ζ| ≤ 1. Now, we define 1 1 (f ◦ ϕn )(ζ) = f (ϕn (ζ)). Fn (ζ) = rn rn Then Fn is harmonic for |ζ| ≤ 1+δ for some δ > 0 (because f and ϕ n have extension to the boundary) and satisfies the assumption of Case (a). Next, we see that 1 1 fz (z)ϕ�n (ζ) and (Fn )ζ (ζ) = fz (z)ϕ�n (ζ). (Fn )ζ (ζ) = rn rn Thus, by applying Case (a) to Fn , we conclude that 1 1 1 |(Fn )ζ (ζ)|2 + |(Fn )ζ (ζ)|2 = 2 |fz (z)|2 |ϕ�n (ζ)|2 + 2 |fz (z)|2 |ϕ�n (ζ)|2 ≥ 2 ; rn rn π or equivalently (17)

|fz (z)|2 + |fz (z)|2 ≥

1 rn2 . |ϕ�n (ζ)|2 π 2

By applying the theorem on normal families of analytic functions (cf. [54, 60]), we see that there exists a subsequence {ϕ nk } of the sequence {ϕn } which converges uniformly on every compact subset of D to an analytic function ϕ such that ϕ(0) = 0. Because of the inclusion (16), for the subsequence {ϕ nk }, Schwarz’s lemma implies that � � 1 |ζ| ≤ |ϕnk (ζ)| ≤ |ζ| for |ζ| < 1 1− nk for large values of k. Consequently, ϕ is analytic and univalent mapping of D onto itself satisfying the conditions ϕ(0) = 0 and ϕ� (0) > 0. Therefore, we must have ϕ(ζ) = ζ. This implies that ϕ�nk (ζ) → ϕ� (ζ) = 1 as k → ∞

on every compact subset of D. As (17) holds with n replaced by n k , allowing k → ∞, we obtain that 1 |fz (z)|2 + |fz (z)|2 ≥ 2 , π and the theorem is established.

302


Remark 8. The conditions f (0) = 0 in the statement can be omitted. Indeed, by • setting f1 (z) = f (z) − f (0), the problem is reduced to the previous case. Heinz’s simple proof indeed gives D f (0) ≥ c, where c≥1−

∞ 4� 1 2π 4 ≈ 0.1788. =1+ − π n=2 n2 π 3

This constant c was improved to 0.32 by J.C.C. Nitsche [46, 47], 0.4345 by H.L. de Vries [67], 0.4476 by J.C.C. Nitsche [48], 0.6411 by H.L. de Vries [68], and 0.65844 by Wegmann [70]. The conjectured value of c by Wegmann was 27 ≈ 0.6839, c= 4π 2 which became known as the Heinz constant. This was finally verified by R.R. Hall [25] in 1982. We remark that the constant 1/π 2 in (13) is clearly smaller than that of Nitsche at z = 0. Improvements in special cases appear in [51, 52]. In [36], Kalaj generalized the result of Heinz under the assumption that the range of f is an arbitrary convex domain rather than the unit disk. We refer to [34] for recent development with regards of the Nitsche conjecture (although we do not discuss it here). Theorem 15 (Rad´ o’s Theorem). There is no univalent harmonic mapping of D onto C. Proof. Let us prove a stronger version. Suppose that f is a univalent harmonic mapping of D onto a domain Ω = f (D) which contains a disk D(z 0 ; R) for some R > 0 and z0 ∈ C. We may without loss of generality assume that f (0) = 0 and z0 = 0. Let ΩR = f −1 (DR ), where ΩR is a subdomain of D, and let ϕ be a conformal mapping of D onto ΩR = ϕ(D) with ϕ(0) = 0. Define 1 F = f ◦ ϕ. R Then F maps D univalently onto D with F (0) = 0 and F is obviously harmonic. By Heinz’s lemma, it follows that |Fζ (0)|2 + |Fζ¯(0)|2 ≥ c,

where c > 0 is an absolute constant. But, as in the proof of Heinz’s lemma, a calculation gives 1 1 Fζ (0) = fz (0)ϕ� (0) and Fζ (0) = fz (0)ϕ� (0). R R Since |ϕ� (0)| ≤ 1 by the Schwarz lemma, cF 2 ≤ |fz (0)|2 + |fz¯(0)|2 , |ϕ� (0)| giving a finite upper bound for R. This shows that the range of f cannot contain disks of arbitrarily large radius centered at the origin. cR2 ≤

Corollary 2. There exists no univalent harmonic mapping of a proper subdomain of the plane C onto C.


303

Proof. Suppose there exists a univalent harmonic function f from Ω � C onto C. Then by the Riemann mapping theorem there is a conformal mapping ϕ of D onto Ω. It follows that the composition f ◦ ϕ is harmonic and maps D univalently onto C, which is a contradiction to Radó’s theorem.

4. Univalent Harmonic Mappings We begin the discussion with the following note. If the second complex dilatation ω of a harmonic mapping f on the domain Ω satisfies |ω(z)| ≤ k < 1 in Ω, then f is quasiconformal with the maximum dilatation K = (1 + k)/(1 − k); i.e. f is K-quasiconformal. Theorem 7 gives the following: Corollary 3. Sense-preserving harmonic mappings are locally quasiconformal. Suppose that f is a harmonic mapping of a proper simply connected domain Ω of C. By the Riemann mapping theorem there is a conformal mapping ϕ of D onto Ω. It follows that the composition f ◦ ϕ maps D harmonically onto Ω. As a consequence of this observation, we may assume that Ω is the unit disk and that f is sense-preserving in D. We next observe that, because J f = |h� |2 − |g � |2 > 0 for the sense-preserving harmonic mapping f = h + g, fz (0) = h� (0) �= 0, and so we arrive at the normalized form of f : f (z) − f (0) . fz (0) Thus, the analytic and co-analytic parts of the harmonic mapping f = h + g in D may be written, respectively, ∞ ∞ � � (18) h(z) = z + an z n and g(z) = bn z n . n=2

n=1

Denote by SH the class of all complex-valued harmonic univalent and sensepreserving mappings f in the unit disk D = {z ∈ C : |z| < 1} with f (0) = 0 = fz (0) − 1. We note that SH reduces to S, the class of normalized univalent analytic functions in D whenever the co-analytic part of f is zero, i.e., g(z) ≡ 0 in D. Furthermore, for f = h + g ∈ SH with g � (0) = b and |b| < 1 (because Jf (0) = 1 − |g � (0)|2 = 1 − |b|2 > 0), the function

f − bf 1 − |b|2 is also in SH . Univalency of F is obtained by applying an affine mapping to f . We observe that Fz (0) = 0. Thus, we may sometimes restrict our attention to the subclass 0 = {f ∈ SH : fz (0) = 0}. SH Note that the condition fz (0) = 0 is equivalent to the assertion that the second complex dilatation ω(z) of f is zero, i.e. ω(0) = 0 or g(z) = O(z 2 ) for z near the origin. Clearly, 0 � SH . S � SH 0 0 is compact Although both SH and SH are known to be normal families, only SH with respect to the topology of locally uniform convergence (see [18, 22]). Indeed, SH is not a compact family because it is not preserved under passage to locally (19)

F =

304


uniform limits. The limit function is necessarily harmonic in D, but it need not be univalent. For example, the sequence of affine mappings f n defined in Example 3 demonstrates that SH is not a compact family.

Identifying z = x + iy ∈ C with (x, y) ∈ R2 and points in R2 as 2 × 1 column matrix, we may rewrite the equation (19) in matrix form as � � � �� Re F 1 − b1 0 Re f = , (1 − |b1 |2 ) Im F Im f 0 1 + b1 and so, pre-multiplying by the inverse of the 2 × 2 matrix gives �� Re F 0 Re f 1 + b1 . = Im F Im f 0 1 − b1

Simplifying the last relation reveals that the transformation (19) is one-to-one and its inverse is given by f = F + b1 F .

0 This observation enables us to derive a number of properties about S H and SH , [18]. 0 4.1. Normality of the families SH and SH . In this section we consider sets of functions, which are called families to make avoiding confusion with sets of points. We always assume that all functions are defined on a fixed domain Ω.

We are usually interested in studying subfamilies of functions defined on a set Ω, e.g., analytic, harmonic or quasiconformal functions. Typical examples of such subfamilies are bounded functions, functions which are non-vanishing, and functions having certain values at given points, in the respective family. Consider a sequence of complex-valued functions {f n }, fn : Ω ⊆ C → C and n ∈ N. For a fixed z0 ∈ Ω, {fn (z0 )} is an ordinary sequence of complex numbers. Definition 10. A sequence {fn } of functions is said to be convergent at z0 ∈ Ω if the sequence {fn (z0 )} converges. We shall call this limit f (z0 ). The sequence {fn } of functions is said to converge pointwise to f on Ω if {fn (z0 )} converges to f (z0 ) at each point z0 ∈ Ω. Since the limit of a sequence, when it exists, is unique, in the case of ‘pointwise’ convergence, we have a uniquely defined function f from Ω into C and we call f , the pointwise limit, or simply the limit function of the sequence {f n } and write f (z) = lim fn (z), n→∞

z ∈ Ω.

An equivalent formulation of the above discussion is given by the following: Definition 11. Let fn and f be functions from Ω into C. We say that fn → f on Ω iff for every � > 0 and every z ∈ Ω, there exists N = N (�, z) such that |fn (z) − f (z)| < � for all n ≥ N .

This convergence is said to be uniform if it is possible that N (�, z) can be chosen independent of z ∈ Ω. That is, one N (�) works for all z ∈ Ω.


305

We often write fn → f uniformly on Ω, or limn→∞ fn (z) = f (z) uniformly on Ω to denote the uniform convergence of f n to f on Ω. Also note that the uniform convergence on Ω implies pointwise convergence. The converse is false, for example consider fn (z) = z n for |z| < 1.

Of course, there is no reason to expect any given sequence of functions to be convergent. In fact, it is possible that a sequence of functions {f n } does not contain a single convergent subsequence. This can be a serious obstacle in many situations, and it is therefore often desirable to rule out this possibility. This leads to discussion of the normality of a given family of functions. Again, a well-known fact is that every bounded sequence of complex numbers {z n } possesses a limit point and so, {zn } has a convergent subsequence. Analogous situation for sequences of functions {fn } leads to the definition of a normal family. Definition 12. Let F be a family of complex-valued functions f defined on a domain Ω. We say that F is a normal family if every sequence of functions contains a subsequence which converges uniformly on compact subsets of Ω (i.e. locally uniformly on Ω). A normal family is said to be compact if every limit function is a member of the family. It is well-known that if the family consists of analytic functions defined on a domain Ω, then the limit of such a subsequence must be analytic on Ω as well. We only consider families of harmonic functions defined on a domain Ω, in particular with Ω = D. Again, we remark from classical results in the function theory that a limit of a sequence of harmonic functions is necessarily harmonic. Definition 13. A family F of harmonic functions defined on Ω is said to be locally uniformly bounded in Ω if for any compact set K ⊂ Ω, there exists a constant M = M (K) such that |f (z)| ≤ M for all f ∈ F, and for all z ∈ K. We say that the family F is uniformly bounded on Ω if there exists an absolute constant M > 0 such that |f (z)| ≤ M for all z ∈ Ω and all f ∈ F. Clearly, uniformly boundedness of a family implies that each member of the family is bounded. However the converse is not true. For instance, consider � � � � 1 1 F1 = : n ∈ N , F2 = : n∈N . z − ein 1 − zn

Then each Fk (k = 1, 2) is a locally uniformly bounded family in D but none of them is uniformly bounded in D. The key ingredient in the proof of the Riemann mapping theorem is Montel’s theorem (for a proof we refer to recent books of Ponnusamy [54], and Ponnusamy and Silverman [60]) which states that a family of analytic functions is normal if and only if it is locally bounded. The same result holds for harmonic functions as well. Theorem 16 (Montel’s theorem). A family of harmonic functions is locally bounded if and only if it is normal.

306


Thus, to prove the normality of SH it suffices to show its local boundedness. The proof is based on the Schwarz lemma for quasiconformal mappings. We write M∞ (r, f ) = max |f (z)|. |z|=r

We are now ready to prove the main result in this section. 0 Theorem 17. The family SH is normal. The family SH is normal and compact. 0 Proof. Recall that for each f ∈ SH there is an f0 ∈ SH such that (see Exercise 27)

f = f 0 + b1 f0 ,

where b1 = fz (0) and |b1 | < 1. Thus,

M∞ (f, r) < 2M∞ (f0 , r).

0 is locally bounded. Hence it is sufficient to show that SH

0 Now we suppose that f = h + g ∈ SH . Then, we may write g � (z) = ω(z)h� (z), where ω : D → D is analytic with ω(0) = 0 (because g � (0) = 0). By the classical Schwarz’s lemma we have |ω(z)| ≤ |z| and thus,

|g � (z)| ≤ |z| |h� (z)|.

Fix R ∈ (0, 1) and consider mapping A(z) = f (Rz). Then (see Definition 7) DA (z) =

1 + |fz (Rz)/fz (Rz)| 1 + |ω(Rz)| 1 + |Rz| 1+R 1 + dA (z) = = ≤ < , 1 − dA (z) 1 − |fz (Rz)/fz (Rz)| 1 − |ω(Rz)| 1 − |Rz| 1−R

and thus the mapping A is K-quasiconformal with K −1 1+R , i.e., R = . K := KR = 1−R K +1 By the Riemann mapping theorem, there exists a conformal (univalent and analytic) mapping F of D onto f (D), normalized by F (0) = 0 and F � (0) > 0. The mapping F is called the conformal associate of f . We note that the mapping G defined by G(z) = F −1 (A(z)) is again K-quasiconformal (being a composition of 1-quasiconformal and K-quasiconformal maps), and it maps the unit disk into itself with G(0) = 0. By the Schwarz lemma for quasiconformal mappings, we have the estimate It follows that (20)

|G(z)| ≤ ϕK (|z|) for all z ∈ D. � |F −1 (f (z))| ≤ ϕK

|z| R

� for |z| ≤ R =

K −1 . K +1

On the other hand, F : D → C is a conformal mapping with F (0) = 0 and F � (0) > 0. We have the well known growth estimate for the class of conformal mappings in D: (21)

|F (z)| ≤

|F � (0)| |z| , (1 − |z|)2

z ∈ D.

The estimates (20) and (21) together yield

|f (z)| = |F (F −1 (f (z)))| ≤

|F � (0)|ϕK (r/R) (1 − ϕK (r/R))2


307

for |z| ≤ r < R.

It follows from the well-known Koebe 1/4-theorem that |F � (0)| ≤ 4|w0 | for any w0 which is an omitted value of F . On the other hand, the proof of Rad´ o’s theorem (see the proof Theorem 15) shows that if the disk D R is contained in the range of a harmonic mapping f , then cR2 ≤ |fz (0)|2 + |fz (0)|2 , 0 , we have fz (0) = 1 and fz (0) = 0, and where c > 0 is a constant. Because f ∈ S√H √ thus f omits a value w0 with |w0 | = 1/ c. Then it follows that |F � (0)| ≤ 4/ c 0 showing that f is locally bounded. We conclude that S H , and SH , are locally bounded. 0 0 is compact, suppose that fn = hn + gn ∈ SH , and fn → f uniformly To see that SH on compact subsets of D. Because f is harmonic, it has the canonical presentation f = h + g. We claim that

hn → h and gn → g

locally uniformly, and g (0) = h� (0) − 1 = 0. In order to present a proof of the claim, we consider a compact subset K of the unit disk D and let �

Mn = max |fn (z) − f (z)|, z∈K

where Mn → 0 as n → ∞. According to Cauchy inequality |∂x (fn − f )| ≤ C1

Mn Mn and |∂y (fn − f )| ≤ C2 , 1 − |z| 1 − |z|

where C1 and C2 are constants. Since

gn� (z) − g � (z) = (fn − f )z (z) =

1 [(fn − f )x (z) + i(fn − f )y (z)], 2

it follows from previous two inequalities that |gn� (z) − g � (z)| ≤ C3

Mn 1 − |z|

for some constant C3 > 0. Thus Gn = maxz∈K |gn� (z) − g � (z)| → 0 as n → ∞ because Mn → 0. The same holds for hn . Further, � z � z (gn − g)� (ζ) dζ = (gn − g)� (ζ) dζ. gn (z) − g(z) = gn (0) − g(0) + 0

0

This implies that gn − g converges to zero as n → ∞, and so g n → g and hn → h locally uniformly, and g � (0) = h� (0) − 1 = 0. In particular, h is not constant, and thus by Hurwitz’ theorem we have h � (z) �= 0 for z ∈ D. On the other hand, the classical Schwarz lemma states that |g � (z)| ≤ |z| |h� (z)|, and thus Jf (z) = |h� (z)|2 − |g � (z)|2 ≥ (1 − |z|2 )|h� (z)|2 > 0 for all z ∈ D, i.e., f is locally univalent. As f is a uniform limit of univalent functions on compact subsets of D, it follows from the argument principle that f is 0 0 , showing that SH is a compact family. univalent in D. Thus, f is in SH

308


0 4.2. Area Minimizing Property of the Class SH . As an example of an 0 extremal problem for the functions of the class S H , we study the question, if there 0 exists a function in SH which maps the unit disk to a domain of the smallest area. It turns out that this problem can be completely solved. Namely, we obtain the following result (see [22, Chapter 6]): 0 . Then the area of f (D) is greater than or Theorem 18. Let f be a function in SH equal to π/2. Furthermore, the minimum is attained only for 1 f0 (z) = z + z 2 2 and its rotations. 0 has the form f = h + g, where h Proof. As noted above, each function f ∈ S H and g are analytic functions in D with the series representation (18). As in the introduction, we may write g � (z) = ω(z)h� (z), where ω : D → D is analytic with ω(0) = 0 (because g � (0) = 0). Thus, by the classical Schwarz’s lemma we have |ω(z)| ≤ |z|, and by the definition of the Jacobian, we have � � Jf (z) = |h� (z)|2 − |g � (z)|2 = (1 − |ω(z)|2 |h� (z)|2 ≥ (1 − |z|2 |h� (z)|2 > 0.

Then the area A of f (D) has the expression �� A = Jf (z) dx dy D �� |h (z)|2 − |zh� (z)|2 dx dy =: B ≥ D

It is a simple exercise to see that �� ∞ � |h� (z)|2 dx dy = π n|an |2 D

n=1

and if we set zh (z) = H (z), then H will be of the form �

�

H(z) =

∞ � n=1

An+1 z n+1 ,

An+1 =

n an n+1

and as a consequence of this observation �� ∞ ∞ � � n2 � 2 2 |an |2 . |zh (z)| dx dy = π (n + 1)|An+1 | = π n + 1 D n=1 n=1 Thus, we have (as a1 = 1) A≥B=

∞ ∞ � � π n � π � 2n +π |an |2 = |an |2 , n 1− 2 n + 1 2 n + 1 n=2 n=1

and therefore, A ≥ π/2. The minimum is attained if we choose a n = 0 for all n ≥ 2.

This happens when the complex dilatation ω is of the form ω(z) = e iα z for some α, which gives 1 g � (z) = eiα zh� (z) = eiα z, i.e., f (z) = z + e−iα z 2 . 2 The proof is complete.


309

The affine mappings of the form f (z) = z + b1 z (0 < |b1 | < 1) show that the class SH contains no area-minimizing functions (see Exercise 32).

Remark 9. We ask whether there exists a univalent harmonic mapping f of D onto a simply connected domain Ω that is the solution of the elliptic partial differential equation fz (z) = ω(z)fz (z), z ∈ Ω,

where we assume that the complex dilatation ω �≡ 0, as ω ≡ 0 is taken care of by the Riemann mapping theorem. Suppose that ω(z) = z, Ω = D and f = h + g, where h and g are analytic functions in D with the series representation (18). Then the differential equation becomes g � (z) = zh� (z) which gives (n + 1)bn+1 = nan for n ≥ 0. Then the area A of f (D) is A=

∞ π � 2n |an |2 2 n=1 n + 1

and the square integrable mean is � 2π ∞ ∞ � � n �2 � � � 1 M2 (f ) = |f (eiθ )|2 dθ = (|an |2 + |bn+1 |2 ) = 1+ |an |2 , 2π 0 n + 1 n=0 n=0 which gives the relation A < (π/2)M2 (f ). If f (D) ⊂ D, then M2 (f ) ≤ 1 and so the area of f (D) is less than π/2. Since the area of D is π, the function f cannot map D onto D. This example shows that some caution is required in the choice of the dilatation. For many related mapping issues, • we refer to the work of Hengartner and Schober [31]. 4.3. Harmonic Univalent Polynomials. Harmonic univalent polynomials form a subtopic of the theory of harmonic mappings. Specifically a harmonic polynomial is a function f = h+g, where h and g are analytic polynomials in z. The degree of f is defined as the larger of the degrees of h and g. Finding a method of constructing sense-preserving univalent harmonic polynomials is another important problem, see [65]. Such polynomials in general have the above form with the normalization h(0) = g(0) = 0 = g � (0) and the dilatation ω is a finite Blaschke product. However, very little is known about harmonic mappings whose dilatation is not a Blaschke product. Very few explicit examples of such mappings are known. In any case, it is appropriate to include one more simple example of a harmonic univalent polynomial (see also Examples 3 and 2) although we would continue our discussion on this topic at a later stage. Example 5. Let f = h + g, where z2 z2 and g(z) = β , 2 2 where α and β are complex constants lying in the closed unit disk |z| ≤ 1 such that |α| + |β| = 1. We see that h(z) = z + α

f (z) = z + α

z2 z2 +β 2 2

310


1 0.5

0.5

0

0

−0.5

−0.5

−1

−1 −1

−0.5

0

0.5

1

Figure 11. Image of D iπ/4 under f (z) = z + e 2 z 2

1.5

−1

−0.5

0

0.5

1

1.5

Figure 12. Image of D iπ/4 under f (z) = z+ 3e 8 z 2 + e−iπ/3 2 z 8

is a harmonic mapping on D. To verify the univalence of f in D, we suppose that f (z1 ) = f (z2 ) for z1 , z2 ∈ D with z1 �= z2 . Then, we have � � β α 0 = (z1 − z2 ) 1 + (z1 + z2 ) + (z1 − z2 )(z1 + z2 ) 2 2 or � � α β z1 − z2 0 = 1 + (z1 + z2 ) + (z1 + z2 ), 2 2 z1 − z2 which is a contradiction. This is because, for z 1 , z2 ∈ D with z1 �= z2 , � � � � � � α β z1 − z2 � 0 = �1 + (z1 + z2 ) + (z1 + z2 )�� > 1 − |α| − |β| = 0, 2 2 z1 − z2

which is false. Thus, for z1 �= z2 , we must have f (z1 ) �= f (z2 ) and hence, f is a univalent harmonic mapping on D. The images of f for various of choices of α and • β are illustrated in Figures 11–15. Harmonic mappings can be considered as close relatives of conformal mappings. However, in contrast to conformal mappings, harmonic mappings are not at all determined (up to normalizations) by their image domains. Thus, it is natural to study the class SK (Ω, Ω� ) of harmonic sense-preserving mappings of a domain Ω onto another domain Ω� . For the case Ω = Ω� = D, we refer to the work of Choquet [17], Heine [29] and Hall [25]. For Ω, a proper sub-domain of C and Ω � , a strip, we refer to the work of Hengartner and Schober [32]. For a recent survey of harmonic mappings, we refer to the paper of Bshouty and Hengartner [11]. However, at a later stage, we shall return to discuss these and other related important questions on harmonic and quasiconformal mappings along with recent investigation on these mappings.

5. Method of Shearing One of the forms of constructing harmonic mappings, introduced by Clunie and Sheil-Small [18], is known as “shear construction.” This method produces certain


311

1 1 0.5

0.5

0

0

−0.5

−0.5 −1

−0.5

0

0.5

1

−1

Figure 13. Image of D iπ/4 under f (z) = z + e 4 z 2 +

−0.5

0

0.5

1

Figure 14. Image of D iπ/4 under f (z) = z + e 8 z 2 +

e−iπ/3 2 z 4

3e−iπ/3 2 z 8

1 0.5 0 −0.5

−1

−0.5

0

0.5

1

1.5

Figure 15. Image of D under f (z) = z +

e−iπ/3 2 z 2

planar harmonic mappings by adjoining functions in S with co-analytic parts that are related to or derived from analytic parts. Moreover, in this method one produces a univalent harmonic mapping of D with a specified dilatation onto a domain in one direction by “shearing” a conformal mapping along parallel lines. Definition 14. A domain Ω is convex in the direction eiα , α ∈ R, if and only if for every complex number a, the set Ω ∩ {a + teiα : t ∈ R} is either connected or empty; i.e. every line parallel to the line through 0 and e iα has a connected (or empty) intersection with Ω. In particular, it is convenient to consider domains convex in the direction of the real axis (denoted by CHD). Thus, a domain Ω is a CHD if every line parallel to the real axis has a connected intersection with Ω. Clearly, a domain Ω in C is convex if and only if it is convex in every direction.

1.5

312


Definition 15. A function f defined in D is said to be CHD if its range is CHD; i.e. if the intersection of f (D) with each horizontal line is connected. For example, the Koebe function k(z) = z/(1 − z)2 , and the convex function c(z) = z/(1 − z) are univalent in D and CHD. 5.1. Shearing Theorem of Clunie and Sheil-Small. The following lemma due to Clunie and Sheil-Small [18] is a basis for the construction of harmonic mappings that are convex in the direction of the real axis, in particular. The proof of next theorem relies on a simple lemma. Lemma 12. Let Ω ⊂ C be a CHD domain, and let p be a real-valued continuous function on Ω. Then the mapping w �→ w + p(w) is univalent in Ω whenever it is locally univalent. When the mapping is univalent, then its range is again a CHD domain. Proof. Assume that the mapping w �→ w + p(w) is locally univalent. Suppose on the contrary that it is not univalent. Then there exist two distinct points w 1 , w2 in Ω such that w1 + p(w1 ) = w2 + p(w2 ). Because p is real-valued, it follows that Im w1 = Im w2 (= c say), and so, by writing w1 = u1 + iv1 and w2 = u2 + iv2 , we have u1 + p(u1 + ic) = u2 + p(u2 + ic). Writing, w = u + iv, we observe that the mapping u �→ ϕ(u) = u + p(u + ic) defined on some interval, which depends on c, is not strictly monotone and hence, is not locally univalent. This is a contradiction and thus, the mapping w �→ w + p(w) must be univalent. Geometrically, this shows that the range of the mapping is CHD. Theorem 19. For analytic functions h and g, assume that f = h + g is harmonic and locally univalent in the unit disk D. Then f is a univalent mapping of D onto a CHD domain if and only if h − g is a conformal mapping of D onto a CHD domain. Proof. Suppose first that f = h + g is univalent and its range Ω = f (D) is a CHD domain. Then the inverse of f exists. We may set w = f (z) so that z = f −1 (z) and, since h − g = h + g − (g + g) = f − 2Re g, it follows that � � h(z) − g(z) = w − 2Re g ◦ f −1 (w) = w + p(w) � � where p(w) = −2Re g ◦ f −1 (w) is a continuous real-valued function on Ω. Since f is locally univalent, Jf (z) = |h� (z)|2 − |g � (z)|2 �= 0 which implies (h − g)� (z) �= 0 in D. This observation gives that h − g is locally univalent and this in turn implies that the mapping w �→ w + p(w) is locally univalent in Ω. From Lemma 12, this mapping is univalent and its range is CHD. This means that h − g is univalent and its range is CHD.


313

Conversely, suppose F = h − g is univalent in D and its range Ω � = F (D) is a CHD. With w = F (z) we now have f = h + g = h − g + (g + g) = F + 2Re g

and as before we obtain that � � (h + g)(z) = f ◦ F −1 (w) = w + 2Re g(F −1 (w)) := w + q(w).

Note that w �→ w + q(w) is locally univalent in Ω� = F (D) (as f is locally univalent by assumption). Finally, by Lemma 12, the mapping f is univalent in D and its range is CHD. Remark 10. 1. From Theorem 19, we note that the imaginary part of the analytic function h − g equals the imaginary part of the harmonic function f = h + g so that the domain is changed or cut or stretched in the real-axis direction. In view of this observation, the method is known as “shearing”. 2. Theorem 19 indeed holds for simply connected domains rather than for func• tions defined on the unit disk D. We can state the following equivalent form of Theorem 19. Theorem 20. For analytic functions h and g, assume that f = h + g is harmonic and locally univalent in the unit disk D. Then f is a univalent mapping of D onto a convex domain if and only if for each choice of α (0 ≤ α < 2π) the analytic function eiα h − e−iα g is univalent and maps D onto a CHD domain. Proof. From the definition, it is clear that the harmonic mapping f = h + g has a convex range if and only if the range is convex in every direction. This is equivalent to saying that the range of every rotation e iα f (= eiα h + e−iα g) is CHD, for 0 ≤ α < 2π. Moreover, as eiα f = eiα h + e−iα g = eiα h − e−iα g + 2Re (e−iα g),

the equivalency of the statements follows.

Corollary 4. Suppose that f = h + g is locally univalent and convex harmonic function on D. Then the analytic function h − e−2iα g is univalent and maps D onto a domain that is convex in the direction eiα for all 0 ≤ α < π. In particular, h + eiγ g is univalent in D for each γ ∈ [0, 2π). Proof. Assume that f = h+g is a convex harmonic mapping on D. By Theorem 20, the analytic function eiα h − e−iα g = eiα (h − e−i2α g)

is (univalent) convex in the direction of real axis. By rotating, we see that h−e −2iα g is convex in the direction of eiα for every α, 0 ≤ α < π.

Corollary 5. Assume that f = h + g is harmonic and locally univalent in D. Then f maps D univalently onto a domain convex in the vertical direction (i.e. convex in the direction of the imaginary axis) if and only if the analytic function h + g is univalent and has the same mapping property.

314


Proof. Observe that f maps D onto a domain convex in the vertical direction if and only if F = eiπ/2 f = i(h + g) = i(h + g − (g − g) = i(h + g) + 2i Im g

maps D onto a CHD domain, i.e. to say that h+g is convex in the vertical direction.

One can use the technique of shearing for constructing interesting examples of harmonic mappings by shearing conformal mappings with a prescribed dilatation ω(z). As a simple demonstration, let φ be a conformal mapping of the unit disk D onto a CHD domain with φ(0) = 0, and let f = h + g be a sense-preserving harmonic function, where h and g are analytic in D. Then the dilatation ω = g � /h� is analytic with |ω(z)| < 1 in D. According to the shear technique, the construction of harmonic mappings proceeds by letting h − g = φ. This gives the pair of linear differential equations h� (z) − g � (z) = φ� (z) and ω(z)h� (z) − g � (z) = 0.

Solving for h� (z) and g � (z), and then integrating with the normalization g(0) = h(0) = 0, we arrive at the formulas for h and g explicitly: � z � z φ� (t) ω(t)φ� (t) h(z) = dt and g(z) = dt = h(z) − φ(z). 0 1 − ω(t) 0 1 − ω(t) So, the shear construction produces the harmonic mapping f defined by f (z) = h(z) + g(z) = 2Re h(z) − φ(z)

and f maps D onto a CHD domain. Various choices of the dilatation ω(z) and the conformal mapping φ(z) produce a number of univalent harmonic mappings, as demonstrated by a number of examples below (see the paper by Greiner [27] for many more interesting examples). Example 6. Note that z −z 3/3 maps D conformally onto a domain which is convex in the direction of the real axis and so is the harmonic function f = h + g, where 1 g � (z) h − g = z − z 3 and � = z 2 (= ω(z)). 3 h (z) As h� (z) − g � (z) = 1 − z 2 and g � (z) = z 2 h� (z), solving these two equations yields that h� (z) = 1 and g � (z) = z 2 . As h(0) − g(0) = 0, i.e. g(0) = h(0), it follows that h(z) = z and g(z) =

1 3 z 3

which gives univalent harmonic mapping 1 f (z) = h(z) + g(z) = z + z 3 . 3 We see that f maps D onto the interior of the region bounded by the hypocycloid with 4 cusps. Also, we note that h − g is a conformal mapping of D onto the interior of an epicycloid with 2 cusps. The geometric behaviors of f (z) = z + 13 z 3 and F (z) = z − 13 z 3 are illustrated in Figures 16 and 17 which are drawn using


1

1

0.5

0.5

0

0

−0.5

−0.5

−1

−1 −1.5

−1

−0.5

0

0.5

1

1.5

−1.5

Figure 16. Image of D under f (z) = z + 13 z 3 .

−1

−0.5

315

0

0.5

1

Figure 17. Image of D under F (z) = z − 13 z 3 .

MATLAB. These figures and many other figures that follow depict the images of concentric circles and radial segments of the unit disk D. • 5.2. Harmonic Koebe Function. The classical Koebe function �� 2 1+z 1 z = −1 k(z) = (1 − z)2 4 1−z maps D conformally onto the slit plane C \{u + iv : u ≤ −1/4, v = 0}. This function plays an important role for many extremal problems for the class S of analytic univalent functions in the unit disk D. It is expected that if we can find a harmonic analog of it, such a function will serve analog role of the classical Koebe function. Using the method of shearing, Clunie and Sheil-Small [18] have actually 0 . constructed harmonic Koebe function which belongs to the class S H We note that the range k(D) is convex in the horizontal direction, so is called a CHD domain. According to the method of shearing there exists a harmonic univalent mapping f = h + g with the same properties if 1. h(z) − g(z) = k(z), 2. g � (z) = ω(z)h� (z) with the dilatation ω(z) = z, for example. Thus, we have the pair of differential equations h� (z) − g � (z) = Solving these, we easily see that h� (z) = and g � (z) =

1+z and g � (z) = zh� (z). (1 − z)3 1+z 2 − (1 − z) = , (1 − z)4 (1 − z)4

z(1 + z) (1 − z)2 − 3(1 − z) + 2 = . (1 − z)4 (1 − z)4

1.5

316


Integrating with g(0) = h(0) = 0, we easily arrive at the formulas for h and g: h(z) = We observe that Also

1 2 1 3 z − 12 z 2 + 16 z 3 2z + 6z , and g(z) = . (1 − z)3 (1 − z)3

Re f = Re (h + g) and Im f = Im(h − g).

z + 13 z 3 z and h(z) − g(z) = . (1 − z)3 (1 − z)2 The function f = h + g obtained in this way is referred to as the harmonic Koebe function (with the dilatation ω(z) = z), denoted usually by K(z). Thus, function K(z) is often written in the following special notation rather than f = h + g: � � � � z + 13 z 3 z K(z) = H(z) + G(z) = Re + iIm . (1 − z)3 (1 − z)2 h(z) + g(z) =

We claim that K(z) maps D onto the slit plane C \{u + iv : u ≤ −1/6, v = 0}. In order to obtain a proof (see also the original article [18]) for the range of K, we may rewrite the last equation as �� 3 �2 1+z 1+z i 1 − 1 + Im −1 . K(z) = Re 6 1−z 4 1−z

Now, consider the M¨ obius transformation 1+z = u + iv w = T (z) = 1−z so that 1 i 1 i K(z) = Re (w3 − 1) + Im (w2 − 1) = (u3 − 3uv 2 − 1) + uv, 6 4 6 2 where u > 0, v ∈ R (as w = T (z) maps D onto the right half-plane {w : Re w > 0}). Now, we observe that:

• The boundary values of K are all −1/6 except for the point 1 which goes to the point at infinity. That is each point ζ = e iθ ∈ ∂D \{1} is mapped onto the point K(eiθ ) = −1/6 as T (ζ) = iv0 and T 3 (ζ) = −iv03 , v0 ∈ R. • If uv = 0, i.e., v = 0 (as u > 0), then x ∈ (−1, 1) is mapped onto K(x) =

1 1+x Re (u3 − 1), where u = > 0, 6 1−x

and so K((−1, 1)) is the interval (− 16 , ∞), as u3 − 1 is monotone and varies between −1 and +∞. Thus, under the mapping K(z), (−1, 1) is mapped monotonically onto the real interval (− 61 , ∞). • Finally the level curve uv = c (0 �= c ∈ R) is carried univalently onto the set � � u3 c2 c − − 1, u > 0 . W =U +i : U = 2 6 2u Note that u > 0, and so the function T (u) =

c2 u3 − −1 6 2u


317

Figure 18. The harmonic Koebe function maps (0, ∞) univalently on the whole real line (−∞, ∞). Thus, the range of the harmonic Koebe function K is the slit plane C \(−∞, − 61 ].

Concerning the behaviour of K(z) near z = 1, we observe the following:

• K(z) → ∞ as z → 1 nontangentially. Indeed on the curve corresponding to the ray v = cu (u > 0, c ∈ R), 1 c K(z) = Re [(1 − 3c2 )u3 − 1] + i u2 6 2 which approaches the point at infinity as u → ∞. • On the other√hand, as z → 1 along the tangential curve corresponding to the curve v = c/ u (u > 0, c ∈ R), one finds that 1 c√ (3c2 + 1) Re (u3 − 3c2 − 1) + i as u → 0. u→− 6 2 6 Because c ∈ R, this amounts to say for all the missing boundary bound (−∞, −1/6]. K(z) =

5.3. Coefficient Conjectures for Univalent Harmonic Mappings. A classical problem for the class S of analytic univalent functions, normalized so that f (0) = 0 and f � (0) = 1, f (z) = z + a2 z 2 + a3 z 3 + · · · ,

|z| < 1,

was to find the sharp bound for the second coefficient a 2 . The solution to this problem, |a2 | ≤ 2, was obtained by Bieberbach and this fact plays a crucial role in the classical function theory, e.g. in the proof of the Koebe one-quarter theorem.

318


It is worth remarking that in the analytic case, the Bieberbach conjecture |a n | ≤ n for each f ∈ S has been proved by de Branges [9]. However, Zalcman’s conjecture asserts that for each f ∈ S, |a2n − a2n−1 | ≤ (n − 1)2 ,

with equality only for the Koebe function. This remarkable conjecture implies the Bieberbach conjecture but it remains an open problem (for n > 3) even after the proof of Bieberbach conjecture by de Branges. This inequality has been proved only for certain special subclasses of S, [10, 44]. However, the case n = 3 has been proved by Krushkal [38] in 1995. Now, it is appropriate to state the Bieberbach type coefficient conjecture for the 0 of the all sense preserving univalent harmonic mappings f = h + g, where class SH h(z) =

∞ �

an z n and g(z) =

n=0

∞ �

bn z n ,

n=1

0 normalized so that a0 = b0 = b1 = 0, and a1 = 1. The family SH is known to be 0 compact and the harmonic Koebe function K = H + G belongs to SH , where

H(z) = and

∞ � z − 12 z 2 + 16 z 3 = An z n (1 − z)3 n=1

G(z) =

∞ � + 16 z 3 = Bn z n . (1 − z)3 n=2

1 2 2z

The coefficients An and Bn are given by 1 1 An = (2n + 1)(n + 1) and Bn = (2n − 1)(n − 1), for all n ≥ 1 6 6 respectively. We remind the reader that the function K has the dilatation ω(z) = z, and 2n2 + 1 An − Bn = n and An + Bn = for all n ≥ 2. 3 Analogous to the analytic univalent case, this suggests the following coefficient conjecture which is due to Clunie and Sheil-Small [18]. 0 Conjecture 1. Let f = h + g belong to SH , where the representation of h and g are as above. Then |an | ≤ An and |bn | ≤ Bn for all n ≥ 2.

Partial results in this direction have been shown by Clunie and Sheil-Small [18, 64], 0 but the general conjecture concerning all the functions in the class S H remains open. Another related conjecture is that � � �|an | − |bn |� ≤ n

0 , see [18]. However, analog of Zalcman’s holds for each n ≥ 2 and for all f ∈ SH conjecture has not been looked at. It would be interesting to consider the harmonic analog of Zalcman’s conjecture.

0 and |b1 | < 1, Finally, as each F ∈ SH has the form F = f0 + b1 f0 for some f0 ∈ SH 0 the representation f0 = h0 + g0 ∈ SH leads to

F = f 0 + b 1 f 0 = h0 + b 1 g 0 + b 1 h0 + g 0 .


319

Thus, if we set F (z) = z + then for n ≥ 2 we have

∞ � n=2

an z n +

∞ �

bn z n ,

n=2

an = a0n + b1 b0n and bn = b1 a0n + b0n ,

where a0n and b0n are the Taylor coefficients of h0 and g0 , respectively. Thus, if Conjecture 1 were true, then |an | < |a0n | + |b0n | ≤ An + Bn and |bn | < |a0n | + |b0n | ≤ An + Bn for n ≥ 2.

This observation leads to:

Conjecture 2. Let f = h + g belong to SH , where the standard representations of h and g are as above. Then we have the sharp bounds (not attained) |an |
|g � (0)| and h + �g is close-toconvex for each �, |�| = 1, then f = h + g is close-to-convex. Next we state and prove a harmonic analog of a well-known theorem of Wolff, Noshiro and Warschawski, see [21, Theorem 2.16]. Although this lemma has been proved by Mocanu [45] in 1981, it is not widely known and therefore, the proof is worth presenting here. Lemma 14. Suppose f = h + g is harmonic mapping in a convex domain Ω such that Re (eiγ h� (z)) > |g � (z)| for all z ∈ Ω, and for some γ ∈ R. Then f is univalent in Ω. Proof. Let z1 , z2 ∈ Ω with z1 �= z2 . Then, because Ω is a convex domain, z(t) = (1 − t)z1 + tz2 ∈ Ω for 0 ≤ t ≤ 1. Now, � � 1 � 1� d z2 − z1 � f (z(t)) dt = (z2 − z1 ) g (z(t)) dt f (z1 ) − f (z2 ) = h� (z(t)) + z2 − z1 0 dt 0 so that � � � � � 1 � f (z2 ) − f (z1 ) � iγ � iγ z2 − z1 � � � ≥ Re e h (z(t)) + e g (z(t)) dt � z2 − z1 � z2 − z1 0 � �� 1� � z2 − z1 � > g � (z(t))�� dt = 0 |g � (z(t))| − �� z2 − z1 0 which proves the univalence of f in Ω. When g(z) ≡ 0, Lemma 14 implies the theorem of Wolff, Noshiro and Warschawski, see [21, Theorem 2.16]. For convenience, we now introduce 1 CH = {f ∈ SH : Re fz (z) > |fz (z)| in D},

and we can improve Lemma 14 as follows:

1 is close-to-convex in D. Theorem 24. [62] Every f ∈ CH 1 and F = h + �g, where |�| = 1. Then, Proof. Let f = h + g ∈ CH

Re F � (z) = Re (h� (z) + �g � (z)) > |g � (z)| + Re (�g � (z)) ≥ |g � (z)| − |�g � (z)| = 0

showing that F is analytic and close-to-convex in D. According to Lemma 13, it follows that the harmonic function f is also close-to-convex in D. It is easy to see that the converse of Theorem 24 is not true. That is, a close-to1 convex function does not necessarily belong to the class C H .


Corollary �∞ 6. [62] Suppose that f = h + g, where h(z) = z + g(z) = n=1 bn z n in a neighborhood of origin with |b1 | < 1. If ∞ �

(22)

n=2

n|an | +

∞ �

n=1

321

�∞

n=2

an z n and

n|bn | ≤ 1,

1 (and, hence f is close-to-convex in D). then f ∈ CH

Proof. Note that the coefficient inequality implies that both h and g are analytic in D. Without loss of generality, we may assume that f is not affine. Then, as fz = h� and fz = g � , it follows from the hypotheses that � �∞ � �� n−1 � nan z Re h (z) ≥ 1 − � � � � n=2

≥1− ≥1− implying that f ∈ inequality).

1 CH

∞ � n=2 ∞ � n=2

n|an | |z|n−1 n|an | ≥

∞ � n=1

n|bn | ≥ |g � (z)|

(since strict inequality occurs either at third or fifth

When the co-analytic part is zero, i.e g ≡ 0, then the coefficient condition (22) is known to imply that f is starlike in D and satisfies the condition that Re f � (z) > 0 in D. This well-known result suggests that in the case of harmonic f , one expects that the condition (22) would imply that f is starlike. This has been shown to be true in [37] (see also [62]). Corollary 7. Assume the hypothesis of Corollary 6 with b 1 = 0. Then f = h + g is starlike in D. For a proof of this corollary we refer to [7]. Univalent harmonic polynomials are 1 are investigated in a recent discussed in [65, 35] and it relation with the class C H paper by Bharanedhar and Ponnusamy [7]. Example 7. Consider f (z) = z + az n + bz m ,

n ≥ 2, m ≥ 1.

By Corollary 6, f is close-to-convex (univalent) in D whenever n|a| + m|b| ≤ 1. On the other hand, by Corollary 7, f is also starlike in D when n|a| + m|b| ≤ 1 with m ≥ 2. Finally, we remark that the univalency of f may be verified by a direct computation. Our next result is a slightly modified version of the theorem due to Mocanu [45] and some of its applications are presented by Ponnusamy and Sairam Kaliraj [59]. Theorem 25. Let G ∈ C 1 (D) be univalent such that G(D) is a convex domain and JG (z) > 0 in D. Suppose that F ∈ C 1 (D) such that (23)

Re I(F, G) > |I(F, G)| for z ∈ D,

322

where


� � F I(F, G) = �� z Gz

� Fz �� . Gz �

Then F is sense preserving and univalent in D. Proof. Set Ω = G(D), ϕ = G−1 and consider � � H(w) = F ϕ(w) . Then it is a simple exercise to see that (23) is equivalent to Re Hw > |Hw |,

w ∈ Ω.

By Lemma 14, we obtain that H is univalent in the convex domain Ω and thus, F = H ◦ G is univalent in D. As � � � � �I(F, G)�2 − �I(F, G)�2 = JF · JG ,

the condition (23) gives that JF · JG > 0, and since JG > 0, it follows that JF > 0. The theorem is proved.

Conjecture 3. Assume the hypothesis of Theorem 25. Then F is sense preserving close-to-convex and univalent in D. Theorem 27 below supports this conjecture. In the special situation, we still have the following theorem due Clunie and Sheil-Small [18] and using this result, one can construct more examples of univalent harmonic functions. We do not present its proof here. Theorem 26. Let f = h + g be a sense-preserving harmonic mapping in the unit disk D, and suppose that h + �g is convex for some |�| ≤ 1. Then f is a univalent harmonic mapping from D onto a close-to-convex domain. Theorem 27. (Ponnusamy and Sairam Kaliraj [59]) Suppose that F = h+ g, where h and g are analytic in D. Suppose further that � � � � � Fz (z) � Fz (z) � � (24) Re > � G� (z) �, z ∈ D, G� (z) for some G which is univalent, analytic and convex in D. Then F is close-to-convex in D. Proof. Proof follows if we use Lemma 13 and the method of proof of Theorem 24. For details, we refer to [59]. Note that Theorem 27 reduces to Lemma 14 if G(z) = z. Corollary 8. Let f = h + g be a sense-preserving harmonic mapping in the unit disk D, and suppose that h is univalent and h(D) is convex. Then f is univalent and close-to-convex in D. Proof. By assumptions Jf = |h� |2 − |g � |2 > 0 in D, and so |h� (z)| > |g � (z)| in D. Set h = G in Theorem 27. Then h� (z) = fz (z) = G� (z), and so (24) is equivalent to Jf > 0. Thus, f is close-to-convex in D and hence, f is univalent in D.


323

However, in this special case, one can present a more direct � proof� for f to be univalent in D. Let D = h(D). Define φ : D → C by φ(w) = g ◦ h−1 (w). Then φ is analytic on the convex domain D and φ� (w) =

g � (z) , h� (z)

where w = φ−1 (z),

so that |φ� (w)| = |g � (z)/h� (z)| < 1 by the hypothesis that f is sense-preserving.

Now, we suppose that z1 , z2 ∈ D, z1 �= z2 , such that f (z1 ) = f (z2 ). Set w1 = h(z1 ) and w2 = h(z2 ). Then h(z1 ) − h(z2 ) = g(z2 ) − g(z1 ), and therefore w1 − w2 = φ(w2 ) − φ(w1 ).

On the other hand, because φ is analytic on the convex domain D, this can be equivalently rewritten as � w1 − w2 = φ� (w) dw, [w1 ,w2 ]

where the integral is taken over a straight-line segment connecting w 1 to w2 . However, this is not possible, because |φ� (w)| < 1 in D. Thus, f (z1 ) �= f (z2 ) and this proves the univalency in D. For many more applications of Theorem 27, we refer to the recent article of Ponnusamy and Sairam Kaliraj [59] (see also [57]). Using these results, one can construct many more interesting examples of univalent harmonic mappings. For example, we refer to the recent works [57, 59, 62]. 6.1. Some Basic Examples. Example 8. Consider h(z) = z +

1 log 2

�

1+z 1−z

� and g(z) = z −

1 log 2

�

1+z 1−z

� .

Then h� (z) = 1 + 1/(1 − z 2 ) and since w = 1/(1 − z 2 ) maps unit disk D into the half-plane Re w > 1/2, it follows that Re h� (z) > 3/2 in D and thus, h is univalent in D and therefore, h� (z) �= 0. Moreover, (h − g)� (z) = so that

�

2 and (h + g)� (z) = 2 1 − z2

� h� (z) + g � (z) Re = Re (1 − z 2 ) > 0. h� (z) − g � (z) showing that |g � (z)/h� (z)| < 1 in D and so, by Lewy’s theorem, f = h + g is locally univalent in D. Note that �

� 1+z . 1−z The geometric behavior of f (z) is illustrated in Figure 22. f (z) = 2Re z + i arg

A computation shows that the function h is non-convex in D (see Figure 20) So, Corollary 8 has its limitation.

•

324


Figure 19. Image of D under f (z) = 12 Re z +

Figure 20. Image of D under h(z) = z +

1 2

i 4

arg

log

�

�

1+z 1−z

1+z 1−z

� .

� .

Example 9. We know that hλ (z) = z + λz 2 is convex whenever |λ| ≤ 1/4. At this place, it worth recalling the fact that if f is a normalized analytic function f such that |f �� (z)| ≤ 1/2 in D, then f is convex in D (see [61] for a general result). In particular, let h(z) = z − (1/4)z 2 . Then h(D) is a convex domain. Next, we consider � � 1+z 2 g(z) = (1/4)(z − 1) exp − . 1−z


325

Figure 21. Image of D under f (z) = z − 14 z 2 + 14 (z − 1)2 exp Then we see that

�

z+1 z−1

�

.

� � 1+z 2−z � 2−z , g (z) = − exp − h (z) = 2 2 1−z �

and therefore, the dilatation is

ω(z) =

� � 1+z g � (z) = − exp − . h� (z) 1−z

Since w = (1 + z)/(1 − z) maps the unit disk D conformally onto the right half-plane Re w > 0, it follows that �� 1+z < 1 in D. |ω(z)| = exp −Re 1−z Thus, f = h + g is sense-preserving. By Corollary 8, 1 1 f (z) = z − z 2 + (z − 1)2 exp 4 4

�

z+1 z−1

�

is harmonic univalent and close-to-convex in D (see Figure 21).

•

326


Example 10. Let p and q be two arbitrary analytic functions in D such that Re p(z) > 0 and Re q(z) > 0 in D with p(0) + q(0) = 1. Define h by �� t � � � z� p(s) + q(s) − 1 exp h(z) = ds dt. s 0 0 Then

��

z

h (z) = exp �

0

and 1+

p(s) + q(s) − 1 s

� ds

zh�� (z) = p(z) + q(z) h� (z)

and hence, h is convex in D (with h(0) = 0 = h� (0) − 1). For instance, consider � � � � 1 1+z 3 1−z p(z) = and q(z) = . 2 1−z 2 3−z

It is easy to see that w = (1 − z)/(3 − z) maps D conformally onto the disk |w − 1/4| < 1/4, and conclude that both p(D) and q(D) lie in the right half-plane Re w > 0 with p(0) + q(0) = 1. (In fact, fixing one of p and q varying the other gives various possibilities, see the discussion on Möbius transformations from [54, 60]). Now, we observe that � � � � 3 1−z 1 3 1 1+z +1 + −1 = − p(z) + q(z) − 1 = 2 1−z 2 3−z 1−z 3−z and so,

p(z) + q(z) − 1 2 1 1 = = − . z (1 − z)(3 − z) 1−z 3−z

Integration and then exponentiation give, h� (z) =

3−z 3(1 − z)

and therefore, h(z) = (z − 2 log(1 − z))/3. Finally, we set � � 1+z g � (z) = − exp − ω(z) = � h (z) 1−z so that

1 g (z) = − 3 �

Solving for g gives

� 1+

2 1−z

�

� � 1+z exp − . 1−z

� � 1+z 1 . g(z) = − (1 − z) exp − 3 1−z

Thus, f = h + g takes the form (see Figure 22) f (z) =

1 1 (z − 2 log(1 − z)) − (1 − z) exp 3 3

�

z+1 z−1

�

and by Corollary 8, f is harmonic univalent and close-to-convex in D.

•


Figure 22. Image � of D under f (z) = � 1 z+1 − 3 (1 − z) exp z−1

1 3 (z

327

− 2 log(1 − z))

Questions and Exercises. 1. Describe the difference between a real-analytic function and an analytic function in an open set Ω. 2. Must every analytic function be harmonic? 3. When can every harmonic function be analytic? Let f (z) = x 2 − y 2 + ix. Is f harmonic on C? Is f 2 harmonic on C? Hint: Set f 2 (z) = (x2 − y 2 )2 − x2 + 2x(x2 − y 2 )i := U + iV . Then ΔV = 8x. 4. Give an example of C 1 -function that is not harmonic. How about the function f (z) = x cos y + iy? 5. Show that neither the reciprocal 1/f nor the inverse f −1 (when they exist) of a harmonic function is in general harmonic. 6. Consider the automorphism of the unit disk: φ a (z) = eiθ (z − a)/(1 − az) (|a| < 1). If f : D → D is a harmonic function, must φ a ◦ f be harmonic in D? 7. Show that the composition of a harmonic mapping with an affine mapping is harmonic, i.e., if f is harmonic, then so is αf + βf + γ, |α| �= |β|. 8. Check whether the precomposition of a harmonic function with a conformal mapping is harmonic. 9. Suppose that f (z) = ln |z| + iy. Determine Ω, analytic functions h and g so that the decomposition f (z) = h(z) + g(z) is possible. Also determine the corresponding second complex dilatation ω(z) which is analytic in Ω with |ω(z)| < 1 for z ∈ Ω. 10. For |α| ≤ 2, let fα (z) = z − z1 + α ln |z|. Show that each fα is sense-preserving harmonic mapping of the exterior of the unit disk onto the punctured plane. Hint: See [32].

328


11. State and prove the composition rule for (f ◦ g) z and (f ◦ g)z . 12. Suppose that f is a harmonic function in an open set Ω ⊆ C such that Jf (z0 ) �= 0 at some point z0 ∈ Ω. Show that f is either sense-preserving or sense-reversing in some neighbourhood of z 0 . 13. Suppose that f is a sense-preserving harmonic mapping of a domain Ω. Must |fz | > 0 on Ω? Must the second complex dilatation ω(z) be an analytic mapping of Ω into the unit disk D? 14. Suppose that f is a sense-preserving (reversing) harmonic diffeomorphism and g is analytic (anti-analytic). Check whether the Jacobian of f ◦ g does not change sign, i.e. Jf ◦g > 0 (< 0). 15. Modify Remark 2 to obtain a harmonic homeomorphism of R n with Jacobian vanishing on a hyperplane. Conclude that Lewy’s theorem is false for all n ≥ 3. 16. If f (z) = z + z 2 − z, determine the critical points of f in C. 17. Verify whether the following functions are harmonic in the upper half-plane Ω = {z = x + iy : y > 0}: (a) f (z) = arg z + iy (b) f (z) = xy + iy. Do these mappings take Ω onto Ω? If not, find f (Ω). 18. Must f (z) = z + Re (ez ) be a harmonic mapping in C? 19. Let f = h + g with h + g = z/(1 − z) and with the dilatation ω(z) = −z (Note that z/(1 − z) is convex in every direction). Compute the formulas for h and g explicitly and determine f (D). 20. Does Example 4 follow from Lemma 1 or Corollary 1? If yes, explain. If not, could you find an analog of Corollary 1 applicable to Example 4? 1 21. Define f (z) = z|z| k −1 for k ≥ 1. Is f a quasiconformal mapping of C with K = k? 22. Let f = h + g with h − g = z. Assume first the dilatation ω(z) = z and then ω(z) = z 2 . Using the shear construction, determine f in each case. Find also f (D). 23. Let f = h + g be a CHD domain. Must fα defined by fα = h + eiα g be convex in the direction eiα ? 24. Suppose that f = h + g ∈ SH . Must h be locally univalent in D? Should h be univalent in D? Hint:. As f ∈ SH is sense preserving, Jf > 0, i.e. |g � (z)| < |h� (z)| in D. Thus, h� (z) �= 0 in D and so the analytic part of every f ∈ SH is necessarily locally univalent in D. 25. Why is not SH a compact family? Hint: See Example 3. 0 . Must the dilatation ω(z) have the property that 26. Suppose that f ∈ SH |ω(z)| ≤ |z| in D? Can it be |ω(z)| ≤ 1 in D? Hint: ω(z) = g � (z)/h� (z), where ω(0) = 0 and |ω(z)| < 1 in D. Schwarz’s lemma gives the result. 27. Let f ∈ SH with fz (0) = b1 . If

φ(w) =

w − b1 w . 1 − |b1 |2


329

0 Show that F = φ ◦ f ∈ SH . Is the mapping f �→ F = φ ◦ f invertible? If so, 0 to f ∈ SH (for a specified value what is its inverse map which sends F ∈ S H of b1 with |b1 | < 1). Hint: f = F + b1 F does �∞the job. 28. Suppose that f (z) = n=0 an z n is analytic and univalent in D. Show that the area A of the range f (D) is given by �� ∞ � |f � (z)|2 dx dy = π n|an |2 . A= D

n=1

Hint: Recall that if Ar equals the area of the range f (Dr ), then �� r � 2π � 2 |f (z)| dx dy = |f � (ρeiθ )|2 ρ dρ dθ Ar = Dr

where

�

|f (ρe )| = �

iθ

2

0

∞ �

0

�� nan ρ

n−1 i(n−1)θ

e

n=1

Using the fact that �

∞ �

� nan ρ

n−1 −i(n−1)θ

e

.

n=1

�

2π

e

inθ

dθ =

0

2π 0

if n = 0; otherwise,

we conclude �

2π

0

|f � (ρeiθ )|2 dθ = 2π

∞ � n=1

n2 |an |2 ρ2(n−1)

and so Ar = 2π

∞ � n+1

� 2

n |an |

2 0

r

ρ2n−1 dρ = π

∞ � n=1

n|an |2 r2n .

Allowing r → 1 gives the desired result. 29. Let f = u + iv be a C 1 -function defined in a domain Ω ⊂ C. Show that −

fθ (reiθ ) ur vθ − uθ vr |fr (reiθ )|2 and |fz (z)|2 + |fz (z)|2 = + . r 2 2r2 Hint: We have

Jf (reiθ ) =

fr = eiθ fz + e−iθ fz and fθ = ir(eiθ fz − e−iθ fz )

so that fz =

e−iθ 2

and eiθ fz = 2

� fr − i �

fθ r

fθ fr + i r

� =

� vθ � uθ �� e−iθ �� ur + + i vr − , 2 r r

=

� vθ � uθ �� eiθ �� ur − + i vr + . 2 r r

�

The conclusion follows from the last three equalities.

330


30. If u is a non-constant harmonic function in an open subset of R 2 , show that the critical points of u are isolated. Verify this fact using particular choices of u. Hint: The differential operator gives � � 1 ∂u ∂u ∂u = −i ∂z 2 ∂x ∂y which vanishes at critical points. But u being harmonic shows that u z (x, y) is analytic and so the critical points of u are isolated unless u z ≡ 0. For instance, if u(x, y) = x2 − y 2 and U (x, y) = 2xy then the only critical point of u and U is (0, 0). We see that uz = x + iy = z and Uz = −iz

so that uz and Uz are analytic. 31. Give three examples of harmonic functions which have no critical points. Hint: Consider u(x, y) = 2ex cos y and observe that uz = ez �= 0 in C 32. Can there exist a function f ∈ SH so that the area of the range f (D) is minimum? Hint: No. Consider f (z) = z + bz, 0 < |b| < 1. We see that Jf (z) = 1 − |b|2 and if A equals the area of f (D), then �� A= Jf dx dx = π(1 − |b|2 ) D

which can be made as small as possible by allowing b → 1. Thus, S H contains no area-minimizing functions. 33. Suppose that h(z) = z/(1 − z) and ω is defined as in Example 10, i.e. � � 1+z g � (z) = − exp − ω(z) = � . h (z) 1−z 34. 35. 36.

37. 38.

39.

Determine g and the harmonic mapping f = h + g. Draw the image of the mapping f using Mathematica or otherwise. Discuss the corresponding mapping properties of f in Example 9 with h λ (z) = z + λz 2 for various values of λ, say λ = 1/5, 1/6, 1/7. Does there exist a univalent harmonic mapping f of the unit disk D onto itself with the dilatation ω(z) = z? Hint: See [31, p.481] and also [71]. Let f = h + g with h(z) − g(z) = z and ω(z) = z. Compute h and g explicitly 0 and determine the range f (D). so that f ∈ SH Remark: Note that φ(z) = z is convex. Does it mean f maps D onto a convex domain? Of course, theorem on shearing implies that |Im f (z)| < 1. Do the previous exercise by replacing ω(z) = z by ω(z) = z 2 . Using Theorem 11 and the Heinz lemma show that there is no harmonic function that maps the unit disk D univalently onto C. More generally, shows that there exists no univalent harmonic mapping of C onto a proper subdomain Ω of C. Prove or disprove the following: For f (z) = z + az n + bz m (n ≥ 2, m ≥ 1) to be univalent in the unit disk D, it is necessary that a and b satisfy the condition n|a| + m|b| ≤ 1 (See Example 7).


331

40. Simplify the statement of Theorem 25 in each of the following cases: (a) F is analytic in D, (b) G is analytic in D, (c) F and G are both analytic in D, (d) F is analytic and G(z) = z in D, (e) G is one of the following functions: � �� 1+z 1 z , − log(1 − z), log . z, 1−z 2 1−z

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Saminathan Ponnusamy

E-mail: [email protected]

Address: Department of Mathematics, Indian Institute of Technology Madras, Chennai-600 036, India Antti Rasila

E-mail: antti.rasila@iki.fi

Address: Institute of Mathematics, Aalto University, P.O.Box 11100, FI-00076 Aalto, Finland