Introduction to Quasiconformal Mappings in the plane with an application to quasiconformal

surgery

Nathan Mercer

Rich Stankewitz

(Nathan Mercer) Department of Mathematics, Holyoke Community College, Holyoke, MA 0104, USA

E-mail address: nmercer@hcc.mass.edu

(Rich Stankewitz) Department of Mathematical Sciences, Ball State University, Muncie, IN 47306, USA

E-mail address: rstankewitz@bsu.edu

Contents

1. Introduction 4 2. Notation 4 3. Differentiability in the Real Sense 6 4. Tentative Definition of Quasiconformal Maps 10 5. Conformal Maps 11 6. Geometric Definition of Quasiconformal Maps 11 7. Metric Definition of Quasiconformal Maps 23 8. Analytic Characterization 24 9. Beltrami Equations 25 10. Riemann Surface Structures 26 11. Fundamental Theorem of Quasiconformal Surgery 30 12. Appendix I 34 13. Appendix II 34

Bibliography 39

1. Introduction

A large portion of this paper is from the Masters thesis of the first author, written under the direction of the second author. It is presented here, not as original work, but as a systematic introduction to the topic at hand. We attempt to present a fairly self-contained paper from which an interested reader could learn many of the aspects needed to understand the background necessary in the statement and proof of the Fundamental Theorem of Quasiconformal Surgery.

This paper will introduce a special kind of complex homeomorphism called a quasiconformal map. Quasiconformal maps have many applications in areas of heat conduction, electrostatic potential, and fluid flow. They are also a valuable tool in the field of complex dynamics. We follow [3] in that we will limit our discussion to the context of the extended complex plane. The study of quasiconformal maps, however, may be extended to higher dimensions.

Quasiconformal maps are generalizations of the well known conformal maps (and it will be shown that every conformal map is a quasiconformal map). Conformal maps impose a very strong condition on the differential (approximating linear map), whereas quasiconformal maps relax this condition considerably. However, quasiconformal maps still retain many aspects (or parallel aspects) of conformal maps. As such these maps become a more flexible tool for such procedures as we will see in the proof of the Fundamental Theorem of Quasiconformal Surgery. This theorem will allow one to glue together, in some sense, two different analytic maps – something prohibited, in the usual sense, by the Identity Theorem (Principle of Analytic Continuation).

The reader is assumed to have a basic understanding of general topology, complex analytic functions, measure theory, and integration theory. Wherever there is an exception, there will be detailed explanation and proof, along with precise references for those results whose proofs are “beyond the scope” of the paper.

The authors would like to thank Xiaosheng Li for his helpful suggestions, as well as for his help in creating the figures in this paper.

2. Notation

We follow Lehto and Virtanen [3] and state that throughout this paper the plane will be taken to mean the extended complex plane, denoted by \overline{ℂ}, which is homeomorphic to the Riemann Sphere. And we will call the usual Euclidean plane the finite plane and denote this by ℂ. We call a subset of the plane a domain if it is both open and connected. Also, all functions will be taken to be complex valued unless otherwise indicated.

We will use the following notation: Δ(z, r) is the Euclidean disk of radius r and center z, C(z, r) is the Euclidean circle of radius r and center z, Ann(z_{0}; r, R) = \{z : r < |z − z_{0}| < R\}, ∂Ω and \overline{Ω} refer to the boundary and closure, respectively, of the set Ω taken in the topology of \overline{ℂ}.

Let U be an open subset of the plane. We will say a complex valued function on U given by f = u + iv where u = Re (f) and v = Im (f) is in the class C^{1}(U) if the partial derivatives f_{x} = u_{x}+iv_{x} and f_{y} = u_{y} +iv_{y} both exist and are continuous on U. Also, when the necessary derivatives of f exist, we will set

(2.1)  ∂f = f_{z} = \frac{1}{2} (f_{x} − if_{y}) and \overline{∂}f = f\overline{z} = \frac{1}{2} (f_{x} + if_{y}) .

2. NOTATION 5

For this paper we will be using a “little-o” notation. So we will make the following definition.

Definition 2.1. Let f and g be complex valued functions defined in a neighborhood U of z_{0} with g(z) ≠ 0 for all z ∈ U \ {z_{0}}. We say f = o(g) as z → z_{0} if lim_{z→z_{0}} \frac{f(z)}{g(z)} = 0. If no confusion can arise, the “ z → z_{0}” will be omitted when the choice of z_{0} is clear.

The following are properties that will be used in what follows. Lemma 2.2. Consider functions f, f_{1}, f_{2}, g, h, and h_{1}, defined in a neighbor-

hood of z_{0}.

a) If f_{1} = o(g) and f_{2} = o(g), then f_{1} + f_{2} = o(g).

b) If f_{1} = o(g) and f_{2} = o(g), then f_{1}f_{2} = o(g^{2}).

c) Suppose i) f = f_{1} + o(g) with lim sup_{z→z_{0}} |f(z)| < +∞, ii) h = h_{1} + o(g) with lim sup_{z→z_{0}} |h_{1}(z)| < +∞.

Then fh = f_{1}h_{1} + o(g).

d) Suppose i) f = f_{1} + o(g), ii) h = h_{1} + o(g) with both h(z) ≠ 0 and h_{1}(z) ≠ 0 for all z ≠ z_{0},

iii) lim sup_{z→z_{0}} \frac{1}{|h(z)|} < +∞, lim sup_{z→z_{0}} |f(z)| < +∞, and lim sup_{z→z_{0}} |g(z)| < +∞.

Then \frac{f}{h} = \frac{f_{1}}{h_{1}} + o(g).

e) If f = o(g), then i) \overline{f} = o(g); ii) |f| = o(g); iii) f = o(|g|); and iv) f = o(ḡ).

Proof. Parts (a), (b), and (e) follow immediately from the definition. Part (c) follows from the following equalities:

lim_{z→z_{0}} \frac{f(z)h(z) − f_{1}(z)h_{1}(z)}{g(z)} = lim_{z→z_{0}} \frac{f(z)h(z) − f(z)h_{1}(z) + f(z)h_{1}(z) − f_{1}(z)h_{1}(z)}{g(z)}
= lim_{z→z_{0}} f(z)\frac{h(z) − h_{1}(z)}{g(z)} + lim_{z→z_{0}} h_{1}(z)\frac{f(z) − f_{1}(z)}{g(z)} = 0,

which shows that fh − f_{1}h_{1} = o(g) and thus fh = f_{1}h_{1} + o(g).

6 CONTENTS

Part (d) is shown as follows:

lim_{z→z_{0}} \frac{\frac{f(z)}{h(z)} − \frac{f_{1}(z)}{h_{1}(z)}}{g(z)} = lim_{z→z_{0}} \frac{f(z)h_{1}(z) − h(z)f_{1}(z)}{h(z)h_{1}(z)g(z)}
= lim_{z→z_{0}} \frac{f(z)h_{1}(z) − f_{1}(z)h_{1}(z) + f_{1}(z)h_{1}(z) − h(z)f_{1}(z)}{h(z)h_{1}(z)g(z)}
= lim_{z→z_{0}} \frac{1}{h(z)} · \frac{f(z) − f_{1}(z)}{g(z)} + lim_{z→z_{0}} \frac{f_{1}(z)}{h(z)h_{1}(z)} · \frac{h_{1}(z) − h(z)}{g(z)} = 0,

and thus \frac{f(z)}{h(z)} = \frac{f_{1}(z)}{h_{1}(z)} + o(g).

Note that we justify that lim_{z→z_{0}} \frac{f_{1}(z)}{h(z)h_{1}(z)} · \frac{h_{1}(z)−h(z)}{g(z)} = 0 from the following argument. Since f = f_{1} + o(g), \frac{f(z)−f_{1}(z)}{g(z)} → 0 as z → z_{0} by definition. But we have that lim sup_{z→z_{0}} |g(z)| < +∞ and so f(z) − f_{1}(z) → 0 as z → z_{0}. But since lim sup_{z→z_{0}} |f(z)| < +∞, we must also have that lim sup_{z→z_{0}} |f_{1}(z)| < +∞. We can similarly show lim sup_{z→z_{0}} \frac{1}{|h_{1}(z)|} < +∞ from the hypotheses lim sup_{z→z_{0}} \frac{1}{|h(z)|} < +∞ and h = h_{1} + o(g). □

3. Differentiability in the Real Sense

In what follows we will sometimes substitute without mention the two-dimensional real vector z = ( \begin{matrix}x \\ y \end{matrix} ) for z = x + iy.

Definition 3.1. Let f : Ω_{1} → Ω_{2} be a function between finite plane domains. We say that f is differentiable in the real sense (also called ℝ-differentiable or real differentiable) at z_{0} = ( \begin{matrix}x_{0} \\ y_{0} \end{matrix} ) when there exists a matrix [ \begin{matrix}α & β \\ γ & δ \end{matrix} ], where α, β, γ, δ ∈ ℝ are such that

f(z) = f(z_{0}) + [ \begin{matrix}α & β \\ γ & δ \end{matrix} ] ( \begin{matrix}x − x_{0} \\ y − y_{0} \end{matrix} ) + E(z)

for all z ∈ Ω_{1} and the function E(z) satisfies \frac{|E(z)|}{|z−z_{0}|} → 0 as z → z_{0}, i.e., E(z) = o(z − z_{0}).

Let f be given as in Definition 3.1. Then using i simply as a place holder, we get

(3.1) f(z) = f(z_{0}) + α(x − x_{0}) + β(y − y_{0}) + iγ(x − x_{0}) + iδ(y − y_{0}) + E(z).

By definition then, we have

f_{x}(z_{0}) = lim_{x→x_{0}} \frac{f(x + iy_{0}) − f(x_{0} + iy_{0})}{x − x_{0}}

= lim_{x→x_{0}} \frac{f(x_{0} + iy_{0}) + α(x − x_{0}) + β(y_{0} − y_{0}) + iγ(x − x_{0}) + iδ(y_{0} − y_{0}) + E(x + iy_{0}) − f(x_{0} + iy_{0})}{x − x_{0}}

= α + iγ + lim_{x→x_{0}} \frac{E(x + iy_{0})}{x − x_{0}} = α + iγ.

Note that lim_{x→x_{0}} \frac{E(x+iy_{0})}{x−x_{0}} = 0 since it is lim_{z→z_{0}} \frac{E(z)}{z−z_{0}} = 0, letting z approach z_{0} along the line z = x + iy_{0}.

Similarly, f_{y}(z_{0}) = β + iδ and so by (3.1) we see

(3.2)  f(z) = f(z_{0}) + f_{x}(z_{0})(x − x_{0}) + f_{y}(z_{0})(y − y_{0}) + E(z).

Note that writing f = u + iv where u = Re f and v = Im f, we then have

u_{x}(z_{0}) = α, u_{y}(z_{0}) = β, v_{x}(z_{0}) = γ, and v_{y}(z_{0}) = δ.

Now, using the fact that if z = x + iy, then

x = \frac{z + z̄}{2} and y = \frac{z − z̄}{2i},

we see from (3.1) that:

f(z) =f(z_{0}) + \frac{α}{2} (z + z̄ − z_{0} − \overline{z_{0}}) + \frac{β}{2i} (z − z̄ − z_{0} + \overline{z_{0}})

+ i\frac{γ}{2} (z + z̄ − z_{0} − \overline{z_{0}}) + \frac{iδ}{2i} (z − z̄ − z_{0} + \overline{z_{0}}) + E(z)

=f(z_{0}) + (\frac{α}{2} + \frac{β}{2i} + \frac{iγ}{2} + \frac{δ}{2}) (z − z_{0}) + (\frac{α}{2} − \frac{β}{2i} + \frac{iγ}{2} − \frac{δ}{2}) (z̄ − \overline{z_{0}}) + E(z)
=f(z_{0}) + ∂f(z_{0})(z − z_{0}) + ¯∂f(z_{0})(z̄ − \overline{z_{0}}) + E(z) by (2.1).

We have thus demonstrated the following theorem.

Theorem 3.2. If a function f is ℝ-differentiable at z_{0}, then it has partial derivatives at z_{0} and can be written in the form

(3.3)  f(z) = f(z_{0}) + A(z − z_{0}) + B(z̄ − \overline{z_{0}}) + E(z)

where A = ∂f(z_{0}) (the z-coefficient) and B = ¯∂f(z_{0}) (the z̄-coefficient) are both complex and E(z) = o(z − z_{0}).

The following theorem is a well known result indicating conditions which imply a function is ℝ-differentiable.

Theorem 3.3 ([5], p. 100). Suppose that a function f = u + iv is defined in an open subset U of the finite plane and that the partial derivatives u_{x}, u_{y}, v_{x}, and v_{y} exist everywhere in U. If each of these partial derivatives is continuous at the point z_{0} of U, then f is differentiable in the real sense at z_{0}.

Equation (3.3) says that a function f(z) that is differentiable in the real sense at z_{0} can be well approximated near z_{0} by a linear function of the form

L(z) = f(z_{0}) + A(z − z_{0}) + B(z̄ − \overline{z_{0}}).

Note that L is just a translation (z ⟼ z−z_{0}), followed by L̃(z) = Az+Bz̄, followed by another translation (z ⟼ z + f(z_{0})). Thus in order to analyze L(z) (and hence locally analyze any real-differentiable function) we first consider L̃(z) = Az + Bz̄. If we let A_{1}, A_{2} be the real and imaginary parts of A and B_{1}, B_{2} be the real and imaginary parts of B we get

(3.4)  L̃(z) =(A_{1} + iA_{2})(x + iy) + (B_{1} + iB_{2})(x − iy)
(3.5)  =(A_{1} + B_{1})x + (B_{2} − A_{2})y + i [(A_{2} + B_{2})x + (A_{1} − B_{1})y] .

We note that in matrix form we see from equation (3.5) that z ↦ L̃(z) is written

(3.6)  ( \begin{matrix}x \\ y \end{matrix} ) ↦ [ \begin{matrix}A_{1} + B_{1} & B_{2} − A_{2} \\ A_{2} + B_{2} & A_{1} − B_{1} \end{matrix} ] ( \begin{matrix}x \\ y \end{matrix} ) .

Hence we immediately see that if a map f has form (3.3) with E(z) = o(z − z_{0}),

then f is real differentiable at z_{0} (as given in Definition 3.1) with [ \begin{matrix}α & β \\ γ & δ \end{matrix} ] =

[ \begin{matrix}A_{1} + B_{1} & B_{2} − A_{2} \\ A_{2} + B_{2} & A_{1} − B_{1} \end{matrix} ]. We also note that the Jacobian of the map f at z_{0} is then

given by

(3.7)  J_{f}(z_{0}) = det [ \begin{matrix}α & β \\ γ & δ \end{matrix} ] = |A|^{2} − |B|^{2} = |∂f(z_{0})|^{2} − |¯∂f(z_{0})|^{2}.

We now discuss the concept of a total differential of a function f. The reader may reference [6], p.55-59 for the details of this matter. For a function f differentiable in the real sense at z_{0}, the total differential is given by df_{z_{0}}(z) = ∂f(z_{0})z +\overline{∂}f(z_{0})\overline{z}. Thus df_{z_{0}}(z) is a linear function. When we write df_{z_{0}}(z) ≠ 0 we mean that the function is not identically the zero function, i.e., not both ∂f(z_{0}) and \overline{∂}f(z_{0}) are zero.

Definition 3.4 ([3], p. 9). A (not necessarily differentiable) homeomorphism f : Ω_{1} → Ω_{2} between arbitrary point sets in the plane is sense-preserving if f preserves the orientation of the boundary of every Jordan domain D such that \overline{D} ⊆ Ω_{1}.

Remark 3.5. If a function f is sense-preserving, then its inverse f^{−1} is sensepreserving. Also if both f and g are sense-preserving, then f ∘ g is also sensepreserving.

We refer the reader to the Orientation Theorem and subsequent discussion in [3], p. 9-10 which gives the following. Let f : Ω_{1} → Ω_{2} be a homeomorphism between domains in \overline{ℂ}. If J_{f}(z_{0}) > 0 at some z_{0} ∈ Ω_{1}, then f is sense-preserving. Conversely, if J_{f}(z_{0}) < 0 at some z_{0} ∈ Ω_{1}, then f is sense-reversing. Thus we see that a sense-preserving map must satisfy J_{f}(z) ≥ 0 for all z ∈ Ω_{1}. We note that it is possible to have J_{f}(z_{0}) = 0 at some z_{0} for a sense-preserving map, as is the case for z_{0} ∈ ℝ with the map (x, y) ↦ (x, y^{3}) defined on all of ℂ.

Note: When df_{z_{0}}(z) ≠ 0 and f is sense-preserving we must have |∂f(z_{0})| > 0 and |∂f(z_{0})| ≥ |\overline{∂}f(z_{0})|.

Returning to our analysis of ℝ-differentiable functions and hence functions of the form L̃(z) = Az + Bz̄, we can use basic linear algebra and (3.7) to obtain the following.

Theorem 3.6. The function L̃(z) = Az + Bz̄ from the finite plane to itself is bijective if and only if |A| ≠ |B|. Furthermore, if |A| ≠ |B|, then L̃ is sensepreserving if and only if |A| > |B|.

Let us take a closer look at the maps of the form L̃(z) by first considering, for 0 < μ < 1, the maps L_{μ}(z) = z + μz̄ = x + iy + μx − iμy = (1 + μ)x + i(1 − μ)y.

We see that there is an expansion in the x-direction and a compression in the ydirection under L_{μ}. We see then that a circle of radius r centered at the origin given by (x, y) = (r cos θ, r sin θ) maps to an ellipse centered at the origin since (x, y) = (r cos θ, r sin θ) ↦ (x̃, ỹ) = ((1 + μ)r cos θ, (1 − μ)r sin θ). Thus the image

of the circle is given by [\frac{x̃}{(1 + μ)r} ]^{2} + [\frac{ỹ}{(1 − μ)r} ]^{2} = 1, an ellipse with major axis

r(1 + μ) and minor axis r(1 − μ).

Our main interest is in sense preserving maps and so let us now consider more general maps of the form L̃(z) = Az + Bz̄ with |A| > |B|. Writing μ = \frac{B}{A}, A = |A|e^{iα}, μ = |μ|e^{iβ} and z = re^{iθ}, we have

(3.8)  L̃(z) =A (z + \frac{B}{A} z̄) = |A|e^{iα} (re^{iθ} + |μ|e^{iβ}re^{−iθ})
(3.9)  =|A|e^{i(α+ \frac{β}{2} )} (re^{i(θ− \frac{β}{2} )} + |μ|re^{−i(θ− \frac{β}{2} )}) .

Thus if we let g(z) = |A| e^{i(α+ \frac{β}{2} )} · z and h(z) = e^{−i \frac{β}{2}} · z, we then have that L̃ = g ∘ L_{μ} ∘ h. But g and h are simply rotations and therefore

(3.10)  L̃ = g ∘ L_{|μ|} ∘ h

also maps circles to ellipses (see Figure 1) for which the ratio of the major to minor axis is \frac{1+|μ|}{1−|μ|}.

re^{iβ/2}  L̃  arg = α + \frac{β}{2} |A|(1 + |μ|)r |A|(1 − |μ|)r

Figure 1

Next we consider a general ℝ-differentiable function f : Ω_{1} → Ω_{2} which is expressed f(z) = f(z_{0})+∂f(z_{0})(z−z_{0})+ ¯∂f(z_{0})(z̄−\overline{z_{0}})+E(z) with E(z) = o(z−z_{0}).

Remark 3.7. Since a function f which is real differentiable at z_{0} is well approximated near z_{0} by the linear function L(z) = f(z_{0})+∂f(z_{0})(z −z_{0})+ ¯∂f(z_{0})(z̄−\overline{z_{0}}) we say that if |∂f(z_{0})| > |¯∂f(z_{0})|, then f maps infinitesimal circles centered at z_{0} to infinitesimal ellipses centered at f(z_{0}). We also note that the ratio of the major axis to the minor axis of the infinitesimal ellipse is

\frac{1 + |μ|}{1 − |μ|} = \frac{1 + |\frac{\overline{∂}f}{∂f} |}{1 − |\frac{\overline{∂}f}{∂f} |} = \frac{|∂f| + |\overline{∂}f|}{|∂f| − |\overline{∂}f|}.

This ratio is, of course, intimately connected with the directional derivative, which we state now.

Definition 3.8. The directional derivative of f at z_{0} in the direction of θ is

∂_{θ}f(z_{0}) = lim_{r→0} \frac{f(z_{0} + re^{iθ}) − f(z_{0})}{r}.

By examining the stretching a ℝ-differentiable map exhibits locally, as in Figure 1, the following lemma is clear. Nevertheless, a formal proof will follow.

Lemma 3.9. Let Ω_{1} and Ω_{2} be domains in \overline{ℂ}. If f : Ω_{1} → Ω_{2} is sense preserving and real differentiable at z_{0}, then

max_{θ} |∂_{θ}f(z_{0})| = |∂f(z_{0})| + |¯∂f(z_{0})| and min_{θ} |∂_{θ}f(z_{0})| = |∂f(z_{0})| − |¯∂f(z_{0})|.

Proof. Letting z = z_{0} + re^{iθ} we see that

∂_{θ}f(z_{0}) = lim_{r→0} \frac{f(z_{0}) + ∂f(z_{0})(re^{iθ}) + ¯∂f(z_{0})(re^{−iθ}) + E(z) − f(z_{0})}{r}

= ∂f(z_{0})e^{iθ} + ¯∂f(z_{0})e^{−iθ} + lim_{z→z_{0}} \frac{E(z)}{|z − z_{0}|} = ∂f(z_{0})e^{iθ} + ¯∂f(z_{0})e^{−iθ}.

Let θ_{0} = \frac{γ−α}{2} where α = arg(∂f(z_{0})) and γ = arg(¯∂f(z_{0})). Then one can

easily show that max_{θ} |∂_{θ}f(z_{0})| = |∂f(z_{0})e^{iθ_{0}} + ¯∂f(z_{0})e^{−iθ_{0}}| = |∂f(z_{0})| + |¯∂f(z_{0})|.

Similarly, if we set θ_{1} = \frac{γ−α−π}{2} , we get min_{θ} |∂_{θ}f(z_{0})| = |∂f(z_{0})e^{iθ_{1}}+¯∂f(z_{0})e^{−iθ_{1}}|

= |∂f(z_{0})|−|¯∂f(z_{0})|, noting that |∂f(z_{0})| ≥ |¯∂f(z_{0})| since f is sense-preserving. □

4. Tentative Definition of Quasiconformal Maps

We now wish to give a “tentative” definition for quasiconformal maps. This definition relates to the local stretching displayed by quasiconformal maps. For a differentiable map to be quasiconformal, there must be a bound to this stretching in one direction versus any other direction. First we must make a preliminary definition.

Definition 4.1. Let f be a function that is real-differentiable at z_{0} such that ∂f(z_{0}) ≠ 0. We call the ratio

μ_{f}(z_{0}) = \frac{¯∂f(z_{0})}{∂f(z_{0})}

the complex dilatation of f at z_{0}.

Definition 4.2. Let f be a sense-preserving function that is real-differentiable at z_{0} such that J_{f}(z_{0}) ≠ 0 (thus |μ_{f}(z_{0})| ≠ 1). We set

K̃_{f}(z_{0}) = \frac{1 + |μ_{f}(z_{0})|}{1 − |μ_{f}(z_{0})|} = \frac{|∂f(z_{0})| + |¯∂f(z_{0})|}{|∂f(z_{0})| − |¯∂f(z_{0})|} = \frac{max_{θ} |∂_{θ}f(z_{0})|}{min_{θ} |∂_{θ}f(z_{0})|}.

According to the discussion at the end of the last section K̃_{f}(z_{0}) is the ratio of the maximum stretch to the minimum stretch of an infinitesimal circle around z_{0} under f.

Definition 4.3. [Tentative Definition] Let f : Ω_{1} → Ω_{2} be a real-differentiable sense-preserving bijection between plane domains with J_{f} > 0 on Ω_{1}. For any K ≥ 1, we say f is K-quasiconformal if K̃_{f} = sup_{z∈Ω_{1}} K̃_{f}(z) ≤ K.

Remark 4.4. Definition 4.3 requires f to be real-differentiable everywhere in Ω_{1}, but in the following pages we will not impose such a strict requirement. It is for this reason that Definition 4.3 is our “tentative” definition.

5. Conformal Maps

In this paper we will rely on many facts from the theory of conformal mappings of the complex plane. Let us now recall the definition of a conformal map.

Definition 5.1. Let Ω_{1} and Ω_{2} be open subsets of the plane \overline{ℂ}. We call a function f : Ω_{1} → Ω_{2} conformal if f is bijective and analytic.

Remark: Other sources may define a map to be conformal on its domain when the complex derivative both exists and is nonzero at each point of its domain. This amounts to f being analytic and \underline{locally} bijective. For example, the map f(z) = e^{z} defined on the finite plane ℂ is such a map, but it is \underline{not} conformal by Definition 5.1.

Note that f is complex differentiable at z_{0} ∈ ℂ with f(z_{0}) ∈ ℂ, exactly when (see [5], p. 63) we can express f(z) = f(z_{0})+f′(z_{0})(z −z_{0})+o(z −z_{0}). Thus, if f is conformal at z_{0}, we then have that f is ℝ-differentiable with \overline{∂}f(z_{0}) = 0, ∂f(z_{0}) = f′(z_{0}) ≠ 0 and μ_{f}(z_{0}) = 0. Here we say conformal f : Ω_{1} → Ω_{2} maps infinitesimal circles centered at z_{0} to infinitesimal circles centered at f(z_{0}), since f would be well approximated near z_{0} by L(z) = f(z_{0}) + f′(z_{0})(z − z_{0}) which maps such circles to such circles. It is clear then that a conformal map f is 1-quasiconformal according to our Tentative Defintion 4.3 since K̃_{f}(z) ≡ 1.

Remark 5.2. When f is injective and real-differentiable in a domain Ω_{1} with ∂f(z) never vanishing and μ_{f}(z) ≡ 0, then \overline{∂}f(z) ≡ 0. Therefore f would be conformal onto its range and ∂f = f′.

6. Geometric Definition of Quasiconformal Maps

In this section we will introduce the notion a quadrilateral in the plane \overline{ℂ} and introduce the concept of a “conformal module” of a quadrilateral. This will help give us a precise definition of a quasiconformal map. The geometric definition will be more flexible than the “tentative” definition in Definition 4.3 since it will not require differentiability. However, as we will see in Section 8, such quasiconformal maps will turn out to be real-differentiable almost everywhere.

Definition 6.1. A quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) is a Jordan domain in \overline{ℂ} with four distinct points on ∂Q, the boundary of Q, with a positive cyclic order. We will call these four points the vertices of Q. We will also denote a quadrilateral more briefly by Q \binom{→}{z}, where z^{→} = (z_{1}, z_{2}, z_{3}, z_{4}), or simply by Q when the boundary points are either unimportant or understood.

Definition 6.2. For a > 0, define R(a) = Q(0, a, a+i, i) where Q is a rectangle with vertices in the usual sense at 0, a, a + i, and i.

Theorem 6.3. Let Q(z_{1}, z_{2}, z_{3}, z_{4}) be a quadrilateral in the plane \overline{ℂ}. Then there exists a unique real number a > 0 and a conformal map f mapping Q onto R(a) such that the map f, (extended to ∂Q as given by the Caratheodory-Osgood Theorem), satisfies f(z_{1}) = 0, f(z_{2}) = a, f(z_{3}) = a + i, and f(z_{4}) = i. We may sometimes denote this correspondence of vertices by f(z_{1}, z_{2}, z_{3}, z_{4}) = (0, a, a+i, i).

Proof. By the Riemann Mapping Theorem there exists a conformal map w : Q(z_{1}, z_{2}, z_{3}, z_{4}) → {z : Imz > 0}. By the Caratheodory-Osgood Theorem we can extend w to a homeomorphism from \overline{Q} to \{z : Im z ≥ 0\} ∪ \{∞\} such that, without loss of generality, w(z_{1}, z_{2}, z_{3}, z_{4}) = (α, β, γ, ∞). Now we can apply the Schwarz-Christoffel Formula ([6], p. 236) which guarantees the existence of a conformal map g : \{z : Im z ≥ 0\} → \overline{R} where R is some rectangle in the plane such that α, β, γ, and ∞ are mapped to the vertices of R. Now we can use a map h which simply rotates, translates, and dilates (hence h is conformal) R to some R(a). Therefore h ∘ g ∘ w(Q(z_{1}, z_{2}, z_{3}, z_{4})) = R(a) and thus existence has been shown.

To show uniqueness, suppose that f_{1} : Q → R(a′) and f_{2} : Q → R(a) are as in the statement of the theorem. Then the map f_{2} ∘ f_{1}^{−1} : R(a′) → R(a) is conformal and fixes i and 0. So we can repeatedly apply the Schwarz Reflection Principle to extend f_{2} ∘ f_{1}^{−1} : R(a′) → R(a) to be conformal from the finite plane ℂ to itself. Therefore f_{2} ∘ f_{1}^{−1} must have the form f_{2} ∘ f_{1}^{−1}(z) = cz + d where c and d are complex number with c ≠ 0 (see [5], p. 388). But since f_{2} ∘ f_{1}^{−1} fixes i and 0, we must have that f_{2} ∘ f_{1}^{−1} is the identity map. Hence a = a′. □

Definition 6.4. We define the conformal module of a quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) to be M (Q(z_{1}, z_{2}, z_{3}, z_{4})) = a where a > 0 is as in Theorem 6.3.

Remark 6.5. If R is a rectangle with vertices (in the usual sense) z_{1}, z_{2}, z_{3}, and z_{4}, then a = M (R(z_{1}, z_{2}, z_{3}, z_{4})) = \frac{|z_{2}−z_{1}|}{|z_{3}−z_{2}|}, i.e., the ratio of (oriented) side lengths of R. This holds since a linear map (rotation, dilation, translation) can conformally map R onto R(a) while preserving this ratio.

The next lemma will display the effect on the modulus of a quadrilateral after having the vertices reordered or having the quadrilateral transformed by a conformal map.

Lemma 6.6. Let Q(z_{1}, z_{2}, z_{3}, z_{4}) be a quadrilateral in the plane. Then:

a) M (Q (z_{2}, z_{3}, z_{4}, z_{1})) = \frac{1}{M (Q (z_{1}, z_{2}, z_{3}, z_{4}))}, and

b) M is conformally invariant. That is, if f is conformal on a domain containing \overline{Q}, then M(f(Q)(f(z_{1}, z_{2}, z_{3}, z_{4}))) = M(Q(z_{1}, z_{2}, z_{3}, z_{4})).

Proof. a) Let f_{1} be the conformal map, as in Theorem 6.3, that defines the conformal module a of Q. Set f_{2}(z) = −iz + ia and f_{3}(z) = \frac{z}{a}. Then, since the composition of conformal maps is conformal (see Figure 2 below), we see that f_{3} ∘ f_{2} ∘ f_{1} : Q(z_{2}, z_{3}, z_{4}, z_{1}) → R (\frac{1}{a}) satisfies Theorem 6.3 above, thus showing M (Q (z_{2}, z_{3}, z_{4}, z_{1})) = \frac{1}{a} = \frac{1}{M(Q(z_{1},z_{2},z_{3},z_{4}))}.

b) Let f : Ω_{1} → Ω_{2} be a conformal map with \overline{Q} ⊆ Ω_{1}. Let a denote the conformal module of Q and let g : Q → R(a) be the corresponding conformal map as in Theorem 6.3. The composition g ∘ f^{−1} : f(Q) → R(a) satisfies Theorem 6.3 and shows M(f(Q)(f(z_{1}, z_{2}, z_{3}, z_{4}))) = a. Thus M is conformally invariant. □

We will now set the stage to develop another useful way of computing and estimating the conformal module of a quadrilateral. The vertices of a quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) divide its boundary into four Jordan arcs (see Figure 3). We will denote these Jordan arcs (z_{1}z_{2}), (z_{2}z_{3}), (z_{3}z_{4}), and (z_{4}z_{1}).

Figure 2^{z_{3} z_{2} i  z_{1} z_{4}  f_{1} f_{3}∘f_{2}∘f_{1}  f_{3}  f_{1}(z_{3}) = a + i f_{1}(z_{4}) = i f_{1}(z_{1}) = 0  f_{1}(z_{2}) = a ai 0 = f^{2}(f^{1}(z^{2}))f_{2} \frac{1}{a} = f_{3}(f_{2}(f_{1}(z_{3})))  1 + ai 1 = f^{2}(f^{1}(z^{3})) 0 = f_{3}(f_{2}(f_{1}(z_{2})))}

Figure 3^{^{z_{2}} ^{z_{3}}  ^{γ}  ^{z_{1}} ^{z_{4}}}

Let Γ(Q)= {locally rectifiable Jordan arcs in Q which join (z_{1}z_{2}) to (z_{3}z_{4})}, and let A(Q)= {Borel measurable ρ : Q → [0, ∞) such that ∫_{γ} ρ(z)|dz| ≥ 1 for

all γ ∈ Γ(Q)}. We will call such ρ ∈ A(Q) admissible (for Q). We will also let ℓ_{ρ}(γ) = ∫_{γ} ρ(z)|dz| and call this the ρ-length of γ. When ρ = 1 we simply write

ℓ(γ), the usual Euclidean length. We call A_{ρ}(Q) = ∫∫_{Q} ρ^{2}(z)dxdy the ρ-area of Q.

When ρ = 1 we write A(Q) (or area(Q)), the usual Euclidean area.

Definition 6.7. We define M_{1} (Q \binom{→}{z}) = inf_{ρ∈A(Q)} ∫∫_{Q} ρ^{2}(z)dxdy.

Lemma 6.8. The function M_{1} is conformally invariant, i.e., M_{1}(f(Q)(f(z_{1}, z_{2}, z_{3}, z_{4}))) = M_{1}(Q(z_{1}, z_{2}, z_{3}, z_{4})) for any quadrilateral Q and any conformal f defined on a domain which contains \overline{Q}.

Proof. Let f : Q_{1} → Q_{2} be conformal. Take ρ_{2} ∈ A(Q_{2}). Set ρ_{1}(z) = ρ_{2} ∘ f(z)|f′(z)| for z ∈ Ω_{1}. Choose any γ_{1} ∈ Γ(Q_{1}) and set γ_{2} = f ∘ γ_{1}. Note that γ_{2} ∈ Γ(Q_{2}). Then, letting γ_{2} be a curve parametrized over the interval [b, c] we

have that 1 ≤ ∫_{γ_{2}} ρ_{2}(z)|dz| = ∫_{b}^{c} ρ_{2}[γ_{2}(t)]|γ_{2}^{•}(t)|dt = ∫_{b}^{c} ρ_{2}{f[γ_{1}(t)]}|f′[γ_{1}(t)]||γ_{1}^{•}(t)|dt

= ∫_{b}^{c} ρ_{1}[γ_{1}(t)]|γ_{1}^{•}(t)|dt = ∫_{γ_{1}} ρ_{1}(z)|dz|. Therefore ρ_{1} ∈ A(Q_{1}). We recall that J_{f} =

|f′|^{2} from equation (3.7) and so we apply the change of variable formula (see [5], p.

427) to get M_{1}(Q_{1}) ≤ A_{ρ_{1}}(Q_{1}) = ∫∫_{Q_{1}} [ρ_{1}(z)]^{2}dxdy = ∫∫_{Q_{1}} {ρ_{2}[f(z)]}^{2}|f′(z)|^{2}dxdy =

∫∫_{Q_{2}} [ρ_{2}(w)]^{2}dudv = A_{ρ_{2}}(Q_{2}).

Therefore, taking infimum over the right hand side, we get M_{1}(Q_{1}) ≤ M_{1}(Q_{2}). By the same argument using f^{−1} we obtain M_{1}(Q_{1}) ≥ M_{1}(Q_{2}). Therefore equality holds. □

Theorem 6.9. For every quadrilateral Q in the plane M_{1}(Q) = M(Q).

Proof. We will first show that M_{1}(R(a)) = a = M(R(a)). Let ρ ∈ A(R(a)). Then for fixed 0 < x < a, invoking the Cauchy-Schwarz inequality (see Appendix

I) we have 1 ≤ ℓ_{ρ}(γ) = ∫_{0}^{1} ρ(x + iy)dy ≤ [∫_{0}^{1} ρ^{2}(x + iy)dy]^{\frac{1}{2}} where γ is the vertical

path γ(y) = x + iy for 0 ≤ y ≤ 1. Then a = ∫_{0}^{a} 1^{2}dx ≤ ∫_{0}^{a} ∫_{0}^{1} ρ^{2}(x + iy)dydx =

∫∫_{R(a)} ρ^{2}(x + iy)dydx = A_{ρ}(R(a)). So, taking infimum, a ≤ M_{1}(R(a)). But if ρ ≡ 1,

which can easily seen to be admissible, then M_{1}(R(a)) ≤ area(R(a)) = a. Thus M_{1}(R(a)) = a = M(R(a)) for the rectangle R(a).

Since every quadrilateral Q is mapped conformally onto some R(a), the general case follows from conformal invariance of both M and M_{1}. □

Theorem 6.10. [Rengel’s Inequality] Consider the Jordan arcs (z_{1}z_{2}) and (z_{3}z_{4}) for a quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) such that \overline{Q} ⊂ ℂ. Let s be the Euclidean distance between (z_{1}z_{2}) and (z_{3}z_{4}),i.e., s = inf |w − z| where w ∈ (z_{1}z_{2}) and z ∈ (z_{3}z_{4}), and let t be the Euclidean distance between (z_{2}z_{3}) and (z_{4}z_{1}). Then

\frac{t^{2}}{area(Q)} ≤ M(Q(z_{1}, z_{2}, z_{3}, z_{4})) ≤ \frac{area(Q)}{s^{2}}.

Also, equality holds if and only if Q is a rectangle.

\begin{aligned}z_{1} & z_{3} & z^{2} & z_{4} & S \\ & Figure 4 \end{aligned}

Proof. Notice that ℓ(γ) ≥ s > 0 for every γ ∈ Γ(Q). Now take ρ(z) ≡ 1/s and note that ℓ_{ρ}(γ) = \frac{ℓ(γ)}{s} ≥ 1 for every γ ∈ Γ(Q). Thus ρ is admissible and so

M_{1} (Q \binom{→}{z}) ≤ ∫∫_{Q} ρ^{2}(z)dxdy = \frac{1}{s^{2}} ∫∫_{Q} dxdy = \frac{1}{s^{2}} (areaQ).

By reordering the vertices and setting Q̃ = Q(z_{2}, z_{3}, z_{4}, z_{1}) we see then that \frac{1}{M(Q)} = M(Q̃) ≤ \frac{areaQ̃}{t^{2}} = \frac{areaQ}{t^{2}}, which gives the lower bound on M(Q).

Furthermore, if Q is a rectangle, then equality holds by the Remark 6.5. For the converse, when equality holds, we refer the reader to ([3], pp. 23) for the proof that Q must be a rectangle. □

We are now able to state the geometric definition of quasiconformal maps, which we will take to be our main definition.

Definition 6.11. Let f : Ω_{1} → f(Ω_{1}) be a sense-preserving homeomorphism of a plane domain and let K ≥ 1. We say f is K-quasiconformal on Ω_{1} (or f ∈ K − QC) if

K_{f} := sup_{\overline{Q}⊆Ω_{1}} \frac{M (f (Q))}{M (Q)} ≤ K,

where \overline{Q} denotes the closure of quadrilateral Q in the plane \overline{ℂ}. We call K_{f} the maximal dilatation of f. A map is called quasiconformal if it is K-quasiconformal for some K.

Lemma 6.12. Let f : Ω_{1} → f(Ω_{1}) be K-quasiconformal. Suppose that quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) ⊆ \overline{Q} ⊆ Ω_{1}. Let Q̃ = Q(z_{2}, z_{3}, z_{4}, z_{1}) and f(Q̃) = f(Q)(f(z_{2}), f(z_{3}), f(z_{4}), f(z_{1})) denote a new cyclic order on Q and f(Q) respectively. Then M(f(Q̃)) ≥ \frac{1}{K} M(Q̃).

Proof. By Lemma 6.6 part (a), we see that M(f(Q̃)) = \frac{1}{M(f(Q))} since this is merely a new cyclic order on the same quadrilateral. Also, M(Q̃) = \frac{1}{M(Q)}, and so M(Q̃) = \frac{1}{M(Q)} ≤ K \frac{1}{M(f(Q))}, which follows from f being K -quasiconformal. Therefore \frac{1}{K} M(Q̃) ≤ M(f(Q̃)). □

Theorem 6.13. Let f : Ω_{1} → Ω_{2} be K-quasiconformal and let quadrilateral Q(z_{1}, z_{2}, z_{3}, z_{4}) ⊆ Ω_{1} with \overline{Q} ⊆ Ω_{1}. Then,

a) \frac{1}{K} M(Q) ≤ M(f(Q)) ≤ K · M(Q).

b) f^{−1} : Ω_{2} → Ω_{1} is K-quasiconformal.

c) If g : Ω_{2} → Ω_{3} is K_{0}-quasiconformal, then g ∘f : Ω_{1} → Ω_{3} is at least K_{0} ·Kquasiconformal, that is, K_{g∘f} ≤ K_{0}K.

Remark 6.14. By Theorem 6.13 (a) we see that a 1-QC map f on Ω satisfies M(f(Q)) = M(Q) for all quadrilaterals Q with \overline{Q} ⊂ Ω.

Proof. a) By the geometric definition of quasiconformal maps we get that \frac{M(f(Q))}{M(Q)} ≤ K. Therefore M(f(Q)) ≤ K · M(Q). The remaining inequality follows directly from Lemma 6.12.

b) Let quadrilateral Q_{2} ⊆ \overline{Q_{2}} ⊆ Ω_{2} and set Q_{1} = f^{−1}(Q_{2}) ⊆ Ω_{1}. Then \frac{M(f^{−1}(Q_{2}))}{M(Q_{2})} = \frac{M(Q_{1})}{M(f(Q_{1}))} ≤ K, by the first inequality in part (a). Thus f^{−1} is K − QC.

c) Since g ∈ K_{0} − QC, we get that for any quadrilateral Q with \overline{Q} ⊆ Ω_{1} we have M(g(f(Q))) ≤ K_{0}M(f(Q)) ≤ K_{0}K · M(Q), since f ∈K -QC. □

Theorem 6.15. Let f : Ω_{1} → Ω_{2} be a function between plane domains. Then f is 1-quasiconformal if and only if f is conformal.

Proof. “ ⇐” Since we have shown that the conformal module is conformally invariant we see that \frac{M(f(Q))}{M(Q)} = 1 for all quadrilaterals Q ⊆ Ω_{1} with \overline{Q} ⊆ Ω_{1}.

“ ⇒” Since f is 1-quasiconformal, it is therefore bijective. So we need only show that f : Ω_{1} → Ω_{2} is analytic. Let z_{0} ∈ Ω_{1} and let quadrilateral Q ⊆ \overline{Q} ⊆ Ω_{1} be such that z_{0} ∈ Q. Let f_{1} : Q → R(a) and f_{2} : f(Q) → R(a′) be conformal maps as in Theorem 6.3. Let g = f_{2} ∘ f ∘ f_{1}^{−1}. By Theorem 6.13(c) and the fact that f_{1} : Q → R(a) and f_{2} : f(Q) → R(a′) are both conformal (and hence 1-QC ), we see that g is also 1-quasiconformal.

We will show that the function g must be the identity function. This proves that f is conformal on Q since f = f_{2}^{−1} ∘ f_{1} is a composition of conformal maps. Since quadrilateral Q was arbitrary in Ω_{1}, we get that f is analytic on all of Ω_{1}.

\begin{aligned}g & 0 & a & i & R_{2} & R_{1} & ζ_{0} & R′_{1} & R′_{2} & ζ′_{0} & 0 & a′ & i \\ & & & & & Figure 5 \end{aligned}

Since f ∈ 1 − QC, we see by Remark 6.14 that a = M(Q) = M(f(Q)) = a′. Take ζ_{0} ∈ R(a). Let R_{1} = R(a)∩{z|Rez < Reζ_{0}} and R_{2} = R(a)∩{z|Rez > Reζ_{0}}. Let ζ0^{′} = g(ζ_{0}) and Rn^{′} = g(R_{n}) for n = 1, 2. Then, since g is 1-QC, we may use continuity of the module ([3], p. 26 and 30) to obtain M(R_{n}) = M(Rn^{′}) for n = 1, 2 even though the rectangles R_{n} are not compactly inside the domain of g. Now Rengel’s Inequality, with s = 1, tells us that M(Rn^{′}) ≤ area(Rn^{′}). So,

a = a′ = area(R(a′)) ≥ area(R1^{′}) + area(R2^{′}) ≥ M(R1^{′}) + M(R2^{′}) = M(R_{1}) +

M(R_{2}) = Reζ_{0} + (a − Reζ_{0}) = a. Therefore, equality must hold at each step and so M(Rn^{′}) = area(Rn^{′}) for n = 1, 2. Then the equality of Rengel’s Inequality tells us that Rn^{′} are rectangles for n = 1, 2. But since M(R_{n}) = M(Rn^{′}) for n = 1, 2, we must have that R_{n} = Rn^{′} for n = 1, 2. In particular, Re ζ_{0} = Re ζ0^{′}.

Similarly, we obtain that Im ζ_{0} = Im ζ0^{′}. Thus g is the identity map, which completes the proof. □

We now state and prove several chain rules that will be used to prove two important properties of quasiconformal maps.

Lemma 6.16. [Chain rule for ∂ and \overline{∂}] Let f be real-differentiable at z_{0} and let g be real-differentiable at w_{0} = f(z_{0}). Then \overline{f} and g ∘ f are also real-differentiable at z_{0}. Further, the following hold at z_{0},

a) ∂\overline{f} = \overline{\overline{∂}f} and \overline{∂} \overline{f} = \overline{∂f};

b) \overline{∂}(g ∘ f) = [(∂g) ∘ f]\overline{∂}f + [(\overline{∂}g) ∘ f]\overline{∂}\overline{f};
c) ∂(g ∘ f) = [(∂g) ∘ f]∂f + [(\overline{∂}g) ∘ f]∂\overline{f}.

Proof. (a) We consider f(z) = f(z_{0}) + A(z − z_{0}) + B(z̄ − \overline{z_{0}}) + o(z − z_{0}) as in equation (4). Then, noting that \overline{o(z − z_{0})} = o(z − z_{0}), we see that \overline{f}(z) = \overline{f(z_{0})} + \overline{A}(\overline{z} − \overline{z_{0}}) + \overline{B}(z − z_{0}) + o(z − z_{0}), which shows that f̄ is ℝ-differentiable. We examine the z and \overline{z} coefficients to then conclude that ∂\overline{f}(z) = \overline{B} = \overline{\overline{∂}f} and \overline{∂} \overline{f}(z) = \overline{A} = \overline{∂f}.

To prove (b) and (c) we first note that since f and g are real-differentiable functions they can be written in the form

(6.1)  f(z) =f(z_{0}) + A_{1}(z − z_{0}) + B_{1}(\overline{z} − \overline{z_{0}}) + α(z)
(6.2)  g(w) =g(w_{0}) + A_{2}(w − w_{0}) + B_{2}(\overline{w} − \overline{w_{0}}) + β(w)

where α(z) = o(z − z_{0}), β(w) = o(w − w_{0}) and w_{0} = f(z_{0}).

Now set L_{1}(z) = A_{1}(z − z_{0}) + B_{1}(\overline{z} − \overline{z_{0}}) and L̃_{1}(z) = df_{z_{0}}(z) = A_{1}z + B_{1}\overline{z} = L_{1}(z +z_{0}). Similarly set L_{2}(z) = A_{2}(w −w_{0})+B_{2}(\overline{w} −\overline{w_{0}}) and L̃_{2}(w) = dg_{w_{0}}(w) = A_{2}w + B_{2}\overline{w} = L_{2}(w + w_{0}).

Note that L̃_{1}(z_{1} + z_{2}) = L̃_{1}(z_{1}) + L̃_{1}(z_{2}) and L̃_{2}(w_{1} + w_{2}) = L̃_{2}(w_{1}) + L̃_{2}(w_{2}). Thus

g(f(z)) =g(w_{0}) + L_{2}(f(z)) + β(f(z))
=g(w_{0}) + L_{2}(f(z_{0}) + L_{1}(z) + α(z)) + β(f(z))
=g(w_{0}) + L̃_{2}(L_{1}(z) + α(z)) + β(f(z))
=g(w_{0}) ˜ +L_{2}(L_{1}(z)) + L̃_{2}(α(z)) + β(f(z))
=g(w_{0}) ˜ +L_{2}(L̃_{1}(z − z_{0})) + L̃_{2}(α(z)) + β(f(z)).

But

L̃_{2}(L̃_{1}(z − z_{0})) =A_{2}[A_{1}(z − z_{0}) + B_{1}(\overline{z} − \overline{z_{0}})] + B_{2}[\overline{A_{1}}(\overline{z} − \overline{z_{0}}) + \overline{B_{1}}(z − z_{0})]
=(A_{2}A_{1} + B_{2}\overline{B_{1}})(z − z_{0}) + (A_{2}B_{1} + B_{2}\overline{A_{1}})(\overline{z} − \overline{z_{0}}).

We then examine the z and z̄ coefficients to get that

(6.3)  \overline{∂}(g ∘ f)(z_{0}) = A_{2}B_{1} + B_{2}\overline{A_{1}}

and

(6.4)  ∂(g ∘ f)(z_{0}) = A_{2}A_{1} + B_{2}\overline{B_{1}},

provided that both L̃_{2}(α(z)) = o(z − z_{0}) and β(f(z)) = o(z − z_{0}), which we will now show.

As z → z_{0} we see that \frac{L̃_{2}(α(z))}{z−z_{0}} = \frac{A_{2}·α(z)}{z−z_{0}} + \frac{B_{2}(\overline{α(z)})}{z−z_{0}} → 0 since α = o(z − z_{0}). Thus L̃_{2}(α(z)) = o(z − z_{0}).

We next consider \frac{β(f(z))}{z−z_{0}} as z → z_{0}. When f(z) = w_{0}, we have that β(f(z)) = β(w_{0}) = 0 and so we only need to consider those z tending to z_{0} such that f(z) ≠ w_{0}. For such z we have

(6.5)  \frac{β(f(z))}{z − z_{0}} = \frac{β(f(z))}{f(z) − w_{0}} · \frac{f(z) − w_{0}}{z − z_{0}} .

However

\frac{f(z) − w_{0}}{z − z_{0}} = \frac{f(z_{0}) + A_{1}(z − z_{0}) + B_{1}(\overline{z} − \overline{z_{0}}) + α(z) − w_{0}}{z − z_{0}} = A_{1}+B_{1} \frac{\overline{z} − \overline{z_{0}}}{z − z_{0}} +\frac{α(z)}{z − z_{0}}

is bounded by |A_{1}| + |B_{1}| as z → z_{0}. Also β = o(w − w_{0}), and so using w = f(z) → w_{0} as z → z_{0} we see that \frac{β(f(z))}{f(z)−w_{0}} → 0. Therefore, by examining Equation (6.5) we may conclude that β(f(z)) = o(z − z_{0}).

Note equations (6.1) and (6.2) where we see that A_{2} = ∂g(w_{0}) = (∂g ∘ f)(z_{0}), A_{1} = ∂f(z), B_{2} = \overline{∂}g(w_{0}) = (\overline{∂}g ∘ f)(z_{0}), and \overline{B_{1}} = \overline{\overline{∂}f(z_{0})} = ∂\overline{f}(z_{0}) by part (a). Parts (b) and (c) now follow from (6.3) and (6.4). □

Remark 6.17 (Chain rule for differentials). The proof of (b) and (c) above shows that the linear approximation of a composition is the composition of the respective linear approximations, as one would expect. In terms of differentials we may express this as d(g ∘ f)_{z_{0}} = dg_{f(z_{0})} ∘ df_{z_{0}}.

Lemma 6.18 (Chain rule for complex dilatations). We have the following:

a) Let μ_{f} and μ_{h} be complex dilatations for function f : Ω_{1} → Ω_{2} and h : Ω_{2} → Ω_{3} which are real-differentiable at z and f(z), respectively, such that ∂f(z), ∂h(f(z)), and ∂(h ∘ f)(z) are all nonzero. Then

(i) μ_{h∘f}(z) =\frac{μ_{f}(z) + μ_{h}(f(z)) · \frac{\overline{∂f(z)}}{∂f(z)}}{1 + μ_{h}(f(z)) · μ_{f}(z) · \frac{\overline{\overline{∂}f(z)}}{\overline{∂}f(z)}} when \overline{∂}f(z) ≠ 0, and

(ii) μ_{h∘f}(z) =μ_{h} ∘ f(z) · \frac{\overline{∂f(z)}}{∂f(z)} when \overline{∂}f(z) = 0.

b) Let μ_{f} be the complex dilatation for homeomorphism f : Ω_{1} → Ω_{2} that is real-differentiable at z such that ∂f(z) is nonzero and |∂f(z)| ≠ |\overline{∂}f(z)| (i.e. J_{f}(z) ≠ 0). Then, assuming f^{−1} is real-differentiable at f(z), we have that

μ_{f^{−1}}(f(z)) = −μ_{f}(z) · \frac{∂f(z)}{\overline{∂f(z)}}.

c) Let μ_{f} and μ_{h} be complex dilatations for homeomorphisms f : Ω_{1} → Ω_{2} and h : Ω_{1} → Ω_{3} that are real-differentiable at z such that ∂f(z), ∂h(z), and ∂(h ∘ f^{−1})(f(z)) are all nonzero and |∂f(z)| ≠ |\overline{∂}f(z)|. Then, if f^{−1} is realdifferentiable at f(z), we have

μ_{h∘f^{−1}}(f(z)) = \frac{μ_{h}(z) − μ_{f}(z)}{1 − μ_{h}(z) · \overline{μ_{f}}(z)} · \frac{∂f(z)}{\overline{∂f(z)}}.

Remark 6.19. When a real-differentiable homeomorphism f satisfies J_{f}(z_{0}) ≠ 0, then it can be shown that f^{−1} must be real-differentiable at f(z_{0}). So the hypotheses in (b) and (c) on the differentiability of f^{−1} is not necessary. However, for our applications to quasiconformal maps, differentiability of f^{−1} will follow from Lemma 8.3, and so the above weaker result will suffice for our purposes.

Proof. (a) By applying Lemma 6.16 in the case when \overline{∂}f ≠ 0, we have

μ_{h∘f} =\frac{\overline{∂}(h ∘ f)}{∂(h ∘ f)} = \frac{[(∂h) ∘ f]\overline{∂}f + [(\overline{∂}h) ∘ f]\overline{∂}\overline{f}}{[(∂h) ∘ f]∂f + [(\overline{∂}h) ∘ f]∂\overline{f}}

=\frac{[(∂h) ∘ f]\overline{∂}f + [(\overline{∂}h) ∘ f]\overline{∂}\overline{f}}{[(∂h) ∘ f]∂f + [(\overline{∂}h) ∘ f]∂\overline{f}} · \frac{\frac{1}{∂f[(∂h)∘f]}}{\frac{1}{∂f[(∂h)∘f]}} =  \frac{μ_{f} + μ_{h} ∘ f · \frac{\overline{∂f}}{∂f}}{1 + μ_{h} ∘ f · \frac{\overline{\overline{∂}f}}{∂f}}

=\frac{μ_{f} + μ_{h} ∘ f · \frac{\overline{∂f}}{∂f}}{1 + μ_{h} ∘ f · \frac{\overline{∂}f}{∂f} · \frac{\overline{\overline{∂}f}}{\overline{∂}f}} = \frac{μ_{f} + μ_{h} ∘ f · \frac{\overline{∂f}}{∂f}}{1 + μ_{h} ∘ f · μ_{f} · \frac{\overline{\overline{∂}f}}{\overline{∂}f}} .

When \overline{∂}f = 0 we obtain the second formula in part (a) also from the above work.

(b) Consider the analytic function f ∘ f^{−1}(z) = z. By Lemma 6.16 (b) and (c) we obtain (at f(z))

1 =∂(f ∘ f^{−1}) = (∂f ∘ f^{−1})(∂f^{−1}) + (\overline{∂}f ∘ f^{−1})(∂\overline{f^{−1}})
0 =\overline{∂}(f ∘ f^{−1}) = (∂f ∘ f^{−1})(\overline{∂}f^{−1}) + (\overline{∂}f ∘ f^{−1})(\overline{∂} \overline{f^{−1}}).

We will let A = ∂f ∘ f^{−1}, B = ∂f^{−1}, C = \overline{∂}f ∘ f^{−1}, and D = ∂\overline{f^{−1}}. Then, by Lemma 6.16 (a) we have that 1 = AB + CD and 0 = A\overline{D} + C\overline{B}. Using the conjugate of the second equation we solve for B and D in terms of A and C obtaining B = \frac{\overline{A}}{|A|^{2}−|C|^{2}} and D = \frac{−\overline{C}}{|A|^{2}−|C|^{2}} . Note that here we use the hypothesis

that |A| = |∂f(z)| ≠ |\overline{∂}f(z)| = |C|. Therefore, again using Lemma 6.16 (a), we get that μ_{f^{−1}}(f(z)) = \frac{\overline{∂}f^{−1}}{∂f^{−1}} = \frac{\overline{D}}{B} = −\frac{C}{\overline{A}} = −\frac{CA}{A\overline{A}} = −μ_{f}(z)\frac{∂f}{\overline{∂f}}, since μ_{f} = \frac{\overline{∂}f}{∂f} = \frac{C}{A}.

(c) Now we use (a) and (b) to compute the following when C = \overline{∂}f(z) ≠ 0 (and hence D ≠ 0).

μ_{h∘f^{−1}}(f(z)) =\frac{μ_{f^{−1}}(f(z)) + μ_{h} ∘ f^{−1}(f(z)) · \frac{\overline{∂f^{−1}}}{∂f^{−1}}}{1 + μ_{h} ∘ f^{−1}(f(z)) · μ_{f^{−1}}(f(z)) · \frac{\overline{\overline{∂}f^{−1}}}{\overline{∂}f^{−1}}}

=\frac{−μ_{f}(z)\frac{∂f}{\overline{∂f}} + μ_{h}(z)\frac{\overline{∂f^{−1}}}{∂f^{−1}}}{1 − μ_{h}(z)μ_{f}(z)\frac{∂f}{\overline{∂f}} \frac{\overline{\overline{∂}f^{−1}}}{\overline{∂}f^{−1}}} = \frac{−μ_{f}(z)\frac{A}{\overline{A}} + μ_{h}(z)\frac{\overline{B}}{B}}{1 − μ_{h}(z)μ_{f}(z)\frac{A}{\overline{A}} \frac{D}{\overline{D}}}

=\frac{(−μ_{f}(z) + μ_{h}(z))\frac{A}{\overline{A}}}{1 − μ_{h}(z)μ_{f}(z)\frac{A}{\overline{A}} \frac{(−\overline{C})}{(−C)}} = \frac{(−μ_{f}(z) + μ_{h}(z))\frac{A}{\overline{A}}}{1 − μ_{h}(z)\overline{μ_{f}}(z)}

=\frac{μ_{h}(z) − μ_{f}(z)}{1 − μ_{h}(z) · \overline{μ_{f}}(z)} · \frac{∂f}{\overline{∂f}} .

When C = \overline{∂}f(z) = 0 (and hence D = 0) we see that μ_{f}(z) = 0 and so our desired formula in (c) readily follows from formula (ii) in (a). □

Corollary 6.20. Let f : Ω_{1} → Ω_{2} and h : Ω_{2} → Ω_{3} be real-differentiable functions with complex dilatations μ_{f}and μ_{h}, respectively. Then we have

a) If h is conformal, then μ_{h∘f} = μ_{f}.

b) If f is conformal, then μ_{h∘f} = (μ_{h} ∘ f)\frac{\overline{f′}}{f′}.

Proof. Both (a) and (b) follow directly from Lemma 6.18 (a) noting that μ_{h} ≡ 0 when h is conformal and noting that ¯∂f(z) = 0 and ∂f(z) = f′(z) when f is conformal. □

Remark 6.21. By Corollary 6.20 we see that the modulus of the complex dilatation is preserved by pre and post compositions with conformal maps. Hence, K̃_{f} is similarly preserved.

We will now investigate how μ_{f} is related to K̃_{f} for certain maps. We begin with an obvious lemma.

Lemma 6.22. Let f : Ω_{1} → Ω_{2} be a sense-preserving function between plane domains that is differentiable in the real sense at z_{0} ∈ Ω_{1} and let K ≥ 1. Then the statements

(i) |μ_{f}(z_{0})| ≤ \frac{K − 1}{K + 1} and (ii) K̃_{f}(z_{0}) = \frac{|∂f(z_{0})| + |\overline{∂}f(z_{0})|}{|∂f(z_{0})| − |\overline{∂}f(z_{0})|} ≤ K

are equivalent.

Proof. Note that if in either statement K = 1 holds, then μ(z_{0}) = 0 and thus \overline{∂}f(z_{0}) = 0. So we only need to consider the case where K > 1. Assuming the first inequality, we find

|\frac{\overline{∂}f(z_{0})}{∂f(z_{0})}| = |μ_{f}(z_{0})| ≤ \frac{K − 1}{K + 1} ⇒ \frac{K + 1}{K − 1} |\frac{\overline{∂}f(z_{0})}{∂f(z_{0})}| ≤ 1

⇒ \frac{K|\overline{∂}f(z_{0})| + |\overline{∂}f(z_{0})|}{K|∂f(z_{0})| − |∂f(z_{0})|} ≤ 1 ⇒ K|\overline{∂}f(z_{0})| + |\overline{∂}f(z_{0})| ≤ K|∂f(z_{0})| − |∂f(z_{0})|

⇒ |\overline{∂}f(z_{0})| + |∂f(z_{0})| ≤ K (|∂f(z_{0})| − |\overline{∂}f(z_{0})|) ⇒ \frac{|∂f(z_{0})| + |\overline{∂}f(z_{0})|}{|∂f(z_{0})| − |\overline{∂}f(z_{0})|} ≤ K,

noting that the denominator in the last inequality is nonzero since |μ_{f}(z_{0})| < 1. We then unravel the implications in reverse order to see the reverse implication. □

Theorem 6.23. Let f : Ω_{1} → Ω_{2} be a K-QC function between plane domains. Suppose that f is differentiable in the real sense at z_{0} ∈ Ω_{1} and df_{z_{0}}(z) ≠ 0. Then

K̃_{f}(z_{0}) = \frac{|∂f(z_{0})| + |\overline{∂}f(z_{0})|}{|∂f(z_{0})| − |\overline{∂}f(z_{0})|} ≤ K, or equivalently, |μ_{f}(z_{0})| ≤ \frac{K − 1}{K + 1}.

Proof. Under the assumptions of the theorem we want to show that |μ_{f}(z_{0})| ≤ \frac{K−1}{K+1}. Since translations are conformal and pre and post composition with conformal maps preserves the modulus of complex dilatation (Corollary 6.20), we can, without loss of generality, assume that f(z_{0}) = 0 = z_{0}. Since f is real-differentiable at z_{0} we can then write f(z) = Az + B\overline{z} + o(z). Therefore |μ_{f}(0)| = \frac{|B|}{|A|} = |μ_{L}(0)| where L(z) = Az + B\overline{z}. Recall that equation 3.10 allows us to write L = g ∘ L_{|μ|} ∘ h where h(z) = e^{−i \frac{β}{2}} · z, g(z) = |A| e^{i(α+ \frac{β}{2} )} · z and L_{|μ|}(z) = z + |μ| \overline{z}, where μ = μ_{f}(0).

Consider H = g^{−1} ∘ f ∘ h^{−1}. Since h and g are just rotations we have that dh^{−1} = h^{−1}, dg^{−1} = g^{−1} and, recalling Remark 6.17, we have that dH = dg^{−1} ∘ df ∘ dh^{−1} = g^{−1} ∘ L ∘ h^{−1} = L_{|μ|}. This shows that H(z) = z + |μ| \overline{z} + o(z). This last expression of H together with the fact that H = g^{−1} ∘ f ∘ h^{−1} ∈ K-QC (since g^{−1} and h^{−1} are conformal) will allow us below to show that |μ_{f}(0)| = |μ_{H}(0)| ≤ \frac{K−1}{K+1} which will finish the proof of the theorem.

O  O z_{1} = (−1 + i)δ  δ(1 + |μ|) + i(1 − |μ|)δ

Figure 6. Illustration of mapping H(z)

For δ > 0 consider the oriented square Q((−1 + i)δ, (−1 − i)δ, (1 − i)δ, (1 + i)δ). Then H((1+i)δ) = δ(1+|μ|)+i(1−|μ|)δ+o(δ) (see Figure 6). By Rengel’s inequality we have that M(H(Q)) ≤ \frac{area(H(Q))}{s^{2}} = \frac{[4δ^{2}(1−|μ|^{2})+o(δ^{2})]}{[2δ(1+|μ|)+o(δ)]^{2}} = \frac{[4(1−|μ|^{2})+o(1)]δ^{2}}{[2(1+|μ|)+o(1)]^{2}δ^{2}} =

\frac{4(1−|μ|^{2})+o(1)}{[2(1+|μ|)]^{2}+o(1)} = \frac{1−|μ|}{1+|μ|} + o(1) as we let δ → 0 (using Lemma 2.2 (d) in the last step). Also, by Theorem 6.13 (a), noting that the conformal module of the square Q is 1, we have \frac{1}{K} = \frac{M(Q)}{K} ≤ M(H(Q)) ≤ \frac{1−|μ|}{1+|μ|} + o(1). Letting δ → 0 we see that

\frac{1+|μ|}{1−|μ|} ≤ K. Therefore \frac{K−1}{K+1} ≥ |μ_{H}(z_{0})|, as desired. □

Theorem 6.24. Let f : Ω_{1} → Ω_{2} be a C^{1} sense-preserving bijection between plane domains with J_{f} > 0 on Ω_{1}. If sup_{z∈Ω_{1}} |μ_{f}(z)| ≤ \frac{K−1}{K+1} (equivalently K̃_{f} ≤ K)

for some K, then f is K-quasiconformal.

Remark 6.25. Theorem 6.24 shows that a C^{1} function that is K−QC under the tentative definition in Definition 4.3 is quasiconformal under our main geometric definition in Definition 6.11. However, not all quasiconformal maps are C^{1} (or even everywhere real differentiable) and so these definitions are not equivalent.

Remark 6.26. The complex dilatation μ_{f} for a real-differentiable function f measures the deviation from conformality. So for a function to be quasiconformal, there must be a bound on μ_{f}. Likewise, the closer μ_{f} is to zero, the closer f is to being conformal.

Proof. (Theorem 6.24) Let quadrilateral Q ⊆ \overline{Q} ⊆ Ω_{1}. Let f_{1} : Q → R(a) and f_{2} : f(Q) → R(a′) be as in Theorem 6.3. Quasiconformality of f is shown if we can show that a′ ≤ Ka. Set g = f_{2} ∘ f ∘ f_{1}^{−1} : R(a) → R(a′).

Since f is a C^{1} bijection between plane domains with J_{f} > 0 and f_{1}, f_{2} are both conformal, g is also a C^{1} bijection between plane domains with with J_{g} > 0 (since J_{g} is the product of J(f_{2}), J(f) and J(f_{1}^{−1}) evaluated at corresponding points according to the chain rule). Therefore the partial derivative \frac{∂g}{∂x} exists and is continuous in R(a). By considering Figure 7 we see that, fixing 0 < y_{0} < 1, we have

a′ ≤ |g(a + iy_{0}) − g(iy_{0})| ≤ ∫_{0}^{a} |\frac{∂g}{∂x}(x + iy_{0})| dx ≤ \{∫_{0}^{a} [\frac{∂g}{∂x}(x + iy_{0})]^{2} dx\}^{\frac{1}{2}} · a^{\frac{1}{2}} ,

where the last inequality follows from the Cauchy-Schwarz inequality (see Appendix I). Since sup_{z∈Ω_{1}} |μ_{f}(z)| ≤ \frac{K−1}{K+1} and f_{1}^{−1}, f_{2} are both conformal, by Corollary 6.20 (a)

Ω_{1}  Ω_{2} f(Q) Q f_{1}  f_{2} f g R(a)  R(a′) 0 0  a′ a y_{0}  i i

Figure 7

and (b), we see that sup_{z∈Ω_{1}} |μ_{g}(z)| ≤ \frac{K−1}{K+1}. But Lemma 6.22 lets us alternatively as-

sume for all z ∈ R(a) that \frac{|∂g(z)|+|\overline{∂}g(z)|}{|∂g(z)|−|\overline{∂}g(z)|} ≤ K which implies that \frac{(|∂g(z)|+|\overline{∂}g(z)|)^{2}}{|∂g(z)|^{2}−|\overline{∂}g(z)|^{2}} ≤

K. But J_{g}(z) = |∂g(z)|^{2} −|\overline{∂}g(z)|^{2} and |\frac{∂g}{∂x}(z)| = |∂g(z)+\overline{∂}g(z)| ≤ |∂g(z)|+|\overline{∂}g(z)|

yields |\frac{∂g}{∂x} |^{2} ≤ J_{g} · K. We use this fact in the following list of inequalities. We have

(a′)^{2} = ∫_{0}^{1}(a′)^{2}dy ≤ a ∫_{0}^{1} ∫_{0}^{a} |\frac{∂g}{∂x} |^{2} (x + iy)dxdy = a ∫∫_{R(a)} |\frac{∂g}{∂x} |^{2} (x + iy)dxdy

≤ a ∫∫_{R(a)} K · J_{g}(x + iy)dxdy = aK ∫∫_{R(a′)} 1dxdy = aKa′. Thus a′ ≤ Ka as desired. □

We now discuss two basic examples of quasiconformal maps.

Example 6.27. a) Let g map the plane onto itself by g(z) = Az + B\overline{z} for some complex numbers A and B such that |A| > |B| ≥ 0. Then, by Theorem 6.24, g is K-quasiconformal with K = \frac{|A|+|B|}{|A|−|B|}.

b) Let f(z) = kx + iy where k ≥ 1 and z = x + iy. Then f(z) = k (\frac{z+\overline{z}}{2} ) +

i (\frac{z−\overline{z}}{2i} ) = (\frac{k}{2} + \frac{1}{2}) z + (\frac{k}{2} − \frac{1}{2}) \overline{z}. So \frac{|∂f(z)|+|\overline{∂}f(z)|}{|∂f(z)|−|\overline{∂}f(z)|} = \frac{\frac{k+1}{2} + \frac{k−1}{2}}{\frac{k+1}{2} − \frac{k−1}{2}} = k. Therefore f

is k-quasiconformal by Theorem 6.24. If 0 < k < 1, then f is \frac{1}{k} -QC by a similar calculation.

Essentially, for a function f to be quasiconformal there must be a bound, locally, on how much f can stretch in any direction compared to any other direction. However this is not a global bound. For example, there are conformal maps f : {z : |z| < 1} → {z : Re z > 0}. So here there is an unbounded stretch in the global sense even though conformality implies no stretch in one direction more than any other locally.

7. Metric Definition of Quasiconformal Maps

Recall from the introduction that there are several equivalent definitions of quasiconformal maps. One of these is the metric definition. We previously mentioned the complex dilatation of f at z_{0}. But for the metric definition we will now introduce the notion of circular dilatation at z_{0}.

Definition 7.1. Let f : Ω_{1} → Ω_{2} be a function between plane domains. For

z_{0} ∈ Ω_{1}, we define H_{f}(z_{0}) = lim sup_{r→0} \frac{sup_{θ}|f(z_{0}+re^{iθ})−f(z_{0})|}{inf_{θ}|f(z_{0}+re^{iθ})−f(z_{0})|} to be the circular dilatation

of f at z_{0}.

Remark 7.2. Intuitively, H_{f} measures the local ratio of the maximum stretch to the minimum stretch of f. Thus, we expect H_{f}(z_{0}) to equal K̃_{f}(z_{0}) = \frac{max_{θ} |∂_{θ}f|}{min_{θ} |∂_{θ}f|}, as calculated in Section 4, when f is real-differentiable at z_{0}. However, we also note that H_{f}(z_{0}) exists whether or not f is differentiable at z_{0}.

Lemma 7.3. If f : Ω_{1} → Ω_{2} between plane domains is differentiable in the real sense at z_{0}, and J_{f}(z_{0}) > 0, then H_{f}(z_{0}) = \frac{|∂f(z_{0})|+|\overline{∂}f(z_{0})|}{|∂f(z_{0})|−|\overline{∂}f(z_{0})|} = \frac{max_{θ} |∂_{θ}f|}{min_{θ} |∂_{θ}f|} =

\frac{1+|μ_{f} (z_{0})|}{1−|μ_{f} (z_{0})|} = K̃_{f}(z_{0}).

The proof follows the same line of calculation as the proof of Lemma 3.9 and so will be omitted.

We will now state two theorems without proof. The reader may refer to [3], p. 177 for the first theorem and [3], p. 178 for the second. We will simply quote these theorems as needed.

Theorem 7.4. Suppose f : Ω_{1} → Ω_{2} is K-quasiconformal. Then,

a) for every z_{0} ∈ Ω_{1}, we have H_{f}(z_{0}) ≤ λ(K) for some positive number λ(K); and b) we have ∥H_{f}∥_{∞} ≤ K.

Theorem 7.5. Suppose f : Ω_{1} → Ω_{2} is a sense-preserving homeomorphism such that H_{f}(z) < +∞ at each point of Ω_{1} and H_{f}(z) ≤ K almost everywhere in Ω_{1}. Then f is K-quasiconformal.

Remark 7.6. It is actually sufficient, in Theorem7.5, to replace the assumption that that H_{f}(z) < +∞ everywhere on Ω_{1} by the assumption that H_{f}(z) < +∞ \underline{off} a set of σ-finite linear measure ([3], p. 178). However, once f is shown to be Kquasiconformal, we see, by Theorem 7.4, that H_{f}(z) must be bounded everywhere.

Corollary 7.7. If f : Ω_{1} → Ω_{2} is K-quasiconformal and has circular dilatation of 1 almost everywhere, then f is conformal.

Proof. By Theorem 7.4 we have that H_{f}(z_{0}) ≤ λ(K) for all z_{0} ∈ Ω_{1}. Also ∥H_{f}(z)∥_{∞} = 1. So by Theorem 7.5 f is 1-quasiconformal. Hence by Theorem 6.15 we see that f is conformal. □

Caution – If a homeomorphism f has circular dilatation equal to 1 almost everywhere, but is not known to be quasiconformal, it is not necessarily true that f is conformal. A counterexample for this is the function f(z) = z + ig(x) defined on the unit square where g(x) is the Cantor function and x = Rez.

8. Analytic Characterization

In this section we address the connections between differentiability conditions and quasiconformality. Since the details are not provided, the reader may wish to refer the discussion beginning on page 162 in [3] for further examination. We begin with a definition.

Definition 8.1. We say that a complex valued continuous map f(x, y) on a domain Ω ⊂ \overline{ℂ} is absolutely continuous on lines if in every rectangle R = \{x + iy : a < x < b, c < y < d\}, with \overline{R} ⊂ Ω \ \{∞, f^{−1}(∞)\}, the map f(x, ·) is absolutely continuous for almost all a < x < b and the map f(·, y) is absolutely continuous for almost all c < y < d.

Clearly any map satisfying Definition 8.1 has both partial derivatives f_{x} and f_{y} existing a.e. in Ω. But it is well known that a function with partial derivatives defined at a point z_{0} is not necessarily real differentiable at z_{0}. However, we do have the following somewhat surprising result whose proof we omit.

Theorem 8.2 ([3], p. 128 and 130). A continuous open map f on a domain Ω in \overline{ℂ} having finite partial derivatives a.e. in Ω is real differentiable a.e. in Ω.

By showing that a quasiconformal map f is indeed absolutely continuous on lines we can conclude a.e. real differentiability and, indeed, in the manner of the proof of Theorem 6.23 obtain a bound on μ_{f}, which in turn implies conditions on J_{f} and df. The details can be found in [3] starting in p. 162. We summarize these results as follows.

Lemma 8.3. Let f : Ω_{1} → Ω_{2} be K-quasiconformal between plane domains. Then we have all of the following

a) f is absolutely continuous on lines;

b) f is differentiable in the real sense almost everywhere;

c) |∂f(z)| ≠ 0 a.e. and | μ_{f}(z) |≤ \frac{K−1}{K+1} almost everywhere.

d) J_{f}(z) > 0 a.e.

e) df_{z_{0}} ≠ 0 a.e.

It turns out that conditions (a) and (c) are enough to ensure that a map f is quasiconformal. This leads us to the so-called analytic definition of quasiconformality which we now state as a lemma.

Lemma 8.4. ([3], p. 168) Suppose a homeomorphism f : Ω_{1} → Ω_{2} between domains in \overline{ℂ} is absolutely continuous on lines and |μ_{f}(z)| ≤ \frac{K−1}{K+1} almost everywhere. Then f is K-quasiconformal.

Theorem 8.5. Suppose f : Ω_{1} → Ω_{2} is quasiconformal with μ_{f}(z) = 0 almost everywhere in Ω_{1}. Then f is conformal.

Proof. Since f is quasiconformal, we see by Lemma 8.3 that f is absolutely continuous on lines. Since μ_{f}(z) = 0 almost everywhere, Lemma 8.4 applies with K = 1. Thus f is 1-quasiconformal and hence is conformal. □

Caution – If a homeomorphism f has complex dilatation μ_{f} equal to 0 almost everywhere, but is not known to be quasiconformal, it is not necessarily true that f is conformal. A counterexample for this is the function f(z) = z + ig(x) defined on the unit square where g(x) is the Cantor function and x = Rez.

9. Beltrami Equations

Lemma 8.3 shows us that a quasiconformal map f satisfies ||μ_{f}||_{∞} < 1 on its domain. We will now ask, and answer (see Theorem 9.8) the question: Given a complex valued function μ with ||μ||_{∞} < 1 on an open set Ω, can we find a quasiconformal map f such that μ_{f} = μ.

We begin with a definition.

Definition 9.1. We call a complex valued function μ on a domain Ω a Beltrami coefficient if μ is Lebesgue measurable and ∥μ∥_{∞} < 1.

Remark 9.2. Given Beltrami coefficients μ and ν on an open set Ω_{1}, we will often write μ = ν to mean μ = ν almost everywhere since in what follows it is only important that equality holds almost everywhere.

Remark 9.3. By Lemma 8.3, a K-quasiconformal map f has μ_{f} defined and satisfying |μ_{f}| ≤ \frac{K−1}{K+1} almost everywhere. Thus μ_{f} is a Beltrami coefficient.

Lemma 9.4. If μ_{f} and μ_{h} are Beltrami coefficients for quasiconformal maps f and h, then,

a) μ_{h∘f}(z) = \frac{μ_{f}(z) + μ_{h} ∘ f(z) · \frac{\overline{∂f(z)}}{∂f(z)}}{1 + μ_{h} ∘ f(z) · μ_{f}(z) · \frac{\overline{\overline{∂}f(z)}}{\overline{∂}f(z)}} where f : Ω_{1} → Ω_{2} and h : Ω_{2} → Ω_{3};

b) μ_{h∘f^{−1}}(f(z)) = \frac{μ_{h}(z) − μ_{f}(z)}{1 − μ_{h}(z) · \overline{μ_{f}}(z)} · \frac{∂f(z)}{\overline{∂f(z)}} where f : Ω_{1} → Ω_{2}, h : Ω_{1} → Ω_{3}.

Remark 9.5. Note that when \overline{∂}f(z) = 0 we have μ_{f}(z) = 0 and so we regard the formula in (a) as collapsing to the formula in Lemma 6.18 (a)(ii) despite the

fact that the term \frac{\overline{\overline{∂}f(z)}}{\overline{∂}f(z)} is not formally defined.

Proof. Noting that since all maps involved are quasiconformal, Lemma 8.3 shows that Lemma 6.18 may be applied, from which (a) and (b) readily follow. However, it should be noted that we inherently used the fact (see [3], p. 165) that the image of a null set under a quasiconformal map is a null set, and so one can safely assume in (a), for example, that the set of points z_{0} where both f is ℝ-differentiable at z_{0} and g is ℝ-differentiable at f(z_{0}) is of full measure in Ω_{1}. Indeed, this set is Ω_{1} \ ({z : f is not differentiable} ∪ {f^{−1}(w) : g is not differentiable at w}). □

Corollary 9.6. Let f : Ω_{1} → Ω_{2}, g : Ω_{1} → Ω_{3}, and h : Ω_{2} → Ω_{4} be quasiconformal maps. Then,

a) We have that μ_{f} = μ_{g} almost everywhere if and only if f ∘ g^{−1} is conformal;

b) If h is conformal, then μ_{h∘f} = μ_{f} almost everywhere;

c) If f is conformal, then μ_{h∘f} = (μ_{h} ∘ f)\frac{\overline{f′}}{f′} almost everywhere.

Proof. a) “ ⇒” As a composition of quasiconformal maps it follows that f ∘ g^{−1} is K -quasiconformal where K = K_{f} · K_{g}. Now by the previous lemma, μ_{f∘g^{−1}} = \frac{μ_{f} −μ_{g}}{1−μ_{f} \overline{μ_{g}}} · \frac{∂g}{\overline{∂g}}. But μ_{f} = μ_{g} almost everywhere implies, then, μ_{f∘g^{−1}} =

\frac{μ_{f} −μ_{g}}{1−μ_{f} \overline{μ_{g}}} · \frac{∂g}{\overline{∂g}} = 0 almost everywhere. Hence, by Theorem 8.5 we get that f ∘ g^{−1} is conformal.

“ ⇐” Assuming f ∘ g^{−1} is conformal, we have that μ_{f∘g^{−1}} = 0 which by Lemma 9.4 (b) is true only when μ_{f} − μ_{g} = 0 (since ∂f ≠ 0 a.e. by Lemma 8.3(c)). Therefore μ_{f} = μ_{g}.

Parts (b) and (c) are direct consequences of Lemma 9.4 (a). □

Theorem 9.7. Let μ be a Beltrami coefficient on the plane \overline{ℂ}. Let z_{1}, z_{2}, and z_{3} be distinct points in the plane and let w_{1}, w_{2}, and w_{3} also be distinct points in the plane. Then there exists a unique quasiconformal map f from the plane to itself such that \overline{∂}f = μ · ∂f almost everywhere and f(z_{n}) = w_{n} for n = 1, 2, 3.

Proof. – We will only show uniqueness. The reader may refer to ([3], p. 194) for existence. Suppose that both f_{1}, f_{2} satisfy the statement of the theorem. Then T = f_{2}^{−1} ∘ f_{1} fixes z_{1}, z_{2}, and z_{3}. But since μ_{f_{1}} = μ = μ_{f_{2}}, we know that T is conformal. Thus T is a Möbius map that fixes 3 distinct points and therefore is the identity map. Hence f_{1} = f_{2}. □

We will state the following theorem without proof. The reader may refer to ([3], p. 194) for the proof of this important theorem.

Theorem 9.8 (Existence Theorem). Let μ be a Beltrami coefficient on an open subset U of the plane \overline{ℂ}. Then there exists a quasiconformal map f on U with μ_{f} = μ almost everywhere in U.

10. Riemann Surface Structures

In order to understand and prove the Fundamental Theorem of Quasiconformal Surgery it will be useful to understand Riemann Surfaces. In particular, we will focus on domains in \overline{ℂ} that have a “conformal structure” induced by Beltrami coefficients. Much like the way the topology on a set determines which functions are continuous, we will see that the conformal structure on a set determines which functions are analytic. We now develop the necessary ideas.

Definition 10.1. A manifold of dimension n is a connected Hausdorff space X for which every point has a neighborhood U that is homeomorphic to an open subset V of ℝ^{n}. Such a homeomorphism f : U → V is called a chart.

Definition 10.2. A two-dimensional manifold is called a surface.

Definition 10.3. Two charts f_{α} : U_{α} → ℂ and f_{β} : U_{β} → ℂ, whose domains intersect, are called compatible if the maps f_{β} ∘ f_{α}^{−1} : f_{α}(U_{α} ∩ U_{β}) → f_{β}(U_{α} ∩ U_{β}) and f_{α} ∘ f_{β}^{−1} : f_{β}(U_{α} ∩ U_{β}) → f_{α}(U_{α} ∩ U_{β}) are holomorphic (i.e., analytic). If A is a collection of compatible charts and if any x in X is in the domain of some f in A, then we say that A is an atlas. When we endow X with an atlas A, we say that (X, A) is a Riemann surface. See Figure below.

Definition 10.4. Let U and V be Riemann Surfaces and Φ(U) and Φ(V) be their respective atlases. We say that f : U → V is analytic between Riemann surfaces if for any φ ∈ Φ(U) and ψ ∈ Φ(V) such that f(U) meets the domain of ψ we have ψ ∘ f ∘ φ^{−1} is analytic (in the usual sense) on its domain. See Figure 9.

We will be using the following notation when talking about Riemann surfaces whose underlying set is an open subset of the plane \overline{ℂ}. Let U be an open subset of the plane, and let μ be a Beltrami coefficient on U. Then U[μ], sometimes called U with the conformal structure μ, denotes the Riemann Surface U with atlas A(μ) = \{φ : U → \overline{ℂ} | φ is quasiconformal with μ_{φ} = μ almost everywhere\}, where \overline{ℂ} denotes the plane. The set A(μ), also called the set of all μ-conformal maps on

f_{β} f_{α}  u_{β} u_{α}S-surface f^{β} ∘f{−1 \atop α} conformal

Figure 8

f analytic ψ∘f ∘φ^{−1} φ ∈ Φ(U)  ψ ∈ Φ(V) U  V z  f(z)

Figure 9

U, is nonempty by the Existence Theorem 9.8. Also, for any two such maps φ and ψ, we see that φ ∘ ψ^{−1} is conformal by Corollary 9.6 (a) and so A(μ) is indeed an atlas. We denote the Riemann surface with the atlas A(0) of all conformal maps on U by U[0], which we will often simply write as U.

Remark 10.5. When checking to see if f : U[η] → V [ν] is analytic between Riemann surfaces, we note that it suffices to use only a single element from each atlas, instead of checking all charts. Indeed, if φ_{1} ∈ Φ(U) and ψ_{1} ∈ Φ(V) are such that ψ_{1} ∘ f ∘ φ_{1}^{−1} is analytic, then for \underline{any} φ_{2} ∈ Φ(U) and ψ_{2} ∈ Φ(V) we must have that ψ_{2} ∘ f ∘ φ_{2}^{−1} is also analytic. This follows since ψ_{2} ∘ f ∘ φ_{2}^{−1} = (ψ_{2} ∘ ψ_{1}^{−1}) ∘ (ψ_{1} ∘ f ∘ φ_{1}^{−1}) ∘ (φ_{1} ∘ φ_{2}^{−1}) is a composition of analytic (in the usual sense) maps.

Lemma 10.6. Let U and V be domains in the plane \overline{ℂ}. Let ν be a Beltrami coefficient on V. If α : U → V is quasiconformal with dilatation μ_{α}, then we can find a unique Beltrami coefficient η on U such that the map α : U[η] → V [ν] is analytic between Riemann surfaces. Here we call η the “pull back” of ν by α and we denote this pull back by η = α^{∗}ν.

Proof. Let φ ∈ A(ν). Since both α, φ are quasiconformal, their composition φ ∘ α is quasiconformal. We set η = μ_{φ∘α} which is a Beltrami coefficient by Remark 9.3. Now to show analyticity between Riemann surfaces we must choose ψ ∈ A(η). Since ψ, φ ∘ α ∈ A(η), we see that φ ∘ α ∘ ψ^{−1} is conformal. Hence α : U[η] → V [ν] between Riemann surfaces is analytic.

The uniqueness follows from Corollary 9.6 (a). □

Remark 10.7. We note that when we pull back a Beltrami coefficient by a quasiconformal map we still get a Beltrami coefficient.

We will use this “pull back” method often throughout the rest of this paper. This method lets us put a structure on a set so that we have the correct dilatation to insure analyticity of a given map. We will next show that we can “pull back” conformal structures by not only quasiconformal maps, but also by analytic (in the usual sense) and quasiregular maps as well. In order to show this we need the following result.

Theorem 10.8. Let f : U[μ] → V [ν] and g : V [ν] → W [η] each be analytic between Riemann surfaces. Then g ∘ f : U[μ] → W [η] is analytic.

Proof. Choose φ ∈ A(μ), ψ ∈ A(ν), and β ∈ A(η) arbitrarily. Since f is analytic between Riemann surfaces, we have that ψ ∘ f ∘ φ^{−1} is analytic. Similarly β ∘ g ∘ ψ^{−1} is analytic. Therefore β ∘ g ∘ f ∘ φ^{−1} = (β ∘ g ∘ ψ^{−1}) ∘ (ψ ∘ f ∘ φ^{−1}) is analytic. Hence g ∘ f : U[μ] → W [η] is analytic between Riemann surfaces. □

Lemma 10.9. ([1], p. 182) Let η and ν be Beltrami coefficients on domains of the plane U and V respectively. Let g : U → V be analytic and nonconstant. Then the following are equivalent:

a) g : U[η] → V [ν] is analytic.

b) ν(g(z)) = \frac{g′(z)}{\overline{g′(z)}}η(z) almost everywhere in U.

Proof. First, let ψ be ν-conformal on V, i.e., μ_{ψ} = ν, and note that μ_{ψ∘g} =

μ_{1} where μ_{1}(z) = \frac{\overline{g′ (z)}}{g′ (z)}ν(g(z)) almost everywhere by Corollary 9.6 (c), since the analytic map g is locally conformal away from the discrete (in U) set E of points where the multiplicity of g is strictly greater than one.

“(a) ⇒(b)” Let φ ∈ A(η). By (a), the map (ψ∘g)∘φ^{−1} is analytic and therefore conformal away from a discrete set in φ(U). Thus, by applying Corollary 9.6 (a)

locally to (ψ ∘ g) ∘ φ^{−1}, we have η(z) = μ_{φ}(z) = μ_{ψ∘g}(z) = \frac{\overline{g′ (z)}}{g′ (z)}ν(g(z)) almost everywhere in U. Therefore (b) holds.

“(b) ⇒(a)” Let φ ∈ A(η). Now (b) implies that ψ ∘ g is η-conformal in a neighborhood of each z in U \ E. Thus we see that (ψ ∘ g) ∘ (φ^{−1}) : φ(U \ E) → \overline{ℂ} is locally conformal. Since (ψ ∘g)∘(φ^{−1}) is continuous on all of φ(U) and E is a set of isolated points (in U), Riemann’s Theorem on removable singularities implies that (ψ ∘ g) ∘ (φ^{−1}) from φ(U) into ψ(V) is analytic. Thus g : U[η] → V [ν] is analytic between Riemann Surfaces. □

Corollary 10.10. Let U and V be domains in ℂ and let g : U → V be surjective and analytic. Let ν be a Beltrami coefficient on V. Then there exists a unique Beltrami coefficient η on U such that g : U[η] → V [ν] is analytic. We denote such η by η = g^{∗}ν and call it the pull back of ν by g.

Proof. Set η(z) = \frac{\overline{g′ (z)}}{g′ (z)}ν(g(z)) where g′(z) ≠ 0, which is almost everywhere. But by Lemma 10.9, this is equivalent to saying that g : U[η] → V [ν] is analytic provided that η is Beltrami, which holds since ∥η(z)∥_{∞} = ∥ν(g(z))∥_{∞} < 1, using the fact that ν is a Beltrami coefficient. Again, the uniqueness follows from applying Corollary 9.6 (a) locally on the set U \ {z : g′(z) = 0}. □

Remark 10.11. We see here that when we pull back a Beltrami coefficient by an analytic map the infinity norm of the Beltrami coefficient does not increase.

Lemma 10.12. Let R be a rational map from the plane \overline{ℂ} to itself, and let φ, mapping the plane to itself, be μ-conformal for some Beltrami coefficient μ. Then

φRφ^{−1} is rational if and only if μ(R(z)) = \frac{R′ (z)}{\overline{R′ (z)}} · μ(z) almost everywhere in the

plane. Furthermore, when this is true, we have deg R = deg φRφ^{−1}.

Proof. “ ⇒” Suppose φRφ^{−1} from the plane to itself is rational. Then φRφ^{−1} must be analytic in the usual sense because it is a rational map from the plane to itself. Therefore, by definition of a function being analytic between Riemann surfaces, noting that φ is μ-conformal, we see R : \overline{ℂ}[μ] → \overline{ℂ}[μ] is analytic. So by

Lemma 10.9, we see that μ(R(z)) = \frac{R′ (z)}{\overline{R′ (z)}} · μ(z) almost everywhere in the plane \overline{ℂ}.

“ ⇐” Now suppose that μ(R(z)) = \frac{R′ (z)}{\overline{R′ (z)}} · μ(z) almost everywhere in the plane.

By Lemma 10.9 we have that R : \overline{ℂ}[μ] → \overline{ℂ}[μ] is analytic. But analyticity between Riemann surfaces implies that φRφ^{−1} is analytic between the plane and itself since φ ∈ A(μ). However, the only analytic maps from the plane to itself are rational maps (see [5], p. 358).

As to the deg φRφ^{−1}, we know that φ and φ^{−1} are injective since they are quasiconformal. Therefore deg R = deg φRφ^{−1}. □

Definition 10.13. A quasi-regular map is a composite map f ∘ φ where φ is quasiconformal on the plane and f is rational.

Lemma 10.14. Let g be a quasi-regular map. If g : U → V is surjective, and ν is a Beltrami coefficient on V, then there exists a unique Beltrami coefficient η on U such that g : U[η] → V [ν] is analytic. We denote η = g^{∗}ν and call η the “pull back” of ν by g.

Proof. Let g = f ∘ φ where φ is quasiconformal on the plane and f is rational. Using Lemma 10.6 and Corollary 10.10, we set η = φ^{∗}(f^{∗}ν). Thus we have U[φ^{∗}(f^{∗}ν)] →^{φ} φ(U)[f^{∗}ν] →^{f} V [ν] and note that g = f ∘ φ : U[η] → V [ν] is analytic by Theorem 10.8.

Uniqueness follows as in the proofs of Lemma 10.6 and Corollary 10.10. □

Remark 10.15. We summarize Lemma 10.6, Corollary 10.10 and Lemma 10.14 by stating that for Beltrami coefficients η and ν on domains U and V, respectively, a function f : U[η] → V [ν] is analytic if and only if η = μ_{ψ∘g} for any ψ ∈ A(ν) if and only if η = f^{∗}ν (almost everywhere), whether f : U → V is quasiconformal, analytic, or quasi-regular. We will use this fact in the following arguments without mention.

Lemma 10.16. Suppose surjective g : \overline{ℂ} → \overline{ℂ} is continuous and locally quasiconformal away from a finite set of points K (but not necessarily a global homeomorphism). If ∥μ_{g}∥_{∞} < 1, then g is quasi-regular.

Proof. Let φ : \overline{ℂ} → \overline{ℂ} be quasiconformal with μ_{φ} = μ_{g}, which is guaranteed by the Existence Theorem. Set A = g ∘ φ^{−1}. Then away from φ(K) we see that A is locally quasiconformal. Now since μ_{φ} = μ_{g}, we have that A is locally conformal away from φ(K). Also A is continuous at each point in φ(K) since both φ^{−1} and g are continuous. Thus by Riemann’s Theorem on removable singularities, A is analytic on the entire plane, and hence A is rational. Therefore the map g = A ∘ φ is quasi-regular. □

Remark 10.17. Lemma 10.16 shows that quasi-regular maps are those which are globally continuous, but locally quasiconformal away from a finite set of points – just like rational maps are those which are globally continuous, but locally conformal away from a finite set of points.

11. Fundamental Theorem of Quasiconformal Surgery

Shishikura developed the idea of quasiconformal surgery “to create from given rational functions a new one preserving their dynamics (in some sense)” ([7], p. 7). The Identity Theorem (Principle of Analytic Continuation) ([5], p. 307) often restricts one from “gluing” analytic functions together to form a globally analytic map. Considering a conjugation by quasiconformal maps, however, the “gluing” process is a possibility ([7], p. 7). In this section we present this result (the Fundamental Theorem of Quasiconformal Surgery) along with an important application.

Theorem 11.1 (Fundamental Theorem of Quasiconformal Surgery). Let the map g : \overline{ℂ} → \overline{ℂ} be quasi-regular. Suppose, for i = 1, ..., m, there are disjoint open subsets E_{i} of the plane \overline{ℂ}, quasiconformal maps Φ_{i} : E_{i} → Ei^{′} and an integer N ≥ 0 satisfying the following conditions:

i) g(E) ⊆ E where E = E_{1} ∪ ... ∪ E_{m};

ii) Φ ∘ g ∘ Φ_{i}^{−1} : Ei^{′} → ∪_{i=1}^{m}Ei^{′} is analytic where Φ : E → \overline{ℂ} is defined by Φ|_{E_{i}} = Φ_{i};

iii) \overline{∂}g = 0 almost everywhere on \overline{ℂ} \ g^{−N}(E) where g^{−N}(E) is inverse image of E under g^{N}, the N^{th} iterate of g.

Then there exists a quasiconformal mapping φ of the plane \overline{ℂ} such that φ∘g∘φ^{−1} is rational. Moreover, φ∘Φ_{i}^{−1} is conformal in Ei^{′} and \overline{∂}φ(z) = 0 almost everywhere on \overline{ℂ} \ ∪_{n≥0}g^{−n}(E).

Proof. Our goal is to define a Beltrami coefficient σ on the plane for which g^{∗}σ = σ. Then the Existence Theorem guarantees that there is quasiconformal map φ which is a solution to the Beltrami equation \overline{∂}φ = σ · ∂φ. As we shall see, this φ will be the quasiconformal map we seek to prove the theorem.

We begin by defining, on E_{i}, σ = Φ_{i}^{∗}σ_{0} where we define σ_{0} = 0 on Ei^{′}, i.e., σ = μ_{Φ_{i}} on E_{i}. This says that Φ_{i} is a chart on E_{i}[σ]. So by definition of the “pull back” we see that Φ_{i} : E_{i}[σ] → Ei^{′}[σ_{0}] is analytic. In this way σ is defined on all of E = ∪_{i=1}^{m}E_{i}.

Next we will show that g^{∗}σ = σ on E, i.e., g : E[σ] → E[σ] is analytic. Without loss of generality, we will assume that E_{i} and E_{j} are components of E and that g(E_{i}) ⊆ E_{j}. Recall that it is enough to check analyticity with respect to one chart from each atlas. Since Φ_{i} is a chart on E_{i} and Φ_{j} is a chart on E_{j}, we have, by assumption (ii) that Φ_{j} ∘ g ∘ Φ_{i}^{−1} : Ei^{′} → Ej^{′} is analytic. Therefore we have shown that g : E_{i}[σ] → E_{j}[σ] is analytic and, in general, g : E[σ] → E[σ] is analytic.

We will now define σ on ∪_{n≥1}g^{−n}(E) by successively pulling back by g. Note that the set ∪_{n≥1}g^{−n}(E) is the set of points that are mapped into E under some iterate of g. Consider a component W_{1} of E. We endow g^{−1}(W_{1}) with the conformal structure g^{∗}σ. We note that for any z_{0} ∈ g^{−1}(W_{1})∩E we have σ(z_{0}) = g^{∗}σ(z_{0}) from above. Thus we may extend σ to g^{−1}(W_{1}) ∪ E by setting σ = g^{∗}σ on g^{−1}(W_{1}) \ E. Performing this procedure on each component of E, we extend σ to all of g^{−1}(E) by g^{∗}σ. With σ now defined on all of g^{−1}(E), we can similarly extend σ to all of g^{−2}(E) by defining σ|_{g^{−2}(E)} = g^{∗}(σ|_{g^{−1}(E)}). In this manner, we can inductively define σ such that g^{∗}σ = σ on ∪_{n≥0}g^{−n}(E) and so g : ∪_{n≥0}g^{−n}(E)[σ] → ∪_{n≥0}g^{−n}(E)[σ] is analytic.

Now we set σ = 0 on the rest of \overline{ℂ}.

We now have σ defined everywhere, but we must now show that σ is a Beltrami coefficient. Note that σ is clearly Lebesgue measurable. So we must show that ∥σ∥_{∞} < 1. We know that on E, σ is a Beltrami coefficient. On g^{−1}(E), since we pulled back by a quasi-regular map, σ is again Beltrami. We continue this up to g^{−N}(E) and since we pulled back a finite number of times, we still have σ Beltrami here. However, we claim that for n > N we have ∥σ |_{g^{−n}(E)}∥_{∞} = ∥σ |_{g^{−N} (E)}∥_{∞}. Indeed, for any z ∈ g^{−(N+1)}(E) \ g^{−N}(E) condition (iii) gives \overline{∂}g(z) = 0. Since σ(z) = μ_{ψ∘g}(z) for any ψ ∈ A(σ |_{g^{−N} (E)}), we then see by Lemma 6.18(a)(ii) that |σ(z)| = |σ(g(z))| and thus the pullback by g of σ |_{g^{−N} (E)} to σ |_{g^{−(N+1)}(E)} does not increase in norm. The claim for general n > N now follows by induction. Hence, noting that σ = 0 off of ∪_{n≥0}g^{−n}(E), we see that σ is a Beltrami coefficient with

∥σ∥_{∞} = max_{k=0,...,N} ∥σ |_{g^{−k}(E)}∥_{∞} = ∥σ |_{g^{−N} (E)}∥_{∞} < 1.

Now the Existence Theorem provides a quasiconformal map φ on the plane such that φ is a solution to \overline{∂}φ = σ · ∂φ on the plane, i.e., σ = φ^{∗}σ_{0} on the plane where σ_{0} ≡ 0.

We will now show that φ ∘ g ∘ φ^{−1} : \overline{ℂ}[0] → \overline{ℂ}[0] is analytic and hence rational. As in Remark 10.15, it suffices to show that μ_{φ∘g} = μ_{φ} = σ a.e. in \overline{ℂ}. Indeed, on ∪_{n≥0}g^{−n}(E), this equality holds since g^{∗}σ = σ there. For z ∉ ∪_{n≥0}g^{−n}(E) we see that g(z) ∉ ∪_{n≥0}g^{−n}(E) and also z ∉ g^{−N}(E), which by (iii), implies \overline{∂}g(z) = 0 for almost all such z. By Lemma 6.18 we then have μ_{φ∘g}(z) = μ_{φ}(g(z))\frac{\overline{∂g(z)}}{∂g(z)} = σ(g(z))\frac{\overline{∂g(z)}}{∂g(z)} = 0 = σ(z) for almost all z ∉ ∪_{n≥0}g^{−n}(E). Hence we have shown that φ ∘ g ∘ φ^{−1} is rational.

Now since φ and Φ_{i} are both σ-conformal on E_{i}, we have that φ ∘ Φ_{i}^{−1} is conformal on Ei^{′}. Lastly we note that \overline{∂}φ = 0 on \overline{ℂ} \ ∪_{n≥0}g^{−n}(E) as σ = 0 there. The proof of the theorem is now complete. □

Definition 11.2. A function f : U → V between domains in \overline{ℂ} is called proper if for any compact K ⊆ V we have that f^{−1}(K) is compact in U.

The following characterization of proper maps will be useful later. The proof is left to the reader.

Lemma 11.3. Let f : U → V be a map between domains in \overline{ℂ}. Then f is proper if and only if f(z_{n}) → ∂V as z_{n} → ∂U.

Note that z_{n} → ∂U means that given any compact K ⊂ U we have that z_{n} ∉ K for all sufficiently large n.

Definition 11.4. Let U_{1} and U_{2} be simply connected bounded domains of the finite plane ℂ such that ∂U_{1} = γ_{1} and ∂U_{2} = γ_{2} are analytic (see Appendix II Definition 13.1 for definition) simple closed curves and \overline{U_{1}} ⊆ U_{2}. We call f : U_{1} → U_{2} polynomial-like of degree d if

i) f is proper,

ii) f has degree d, i.e., every point in U_{2} is taken d times counting multiplicity by points in U_{1} under f,

iii) f is analytic.

Remark 11.5. By Lemma 13.12 in Appendix II, a polynomial-like map f extends analytically to domain U1^{′} ⊇ \overline{U_{1}} and this extension maps γ_{1} onto γ_{2} precisely d times.

In the proof of the next theorem we will require the following fact.

Fact: Let d be a positive integer. For any C^{∞}-smooth curve Γ(t) for 0 ≤ t ≤ 1 with Γ(t) ∈ C(0, R^{d}), there exists a C^{∞}-smooth curve γ(t) with γ(t) ∈ C(0, R) such that [γ(t)]^{d} = Γ(t). Such can easily be accomplished by noting that locally γ(t) will be given by h(Γ(t)) where h is a suitably chosen branch of z ↦ z^{1/d}. Note that if Γ(t) traverses C(0, R^{d}) exactly d times, then γ(t) will traverse C(0, R) exactly once.

Theorem 11.6. If f : U_{1} → U_{2} is polynomial-like of degree d, then there exists a quasiconformal map of the plane φ such that f = φ ∘ P ∘ φ^{−1} on U_{1} where P is a polynomial of degree d.

Proof. Fix r > 1. Let Φ_{1} map \overline{ℂ} \ \overline{U_{2}} conformally onto \overline{ℂ} \ \overline{Δ(0, r^{d})} such that Φ_{1}(∞) = ∞. The existence of this map is guaranteed by the Riemann Mapping Theorem. Now by the Caratheodory-Osgood Theorem we extend Φ_{1} to ∂U_{2} so that Φ_{1}(∂U_{2}) = ∂(Δ(0, r^{d})), and by Corollary 13.6 in Appendix II, Φ_{1} extends analytically to a domain containing \overline{ℂ} \ U_{2}.

\begin{aligned}r & r^{d} & U_{2} & U_{1} & f & Φ & ∞ & ∞ \\ & & & & Figure 10 \end{aligned}

Next we define a map Φ_{2} on ∂U_{1} = γ_{1} as follows. We note that f(z) will d times traverse γ_{2} as z traverses γ_{1} once. Setting Γ(t) = Φ_{1}(f(γ_{1}(t))), which is C^{∞} smooth and d times traverses the circle C(0, r^{d}), we see by the Fact given just before the statement of this theorem, that there exists a C^{∞} smooth parametrization γ(t) of C(0, r) such that [γ(t)]^{d} = Γ(t). We then define Φ_{2} on γ_{1} by Φ_{2}(γ_{1}(t)) = γ(t) and note that Φ_{2} is C^{∞} smooth here. Thus [Φ_{2}(z)]^{d} = Φ_{1}(f(z)) on γ_{1} = ∂U_{1}. Setting Φ_{2} equal to Φ_{1} on γ_{2} we may then extend Φ_{2}, in a C^{∞} smooth way, such that Φ_{2} : \overline{U_{2} \ U_{1}} → \overline{Δ(0, r^{d}) \ Δ(0, r)} as in Theorem 13.9 from Appendix II. We note that on U_{2} \ \overline{U_{1}} the map Φ_{2} is K -QC for some K by Remark 13.10 from Appendix II. We now sew Φ_{1} and Φ_{2} together to obtain a map Φ on \overline{ℂ}\U_{1} defined by Φ = Φ_{1} on \overline{ℂ} \ U_{2} and Φ = Φ_{2} on \overline{U_{2} \ U_{1}}. By Lemma 13.13, the map Φ is K − QC on \overline{ℂ} \ \overline{U_{1}}.

We now define a map g(z) on \overline{ℂ} and show that it is quasiregular. Set

g(z) = \{ \begin{matrix}f(z) & for z ∈ U_{1} \\ Φ^{−1}[(Φ(z))^{d}] & for z ∈ \overline{ℂ} \ U_{1}. \end{matrix}

On U_{1}, g is analytic and so conformal away from the finite set {z : f′(z) = 0}. Therefore ∥μ_{g}∥_{∞} = 0 on U_{1}. On \overline{ℂ} \ \overline{U_{1}} we see that g is a composition of K -QC Φ, followed by analytic z ⟼ z^{d} (which is locally conformal), followed by conformal Φ^{−1} (on \overline{ℂ} \ Δ(0, r^{d})). Hence ∥μ_{g}∥_{∞} ≤ ∥μ_{Φ}∥_{∞} < 1 on \overline{ℂ} \ \overline{U_{1}}. (Upon closer inspection we can actually see that μ_{g} = 0 on \overline{ℂ} \ \overline{U_{2}}.) We now show that g is locally quasiconformal away from the set \{z : f′(z) = 0\} ∪ \{∞\}. On U_{1} and \overline{ℂ} \ \overline{U_{1}}, this is clear. For z_{0} ∈ γ_{1}, we see g defined on Δ(z_{0}, ϵ) for a sufficiently small ϵ is K − QC since g is the map created when the conformal map f and the K − QC map Φ^{−1}[(Φ(z))^{d}] are sewn together (see Lemma 13.13 in Appendix II). We may now apply Lemma 10.16 to see that g is quasi-regular.

One can see that the Fundamental Theorem of Quasiconformal Surgery 11.1 can be applied, with E = \overline{ℂ} \ \overline{U_{1}} and N = 1, to see that there exists a quasiconformal φ such that P = φ∘g ∘φ^{−1} is rational. But since only ∞ maps to ∞ with multiplicity d we see that P is actually a polynomial of degree d. □

Remark 11.7. We have, in a sense, glued g|_{\overline{ℂ}\E} = f|_{U_{1}} to Φ^{−1} ∘ z^{d} ∘ Φ|_{\overline{ℂ}\U_{2}} which is z ⟼ z^{d}, at least up to conjugation by a \underline{conformal} map. Now pasting them together did not make a polynomial. However, when we conjugated this “pasting” by a quasiconformal map, we then have a true polynomial.

12. Appendix I

Here we present the Cauchy-Schwarz inequality.

If f and g are real-valued functions that are Lebesgue Measurable over the

interval [c, d], then ∫_{c}^{d} |f(y)g(y)| dy ≤ \{∫_{c}^{d}[f(y)]^{2}dy\}^{\frac{1}{2}} \{∫_{c}^{d}[g(y)]^{2}dy\}^{\frac{1}{2}} .

Furthermore, equality holds if and only if there exist constants α and β (not both zero) such that α |f|^{2} = β |g|^{2} almost everywhere.

13. Appendix II

Definition 13.1 ([4], p. 186). We say that a curve γ(t) = x(t)+iy(t) from the interval [0, 1] into ℂ is analytic if for each t_{0} ∈ (0, 1) we have that both the real part x(t) and the imaginary part y(t) can be represented by a power series centered at t_{0}; we further assume that x′(t) and y′(t) do not vanish at the same time.

Remark 13.2. We note that analytic curves are C^{∞}-smooth.

Remark 13.3. We note that an open analytic curve γ defined on the open interval (0, 1) in ℝ is the restriction of an analytic function f, and thus the set of image points of this curve (which, by an abuse of notation, we also call γ when no confusion will arise) is the image of an open interval under a complex analytic map. To see this we first fix a point t_{0} of (0, 1). We may express x(t) = ∑_{n=0}^{∞} a_{n}(t−t_{0})^{n} and y(t) = ∑_{n=0}^{∞} b_{n}(t − t_{0})^{n} by convergent power series in some real neighborhood of t_{0}. Thus the power series f(z) = ∑_{n=0}^{∞}(a_{n} + ib_{n})(z − t_{0})^{n} is then convergent in some complex neighborhood of t_{0} and therefore defines an analytic function there. Also note that f′(t_{0}) = x′(t_{0})+iy′(t_{0}) ≠ 0. Doing this at each point in (0,1) clearly defines an analytic map f in a complex neighborhood of (0, 1).

The Schwarz reflection principle then allows one to extend analytic maps across analytic curves as well. The details can be seen in the cited reference of the following.

Lemma 13.4. [[4], p. 184] Let U and V be domains in ℂ and let g : U → V be an analytic function. Suppose ∂U includes an open analytic arc α, ∂V includes an open analytic arc β, and that g extends continuously to U ∪ α where g(z) ∈ β for all z ∈ α. Then g analytically extends across α, i.e., g can be extended analytically in a neighborhood U′ which contains α.

Remark 13.5. By considering the details of the proof one can see that in Lemma 13.4 we have that g′ ≠ 0 on α.

Corollary 13.6. If f : D_{1} → D_{2} is a conformal map between simply or multiply connected domains in ℂ each with boundaries that consist of a finite number of analytic curves, then f analytically extends to a domain containing \overline{D_{1}}. Furthermore, f^{−1} analytically extends to a domain containing \overline{D_{2}}.

Proof. Suppose that D_{1} has analytic boundary arcs γ_{1}, …, γ_{n} and D_{2} has analytic boundary arcs Γ_{1}, …, Γ_{m}. Then by applying the techniques used in the proof of Theorem 4.7 in [5] (which is used in the proof of the Carathéodory-Osgood Theorem), we see that f continuously extends to each γ_{j}, mapping it into some Γ_{k}. We then apply Lemma 13.4 to show the existence of an analytic extension across each γ_{j} and thus to a domain containing \overline{D_{1}}.

We note that we may apply the same argument to f^{−1} to conclude that f^{−1} analytically extends to a domain containing \overline{D_{2}}. □

We note that we get the following similar result if our boundary curves are merely C^{∞}-smooth instead of analytic.

Theorem 13.7 ([2], p. 23 and 24). If f : D_{1} → D_{2} is a conformal map between simply or multiply connected domains in ℂ each with C^{∞}-smooth boundaries, then f and all of its derivatives extend continuously to \overline{D_{1}}. Furthermore, f^{−1} and all of its derivatives have continuous extensions to \overline{D_{2}}.

By employing the same techniques as the proof of Lemma 13.4 and using the harmonic reflection principle stated on p. 234 in [5], we see that harmonic functions can also be reflected over analytic curves. Specifically we have the following:

Lemma 13.8. Let U be a domain in ℂ and let g : U → ℝ be a harmonic function. Suppose ∂U includes an open analytic arc α such that g extends continuously to U ∪ α where g(z) = 0 for all z ∈ α. Then g harmonically extends across α, i.e., g can be extended harmonically in a neighborhood U′ which contains α.

Theorem 13.9. Let U_{1} and U_{2} be simply connected domains in ℂ bounded by closed analytic Jordan curves γ_{1} and γ_{2}, respectively. Suppose further that \overline{U_{1}} ⊂ U_{2}. Given C^{∞}-smooth homeomorphisms g_{j} : γ_{j} → C(0, r_{j}), for j = 1, 2 and r_{1} < r_{2}, each which preserves orientation, there exists a C^{∞}-smooth homeomorphism G from \overline{U_{2} \ U_{1}} onto the closed true annulus \overline{Ann(0; r_{1}, r_{2})} which extends both g_{1} and g_{2}. Furthermore, G and all of its derivatives extend continuously to \overline{U_{2} \ U_{1}}. We also note that there exists η > 0 (which depends on r_{1}, r_{2}, g_{1}, g_{2}, U_{1} and U_{2}) such that the Jacobian J_{G}(z) ≥ η for all z ∈ \overline{U_{2} \ U_{1}}.

Remark 13.10. We note that the map G in Theorem 13.9 satisfies ∥μ_{G}∥_{∞} < 1 on U_{2} \ \overline{U_{1}} and is therefore quasiconformal on U_{2} \ \overline{U_{1}}. Since |∂G(z)|^{2} − |¯∂G(z)|^{2} = J_{G}(z) ≥ η > 0, we have that |∂G(z)| ≥ \sqrt{η} > 0. Using these two facts then allows us to note that |μ_{G}(z)| = |¯∂G(z)/∂G(z)| is bounded above by 1 and continuous on the compact set \overline{U_{2} \ U_{1}}. Thus it follows that ∥μ_{G}∥_{∞} < 1.

Lemma 13.11 ([4], p. 333). Every doubly connected domain A in \overline{ℂ} whose boundary components each contain more than one point can be mapped conformally onto a true annulus Ann(0; 1, R) for some 0 < R < ∞. We call log R the modulus of A.

proof of Theorem 13.9. We begin by first proving Theorem 13.9 in the special case that γ_{1} = C(0, r_{1}) and γ_{2} = C(0, r_{2}). Under the given assumptions on g_{1} and g_{2} we may now express g_{1}(r_{1}e^{iθ}) = r_{1}e^{if_{1}(θ)} and g_{2}(r_{2}e^{iθ}) = r_{2}e^{if_{2}(θ)} where for j = 1, 2 we have

(i) f_{j} maps ℝ into ℝ and is a C^{∞}-smooth increasing function of θ (since g_{j} is C^{∞}-smooth and orientation preserving), and

(ii) f_{j}(θ + 2π) = f_{j}(θ) + 2π (since g_{j} is a homeomorphism of a circle, f_{j} must map any interval of length 2π onto an interval of length 2π).

We now define f : [r_{1}, r_{2}] × ℝ → ℝ to be an interpolation of f_{1} and f_{2} given by

f(r, θ) = \frac{r − r_{2}}{r_{1} − r_{2}} f_{1}(θ) + (1 − \frac{r − r_{2}}{r_{1} − r_{2}} ) f_{2}(θ).

Note that for fixed r, the map f(r, θ) is a C^{∞}-smooth function which is increasing in θ (by (i)) and f(r, θ+2π) = f(r, θ)+2π (by (ii)). We now see that G(re^{iθ}) = re^{if(r,θ)} will satisfy the conclusion of our theorem.

Now we return to considering the general case without the assumptions on γ_{1} and γ_{2} given above. By Lemma 13.11 there exists a conformal map h from U_{2} \ \overline{U_{1}} onto an annulus of the form Ann(0; 1, R), which by Corollary 13.6 extends C^{∞}- smoothly to all of \overline{U_{2} \ U_{1}}. Note also that Corollary 13.6 allows one to conclude that h^{−1} is also a C^{∞}-smooth homeomorphism on \overline{Ann(0; 1, R)}.

Now consider the C^{∞}-smooth homeomorphic mapping k from \overline{Ann(0; r_{1}, r_{2})} onto \overline{Ann(0; 1, R)} given by re^{iθ} ↦ [\frac{R−1}{r_{2}−r_{1}} (r − r_{1}) + 1]e^{iθ}, which merely stretches radially. Hence we see that the maps g_{1} ∘ h^{−1} ∘ k and g_{2} ∘ h^{−1} ∘ k are C^{∞}-smooth homeomorphisms from C(0, r_{1}) and C(0, r_{2}) into themselves, respectively. This case was already handled above, and so we let G̃ be the C^{∞}-smooth homeomorphism from \overline{Ann(0; r_{1}, r_{2})} to itself, which extends the maps g_{1} ∘ h^{−1} ∘ k and g_{2} ∘ h^{−1} ∘ k. We then see that G = G̃ ∘ k^{−1} ∘ h has the required properties.

Finally, we conclude the proof by noting that since each map G̃, k^{−1} and h has Jacobian bounded below by some positive constant (which can be shown through direct calculation), so does G since J_{G}(z_{0}) = JG̃(k^{−1}(h(z_{0}))J_{k^{−1}}(h(z_{0}))J_{h}(z_{0}) for any z_{0} ∈ \overline{U_{2} \ U_{1}}. □

Lemma 13.12. Let f : U → V be a proper analytic map of degree d where U and V are simply connected domains in ℂ bounded by closed analytic Jordan curves γ_{1} and γ_{2}, respectively. Then f can be extended analytically to a map f : U′ → ℂ such that domain U′ ⊃ \overline{U} and f(U′) contains \overline{V}. We also note that this extension of f then maps γ_{1} onto γ_{2} precisely d times.

Proof. We first consider the case where U = V = Δ(0, 1). Fix a point z_{0} ∈ ∂Δ(0, 1). Since f is proper, there exists a small disk Δ = Δ(z_{0}, δ) such that f(z) ≠ 0 on Δ ∩ Δ(0, 1). Consider the harmonic function h(z) = log |f(z)| = Re Logf(z) defined on Δ ∩ Δ(0, 1), where Log(z) denotes the principal logarithm. Since |f(z)| → 1 as |z| → 1 (f is proper), we see that h(z) → 0 as |z| → 1. Thus we may reflect h(z) across |z| = 1 to extend h(z) to be harmonic on all of Δ (Theorem 3.4 in [5] states this reflection principle for real line segments, but it also holds for arcs of circles by a simple application of a Möbius transformation). Letting v(z) be a harmonic conjugate of h(z) in Δ, which on Δ∩Δ(0, 1) must equal Im Logf(z)+c for some real constant c, we see that the analytic function e^{h(z)+i(v(z)−c)} defined on all of Δ must equal f(z) on Δ ∩ Δ(0, 1). Thus extending f in this way in a disk about each point in ∂Δ(0, 1) we see that f can be extended to a domain U′ containing \overline{Δ(0, 1)}. Clearly, since f is an open continuous map, we must then have that f(U′) contains the compact set f(\overline{Δ}) = \overline{Δ(0, 1)}.

Note that by using the Argument Principle and the hypothesis that f maps U onto V in a d-to-1 fashion, we see that as the curve γ(t) traverses C(0, 1) once, f(γ(t)) traverses C(0, 1) exactly d times.

Now we address the general case. Let ϕ : U → Δ(0, 1) and ψ : V → Δ(0, 1) be Riemann maps. Since ∂U and ∂V are analytic curves we can see by Corollary 13.6 that ϕ, ψ, ϕ^{−1}, and ψ^{−1} all analytically extend across the boundary of their respective domains. Note that we may assume that these extensions are also conformal, even on these extended domains. From the special case above, we see that g = ψ ∘ f ∘ ϕ^{−1} : Δ(0, 1) → Δ(0, 1) extends across the boundary, and so one can easily show that ψ^{−1} ∘g ∘ϕ (using the extensions of each of these maps) extends f to a domain containing \overline{U}. □

The following lemma expresses the fact that we may sometimes “sew” quasiconformal maps together along a common boundary arc (where the maps agree) and obtain a resulting map which is also quasiconformal.

Lemma 13.13. Let Δ be an open disk which is met by an analytic arc γ such that there are two components D_{1} and D_{2} of Δ\γ. Suppose f_{1} and f_{2} are quasiconformal in D_{1} and D_{2}, respectively, such that f_{1}(D_{1}) and f_{2}(D_{2}) are disjoint and such that each map extends continuously and injectively to Δ ∩ γ on which f_{1} = f_{2}. Then the map F (z) defined by f_{1} on D_{1} and f_{2} on D_{2} and f_{1} = f_{2} on Δ∩γ is quasiconformal on all of Δ.

Proof. Since f_{1} and f_{2} are each absolutely continuous on lines, one can easily show, by continuity and the fact that f_{1} = f_{2} on Δ ∩ γ, that so is F. Since F is a continuous injective map of an open set Δ, it is a homeomorphism (see [3], p. 6). Since γ has measure zero, it is easy to see that by using Lemma 8.3 we have that |μ_{F} | is bounded almost everywhere by \frac{K−1}{K+1} where K = max\{K_{f_{1}}, K_{f_{2}}\}. Hence by Lemma 8.4 the map F is K-quasiconformal. □

Remark 13.14. One could also apply Theorem I.8.3 in [3] to obtain the above result.

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