On the Argument Oscillation of Conformal Maps - Project Euclid

On the Argument Oscillation of Conformal Maps J. J. Car m ona & C h. P om m e r e n k e

1. Introduction Let D = { z ∈ C : |z| < 1 } be the unit disk and T its boundary. We shall consider (injective) conformal maps f of D into C. For ζ ∈ T we denote by f (ζ) the angular (= radial) limit if it exists and is finite. This holds for almost all ζ ∈ T; even the exceptional set has zero logarithmic capacity, by the well-known Beurling theorem (see [Be; Po2, p. 215]). Furthermore, the set { ζ ∈ T : f (ζ) = a } has zero capacity for every a ∈ C [Du; Po2, p. 219]. A stronger condition is that f is continuous at ζ ; that is f (z) → f (ζ)

as z → ζ, z ∈ D.

(1.1)

Suppose now that the angular limit f (ζ) 6= ∞ exists at ζ ∈ T. The function gζ (z) = log[f (z) − f (ζ)],

z ∈ D,

(1.2)

is analytic and univalent in D for any branch of the logarithm. It therefore has finite angular limits at all points except a set of zero capacity. For convenience, throughout this paper we will write E(ζ) = { ζ 0 ∈ T : f (ζ 0 ) and gζ (ζ 0 ) exist and are finite }. Thus T \ E(ζ) has zero capacity. Then we define the argument by ( for z ∈ D, Im gζ (z) arg[f (z) − f (ζ)] = it lim Im gζ (re ) for z = e it ∈ E(ζ).

(1.3)

r→1

We also consider the analytic function h(z) = log

f (z) − f (ζ) = gζ (z) − log(z − ζ), z ∈ D ∪ E(ζ). z−ζ

(1.4)

In the unit disk, we always use the branch of the logarithm determined by θ + π/2 < arg(z − ζ) < θ + 3π/2,

z ∈ D ∪ E(ζ), ζ = e iθ .

The function h is a Bloch function [Po2, p. 173]. We will make frequent use of the harmonic function Received May 8, 1998. Revision received February 25, 1999. The work of first author is partially supported by the grants PB95-0956-C02-02 and 1995SGR-00482. Michigan Math. J. 46 (1999).

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J. J. Car m ona & C h. P om m e r e n k e

298

arg

f (z) − f (ζ) = Im h(z) z−ζ = arg[f (z) − f (ζ)] − arg(z − ζ), z ∈ D ∪ E(ζ).

(1.5)

The McMillam twist theorem [Mc; Po2, p. 142] shows that, for almost all ζ ∈ T, there are only two alternatives: (ζ) has a finite nonzero angular limit as z → ζ ; (i) log f (z)−f z−ζ

(ζ) oscillates between −∞ and +∞ along any curve in D ending (ii) arg f (z)−f z−ζ at ζ. This gives the definitive answer to the problem of argument oscillation almost everywhere as far as angular approach is concerned. See [CaPo] for a discussion of the behavior that is possible on subsets of T of zero measure. In this paper we study the oscillation of the argument for unrestricted approach. It turns out that the exceptional sets tend to be either countable or of zero measure.

2. Results Generically we define

osc = lim sup − lim inf,

provided that both limits are finite; otherwise, we define osc = +∞. Theorem 1. Let f map D conformally into C. Except possibly for countably many ζ ∈ ∂T, we have: If the angular limit f (ζ) exists and if osc

z→ζ, z∈D

arg[f (z) − f (ζ)] < 2π,

(2.1)

then f is continuous at ζ. The constant 2π is best possible. We shall derive this theorem in Section 7 from two results of a topological nature. The condition that f be continuous at ζ is important for the following reason. There are two causes that may contribute to the oscillation of arg[f (z) − f (ζ)], namely: (i) nearby oscillation—that is, for f (z) − f (ζ) small; (ii) faraway oscillation—that is, for |f (z) − f (ζ)| > c > 0 (e.g., if f (z) winds around the entire domain). If f is continuous at ζ then only case (i) is possible. Now we consider the oscillation of the difference quotient. If ζ ∈ T and f (ζ) 6= ∞ exists, then we define f (z) − f (ζ) . (2.2) 1(ζ) = osc arg z→ζ, z∈D z−ζ The quantity 1(ζ) is closely related to tangency properties of the boundary of G = f (D). We say that ∂G has a tangent at the prime end corresponding to ζ = e iθ if β as t → θ + , e it ∈ E(ζ), arg[f (e it ) − f (ζ)] → (2.3) β + π as t → θ − , e it ∈ E(ζ).

On the Argument Oscillation of Conformal Maps

299

Observe that this definition is completely different, in the case of an arbitrary conformal map, from the usual concept (see [Fa, p. 31]) that the planar set ∂f (D) has a geometric tangent at f (ζ). However, if ∂G is a Jordan curve then the definitions are equivalent. We have the following result, which should be compared to the classical Lindelöf theorem (see [Po2, p. 51]) and to the theorem of Wolff [Wo]. Its proof will be an application of Theorem 3. Theorem 2. Let f map D conformally into C. If f (ζ) 6= ∞ exists at ζ ∈ T, then ∂f (D) has a tangent at f (ζ) if and only if 1(ζ) = 0. Theorem 3.

Let f map D conformally into C. If f (ζ) 6= ∞ exists at ζ ∈ T, then 1(ζ) =

osc

η→ζ, η∈E(ζ)

arg

f (η) − f (ζ) . η−ζ

(2.4)

If f is continuous at ζ, then osc arg[f (rζ) − f (ζ)] ≤ 1(ζ) ≤ 3π +

r→1

osc

z→ζ, z∈C

arg[f (z) − f (ζ)],

(2.5)

where C is any curve in D ∪ {ζ } ending at ζ. The constant 3π is best possible. Note that the right-hand side of (2.4) is essentially a geometric quantity (see Section 3 for details). The radial oscillation in (2.5) has been studied previously (see e.g. [CaPo]). Theorem 4. Let f map D conformally into C. Then the sets { ζ ∈ T : 0 < 1(ζ) < π } and { ζ ∈ T : f is continuous at ζ, 2π < 1(ζ) < ∞ } have zero Lebesgue measure on T. If ∂f (D) is locally connected (see the definition in [Po2, p. 19]) then it follows from Theorem 4 that 1(ζ) = 0

or

π ≤ 1(ζ) ≤ 2π

or

1(ζ) = +∞

for almost all ζ ∈ T. However “linear measure zero” cannot be replaced by “Hausdorff dimension < 1”, as the following theorem shows. Theorem 5.

There exists a conformal map f onto a Jordan domain such that dim{ ζ ∈ T : 0 < 1(ζ) < π } = 1.

(2.6)

3. One-Sided Oscillation Throughout this section we assume that f maps D conformally into C, that ζ ∈ T, and that the angular limit f (ζ) 6= ∞ exists. We write G = f (D), We define

ζ = e iθ (0 ≤ θ < 2π),

w = f (ζ).


300

α ± (ζ) = β ± (ζ) =

lim inf

arg[f (e it ) − w],

lim sup

arg[f (e it ) − w].

t→θ ± , e it ∈E(ζ)

t→θ ± , e it ∈E(ζ)

(3.1)

Also, if C is a Jordan arc such that ζ ∈ C and C \ {ζ } ⊂ D, we put α C (ζ) = lim inf arg[f (z) − w], z→ζ, z∈C

β C (ζ) = lim sup arg[f (z) − w].

(3.2)

t→ζ, z∈C

Note that α ± (ζ) and β ± (ζ) can essentially be determined from the domain G, whereas one must know the function f in order to find α C (ζ) and β C (ζ). When there is no possibility of confusion, we will omit the point ζ and simply write α ± and β ± . Now we restate Theorem 3. Theorem 6.

If f (ζ) 6= ∞ exists, then

1(ζ) = max(β + − α + , β + − α − + π, β − − α + − π, β − − α − ),

(3.3)

and if f is continuous at ζ then βR − αR ≤ 1(ζ) ≤ 3π + β C − α C ,

(3.4)

where R = [0, ζ) and C is any curve in D ∪ {ζ } ending at ζ. The constant 3π is best possible. To see that (2.4) and (3.3) are equivalent, we first observe that θ + π/2 as t → θ + , arg(e it − ζ) → θ + 3π/2 as t → θ − . Therefore, f (η) − w π 3π = max β + − θ − , β − − θ − lim sup arg η−ζ 2 2 η→ζ, η∈E(ζ) and lim inf arg

η→ζ, η∈E(ζ)

(3.5)

f (η) − w π 3π = min α + − θ − , α − − θ − . η−ζ 2 2

Then the foregoing identities give the equivalence. Proposition 1.

If f is continuous at ζ, then α + (ζ) ≤ α C (ζ) ≤ α − (ζ) ≤ α + (ζ) + 2π,

(3.6)

β − (ζ) ≥ β C (ζ) ≥ β + (ζ) ≥ β − (ζ) − 2π.

(3.7)

Proof. Let g = gζ , where gζ is defined by (1.2). By (1.1) we have Re g(z) = log|f (z) − w| → −∞ as z → ζ, z ∈ D.

(3.8)


301

Let us consider a parameterization p(t), t ∈ [0, 1], of C with p(1) = ζ. For each b ∈ (−∞, Re g(0)), let t b be the last value on the curve C such that Re g(p(t b )) = b. Let now L(b) be the connected component of { s : Re s = b } ∩ g(D) that contains g(p(t b )). Then g −1(L(b)) is a crosscut in D. Say that Q(b) is the component of D \ g −1(L(b)) for which ζ belongs to its boundary. The definition of t b gives that Q(b) \ C is the union of two domains U + (b) and U − (b). We have R = { g(p(t)) : t b < t < 1 } ⊂ g(D) and therefore B = { g(p(t)) − 2πi : t b ≤ t < 1 } ⊂ C \ g(D), because f is univalent. If B − denotes the Jordan curve B with the reversed orientation, then J = B − ∪ [g(p(t b )) − 2πi, g(p(t b ))] ∪ R ∪ {∞} is a positively oriented Jordan curve in C ∞ . Let H + be its interior. Since the tangent vector to g(C ) at the point g(p(t b )) has negative real part and g(U + (b)) ∩ J = ∅, we conclude that

g(U + (b)) ⊂ H + .

Then g(e it ) ∈ H + for some small interval t ∈ (θ, θ + δ) with e it ∈ E(ζ). Letting b → −∞, we infer from (3.1) and (3.2) that α C (ζ) − 2π ≤ α + (ζ) ≤ α C (ζ), β C (ζ) − 2π ≤ β + (ζ) ≤ β C (ζ).

(3.9)

Similarly, if we consider the Jordan arc B − + 4πi then we have α C (ζ) ≤ α − (ζ) ≤ α C (ζ) + 2π, β C (ζ) ≤ β − (ζ) ≤ β C (ζ) + 2π.

(3.10)

If we apply the inequalities (3.9) to curves that are sufficiently tangential to { e it : t < θ } or to { e it : t > θ }, we obtain α − − 2π ≤ α + ≤ α − and β − − 2π ≤ β + ≤ β − . This and (3.10) give (3.6) and (3.7). It is not difficult to see that the hypothesis of continuity of f at ζ is essential in each inequality of (3.6) and (3.7). Also, the constant 2π is best possible, as will be shown by Example 1. However, if we are considering only Stolz angles, then some improvement of Proposition 1 is possible; see [Wa]. We recall that the function f is isogonal at ζ (see [Po2, p. 80]) if, for some γ , arg

f (z) − w →γ z−ζ

as z → ζ in every Stolz angle.

It follows from this definition that, for every ε > 0, there exists a δ > 0 such that { z : 0 < |z − w| < δ, γ 1 + ε < arg(z − w) < γ 2 − ε } ⊂ G, with γ 1 = γ + θ + π/2 and γ 2 = γ 1 + π.

(3.11)

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Proposition 2.


If f is continuous and isogonal at ζ, then α − (ζ) = β + (ζ) + π.

Proof. Since f is continuous at ζ, it follows from (3.11) (for ε → 0) that β + (ζ) ≤ γ 1,

α − (ζ) ≥ γ 2 .

Then α − − β + ≥ γ 2 − γ 1 = π. However, by Ostrowski’s theorem [Po2, p. 252], the domain does not contain any sector of vertex f (ζ) of angle larger than π. Hence α − − β + = π. It is interesting to point out that the inequalities α − ≤ α + +2π and β − ≤ β + +2π of Proposition 1 cannot be improved even in the case when ζ is an isogonal point, as Example 1 will show. On the other hand, if ∂G has a tangent at f (ζ) then one has α + = β + , α − = β − , β − = β + + π. Example 1. Put D1 = { z : Re z > 0, |z| < 2 }. For each natural number n ≥ 1, consider { z : z = re is , π2 ≤ s < 3π , n1 < r < n1 + 31n } if n is an odd number, 2 Cn = , n1 − 31n < r < n1 } if n is an even number. { z : z = re is , π2 < s ≤ 3π 2 S Now G = D1 ∪ n≥1 Cn is a simply connected domain. Let f a conformal map from D onto G with f (1) = 0. Because, at point 1, we have β − = 3π/2, α − = π/2, β + = −π/2, and α + = −3π/2, it follows that α − = α + + 2π and β − = β + + 2π. The function f is continuous at 1 and so the fact that 1 is isogonal can be proved, with a little effort, by means of Ostrowski’s theorem (mentioned in the proof of Proposition 2). Corollary 1. Let f map D conformally onto G, and assume that G is starlike with respect to f (0) = 0. If ζ ∈ T and f (ζ) 6= ∞, then 1(ζ) ≤ π. Proof. We write β 1(t) = lim r→1 arg f (re it ). Then β 1(t) exists for all t and β 1 is an increasing function [Po, p. 66]. Therefore, β1 ≤ α+ ≤ β + ≤ β1 + π and

β 1 + π ≤ α − ≤ β − ≤ β 1 + 2π.

Hence, by (3.3), we obtain 1(ζ) ≤ π.

4. Proofs of Theorem 3 and Theorem 2 Proof of Theorem 3. (a) First we recall [Po2, p. 52] that Z 2π it i e +z arg(f (e it ) − a) dt log(f (z) − a) = log|f (0) − a| + 2π 0 e it − z if f maps D conformally onto G and a ∈ / G. Therefore, if we apply this formula to the functions f and g(z) = z with the point a = f (ζ), ζ ∈ T, and afterwards take its imaginary part, we obtain the Poisson representation


arg

1 f (z) − f (ζ) = z−ζ 2π

Z E(ζ)

f (e it ) − f (ζ) 1 − |z|2 dt, z ∈ D. arg it 2 |e − z| e it − ζ

303

(4.1)

Now the proof is a consequence of a general fact about Poisson integrals. For the sake of completeness we will present it. Let q(z) = Im h(z), z ∈ D ∪ E(ζ) with h the function defined in (1.4), and assume that β = lim supη→ζ, η∈E(ζ) q(η) is finite. Given ε > 0, there exists 0 < δ < 1 such that q(η) < β + ε

for all η ∈ E(ζ), |η − ζ| < δ.

Now we decompose the integral in (4.1) as the sum of two integrals over the sets η ∈ E(ζ), |η − ζ| > δ, and η ∈ E(ζ), |η − ζ| ≤ δ, respectively. Estimates of each term give δ (1 − |z|2 ) c + β + ε for z ∈ D and |z − ζ| < . δ2 2 Hence (4.2) implies that lim sup q(z) ≤ β. q(z) ≤

z→ζ, z∈D

(4.2)

(4.3)

The reverse inequality in (4.3) is almost trivial because q(ξ) is the radial limit of q(rξ) as r → 1 in each point ξ ∈ E(ζ). An analogous argument gives lim inf q(z) = lim inf q(η),

z→ζ, z∈D

η→ζ, η∈E(ζ)

so (2.4) holds. (b) If f is continuous at ζ then we obtain from (3.3) and Proposition 1 that 1(ζ) ≤ max[β C − (α C − 2π), β C − α C + π, β C + 2π − (α C − 2π) − π, β C + 2π − α C ] = β C − α C + 3π, which proves the right-hand inequality (3.4) and thus assertion (2.5). The left-hand inequality (3.4) is immediate from (2.2). (c) Finally, we construct an example where equality on the right side of (2.5) holds for C = [0, ζ). Example 2. Consider the parabolas P ± = { t ± it 2 : 0 ≤ t < +∞ } and let H be the domain between P + and P − . Let 0 < ε n < 2−n and A n = { w : 2−n < |w| < 2−n + ε n }. Then [ [ [(A n \ H¯ ) ∪ (P − ∩ A n )] ∪ [(A n \ H¯ ) ∪ (P + ∩ A n )] G=H∪ n even

n odd

is a simply connected domain. It has a cusp at 0 and moreover narrow annular corridors whose entrances accumulate at 0 and that go almost completely around 0 in the positive (n even) and negative (n odd) directions. Let f map D conformally onto G such that f (1) = 0. By (2.4) and (3.5), our construction gives


304

π 3π − −2π − = 3π. 1(1) = 2π − 2 2 Now { f (r) : 0 < r < 1 } has, up to multiplicative bounds, the same distance from the two parts { f (e it ) : 0 < t < π } and { f (e it ) : −π < t < 0 } of the boundary (see e.g. [PoRo, Thm. 3.3]). If ε n tends to zero rapidly enough, it follows that f ([0, 1)) cannot cut infinitely many segments P − ∩ A n or P + ∩ A n . So { f (r) : 1 − ε < r < 1 } ⊂ H for some ε > 0. Thus arg f (r) → 0 as r → 1. As in Proposition 1, the hypothesis of continuity of f at ζ is essential in (2.5). To prove that we make a little modification of Example 2; we will sketch the idea. Consider P ± (1) = { t ± it 2 : 0 ≤ t < 1 } and let H be the domain between P + (1) and P − (1). The domain G is defined as before, but now A n is always a narrow corridor that starts in { w : 2−n < |w| < 2−n + ε n } ∩ P + (1) with A n ⊂ { z : |z| ≤ 2 } and that goes around H at least n times in the positive direction. In this situation one actually has α − = 0 and β − = +∞, and the radial oscillation is zero. Proof of Theorem 2. By (3.5) and by our hypothesis (see (2.3)), we have arg

f (e it ) − f (ζ) → β − θ − π/2 e it − ζ

as e it → ζ, e it ∈ E(ζ).

Now the representation formula (4.1) and the properties of the Poisson kernel give (ζ) exists when z → ζ (z ∈ D); hence 1(ζ) = 0. that lim arg f (z)−f z−ζ Conversely, if 1(ζ) = 0 then Theorem 3 implies that there exists lim arg

f (e it ) − f (ζ) =α e it − ζ

as e it → ζ, e it ∈ E(ζ).

Now considering t → θ ± and (3.5), we conclude that α+ = β +,

α− = β −,

α = β + − (θ + π/2) = β − − (θ + 3π/2).

From this it follows that β − = β + + π. Hence (see (2.3)), ∂G has a tangent at f (ζ).

5. Proof of Theorem 4 We say that f is “twisting” at ζ ∈ T [Po2, p. 141] if αR (ζ) = −∞ and βR (ζ) = +∞, which implies that 1(ζ) = ∞ (recall that αR = α [0,ζ ] in (3.2)). The McMillan twist theorem states that f is isogonal or twisting at almost all ζ ∈ T. Hence we may assume that f is isogonal at ζ. Let 3 denote the linear measure in C [Po2, p. 129]. (i) First we consider the case 0 < 1(ζ) < π. It follows from (1.5) and (3.5) that (5.1) osc arg[f (z) − f (ζ)] ≤ π + 1(ζ) < 2π. z→ζ, z∈D

Hence Theorem 1 shows that f is continuous at ζ with at most countably many exceptions. Thus it suffices to show that 3(L) = 0, where L = { ζ ∈ T : f is isogonal and continuous at ζ and 0 < 1(ζ) < π }.


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If E is an arbitrary subset of C and w ∈ E, let TE (w) denote the union of all rays { w + te iλ : 0 ≤ t < ∞ } such that there exist wn ∈ E with wn → w and arg(wn − w) → λ as n → ∞. The Kolmogoroff–Ver˘cenko theorem (see [Sa, Chap. 9; Po2, p. 127]) states that E = E 0 ∪ { w ∈ E : TE (w) is C or a half-plane or a full line }, where 3(E 0 ) = 0. Let (I n ) be the countable collection of all different intervals { e it : q < t < q 0 } with q, q 0 rational numbers. Fix n ≥ 1 and put \ [ { f (se it ) : t ∈ I n }. (5.2) Bn = r arg ζ }, Uδ− (ζ) = { z ∈ Uδ (ζ) : arg z < arg ζ }. The one-sided cluster sets at ζ are defined by \ f (Uδ± (ζ)) ⊂ ∂G C ± (f, ζ) =

(7.1)

δ>0

and the total cluster set by C(f, ζ) = C + (f, ζ) ∪ C − (f, ζ). The prime end fˆ(ζ) of G is called symmetric if C + (f, ζ) = C − (f, ζ). The Collingwood symmetry theorem (see [Po2, p. 38]) states that there are at most countably many nonsymmetric prime ends. The symmetry of prime ends has some curious consequences. Theorem 7. Let f map D conformally onto G, let ζ, ζ 0 ∈ T (ζ 6= ζ 0 ), and let E be a continuum. Assume that: (i) the angular limits f (ζ), f (ζ 0 ) exist and f (ζ) ∈ E, f (ζ 0 ) ∈ E; (ii) there exists a neighborhood V of ζ such that f (V ) ∩ E = ∅; and (iii) the prime end fˆ(ζ) is symmetric. Then C(f, ζ) ⊂ E ∩ ∂G. ¯ Let z 0 = / V. Proof. We may assume that V = { z ∈ D : |z − ζ| < ρ } and ζ 0 ∈ 0 (1 − ρ)ζ and 00 = [z 0 , ζ). Now take a circular arc 0 from z 0 to ζ such that 0 ⊂ D \ V. By (i) and (ii) we have / E, f (z 0 ) ∈

f (ζ 0 ) ∈ E.

Hence there exists a first point z1 ∈ 0 where f (0) meets E (it is possible that z1 = ζ 0 ). We consider the open arc 01 between z 0 and z1 . Then the Jordan arc C = C 0 ∪ C1, where C 0 = f (00 ) and C1 = f (01 ), satisfies C ∩ E = ∅ and C¯ ∩ E = {f (ζ), f (z1 )}. The components of C \ E are simply connected domains. By (ii) there exists a component H of C \ E such that f (V ) ⊂ H. The Jordan arc C lies in H, so C is a crosscut of H and thus H \ C has exactly two components H + and H − . If we write V + = { z ∈ Uρ (ζ) : arg z > arg ζ }, V − = { z ∈ Uρ (ζ) : arg z < arg ζ }, then we may assume that

f (V ± ) ∩ H ± 6= ∅.

Since f is univalent, we have C 0 ∩ f (V ± ) = ∅, by (ii). Therefore,

C1 ∩ f (V ± ) = ∅,

(E ∪ C ) ∩ f (V ± ) = ∅

and so f (V ± ) ⊂ H ± , which implies

E ∩ f (V ± ) = ∅

308


C ± (f, ζ) ⊂ f (V ± ) ∩ ∂G ⊂ H ± ∩ ∂G. Using (iii) and the fact that C ∩ ∂G = ∅, we obtain C(f, ζ) = C + (f, ζ) ∩ C − (f, ζ) ⊂ (H + ∩ H − ) ∩ ∂G ⊂ (C ∪ E ) ∩ ∂G = (C ∩ ∂G) ∪ (E ∩ ∂G) = E ∩ ∂G. Corollary 2. Suppose that fˆ(ζ) is symmetric and that f is not continuous at ζ. Let H be a domain and A a Jordan arc such that A \ {w} ⊂ H and that A begins at w = f (ζ). If f (Uδ (ζ)) ⊂ C \ H for some δ > 0 then f (D) = G ⊂ C \ H. Proof. Suppose that f (Uδ (ζ)) ⊂ C \ H but there exists z ∈ D with f (z) ∈ H. We may assume that A ends at f (z). Let w1 be the last point where A meets C \ f (D). By [Po2, p. 29] we know that w1 = f (ζ 0 ) for some ζ 0 6= ζ ; it is possible that w1 = w. We consider the subarc A1 of A between w and w1 . Now E = A1 satisfies all hypotheses of Theorem 7, so C(f, ζ) ⊂ E ∩ ∂G ∩ f (Uδ (ζ)) = {w} and f is continuous [Po2, p. 35] at ζ, which contradicts our assumption. Let H be a connected component of C \ G. The angular limit w = f (ζ) is called a transition point of G with respect to H if there exists a Jordan arc A that begins at w and such that A \ {w} ⊂ H. It is easy to see that is equivalent to saying there exists a continuum K (i.e., a compact connected set with more than one point) with K ⊂ H ∪ {w}. A point b ∈ ∂G is accessible from G if there exists a Jordan arc L that lies in G except for the endpoint b. We shall now study the relationship between the concepts of transition point, continuity point, and symmetric prime end. Proposition 3. Let f map D conformally onto G, and let w = f (ζ) be a transition point of G with respect to H. Let H˜ = { b ∈ ∂G : there exists a continuum F ⊂ C \ G, ∂H * F, and b, w ∈ F }. Assume that f is not continuous at ζ and that fˆ(ζ) is symmetric. Then ∂H ∪ H˜ ⊂ C(f, ζ).

(7.2)

Proof. First we will show that ∂H ⊂ C(f, ζ). Assume this inclusion is not true; then there exists b ∈ ∂H with b ∈ / C(f, ζ). Choose a disc V of center b such that V ∩ C(f, ζ) = ∅. Since b ∈ ∂H ⊂ ∂G, we can select two points c∈G∩V

and

c 0 ∈ H ∩ V.

Let A ⊂ H be the arc of transition and let a 6= w be the other endpoint of A. Consider an arc L ⊂ H joining a with c 0 . Now take c 00 as the first point of [c, c 0 ] ∈ V that meets G c . The set E = A ∪ L ∪ [c 00 , c 0 ] allows us to apply Theorem 7, and we conclude that C(f, ζ) ⊂ E ∩ V c ∩ ∂G ⊂ {w}, in contradiction to the hypothesis that f is not continuous at ζ. Thus ∂H ⊂ C(f, ζ).


309

Assume now that there exists a b ∈ H˜ and that b ∈ / C(f, ζ). Consider the continuum E = F ∪ [b, c 00 ], where F is as in the definition of H˜ and where c 00 is chosen as before. It follows from Theorem 7 that C(f, ζ) ⊂ (F ∩ C(f, ζ)) ∪ ([c 00 , b] ∩ C(f, ζ)) ⊂ F. But this contradicts the facts that ∂H ⊂ C(f, ζ), b ∈ H˜ , and ∂H * F, so H˜ ⊂ C(f, ζ) and therefore (7.2) is proved. We have the following consequence. Corollary 3. Except for at most countable many points ζ ∈ T, if f (ζ) is a transition point then f is continuous at ζ. Proof. By the Collingwood symmetry theorem and the fact that there are only countably many components H of C \ G, it is enough to show that in each component H there is at most a transition point w = f (ζ), with fˆ(ζ) symmetric and f not continuous at ζ, where for this point ζ the set f −1(ζ) is a singleton. Assume that w = f (ζ) and w 0 = f (ζ 0 ), w 6= w 0 , are two transition points with respect to H. Then let B be an arc joining the endpoints of the corresponding arcs A, A0 and put E = A ∪ B ∪ A0 . We can apply Theorem 7 to conclude that f must be continuous at ζ and ζ 0 . If w = w 0 and ζ 6= ζ 0 , we can take E = {w} to infer that f would be continuous at ζ ; hence ζ = ζ 0 . Proof of Theorem 1. By the Collingwood symmetry theorem, we may assume that fˆ(ζ) is symmetric. It follows from (2.1) that there is a sector H(ζ) of vertex f (ζ) with f (Uδ (ζ)) ⊂ C \ H(ζ) for some δ = δ(ζ) > 0. If f is continuous at ζ then we are finished. Otherwise, G ⊂ C \ H(ζ) by Corollary 2. We conclude that f (ζ) is a transition point (with A the midline of H(ζ)), and Corollary 3 implies that f is continuous at ζ with at most countably many exceptions. In order to show that the constant 2π is best possible we present an example in which there are uncountable many points where the oscillation (2.1) equals 2π and where the function is not continuous. Let K be the usual Cantor set and let Q be the square { x + iy : 0 < x < 1, 0 < y < 2 }. Now take any conformal map f from D onto the simply connected domain Q \ (K + i[0, 1]). For each point w = a + i (a ∈ K) there exists a point ζ ∈ T such that f (ζ) = w [Po2, p. 29]; let us denote by E the set of such points. Now we have finished because the set E is uncountable and in each point ζ ∈ E one has lim inf Im f (z) = 0 z→ζ

and

osc

z→ζ, z∈D

arg[f (z) − f (ζ)] = 2π.

8. Additional Topological Results In the proof of Corollary 3, we have seen what occurs if there are two transition points with respect to the same component H. It is interesting to study what happens if we relax this hypothesis and assume that one of those points is accessible only from G. Before stating the next result, we will need the following definition:

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A continuum E is indecomposable if it cannot be written as the union of two proper subcontinua (see [Na, pp. 7–14] for more information). Theorem 8. Let f map D conformally onto G, and let ζ ∈ T. Assume that: (a) the radial limit f (ζ) = w exists and is a transition point with respect to the ¯ component H of C \ G; ˆ (b) the prime end f (ζ) is symmetric; (c) f is not continuous at ζ ; and (d) there exists a point w 0 ∈ ∂H (w 0 6= w) that is accessible from G. Then C(f, ζ) = ∂H and ∂H is indecomposable. Proof. We already know that ∂H ⊂ C(f, ζ).

(8.1)

Assume that ∂H is decomposable; then there exist continua A, B such that ∂H = A ∪ B

(A 6= ∂H, B 6= ∂H, and w 0 ∈ A).

(8.2)

First we will prove the following. Claim: If b ∈ ∂H is accessible from H then b ∈ B. In order to prove the claim, assume that b ∈ / B; thus b ∈ A by (8.2). There exists a curve C ⊂ H ∪ {w, b} joining w and b. Then E = A ∪ C ⊂ C \ G is a continuum that contains two accessible points w and w 0 , so Theorem 7 entails C(f, ζ) ⊂ E ∩ ∂G = (A ∩ ∂G) ∪ (C ∩ ∂G) = A ∪ {w}.

(8.3)

Recall that w ∈ C(f, ζ). Therefore, if w ∈ / A then (8.3) implies C(f, ζ) = {w}, which is impossible by (c). If w ∈ A then (8.3) and (8.2) imply that C(f, ζ) ⊂ A ∪ {w} = A Ã ∂H, which contradicts (8.1). Therefore b ∈ B and the claim is proved. From the claim we see that the set of points accessible from H lies in B. Since this is a dense subset of ∂H, it follows that B = B¯ = ∂H, which contradicts (8.2). Therefore ∂H is indecomposable. Because ∂H is a continuum that contains f (ζ) and another point w 0 ∈ ∂H, accessible from G and with ∂H ⊂ C(f, ζ), the minimality property of cluster sets (see Proposition 2.22 of [Po2, p. 38]) implies that ∂H = C(f, ζ). It is a natural question to ask whether there exists a conformal map under the hypotheses of the previous theorem. The answer is yes, as the following result shows. Proposition 4. There exists a continuum E with the following properties: (a) C \ E = G ∪ H, where G and H are disjoint domains and G is simply connected; (b) ∂G = ∂H = E;


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(c) E is indecomposable; (d) there is a point ζ ∈ T where the conformal map from D onto G has a radial limit f (ζ) = w and C + (f, ζ) = C − (f, ζ) = E; (e) w is a transition point of f with respect to H. Proof. To fix ideas, let us take a k = (−1) k+1π/2 k+1 and put z k = e ia k . Also, we consider a sequence of points (wk ) such that, for all k ≥ 1, |wk | > 1,

Im w 2k+1 > Im w 2k+3 > 0,

Im w 2k < Im w 2k+2 < 0

with lim k→∞ wk = 1. Write G0 = D and H 0 = { z : |z| > 1 }. To start the construction, select ε1 > 0 such that z 3 does not belong to the arc in T between z1e iε1 and z1e−iε1. Then consider an open strip B1 with B1 ⊂ H 0 of width 2ε1 joining the arc (z1e iε1, z1e−iε1 ) in T with the point w 2 and satisfying, moreover, B1 ∩ [1, ∞) = ∅ and

d(x, ∂B1 ) < 1 for all points x ∈ T.

Now we put G1 = G0 ∪ B1 and H1 = C \ G1 . Assume that we have already constructed the domains Gn and H n . Consider an open strip B n+1 with B n+1 ⊂ H n of width 2ε n+1, for suitably chosen ε n+1 < ε n , joining the arcSI n+1 = / s≤n Bs (z n+1e iε n+1 , z n+1e−iε n+1 ) with the first point wk (k ≥ n+2) such that wk ∈ and with the additional restrictions B n+1 ∩ [1, ∞) = ∅, d(x, ∂B n+1) < ε n

z n+3 ∈ / I n+1 and for x ∈ ∂Gn .

(8.4)

We must show that this construction is always possible. One way to see this is to consider a continuous conformal map from D onto H n , mapping 0 to ∞, and then apply its uniform continuity. As before, weSdenote Gn+1 = Gn ∪ B n+1 and H n = C \ Gn+1. Then the do∞ ¯ A conformal map from D onto G with mains are G = n=1 Gn and H = C \ G. f (1) = 1 gives our example. The point 1 is a transition point, since [0, 1) ⊂ G and (1, ∞) ⊂ H. The prime end fˆ(1) is symmetric and C(f, 1) = E because members of the family of crosscuts (z n e iε n , z n e−iε n ) are alternatively above and below the real axis and, by (8.4), each point of E is an accumulation point in ∂B n ; hence (d) holds. The continuum E = ∂H = ∂G is indecomposable. A direct proof is possible, but it is enough to apply Theorem 8 because E has a dense set of accessible points from the component H.

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[Du] J. Dufresnoy, Sur les fonctions méromorphes et univalentes dans le cercle unité, Bull. Sci. Math. 69 (1945), 21–36. [Fa] K. Falconer, The geometry of fractal sets, Cambridge Univ. Press, Cambridge, 1985. [Ha] D. H. Hamilton, Conformal distortion of boundary sets, Trans. Amer. Math. Soc. 308 (1988), 69–81. [Mc] J. E. McMillan, Boundary behaviour of a conformal mapping, Acta Math. 123 (1969), 43–67. [Na] S. B. Nadler, Jr., Continuum theory, an introduction, Marcel Dekker, New York, 1992. [Po1] Ch. Pommerenke, Univalent functions, Vandenhoeck & Ruprecht, Göttingen, 1975. [Po2] , Boundary behaviour of conformal maps, Springer-Verlag, Berlin, 1992. [PoRo] Ch. Pommerenke and S. Rohde, The Gehring–Hayman inequality in conformal mapping, Quasiconformal mappings and analysis (P. Duren, J. Heinonen, B. Osgood, B. Palka, F. W. Gehring, eds.), pp. 309-319, Springer-Verlag, New York, 1998. [Sa] S. Saks, Theory of the integral, 2nd ed., Hafner, New York, 1964. [Wa] S. E. Warschawski, On the preservation of angles at a boundary point in conformal mapping, Bull. Amer. Math. Soc. 42 (1936), 675–680. [Wo] J. Wolff, Sur la conservation des angles dans la représentation conforme d’un domaine au voisinage d’un point frontière, Comptes Rendus 200 (1935), 42–43.

J. J. Carmona Departament de Matemàtiques Universitat Autònoma de Barcelona 08193 Bellaterra (Barcelona) Spain

Ch. Pommerenke Fachbereich Mathematik MA 8-2 Technische Universität D-10623 Berlin Germany

[email protected]

[email protected]