A condition of quasiconformal extendability

Y. Gotoh and M. Taniguchi

1999/2

Recently, Heinonen and Koskela showed, as a corollary of their deep result, the following extension theorem.

Proposition 1 ([3],4.2 Theorem) Suppose that f is a quasiconformal map of the complement of a closed set E in R^{n} into R^{n}, n ≥ 2, and suppose that each point x ∈ E has the following property: there is a sequence of radii r_{j}, r_{j} → 0 as j → ∞, such that the annular region B(x, ar_{j}) − B(x, r_{j}/a) does not meet E for some a > 1 independent of x. Then f has a quasiconformal extension to R̂^{n} = R^{n} ∪ {∞}. Moreover, the dilatation of the extension agrees with the dilataiton of f.

There, they remarked that this result may be new even for conformal maps in the plane. So it is noteworthy to give a different proof of a more general extension theorem on 2-dimensional quasiconformal maps of the plane based on some classical results in the function theory.

We begin with the following definition, which weakens the condition in the above theorem to a conformally invariant one.

Definition We say that a closed set E in the complex plane is annularly coarse if each point x ∈ E has the following property: there is a sequence of mutually disjoint nested annuli {R_{k}}_{k=1}^{∞} such that the modulus m(R_{k}) of R_{k} satisfies

m(R_{k}) ≥ c

with a positive c. Here we say that a sequence of annuli {R_{k}}_{k=1}^{∞} is nested if every R_{k} (k > 1) separates R_{k−1} from x.

Also note that the positive constant c can depend on x.

Now we will prove the following

Theorem 2 Suppose that f is a quasiconformal map of the complement of a closed set E in the complex plane C into C and suppose that E is annularly coarse. Then f has a quasiconformal extension to Ĉ. Moreover, the dilatation of the extension agrees with the dilataiton of f.

1 Known facts and basic lemmas

In 2-dimensional case, we have the following

Proposition 3 Let E be a compact set in C. Then the following conditions are mutually equivalent.

1. Every conformal map of D = C − E is the restriction of a Möbius transformaiton.

2. Every quasiconformal map of D = Ĉ − E has a quasiconformal extension to the whole Ĉ.

3. For every relatively compact neighborhood U of E, every quasiconformal map of U − E has a quasiconformal extension to U.

Proof. First assume the condition 1) and take any quasiconformal map f of D = C − E. Here we may assume that f(∞) = ∞. Let μ be the Beltrami coefficient of f^{−1} on f(C − E). Set μ = 0 on C − f(C − E), and we have a quasiconformal map g of Ĉ with the complex dilatation μ (cf. [1] and [4]). Then, g ∘ f has vanishing complex dilatation on C − E, and hence the assumption implies that it is a Möbius transformation T. Thus f can be extended a quasiconformal map g^{−1} ∘ T of the whole Ĉ.

Next assume the condition 2) and take a relatively compact neighborhood U of E and a quasiconformal map f of U −E arbitrarily. Since E is compact, the famous extention theorem ( [6] II Theorem 8.1) gives a neighborhood V of E in U and a quasiconformal map g of Ĉ−E which coincides with f on V −E. Then the assumption implies that g can be extended to a quasiconformal map of Ĉ, which clearly gives a quasiconformal extension of f to U.

Finally, assume the condition 3) and take any conformal map f of D = C − E. Then f can be extended to a quasiconformal map g of C. Hence if E has vanishing area, then this g is actually conformal, and hence is a Möbius transformation. If not, consider the extremal (horizontal) slit map h of C − E. Then h should be extended a quasiconformal map of C. But this is impossible, for C−f(C−E) has vanishing area by Koebe’s uniformization theorem.

Remark Koebe’s uniformization theorem asserts that every planar domain Ω can map conformally onto the complement of some union of horizontal slits and points whose total area vanishes. And an example of such univalent holomorphic maps are the extremal slit maps. (See for instance, [5].)

As a condition which assures these extension properties, we know the following; we say that a compact set E has absolutely vanishing area if C − g(C−E) has vanishing area for every univalent holomorphic map g of C−E.

Actually, the following fact is classically well-known.

Lemma 4 Let E be a compact set in C with absolutely vanishing area. Then every conformal map of D = C − E is the restriciton of a Möbius transformaiton.

Proof. Ahlfors and Beurling ([2] showed that D belongs to O_{AD} if and only if E has absolutely vanishing area, which is also equivalent the condition 1) in Proposition 3. (Also see [8] VI and [7] I §2.)

Now, it is clear from the definition that an annularly coarse compact set is totally disconnected (or even absolutely disconnected). Furthermore, we see the following

Lemma 5 Every annularly coarse compact set E has absolutely vanishing area.

Proof. It suffices to show that E has vanishing area. For this purpose, fix a point a ∈ E arbitrarily. Then there is a sequence of mutually disjoint nested annuli R_{k} such that m(R_{k}) ≥ c for every k with a positive constant c.

Let d_{k} be the diameter of the bounded component F_{k} of C−R_{k}. Then we can find a positive constant η (depending only on c) such that R_{k} ∩B(a, 2d_{k}) contains a ball B_{k} with radius ηd_{k}, where and in the sequel, we set B(a, r) = {|z − a| ≤ r}.

For the sake of convenience, we include a direct proof of this assertion. Let A_{k} be the distance between F_{k} and F′_{k} = C − (F_{k} ∪ R_{k}), and z_{k} ∈ F_{k} and z′_{k} ∈ F′_{k} satisfy |z_{k} − z′_{k}| = A_{k}. Also take two points w_{k}, w′_{k} ∈ F_{k} satisfing |w_{k} − w′_{k}| = d_{k}. Further we may assume that |w_{k} − z_{k}| ≥ d_{k}/2. Then R_{k} separates w_{k} and z_{k} from z′_{k} and ∞, and hence

T_{k}(z) = −\frac{z − z_{k}}{w_{k} − z_{k}}

maps R_{k} onto a region admissible to the extremal problem of Teichmüller (see [1]). Under the notation of [1], we have

c < M(R_{k}) ≤ \frac{1}{2π} log Ψ(|(z_{k} − z′_{k})/(z_{k} − w_{k})|)
≤ \frac{1}{2π} log Ψ(2A_{k}/d_{k}),

and since log Ψ(x) → 0 as x → 0, we can find a positive η = η(c) such that

A_{k} ≥ η2d_{k}.

for every k, which gives the assertion.

Now set r_{k} = 2d_{k} for every k, and we have

\frac{Area(E ∩ B(a, r_{k}))}{Area(B(a, r_{k}))} < 1 − \frac{π(ηd_{k})^{2}}{πr_{k}^{2}} = 1 − \frac{η^{2}}{4} .

This implies that a is not a density point of E. Since a is arbitrary, we conclude that the area of E vanishes.

2 Proof of Theorem 2

First fix an annularly coarse closed set E arbitrarily. For every n, set E_{n} = E ∩ B(0, n). Then every E_{n} is compact and the assumption implies that there is a neighborhood U_{n} of E_{n} such that the boundary of U_{n} is a comapct set in D = C − E.

Let f be a quasiconformal map of C−E. Then Proposition 3 implies that f can be extended uniquely to a quasiconformal map, say f_{n}, of C − E ∩ U_{n} and the maximal dilataiton of f_{n} is the same as that of f by Lemma 5.

Since K-quasiconformal maps are sequencially compact, we conclude that f has a desired extension to the whole C.

References

[1] L. Ahlfors Lectures on Quasiconformal mappings Van-Nostland, 1966.

[2] L. Ahlfors and A. Beurling Conformal invariants and function-theoretic null-sets Acta Math., 83, (1950), 101-129.

[3] J. Heinonen and P. Koskela Definition of quasiconformality Invent. math., 120, (1995), 61–79.

[4] Y. Imayoshi and M. Taniguchi An Introduction to Teichmüller Spaces Springer-Verlag, 1991.

[5] Y. Kusunoki Theory of Abelian integrals and its applications to conformal mappings Mem. Coll. Sci. Univ. Kyoto (Math.), 32, (1959), 235–258; 33. (1961), 429–433.

[6] O. Lehto and K. Virtanen Quasiconformal Mappings in the Plane Springer-Verlag, 1973.

[7] L. Sario and M. Nakai Classification Theory of Riemann Surfaces Springer-Verlag, 1970.

[8] L. Sario and K. Oikawa, Capacity Functions Springer-Verlag, 1969.