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The radial stretch is quasiconformal with K equal to max of alpha and one over alpha

Statement

Assume the Axiom of Choice. For α>0 define fα(0)=0 and fα(z)=∣z∣α−1z for z≠0. Verify:

(a) fα is a homeomorphism of C onto itself, orientation-preserving, with inverse f1/α, and belongs to Wloc1,2. For z≠0, (fα)z=α+12∣z∣α−1,(fα)zˉ=α−12∣z∣α−1zzˉ. Consequently μfα(z)=α−1α+1zzˉ(z≠0),∣μfα∣=∣α−1∣α+1,Kfα=max⁡(α,1α), and fα is analytically and geometrically Kfα-quasiconformal.

(b) fα maps ∣z∣=r onto ∣w∣=rα and each ray onto itself. It is conformal exactly when α=1.

(c) The extremal-length distortion bound is sharp on every annular connecting family. For A(r,R) (Annuli in the complex plane) with 0<r<R<∞, let Γr,R be the paths joining its boundary circles. Then λ(fαΓr,R)=12πlog⁡Rαrα=αλ(Γr,R). If α≥1, then Kfα=α and this attains the upper extremal-length bound; if 0<α<1, then Kfα=1/α and the ratio α=1/Kfα attains the lower bound. The reciprocal modulus bounds are attained at the corresponding opposite endpoints (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K).

Facts & Assumptions

Given: Choice, α>0, the ACL/Sobolev convention, and the annulus path-family conventions.

[F1]

The map has polar form fα(reit)=rαeit. The function r↦rα is a strictly increasing homeomorphism of [0,∞) with inverse s↦s1/α, so fα is a homeomorphism with inverse f1/α.

[F2]

On C∖{0}, direct Wirtinger differentiation gives the derivatives in the Statement. In particular the real Jacobian is Jfα=∣(fα)z∣2−∣(fα)zˉ∣2=α∣z∣2α−2>0.

[F3]

On each horizontal or vertical line not passing through 0, fα is smooth. On the two coordinate lines through 0, its components are constant multiples of g(t)=sgn⁡(t)∣t∣α, which is absolutely continuous on compact intervals because g′(t)=α∣t∣α−1∈Lloc1 for α>0.

[F4]

The function is locally bounded, and its classical first partial derivatives off 0 are bounded by Cα∣z∣α−1. Since ∫∣z∣<R∣z∣2α−2 dA=2π∫0Rr2α−1 dr<∞ for α>0, they are locally square-integrable (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma). The excluded point 0 is null because it lies in boxes of arbitrarily small area (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included). The ACL characterization therefore gives fα∈Wloc1,2 and identifies these almost-everywhere classical derivatives with its weak derivatives (Absolute continuity on almost every coordinate line, The ACL characterisation of W1,p).

[F5]

The ratio ∣(fα)zˉ∣/∣(fα)z∣=∣α−1∣/(α+1)<1 off 0, and 1+∣α−1∣/(α+1)1−∣α−1∣/(α+1)=max⁡(α,1α). The modulus-distortion lemma gives the quadrilateral inequalities for analytic maps; together with orientation preservation this is the geometric definition (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K, Orientation-preserving homeomorphisms and the geometric definition of quasiconformality).

[F6]

For every finite round annulus, λ(Γr,R)=(2π)−1log⁡(R/r) and μ(Γr,R)=1/λ(Γr,R) (Extremal length of the rectangle and of the round annulus).

[F7]

At a point where the real derivative is invertible, the inverse-function theorem makes the map a local diffeomorphism; for a smooth local diffeomorphism its local-homology orientation multiplier is the sign of its determinant (The Euclidean inverse function theorem, Smooth orientation sign is the local integral homology multiplier, R-orientation of a topological manifold).

Proof

technique · use the polar form for the homeomorphism and annulus images, compute the Wirtinger derivatives off the origin, establish local Sobolev regularity by ACL, then compare the resulting constants
1.1F1F2F7given

By [F1], fα is a homeomorphism with inverse f1/α. At every z≠0, [F2] gives a positive Jacobian, so the Euclidean inverse function theorem makes fα a local diffeomorphism there; the smooth-to-local-homology orientation lemma identifies its local orientation multiplier with this positive determinant sign. The local orientation sign of the homeomorphism is locally constant on connected C by Orientation-preserving homeomorphisms and the geometric definition of quasiconformality, so the sign at 0 is positive as well.

1.2F2F4givenalgebra

Write fα(z)=z∣z∣α−1 for z≠0. Using ∂z∣z∣=zˉ/(2∣z∣) and ∂zˉ∣z∣=z/(2∣z∣) gives (fα)z=∣z∣α−1+α−12∣z∣α−3zzˉ=α+12∣z∣α−1, (fα)zˉ=α−12∣z∣α−3z2=α−12∣z∣α−1zzˉ. Thus [F2] and [F4] provide the stated almost-everywhere derivatives and Wloc1,2 regularity; the point 0 is a null set.

2.1F4F5step 1.1givenalgebra

Since ∣(fα)z∣=(α+1)∣z∣α−1/2 and ∣(fα)zˉ∣=∣α−1∣∣z∣α−1/2, the Beltrami coefficient has constant modulus ∣α−1∣/(α+1). If α≥1, the quotient ∣μ∣=(α−1)/(α+1) gives Kfα=α; if 0<α<1, it gives Kfα=1/α. The analytic inequality holds almost everywhere; [F5] and step 1.1 then give geometric Kfα-quasiconformality. If α≠1, its ∂ˉ derivative is nonzero on C∖{0}, so it is not holomorphic; if α=1, it is the identity. This proves (a) and the conformality claim in (b).

3.1F1F6step 2.1givenalgebra∎

The polar formula in [F1] gives ∣fα(z)∣=∣z∣α and preserves the argument, so A(r,R) maps to A(rα,Rα) and its connecting family maps onto the target connecting family. By [F6], λ(fαΓr,R)=12πlog⁡Rαrα=α12πlog⁡Rr=αλ(Γr,R). The two cases in step 2.1 show this is the upper endpoint for α≥1 and the lower endpoint for 0<α<1. Taking reciprocals shows the corresponding modulus endpoint is also attained.

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