Conformally Invariant Variational Problems. - SAM

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Proof of Lemma II.4
Since u(0) = 0 we can define v(z) = u(z)
z
which is holomorphic
and hence harmonic. On the disc D 2 we have
∆ | v | 2 = 2 | ∇v | 2 0
(II.25)
Moreover since u(D 2 )CD 2
| u || on D 2 thus
| v | 2 1 on ∂D 2 (II.26)
Combining (II.25), (II.26) together with the maximum principle
gives
| v | 2 1 on D 2
which implies (II.24) and Lemma II.8 is proved.
Proof of theorem II.2
Since u is a conformal positive diffeomorphism of the disc D 2
we obtain from proposition (II.1) that u is holomorphic. u −1
satisfies also the positivity and conformality condition therefore
it is also holomorphic.
First we assume that u(0) = 0 (and hence u −1 (0) = 0) we
can then apply Lemma II.8 to both u and u −1 . This gives
∀z ∈ D 2 | z |=| u◦u −1 (z) || u −1 (z) || z |
This series of inequalities is then an equality and this implies
∀z ∈ D 2 | u(z) |=| z | (II.27)
v(z) := u(z)/z is holomorphic and has a constant modulus.
Therefore there exists θ ∈ R such that
u(z)
z
= v(z) ≡ e iθ
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