Conformally Invariant Variational Problems. - SAM

neighborhood of ∂D 2 (the boundary condition (V.17) is clearly
satisfied with this choice). Observe that the restriction of dz =
d(re iθ ) to ∂Br(0) 2 is equal to ie iθ rdθ = izdθ. For this choice
of X, (V.23) becomes
(∫
)
lim I α(θ) H(u)(z)z 2 dθ = 0 . (V.24)
r→1− ∂Br 2(0) Let z 0 ∈ D 2 and choose for α := G(θ,z 0 ) the Poisson Kernel
such that, for any harmonic function f
f(z 0 ) =
∫ 2π
0
G(θ,z 0 ) f(e iθ ) dθ .
Since H(u)isholomorphic,H(u)(z)z 2 isholomorphicandhence
harmoniconD 2 . ThisisalsoofcoursethecaseforH(u)(rz)r 2 z 2
for any 0 < r < 1. We then obtain from (V.24)
(∫ 2π
0 = lim I G(θ,z 0 ) H(u)(re iθ )r 2 e 2iθ dθ
r→1−
0
= lim
r→1− I(H(u)(rz 0)r 2 z 2 0) = I(H(u)(z 0 )z 2 0) .
This holds for any z 0 and therefore the holomorphic function
H(u)(z)z 2 , whose imaginarypart vanishes, has to be identically
equal to a real constant :
H(u)(z)z 2 ≡ c .
Since H(u) is holomorphic without pole at the origin, c has to
be equal to zero. We have then established that H(u) ≡ 0 on
D 2 which concludes the proof of proposition V.3. ✷
We will now concentrate our efforts for proving the existence
of a minimizer of E in C(Γ). While taking a minimizing sequence
u n we have already explained the risk for the boundary
requirement u n ∈ C 0 (∂D 2 ,Γ) not to be preserved at the limit
do to the lack of Sobolev embedding (V.1).
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