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5.16. Lemma. Let p > 1 and α ∈Ê\(π/2)�. If v ∈ Xq (Σ), 1/p+1/q = 1, belongs to the cokernel<br />

of<br />

∂α : X 1,p (Σ) → X p ∈Ê<br />

(Σ),<br />

then v is smooth on Σ \ {0}, it solves the equation ∂v − αv = 0 with boundary conditions<br />

v(s) ∈Ê, v(−s) ∈ iÊ∀s > 0,<br />

(102)<br />

v(s + i) ∀s ∈Ê,<br />

and the function (T v)(ζ) = 2ζv(ζ2 ) is smooth on Cl(À+ ) ∩�1. In particular, v(z) = O(|z| −1/2 )<br />

and Dv(z) = O(|z| −3/2 ) for z → 0.<br />

Proof. Since v ∈ Xq (Σ) annihilates the image of ∂α, there holds<br />

�<br />

Re 〈v(z), ∂u(z) + αu(z)〉ds dt = 0, (103)<br />

Σ<br />

for every u ∈ X1,p (Σ). By letting u vary among all smooth functions in X1,p (Σ) which are<br />

compactly supported in Σ \ {0}, the regularity theory for weak solutions of ∂ (the analogue of<br />

Theorem 5.4) and a bootstrap argument show that v is smooth on Σ\{0} and it solves the equation<br />

∂v − αv = 0 with boundary conditions (102). There remains to study the regularity of v at 0.<br />

Set w(ζ) := (T v)(ζ) = 2ζv(ζ2 ). By (89), the function w is in Lq (À+ ∩�1). Let ϕ ∈<br />

C∞ c (Cl(À+ ) ∩�1) be real onÊ+ and purely imaginary on iÊ+ . Then the function u defined by<br />

u(ζ2 ) = ϕ(ζ) belongs to X1,p (Σ), and by (103) we have<br />

�<br />

0 = Re 〈v, ∂u + αu〉dsdt = 4Re<br />

Σ<br />

The above identity can be rewritten as<br />

Re<br />

�À+ ∩�1<br />

�À+ ∩�1<br />

= Re<br />

|ζ 2 |〈 1 1<br />

w(ζ), ∂ϕ(ζ) + αϕ(ζ)〉dσdτ<br />

2ζ 2ζ<br />

�À+ ∩�1<br />

〈w(ζ), ∂ϕ(ζ)〉dσdτ = −Re<br />

〈w(ζ), ∂ϕ(ζ) + 2αζϕ(ζ)〉dσdτ.<br />

�À+ ∩�1<br />

〈2αζw(ζ), ϕ(ζ)〉dσdτ,<br />

so w is a weak solution of ∂w = 2αζw on Cl(À+ )∩�1 with real boundary conditions. Since w is in<br />

L q (À+ ∩�1), Lemma 5.7 implies that w is in W 1,q (À+ ∩�1). In particular, w is square integrable<br />

onÀ+ ∩�1, and so is the function<br />

f(ζ) := e −αζ2 /2 w(ζ).<br />

The function f is anti-holomorphic, it takes real values onÊ+ and on iÊ+ , so by a double Schwarz<br />

reflection it can be extended to an anti-holomorhic function on�1\{0}. Since f is square integrable,<br />

the singularity 0 is removable and f is anti-holomorphic on�1. Therefore<br />

is smooth on Cl(À+ ) ∩�1, as claimed.<br />

(T v)(ζ) = w(ζ) = e αζ2 /2 f(ζ)<br />

We can finally prove the following Liouville type result:<br />

5.17. Proposition. If 0 < α < π/2, the operator<br />

is an isomorphism, for every 1 < p < ∞.<br />

∂α : X 1,p (Σ) → X p (Σ)<br />

65

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