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Proof. We give a sketch of the proof, details are left to the reader.
First, given a sequence αn → 0 and associated solutions un: [0, αn]×Ì→T ∗ M of (74), one can
show that un → (c, 0) ∈ ΛT ∗ M uniformly. Here, it is again essential to make a case distinction for
the three cases of potential gradient blow-up, Rn = |∇un(zn)| → ∞, namely, modulo subsequence,
αn · Rn → 0 or → ∞ or → k > 0.
The most interesting case is αn ·Rn → k > 0 which is dealt with by rescaling vn = un(αn·, αn·)
as in the proof of Lemma 6.6.
For the converse, we need a Newton type method which is hard to implement for the shrinking
domains [0, α] ×Ìwith α → 0. Instead, we consider the conformally rescaled equivalent problem.
Let v(s, t) = u(αs, αt) and consider the corresponding problem for α → 0,
v: [0, 1] ×Ìα −1 → T ∗ M, ∂J,Hαv = 0,
π(v(0, t)) = c(αt), v(1, t) ∈ÇM ∀t ∈Ìα −1,
where Hα(t, ·) = αH(αt, ·) andÌα −1 =Ê/α −1�. The proof is now based on the Newton method
which requires to show that:
(a) for vo(s, t) = 0 ∈ T ∗ c(αt) M we have ∂J,Hvo → 0 as α → 0, which is obvious, and
(b) the linearization Dα of ∂J,Hα at vo is invertible for small α > 0 with uniform bound on
�D −1
α �Op as α → 0.
We sketch now the proof of this uniform bound.
After suitable trivializations, the linearization Dα of ∂J,Hα at vo with the above Lagrangian
boundary conditions can be viewed as an operator Dα on
H 1,p
iÊn ,Ên(α) := � v: [0, 1 ×Ìα −1 →�n | v(0, ·) ∈ iÊn , v(1, ·) ∈Ên � ,
with norm � · �1,p;α. Assuming that D −1
α is not uniformly bounded as α → 0 means that we would
have αn → 0 and vn ∈ H 1,p
iÊn ,Ên(αn) with �vn�1,p;αn = 1 such that �Dαnvn�0,p;αn → 0. The limit
operator to compare to is the standard ∂-operator on maps
v: [0, 1] ×Ê→�n , s.t. v(0, t) ∈ iÊn , v(1, t) ∈Ên ∀t ∈Ê.
This comparison operator is clearly an isomorphism (from which one easily shows that Dαn has
to be invertible for αn small).
Let β ∈ C∞ (Ê, [0, 1]) be a cut-off function such that
�
1, t ≤ 0,
β(t) =
β
0, t ≥ 1,
′ ≤ 0,
and set βn(t) = β(αnt − 1) · β(−αnt), hence
We have
Since the linearization Dα is of the form
with some matrix A(s, t), we observe that
βn −1 |[0, αn ] ≡ 1 and suppβn ⊂ [−α −1
n , 2α −1
n ] .
�vn�1,p;αn ≤ �βnvn�1,p;Ê≤c1�vn�1,p;αn .
Dαv = ∂sv + i∂tv + αA(s, t)v
�∂(βnvn) − Dαn(βnvn)�0,p;Ê= αn�Aβnvn�0,p;Ê→ 0 .
53
(75)