Conformally Invariant Variational Problems. - SAM

Exchangingtheroleofiandj assumingmoreoverthatj ≠ k and
subtracting to (II.14) the corresponding identity for the triplet
(j,i,k) instead of (i,j,k) gives
∂ 2 e −λ
∂x j ∂x k
∂u
− ∂2 e −λ
∂x i ∂x i ∂x k
∂u
∂x j
= 0
(II.15)
We are assuming that n 3 hence, for any pair i ≠ k there
exists a j ≠ k and j ≠ i. Applying (II.15) to this triplet and
since ∂u
∂x i
and ∂u
∂x j
arelinearlyindependentweobtaininparticular
∂ 2 e −λ
∀i ≠ k = 0 (II.16)
∂x i ∂x k
We can exchange the role playedby the canonicalbasis withany
orthonormal basis of R n . Hence, considering the bilinear form
given by the Hessian of e −λ , we have proved that
U⊥V =⇒ d 2 e −λ (U,V) = 0
For any U ≠ 0 we introduce L U ∈ (R n ) ∗ given by
(II.17)
∀V ∈ R n L U (v) : = d 2 e −λ (U,V)− d2 e −λ (U,U)
| U | 2 < U,V >
From (II.16) we have that L U is identically zero on U ⊥ and that
L U (U) = 0 thus for any U ∈ R n L U ≡ 0 i.e.
∀U ∈ R n ∀V ∈ R n d 2 e −λ (U,V) = d2 e −λ (U,U)
| U | 2 < U,V >
Arguing as in the end of the proof of Lemma II.3 gives the
existence of f ∈ C 1 such that
and
∀U,V ∈ R n d 2 e −λ (U,V) = f < U,V >
f = ∂2 e −λ
∂x 2 i
10
∀i = 1...m