Constructions of harmonic maps between Hadamard manifolds

4.1. PROOF OF THEOREM 77
Proposition 4.1.7. Let m, n ≥ 2 and u ∈ C 2 (B m , D n ) ∩ C 1 (B m , D n ). If u is a proper
harmonic map, then the boundary value of u is a constant map.
Proof.
Since u α ∈ R and τ(u) α =0, we have
0=a(u)(z)(1 −|z| 2 )〈Lu, u〉 + 2(1 −|z| 2 )(2|〈Nu,u〉| 2 −|u| 2 |Nu| 2 )
m∑
+2 (2|〈X j u, u〉| 2 −|u| 2 |X j u| 2 ).
j=1
For p ∈ ∂B m , let {z j } ∞ j=1 be a sequence satisfying the conditions in Corollary 4.1.6. Then,
since u ∈ C 1 (B m , D n ), it follows from Corollary 4.1.6 that
Thus we conclude that
and hence
lim a(u)(z j)(1 −|z j | 2 )〈Lu, u〉(z j )=0,
j→∞
lim
j→∞ 2(1 −|z j| 2 )(2|〈Nu,u〉| 2 −|u| 2 |Nu| 2 )(z j )=0.
2
2
m∑
|〈X j u, u〉| 2 =
j=1
m∑
|〈X j u, u〉| 2 =
j=1
On the other hand, on ∂B m , it holds that
m∑
|X j u| 2 at p,
j=1
m∑
|X j u| 2 on ∂B m .
j=1
0=X j |u| 2 =2〈X j u, u〉.
Therefore
which asserts that
m∑
|X j u| 2 =0,
j=1
Since u α ∈ R, we obtain
X j u α =0 on∂B m for 1 ≤ j ≤ m, 1 ≤ α ≤ n.
¯X j u α =0 on∂B m for 1 ≤ j ≤ m, 1 ≤ α ≤ n.