Constructions of harmonic maps between Hadamard manifolds

82 CHAPTER 4. NON-EXISTENCE OF PROPER HARMONIC MAPS
Hence the function
‖u(z 1 ,z 2 ) − (0, 1)‖
‖(z 1 ,z 2 ) − (0, 1)‖
is bounded on B 2 −{(0, 1)}, which implies that u is Lipschitz continuous at (0, 1).
Claim 3. u is not C 1 at (0, 1).
Let v be the second component of u, that is,
v(z 1 ,z 2 )= (1 −|z|2 ) 2 − 4|1 − z 2 | 4 +8(Iz 2 ) 2
(1 −|z| 2 +2|1 − z 2 | 2 ) 2 +8(Iz 2 ) 2 .
Then
∂v
v(0, 1) − v(0, 1 − h) 4(2 − h)
(0, 1) = lim
= lim
∂z2 h→+0 h
h→+0 (2 + h) =2. 2
On the other hand, from a straightforward calculation, we obtain
v z 2 ×{(1 −|z| 2 +2|1 − z 2 | 2 ) 2 +8(Iz 2 ) 2 } 2
= 16(Iz 2 ) 2 [ ¯z 2 {2|1 − z 2 | 2 − (1 −|z| 2 ) − (1 −|z| 2 ) 2 } + 2(1 −|z| 2 )]
−32 √ −1Iz 2 |1 − z 2 | 2 {2|1 − z 2 | 2 +1−|z| 2 }
−2(1 −|z| 2 +2|1 − z 2 | 2 ) 2 { ¯z 2 (1 −|z| 2 ) 2 +4|1 − z 2 | 2 (1 − ¯z 2 )}
+2(2 − ¯z 2 ){(1 −|z| 2 ) 2 − 4|1 − z 2 | 4 }(1 −|z| 2 +2|1 − z 2 | 2 ).
Here we have
{(1 −|z| 2 +2|1 − z 2 | 2 ) 2 +8(Iz 2 ) 2 } 2 = {(2θ 3 + r − 2r‖θ‖ 2 ) 2 +8(θ 4 ) 2 }r 4 .
Hence
v z 2(0, 1 − r) = −2r(2 + r){(1 − r)(2 − r)2 +4r} + 2(1 + r){(2 − r) 2 − 4r 2 }
r 4 (r +2) 4
−→ +∞ (r → 0).
Therefore, u is not C 1 at (0, 1).