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n
n
n
(
QN u
)
=−
c x
+
(
a
+
αβ
+
)
f x
(
)
+∧
T u
+∨
H u
+
I
i
i
i
ij
ij
ij
j
ij
j
ij
j
i
j
=
1
j
=
1
j
=
1
We claim that
(
QN u
)
>=
0,
i
1, 2,
(7)
,
n
.
i
On the contrary, suppose that there are some i such that (
QN u
)
=
0
, namely
i
n
n
n
cx
=
(
a
+
αβ
+
)
f x
(
)
+ ∧
Tu
+ ∨
Hu I
+
i
i
ij
ij
ij
j
ij
j
ij
j
i
j
=
1
j
=
1
j
=
1
Then we have
1
n
n
n
hx a f x Tu HuI
c
==
|
|
|
(
++
αβ
)()
+∧
+∨
+
|
i
i
ij
ij
ij
j
ij
j
ij
j
i
j
=
1
j
=
1
j
=
1
i
1 {(|
n
n
a
+
αβ
+
p h
+
a
+
αβ
+
q
|
|
|
|
|)
(|
|
|
|
|
|)
ij
ij
ij
j
j
ij
ij
ij
j
c
=
1
=
1
j
j
i
n
n
+∧
|
Tu Hu I
||
|
+
+∨
|
||
|
+
+
|
|}
ij
j
ij
j
i
j
=
1
j
=
1
1
n
1
n
+
≤+
(
ω
{ |
a
| |
+ +
α
|
|
β
|)|
p h
| }(
++
ω
{ |
a
| |
+ +
α
|
|
β
|)
q
ij
ij
ij
j
j
ij
ij
ij
j
c
c
j
=
1
j
=
1
i
i
n
n
+
+
+∧
|
Tu Hu I
||
|
+∨
|
||
|
+
|
|}
ij
j
ij
j
i
j
=
1
j
=
1
n
=
kh F
+
(8)
ij
j
j
j
=
1
Which is a contradiction. Therefore (7) holds and Lu Nu
=
n
QNu
≠∈ Ω
0,
u
KerL
= Ω
(9)
R
ΦΩ
:(
KerL
) [0,1]
×
Ω
KerL
Consider the homotopy
defined by
Φ=
(, )
u
μμ
diag
(
c
,
c
, ,
c u
)
+
(1
μ
)
QNu
1
2
n
Φ⋅
(,0)
=
JQN
,
Φ=
(, ) 0
u
Note that
if
, then we have
 
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