It turns out from 6 the euler relation and the

It turns out from (6), the Euler
relation, and the inequality d
0
< 0 that f
t
0
, Δ
0
( x 0 ) = 0 , which contradicts the non-degeneracy
condition. Therefore, J negationslash = ∅ . For j ∈ J , we can write
λ
j
( s ) = c
j
s
m
j
+ higher-order terms in s,
where c
j
negationslash = 0 and m
j
∈ Q .
Put J
1
:= { j ∈ J : g
j
|
C I
negationslash≡ 0 } . If J
1
= ∅ , then
∂g
j
∂x
i
( φ ( s )) ≡ 0 for all i ∈ I and all j ∈ J.
We deduce from the condition (a5) that
∂f
t ( s )
∂x
i
( φ ( s )) ≡ 0 for all i ∈ I.
Consequently,
∂f
t
0
, Δ
0
∂x
i
( x
0
) = 0 for all i ∈ I.
It follows from (6), the Euler relation, and the inequality d
0
< 0 that f
t
0
, Δ
0
( x 0 ) = 0 , which
contradicts the non-degeneracy condition. Hence J
1
negationslash = ∅ . For each j ∈ J
1
, let d
j
be the
13

minimal value of the linear function
∑
i ∈ I
α
i
q
i
on R I ∩ Γ( g
j
) and Δ
j
be the maximal face of
R I ∩ Γ( g
j
) where this linear function takes its minimum value. We can write
g
j
( φ ( s )) = g
j, Δ
j
( x
0
) s
d
j
+ higher-order terms in s,
where g
j, Δ
j
is the face function associated with g
j
and Δ
j
. By the condition (a4), then
g
j, Δ
j
( x
0
) = 0 for all j ∈ J
1
. (7)
On the other hand, for i ∈ I and j ∈ J
1
,
∂g
j
∂x
i
( φ ( s )) =
∂g
j, Δ
j
∂x
i
( x
0
) s
d
j
− q
i
+ higher-order terms in s.
For i / ∈ I and j ∈ J
1
, the function g
j, Δ
j
does not depend on the variable x
i
, and hence,
∂g
j, Δ
j
∂x
i
( x
0
) = 0 . (8)
The condition (a5) implies that for all i ∈ I ,
∂f
t
0
, Δ
0
∂x
i
( x
0
) s
d
0
− q
i
+ · · · +
summationdisplay
j ∈ J
2
c
j
∂g
j, Δ
j
∂x
i
( x
0
) s
ℓ − q
i
+ · · · = 0 , (9)
where ℓ := min
j ∈ J
1
( m
j
+ d
j
), J
2
:= { j ∈ J
1
: ℓ = m
j
+ d
j
} and the dots stand for the
higher-order terms in s . There are three cases to be considered.
Case 1: ℓ > d
0
. By (6) and (9), we have
∂f
t
0
, Δ
0
∂x
i
( x
0
) = 0 for i = 1 , . . . , n.
This, together with the Euler relation, implies that
d
0
f
t
0
, Δ
0
= 0 .
Hence, f
t
0
, Δ
0
( x 0 ) = 0 because of d
0
< 0 . This contradicts the non-degeneracy condition.
Case 2: ℓ = d
0
. We deduce from (6), (8) and (9) that
∂f
t
0
, Δ
0
∂x
i
( x
0
) +
summationdisplay
j ∈ J
2
c
j
∂g
j, Δ
j
∂x
i
( x
0
) = 0 for i = 1 , . . . , n.
14

Consequently,
0 =
n summationdisplay
i =1
q
i
x
0
i
∂f
t
0
, Δ
0
∂x
i
( x
0
) +
n summationdisplay
i =1
summationdisplay
j ∈ J
2
c
j
q
i
x
0
i
∂g
j, Δ
j
∂x
i
( x
0
)
=
n summationdisplay
i =1
q
i
x 0
i
∂f
t
0
, Δ
0
∂x
i
( x 0 ) +
summationdisplay
j ∈ J
2
c
j
n summationdisplay
i =1
q
i
x 0
i
∂g
j, Δ
j
∂x
i
( x 0 )
= d
0
f
t
0
, Δ
0
( x
0
) +
summationdisplay
j ∈ J
2
c
j
d
j
g
j, Δ
j
( x
0
)
= d
0
f
t
0
, Δ
0
( x 0 ) ,
where the last equation follows from (7). Since d
0
< 0 , we get f
t
0
, Δ
0
( x 0 ) = 0 , which contra-
dicts the non-degeneracy condition.
Case 3: ℓ < d
0
. By (8) and (9), we obtain
summationdisplay
j ∈ J
2
c
j
∂g
j, Δ
j
∂x
i
( x
0
) = 0 for i = 1 , . . . , n.
This fact and (7) combined give a contradiction with the non-degeneracy condition. square
Lemma 4.2 (Boundedness of singularities at infinity) . There exists a real number r > 0
such that
Σ
∞
( f
t
|
S
) ⊂ D
r
for all t ∈ [0 , 1] .
Proof. Suppose the assertion of the lemma is false. By the Curve Selection Lemma at infinity
(see [25] or [17]), we can find a nonempty set I ⊂ { 1 , . . . , n } with f
t
|
C I
negationslash≡ 0 , a (possibly
empty) set J ⊂ { j ∈ { 1 , . . . , p } : g
j
|
C I
negationslash≡ 0 } , a vector q ∈ R
n
with min
i ∈ I
q
i
< 0 and
0 = d ( q, Γ( f
t
|
C I
)), and analytic curves
φ : (0 , ǫ ) → ( C
∗
)
I
, t : (0 , ǫ ) → [0 , 1] , and λ
j
: (0 , ǫ ) → C , j ∈ J,
such that the following conditions hold
(a1) bardbl φ ( s ) bardbl → ∞ as s → 0;
(a2) t ( s ) → t
0
∈ [0 , 1] as s → 0;
(a3) f
t ( s ) , Δ
0
( φ ( s )) → ∞ as s → 0;
(a4) g
j, Δ
j
( φ ( s )) = 0 for all j ∈ J and all s ∈ (0 , ǫ );
(a5) ∇ f
t ( s ) , Δ
0
( φ ( s )) +
∑
j ∈ J
λ
j
( s ) ∇ g
j, Δ
j
( φ ( s )) = 0 for all s ∈ (0 , ǫ ),
where Δ
0
:= Δ( q, Γ( f
t
|
C
I
) and Δ
j
:= Δ( q, Γ( g
j
|
C
I
)) for j ∈ J.