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since Tr u 2 e v = Tr v 2 e u = Tr ( v ) = 0. Now put s d (0) := v∈K (
1) Tr ( v d ) .
Then W f (1) = W f (1) s d (0) , hence s d (0) = 2 m 2 k ,since m = k k .
Now for α ∈ K ,weget
W f ( α )=
x∈ F
1) Tr ( x 2 e +1 + αx )
(
=
u∈K
1) Tr ( u d + αu )
1) Tr ( v d )
(
(
v
K
2
k
.
= W f ( α )2 m k
2
K we get:
W f ( α + γ )=
x∈ F
For γ
1) Tr ( x 2 e +1 +( α + γ ) x )
(
=
u∈K
1) Tr ( u d + αu )
1) Tr ( v d + γv )
(
(
v∈K
= W f ( α )
v∈K
1) Tr ( v d + γv ) .
(
Putting α =1wehave s d ( γ ):= v∈K (
1) Tr ( v d + γv )
W f (1+ γ )
W f (1)
2 m k
=
∈{±
}
2
since 1 + γ
F \
H ,where H is the Trace-0 hyperplane of
F
(if d
Gold then
H cf.[10]).Letting δ ( γ ):= sign W f (1+ γ )
W f (1)
,the
we have W f ( α )=0
⇐⇒
α
proof is finished.
Remark 1. Equality (6) is in general not true if d
Kasami Welch Niho .We
expect, however, (7) is a property of AB exponents.
The following corollary shows that a Gold exponent can be decomposed into bent
functions. We note here that a bent function is a boolean function defined on
a vector space V with a 2-valued Walsh spectrum (i.e.
{ x∈V (
1) f ( x )+ u·x
:
u
V
}
=
|
V
|}
).
Corollary 2. Let d
Gold and f
| V denote the restriction of f toasubspace V .
Then
1. f
| K is AB.
2. Tr ( f )
| K is bent.
Proof. The first item is already proved. For the second, note that all the lin-
ear functionals on K are of the form Tr ( γx ), where γ
K ,asaresultof
nondegeneracy of the trace bilinear form (cf. [8]). Noting that s d ( γ )areWalsh
coecients of Tr x d on K completes the proof.
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