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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.