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f
W
is bent. By direct calculation,
Proof.
We first prove that
2
x
1)
f
W
(
x,v
)+
u·x
+
v·w
(
f
W
)(
u, w
)=2
−
n
H
(
−
∈
U
v
∈
W
n
+
k
2
=2
−
1)
f
(
x,y
)+
v·y
+
u·x
+
v·w
(
−
x
∈
U
v
∈
W
y
∈
W
1)
f
(
x,y
)+
u·x
v
n
+
k
2
=2
−
1)
v·
(
y
+
w
)
.
(
−
(
−
x
∈
U
y
∈
W
∈
W
The inner sum is zero unless
y
=
w
, in which case it is 2
k
. Hence,
(
f
W
)(
u, w
)=2
−
n−k
2
1)
f
(
x,w
)+
u·x
H
−
(
x∈U
=
H
U
(
f
)(
u, w
)
.
|H
U
(
f
)(
u, w
)
|
=1foreach
u
∈
U
and each
w
∈
W
. Therefore,
By assumption,
f
W
is bent.
Next we prove that
f
W
is negabent. We have
(
f
W
)(
u, w
)=2
−
2
x∈U
1)
f
W
(
x,v
)+
u·x
+
v·w
i
wt(
v
)+wt(
x
)
N
(
−
v∈W
=2
−
n
+
k
1)
f
(
x,y
)+
v·y
+
u·x
+
v·w
i
wt(
v
)+wt(
x
)
(
−
2
x
∈
U
v
∈
W
y
∈
W
1)
f
(
x,y
)+
u·x
i
wt(
x
)
v∈W
=2
−
n
+
k
1)
v·
(
y
+
w
)
i
wt(
v
)
.
(
−
(
−
2
x∈U
y∈W
The inner sum can be computed with Lemma 1. We therefore obtain
2
ω
k
x
(
f
W
)(
u, w
)=2
−
n
1)
f
(
x,y
)+
u·x
i
wt(
x
)
i
−
wt(
y
+
w
)
N
(
−
∈
U
y
∈
W
=2
−
2
ω
k
i
−
wt(
w
)
x∈U
1)
f
(
x,y
)+
u·x
+
y·w
i
wt(
x
)
−
wt(
y
)
(
−
y∈W
=
ω
k
i
−
wt(
w
)
(
f
)(
u, w
)
,
where
ω
=(1+
i
)
/
√
2and
w
is the complement of
w
.Since
f
is negabent, this
shows that
N
f
W
is also negabent.
4Con ru ons
Throughout this section we use the following notation. Define
V
to be an
mn
-
dimensional vector space over
F
2
,sothat
V
=
V
n
⊕
V
n
⊕···⊕
V
n
.
m
times
