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Theorem 2.
Let
f, g
:
V
n
→
F
2
be two Boolean functions. Suppose that
f
and
g
are related by
g
(
x
)=
f
(
Ax
+
b
)+
c
·
x
+
d,
where
A
∈
O
(2
,n
)
,b,c
∈
V
n
,d
∈
V
1
.
Then, if
f
is bent-negabent,
g
is also bent-negabent.
Proof.
As discussed above,
g
is bent if
f
is bent. It remains to show that
g
is
negabent. From [3, Lem. 2] we know that, if
f
(
Ax
) is negabent, so is
f
(
Ax
+
b
)+
c
x
+
d
. It is therefore sucient to assume that
b
and
c
are all-zero vectors
and
d
=0.Observethat
·
wt(
x
)=
x
T
Ix,
where
I
is the
n
×
n
identity matrix and the matrix operations are over
Z
.We
therefore have
(
g
)(
u
)=2
−
2
x∈V
n
1)
f
(
Ax
)+
u·x
i
x
T
Ix
.
N
(
−
Now, since
A
is invertible by assumption, there exists
B
such that
AB
=
I
.
Moreover, when
x
ranges over
V
n
,sodoes
Bx
.Thus,
(
g
)(
u
)=2
−
2
x∈V
n
1)
f
(
x
)+
u·Bx
i
(
Bx
)
T
I
(
Bx
)
.
N
(
−
O
(2
,n
)andso
B
T
IB
=
I
. Hence,
Since
A
∈
O
(2
,n
), we have
B
∈
(
Bx
)
T
I
(
Bx
)=
x
T
(
B
T
IB
)
x
=
x
T
Ix.
We conclude
(
g
)(
u
)=2
−
2
x
1)
f
(
x
)+
u·Bx
i
x
T
Ix
N
(
−
∈
V
n
=2
−
2
x∈V
n
1)
f
(
x
)+
B
T
u·x
i
wt(
x
)
(
−
(
f
)(
B
T
u
)
,
=
N
which proves the theorem.
It is known that the dual of a bent function is again a bent function, and it
was proved in [3, Thm. 11] that the dual of a bent-negabent function is also
bent-negabent. The following theorem generalizes this concept by showing that,
if a bent-negabent function is bent with respect to certain subspaces, then the
corresponding partial duals are also bent-negabent.
Theorem 3.
Write
V
n
=
U
⊕
W
,where
dim
W
=
k
and
k
≤
n
, so that
dim
U
=
n
k
.Let
f
:
V
n
→
F
2
be a bent-negabent function that is bent with respect to
U
and bent with respect to
W
.Then
f
W
is also bent-negabent.
−
