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s
:
s
i
are integers, s
i
≥
i
=
1
s
i
=
s
n
B
=
0
,
Then A
⊂
B, and the following equality
min
s
f
(
s
)=
min
s
f
(
s
)
(3.41)
∈
B
∈
A
is satisfied. That is to say, the minimum of (
3.32
) subject to (
3.33
) is obtained with
an s
∈
A.
Proof.
The relation
A
⊂
B
is clear. To proof equality (
3.41
) first let us suppose
d
>
0
.
n
∑
i
=
Let
s
∈
B
be given, then
s
i
=
s
and
s
i
=
a
i
+
δ
i
is satisfied, where
δ
i
,
i
=
1
,
2
,...,
n
,
1
are integers. If every
δ
i
different from zero has the same sign (that is
δ
i
≥
0for
=
,
,...,
i
1
2
n
)then
n
i
=
1
s
i
=
n
i
=
1
a
i
+
n
i
=
1
δ
i
=
s
−
d
+
n
i
=
1
δ
i
n
i
=
1
δ
i
=
d
⇒
and so
s
∈
A
. If, on the contrary, not all the
δ
i
have the same sign, let us call
H
1
=
{
i
:
δ
i
>
0
},
H
2
=
{
i
:
δ
i
<
0
}.
We can write
n
i
=
1
s
i
=
n
i
=
1
a
i
+
∑
H
1
δ
i
+
∑
+
∑
i
H
1
δ
i
+
∑
H
2
δ
i
=
s
−
d
H
2
δ
i
=
s
.
i
∈
i
∈
∈
i
∈
δ
i
for
i
Then
i
∈
H
1
δ
i
=
d
−
∑
i
∈
H
2
δ
i
>
d
, and we can select
K
⊂
H
1
and
∈
K
such that
∑
n
∑
i
=
δ
i
are integers, 1
≤
δ
i
≤
δ
i
and
1
δ
i
=
d
. Now we can build
s
∈
A
as follows
s
i
=
a
i
+
δ
i
for
i
∈
K
⊂
H
1
,
s
i
=
a
i
for
i
∈
K
.
We h ave
n
i
=
1
c
i
(
s
i
−
r
i
)
2
=
i
∈
K
c
i
(
s
i
−
r
i
)
2
+
i
∈
K
c
i
(
s
i
−
r
i
)
2
f
(
s
−
r
)=
≥
i
∈
K
c
i
(
s
i
−
r
i
)
2
+
i
∈
K
c
i
(
a
i
−
r
i
)
2
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