Take
h
i
=
ϕ
i
2
in the above we see that the second term and the second
from the end are zero. For computing the rest terms in the exposition (19)
we invoke the sufficient conditions for
f
to achieve its maximum at
ϕ
. If
the dimension of the underlying space,
R
n
,
n
is even, then the following
equations must be satisfied [9, 11],
D
k
f
α
ϕ
α
= 0
,
k
= 1
,
2
,
· · ·
, n
−
1
.
(20)
A direct calculation shows
D
2
f
α
ϕ
α
1
= 2
f
i
=
j
h
i
1
h
i
2
ˆ
i
ˆ
j
ϕ
α
1
−
i
h
i
1
, h
i
2
λ
= 0
for arbitrary
h
i
1
and
h
i
2
taken from
L
(
i
)
2
(0
,
1)
. Let
h
i
1
=
h
i
2
=
ϕ
i
2
. We get
the value of the third term,
f
N
i
=
j
ϕ
i
2
ϕ
i
2
ˆ
i
ˆ
j
ϕ
α
1
=
n
2
λ
N
+1
.
The third term from the end of (19) is in complete symmetry with the
above, and, due to (20), all the rest terms are zero except the first and last
ones. Then,
f
N
(
ϕ
3
, t
) =
t
n
+
t
n
−
2
(1
−
t
)
2
n
2
+
t
2
(1
−
t
)
n
−
2
n
2
+ (1
−
t
)
n
λ
N
+1
.
It follows from the assumption,
f
N
(
ϕ
3
,
0) =
f
N
(
ϕ
3
,
1) =
λ
N
+1
, and it
reaches minimum at
t
= 1/2. Let
t
= 1
/
2
. It yields
f
N
ϕ
3
,
1
2
=
f
N
α
ϕ
α
1
+
ϕ
α
2
2
= (2 +
n
)2
−
n
λ
N
+1
.
Since
ϕ
α
1
+
ϕ
α
2
=
√
2
,
ϕ
α
1
+
ϕ
α
2
√
2
∈
B
α
, we have
f
N
ϕ
3
,
1
√
2
=
f
N
α
ϕ
α
1
+
ϕ
α
2
√
2
= (2 +
n
)
·
2
−
n
2
λ
N
+1
=
ρ
(
n
)
λ
N
+1
.
It is evident, if
ρ
(
n
)
≥
1
then
(
ρ
(
n
))
1
n
≥
1
, so that
α
ϕ
α
1
+
ϕ
α
2
√
2
ρ
1
n
∈
B
n
.
A direct computation shows this is possible only for
n
= 2
,
4
and 6. In
these cases if
ϕ
=
1
α
ϕ
α
N
+1
,
1
, and
ϕ
=
2
α
ϕ
α
N
+1
,
2
both render the supremum
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2005. № 4
45
