Information Technology Reference
In-Depth Information
→
Note that, as
k
0,
V
∗
strongly
.
G
0
→
u
0
∈
H
,H
0
→
0in
(B.19)
Since 0
∈
K
, we may choose in (
B.17
)
v
=
0 and get
2
2
2
2
∗
u
1
H
+
k
u
1
V
≤
C(
G
0
H
+
H
0
),
(B.20)
V
∗
. We claim that
u
1
−
P
u
0
where
·
∗
denotes the norm in
2
2
H
+
k
u
1
V
→
0as
k
→
0
.
(B.21)
To prove (
B.21
), we note that
u
1
∈
K
. Also, by the definition of
P
in Remark
B.1.1
,
we have
(u
0
−
P
u
0
,u
1
−
P
u
0
)
≤
0
,
hence
2
u
1
−
P
u
0
H
=
(u
1
−
P
u
0
,u
1
−
P
u
0
)
≤
(u
1
−
u
0
,u
1
−
P
u
0
).
(B.22)
By (
B.4b
) with
λ
=
0, (
B.9c
) and (
B.9e
), (
B.17
), (
B.22
), it follows for every
v
∈
K
2
2
u
1
−
P
u
0
H
+
u
1
−
V
≤
(u
1
−
u
0
,u
1
−
P
αk
v
u
0
)
+
A
(u
1
−
v),u
1
−
V
∗
,
V
k
v
≤
(u
1
−
u
0
)
+
k
u
1
−
u
0
+
k
A
u
1
,u
1
−
v
V
∗
,
V
−
k
A
v,u
1
−
v
V
∗
,
V
u
0
,v
−
P
(
B.17
)
≤
u
1
−
u
0
H
v
−
P
u
0
H
+
(G
0
−
u
0
,u
1
−
v)
√
k
√
k
+
H
0
−
A
v,u
1
−
v
V
∗
,
V
.
Therefore, it holds for all
v
∈
K
that
2
2
V
u
1
−
P
u
0
H
+
k
u
1
−
u
0
C
≤
u
1
−
u
0
H
−
P
u
0
H
v
.
(G
0
−
√
k
H
0
−
√
k
v
+
u
0
,u
1
−
v)
+
A
v,u
1
−
(B.23)
V
∗
,
V
∈
K
·
H
, there is a sequence
Since
P
u
0
{
v
k
}
k>
0
⊂
K
such that, as
k
→
0,
2
k
v
k
and obtain, after
passing to the limit and using (
B.19
), (
B.20
), the assertion (
B.21
). We have
v
k
V
→
0, and
v
k
−
P
u
0
H
→
0. We choose in (
B.23
)
v
=
Lemma B.3.1
For any fixed n>
0,
there exists C
n
>
0
such that
,
as k
→
0,
C
n
S(
0
,nk)
.
2
2
2
2
u
k,n
−
P
u
0
H
+
k
u
k,n
V
≤
u
0
H
+
f
(B.24)
Moreover
,
as k
→
0,
2
2
u
k,n
−
P
u
0
H
+
k
u
k,n
V
→
0
(non-uniformly in n) .
(B.25)
Search WWH ::
Custom Search