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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)
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