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=
Proof
We proceed by induction on
n
. Inequality (
B.24
)for
n
1 follows from
(
B.20
), (
B.21
), if we note that the right hand side of (
B.20
) is bounded by
C
{
u
0
2
2
L
1
(
0
,k
2
L
2
(
0
,k
for
f
∈
S(
0
,T)
.
Assume now the assertion (
B.24
)istrueforsome
n
. Then, arguing as in the proof
of (
B.20
), (
B.21
), we obtain (
B.24
)for
n
H
+
g
;
H
)
+
h
;
V
∗
)
}
+
1.
Next, we address the case (iii) in Theorem
B.2.2
.
Lemma B.3.2
If
(
B.15a
), (
B.15b
)
holds
,
as k
→
0
we have
2
2
V
≤
Ck
2
.
u
1
−
P
u
0
H
+
k
u
1
−
P
u
0
(B.26)
·
H
into (
B.23
) to obtain
=
P
u
0
∈
K
Proof
We use (
B.18
) and insert
v
k
0
C
k
2
2
u
1
−
P
u
0
H
+
k
u
1
−
P
u
0
V
≤
u
1
−
P
u
0
H
g(τ)
H
d
τ
k
(h(
0
)
u
0
)
+
−
AP
u
0
,u
1
−
P
k
0
d
τ
.
(B.27)
+|
u
1
−
P
u
0
|
V
h(τ )
−
h(
0
)
∗
Theassertion(
B.26
) follows from (
B.27
) with the estimates
k
0
g(τ)
H
d
τ
≤
k
g
L
∞
(
0
,k
;
H
)
,
k
0
Ck
3
/
2
h
L
2
(
0
,k
;
V
∗
)
.
h(τ )
−
h(
0
)
∗
d
τ
≤
Remark B.3.3
Inserting
v
=
u
k,
1
into (
B.12b
), we obtain by similar arguments
2
2
V
u
k,
2
−
u
k,
1
H
+
k
u
k,
2
−
u
k,
1
≤
(f
k,
1
−
k
A
u
k,
1
,u
k,
2
−
u
k,
1
)
≤
Ck
2
d
τ
.
2
k
2
2
h
(τ )
2
∗
0
<τ <
2
k
g(τ)
sup
H
+
h(
0
)
−
AP
u
0
H
+
(B.28)
0
For the proof of Theorem
B.2.2
as well as for a priori error estimates, we require
the following perturbation results.
Lemma B.3.4
Assume
0
∈
K
and we are given sequences
{
w
m
}
,
{
p
m
}
,
{
q
m
}
which
satisfy
,q
m
∈
V
∗
for m
w
0
∈
H
,w
m
+
1
∈
K
,p
m
∈
H
≥
0
,
(B.29)
≥
and are such that
,
for m
0,
it holds
w
m
+
1
−
w
m
+
k
A
w
m
+
1
,w
m
+
1
V
∗
,
V
≤
k(p
m
+
q
m
,w
m
+
1
).
(B.30)
Define
,
for M
≥
1,
with α as in
(
B.4b
),
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