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Adding these inequalities together gives an inequality of the type ( B.30 ), where
p m +
q m =
f k,m + 2
f k,m + 1 , w m =
u k, m + 1
u k,m ,m
1. We apply Lemma B.3.9
to w k (t) defined in ( B.52 ), with s(t)
=
f k (t
+
k) and w 0 :=
u k, 1
Pu 0 . Observing
kU k (t) , we get ( B.61 ) with ( B.60 ) from ( B.53 ), ( B.54 ).
that w k (t)
=
From ( B.61 ) the size of E(k,T ) as k
0 for fixed T> 0 is important. We have
Lemma B.3.14 Assume 0 ,u 0 K · H and ( B.11 ). Then
E(k,T ) = o( 1 ) as k
0 ,
(B.62a)
and , for compatible data satisfying ( B.15a ), ( B.15b ), it holds
E(k,T )
Ckas k
0 .
(B.62b)
Proof We show ( B.62b ). The terms in the second row of the definition ( B.60 )of
E(k,T ) can be bounded as in ( B.62b ), by ( B.26 ), ( B.28 ). The terms in the first row
of ( B.60 ) are boun d u sing ( B.15a ), ( B.15b ) as follows: we decompose f
=
g
+
h
S( 0 ,T) and define g k (t), h k (t) as in ( B.55 ). Assume k
=
T/M for M
∈ N
. Then
h(t + k) h(t) L 2 (J ; V ) k h (t) L 2 ( 0 ,T + k ; V ) ,
h k (t) h(t) L 2 (J ; V ) k h (t) L 2 ( 0 ,T + k ; V ) ,
as well as
M
1
g(t
+
k)
g(t)
L 2 (J ; H ) =
J k,m
g(τ
+
k)
g(τ)
H
d τ
m =
0
M
1
k
=
0 g(τ + (m +
1 )k) g(τ + mk) H
d τ
0
m
=
M
1
k
Va r (g, J k,m ; H
)
m =
0
; H
k Va r (g, J
).
Furthermore,
g(τ) d τ g(t) H
M
1
1
k
g k (t) g(t) L 1 (J ; H ) =
d t
J k,m
J k,m
m
=
0
M
1
1
k
J k,m
g(s)
g(t)
H
d t d s
J k,m
m =
0
M
1
k
Va r (g, J k,m ; H
)
m
=
0
k Va r (g, J
; H
).
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