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αk
M
1
2
2
V
X M :=
m M { w m H } ,
max
M :=
1 w m
(B.31a)
1
m
=
M
1
M
1
2
k
α
2
2
P M :=
k
0
p m H
,
:=
w 0
H +
q m
.
M
m =
m =
0
(B.31b)
Then it holds
Q 2 M ) 2 .
(P M +
max (X M ,Y M )
P M +
(B.32)
Proof By ( B.30 ) and ( B.4b ) with λ
=
0,
2
V
2 (w m + 1 w m ,w m + 1 ) +
2 w m + 1
2 k p m H w m + 1 H +
2 k q m w m + 1 H
k
α
2
2
H
2 k
p m H
w m + 1 H +
q m
+
αk
w m + 1
.
Using here
2
2
H
(w m + 1
w m ,w m + 1 )
w m + 1
H
w m
(B.33)
and summing from m
=
0 ,...,M
1, we get
M
1
2
Y M
Q 2 M
Q 2 M .
w M
H +
2 k
0
w m + 1 H
p m H +
2 X M P M +
(B.34)
m
=
For M
1, we infer from ( B.34 ) the bound
Y M
Q 2 M ,
2 X M P M +
(B.35)
and also, for 1
m
M ,
2
Q 2 M .
w m
H
2 X M P M +
(B.36)
Hence, from ( B.36 ),
X 2 M =
2
Q 2 M ,
M
w m
H
2 X M P M +
max
1
m
and, combining with ( B.35 ), we get ( B.32 ).
Remark B.3.5 We can also bound the jumps w m + 1
w m :
M
1
Q 2 M ) 2 .
2
(P M +
0
w m + 1
w m
H
P M +
(B.37)
m
=
To obtain this, replace ( B.33 )by
2
2
2
H
2 (v
w,v)
=
v
w
H +
v
H
w
.
We also need a continuous analogue of Lemma B.3.4 .
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