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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
kα
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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