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Let
ρ>
0 be a parameter. Then (
A.22
) is equivalent to
Find
u
∈
K
such that
(A.23)
ρf
−
ρBu
+
u
−
u, v
−
u
V
≤
0
∀
v
∈
K
,
or to the fixed point problem
u
=
P
K
(ρf
−
ρBu
+
u)
in
K
.
(A.24)
Define, for
v
∈
K
,
S
v
=
P
(v
−
ρ(Bv
−
f))
. Then, by (
A.6
), for every
K
v
1
,v
2
∈
K
,
S
v
1
−
S
v
2
V
≤
(v
1
−
v
2
)
−
ρB(v
1
−
v
2
)
V
,
and hence
2
2
ρ
2
2
V
S
v
1
−
S
v
2
V
≤
v
1
−
v
2
V
−
2
ρ
B(v
1
−
v
2
), v
1
−
v
2
V
+
B(v
1
−
v
2
)
2
V
2
ρC
2
+
ρ
2
C
1
).
≤
v
1
−
v
2
(
1
−
2
ρC
2
+
ρ
2
C
1
)
2
<
1. Then
We fix
ρ>
0 such that
κ
:=
(
1
−
S
is a contraction on
K
and (
A.24
) has a unique solution.
Remark A.3.4
The proof is constructive—the contraction
S
converges with rate
κ<
1. The rate is maximal, if
ρ
2
C
1
}=
C
2
/C
1
,
ρ
=
ρ
∗
=
arg min
{
1
−
2
ρC
2
+
(A.25)
then
C
2
/C
1
)
2
<
1
.
κ
=
κ
∗
=
(
1
−
(A.26)
In particular, if
v
(
0
)
∈
K
is arbitrary, the sequence
v
(i
+
1
)
v
(i)
f)
v
(i)
ρ(Bv
(i)
=
S
=
P
−
−
(A.27)
K
converges to
u
∈
K
, the solution of (
A.17
), and
v
(i)
Cκ
i
v
(
0
)
u
−
V
≤
∗
u
−
V
,i
≥
0
.
(A.28)
Corollary A.3.5
If
M
⊆
V
is a closed
,
linear subspace
,
the problem
∈
M
≥
∀
∈
M
Find u
such that
b(u, v)
(v)
v
admits a unique solution
.
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