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In the following, we prove the convergence property of Algorithm 2.1. We first
assume that
{}
k
x
is an infinite sequence generated by Algorithm 2.1. The following
assumptions are satisfied.
(H
5
) The objective function
f
(
x
)
has a lower bound on
n
R
.
(H
6
) The gradient
g
(
x
)
is uniformly continuous on an open convex set
D
that
{}
contains
k
x
.
(H
7
) There exists
2
m
>
0
, for any
k
,
P
T
B
P
≥
m
P
.
k
k
k
k
The assumption (H
5
) is very mild. Because the objective function
f
(
x
)
can be
replaced by
f
e
, if the assumption is not satisfied.
According to the related reference, we have the following Lemma.
(
x
)
Lemma 2.1
[5]
.
Suppose that (H
2
), (H
3
) and (H
4
) hold for
ω
γ
s
,
and
. Then when
k
k
k
k
∈
I
is sufficiently large, we have
g
≤
c
γ
,
k
3
k
where
c
is a constant.
Analogous to the proof of the corresponding Lemmas in [5], we can obtain the
following Lemmas.
>
0
3
{}
Lemma 2.2.
Let
k
x
be an infinite iteration sequence generated by Algorithm 2.1. If
{}
g
(
x
)
is uniformly continuous on an open convex set
D
containing
k
x
, then there
{
}
exists a constant
σ
>
0
such that
f
+
σ
T
is monotone and non-increasing.
1
l
(
k
)
1
k
{}
)
Further, the sequence
is convergent (which converges to a finite value or
−∞
).
f
l
(
k
Without loss of generality, we assume that
lim
f
k
l
=
f
*
.
(
)
k
→
∞
{}
Lemma 2.3.
Let
k
x
be an infinite iteration sequence generated by Algorithm 2.1. If
{}
k
g
(
x
)
is uniformly continuous on an open convex set
D
containing
x
and
lim
f
k
=
f
*
(
0
≤
i
≤
M
−
1
),
then we have
k
∈
K
,
k
→
∞
i
−
1
(1)
li
,
1
g
=
0
k
k
∈
K
k
→
∞
i
−
(2)
lim
f
k
=
f
*
.
k
∈
K
,
k
→
∞
i
Proof.
Now we show that (1) holds in two cases.
Case 1:
When
I
∩
K
is an infinite index set, it follows from Lemma 2.1 that
i
lim
,
g
=
0
.
k
k
∈
I
∩
K
k
→
∞
i
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