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1500 JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013
In view of the boundedness of { x k } and J
k
being a
subset of the finite set I = {1, 2, , m}
as well as
Lemma 2.1, we know that there exists an infinite index
'
set K ⊆ K such that
' '
T
k k k k
k
x → x, J ≡ J , ∀k∈K , det( A A ) ≥ε, ε ≥ ε.
As a result,
T
lim( ) ( T
A A =∇g x) ∇g ( x),
'
k∈K
k k '
'
J J
T
det( ∇g
'( x) ∇g '( x)) ≥ ε > 0.
J
J
T −1
Hence, we obtain T
|| ( A ) || ||
k
Ak → ∇g '( x) ∇ g '( x) ||,
J J
T −1
this contradict || ( Ak
Ak) || →∞, ( k∈ K).
So the first
conclusion 1) follows.
k
2) We firstly show that lim d0
= 0 .
k →∞
k
We suppose by contradiction that lim d0
≠ 0, then
k →∞
there exist an infinite index set K and a constant σ > 0
k
such that || d0
|| > σ holds for all k∈
K.
Taking notice
of the boundedness of { x
k } , by taking a subsequence if
necessary, we may suppose that
k
'
x → x, Jk
≡ J , ∀k∈
K.
Using Taylor expansion, we analyze the first search
inequality of Step 5, combining the proof of Theorem 3.2,
k
the fact that x → x * ,( k →∞ ) implies that it is true.
k
k
The proof of limd
= 0; limd = 0 are elementary
k→∞
k→∞
from the result of 1) as well as formulas (9) and (10).
3) The proof of 3) is elementary from the formulas (9),
(10) and assumption H2.1.
Lemma 4.2. Let H2.1 to H4.1 holds,
k+
1 k
lim || x − x || = 0 . Thereby, the entire sequence { x
k }
k →∞
* k
converges to x i.e. x → x * , k →∞ .
Proof.
From the Lemma 4.1, it is easy to see that
k+
1 k k 2 k
lim || x − x || = lim(|| tkd + tkd
||)
k→∞
k→∞
k
k
≤ lim(|| d || + || d
||) = 0
k →∞
Moreover, together with Theorem 1.1.5 in [4], it shows
k
that x → x * , k →∞
Lemma 4.3 It holds, for k large enough, that
* k
* k *
1) J
k
≡ I( x ) I* , b → uI = ( u ,
* j
j∈I* ), v →( uj, j∈I*
)
k k T k
2) I + ⊆ Lk = { j∈ Jk : g
j( x ) +∇ g
j( x ) d0
= 0} ⊆ Jk.
Proof.
1) Prove J
k
≡ I*
.
On one hand, from Lemma 2.1, we know, for k large
enough, that I *
⊆ J k
. On the other hand, if it doesn’t
hold that J
k
⊆ I*
, then there exist constants j 0
and
β > 0 , such that
*
g
j
( x ) ≤− β < 0, j
0
0
∈ Jk.
k
So, according to d 0
→ 0 and the functions g ( x ),
j
( j∈
I ) are continuously differentiable, for k large
k
enough, if v < 0 , we have
j0
k * T k k * T k
+∇
j0 0
=−
j
+∇
0 j0
0
p ( x ) g ( x ) d v g ( x ) d
j0
1 k
≥− v
j
> 0.
0
2
Otherwise,
k * T k
p ( x ) +∇g ( )
j
j
x d
0
0
0
k * T k 1
k
= g
j
( x ) +∇g ( )
0 j
x d
0 0
≤− β < 0, ( vj
≥0)
0
2
which is contradictory with (8) and the fact j 0
∈ J k
. So,
J
k
≡ I*
(for k large enough).
k
* k *
Prove that b → uI = ( u ,
* j
j∈I* ), v →( uj, j∈ I*
).
k *
For the v →( uj
, j∈I*
) statement, we have the
k
following results from the definition of v ,
k * T −1 T *
v →−−B∇ f( x ) =−( A A ) A ∇ f( x )
* * * *
In addition, since x * is a KKT point of (1), it is
evident that
* *
∇ f( x ) + Au .
* I
= 0, u
* I
= −B * *
∇ f( x )
T −1 T *
i.e. uI
=−( A
* *
A* ) A*
∇ f( x ).
k
Otherwise, from (8), the fact that d0 → 0 implies that
k k k k
*
∇ f ( x ) + Hkd0 + Akb = 0, b →−B*
∇ f( x ) = uI
.
*
The claim holds.
2) For
Furthermore, it has
*
lim( k k
x , d ) ( x , 0)
0
k →∞
k *
uI+ uI+
x→∞
= , we have
Lk
*
⊆ I( x ) .
lim = > 0 , so the proof is
finished.
In order to obtain super-linear convergence, a crucial
requirement is that a unit step size is used in a
neighborhood of the solution. This can be achieved if the
following assumption is satisfied.
H4.2 Let
2 k k k k
|| (
xxLx ( , uJ ) H) || (|| ||)
k k
d o d
∇ − = , where
= +∑ .
k
k
Lxu ( , ) f( x) u g( x)
Jk
Jk
j
j∈Jk
Lemma 4.4 Suppose that Assumption H 2.1 to H 4.2
are all satisfied. Then, the step size in Algorithm A
always one, i.e. tk
≡ 1, if k is large enough.
Proof.
It is only necessary to prove that
f ( x k + d k + d k ) ≤ f( x k ) + α∇f( x k ) T d
k , (19)
k k k
g ( x + d + d ) ≤0, j∈I.
(20)
j
*
For (12) if j ∈ I \ I we have g ( ) 0
*
j
x < ,
k k
k
*
( x , d , d ) →( x , 0, 0)( k →∞ ), then, it is easy to
obtain g ( k k k
j
x + d + d ) ≤0
holds.
If j ∈ I*
we have
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