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JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1501
g ( x + d + d ) = g ( x + d ) +∇ g ( x + d ) d + O(|| d
|| )
k k k k k k k T k k 2
j j j
k k k T k k
gj
x + d +∇ gj
x d + O d
k
d k
+ O d
2
k k k T k k
gj
x + d +∇ gj
x d
+ O d
k
d
= ( ) ( ) (|| |||| ||) (|| || )
= ( ) ( ) (|| |||| ||).
In addition, from (9) and (10),
k T k k T k
∇ g
j
( x ) d =∇g j( x ) d0
− δ
k,
k T
k
( )
k τ
k k
∇ g
j
x d = −|| d0
|| − g
j( x + d )
k k T k
+ g
j( x ) +∇g j( x ) d ,
so, for τ ∈ (2,3) we have
k k k
gj
( x + d + d
)
= − || d || + g ( x ) +∇g ( x ) d − δ + O(|| d |||| d
||)
k τ k k T k k k
0 j j 0 k
k τ
k k
0
(21)
≤− || d || + O(|| d |||| d ||) ≤0.
Hence, the second inequalities of (20) hold for t = 1
and k is sufficiently large.
The next objective is to show the first inequality of (19)
holds. From Taylor expansion and taking into account
Lemma 4.1 and Lemma 4.3, we have
k k k k k T k
s f( x + d + d
) − f( x ) + α∇f( x ) d
k T k k 1 k T 2 k T k
= ∇ f ( x ) ( d + d ) + ( d ) ∇ f( x ) d (22)
2
k T k k 2
−α∇ f( x ) d + o(|| d || ).
On the other hand, from the KKT condition of (8) and
the active set L
k
defined by Lemma 4.3 one has
∑
∇ f( x ) =−H d − u ∇g ( x )
k k k k
k 0
j j
j∈Lk
k k k k 2
=−Hd k
−∑
uj∇ g
j( x) + o(|| d || ),
j∈Lk
So, from (23) and Lemma 4.3, we have
∇f( x ) d
k T k
∑
k T k k k T k
( d ) Hkd uj g
j( x ) d
k 2
o(|| d || )
j∈Lk
k
d
T k k k T k
Hkd ∑ uj g
j
x d0
o
k
d
2
j∈Lk
=− − ∇ +
= −( ) − ∇ ( ) + (|| || )
k T k
k
∇ f( x ) ( d + d
)
∑
k T k k k T k
k
k 2
k j j
j∈Lk
=−( d ) H d − u ∇ g ( x ) ( d + d ) + o(|| d || ),
Again, from (21) and Taylor expansion, it is clear that
k 2
k k k
o(|| d || ) = gj( x + d + d
)
1
g x +∇ g x d + d
+ d ∇ g x d + o d
2
where j∈
L , then, we obtain
(23)
(24)
(25)
k k T k k k T 2 k T k k 2
=
j( )
j( ) ( ) ( )
j( ) (|| || )
k
∑
k k T k
k
− u ∇ g ( x ) ( d + d
)
j j
j∈Lk
k k T
∑ uj
g
j( x )
j∈Lk
1 ⎛ ( )
2 T ⎞
k T k (
k )
k (||
k ||
2
+ d uj
g
j
x d o d ),
2 ⎜∑
∇ ⎟
+
j∈Lk
= ∇
From (25) and (26), we have
⎝
k T k
k
∇ f( x ) ( d + d
)
k T k k k
=−( d ) H d − u ∇g ( x )
1
k j j
j∈Lk
⎛
k T k 2 k T k k 2
+ ( d ) j j( ) (|| || )
2 ⎜
u ∇ g x ⎟
d + o d
j∈Lk
⎝
∑
∑
Substituting (27) and (24) into (22), it holds that
1 k T k k k
s ( α − )( d ) Hkd + (1 −α) ∑ ujg j( x )
2
1 ⎛
⎞
2
⎜ ∑
⎟
⎝
j∈Lk
⎠
1 k T k k k
=( α − )( d ) Hkd + (1 −α) ∑ ujg j( x )
2
⎠
⎞
⎠
j∈Lk
(26)
(27)
+
k T
( d )
T
2 k k 2 k
k k 2
∇ f ( x ) + u ∇ g ( x ) − H d + o(|| d || )
j j k
j∈Lk
1 k T 2 k k k k 2
+ ( d ) ( ∇ L( x , uJ
) − H ) (|| || ).
k k
d + o d
2
Then, together assumption H 3.1 and H 4.2 as well as
k k
ug( x) ≤ 0, shows that
j
j
1 1
s ≤ α − a d + o d ≤ α∈
2 2
Hence, the inequality of (19) holds.
k 2 k 2
( ) || || (|| || ) 0. ( (0, )).
Furthermore, in a way similar to the proof of Theorem
5.2 in [5] and in [19, Theorem 2.3], we may obtain the
following theorem:
Theorem 4.5 Under all above-mentioned assumptions,
the algorithm is superlinearly convergent, i.e., the
sequence { x k } generated by the algorithm satisfies that
1 * *
|| k
k
x + − x || = o(|| x − x ||).
Proof.
From Lemma 4.1 and Lemma 4.4, we can know that
the sequence { x
k } yielded by Algorithm A has the form
of
k 1 k k
k
x + = x + d + d
k k k
k
k
= x + d0 + ( d + d −d0)
k
k k
x + d + d
.
0
k
where k
k
( k
d = d + d − d ) (for k large enough) and
k
k 3
d O d0
0
|| || = (|| || ) . Consequently, we can obtain the
result together with Ref. [5] and [19].
V. NUMERICAL RESULTS
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