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Convergence analysis of a method for
variational inclusions
M.H. Rashid
a b
c
, J.H. Wang & C. Li
a d
a
Department of Mathematics, Zhejiang University, Hangzhou
310027, P.R. China
b
Department of Mathematics, Faculty of Science, University of
Rajshahi, Rajshahi – 6205, Bangladesh
c
Department of Applied Mathematics, Zhejiang University of
Technology, Hangzhou 310032, P.R. China
d
Department of Mathematics, College of Sciences, King Saud
University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
Version of record first published: 26 Sep 2011.
To cite this article: M.H. Rashid , J.H. Wang & C. Li (2012): Convergence analysis of a method for
variational inclusions, Applicable Analysis: An International Journal, 91:10, 1943-1956
To link to this article: http://dx.doi.org/10.1080/00036811.2011.618127
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Applicable Analysis
Vol. 91, No. 10, October 2012, 1943–1956
Convergence analysis of a method for variational inclusions
M.H. Rashidab, J.H. Wangc* and C. Liad
a
Department of Mathematics, Zhejiang University, Hangzhou 310027, P.R. China;
Department of Mathematics, Faculty of Science, University of Rajshahi, Rajshahi – 6205,
Bangladesh; cDepartment of Applied Mathematics, Zhejiang University of Technology,
Hangzhou 310032, P.R. China; dDepartment of Mathematics, College of Sciences,
King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
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b
Communicated by J.C. Yao
(Received 10 July 2011; final version received 24 August 2011)
Consider the following variational inclusion problem:
0 2 f ðxÞ þ gðxÞ þ FðxÞ,
where f is differentiable in a neighbourhood of a solution x and g is
and F is a set-valued mapping, and the method
differentiable at x,
introduced in Jean-Alexis and Pietrus [C. Jean-Alexis and A. Pietrus, On
the convergence of some methods for variational inclusions, Rev. R. Acad.
Cien. serie A. Mat. 102(2) (2008), pp. 355–361] for solving this problem:
0 2 f ðxk Þ þ gðxk Þ þ ðrf ðxk Þ þ ½2xkþ1 xk , xk ; gÞðxkþ1 xk Þ þ Fðxkþ1 Þ,
where rf(x) denotes the Fréchet derivative of f at x and [x, y; g] the firstorder divided difference of g on the points x and y. Local converge analysis
are provided for the method under the weaker conditions than Jean-Alexis
and Pietrus (2008). Moreover, if rf and the first-order divided difference of
g are p-Hölder continuous at a solution, then we show that this method
converges superlinearly. In particular, our results extend the corresponding
ones Jean-Alexis and Pietrus (2008), and fix a gap in the proof in (JeanAlexis and Pietrus (2008), Theorem 1).
Keywords: set-valued mapping; variational inclusions; pseudo-Lipschitz
continuity; local convergence; divided difference
AMS Subject Classifications: Primary: 47H04, 90C30; Secondary: 49J53,
65K10
1. Introduction
! 2Y be a setLet X and Y be Banach spaces. Let f : X ! Y, g : X ! Y and let F : X !
valued mapping with closed graph. We consider the following variational inclusion
problem:
0 2 f ðxÞ þ gðxÞ þ FðxÞ,
ð1:1Þ
*Corresponding author. Email: wjh@zjut.edu.cn
ISSN 0003–6811 print/ISSN 1563–504X online
ß 2012 Taylor & Francis
http://dx.doi.org/10.1080/00036811.2011.618127
http://www.tandfonline.com
1944
M.H. Rashid et al.
where f is differentiable in a neighbourhood of a solution x of (1.1), g is continuous
on and differentiable at x but may be not differentiable in .
As pointed out in [1], the problem (1.1) can be viewed as a perturbed problem of
the following one:
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0 2 f ðxÞ þ FðxÞ,
ð1:2Þ
with g being the perturbation function. Variational inclusion (1.2) is a very general
framework and many problems from applied mathematical areas, such as variational
inequality problems including linear and nonlinear complementary problems,
systems of nonlinear equations, abstract inequality systems and etc, can be cast as
problem (1.2) (see [2–8] and references therein).
In particular, in the special case when F ¼ {0}, the variational inclusion problem
(1.1) is reduced to the classical problem of solving systems of nonlinear equations:
f ðxÞ þ gðxÞ ¼ 0:
ð1:3Þ
When f is differentiable and g is a continuous function admitting first- and secondorder divided differences, Cătinas [9] proposed a method for solving (1.3), which is a
combination of Newton’s method with the secant’s method:
xkþ1 ¼ xk ðrf ðxk Þ þ ½xk1 , xk ; gÞ1 ð f ðxk Þ þ gðxk ÞÞ
k ¼ 1, 2, . . . ,
where rf(x) denotes the Fréchet derivative of f at x, and [x, y; g] the first-order
divided difference of g on the points x and y. Geoffroy and Piétrus [10] extended this
method to solve the variational inclusions (1.1)
0 2 f ðxk Þ þ gðxk Þ þ ðrf ðxk Þ þ ½xk1 , xk ; gÞðxkþ1 xk Þ þ Fðxk Þ:
They proved that the sequence generated by this method converges superlinearly
under certain conditions.
In a recent paper [1], Jean-Alexis and Pietrus presented the following method for
solving the variational inclusion problem (1.1):
0 2 f ðxk Þ þ gðxk Þ þ ðrf ðxk Þ þ ½2xkþ1 xk , xk ; gÞðxkþ1 xk Þ þ Fðxkþ1 Þ,
ð1:4Þ
where rf (x) denotes the Fréchet derivative of f at x and [x, y; g] the first-order
divided difference of g on the points x and y. Under the assumption that rf and the
first-order divided difference of g are p-Hölder continuous around a solution x (i.e.
there exists a neighbourhood of x such that rf and the first-order divided difference
of g are p-Hölder continuous on this neighbourhood) and that (f þ g þ F)1 is
with F having closed graph, they proved that the
pseudo-Lipschitz around ð0, xÞ
sequence generated by (1.4) converges superlinearly. Unfortunately, the proof of the
main theorem (Theorem 1) contains a gap, as we detailed in Section 3.
In this article, we continue to study the local convergence of the sequence
generated by (1.4). However, we only assume that rf and the first-order divided
difference of g are continuous at a solution x of (1.1). Then we show that the
sequence obtained by method (1.4) is convergent, provided that ( f þ g þ F )1 is
with F having closed graph. Furthermore, if rf and
pseudo-Lipschitz around ð0, xÞ
the first-order divided difference of g are additionally p-Hölder continuous at the
then we show that the sequence generated by (1.4) converges
solution x,
Applicable Analysis
1945
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superlinearly, which, in particular, extends [1, Theorem 1], and fix the gap in the
proof of the main theorem [1, Theorem 1].
The outline of this article is as follows. Section 2 presents some necessary notions
and preliminary results on fixed point lemma. The main theorems are presented in
Section 3, where the local convergence and superlinear convergence results of the
sequence generated by (1.4) are established.
2. Notations and preliminary results
In this section we recall some standard notations and notions. Let X, Y be Banach
spaces. Let x 2 X and r 4 0. We use Br(x) to denote the closed ball centred at x with
radius r, while L(X, Y) stands for the set of all bounded linear operators from X to Y.
Let A X and C X. The distance from x to A is defined by
distðx, AÞ :¼ inffkx ak : a 2 Ag,
while the excess from A to C is defined by
eðC, AÞ :¼ supfdistðx, AÞ : x 2 Cg:
Consider a set-valued mapping F : X ! 2Y. The inverse F1 of F is defined by
F 1 ð yÞ :¼ fx 2 X : y 2 FðxÞg
and the graph of F is denoted by
gph F :¼ fðx, yÞ 2 X Y : y 2 FðxÞg:
We begin with the definition of the first-order divided difference operators (see [11]
for details).
Definition 2.1 Let g : X ! Y and x, y 2 X. An operator [x, y; g] 2 L(X, Y) is called
the first-order divided difference of the function g : X ! Y on the points x, y 2 X if
[x, y; g] satisfies
gð yÞ gðxÞ x 6¼ y,
½x, y; gð y xÞ ¼
rgðxÞ
y ¼ x,
and in the sequel, rg denotes the Fréchet derivative operator of g.
The notion of pseudo-Lipschitz was introduced by Aubin [12,13] and has been
studied extensively (see, e.g. [1,6,8,14,15]).
! 2X be a set-valued mapping and let ðy,
xÞ
2 gph . Then
Definition 2.2 Let : Y !
xÞ
if there exist constants ry , rx and M
is said to be pseudo-Lipschitz around ðy,
such that the following inequality holds:
ð y2 ÞÞ Mk y1 y2 k
eðð y1 Þ \ Brx ðxÞ,
for every y1 , y2 2 Bry ðyÞ:
ð2:1Þ
Remark 2.1 The pseudo-Lipschitz property of a set-valued mapping is equivalent
to the openness with linear rate of 1 ( the covering property) and to the metric
regularity of 1 ( a basic well-posedness property in optimization) (see [12,13,16–20]
for more details).
1946
M.H. Rashid et al.
The following fixed point lemma for set-valued mappings, which was proved in
[21, Lemma (fixed point)], is a generalization of the fixed point theorem [22].
LEMMA 2.1 Let be a set-valued mapping from X into the closed subsets of X.
Let r 2 (0, þ1), 2 (0, 1) and 0 2 X. Suppose that
(a) dist (0, (0)) 5 r(1 ),
(b) e((x1) \ Br(0), (x2)) kx1 x2k, 8x1, x2 2 Br(0).
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Then has a fixed point in Br(0), that is, there exists x 2 Br(0) such that x 2 (x).
3. Convergence analysis
This section is devoted to the study of the local convergence of the sequence
generated by method (1.4). Let x be a solution of (1.1). Consider the following
assumptions:
(I) The function f is Fréchet differentiable and its derivative rf is continuous
at x.
(II) g is Fréchet-differentiable at x and the first-order divide difference of g is
that is, for any " 4 0, there exists 4 0 such that
continuous at x,
"
k½x, y; g rgðxÞk
ð3:1Þ
þ k y xk
.
for all x, y 2 X with x 6¼ y and kx xk
(III) F have closed graph and (f þ g þ F)1 is pseudo-Lipschitz around ð0, xÞ.
Fix x 2 X. For simplicity, we define the mapping Zx: X ! Y by
8
f ðxÞ þ gð yÞ gðxÞ þ rf ðxÞð
y xÞ
ðrf ðxÞ
>
< f ðxÞ
þ ½2y x, x; gÞð y xÞ,
y 6¼ x,
Zx ð yÞ :¼
>
:
f ðxÞ þ rf ðxÞð
y xÞ,
f ðxÞ
y ¼ x:
ð3:2Þ
The following lemma about Zx() plays a crucial role in this section.
Let x 2 X be fixed. Then, for any x0 , x00 2 X with x0 6¼ x00 , one has that
rf ðxÞk þ k½x00 , x0 ; g ½2x00 x, 2x0 x; gk kx0 x00 k:
kZx ðx0 Þ Zx ðx00 Þk ðkrf ðxÞ
LEMMA 3.1
ð3:3Þ
Proof
Let x0 , x00 2 X with x0 6¼ x00 . We divide the proof into two cases.
Case 1. x00 6¼ x and x0 6¼ x. Then by the definition of first-order divided difference of g,
one concludes that
½2x0 x, x; gðx0 xÞ þ ½2x00 x, x; gðx00 xÞ
1
1
¼ ½2x0 x, x; gðx ð2x0 xÞÞ ½2x00 x, x; gðx ð2x00 xÞÞ
2
2
1
¼ ð gð2x00 xÞ gð2x0 xÞÞ
2
¼ ½2x00 x, 2x0 x; gðx0 x00 Þ
ð3:4Þ
Applicable Analysis
1947
gðx0 Þ gðx00 Þ ¼ ½x00 , x0 ; gðx0 x00 Þ:
ð3:5Þ
and
Note that
Zx ðx0 Þ Zx ðx00 Þ ¼ gðx0 Þ gðx00 Þ ½2x0 x, x; gðx0 xÞ þ ½2x00 x, x; gðx00 xÞ
rf ðxÞÞðx0 x00 Þ:
þ ðrf ðxÞ
This, together with (3.4) and (3.5), gives (3.3).
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Case 2 x00 ¼ x or x0 ¼ x. Without loss of generality, we assume that x00 ¼ x. Then
rf ðxÞÞðx0 xÞ:
Zx ðx0 Þ Zx ðx00 Þ ¼ gðx0 Þ gðxÞ ½2x0 x, x; gðx0 xÞ þ ðrf ðxÞ
ð3:6Þ
Note that
1
½2x0 x, x; gðx0 xÞ ¼ ½2x0 x, x; gðx ð2x0 xÞÞ
2
1
¼ ð gð2x0 xÞ gð2x xÞÞ
2
¼ ½2x x, 2x0 x; gðx0 xÞ:
Combing this with (3.6) yields (3.3) for the case when x00 ¼ x. The proof is complete.
g
! 2Y by
Define a set-valued mapping Qx : X !
þ gðÞ þ rf ðxÞð
xÞ
þ FðÞ:
Qx ðÞ ¼ f ðxÞ
ð3:7Þ
LEMMA 3.2 Let x be a solution of (1.1). Suppose that assumptions (I)–(III) hold.
there is x^ 2 B ðxÞ
satisfying
Then, there exists 4 0 such that for each x 2 B ðxÞ,
^
0 2 f ðxÞ þ gðxÞ þ ðrf ðxÞ þ ½2x^ x, x; gÞðx^ xÞ þ FðxÞ
ð3:8Þ
1
kx^ x k kx x k:
2
ð3:9Þ
and
Proof From [21, Corollary 2], the assumption (III) implies that the mapping Q1
x ðÞ
Hence, there exist rx 4 0, r0 4 0 and M such that
is pseudo-Lipschitz around ð0, xÞ.
Q1
e Q1
x ð y1 Þ \ Brx ðxÞ,
x ð y2 Þ Mk y1 y2 k
for all y1 , y2 2 Br0 ð0Þ:
ð3:10Þ
Let " 4 0 be such that
1
M" :
6
ð3:11Þ
1948
M.H. Rashid et al.
Assumptions (I) and (II) imply that there exists r 4 0 such that
krf ð yÞ rf ðxÞk
"
2
for all y 2 Br ðxÞ
ð3:12Þ
and
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k½z, y; g rgðxÞk
"
2
ð3:13Þ
such that z 6¼ y and kz xk
þ k y xk
r. Let 4 0 be such that
for all z, y 2 Br ðxÞ
(
)
rx r
r0
, , 1,
min
:
ð3:14Þ
3 6
5"
such that x 6¼ x,
and define
Fix x 2 B ðxÞ
rx :¼ 3M"kx xk:
ð3:15Þ
Then, from (3.11) we have
rx :
2
ð3:16Þ
! 2X by
Define a set-valued map x: X !
x ðÞ ¼ Q1
x ½Zx ðÞ:
ð3:17Þ
r :¼ rx and :¼ 13 to
Below, we will apply Lemma 2.1 to the map x with 0 ¼ x,
^ that is,
^
conclude that there exists a fixed point x 2 Brx ðxÞ such that x^ 2 x ðxÞ,
^ 2 Qx ðxÞ,
^ which implies that
Zx ðxÞ
^
0 2 f ðxÞ þ gðxÞ þ ðrf ðxÞ þ ½2x^ x, x; gÞðx^ xÞ þ FðxÞ,
B ðxÞ
and so
i.e. (3.8) holds. Furthermore, x^ 2 Brx ðxÞ
1
kx xk,
rx ¼ 3M"kx xk
kx^ xk
2
i.e. (3.9) holds.
Hence, to complete the proof, it is sufficient to show that Lemma 2.1 is applicable
r :¼ rx and :¼ 13. To do this, it remains to prove that
for the map x with 0 ¼ x,
both conditions (a) and (b) of Lemma 2.1 hold. It is obvious that x 2 Q1
x ð0Þ \ Brx ðxÞ
and according to the definition of the excess e, we obtain
e Q1
x ðxÞ
e Q1
Q1
: ð3:18Þ
x ðxÞÞ
distðx,
x ð0Þ \ Brx ðxÞ,
x ð0Þ \ B ðxÞ,
x ½Zx ðxÞ
with u 6¼ x, (3.2) gives
For all u 2 B ðxÞ
þ gðuÞ þ rf ðxÞðu
xÞ
f ðxÞ gðxÞ ðrf ðxÞ þ ½2u x, x; gÞðu xÞk
kZx ðuÞk ¼ k f ðxÞ
f ðxÞ rf ðxÞð
x xÞk þ kðrf ðxÞ
rf ðxÞÞðu xÞk
k f ðxÞ
þ kð½x, u; g ½2u x, x; gÞðu xÞk:
ð3:19Þ
1949
Applicable Analysis
f ðxÞ rf ðxÞð
x xÞ ¼
Since f ðxÞ
we have
R1
0 ½rf ðx þ tðx xÞÞ rf ðxÞðx xÞdt,
from (3.12)
f ðxÞ rf ðxÞð
x xÞk "kx xk:
k f ðxÞ
ð3:20Þ
þ ku xk
2 r and k2u x xk
þ
On the other hand, since u 6¼ x, kx xk
4 r, (3.13) is applicable to concluding that
kx xk
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k½x, u; g ½2u x, x; gkku xk ðk½x, u; g rgðxÞk
þ k½2u x, x; g rgðxÞkÞku
xk "ku xk:
þ 2ku xkÞ This, together with (3.19) and (3.20), gives that kZx ðuÞk "ðkx xk
with u 6¼ x,
5": By (3.14), we have 5" r0 and deduce that for all u 2 B ðxÞ
f ðxÞ
Zx ðuÞ 2 Br0 ð0Þ. Furthermore, in the case when u ¼ x, we have Zx ðxÞ ¼ f ðxÞ
x xÞ. Hence, it follows from (3.20) that kZx(x)k " r0. Consequently,
rf ðxÞð
we obtain
Zx ðuÞ 2 Br0 ð0Þ 8u 2 B ðxÞ:
ð3:21Þ
On the other hand, letting u ¼ x in (3.19), we obtain
k f ðxÞ
f ðxÞ rf ðxÞðx xÞk þ k½x, x;
g ½2x x, x; gkkx xk
kZx ðxÞk
2"kx xk:
implies that
This, together with (3.18) and (3.10) (with y1 ¼ 0 and y2 ¼ Zx ðxÞ),
1
MkZx ðxÞk
2M"kx xk ¼ 1 rx ¼ ð1 Þr:
x ðxÞÞ
distðx,
3
This shows that the condition (a) of Lemma 2.1 is satisfied.
Now, we show that the condition (b) of Lemma 2.1 is satisfied. To prove this, let
Then we have that x0 , x00 2 Brx ðxÞ
B ðxÞ
by (3.16) and hence
x0 , x00 2 Brx ðxÞ.
0
Zx ðx Þ, Zx ðx00 Þ 2 Br0 ð0Þ by (3.21). This together with (3.10) (with y1 ¼ Zx(x0 ) and
y2 ¼ Zx(x00 )) implies that
x ðx00 ÞÞ eðx ðx0 Þ \ B ðxÞ,
x ðx00 ÞÞ
eðx ðx0 Þ \ Brx ðxÞ,
0
00
Q1
¼ eðQ1
x ½Zx ðx Þ \ B ðxÞ,
x ½Zx ðx ÞÞ
MkZx ðx0 Þ Zx ðx00 Þk:
ð3:22Þ
þ kx00 xk
2 r and k2x00 x xkþ
In the case when x0 6¼ x00 , since kx0 xk
0
6 r, (3.13) is applicable to concluding that
k2x x xk
k½x00 , x0 ; g ½2x00 x, 2x0 x; gk kx0 x00 k
0
þ k½2x00 x, 2x0 x; g rgðxÞkÞkx
ðk½x00 , x0 ; g rgðxÞk
x00 k
"kx0 x00 k:
This, together with (3.3) and (3.12), yields
rf ðxÞk þ k½x00 , x0 ; g ½2x00 x, 2x0 x; gk kx0 x00 k
kZx ðx0 Þ Zx ðx00 Þk krf ðxÞ
2"kx0 x00 k:
1950
M.H. Rashid et al.
Combining this with (3.22) and (3.11) gives
1
x ðx00 ÞÞ 2M"kx0 x00 k kx0 x00 k ¼ kx0 x00 k:
eðx ðx0 Þ \ Brx ðxÞ,
3
ð3:23Þ
Thus the condition (b) of Lemma 2.1 is satisfied. The proof of the lemma is
complete.
g
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The following theorem shows that under the assumptions (I)–(III), the sequence
{xn} generated by (1.4) is convergent.
THEOREM 3.1 Let x be a solution of (1.1). Then under the assumptions (I)–(III), there
there is a sequence {xn} generated by (1.4)
exists 4 0 such that for each x0 2 B ðxÞ,
with initial point x0 satisfying
1
kxk xk
kxkþ1 xk
2
Proof
for each k ¼ 0, 1, . . . :
ð3:24Þ
By Lemma 3.2, there exists 4 0 such that
) there is x^ 2 B ðxÞ
such that (3.8) and (3.9) hold.
x 2 B ðxÞ¼
ð3:25Þ
It follows from (3.25) that there exists x1 2 B ðxÞ
such that
Let x0 2 B ðxÞ.
0 2 f ðx0 Þ þ gðx0 Þ þ ðrf ðx0 Þ þ ½2x1 x0 , x0 ; gÞðx1 x0 Þ þ Fðx1 Þ
and
1
kx0 xk,
kx1 xk
2
and so (3.24) holds for k ¼ 0. Now assume that (1.4) generated x0, x1, . . . , xk
such that
satisfying (3.24). Then, by (3.25), we can choose xkþ1 2 B ðxÞ
0 2 f ðxk Þ þ gðxk Þ þ ðrf ðxk Þ þ ½2xkþ1 xk , xk ; gÞðxkþ1 xk Þ þ Fðxkþ1 Þ
and
1
kxk xk,
kxkþ1 xk
2
and so (3.24) holds for k. This completes the proof of the Theorem 3.1.
g
The remainder of this section is devoted to the study of the superlinear
convergence of the sequence generated by the method (1.4). For this purpose, let
p 2 (0, 1]. Consider the following assumption:
that is,
(Ip) f is Fréchet-differentiable and its derivative is p-Hölder continuous at x,
there exist " 4 0 and r 4 0 such that
"kx xk
p
krf ðxÞ rf ðxÞk
for all x 2 Br ðxÞ:
(IIp) g is Fréchet-differentiable at x and the first-order divided difference of g is
that is, there exist " 4 0 and r 4 0 such that
p-Hölder continuous at x,
p Þ for all x, y 2 Br ðxÞ
with x 6¼ y:
"ðkx xk
p þ k y xk
k½x, y; g rgðxÞk
1951
Applicable Analysis
The following lemma is an analogue of Lemma 3.2. However, the proof
technique is slightly different.
LEMMA 3.3 Let x be a solution of (1.1). Let p 2 (0, 1]. Suppose that assumptions (Ip),
there
(IIp) and (III) hold. Then there exist 4 0 and 4 0 such that for each x 2 B ðxÞ,
satisfying
is x^ 2 B ðxÞ
^
0 2 f ðxÞ þ gðxÞ þ ðrf ðxÞ þ ½2x^ x, x; gÞðx^ xÞ þ FðxÞ
ð3:26Þ
kx xk
pþ1 :
kx^ xk
ð3:27Þ
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and
Proof From [21, Corollary 2], the assumption (III) implies that Q1
x ðÞ is pseudo Then there exist constants rx 4 0, r0 4 0 and M such that
Lipschitz around ð0, xÞ.
Q1
e Q1
x ð y1 Þ \ Brx ðxÞ,
x ð y2 Þ Mk y1 y2 k
for all y1 , y2 2 Br0 ð0Þ:
ð3:28Þ
Assumptions (Ip) and (IIp) imply that there exist L 4 0 and r 4 0 such that
krf ðxÞ rf ðxÞk
L
p
kx xk
2
for all x 2 Br ðxÞ
ð3:29Þ
and
k½x, y; g rgðxÞk
L
p þ k y xk
pÞ
ðkx xk
2
with x 6¼ y:
for all x, y 2 Br ðxÞ
ð3:30Þ
Write
s :¼ Lð1 þ ð4 þ 3p Þð p þ 1ÞÞ
and
q :¼ Lð2 þ 3p Þ:
ð3:31Þ
Take
4
Mð2p þ 5ÞLð1 þ 2ð p þ 1ÞÞ
:
3ð p þ 1Þ
Let 4 0 be such that
(
min
1 1 )
rx r
ð p þ 1Þr0
2ð p þ 1Þ p 1 p
, , 1,
,
,
:
Mqð2p þ 5Þ
3 6
s
ð3:32Þ
such that x 6¼ x,
and define
Fix x 2 B ðxÞ
pþ1 :
rx :¼ kx xk
ð3:33Þ
Then, it follows from (3.32) that
rx pþ1 ¼ p :
! 2X by
Define a set-valued map x: X !
x ðÞ ¼ Q1
x ½Zx ðÞ:
ð3:34Þ
1952
M.H. Rashid et al.
pþ1Þ
r :¼ rx and :¼ 2ð2pþ5
Below, we will apply Lemma 2.1 to the map x with 0 ¼ x,
to
such that x^ 2 x ðxÞ
^ which implies
conclude that there exists a fixed point x^ 2 Brx ðxÞ
^ 2 Qx ðxÞ,
^ that is,
that Zx ðxÞ
^
0 2 f ðxÞ þ gðxÞ þ ðrf ðxÞ þ ½2x^ x, x; gÞðx^ xÞ þ FðxÞ,
B ðxÞ,
one has
and so (3.26) holds. Furthermore, since x^ 2 Brx ðxÞ
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pþ1 ,
rx ¼ kx xk
kx^ xk
i.e. (3.27) holds.
Hence, to complete the proof it is sufficient to show that Lemma 2.1 is applicable.
To this end, it remains to prove that both conditions (a) and (b) of Lemma 2.1 hold.
and according to the definition of the excess e, we
It is trivial that x 2 Q1
x ð0Þ \ Brx ðxÞ
obtain
ð3:35Þ
eðQ1
x ðxÞÞ
eðQ1
Q1
x ðxÞÞ
distðx,
x ð0Þ \ Brx ðxÞ,
x ð0Þ \ B ðxÞ,
x ½Zx0 ðxÞÞ:
with u 6¼ x, by definition,
Moreover, for each u 2 B ðxÞ
þ gðuÞ þ rf ðxÞðu
xÞ
f ðxÞ gðxÞ ðrf ðxÞ þ ½2u x, x; gÞðu xÞk
kZx ðuÞk ¼ k f ðxÞ
f ðxÞ rf ðxÞð
x xÞk þ kðrf ðxÞ
rf ðxÞÞðu xÞk
k f ðxÞ
þ kð½x, u; g ½2u x, x; gÞðu xÞk:
ð3:36Þ
R1
rf ðxÞðx
xÞ
¼
Since f ðxÞ f ðxÞ
we have
0 ½rf ðx þ tðx xÞÞ rf ðxÞðx xÞdt,
f ðxÞ rf ðxÞð
x xÞk k f ðxÞ
from (3.29)
L
kx xkpþ1 :
pþ1
ð3:37Þ
Note that from (3.29) we have
rf ðxÞÞðu xÞk kðrf ðxÞ
L
p ku xk:
kx xk
2
þ ku xk
2 r and k2u x xk
þ
On the other hand, since u 6¼ x, kx xk
4 r, (3.30) is applicable to concluding that
kx xk
k½x, u; g ½2u x, x; gkku xk
þ k½2u x, x; g rgðxÞkÞku
ðk½x, u; g rgðxÞk
xk
L
p þ ku xk
p þ k2u x xk
p þ kx xk
p Þku xk:
ðkx xk
2
ð3:38Þ
Consequently, it follows from (3.36) that
kZx ðuÞk Lð1 þ ð p þ 1Þð4 þ 3p ÞÞ pþ1
pþ1
ð3:39Þ
Note that s (p þ 1)r0 and 1 because of (3.32). It follows from (3.39) that
kZx ðuÞk s
s
pþ1 r0 :
pþ1
pþ1
Applicable Analysis
1953
with u 6¼ x, we have Zx ðuÞ 2 Br0 ð0Þ. Furthermore, in the case
Hence, for all u 2 B ðxÞ
f ðxÞ rf ðxÞð
x xÞ. Then, it follows from (3.37)
when u ¼ x, we have Zx ðxÞ ¼ f ðxÞ
L pþ1
r0 . Therefore, we get
that kZx ðxÞk pþ1
Zx ðuÞ 2 Br0 ð0Þ 8u 2 B ðxÞ:
ð3:40Þ
On the other hand, letting u ¼ x in (3.36), from (3.38) and (3.37) we obtain
kZx ðxÞk
Lð1 þ 2ð p þ 1ÞÞ
kx xkpþ1 :
pþ1
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This, together with (3.28), (3.33) and (3.35), gives
x ðxÞÞ
MkZx ðxÞk
distðx,
MLð1 þ 2ð p þ 1ÞÞ
kx xkpþ1
pþ1
3
kx xkpþ1
5
2p þ 5
2ð p þ 1Þ
¼ 1
rx
2p þ 5
¼ ð1 Þr,
that is, the condition (a) of Lemma 2.1 is satisfied.
Now, we show that condition (b) of Lemma 2.1 holds. To do this, let x0 , x00 2 Brx ðxÞ.
B ðxÞ
by (3.34) and Zx ðx0 Þ, Zx ðx00 Þ 2 Br0 ð0Þ by (3.40).
Then we have x0 , x00 2 Brx ðxÞ
This together with (3.28) implies that
x ðx00 ÞÞ eðx ðx0 Þ \ B ðxÞ,
x ðx00 ÞÞ
eðx ðx0 Þ \ Brx ðxÞ,
0
00
Q1
¼ eðQ1
x ½Zx ðx Þ \ B ðxÞ,
x ½Zx ðx ÞÞ
MkZx ðx0 Þ Zx ðx00 Þk:
ð3:41Þ
þ kx00 xk
2 r and k2x00 x xkþ
In the case when x0 6¼ x00 , since kx0 xk
0
6 r, (3.30) is applicable to concluding that
k2x x xk
k½x00 , x0 ; g ½2x00 x, 2x0 x; gk kx0 x00 k
þ k½2x00 x, 2x0 x; g rgðxÞk
Þkx0 x00 k
ðk½x00 , x0 ; g rgðxÞk
L
p þ k2x0 x xk
p Þkx0 x00 k:
ðkx x00 kp þ kx x0 kp þ k2x00 x xk
2
This, together with (3.3) and (3.29) yields that
kZx ðx0 Þ Zx ðx00 Þk
rf ðxÞk þ k½x00 , x0 ; g ½2x00 x, 2x0 x; gk kx0 x00 k
krf ðxÞ
1
p þ k2x0 x xk
p Þ kx0 x00 k
p þ ðkx x00 kp þ kx x0 kp þ k2x00 x xk
L kx xk
2
Lð2 þ 3p Þp kx0 x00 k
¼ qp kx0 x00 k:
1954
M.H. Rashid et al.
Combing this with (3.41) gives
x ðx00 ÞÞ Mqp kx0 x00 k eðx ðx0 Þ \ Brx ðxÞ,
2ð p þ 1Þ 0
kx x00 k ¼ kx0 x00 k,
2p þ 5
ð3:42Þ
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where the second inequality holds because of (3.32). This implies that the condition
(b) of Lemma 2.1 is satisfied. The proof of the lemma is complete.
g
The following theorem, which was given in [1] (i.e. [1, Theorem 1)) in the case
when r f and the first-order divided difference of g are p-Hölder continuous around
shows that the sequence generated by (1.4) converges superlinearly.
a solution x,
However, the proof given in [1] contains a gap as we explain in the following. In fact,
one key step in their proof is to show that condition (b) of Lemma 2.1 is satisfied. To
do this, they used the following estimate:
kx0 x00 kp ¼ kx0 x00 kp1 kx0 x00 k ð2Þp1 kx0 x00 k
and p 2 (0, 1], the last inequality of which is clearly not true
for all x0 , x00 2 B ðxÞ
because p 2 (0, 1].
THEOREM 3.2 Let x be a solution of (1.1). Let p 2 (0, 1]. Suppose that assumptions
(Ip), (IIp) and (III) hold. Then, there are 4 0 and 4 0 such that for each x0 2 B ðxÞ,
there exists a sequence {xn} generated by (1.4) with initial point x0, which converges to
x and satisfies that
kxk xk
pþ1
kxkþ1 xk
Proof
for each k ¼ 0, 1, . . . :
ð3:43Þ
By Lemma 3.3, there exist 4 0 and 1 4 0 such that
) there is x^ 2 B1 ðxÞ
satisfying 1.3 and 1.4.
x 2 B1 ðxÞ¼
ð3:44Þ
Take
(
1p )
1
0 5 5 min 1 ,
:
ð3:45Þ
Then, it follows from (3.44) that there exists x1 2 B ðxÞ
such that
Let x0 2 B ðxÞ.
0 2 f ðx0 Þ þ gðx0 Þ þ ðrf ðx0 Þ þ ½2x1 x0 , x0 ; gÞðx1 x0 Þ þ Fðx1 Þ
and
kx0 xk
pþ1 ,
kx1 xk
which implies that (3.43) holds for k ¼ 0. We proceed by induction. To do this,
assume that x0, x1, . . . , xk satisfies (3.43), Then, by (3.44) again, we can obtain
such that
xkþ1 2 B ðxÞ
0 2 f ðxk Þ þ gðxk Þ þ ðrf ðxk Þ þ ½2xkþ1 xk , xk ; gÞðxkþ1 xk Þ þ Fðxkþ1 Þ
Applicable Analysis
1955
kxk xk
pþ1 ,
kxkþ1 xk
ð3:46Þ
and
we have from (3.43) that
so (3.43) holds for k. On the other hand, since fxk g B ðxÞ,
p kxk xk
kxkþ1 xk
for each k:
ð3:47Þ
As p 5 1 by (3.45), it follows that {xk} converges to x and the proof is complete. g
In particular, in the case when g ¼ 0, method (1.4) is reduced to
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0 2 f ðxk Þ þ rf ðxk Þðxkþ1 xk Þ þ Fðxkþ1 Þ,
ð3:48Þ
and we immediately have the following two corollaries: the first one extends
[5, Theorem 1] which was proved in the stronger assumption that rf is continuous on
a neighbourhood of a solution; while the second one extends [6, Theorem] which is in
the stronger assumption that r f is p-Hölder continuous on a neighbourhood of a
solution.
COROLLARY 3.1 Let x be a solution of (1.2). Then under the assumptions (I) and (III),
there is a sequence {xn} generated by
there exists 4 0 such that for each x0 2 B ðxÞ,
(3.48) with initial point x0 satisfying
1
kxk xk
kxkþ1 xk
2
for each k ¼ 0, 1, . . . :
COROLLARY 3.2 Let x be a solution of (1.2). Let p 2 (0, 1]. Suppose that assumptions
there
(Ip) and (III) hold. Then, there are 4 0 and 4 0 such that for each x0 2 B ðxÞ,
exists a sequence {xn} generated by (3.48) with initial point x0, which converges to x
and satisfies that
kxk xk
pþ1
kxkþ1 xk
for each k ¼ 0, 1, . . . :
Acknowledgements
M.H. Rashid was fully supported by China Scholarship Council. Research of J.H. Wang was
partially supported by the National Natural Science Foundation of China (grant 11001241).
C. Li was supported in part by the National Natural Science Foundation of China (grant
11171300) and by Zhejiang Provincial Natural Science Foundation of China (grant
Y6110006).
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