A regularized parameter choice in regularization for a common solution of a finite system of...

ISSN 0965�5425, Computational Mathematics and Mathematical Physics, 2014, Vol. 54, No. 3, pp. 397–406. © Pleiades Publishing, Ltd., 2014.

397

1. INTRODUCTION AND PRELIMINARIES

Let E be a real Banach space and E* be its dual space. For the sake of simplicity, the norms of E andE* are denoted by the symbol ||.||. The symbol ⟨x, x*⟩ denotes the value of x* ∈ E* at x ∈ E.

Let A be an m�accretive mapping in E, i.e., A : E E has the following properties:

(i) ⟨A(x) – A(y), (x – y)⟩ ≥ 0, for all x, y ∈ �(A), the domain of A, j(x – y) ∈ J(x – y), the normalizedduality mapping of E, the mapping from E onto E*, satisfying the condition

and

(ii) �(A + λI) = E for each λ > 0, where �(A) denotes the range of A and I is the identity mapping in E.

Note that if there exist two positive constants α and γ such that

and

then A is said to be α�strongly accretive and γ�cocoercive, respectively. Obviously, every γ�cocoercive map�ping is (1/γ)�Lipschtiz continuous. Here, A is called L�Lipschitz continuous, if there exists a positive con�stant L such that, for all x, y ∈ �(A), we have ||A(x) – A(y)|| ≤ L||x – y||. In the case that L = 1, A is said tobe nonexpansive. When γ or α = 0, i.e., A satisfies condition (i), A is called accretive. It is well�known in[1] that if A is accretive and Lipschitz continuous with �(A) = E, a Banach space, then A is m�accretive.

Recall also that a mapping T in E is said to be pseudocontractive, if

x j x( ),⟨ ⟩ x2, j x( ) x , j x( ) J x( ), x∀ E,∈ ∈= =

A x( ) A y( )– j x y–( ),⟨ ⟩ α x y–2,≥

A x( ) A y( )– j x y–( ),⟨ ⟩ γ A x( ) A y( )–2, x∀ y � A( ),∈,≥

Tx Ty– j x y–( ),⟨ ⟩ x y–2,≤

A Regularized Parameter Choice in Regularization for a Common Solution of a Finite System of Ill�Posed Equations Involving

Lipschitz Continuous and Accretive Mappings1

Nguyen Buonga and Nguyen Dinh Dungb

a Vietnamese Academy of Science and Technology, Institute of Information Technology, 18, Hoang Quoc Viet, Hanoi, Vietnam

e�mail: [email protected] Thainguyen National University, Thainguyen City, Vietnam

e�mail: [email protected] May 22, 2013; in final form, October 3, 2013

Abstract—In this paper, we present a regularized parameter choice in a new regularization method ofBrowder–Tikhonov type, for finding a common solution of a finite system of ill�posed operator equa�tions involving Lipschitz continuous and accretive mappings in a real reflexive and strictly convexBanach space with a uniformly Gâteaux differentiate norm. An estimate for convergence rates of reg�ularized solution is also established.

DOI: 10.1134/S0965542514030130

Keywords: accretive and α�strong accretive mapping, γ�co�coercive mapping, reflexive Banach space,Fréchet differentiable and the Browder–Tikhonov regularization.

1 The article is published in the original.

398

COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS Vol. 54 No. 3 2014

NGUYEN BUONG, NGUYEN DINH DUNG

for all x, y ∈ �(T). Obviously, every nonexpansive mapping is pseudocontractive, and if A is accretive, thenT = I – A is also pseudocontractive.

In this paper, we consider the problem of finding a common solution for a finite system of followingequations

(1)

where each fi is an element in E, A0 is L0�Lipschitz continuous and accretive and the other Ai is γi�cocoer�cive such that �(Ai) = E for i = 0, 1, …, N. Denote by Si the set of solutions for i�equation in (1), Through�

out this paper, we assume that S := ≠ .

We are specially interested in the situation where the data fi is not exactly known, i.e., we have only the

approximations , satisfying

0, i = 0, 1, …, N. (2)

It is well�known in [2] that each equation in (1), in general, is ill�posed, by this we mean that the solu�tions do not depend continuously on the data fi. Consequently, the system of equations is ill�posed, in gen�eral. Many practical inverse problems arising in mathematics, mechanics and physics, are naturally for�mulated in such a way and some methods are studied for solving (1), when Ai are continuous and weaklyclosed, see, for example, [3–6].

In 2006, to solve (1) with fi = θE*, the zero element in E*, when Ai : E E*, i = 0, 1, …, N, are N + 1potential, monotone and hemicontinuous mappings, in [7], Buong proposed the following regularizationmethod of Browder–Tikhonov’s type:

(3)

where also are the hemi�continuous and monotone approximations. When Ai are continuous andweakly closed in Hilbert spaces, a method of regularization, that is different from ones in [3–6], was pro�posed by in [8].

The first result in the case that N = 0 and A := A0, an m�accretive mapping with domain �(A) = E, aBanach space possesses the ES and approximation properties [9], was firstly studied by Alber in [2], Heproved that, if A is a demicontinuous and accretive mapping, then the equation

(4)

for each α > 0, has a unique solution ; if J is continuous and at the same time weak to weak continuous

in E and δ/α 0, then x∗, a solution of the equation

(5)

In [10], Alber and Ryazantseva showed the residual principle to choose α = α(δ), based on the function

ρ(α) = , for the case x+ = θE, the null element in E. They proved the following theorem.

Theorem 1 (see [10, Theorem 3.2.3]). Let E be a real, reflexive and strictly convex Banach space withstrictly convex dual space E*, J be a continuous and weak�to�weak continuous duality mapping in E and letA be a hemicontinuous and accretive mapping with D(A) = E, that possesses an approximation. Assume thatδ ∈ (0, 1] and

Then, there exists at least a value such that

(6)

Ai x( ) fi, i 0 1 … N,, , ,= =

∩i 0=N

Si 0

fi δ

fi fi δ

– δ, δ≤

αμiAi

hx( ) αJ x( )+

i 0=

N

∑ θE*,=

μ0 0 μi μi 1+ 1, i< < < 1 2 … N 1,–, , ,= =

Aih

A x( ) α x x+

–( )+ fδ,=

xα

δ

xα

δ

A x( ) f.=

α xα

δ

A θE( ) fδ– Kδp, K 1, 0 p 1.≤<> >

α

ρ α( ) Kδp,=


A REGULARIZED PARAMETER CHOICE IN REGULARIZATION 399

where is the solution of Eq. (4) with α = and x+ = θE. Furthermore, let δ 0. One has: (1) 0;

(2) if p ∈ (0, 1) then δ/ 0 and x∗, the unique solution of the following inequality

where denotes the set of all solutions of the Eq. (5); (3) if p = 1 and = {x∗} then converges weakly to x∗and δ/α(δ) ≤ C, a positive constant.

In [11], Buong considered the generalized discrepancy principle for selecting α by the Eq. (6), underthe conditions: A is weak continuous and satisfies the following condition

(7)

where A'(x) denotes the derivative of A at x ∈ E, A'(x∗)* is the adjoint of A'(x∗), τ is some positive constant

and Q is the normalized duallity mapping of E*.In this paper, for the stated problem, we propose a new regularization method, that consists of the fol�

lowing operator equation

(8)

where μ ∈ (0, 1) is a fixed number. In the next section, we consider: (1) the strong convergence of the solu�

tion of (8); (2) the problem of selecting the value = α(δ), and (3) convergence rates for the regular�

ized solution with condition (7) only for A0.

All terminologies and some facts, that will be used in this paper, are given in [12] (see, also [13]).

2. MAIN RESULTS

For an m�accretive mapping A and any fixed element f in E, we can define a mapping u = Tf(x) by

(9)

for each x ∈ E. Since A is m�accretive, the existence of Tf is asserted. It is not difficult to verify that Tf hasthe following properties:

(i) D(Tf) = E;

(ii) Tf is nonexpansive;

(iii) Fix(Tf) = where Fix(Tf) denotes the set of fixed points of Tf, i.e.,

Clearly, Af := A – f is also an m�accretive mapping in E.

Theorem 2. Let E be a real, reflexive and strictly convex Banach space with a uniformly Gâteaux differen�tiable norm, let A0 be a Lipschtiz continuous and accretive mapping and let Ai be a γi�cocoercive mapping on Efor i = 1, 2, …, N. Then, we have:

(i) for each α > 0 and any ∈ E, Eq. (8) has a unique solution ;

(ii) if S ≠ , the elements satisfy (2) for i = 0, 1, …, N, and the parameter α is chosen, such that α,

δ/α 0, then converges strongly to an element x∗ ∈ S, satisfying

(10)

Proof. (i) Since Ai is accretive and Lipschitz continuous on E for i = 0, 1, …, N, the mapping A := A0 +

also is an accretive and Lipschitz continuous mapping on E. So, it is m�accretive, and hence,

xα

δ α α

α xα

δ

x* J x* y–( ),⟨ ⟩ 0 y∀ S̃,∈≤

S̃ S̃ xα

δ

A y( ) A x*( )– QA' x*( )*J y x*–( )– τ A y( ) A x*( )– , y E,∈≤

A0 x( ) αμAi x( ) fi

δ–( )

i 1=

N

∑ α x x+

–( )+ + f0 δ

,=

xα

δ α

xα δ( )

δ

A u( ) u+ f x,+=

S̃

Fix Tf( ) x E : x∈ Tf x( )={ }.=

fi δ

xα

δ

0 fi δ

xα

δ

x* x+

– J x* z–( ),⟨ ⟩ 0 z∀ S.∈≤

αμAii 1=

N∑

400



Eq. (8) has a solution, denoted by , for each α > 0 and any ∈ E. This solution is unique, because the

mapping [A + α(I – x+)](.) is α�strongly accretive.

(ii) Without any loss of generality, we assume that Nαμ ≤ 1. From (8), we have

for any z ∈ S, and hence

because Ai is accretive. Therefore,

which implies the boundedness of { }. Then, there exists a positive constant M1 such that ≤ M1, forall α, δ > 0 with α, δ/α 0. Consequently,

(11)

Again, from (8) and that Ai is (1/γi)�Lipschitz continuous for i = 1, 2, …, N, we also have that

which implies

(12)

Also by virtue of (8) and that A0 is accretive and Ai is γi�cocoercive, we obtain

Thus,

(13)

Now, we consider the mapping Ti = I – Ai and = Ti + fi. Clearly, z ∈ S if and only if z ∈

. Since Ai is accretive, the mapping Ti is pseudocontractive. So, is the mapping . On the

other hand, form (12) and (13), we obtain that ||(I – ) || 0 as α, δ/α 0 for i = 0, 1, …, N.

We first observe that the mapping 2I – has a nonexpansive inverse, denoted by �i = (2I – )–1

.

xα

δfi

δ

A0 xα

δ( ) A0 z( )– αμAi xα

δ( ) Ai z( )–( )i 1=

N

∑ α xα

δx

+–( ) J xα

δz–( ),+ +

= f0 δ

f0– J xα

δz–( ),⟨ ⟩ αμ

fi δ

fi–( )i 1=

N

∑ J xα

δz–( ), ,+

xα

δx

+– J xα

δz–( ),⟨ ⟩ 1

α�� f0

δf0– J xα

δz–( ),⟨ ⟩ αμ

α�� fi

δfi– J xα

δz–( ),⟨ ⟩ ,

i 1=

N

∑+≤

xα

δz–⟨ ⟩

2x

+z– J xα

δz–( ),⟨ ⟩ 2 δ

α�� xα

δz– z∀ S,∈+≤

xα

δxα

δ

xα

δz–

2

x+

z– J xα

δz–( ),⟨ ⟩ 2 δ

α�� M1 z+( ) z∀ S.∈+≤

A0 xα

δ( ) f0– α xα

δx

+– αμ

Ai xα

δ( ) Ai z( )–i 1=

N

∑ 2δ α xα

δx

+– αμ 1

γi

�� M1 z+( )i 1=

N

∑ 2δ,+ +≤+ +≤

A0 xα

δ( ) f0–α δ/α 0→,

lim 0.=

γi Ai xα

δ( ) fi–2

i 1=

N

∑ Ai xα

δ( ) fi– J xα

δz–( ),⟨ ⟩

i 1=

N

∑≤

≤ α1 μ–x

+xα

δ– J xα

δz–( ),⟨ ⟩ δ/αμ( ) Nδ+[ ] J xα

δz–( )+

≤ α1 μ–x

+z– α1 μ– δ/α Nδ+[ ]+( ) M1 z+( ).

Ai xα

δ( ) fi–α δ/α 0→,

lim 0, i 1 2 … N., , ,= =

T fi

∩i 0=N

Fix T fi( ) T

fi

T fi xα

δ

T fi T

fi



Indeed, 2I – = I + I – = I + Ai – fi, that is 1�strongly accretive on E. So, �(2I – ) = E. Next,

from (9),

the mapping Ai(.) = Ai(.) – fi is m�accretive, and (I + Ai)–1 is nonexpansive, it follows that �i is nonexpan�

sive. Clearly, Fix(�i) = Fix( ) = Si. Thus,

Consequently,

and hence, || – || 0 as α, δ/α 0.

Next, let {xk} be any subsequence of { } with αk, δk/αk 0 as k ∞. We consider the functional

ϕ(x) = μk||xk – x||2 for all x ∈ E. We see that ϕ(x) ∞ as ||x|| ∞ and ϕ is continuous and convex, soas E is reflexive, there exists ∈ E such that ϕ( ) = minx ∈ Eϕ(x). Hence, the set

≠ .

It is easy to see that C* is a bounded, closed, and convex subset of E (see, [14]). Since ||xk – �ixk|| 0, wehave that

which implies that �iC* ⊂ C*, that is C* is �i�invariant, for i = 0, 1, …, N. Now, we show that C* contains

a common fixed point of . Since E is a strictly convex and reflexive Banach space, C* is a Che�

byshev set (see [14]) in E, and hence, for a point z ∈ , there exists a unique ∈ C* such that

By z = �iz and ∈ C*, we have

and hence = . So, there exists a point ∈ ∩ C*. Now, from Lemma 1 in [13], it fol�lows that

(14)

Taking z = in (11) and x = x+ in (14), we obtain that μk||xk – ||2 = 0. Hence, there exists a subse�quence { } of {xk} which strongly converges to as l ∞. Again, from (11) and the norm to weak star

continuous property of the normalized duality mapping J on bounded subsets of E, we obtain that

(15)

Since z and belong to S, a closed and convex subset, by replacing z in (15) by sz + (1 – s) for s ∈ (0, 1),using the well�known property J(s( – z)) = sJ( – z) for s > 0, dividing by s and taking s 0, we obtain

The uniqueness of x∗ in (10) guarantees that = x∗. So, all the net { } converges strongly to x∗ as α,

δ/α 0. This completes the proof.

T fi T

fi Tfi

2I T fi–( )x I I T

fi–+( )x I Ai+( )x fi,–= =

T fi

xα

δT

fixα

δ– 2I T

fi–( )xα

δxα

δ– �i

1–xα

δxα

δand �i�i

1–xα

δ– xα

δ.= = =

xα

δ �ixα

δ– �i�i

1–xα

δ �ixα

δ– �i

1–xα

δxα

δ–≤ I T

fi–( )xα

δ,= =

xα

δ �ixα

δ

xα

δ

z̃ z̃

C* := u E : ϕ u( )∈ ϕ x( )x E∈

min={ } 0

ϕ �i z̃( ) μk xk �i z̃–2

μk �ixk �i z̃–2

μk xk z̃–2≤ ϕ z̃( ),= = =

�i{ }i 0=N

∩i 0=N

Fix �i( ) z̃

z z̃– z x–x C*∈

inf .=

�i z̃

z �i z̃– �iz �i z̃– z z̃– ,≤=

�i z̃ z̃ z̃ ∩i 0=N

Fix �i( )

μk x z̃– J xk z̃–( ),⟨ ⟩ 0 x∀ E.∈≤

z̃ z̃xkl

z̃

z x+

– J z̃ z–( ),⟨ ⟩ 0 z∀ S∈≤ Fix �i( ).i 0=

N

∩=

z̃ z̃z̃ z̃

z̃ x+

– J z̃ z–( ),⟨ ⟩ 0 z∀ S.∈≤

z̃ xα

δ

402



Next, we consider the function ρ(α) = α|| – x+||.

Lemma 1.(i) The function ρ(α) is continuous on (α0, +∞), for each α0 > 0;

(ii) If

(16)

for all δ ≥ 0, where = fi, then

Proof. Let α and β be two arbitrary numbers in (α0, +∞). From (8) it follows that

Therefore,

Consequently,

that implies the continuity of ρ(α) at any α ∈ (α0, +∞). Now, from (8), we also have that

and hence,

Thus,

that together with

and the continuity of Ai implies the conclusion of the Lemma.

Theorem 3. Let x+ ∉ S be a point in E such that (16) holds. Then, there exists at least a value = α(δ)such that

(17)

xα

δ

Ai x+( ) fi

δ–( )

i 1=

N

∑ 0,>

fi 0

ρ α( )α +∞→lim +∞.=

A0 xα

δ( ) A0 xβ

τ( )– αμAi xα

δ( ) fi δ

–( )i 1=

N

∑ βμAi xβ

δ( ) fi δ

–( )i 1=

N

∑– α xα

δx

+–( ) β xβ

δx

+–( )–+ + 0.=

α xα

δxβ

δ– J xα

δxβ

δ–( ),⟨ ⟩ β α–( ) xβ

δJ xα

δxβ

δ–( ),⟨ ⟩– αμ

Ai xα

δ( ) Ai xβ

δ( )– J xα

δxβ

δ–( ),( )

i 1=

N

∑+

– βμ αμ–( ) Ai xβ

δ( ) fi δ

– J xα

δxβ

δ–( ),(⟨ ⟩

i 1=

N

∑ 0.=

xα

δx

+– xβ

δx

+–– xα

δxβ

δ– α β–

α0

�� xβ

δ αμ βμ–

α0

�� Ai xβ

δ( ) fi δ

– ,

i 1=

N

∑+≤ ≤

A0 xα

δ( ) A0 x+( )– αμ

Ai xα

δ( ) Ai x+( )–( )

i 1=

N

∑ α xα

δx

+–( )+ + f0

δA0 x

+( )– αμfi

δAi x

+( )–( ),

i 1=

N

∑+=

xα

δx

+–

f0 δ

A0 x+( )–

α�� 1

α1 μ–�� fi

δAi x

+( )– .

i 1=

N

∑+≤

xα

δx

+–

α +∞→lim 0,=

ρ α( ) αμAi xα

δ( ) fi δ

–( )i 1=

N

∑ A0 xα

δ( ) f0 δ

– ,–≥

α

α K 2–( )δp/ 2 z x

+–( ), z S,∈≥



and

(18)

where is the solution of (8) with α = . Furthermore, let δ 0. One has: (1) 0; (2) if p ∈ (0, 1) then

δ/α(δ) 0 and x∗, solving (10); (3) if p = 1, J is weak�to�weak continuous and S = {x∗} then

converges weakly to x∗ and δ/α(δ) ≤ C, a positive constant.

Proof. From (8), the accretive property of Ai and the definition of J it follows

(19)

for any z ∈ S. With every fixed δ > 0, we have

for sufficiently small α. Then, for these values of α, we can assume that Nαμ ≤ 1, 0 < δ < 1, and hence,

(20)

Now, consider the function

for α > α0 > 0. Since the function d(α) is continuous, by Lemma 1,

From (20) it implies that there is a value of α > 0 such that d(α) < 0. Hence, there exists = α(δ),satisfying (17) and (18). The first conclusion of the theorem is proved.

Further, from (17) with x+ ∉ S, it follows that

(21)

Therefore, in the case that 0 < p < 1, we have δ/ 0, as δ 0. Now, we need to prove thatα(δ) 0, as δ 0. If it is not true, then considerably two cases: (1) there exists a sequence =

α(δk) C0, some positive constant, with δk 0 as k ∞; (2) there exists a sequence = α(δk) ∞

with δk 0 as k ∞; In the case 1, firstly we prove that { } is bounded. Without any loss of generality,

we assume that || – x+|| ∞ as k ∞. From (8) and the property of Ai, we can write that

which is impossible. So, from (18) and ρ( ) = , we also obtain

ρ α( ) α δ( ) xα δ( )

δx

+– Kδp

, K 2, 0 p 1,≤<>= =

xα δ( )

δ α α

xα δ( )

δ

xα δ( )

δ

α xα

δx

+– 2α z x

+– δ 1 Nαμ

+( ),+≤

2α z x+

– K 2–( )δp,<

ρ α( ) K 2–( )δp2δ K 2–( )δp

2δp+<+< Kδp

.=

d α( ) ρ α( ) Kδp–=

d α( )α +∞→lim +∞ Kδp

– +∞.= =

α

δ/α 2δ1 p–z x

+– / K 2–( ).≤

ααk

αk

xαk

δk

xαk

δk

0 A0 xαk

δk( ) αkμ

Ai xαk

δk( ) fi

δk–( )i 1=

N

∑ αk xαk

δk x+

–( ) f0

δk–+ + J xαk

δk x+

–( ),=

≥ A0 x+( ) f0

δk– J xαk

δk x+

–( ),⟨ ⟩ αkμ

Ai x+( ) fi

δk– ) αk xαk

δk x+

–( ) J xαk

δk x+

–( ),+⟨ ⟩i 1=

N

∑+

≥ αk xαk

δk x+

– A0 x+( ) f0

δk– αkμ

Ai x+( ) fi

δk–i 1=

N

∑+⎝ ⎠⎜ ⎟⎛ ⎞

– xαk

δk x+

– +∞,

αk αk xαk

δk x+

–

C0 xαk

δk x+

–k ∞→lim K δk

p

k ∞→lim 0.= =

404



It means that limk →∞|| – x+|| = 0. Again, from (8) with δ, α and x replaced, respectively, by δk, αk

and , and after passing k ∞, we obtain that

(22)

for any z ∈ S. Acting on the last equality by J(x+ – z) and using the accretive property of A0 and λi�coco�ercive property of Ai for i = 1, 2, …, N, we have

Consequently, ||Ai(x+) – Ai(z)|| = 0 for i = 1, 2, …, N. It means that x+ ∈ . Now, from (22) we

obtain that x+ ∈ S0. Hence, x+ ∈ , that contradicts the assumption. In the case (2), from ρ( ) =

and (18), it follows that

(23)

Now, from (8) we have that

Tending k ∞ and using the continuity of Ai for i = 0, 1, …N, (16), (23) and the facts that ∞and δk 0 in the last inequality, we obtain the contradiction +∞ ≤ 0. In final, we have that = α(δ) 0,

as δ 0. By Theorem 2, x∗, solving (10). In the case that p = 1, we have, from (21) that δ/α(δ) ≤

C := 2||z – x+||/(K – 2). Moreover, by the similar argument as above, α(δ) 0 as δ 0. Then, from (19), it

follows the boundedness of { }. Since E is reflexive, there exists a subsequence {xk := }, that con�

verges weakly to some element x∞ ∈ E as k ∞. By (8) with α, δ and x replaced by αk, δk and xk, respec�tively, we obtain that

Thus, ||A0(xk) – || 0 as k ∞. By Lemma 1.15.12 in [10], A0(x∞) = f0. By the similar argument,from (8) and the accretive property of A0, it also follows that

Therefore, ||Ai(xk) – fi|| 0 as k ∞, for i = 1, 2, …, N. Again, by Lemma 1.15.12 in [10], Ai(x∞) = fi.

It means that x∞ ∈ S, i.e., x∞ = x∗ and all net { } converges weakly to x∗, because S contains only one

element x∗. Theorem is proved.

Theorem 4. Let Ai, E and be as in Theorem 2. In addition, assume that the following conditions hold:

(i) A0 is Fréchet differential with (7);

(ii) There exists an element ω ∈ E such that (x∗)ω = x+ – x∗;

(iii) The parameter α = α(δ) can be chosen by (18).

xαk

δk

xαk

δk

A0 x+( ) A0 z( )– C0

μAi x

+( ) Ai z( )–( )i 1=

N

∑+ 0,=

λi Ai x+( ) Ai z( )–

2

i 1=

N

∑ 0.≤

∩i 1=N

Si

∩i 0=N

Si α

α xα

δx

+–

xαk

δk x+

–k ∞→lim

ρ αk( )αk

��k +∞→lim K

δkp

αk

��k +∞→lim 0.= = =

αkμ

Ai xαk

δk( ) fi

δk–i 1=

N

∑ A0 xαk

δk( ) f0

δk–– αk xαk

δk x+

–≤ ρ αk( ) Kδkp.= =

αk

α

xα δ( )

δ

xα δ( )

δxα δk( )

δk

A0 xk( ) f0

δk– αkμ

Ai xk( ) fi

δk–i 1=

N

∑ αk xk x+

– .+≤

f0

δk

λi Ai xk( ) fi–2

i 1=

N

∑ fi

δk fi–( )i 1=

N

∑ αk1 μ–

xk x+

–( ) J xk x∞–( ),+ Nδk αk1 μ–

xk x+

–+( ) xk x∞– .≤ ≤

xα δ( )

δ

fi δ

A0'



Then, for 0 < p < 1, we have

Proof. From (8), the accretive property of Ai and the conditions of the theorem it follows that

(24)

Now, we have

and, from (7), that

Therefore, from (24) it implies that

Because = α(δ) is chosen by (18) with 0 < p < 1, we have Nαμ(δ) ≤ 1 and there exists a positive con�

stant c such that || – x∗|| ≤ c ≤ || – x+||, for sufficiently small δ. Thus, from (21) and the last ine�

quality it follows that

for sufficiently small δ. Now, by using the implication

we obtain

Theorem is proved.

ACKNOWLEDGMENTS

The authors are extremely grateful to the referees for their useful comments and suggestions whichhelped to improve this paper.

This work was supported by the Co�Operative Programm with the Belarussian Republican Foundationfor Fundamental Research, VAST.Belarus03/13�14.

xα δ( )

δx*– O δθ( ), θ min 1 p– μp/2,{ }.= =

xα

δx*–

2

xα

δx*– J xα

δx*–( ),⟨ ⟩ 1

α�� f0

δA0 xα

δ( )– αμfi

δAi xα

δ( )–( )i 1=

N

∑ J xα

δx*–( ),+= =

+ x+

x*– J xα

δx*–( ),⟨ ⟩ δ

α�� 1 Nαμ

+( ) xα

δx*– ω A0' x*( )*J xα

δx*–( ),⟨ ⟩ .+≤

ω A0' x*( )*J xα

δx*–( ),⟨ ⟩ ω A0' x*( )*J xα

δx*–( )≤

A0' x*( )*J xα

δx*–( ) QA0' x*( )*J xα

δx*–( ) τ 1+( ) A0 xα

δ( ) f0– τ 1+( ) δ A0 xα

δ( ) f0 δ–+( )≤ ≤=

≤ τ 1+( ) δ α xα

δx

+– αμ

Ai xα

δ( ) fi δ

–i 1=

N

∑+ + .

xα

δx*–

2 δα�� 1 Nαμ

+( ) xα

δx*–≤

+ ω τ 1+( ) δ α xα

δx

+– αμ

Nδ xα

δx*– 1

λi

��i 1=

N

∑+⎝ ⎠⎜ ⎟⎛ ⎞

+ + .

α

xα δ( )

δxα δ( )

δ

xα δ( )

δx*–

2 4 z x+

–K 2–

��δ1 p–xα δ( )

δx*– ω τ 1+( ) 2δ Kδp

c1 μ–

Kδp( )μ 1

λi

��i 1=

N

∑+ +⎝ ⎠⎜ ⎟⎛ ⎞

+≤

≤ 4 z x+

–K 2–

��δ1 p–xα δ( )

δx*– ω τ 1+( ) 2 K c

1 μ–K

μ 1λi

��i 1=

N

∑+ +⎝ ⎠⎜ ⎟⎛ ⎞

δμp,+

a b c 0 s t as

bat

c as⇒+≤,>,≥, , O b

s/ s t–( )c+( ),=

xα δ( )

δx*– O δθ( ).=

406



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