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Strongly Pseudomonotone Nonlinear Variational

Inequalities R. U. Verma vol. 8, iss. 1, art. 6, 2007

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GENERAL SYSTEM OF STRONGLY

PSEUDOMONOTONE NONLINEAR VARIATIONAL INEQUALITIES BASED ON PROJECTION SYSTEMS

R. U. VERMA

Department of Mathematics, University of Toledo Toledo, Ohio 43606, USA

EMail:verma99@msn.com

Received: 26 February, 2006

Accepted: 11 December, 2006

Communicated by: R.N. Mohapatra 2000 AMS Sub. Class.: 49J40, 65B05, 47H20.

Key words: Strongly pseudomonotone mappings, Approximation solvability, Projection methods, System of nonlinear variational inequalities.

Abstract: Let K1 and K2, respectively, be non empty closed convex subsets of real Hilbert spacesH1 andH2.TheApproximationsolvabilityof a generalized system of nonlinear variational inequality(SN V I)problems based on the convergence of pro- jection methods is discussed. The SNVI problem is stated as follows: find an element (x, y)K1×K2such that

hρS(x, y), xxi ≥0, ∀xK1and forρ >0, hηT(x, y), yyi ≥0, ∀yK2and forη >0, whereS :K1×K2H1andT :K1×K2H2are nonlinear mappings.

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Contents

1 Introduction 3

2 General Projection Methods 7

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1. Introduction

Projection-like methods in general have been one of the most fundamental tech- niques for establishing the convergence analysis for solutions of problems arising from several fields, such as complementarity theory, convex quadratic programming, and variational problems. There exists a vast literature on approximation-solvability of several classes of variational/hemivariational inequalities in different space set- tings. The author [6,7] introduced and studied a new system of nonlinear variational inequalities in Hilbert space settings. This class encompasses several classes of non- linear variational inequality problems. In this paper we intend to explore, based on a general system of projection-like methods, the approximation-solvability of a system of nonlinear strongly pseudomonotone variational inequalities in Hilbert spaces. The obtained results extend/generalize the results in [1], [5] – [7] to the case of strongly pseudomonotone system of nonlinear variational inequalities. Approximation solv- ability of this system can also be established using the resolvent operator technique but in the more relaxed setting of Hilbert spaces. For more details, we refer the reader to [1] – [10].

LetH1 andH2 be two real Hilbert spaces with the inner producth·,·iand norm k · k. Let S : K1 × K2 → H1 and T : K1 × K2 → H2 be any mappings on K1 × K2, where K1 and K2 are nonempty closed convex subsets of H1 and H2, respectively. We consider a system of nonlinear variational inequality (abbreviated as SNVI) problems: determine an element(x, y)∈K1 ×K2 such that

(1.1) hρS(x, y), x−xi ≥0∀x∈K1

(1.2) hηT(x, y), y −yi ≥0∀y∈K2, whereρ, η >0.

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The SNVI(1.1)−(1.2)problem is equivalent to the following projection formulas x =Pk[x−ρS(x, y)] forρ >0

y =Qk[y −ηT(x, y)] forη >0,

wherePk is the projection ofH1ontoK1 andQK is the projection ofH2 ontoK2. We note that the SNVI(1.1)−(1.2)problem extends the NVI problem: determine an elementx ∈K1 such that

(1.3) hS(x), x−xi ≥0, ∀x∈K1.

Also, we note that the SNVI (1.1)− (1.2) problem is equivalent to a system of nonlinear complementarities (abbreviated as SNC): find an element(x, y)∈K1× K2such thatS(x, y)∈K1, T(x, y)∈K2, and

(1.4) hρS(x, y), xi= 0 for ρ >0,

(1.5) hηT(x, y), yi= 0 for η >0,

whereK1 andK2, respectively, are polar cones toK1andK2defined by K1 ={f ∈H1 :hf, xi ≥0, ∀x∈K1}.

K2 ={g ∈H2 :hg, yi ≥0, ∀g ∈K2}.

Now, we recall some auxiliary results and notions crucial to the problem on hand.

Lemma 1.1. For an elementz ∈H, we have

x∈K and hx−z, y−xi ≥0, ∀y∈K if and only if x=Pk(z).

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Lemma 1.2 ([3]). Letk},{βk}, and {γk}be three nonnegative sequences such that

αk+1 ≤(1−tkkkk for k = 0,1,2, ..., wheretk ∈ [0,1],P

k=0tk = ∞, βk =o(tk),andP

k=0γk < ∞.Thenαk → 0as k → ∞.

A mapping T : H → H from a Hilbert space H into H is called monotone if hT(x)−T(y), x−yi ≥0for all x, y ∈H.The mappingT is(r)−strongly monotone if for eachx, y ∈H,we have

hT(x)−T(y), x−yi ≥r||x−y||2 for a constantr >0.

This implies that kT(x) − T(y)k ≥ rkx − yk, that is, T is (r)-expansive, and when r = 1, it is expansive. The mapping T is called (s)-Lipschitz continu- ous (or Lipschitzian) if there exists a constant s ≥ 0 such thatkT(x)−T(y)k ≤ skx−yk, ∀x, y ∈H. T is called(µ)-cocoercive if for eachx, y ∈H,we have

hT(x)−T(y), x−yi ≥µ||T(x)−T(y)||2 for a constantµ >0.

Clearly, every(µ)-cocoercive mappingT is(1µ)-Lipschitz continuous. We can eas- ily see that the following implications on monotonicity, strong monotonicity and expansiveness hold:

strong monotonicity

⇓ monotonicity

⇒expansiveness

T is called relaxed(γ)-cocoercive if there exists a constantγ >0such that hT(x)−T(y), x−yi ≥(−γ)kT(x)−T(y)k2, ∀x, y ∈H.

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T is said to be(r)-strongly pseudomonotone if there exists a positive constantrsuch that

hT(y), x−yi ≥0⇒ hT(x), x−yi ≥rkx−yk2, ∀x, y ∈H.

T is said to be relaxed(γ, r)-cocoercive if there exist constantsγ,r >0such that hT(x)−T(y), x−yi ≥(−γ)kT(x)−T(y)k2+rkx−yk2.

Clearly, it implies that

hT(x)−T(y), x−yi ≥(−γ)kT(x)−T(y)k2, that is,T is relaxed(γ)-cocoercive.

T is said to be relaxed(γ, r)-pseudococoercive if there exist positive constantsγ andrsuch that

hT(y), x−yi ≥0⇒ hT(x), x−yi ≥(−γ)kT(x)−T(y)k2+rkx−yk2, ∀x, y ∈H.

Thus, we have following implications:

(r)-strong monotonicity

relaxed(γ, r)-cocoercivity

relaxed(γ, r)-pseudococoercivity

⇒ strong(r)-pseudomonotonicity

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2. General Projection Methods

This section deals with the convergence of projection methods in the context of the approximation-solvability of the SNVI(1.1)−(1.2)problem.

Algorithm 1. For an arbitrarily chosen initial point(x0, y0) ∈ K1 ×K2,compute the sequences{xk}and{yk}such that

xk+1 = (1−ak−bk)xk+akPk[xk−ρS(xk, yk)] +bkuk yk+1 = (1−αk−βk)ykkQK[yk−ηT(xk, yk)] +βkvk,

wherePK is the projection of H1 ontoK1, QK is the projection ofH2 ontoK2, ρ, η > 0 are constants, S : K1 ×K2 → H1 and T : K1 ×K2 → H2 are any two mappings, anduk andvk,respectively, are bounded sequences in K1 andK2.The sequences{ak},{bk},{αk},andk}are in[0,1]with(k ≥0)

0≤ak+bk ≤1, 0≤αkk≤1.

Algorithm 2. For an arbitrarily chosen initial point(x0, y0) ∈ K1 ×K2,compute the sequences{xk}and{yk}such that

xk+1 = (1−ak−bk)xk+akPk[xk−ρS(xk, yk)] +bkuk yk+1 = (1−ak−bk)yk+akQK[yk−ηT(xk, yk)] +bkvk,

wherePK is the projection of H1 ontoK1, QK is the projection ofH2 ontoK2, ρ, η > 0 are constants, S : K1 ×K2 → H1 and T : K1 ×K2 → H2 are any two mappings, anduk andvk,respectively, are bounded sequences in K1 andK2.The sequences{ak}and{bk},are in[0,1]with(k≥0)

0≤ak+bk ≤1.

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Algorithm 3. For an arbitrarily chosen initial point(x0, y0) ∈ K1 ×K2,compute the sequences{xk}and{yk}such that

xk+1 = (1−ak)xk+akPk[xk−ρS(xk, yk)]

yk+1 = (1−ak)yk+akQK[yk−ηT(xk, yk)],

wherePK is the projection of H1 ontoK1, QK is the projection ofH2 ontoK2, ρ, η > 0 are constants, S : K1 ×K2 → H1 and T : K1 ×K2 → H2 are any two mappings. The sequence{ak} ∈[0,1]fork ≥0.

We consider, based on Algorithm 2, the approximation solvability of the SNVI (1.1)−(1.2)problem involving strongly pseudomonotone and Lipschitz continuous mappings in Hilbert space settings.

Theorem 2.1. LetH1 andH2 be two real Hilbert spaces and, K1 andK2, respec- tively, be nonempty closed convex subsets ofH1 andH2.LetS :K1 ×K2 →H1 be strongly (r)− pseudomonotone and (µ)−Lipschitz continuous in the first variable and letS be(ν)−Lipschitz continuous in the second variable. LetT : K1×K2 → H2 be strongly (s)−pseudomonotone and (β)−Lipschitz continuous in the second variable and letT be(τ)−Lipschitz continuous in the first variable. Letk · kdenote the norm onH1×H2 defined by

k(x, y)k = (kxk+kyk)∀(x, y)∈H1×H2. In addition, let

θ = s

1−2ρr+ρ+ ρµ2

2

2µ2 +ητ <1

σ= s

1−2ηr+η+ ηβ2

2

2β2+ρν <1,

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let(x, y) ∈ K1×K2 form a solution to the SNVI(1.1)−(1.2)problem, and let sequences{xk},and{yk}be generated by Algorithm2. Furthermore, let

(i) hS(x, yk), xk−xi ≥0;

(ii) hT(xk, y), yk−yi ≥0;

(iii) 0≤ak+bk ≤1;

(iv) P

k=0ak =∞,andP

k=0bk <∞;

(v) 0< ρ < µ2r2 and0< η < 2s

β2.

Then the sequence{xk, yk}converges to(x, y).

Proof. Since(x, y)∈K1×K2forms a solution to the SNVI(1.1)−(1.2)problem, it follows that

x =PK[x−ρS(x, y)] and y =QK[x−ηT(x, y)].

Applying Algorithm2, we have kxk+1−xk

(2.1)

=k(1−ak−bk)xk+akPK[xk−ρS(xk, yk)] +bkuk

−(1−ak−bk)x −akPK[x−ρS(x, y)]−bkxk

≤(1−ak−bk)kxk−xk

+akkPK[xk−ρS(xk, yk)]−PK[x−ρS(x, y)]k+M bk

≤(1−ak)kxk−xk+akkxk−x−ρ[S(xk, yk)−S(x, yk) +S(x, yk)−S(x, y)]k+M bk

≤(1−ak)kxk−xk+akkxk−x−ρ[S(xk, yk)−S(x, yk)]k +ρk[S(x, yk)−S(x, y)]k+M bk,

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where

M = max{supkuk−xk,supkvk−yk}<∞.

SinceSis strongly(r)−pseudomonotone and(µ)−Lipschitz continuous in the first variable, andSis(ν)−Lipschitz continuous in the second variable, we have in light of (i) that

kxk−x−ρ[S(xk, yk)−S(x, yk)]k2 (2.2)

=kxk−xk2−2ρhS(xk, yk)−S(x, yk), xk−xi +ρ2kS(xk, yk)−S(x, yk)k2

=kxk−xk2−2ρhS(xk, yk), xk−xi+ 2ρhS(x, yk), xk−xi +ρ2kS(xk, yk)−S(x, yk)k2

≤ kxk−xk2−2ρrkxk−xk22µ2kxk−xk2 + 2ρhS(x, yk), xk−xi

≤ kxk−xk2−2ρrkxk−xk22µ2kxk−xk2 + 2ρhS(x, yk), xk−xi

= [1−2ρr+ρ2µ2]kxk−xk2+ 2ρhS(x, yk), xk−xi.

On the other hand, we have

2ρhS(x, yk), xk−xi ≤ρ[kS(x, yk)k2+kxk−xk2] and

kS(x, yk)k2 (2.3)

= 1 2

kS(x, yk)−S(xk, yk)k2−2[kS(xk, yk)k2

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−1

2kS(xk, yk) +S(x, yk)k2

≤ 1 2

kS(x, yk)−S(xk, yk)k2−2[kS(xk, yk)k2

−1

2 | kS(xk, yk)−S(x, yk)k |2]

≤ 1

2kS(x, yk)−S(xk, yk)k2

≤ µ2

2 kxk−xk2, where

kS(xk, yk)k2− 1

2 | kS(xk, yk)−S(x, yk)k |2>0.

Therefore, we get

2ρhS(x, yk), xk−xi ≤

ρ+ ρµ2

2

kxk−xk2.

It follows that

kxk−x−ρ[S(xk, yk)−S(x, yk)]k2

1−2ρr+ρ+ ρµ2

2

+ (ρµ)2

kxk−xk2.

As a result, we have

(2.4) kxk+1−xk ≤(1−ak)kxk−xk+akθkxk−xk+akρνkyk−yk+M bk,

whereθ = r

1−2ρr+ρ+

ρµ2 2

2µ2.

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Similarly, we have (2.5)

yk+1−y

≤(1−ak)kyk−yk+akσkyk−yk+akητkxk−xk+M bk, whereσ =

r

1−2ηr+η+

ηβ2 2

2β2. It follows from(2.4)and(2.5)that

kxk+1−xk+kyk+1−yk (2.6)

≤(1−ak)kxk−xk+akθkxk−xk+akητkxk−xk+M bk

+ (1−ak)kyk−yk+akσkyk−yk+akρνkyk−yk+M bk

= [1−(1−δ)ak](kxk−xk+kyk−yk) + 2M bk,

whereδ = max{θ+ητ , σ+ρν}andH1 ×H2 is a Banach space under the norm k · k.

If we set

αk=kxk−xk+kyk−yk, tk = (1−δ)ak, βk = 2M bk fork = 0,1,2, ...,

in Lemma1.2, and apply(iii)and(iv),we conclude that kxk−xk+kyk−yk →0 ask → ∞.

Hence,

kxk+1−xk+kyk+1−yk →0.

Consequently, the sequence {(xk, yk)}converges strongly to(x, y),a solution to the SNVI(1.1)−(1.2)problem. This completes the proof.

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Note that the proof of the following theorem follows rather directly without using Lemma1.2.

Theorem 2.2. LetH1 andH2 be two real Hilbert spaces and, K1 andK2, respec- tively, be nonempty closed convex subsets ofH1 andH2.LetS :K1 ×K2 →H1 be strongly(r)− pseudomonotone and(µ)−Lipschitz continuous in the first variable and letS be(ν)−Lipschitz continuous in the second variable. LetT : K1×K2 → H2 be strongly (s)−pseudomonotone and (β)−Lipschitz continuous in the second variable and letT be(τ)−Lipschitz continuous in the first variable. Letk · kdenote the norm onH1×H2 defined by

k(x, y)k = (kxk+kyk) ∀(x, y)∈H1×H2. In addition, let

θ= s

1−2ρr+ρ+ ρµ2

2

2µ2+ητ <1,

σ= s

1−2ηr+η+ ηβ2

2

2β2+ρν <1,

let(x, y) ∈ K1×K2 form a solution to the SNVI(1.1)−(1.2)problem, and let sequences{xk},and{yk}be generated by Algorithm3. Furthermore, let

(i) hS(x, yk), xk−xi ≥0 (ii) hT(xk, y), yk−yi ≥0 (iii) 0≤ak ≤1

(iv) P

k=0ak =∞

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(v) 0< ρ < µ2r2 and0< η < 2s

β2.

Then the sequence{xk, yk)}converges strongly to(x, y).

Proof. Since(x, y)∈K1×K2forms a solution to the SNVI(1.1)−(1.2)problem, it follows that

x =PK[x−ρS(x, y)] and y =QK[x−ηT(x, y)].

Applying Algorithm3, we have kxk+1−xk

(2.7)

=k(1−ak)xk+akPK[xk−ρS(xk, yk)]−(1−ak)x

−akPK[x−ρS(x, y)]k

≤(1−ak)kxk−xk+akkPK[xk−ρS(xk, yk)]−PK[x−ρS(x, y)]k

≤(1−ak)kxk−xk

+akkxk−x−ρ[S(xk, yk)−S(x, yk) +S(x, yk)−S(x, y)]k

≤(1−ak)kxk−xk+akkxk−x−ρ[S(xk, yk)−S(x, yk)]k +ρk[S(x, yk)−S(x, y)]k.

SinceSis strongly(r)−pseudomonotone and(µ)−Lipschitz continuous in the first variable, andSis(ν)−Lipschitz continuous in the second variable, we have in light of(i)that

kxk−x −ρ[S(xk, yk)−S(x, yk)]k2 (2.8)

=kx−xk2 −2ρhS(xk, yk)−S(x, yk), xk−xi +ρ2kS(xk, yk)−S(x, yk)k2

=kx−xk2 −2ρhS(xk, yk), xk−xi+ 2ρhS(x, yk), xk−xi

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2kS(xk, yk)−S(x, yk)k2

≤ kxk−xk2−2ρrkxk−xk22µ2kxk−xk2 + 2ρhS(x, yk), xk−xi

≤ kxk−xk2−2ρrkxk−xk22µ2kxk−xk2 + 2ρhS(x, yk), xk−xi

= [1−2ρr+ρ2µ2]kxk−xk2+ 2ρhS(x, yk), xk−xi.

On the other hand, we have

2ρhS(x, yk), xk−xi ≤ρ[kS(x, yk)k2+kxk−xk2] and

kS(x, yk)k2 (2.9)

= 1 2

(

kS(x, yk)−S(xk, yk)k2

−2

kS(xk, yk)k2− 1

2kS(xk, yk) +S(x, yk)k2 )

≤ 1 2

(

kS(x, yk)−S(xk, yk)k2

−2

kS(xk, yk)k2− 1

2 | kS(xk, yk)−S(x, yk)k |2 )

≤ 1

2kS(x, yk)−S(xk, yk)k2 ≤ µ2

2 kxk−xk2,

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where

kS(xk, yk)k2− 1 2

S(xk, yk)−S(x, yk)

2 >0.

Therefore, we get

2ρhS(x, yk), xk−xi ≤

ρ+ ρµ2

2

kxk−xk2.

It follows that

kxk−x−ρ[S(xk, yk)−S(x, yk)]k2

1−2ρr+ρ+ ρµ2

2

+ (ρµ)2

kxk−xk2.

As a result, we have

(2.10) kxk+1−x|| ≤(1−ak)kxk−xk+akθkxk−xk+akρνkyk−yk,

whereθ = r

1−2ρr+ρ+

ρµ2 2

2µ2. Similarly, we have

(2.11) kyk+1−yk ≤(1−ak)kyk−yk+akσkyk−yk+akητkxk−xk,

whereσ = r

1−2ηr+η+

ηβ2 2

2β2. It follows from(2.9)and(2.10)that

kxk+1−xk+kyk+1−yk (2.12)

≤(1−ak)kxk−xk+akθkxk−xk+akητkxk−xk + (1−ak)kyk−yk+akσkyk−yk+akρνkyk−yk

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= [1−(1−δ)ak](kxk−xk+kyk−yk)

k

Y

j=0

[1−(1−δ)aj](kx0−xk+ky0−yk),

whereδ = max{θ+ητ , σ+ρν}andH1 ×H2 is a Banach space under the norm k · k.

Sinceδ <1andP

k=0akis divergent, it follows that

k→∞lim

k

Y

j=0

[1−(1−δ)aj] = 0 as k→ ∞.

Therefore,

kxk+1−xk+kyk+1−yk →0,

and consequently, the sequence{(xk, yk)}converges strongly to(x, y),a solution to theSN V I(1.1)−(1.2)problem. This completes the proof.

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Strongly Pseudomonotone Nonlinear Variational

Inequalities R. U. Verma vol. 8, iss. 1, art. 6, 2007

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[9] E. ZEIDLER, Nonlinear Functional Analysis and its Applications I, Springer- Verlag, New York, New York, 1986.

[10] E. ZEIDLER, Nonlinear Functional Analysis and its Applications II/B, Springer-Verlag, New York, New York, 1990.

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