System of split variational inequality problems in semi-inner product spaces

0.13 MB·

by  K. R. Kazmi

System of split variational inequality problems in semi-inner product spaces
K.R. Kazmi1 and Mohd Furkan2
Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India
Abstract: Weintroduceanewsystemofsplitvariationalinequalityproblemswhichisanaturalexten-
sionofsplitvariationalinequalityprobleminsemi-innerproductspaces. We usetheretractiontechnique
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to propose an iterative algorithm for computing the approximate solution of the system of split varia-
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0 tional inequality problems. Further, the convergenceanalysis of the iterative algorithmis also discussed.
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Several special cases which can be obtained from the main result are also discussed.
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J

9 Keywords: System of split variational inequality problems, sunny retraction mapping, semi-inner prod-
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uct, generalized adjoint operator,uniformly convex smooth Banach space.

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F AMS classifications: Primary 47J53;Secondary 90C25.
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1 Preliminaries
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We recall the following concepts and results, which are needed to define the problem and to prove the
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main result:
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5 Definition 1.1. [11] Let X be a vector space over the field K=R (or C) of real (or complex) numbers.
0
. A functional [·,·]:X ×X →K is called a semi-inner product if it satisfies the following conditions:
1
0
(1) [x+y,z]=[x,z]+[y,z],∀x,y,z ∈X;
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1
: (2) [λx,y]=λ[x,y],∀λ∈K and ∀x,y ∈X;
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i
X
(3) [x,x]>0, for x6=0;
r
a
(4) |[x,y]|2 ≤[x,x][y,y],∀x,y ∈X.
Thepair(X,[·,·])iscalledasemi-innerproductspace. Asitisobservedin[11]thatkxk=[x,x]1/2,∀x∈
X,isanormonX. Henceeverysemi-innerproductspaceisanormedlinearspace. Ontheotherhand,in
a normedlinear space,one cangeneratesemi-inner product in infinitely many different ways. Further, it
is notedthataHilbertspace H canbe madeinto asemi-inner productspace,while asemi-inner product
is an inner product if and only if the norm it induces verifies the parallelogramlaw.
1Correspondingauthor;E-mail: [e-mail](K.R.Kazmi)
2E-mail: [e-mail] (MohdFurkan)
Let Y be a semi-inner product space and let T :X →Y be an arbitrary operator.
Definition 1.2. [13] The generalized adjoint operator T+ of an operator T is defined as follows: The
domain D(T+) of T+ consists of those y ∈Y for which there exists z ∈X such that
[Tx,y] =[x,z]
Y X
for each x∈X and z =T+y.
Remark 1.1. T+ is an operator from D(T+) into X with the nonempty domain D(T+), since 0 ∈
D(T+). Hence T+(0)=0. As it is observed in [3] that if X and Y are Hilbert spaces then the generalized
adjoint operator is the usual adjoint operator. In general, T+ is not linear even for T is a bounded linear
operator.
Let C be a nonempty closed and convex subset of a Banach space E. Let E∗ be the dual space of E
and h·,·i denote the pairingbetween E andE∗. The normalizedduality mapping J :E →2E∗ is defined
by
J(x)={f ∈E∗ :hx,fi=kxk2,kfk=kxk}
for all x∈E. We denote by j the single normalized duality mapping, i.e., j(x)∈J(x), x∈E.
Definition 1.3. [16] Let U ={x∈E :kxk=1}. A Banach space E is said to be:
(1) uniformly convex if, for any ǫ∈(0,2], there exists δ >0 such that for any x,y ∈U,
x+y
kx−yk≥ǫ implies ≤1−δ;
2
(cid:13) (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
(2) smooth if the limit lim kx+tyk−kxk exists for all x,y ∈U;
t→0 t
(3) uniformly smooth if the limit is attained uniformly for x,y ∈U.
Definition 1.4. The modulus of smoothness of a Banach space E is defined by
1
ρ(τ)=sup (kx+yk+kx−yk)−1:x,y ∈X,kxk=1,kyk=τ ,
2
(cid:26) (cid:27)
where ρ:[0,∞)→[0,∞) is a function.
Remark 1.2. E is uniformly smooth if and only if lim ρ(τ) = 0. If E is smooth then normalized
t→0 τ
duality mapping J is single-valued and if E is uniformly smooth then J is uniformly norm to norm
continuous on bounded subsets of E. If E is a Hilbert space then J =I, where I is the identity mapping.
In 1967, Giles [5] proved that if the underlying semi-inner product space X is a uniformly convex
smooth Banachspace then it is possible to define a semi-inner product uniquely which has the following
properties:
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(i) [x,y]= 0 for some x,y ∈ X if and only if y is orthogonal to x, i.e., if and only if kyk≤ ky+λxk,
for all scalars λ.
(ii) The semi-inner product is continuous, i.e., for each x,y ∈ X, we have Re[y,x+λy] → Re[y,x] as
λ→0.
(iii) The semi-inner product is with the homogeneity property, i.e.,
[x,λy]=|λ|[x,y],∀λ∈K and ∀x,y ∈X.
(iv) GeneralizedRiesz representationtheorem: If f is continuous linear functional on X then there is a
unique vector y ∈X such that f(x)=[x,y],∀x∈X.
The sequence space lp,p > 1 and the function space Lp,p > 1 are uniformly convex smooth Banach
spaces. More precisely, Lp is min{p,2}-uniformly smooth for every p > 1. So one can define semi-inner
product on these spaces uniquely.
Example 1.1. [5] The real sequence space lp for 1 < p < ∞ is a semi-inner product space with the
semi-inner product defined by
1
[x,y]= x y |y |p−2,∀x,y ∈lp.
kykp−2 i i i
p i
X
Example 1.2. [5] The real Banach space Lp(X,µ) for 1<p<∞ is a semi-inner product space with the
semi-inner product defined by
1
[f,g]= fx|gx|p−1sgn(gx)dµ,∀f,g ∈Lp.
kgkp−2
p ZX
Now,wesummarizethefollowingpropertiesofthegeneralizedadjointoperatorfromtheresultsgiven
in [13].
Proposition 1.1. Let X and Y be 2-uniformly convex smooth Banach spaces and let T : X → Y be a
bounded linear operator. Then
(i) D(T+)=Y;
(ii) T+ is bounded, and it holds that
kT+yk≤kTkkyk,∀y∈Y.
Definition 1.5. [14] Let D be a subset of C and Q be a mapping of C into D. Then Q is said to be
C C
sunny if
Q (Q x+t(x−Q x))=Q x,
C C C C
whenever Q x+t(x−Q x)∈C for x∈C and t≥0.
C C
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Definition 1.6. [14] A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny
nonexpansive retraction from C into D.
The following result describes a characterization of sunny nonexpansive retractions on a smooth
Banach space.
Proposition 1.2. [14] Let E be a smooth Banach space and let C be a nonempty subset of E. Let
Q :E →C be a retraction. Then the following are equivalent:
C
(i) Q is sunny and nonexpansive;
C
(ii) kQ x−Q yk2 ≤hx−y,J(Q x−Q y)i,∀x,y ∈E;
C C C C
(iii) hx−Q x,J(y−Q x)i≤0,∀x∈E,y ∈C.
C C
Lemma 1.1. [16] Let p > 1 be a real number and E be a smooth Banach space. Then the following
statements are equivalent:
(i) E is 2-uniformly smooth;
(ii) There is a constant c>0 such that for every x,y ∈E, the following inequality holds
kx+yk2 ≤kxk2+2hy,J(x)i+ckyk2.
Remark 1.3. 1. [5, 11, 15]: Every normed linear space is a semi-inner product space. In fact by
Hahn Banach theorem, for each x ∈ E there exists at least one functional f ∈ E∗ such that
x
hx,f i = kxk2. Given any such mapping f from E into E∗, we can verify that [y,x] = hy,f i
x x
defines a semi-inner product. Hence, we can write the inequality given in Lemma 1.1 as
kx+yk2 ≤kxk2+2[y,x]+ckyk2,∀x,y ∈E.
The constant c is chosen with best possible minimum value. Wecall c as theconstant of smoothness
of E.
2. The inequalities given in Proposition 1.2 (ii) & (iii) can be written as
(ii) kQ x−Q yk2 ≤[x−y,Q x−Q y],∀x,y ∈E;
C C C C
(iii) [x−Q x,y−Q x]≤0,∀x∈E,y ∈C.
C C
Example 1.3. [15] The function space Lp is 2-uniformly smooth for p≥2 and it is p-uniformly smooth
for 1<p<2. If 2≤p<∞, then we have for all x,y ∈Lp,
kx+yk2 ≤kxk2+2[y,x]+(p−1)kyk2,
where (p−1) is the constant of smoothness.
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Let E and E be 2-uniformly convex, smooth Banach spaces and for each i∈{1,2}; let C ⊂E be
1 2 i i
a nonempty, closed and convex set and let J1 : E1 → 2E1∗ and J2 : E2 →2E2∗ be the normalized duality
mappings. Let F,G : C → E and f,g : C → E be nonlinear mappings, and let A : E → E be a
1 1 2 2 1 2
bounded linear operator. We introduce the following system of split variational inequality problems (in
short, SSpVIP): Find (x ,y )∈C ×C such that
1 1 1 1
hλFy +x −y ,J (z −x )i≥0, ∀z ∈C ,
1 1 1 1 1 1 1 1
and such that (x ,y ) with x =Ax ∈C ,y =Ay ∈C solves
2 2 2 1 2 2 1 2
hγfy +x −y ,J (z −x )i≥0, ∀z ∈C ;
2 2 2 2 2 2 2 2
hλGx +y −x ,J (z −y )i≥0, ∀z ∈C ,
1 1 1 1 1 1 1 1
and such that (x ,y ) solves
2 2
hγgx +y −x ,J (z −y )i≥0, ∀z ∈C ,
2 2 2 2 2 2 2 2
for any λ,γ >0.
Above SSpVIP is equivalent to find (x ,y )∈C ×C such that
1 1 1 1
[λFy +x −y ,z −x ]≥0, ∀z ∈C , (1.1)
1 1 1 1 1 1 1
and such that (x ,y ) with x =Ax ∈C ,y =Ay ∈C solves
2 2 2 1 2 2 1 2
[γfy +x −y ,z −x ]≥0, ∀z ∈C ; (1.2)
2 2 2 2 2 2 2
[λGx +y −x ,z −y ]≥0, ∀z ∈C , (1.3)
1 1 1 1 1 1 1
and such that (x ,y ) solves
2 2
[γgx +y −x ,z −y ]≥0, ∀z ∈C , (1.4)
2 2 2 2 2 2 2
for any λ,γ >0.
Some special cases:
1. IfwesetE =H , E =H ,whereH , H areHilbertspaces,thenSSpVIP (1.1)-(1.4)reducesto
1 1 2 2 1 2
the following system of split variationalinequality problems (SSpVIP) in Hilbert spaces: Find (x ,y )∈
1 1
C ×C such that
1 1
hλFy +x −y ,z −x i≥0, ∀z ∈C , (1.5)
1 1 1 1 1 1 1
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and such that (x ,y ) with x =Ax ∈C ,y =Ay ∈C solves
2 2 2 1 2 2 1 2
hγfy +x −y ,z −x i≥0, ∀z ∈C ; (1.6)
2 2 2 2 2 2 2
hλGx +y −x ,z −y i≥0, ∀z ∈C , (1.7)
1 1 1 1 1 1 1
and such that (x ,y ) solves
2 2
hγgx +y −x ,z −y i≥0, ∀z ∈C , (1.8)
2 2 2 2 2 2 2
for any λ,γ >0.
2. If we set F = G, f = g, λ = γ, y = x , then y = x and hence SSpVIP (1.1)-(1.4) reduces to
1 1 2 2
the following split variational inequality problem (in short, SpVIP): Find x ∈C such that
1 1
[Fx ,z −x ]≥0, ∀z ∈C , (1.9)
1 1 1 1 1
and such that x =Ax ∈C solves
2 1 2
[fx ,z −x ]≥0, ∀z ∈C , (1.10)
2 2 2 2 2
3. InCase2,ifE =H , E =H ,thenSpVIP(1.9)-(1.10)reducestothesplitvariationalinequality
1 1 2 2
problem considered and studied by Censor et al. [4]. It is worth mentioning that the SpVIP is quite
general and permit split minimization between two spaces so that the image of a minimizer of a given
function, under a bounded linear operator, is a minimizer of another function. It includes as a special
case, the variational inequality problem, the split zero problem and the split-feasibility problem which
have already been studied and used in practice as a model in the intensity-modulated radiation therapy
planning, see [2, 3]. For a further related work, see [1, 6, 7, 9, 10, 12].
Further, it is worthmentioning that so far the iterative approximationsof split variationalinequality
problem and its generalizations have been studied in the setting of Hilbert spaces. Therefore, a natural
question appears as to whether or not one can study these problems in setting of Banach spaces.
In this paper, we use the retractiontechnique to propose and analyze an iterative algorithmfor com-
puting the approximate solution of SSpVIP (1.1)-(1.4) in 2-uniformly convex smooth Banach spaces.
Further, convergence analysis of the iterative algorithm is discussed. Several special cases which can be
obtained from the main result are also discussed. The problems and the results discussed in this paper
are new and different from the existing problems and results in the literature.
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2 Iterative Algorithms
BymakinguseofProposition1.2,weeasilyobservethatSSpVIP(1.1)-(1.4)canbeformulatedasfollows:
Find (x ,y )∈C ×C with (x ,y )=(Ax ,Ay )∈C ×C such that
1 1 1 1 2 2 1 1 2 2
x =Q (y −λFy ), (2.1)
1 C1 1 1
x =Q (y −γfy ), (2.2)
2 C2 2 2
y =Q (x −λGx ), (2.3)
1 C1 1 1
y =Q (x −γgx ), (2.4)
2 C2 2 2
for λ,γ >0.
Basedonabovearguments,weproposethe followingiterativealgorithmforapproximatinga solution
to SSpVIP (1.1)-(1.4).
∞
Let {α }⊆(0,1) be a sequence such that =∞.
n
n=1
P
Iterative Algorithm 2.1. Given (x0,y0)∈C ×C , compute the iterative sequence {(xn,yn)} defined
1 1 1 1 1 1
by the iterative schemes:
an =Q (yn−λFyn), (2.5)
1 C1 1 1
an =Q (yn−γfyn), (2.6)
2 C2 2 2
bn =Q (xn−λGxn), (2.7)
1 C1 1 1
bn =Q (xn−γgxn), (2.8)
2 C2 2 2
xn+1 =(1−αn)xn+αn an+ρA+(an−Aan) , (2.9)
1 1 1 2 1
(cid:0) (cid:1)
yn+1 =(1−αn)yn+αn bn+ρA+(bn−Abn) , (2.10)
1 1 1 2 1
(cid:0) (cid:1)
for all n=0,1,2,.... and λ, γ, ρ>0, where A+ is the generalized adjoint operator of A, and xn =Axn
2 1
and yn =Ayn for all n.
2 1
If we set E =H , E =H , where H , H are Hilbert spaces, then Iterative Algorithm 2.1 reduces
1 1 2 2 1 2
to the following iterative algorithm for computing the approximate solution of SSpVIP (1.5)-(1.8):
Iterative Algorithm 2.2. Given (x0,y0)∈C ×C , compute the iterative sequence {(xn,yn)} defined
1 1 1 1 1 1
by the iterative schemes:
an =P (yn−λFyn), (2.11)
1 C1 1 1
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an =P (yn−γfyn), (2.12)
2 C2 2 2
bn =P (xn−λGxn), (2.13)
1 C1 1 1
bn =P (xn−γgxn), (2.14)
2 C2 2 2
xn+1 =(1−αn)xn+αn(an+ρA∗(an−Aan)), (2.15)
1 1 1 2 1
yn+1 =(1−αn)yn+αn(bn+ρA∗(bn−Abn)), (2.16)
1 1 1 2 1
for all n=0,1,2,.... and λ, γ, ρ>0, where A∗ is the adjoint operator of A with kA∗k=kAk, and P
Ci
is the metric projection of H onto C for each i∈{1,2}.
i i
If we set F =G, f =g, λ= γ, y =x , then y = x and hence Iterative Algorithm 2.1 reduces to
1 1 2 2
the following iterative algorithm for computing the approximate solution of SpVIP (1.9)-(1.10):
Iterative Algorithm 2.3. Givenx0 ∈C , compute the iterativesequence {xn} definedby the iterative
1 1 1
schemes:
an =Q (xn−λFxn),
1 C1 1 1
an =Q (xn−λfxn),
2 C2 2 2
xn+1 =(1−αn)xn+αn an+ρA+(an−Aan) ,
1 1 1 2 1
(cid:0) (cid:1)
for all n=0,1,2,.... and λ, γ, ρ>0.
3 Main Result
First, we define the following concepts.
Definition 3.1. A mapping F :E →E is said to be
1 1
(1) α-strongly monotone if there exists a constant α>0 such that
[Fx −Fy ,x −y]≥αkx −y k2, ∀x ,y ∈E ;
1 1 1 1 1 1 1 1
(2) β-Lipschitz continuous, if there exists a constant β >0 such that
kFx −Fy k≤βkx −y k, ∀x ,y ∈E .
1 1 1 1 1 1 1
Now,weprovethatthesequenceofapproximatesolutionsofSSpVIP(1.1)-(1.4)generatedbyIterative
Algorithm 2.1 converges strongly to the solution of SSpVIP (1.1)-(1.4).
Theorem 3.1. For each i∈{1,2},let C be a nonempty, closed and convex subset of 2-uniformly convex
i
smooth Banach space E with constant of smoothness c . Let F : C → E be α -strongly monotone
i i 1 1 1
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and β -Lipschitz continuous; let G : C → E be α -strongly monotone and β -Lipschitz continuous;
1 1 1 2 2
let f : C → E be σ -strongly monotone and η -Lipschitz continuous, and let g : C → E be σ -
2 2 1 1 2 2 2
strongly monotone and η -Lipschitz continuous. Let A : E → E be bounded linear operator. Suppose
2 1 2
(x ,y )∈C ×C is a solution to SSpVIP(1.1)-(1.4) then the sequence {(xn,yn)} generated by Iterative
1 1 1 2 1 1
Algorithm 2.1 converges strongly to (x ,y ) provided that the constant λ>0 satisfies the condition:
1 1
α − α2−c β2(1−p2) α + α2−c β2(1−p2)
max i i 1 i i <λ< min i i 1 i i (3.1)
1≤i≤2( p c1βi2 ) 1≤i≤2( p c1βi2 )
1−mθ
α >β c (1−p2); p = i+2; m=ρkA+kkAk;
i i 1 i i 1+m
q
θ = 1−2γσ +c γ2η2; γ >0.
i+2 i 2 i
q
Proof. Given that (x ,y ) is a solution of SSpVIP (1.1)-(1.4), that is, x ,y satisfy the relations (2.1)-
1 1 1 1
(2.4). Since F :C →E is α -stronglymonotoneandβ -Lipschitzcontinuous,fromIterativeAlgorithm
1 1 1 1
2.1 (2.5) and (2.1), we estimate
kan−x k=kQ (yn−λFyn)−Q (y −λFy )k
1 1 C1 1 1 C1 1 1
≤kyn−y −λ(Fyn−Fy )k
1 1 1 1
1
≤ kyn−y k2−2λ[Fyn−Fy ,yn−y ]+cλ2kFyn−Fy k2 2
1 1 1 1 1 1 1 1
(cid:0) (cid:1)
≤θ kyn−y k, (3.2)
1 1 1
where θ1 =(1−2λα1+c1λ2β12)12.
Next, since G : C → E is α -strongly monotone and β -Lipschitz continuous, from Iterative Algo-
1 1 2 2
rithm 2.1 (2.7) and (2.3), we have
kbn−y k=kQ (xn−µGxn)−Q (x −µGx )k
1 1 C1 1 1 C1 1 1
≤θ kxn−x k, (3.3)
2 1 1
where θ2 =(1−2λα2+c1λ2β22)12.
Again, since f : C →E is σ -strongly monotone and η -Lipschitz continuous, from Iterative Algo-
2 2 1 1
rithm 2.1 (2.6) and (2.2), we have
kan−x k≤θ kyn−y k, (3.4)
2 2 3 2 2
where θ3 =(1−2γσ1+c2γ2η12)12.
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Since g :C →E is σ -stronglymonotone andη -Lipschitz continuous,fromIterativeAlgorithm2.1
2 2 2 2
(2.8) and (2.4), we have
kbn−y k≤θ kxn−x k, (3.5)
2 2 4 2 2
where θ4 =(1−2γσ2+c2γ2η22)21.
Now, using the fact that A+ is bounded, we have
kxn+1−x k ≤ (1−αn)kxn−x k+αnkan−x +ρA+(an−Aan)k
1 1 1 1 1 1 2 1
≤ (1−αn)kxn−x k+αnkan−x k+αnρkA+kkan−Aank
1 1 1 1 2 1
≤ (1−αn)kxn−x k+αnkan−x k+αnρkA+k(kan−x −Aan+x k)
1 1 1 1 2 2 1 2
≤ (1−αn)kxn−x k+αnkan−x k+αnρkA+k(kan−x k+kAkkan−x k)
1 1 1 1 2 2 1 1
= (1−αn)kxn−x k+αnθ kyn−y k+αnρkA+k(θ kyn−y k+kAkθ kyn−y k)
1 1 1 1 1 3 2 2 1 1 1
≤ (1−αn)kxn−x k+αnθ kyn−y k+αnρkA+kkAk(θ kyn−y k+θ kyn−y k)
1 1 1 1 1 3 1 1 1 1 1
= (1−αn)kxn−x k+αn θ +ρkA+kkAk(θ +θ ) kyn−y k. (3.6)
1 1 1 1 3 1 1
(cid:0) (cid:1)
Similarly, we obtain
kyn+1−y k≤(1−αn)kyn−y k+αn θ +ρkA+kkAk(θ +θ ) kxn−x k. (3.7)
1 1 1 1 2 2 4 1 1
(cid:0) (cid:1)
Now, define the norm ||·|| on E ×E by
⋆ 1 2
||(x,y)|| =||x||+||y||, (x,y)∈E ×E .
⋆ 1 2
We can easily show that (E ×E ,||·|| ) is a Banach space.
1 2 ⋆
By making using of (3.6) and (3.7), we have the following estimate:
k(xn+1,yn+1)−(x ,y )k =kxn+1−x k+kyn+1−y k
1 1 1 1 ∗ 1 1 1 1
≤(1−α )(kxn−x k+kyn−y k)
n 1 1 1 1
+α θ +ρkA+kkAk(θ +θ ) kyn−y kkyn−y k
n 1 1 3 1 1 1 1
(cid:0) (cid:1)
+α θ +ρkA+kkAk(θ +θ ) kxn−x k
n 2 2 4 1 1
(cid:0) (cid:1)
≤(1−α )(kxn−x k+kyn−y k)
n 1 1 1 1
+α max{k ,k }(kxn−x k+kyn−y k)
n 1 2 1 1 1 1
=(1−α (1−θ))k(x ,y )−(x,y)k , (3.8)
n n n ∗
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