Split equality variational inequality problems for pseudomonotone mappings in Banach spaces

DOI: 10.24193/subbmath.2021.1.13

Split equality variational inequality problems for pseudomonotone mappings in Banach spaces

Oganeditse A. Boikanyo and Habtu Zegeye

Dedicated to Professor Gheorghe Moro¸sanu on the occasion of his 70th anniversary.

Abstract. A new algorithm for approximating solutions of the split equality varia- tional inequality problems (SEVIP) for pseudomonotone mappings in the setting of Banach spaces is introduced. Strong convergence of the sequence generated by the proposed algorithm to a solution of the SEVIP is then derived without assuming the Lipschitz continuity of the underlying mappings and without prior knowledge of operator norms of the bounded linear operators involved. In ad- dition, we provide several applications of our method and provide a numerical example to illustrate the convergence of the proposed algorithm. Our results im- prove, consolidate and complement several results reported in the literature.

Mathematics Subject Classification (2010):47H09, 47J20, 65K15, 47J05, 90C25.

Keywords:Pseudomonotone mapping, split equality variational inequality prob- lem, strong convergence, variational inequality.

1. Introduction

Let K be a nonempty, closed and convex subset of a real Hilbert space H, and T :C →H be a nonlinear mapping. The variational inequality problem (VIP), first introduced by Stampacchia [31] and Fichera [19] in 1964, is a problem that consist of finding an elementx ∈C such that hT x, y−xi ≥ 0 for all y ∈ C. For a nonlinear mappingT :C→H, we denote the solution of the VIP byV I(C, T) if it is nonempty. It is known thatxsolves the VIP if and only ifxis a fixed point of the map P_C(I−λT) :C →C. Variational inequality problems have been studied extensively by several authors, thanks to their relevance in various applications in areas such as mechanics, physics, engineering, convex programming and control theory. Among these studies, VIPs for continuous and pseudomonotone maps will be of particular interest to us. Let us remember that if T is continuous and pseudomonotone, then V I(C, T) is closed and convex [26]. In [30, 33, 34], the authors studied algorithms for

solving uniformly continuous and weakly sequentially continuous pseudomonotone VIPs in Hilbert spaces. The distinctive feature of the algorithms constructed and analyzed in [33, 34] is mainly on the different Armijo-type line search rules used. For further reading on the VIP, particularly iterative methods for finding solutions of VIPs, the interested reader is referred to articles [2, 7, 10, 21–23, 25, 29, 38, 41], and their references.

Let K₁ and K₂ be nonempty, closed and convex subsets of real Hilbert spaces H₁ and H₂, respectively. Also let A : H₁ → H₃ and B : H₂ → H₃ be bounded linear mappings, whereH₃ is another real Hilbert space. Consider two nonlinear mappings T : H₁ → H₁ and S : H₂ → H₂. The split equality variational inequality problem (SEVIP) is formulated as a problem of finding:

(x, y)∈K1×K2 such that (x, y)∈V I(K1, T)×V I(K2, S) andAx=By. (1.1) The SEVIP is quite general and it includes as special cases, split equality zero point problem (see, [18]), common solutions of the variational inequality problem [12], com- mon zeros of mappings [16], split equality feasibility problem [27], has been studied extensively by many authors and applied to solving many real life problems such as in modelling intensity-modulated radiation therapy treatment planning [8, 9], modelling of inverse problems arising from phase retrieval, and in sensor networks in comput- erised tomography and data compression [5, 17].

If, in (1.1), we considerH2=H3, andB =I, the identity mapping onH2, the SEVIP reduces to the split varitional inequality problem (SVIP) that was recently introduced by Censoret al.[10]. The SVIP consists of finding:

(x, y)∈K₁×K₂ such that (x, y)∈V I(K₁, T)×V I(K₂, S) andy=Ax, (1.2) that is, the SVIP constitutes a pair of VIPs, which have to be solved so that the image y=Ax, under a given bounded linear operatorAof the solutionxof the VIP inH1, is a solution of another VIP in another spaceH₂. In Moudafi [27], it was noted that the SVIP generalizes the split fixed point problem, split variational inequality problem, split zero point problem and split feasibility problem (see also [3, 4, 6, 13–15, 35, 40], and the references therein). Many of the results cited above were obtained in the setting of real Hilbert spaces. In [11], Censoret al.studied an iterative algorithm that approximates a solution of the SVIP for a monotone mapping in Hilbert spaces and proved weak convergence results of the algorithm. In [6], Byrne et al.constructed a scheme which approximates the solution of the SVIP for monotone type mappings in Hilbert spaces and proved weak and strong convergence results of the scheme under certain assumptions.

Motivated by the work of Censoret al.[11], Byrneet al.[6] and Thonget al [33], we introduce and study a new algorithm for solving the SEVIP for uniformly continu- ous and weakly sequentially continuous pseudomonotone mappings in the setting of Banach spaces. Strong convergence of the proposed algorithm is proved under mild assumptions and without prior knowledge of operator norms of bounded linear map- pings involved. Some applications of the main results are also provided. A numerical example is given to illustrate the convergence of the proposed algorithm. Our results improve, consolidate and complement several results in the literature.

2. Preliminaries

Let E be a reflexive, strictly convex and smooth Banach space and let C be a nonempty, closed and convex subset of E. Consider the function φ : E×E → R, introduced by Alber [1], defined by

φ(y, x) =||y||²−2hy, J xi+||x||², forx, y∈E, (2.1) whereJ :E→E^∗ is thenormalized duality mappingdefined by

J x:={x^∗∈E^∗:hx, x^∗i=||x||²,||x||=||x^∗||},∀x∈E.

It is known that ifE is uniformly smooth, then J is uniformly norm to norm con- tinuous on each bounded subset ofE (see, [32]). Furthermore, if Eis a reflexive and strictly convex Banach space with a strictly convex dual, thenJ⁻¹ is a duality map- ping from E^∗ into E which satisfies J J⁻¹ = IE^∗ and J⁻¹J = IE (see, [32]). The generalized projection mapping, introduced by Alber [1], is a mapping ΠC :E →C that assigns an arbitrary pointx∈E to the minimizer, ¯x, ofφ(., x) overC.

Lemma 2.1. [1] Let C be a nonempty, closed and convex subset of a real reflex- ive, strictly convex, and smooth Banach space E and let x∈ E. Then φ(y,Π_Cx) + φ(Π_Cx, x)≤φ(y, x)for all y∈C.

Lemma 2.2. [20] Let E be a real smooth and uniformly convex Banach space and let (x_n)and(y_n)be two sequences inE. If either(x_n)or(y_n)is bounded andφ(x_n, y_n)→ 0 asn→ ∞, thenx_n−y_n→0, asn→ ∞.

Lemma 2.3. [1]LetCbe a convex subset of a real smooth Banach spaceE. Letx∈E.

Thenx0= ΠCxif and only if hz−x0, JEx−JEx0i ≤0,∀z∈C.

Consider the functionV :E×E^∗→R, studied by Alber [1], defined by V(x, x^∗) =kxk²−2hx, x^∗i+kx^∗k², for allx∈E andx^∗∈E^∗.

Lemma 2.4. [1]Let E be reflexive, strictly convex and smooth Banach space with E^∗ as its dual. Then for allx∈E andx^∗, y^∗∈E^∗,

V(x, x^∗) + 2hJ_E⁻¹x^∗−x, y^∗i ≤V(x, x^∗+y^∗).

Lemma 2.5. [28] If E is a smooth Banach space and {t_i} ∈(0,1) with PN

i=1t_i = 1, then

φ

z, J_E⁻¹X^N

i=1

tiJExi

≤

N

X

i=1

tiφ(z, xi).

Lemma 2.6. [37]Let(an)be a sequence of nonnegative real numbers such thatan+1≤ (1−βn)an +βnδn, for all n ≥ 1, where (βn) ⊂ (0,1) and (δn) ⊂ R satisfying

∞

X

n=1

βn=∞, andlim sup

n→∞

δn ≤0. Then, lim

n→∞an= 0.

Lemma 2.7. [24] Let (a_n) be a sequence of real numbers such that there exists a subsequence (n_i) of (n) such that a_ni < a_ni+1 for all i ∈ N. Then there exists a nondecreasing sequence (mk) ⊂N such that mk → ∞ and max{am_k, ak} ≤am_k+1. In fact, mk= max{j≤k:aj< aj+1}.

Lemma 2.8. [39]LetEbe a reflexive and smooth real Banach space. Then, there exists α >0such that hx−y, JEx−JEyi ≥α||x−y||² for allx, y∈E.

Lemma 2.9. [36] Let E be a reflexive and smooth real Banach space. Then for each x, y∈E, we have φ(y, x)≥ ¹₂||x−y||².

Lemma 2.10. LetCbe a closed and convex set in a reflexive real Banach spaceE,hbe a real-valued function on E, and K be the set {x∈C :h(x)≤0}. If K is nonempty andhis Lipschitz continuous onC with constantL >0, then

φ(x,Π_Kx)≥ 1

2L²(h(x))², for all x∈C. (2.2) Proof. Clearly (2.2) holds for all x∈K. Hence, it suffices to show that (2.2) holds for every x ∈ C\K. Let x ∈ C but x /∈ K. Since K is closed, there exists y ∈ K such that φ(x, y) = φ(x,ΠKx). It follows from the Lipschitz continuity of h that

|h(x)−h(y)| ≤Lkx−yk. Sincex /∈K andy ∈K, we have h(x)>0 andh(y)≤0.

Thus, from Lemma 2.9, we have

h(x)≤h(x)−h(y) =|h(x)−h(y)| ≤L||x−y|| ≤L

2φ(x,ΠKx)¹₂ ,

and hence the conclusion follows.

Definition 2.11. LetT :C→E^∗ be a mapping. ThenT is called

(a) sequentially weakly continuous on C if for each sequence (x_n) ⊆ C converging weakly tox∈C, the sequence (Ax_n) converges weakly toAx;

(b) monotone ifhx−y, T x−T yi ≥0 for eachx, y∈C;

(c) pseudomonotone onC if for allx, y∈C,

hy−x, T xi ≥0 implies hy−x, T yi ≥0. (2.3) Remark 2.12. In [30], Shehuet al. asserted that using the Monte-Carlo approach, it can be shown that the mapT :R²→R² defined by

T(x, y) =

x²+ (y−1)²

(1 +y),−x³−x(y−1)²

is pseudomonotone onR². The correctness of this method/approach in verifying pseu- domonotonicity of an operator is questionable. We claim that there could still be a pair of points, say (x, y),(u, v)∈R², such that the implication (2.3) does not hold.

Indeed, (2.3) fails to hold for a pair of points (0,1) and (−1,2) in R², as shown by simple computations below

hT(0,1),(−1,2)−(0,1)i= 0≥0 and hT(−1,2),(−1,2)−(0,1)i=−4<0.

Example 2.13. Let the mapS:R²→R²be defined by S(x, y) =

x²+ 1 + (y−1)²

(1 +y),−x³−x

1 + (y−1)² . Claim 1:S is not monotone. Indeed, for the pair (1,0) and (−1,−1), we have

hS(1,0)−S(−1,−1),(1,0)−(−1,−1)i=−3<0.

Claim 2:Sis pseudomonotone. To this end, we assume thathS(x, y),(u, v)−(x, y)i ≥0 is true for each pair (x, y),(u, v)∈R². This means that

x²+ 1 + (y−1)²

(1 +y)(u−x) +

−x³−x

1 + (y−1)²

(v−y)≥0

which implies that

x²+ 1 + (y−1)²

[u+uy−x−xv]≥0 for all (x, y),(u, v)∈R². Therefore,u(1 +y)−x(1 +v)≥0 for all (x, y),(u, v)∈R². Sinceu²+ 1 + (v−1)²>0 for anyu, v∈R, we have for any (x, y),(u, v)∈R²,

hS(u, v),(u, v)−(x, y)i=

u²+ 1 + (v−1)²

[u(1 +y)−x(1 +v)]≥0.

Lemma 2.14. [26]LetK be a nonempty, closed and convex subset of a smooth, strictly convex and reflexive real Banach space E. Let A be a continuous pseudomonotone mapping fromK intoE^∗. Then,V I(K, A)is closed and convex, andp∈V I(K, A)if and only ifhx−p, Axi ≥0, for allx∈K.

3. Main results

In the sequel, we shall make use of the following assumptions:

Assumption 1:

(A1) LetE1 andE2 be uniformly smooth and uniformly convex real Banach spaces with dual spaces E₁^∗ and E₂^∗, respectively, and let E₃ be a real Banach space with dual spaceE₃^∗.

(A2) LetA :E₁→E₃ andB :E₂ →E₃ be bounded linear mappings with adjoints A^∗:E^∗₃→E^∗₁ andB^∗ :E₃^∗→E₂^∗, respectively.

(A3) LetC⊆E₁andD⊆E₂ be nonempty, closed and convex subsets.

(A4) Let T :E1 →E₁^∗ and S :E2 → E₂^∗ be uniformly continuous pseudomonotone mappings that are sequentially weakly continuous on bounded subset of C and D, respectively.

(A5) Let Γ := {(p, q) ∈ C ×D : hx−p, T pi ≥ 0,∀x ∈ C,hy−q, Sqi ≥ 0,∀y ∈ D, andAp=Bq} 6=∅.

Assumption 2:

(B1) Letξ= min{ξ1, ξ2}, whereξ1andξ2are constants given in Lemma 2.8 associated withJE₁ andJE₂, respectively.

(B2) Letl∈(0,1),µ >0 andλ∈(0,_µ^ξ).

(B3) Let (αn) ⊂(0, e] ⊂(0,1), for some constante > 0, be such that lim

n→∞αn = 0 andP∞

n=1αn=∞.

(B4) Let 0< γ≤γn ≤ ξkAxn−Bynk²

2[kA^∗JE3(Axn−Byn)k²+kB^∗JE3(Axn−Byn)k²] forn∈Ω ={n∈N:Ax_n−By_n 6= 0}, otherwiseγ_n=γ >0.

Now, we introduce our algorithm for the SEVIP.

Algorithm 3.1

For arbitraryx0, u∈Candy0, v∈D, define an iterative algorithm by 1. Step 1. Compute:un= ΠCJ_E⁻¹

1

JE₁xn−γnA^∗JE₃(Axn−Byn) and r1(xn, un) =xn−un.

Computevn= ΠDJ_E⁻¹

2

JE₂yn+γnB^∗JE₃(Axn−Byn)

ands1(yn, vn) =yn−vn. 2. Step 2. Compute:zn= ΠCJ_E⁻¹

1

JE1un−λT un

andr2(un, zn) =un−zn. Computewn= ΠDJ_E⁻¹

2

JE₂vn−λSvn

ands2(vn, wn) =vn−wn.

3. Step 3.Computefn=un−τnr2(un, zn), whereτn =l^jn andjn is the smallest non-negative integerj satisfying

hr2(u_n, z_n), T u_n−T(u_n−l^jr₂(u_n, z_n))i ≤µkr2(u_n, z_n)k² (3.1) andgn=vn−κns2(vn, wn), whereκn=l^hn andhnis the smallest non-negative integerhsatisfying

hs₂(v_n, w_n), Sv_n−S(v_n−l^hs₂(v_n, w_n))i ≤µks₂(v_n, w_n)k². (3.2) 4. Step 4.Compute:x_n+1=J_E⁻¹

1

α_nJ_E1u+ (1−α_n)J_E1Π_Cnu_n , where

C_n:={x∈C:hx−f_n, T f_ni ≤0}, (3.3) andy_n+1=J_E⁻¹

2

α_nJ_E2v+ (1−α_n)J_E2Π_Dnv_n , where

Dn:={y∈D:hy−gn, Sgni ≤0}. (3.4) 5. Step 5. Setn:=n+ 1 and go toStep 1.

Lemma 3.1. Assume that Conditions (A1) - (A5) and (B1) - (B4) are satisfied. Then, the sequences (x_n)and(y_n)generated by Algorithm 3.1 are well defined.

Proof. It is enough to show that the search rules in (3.1) and (3.2) are well defined, and the setsCn andDn are nonempty.

Sincel∈(0,1) andT is continuous onC, it follows that

hr2(un, zn), T un−T(un−l^jr2(un, zn))i →0, as j→ ∞.

On the other hand, since kr₂(u_n, z_n)k > 0, there exists a non-negative integer j_n satisfying inequality (3.1). Similarly, from the continuity of the mapping S on D, there exists a non-negative integerhn satisfying inequality (3.2).

Furthermore, since Γ6=∅, choose (p, q)∈Γ. Then by Step 3 of the algorithm,fn∈C and gn ∈ D for each n ≥ 0, and hence by Lemma 2.14, hp−fn, T fni ≤ 0 and hq−gn, Sgni ≤0 for eachn≥0. Hence,p∈Cn andq∈Dn for eachn≥0, showing

thatCn6=∅ andDn6=∅ for eachn≥0.

Lemma 3.2. Assume that Conditions (A1) - (A5) and (B1) - (B4) are satis- fied. If (un), (zn), (vn) and (wn) are sequences generated by Algorithm 3.1, then ξλ⁻¹kr2(un, zn)k²≤ hr2(un, zn), T uniandξλ⁻¹ks2(vn, wn)k²≤ hs2(vn, wn), Svni.

Proof. Using Lemma 2.3 and the definition ofz_n, we have hz−zn, JE₁un−λT un−JE₁zni ≤0, ∀z∈C.

In particular, forz=u_n ∈C, we obtainhun−z_n, J_E1u_n−J_E1z_ni ≤λhun−z_n, T u_ni.

Using Lemma 2.8, we obtain

ξkr2(un, zn)k²≤ξ1kun−znk²≤λhun−zn, T uni.

The second inequality of the lemma can be proved in a similar way.

Lemma 3.3. Assume that Conditions (A1) - (A5) and (B1) - (B4) are met.

Let (p, q)∈Γ,Fn(x) =hx−fn, T fniandGn(y) =hy−gn, Sgni. Then (i)Fn(p)≤0 andFn(un)≥τn ξλ⁻¹−µ

kr2(un, zn)k², and (ii) Gn(q)≤0 andGn(vn)≥κn ξλ⁻¹−µ

ks2(vn, wn)k².

In particular, ifr2(un, zn)6= 0ands2(vn, wn)6= 0, then Fn(un)>0andGn(vn)>0, respectively.

Proof. (i) Since (p, q)∈Γ, it follows thatp∈VI(C, T) andq∈VI(D, S).

By Lemma 2.14,Fn(p) =hp−fn, T fni ≤0 for eachn≥0.

Next, we observe from Step 3 of the algorithm and the definition ofFn that Fn(un) =hun−fn, T fni=hτnr2(un, zn), T fni=τnhr2(un, zn), T fni.

But from the search rule (3.1), hr2(un, zn), T un−T fni ≤µkr2(un, zn)k², which to- gether with Lemma 3.2 imply that

Fn(un) =τnhr2(un, zn), T fni ≥ τn

hhr2(un, zn), T uni −µkr2(un, zn)k²i

≥ τn

h

ξλ⁻¹kr2(un, zn)k²−µkr2(un, zn)k²i . Obviously, ifr₂(u_n, z_n)6= 0, then from Condition (B2), we haveF_n(u_n)>0.

(ii) The proof is similar to the proof of part (i) above.

Lemma 3.4. Assume that Conditions (A1) - (A5) and (B1) - (B4) hold.

(a). If there exist (u_nk)⊂(u_n) and(z_nk)⊂(z_n)such that (u_nk)converges weakly to x∈E1 andτn_kkun_k−zn_kk²→0as k→ ∞, thenx∈VI(C, T).

(b). If there exist (vn_i)⊂(vn) and(wn_i)⊂(wn) such that (vn_i)converges weakly to y∈E₂ andκ_nikv_ni−w_nik²→0 asi→ ∞, theny ∈VI(D, S).

Proof. (a). By considering two possible cases onτ_nk, we first show that

k→∞lim kun_k−zn_kk = 0. (3.5) Case I: Assume that lim inf

k→∞ τn_k>0.

Then there exists a constantτ >0 such thatτ_nk≥τ >0 for allk∈N. Then kun_k−zn_kk²=τ_n⁻¹

k

h

τn_kkun_k−zn_kk²i

≤τ⁻¹h

τn_kkun_k−zn_kk²i

. (3.6)

Therefore, (3.5) follows from (3.6) and the assumption in the lemma.

Case II: Assume that lim inf

k→∞ τn_k= 0.

In this case, we take a subsequence (nk_j) of (nk) if necessary, we assume without loss of generality that

k→∞lim τn_k = 0 and lim

k→∞kun_k−zn_kk=a >0. (3.7) Letfk =¹_lτn_kzn_k+ 1−¹_lτn_k

un_k. Using (3.7), we get

k→∞lim kfk−un_kk= lim

k→∞l⁻¹τn_kkun_k−zn_kk= 0. (3.8) SinceT is uniformly continuous on bounded subsets ofC, it follows from (3.8) that kT fk−T un_kk →0 ask→ ∞.

From (3.1), we have hT un_k−T fk, un_k −zn_ki> µkun_k−zn_kk², and it follows that kun_k−zn_kk →0 ask→ ∞. This contradicts (3.7), hence the limit in (3.5) must hold.

Finally, we show thatx∈VI(C, T).

Since C is weakly closed, we have x ∈ C. Furthermore, from the fact that JE₁ is uniformly continuous on bounded subsets ofE1, we have

lim

k→∞kJ_E1u_nk−J_E1z_nkk = 0. (3.9) From Lemma 2.3 andzn ∈C, we gethz−zn_k, JE₁un_k−λT un_k−JE₁zn_ki ≤0 for all z∈C, which implies that

hz−zn_k, JE₁un_k−JE₁zn_ki −λhun_k−zn_k, T un_ki ≤λhz−un_k, T un_ki. (3.10) Taking the limit inferior ask→ ∞and using (3.9), we get

lim inf

k→∞hz−un_k, T un_ki ≥0 ∀z∈C. (3.11) Thus the inequality in (3.11) implies that we can choose a decreasing sequence of positive real numbers (δk) such that (δk) converges to zero ask → ∞, and for each δk there existsNk, the smallest positive integer, such that

hz−u_nj, T u_nji+δ_k≥0 ∀j≥N_k and ∀z∈C. (3.12) Since (δk) is decreasing, the sequence (Nk) is increasing.

Note that if there exists N >0 such that T u_Nk = 0 for all k ≥N, then it can be shown easily thatx∈VI(C, T).

On the other hand, if there exists a subsequence (Nk_i) of (Nk), again denoted by (Nk), such thatT uN_k 6= 0 for allk∈N, thenhaN_k, T uN_ki= 1 for eachk∈N, where

a_Nk= J_E⁻¹

1T uN_k

kT uN_kk².

From (3.12), we deduce thathz+δkaN_k−uN_k, T uN_ki ≥0 for eachk∈Nandz∈C.

SinceT is pseudomonotone, it follows that

hz+δ_ka_Nk−u_Nk, T(z+δ_ka_Nk)i ≥0 ∀k∈N and ∀z∈C. (3.13) But by our assumption, (uN_k) converges weakly to x ∈ C. Also T is sequentially weakly continuous onE₁implies that (T u_Nk) converges weakly to T x. Moreover, we can suppose thatT x6= 0 (otherwise,xis in VI(C, T)) and so

0≤ kT xk ≤lim inf

k→∞ kT uN_kk.

Since (u_Nk)⊂(u_nk) and (δ_k) converges to zero ask→ ∞, we obtain that 0≤lim sup

k→∞

kδkaN_kk= lim sup

k→∞

δk

kT uN_kk

≤ lim sup_k→∞δk

lim inf_k→∞kT uN_kk ≤ 0 kT xk, and hence kδkaN_kk →0 as k→ ∞. Therefore, taking the limit in (3.13) ask→ ∞, we get hz−x, T zi ≥ 0 for all z ∈ C. In view of Lemma 2.14, we conclude that x∈VI(C, T).

Part (b) of the Lemma can be proved in a similar way.

Remark 3.5. If in Lemma 3.4,T :E1→E₁^∗andS:E2→E₂^∗are uniformly continuous and monotone mappings, then for allz∈C, we have from (3.10)

hz−zn_k, JE1un_k−JE1zn_ki + λhzn_k−un_k, T un_ki ≤λhz−un_k, T un_k−T zi + λhz−un_k, T zi ≤λhz−un_k, T zi.

Taking the limit ask→ ∞, we get 0≤ hz−x, T zi. It then follows from Lemma 2.14 thatx∈VI(C, T). Similarly, we gety∈VI(D, S).

Lemma 3.6. Let (x_n)and(y_n)be sequences generated by Algorithm 3.1. Assume that the Conditions (A1) - (A5) and (B1) - (B4) hold. Then (x_n) and(y_n) are bounded.

Hence,(u_n),(v_n),(z_n) and(w_n)are bounded sequences.

Proof. Let (p, q)∈Γ. Thenp∈VI(C, T),q∈VI(D, S) andAp=Bq.

Denote qn = J_E⁻¹

1

JE₁xn−γnA^∗JE₃(Axn −Byn)

. Then un = ΠCqn and so from Lemmas 2.1 and 2.4, and the properties of the mappingV, we obtain

φ(p, u_n) ≤ φ p, J_E⁻¹

1

J_E1x_n−γ_nA^∗J_E3(Ax_n−By_n)

= V(p, J_E1x_n−γ_nA^∗J_E3(Ax_n−By_n))

≤ V(p, JE₁xn)−2hqn−p, γnA^∗JE₃(Axn−Byn)i

= φ(p, xn)−2γnhqn−p, A^∗JE₃(Axn−Byn)i

= φ(p, xn)−2γnhAqn−Ap, JE3(Axn−Byn)i. (3.14) Furthermore, from Lemma 2.5, Lemma 2.1 and (3.14), we have

φ(p, x_n+1) ≤ α_nφ(p, u) + (1−α_n)φ(p,Π_Cnu_n) (3.15)

≤ αnφ(p, u) + (1−αn)φ(p, un)

≤ α_nφ(p, u) + (1−α_n) [φ(p, x_n)

− 2γnhAqn−Ap, JE₃(Axn−Byn)i]. (3.16) Similarly, if we denotetn=J_E⁻¹

2

JE₂yn+γnB^∗JE₃(Axn−Byn) , then

φ(q, vn)≤φ(q, yn) + 2γnhBtn−Bq, JE₃(Axn−Byn)i, (3.17) and therefore, from Lemma 2.5, Lemma 2.1 and (3.17)

φ(q, yn+1) ≤ αnφ(q, v) + (1−αn)φ(q,ΠD_nvn) (3.18)

≤ (1−αn) [φ(q, yn) + 2γnhBtn−Bq, JE₃(Axn−Byn)i]

+ αnφ(q, v). (3.19)

Denote Υ =φ(p, u) +φ(q, v) and Θn =φ(p, xn) +φ(q, yn). Then adding (3.16) and (3.19), we get

Θ_n+1≤(1−α_n)[Θ_n−2γ_nhAqn−Bt_n, J_E3(Ax_n−By_n)i] +α_nΥ. (3.20) Now observe that

−hAq_n−Bt_n, J_E3(Ax_n−By_n)i=−hAx_n−By_n, J_E3(Ax_n−By_n)i

−hAqn−Axn, JE₃(Axn−Byn)i − hByn−Btn, JE₃(Axn−Byn)i

≤ kqn−xnk kA^∗JE₃(Axn−Byn)k − kAxn−Bynk²

+ kyn−tnk kB^∗JE₃(Axn−Byn)k. (3.21)

From Lemma 2.8 and the definition ofqn, we obtain kq_n−x_nk ≤ 1

ξ₁kγ_nA^∗J_E3(Ax_n−By_n)k ≤ γn

ξ kA^∗J_E3(Ax_n−By_n)k. (3.22) Similarly, from Lemma 2.8 and the definition oftn, we obtain

kyn−tnk ≤ γnξ⁻¹kB^∗JE₃(Axn−Byn)k. (3.23) Combining (3.21), (3.22) and (3.23), we obtain

− 2γnhAqn−Btn, JE₃(Axn−Byn)i ≤ −2γnkAxn−Bynk² + 2γ²_nξ⁻¹[kA^∗JE₃(Axn−Byn)k²+kB^∗JE₃(Axn−Byn)k²]

≤ −γkAxn−Bynk² (3.24)

for all n ∈ Ω, where the last inequality follows from Assumption (B4). If n /∈ Ω, thenAxn−Byn= 0, and in this case inequality (3.24) follows trivially. Finally, using (3.24) in (3.20), we obtain Θn+1 ≤(1−αn)Θn+αnΥ. By mathematical induction, Θn ≤max{Θ0,Υ} for all n ≥0, showing that the sequence (φ(p, xn) +φ(q, yn)) is bounded, which implies that (φ(p, xn)) and (φ(q, yn)) are bounded. By the properties ofφ, we conclude that (x_n) and (y_n) are bounded. Consequently, (u_n), (v_n), (z_n) and

(w_n) are bounded.

Theorem 3.7. Suppose the Assumptions (A1) - (A5) and (B1) - (B4) hold. Then the sequence ((xn, yn)) generated by Algorithm 3.1 converges strongly to (x^∗, y^∗) ∈ Γ, where(x^∗, y^∗) =Q

Γ(u, v).

Proof. Let (x^∗, y^∗)∈Γ be such that (x^∗, y^∗) =Q

Γ(u, v). Denote

Λn = 2h(xn, yn)−(x^∗, y^∗),(JE₁u, JE₂v)−(JE₁x^∗, JE₂y^∗)i.

Then for someM1>0, we have

Λn+1 = 2 [hxn−x^∗, JE1u−JE1x^∗i+hyn−y^∗, JE2v−JE2y^∗i]

+ 2 [hxn+1−xn, JE₁u−JE₁x^∗i+hyn+1−yn, JE₂v−JE₂y^∗i]

≤ Λ_n+M₁[kxn+1−x_nk+kyn+1−y_nk]. (3.25) Since the sequences (u_n) and (z_n) are bounded by Lemma 3.6, the sequence (f_n) is bounded. But T is uniformly continuous implies that there exists L > 0 such that kT fnk ≤Lfor alln≥0. We can then deduce that for eachn≥0, the mappingFn is Lipschitz continuous with Lipschitz constantL >0.

Now, from Lemma 2.1,

φ(x^∗,ΠC_nun) ≤ φ(x^∗, un)−φ(ΠC_nun, un). (3.26) Using Lemmas 2.10 and 3.3, we obtain

φ(Π_Cnu_n, u_n)≥L₁τ_n²kr2(u_n, z_n)k⁴ (3.27) for someL₁>0. Combining (3.26), (3.27) and (3.14), we get

φ(x^∗,ΠC_nun) ≤ φ(x^∗, xn)−2γnhAqn−Ax^∗, JE₃(Axn−Byn)i

− L1τ_n²kun−znk⁴. (3.28)

From the definition ofxn, the properties of the mapV and Lemma 2.4, φ(x^∗, xn+1) = V(x^∗, αnJE₁u+ (1−αn)JE₁ΠC_nun)

≤ V(x^∗, αnJE₁x^∗+ (1−αn)JE₁ΠC_nun) + 2α_nhxn+1−x^∗, J_E1u−J_E1x^∗i

= φ(x^∗, J_E⁻¹

1

α_nJ_E1x^∗+ (1−α_n)J_E1Π_Cnu_n )

+ 2α_nhx_n+1−x^∗, J_E1u−J_E1x^∗i. (3.29) Using Lemma 2.5, (3.29) and (3.28), we obtain for someKb1>0,

φ(x^∗, xn+1) ≤ (1−αn)φ(x^∗,ΠC_nun) + 2αnhxn+1−x^∗, JE₁u−JE₁x^∗i

≤ (1−αn) [φ(x^∗, xn)−2γnhAqn−Ax^∗, JE3(Axn−Byn)i]

+ 2αnhxn+1−x^∗, JE₁u−JE₁x^∗i −Kb1τ_n²kun−znk⁴. (3.30) Similarly, we deduce that for eachn≥0,

φ(y^∗,ΠD_nvn) ≤ φ(y^∗, vn)−φ(ΠD_nvn, vn), (3.31) and also, for someKb2>0, we derive

φ(y^∗, y_n+1) ≤ (1−α_n) [φ(y^∗, y_n) + 2γ_nhBtn−By^∗, J_E3(Ax_n−By_n)i]

+ 2αnhyn+1−y^∗, JE₂v−JE₂y^∗i −Kb2κ²_nkvn−wnk⁴. (3.32) Denote Θ^∗_n=φ(x^∗, x_n) +φ(y^∗, y_n) and Υ^∗=φ(x^∗, u) +φ(y^∗, v). Then combining (3.30), (3.32) and (3.24), we get

Θ^∗_n+1≤(1−αn)Θ^∗_n+αnΛn+1−L^∗h

τ_n²kun−znk⁴+κ²_nkvn−wnk⁴i

(3.33) for someL^∗>0. Furthermore, from (3.15), (3.26) and (3.14), we obtain

φ(x^∗, x_n+1) ≤ α_nφ(x^∗, u) + (1−α_n) [φ(x^∗, x_n)−φ(Π_Cnu_n, u_n)]

− 2(1−αn)γnhAqn−Ax^∗, JE₃(Axn−Byn)i. (3.34) Similarly, from (3.18), (3.31) and (3.17), we obtain

φ(y^∗, yn+1) ≤ αnφ(y^∗, v) + (1−αn) [φ(y^∗, yn)−φ(ΠD_nvn, vn)]

+ 2(1−α_n)γ_nhBt_n−By^∗, J_E3(Ax_n−By_n)i. (3.35) Adding (3.34) and (3.35), and using (3.24), we obtain for someM >0

Θ^∗_n+1≤Θ^∗_n+αnM−γkAxn−Bynk²−[φ(ΠC_nun, un) +φ(ΠD_nvn, vn)], which implies that

φ(Π_Cnu_n, u_n) +φ(Π_Dnv_n, v_n) +γkAx_n−By_nk² ≤ Θ^∗_n−Θ^∗_n+1

+ αnM. (3.36) Finally, we show that the sequence (Θ^∗_n) converges strongly to zero asn → ∞. For this, we consider two possible cases on (Θ^∗_n).

Case I. Assume that there existsn0∈Nsuch that the sequence of real numbers (Θ^∗_n) is decreasing for alln≥n0. It then follows that (Θ^∗_n) is convergent. Taking the limit in (3.36) asn→ ∞, we get

n→∞lim kAxn−By_nk= 0. (3.37)

From the definition ofq_n, Lemma 2.1, Lemma 2.4 and (3.22), we have φ(xn, un) ≤ φ(xn, J_E⁻¹

1

JE₁xn−γnA^∗JE₃(Axn−Byn) )

= V(xn, JE₁xn−γnA^∗JE₃(Axn−Byn))

≤ V(xn, JE1xn)−2hqn−xn, γnA^∗JE3(Axn−Byn)i

≤ φ(xn, xn) + 2γ_n²ξ⁻¹kAk²kAxn−Bynk².

Taking the limit asn→ ∞and noticing (3.37), yieldφ(xn, un)→0 asn→ ∞. Using Lemma 2.2, we obtain

n→∞lim kxn−u_nk= 0. (3.38)

Similarly, starting with the definition oft_n, we obtain

n→∞lim kyn−vnk= 0. (3.39)

Moreover, we also obtain from (3.36) and Lemma 2.2

n→∞lim kΠC_nun−unk= 0 and lim

n→∞kΠD_nvn−vnk= 0. (3.40) From the definition ofxn and Lemma 2.8,

kxn+1−unk ≤ ξ₁⁻¹kαnJE₁u+ (1−αn)JE₁ΠC_nun−JE₁unk

≤ αnK1+ξ₁⁻¹kJE₁ΠC_nun−JE₁unk, (3.41) for some constant K1 > 0. Since JE1 is norm to norm uniformly continuous on bounded subsets ofE₁, we conclude from (3.40) and (3.41) that

n→∞lim kxn+1−unk = 0. (3.42)

Therefore, combining (3.38) and (3.42) yield

n→∞lim kxn+1−xnk= 0. (3.43)

Similarly, one can show that

n→∞lim ky_n+1−y_nk= 0. (3.44)

Since (xn) and (yn) are bounded by Lemma 3.6, we obtain from (3.33) L^∗h

τ_n²kun−znk⁴+κ²_nkvn−wnk⁴i

≤Θ^∗_n−Θ^∗_n+1+αnΛn+1, (3.45) But the convergence of (Θ^∗_n) and Assumption (B3) imply that

n→∞lim τnkun−znk²= 0 and lim

n→∞κnkvn−wnk²= 0. (3.46)

Now, we deduce from Lemma 3.6 that ((xn, yn)) is a bounded sequence in C×D.

Therefore, there exists a subsequence ((xn_k, yn_k)) of ((xn, yn)) such that ((xn_k, yn_k)) converges weakly to (x, y) inE1×E2and

lim sup

n→∞

Λ_n= lim

k→∞Λ_nk. (3.47)

It then follows that (x_nk) converges weakly toxinE₁and (y_nk) converges weakly toy inE2. From (3.38), (un_k) converges weakly toxinE1and from (3.39), (vn_k) converges weakly toy in E2. Using (3.46) and Lemma 3.4, we conclude thatx∈VI(C, T) and y∈VI(D, S), respectively. Moreover,

kAx−Byk²≤2hAx−Axn_k+Byn_k−By, JE₃(Ax−By)i+kAxn_k−Byn_kk². Since (x_nk) converges weakly to x, it follows that (Ax_nk) converges weakly to Ax.

Similarly, (y_nk) converges weakly to y implies that (By_nk) converges weakly to By.

Using (3.37), we getAx=By. Consequently, (x, y)∈Γ.

From (3.25), (3.47), (3.43), (3.44) and Lemma 2.3, we obtain lim sup

n→∞

Λn+1 ≤ lim sup

n→∞

Λn+M1lim sup

n→∞

[kxn+1−xnk+kyn+1−ynk]

= lim

k→∞Λ_nk+M₁ lim

k→∞[kx_nk+1−x_nkk+ky_nk+1−y_nkk]

= 2h(x, y)−(x^∗, y^∗),(JE₁u, JE₂v)−(JE₁x^∗, JE₂y^∗)i

≤ 0. (3.48)

Finally, from (3.33), we have Θ^∗_n+1≤(1−αn)Θ^∗_n+αnΛn+1. Therefore, from (3.48) and Lemma 2.6, we conclude that (Θ^∗_n) converges to zero asn→ ∞. That is,φ(x^∗, xn)→0 andφ(y^∗, yn)→0 asn→ ∞. Hence by Lemma 2.2, we have (xn) and (yn) converges tox^∗ andy^∗, respectively.

Case II. Assume that there exists a subsequence (Θ^∗_ni) of (Θ^∗_n) such that Θ^∗_ni<Θ^∗_ni+1 for all i ≥ 0. Then in view of Lemma 2.7, we can define a nondecreasing sequence (mk)⊂Nsuch thatmk → ∞ask→ ∞and

Θ^∗_m

k≤Θ^∗_m

k+1 and Θ^∗_k ≤Θ^∗_m

k+1 (3.49)

for allk∈N. Following similar steps as in Case I, we derive lim sup

k→∞

Λm_k+1≤0. (3.50)

From (3.33) and (3.49), we obtain α_mkΘ^∗_m

k+1 ≤ α_mkΛ_mk+1, which reduces to Θ^∗_m

k+1 ≤ Λ_mk+1. Taking the limit as k → ∞ and using (3.50), we conclude that Θ^∗_m

k+1→0 as k→ ∞. Again from (3.49), it follows that Θ^∗_k →0 ask→ ∞. There- fore, φ(x^∗, xk) → 0 and φ(y^∗, yk) → 0 as k → ∞. Hence by Lemma 2.2, we have

xk→x^∗ andyk→y^∗ as k→ ∞.

If u = 0 and v = 0, then Algorithm 3.1 can be used to locate an element of the solution with the minimum norm.

Corollary 3.8. Let the Assumptions (A1) – (A5) and (B1) – (B4) hold. Then, the sequence ((xn, yn))generated by Algorithm 3.1 withu= 0 =v converges strongly to the minimum norm point(x^∗, y^∗)∈Γ, that is,(x^∗, y^∗) =Q

Γ(0,0).

Corollary 3.9. Assume thatT :E1→E^∗₁ andS:E2→E₂^∗ are uniformly continuous and monotone mappings. Let the Assumptions (A1) – (A3), (A5) and (B1) – (B4) be satisfied. Then the sequence((xn, yn))generated by Algorithm 3.1 converges strongly to(x^∗, y^∗)∈Γ, where(x^∗, y^∗) =Q

Γ(u, v).

Proof. The mappings T andS are pseudomonotone, hence by Lemma 3.6, (xn) and (yn) are bounded. It then follows from (3.46) and Remark 3.5 thatx∈VI(C, T) and y ∈VI(D, S), where xand y are weak cluster points of (xn) and (yn), respectively.

The rest of the proof is similar to the proof of Theorem 3.7.

4. Applications

In this section, we apply our main result to solve the following problems: split equality zero point problem (SEZPP), common solutions of the variational inequality problem, common zeros of pseudomonotone mappings, split variational inequality problem, split zero point problem (SZPP), split equality feasibility problem (SEFP) and split feasibility problem (SFP).

4.1. Split equality zero point problem

If C = E1 and D = E2, then the SEVIP reduces to the SEZPP, which is to findx ∈ T⁻¹(0) andy ∈ S⁻¹(0) such thatAx = By, where T⁻¹(0) = {p ∈ E1 : 0 =T p} and S⁻¹(0) = {q ∈ E2 : 0 = Sq}. Denote the solution of this problem by z={(p, q)∈E1×E2:p∈T⁻¹(0), q∈S⁻¹(0) andAp=Bq}.

Corollary 4.1. Assume that z6=∅. Let the Assumptions (A1), (A2), (A4) and (B1) – (B4) be satisfied withC=E₁ andD=E₂. Then the sequence ((x_n, y_n))generated by Algorithm 3.1 converges strongly to(x^∗, y^∗)∈z, where(x^∗, y^∗) =Q

z(u, v).

4.2. Common solutions of the variational inequality problem

LetE =E1 =E2 =E3,A=I andB =I. In this case, the SEVIP reduces to finding common solutions of two variational inequality problems for pseudomonotone mappings. DenoteF ={(p, q)∈C×D:hx−p, T pi ≥0,∀x∈Candhy−q, Sqi ≥0,

∀y∈D such that p=q}.

Corollary 4.2. Assume that F 6= ∅. Let the Assumptions (A1), (A3), (A4) and (B1) – (B4) be satisfied with E = E1 = E2 = E3 and A = I = B. Then the sequence ((xn, yn)) generated by Algorithm 3.1 converges strongly to (x^∗, y^∗) ∈ F, where(x^∗, y^∗) =Q

F(u, v).

4.3. Common zeros of pseudomonotone mappings

Let E = E1 = E2 = E3, A = I and B = I. If C = E and D = E, then the SEVIP reduces to finding common zeros of pseudomonotone mappings. Denote F⁰ ={(p, q)∈E×E:p∈T⁻¹(0) and q∈S⁻¹(0) such that p=q}.

Corollary 4.3. Let the Assumptions (A1), (A4), (B1) – (B4) be satisfied withC=D= E=E1=E2=E3 andA=I=B. IfF⁰ 6=∅, then the sequence((xn, yn))generated by Algorithm 3.1 converges strongly to(x^∗, y^∗)∈ F⁰, where (x^∗, y^∗) =Q

F⁰(u, v).

4.4. Split variational inequality problem

IfE₂=E₃andB=I, then the SEVIP reduces to the split variational inequality problem (SVIP) which is to find x∈V I(C, T) andy ∈V I(D, S) such thatAx=y.

DenoteT ={(p, q)∈C×D:p∈V I(C, T), q∈V I(D, S) andAp=q}.

Corollary 4.4. Assume that the Assumptions (A1) – (A4) and (B1) – (B4) hold with E3 =E2 and B =I. IfT 6=∅, then the sequence ((xn, yn))generated by Algorithm 3.1 converges strongly to (x^∗, y^∗)∈ T, where (x^∗, y^∗) =Q

T(u, v).

4.5. Split zero point problem

LetE =E2 =E3 and B =I. IfC =E andD =E, then the SEVIP reduces to the SZPP which is to findx∈T⁻¹(0) and y∈S⁻¹(0) such thatAx=y. Denote S={(p, q)∈E×E:p∈T⁻¹(0), q∈S⁻¹(0) andAp=q}.

Corollary 4.5. Assume that the Assumptions (A1), (A2), (A4), (B1) – (B3) and (B4) with B =I hold. IfS 6=∅, then the sequence ((x_n, y_n)) generated by Algorithm 3.1 converges strongly to (x^∗, y^∗)∈ S, where (x^∗, y^∗) =Q

S(u, v).

Remark 4.6. (a). If E=E1=E2=E3,S= 0,A= 0 andB = 0, then Theorem 3.7 can be used to find solutions of variational inequality problems for uniformly contin- uous pseudomonotone mappings that are sequentially weakly continuous on bounded subsets ofE as well as for uniformly continuous monotone mappings. If in addition, we take C=E andD =E, then Corollary 4.1 will approximate zeros of uniformly continuous pseudomonotone mappings that are sequentially weakly continuous on bounded subsets ofE and also zeros of uniformly continuous monotone mappings.

(b). In view of Corollary 3.9, and the discussion in this section, one can use the results of this section to find solutions of split equality zero point problem for uniformly con- tinuous monotone mappings, common solutions of the variational inequality problem for uniformly continuous monotone mappings, common zeros of uniformly continuous monotone mappings, split variational inequality problems for uniformly continuous monotone mappings, split zero point problem for monotone mappings.

(c). The special cases of the above results can be obtained by taking E₁ = H₁, E2=H2 andE3=H3 to be real Hilbert spaces.

Note that ifE=H, a real Hilbert space, thenJE=I, the identity mapping onH, and ΠC =PC, the metric projection onto C. A well known example of a uniformly con- tinuous, monotone and hence pseudomonotone map isI−PC. Henceforth,E1=H1, E2 =H2 andE3 =H3 are real Hilbert spaces,C ⊂H1 andD ⊂H2 are nonempty, closed and convex subsets. Also A:H1→H3 and B :H2→H3 are bounded linear mappings with adjointsA^∗andB^∗, respectively,T =I−PC andS=I−PD. Thus, we obtain the following applications in Hilbert spaces.

4.6. Split equality feasibility problem

Replacing T with I−PC and S withI−PD, then the SEZPP reduces to the SEFP which seeks to findx∈Candy∈Dsuch thatAx=By. Denote

Γ⁰ ={(p, q)∈C×D:Ax=By}.