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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(2): 257-267

Published online June 30, 2021

### Algorithm of Common Solutions to the Cayley Inclusion and Fixed Point Problems

Department of Mathematical Sciences, Baba Ghulam Shah Badshah University, Rajouri, Jammu and Kashmir, 185234, India

Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India and Department of Mathematics, Faculty of Science, Islamic University of Madinah, Saudi Arabia

Department of Mathematical Sciences, Baba Ghulam Shah Badshah University, Rajouri, Jammu and Kashmir, 185234, India
e-mail : javid2iqbal@gmail.com and waseemm181@gmail.com

Received: July 29, 2020; Revised: October 16, 2020; Accepted: November 16, 2020

### Abstract

In this paper, we develop an iterative algorithm for obtaining common solutions to the Cayley inclusion problem and the set of fixed points of a non-expansive mapping in Hilbert spaces. A numerical example is given for the justification of our claim.

Keywords: resolvent operator, Cayley operator, Cayley inclusion, fixed point, algorithm

### 1. Introduction

Everywhere in the paper V is assumed to be a real Hilbert space with inner product and norm , and , respectively and F(S)={vV:Sv=v} to be the fixed point set of the mapping S. If A:VV and N:V2V are single and multi-valued mappings, respectively then the variational inclusion problem consists of obtaining vV such that

0A(v)+N(v).

Problem (1.1) and related problems have been considered by many authors in papers such as [1, 2, 3, 7, 8, 10, 11], and has applications in economics, physics, and structural analysis. Another problem known as the fixed point problem is the problem of obtaining v*V such that

v*=S(v*).

Here S:VV. This problem 1.2 was considered in [6, 9, 12, 14], and is used for mathematical models of real problems. For the last several years, many researchers, see for example [12, 13, 14], have found common solutions to the problems (1.1) and (1.2). In this paper we find a common solution to the Cayley inclusion problem and the fixed point problem.

### 2. Preliminaries

In this section, we go through the basic definitions and results used in the paper.

Definition 2.1.([14]) A mapping S:VV is called non-expansive if

S(v)S(w)vw   v,wV.

Definition 2.2.([14]) A mapping A:VV is called α-inverse strongly monotone if there exists α∈ R+ such that

A(v)A(w),vwαA(v)A(w)2  v,wV.

Definition 2.3.([14]) Let N:V2V be a multi-valued mapping, then it is said to be

• (i) monotone if for all v,wV, xN(v), yN(w) such that 0vw,xy;

• (ii) strongly monotone if for all v,wV, xN(v), yN(w) there exists &#_120579; ∈ R+ such that θvw2vw,xy;

• (iii) maximal monotone if N is monotone and (I+ηN)(V)=V for all η>0, where I is the identity mapping on V.

Lemma 2.1.([14]) Let {em},{fm} and {gm} be three non-negative real sequences satisfying the following condition:

em+1(1λm)em+fm+gm     mm0,

where m0 is some non-negative integer, {λm} is a sequence in (0, 1) with m=0λm=, fm=o(λm) and m=0gm<, then limmem=0.

Lemma 2.2.([5]) Let E be a real Banach space, J:E2E* be the normalized duality mapping, then for any x,yE, the following conclusion holds:

x+y2x2+2y,j(x+y),  j(x+y)J(x+y).

In particular, If E=V is a real Hilbert space, then

x+y2x2+2y,x+y,  x,yV.

Definition 2.4. ([14]) Let K be a nonempty closed and convex subset of a Hilbert space V, then for any vV, there exists a unique nearest point in K, designated by PK(v), such that

vPK(v)vw,  wK.

This mapping PK from V to K is known as metric projection.

Remark 2.1. The metric projection PK has the following properties:

• (i) PK:VK is non-expansive, i.e., PK(v)PK(w)vw,  v,wV;

• (ii) PK is firmly non-expansive, i.e., PK(v)PK(w)2PK(v)PK(w),vw  v,wV;

• (iii) for each vV, u=PK(v)vu,uw0,  wK.

Definition 2.5.([14]) Let N:V2V be a multi-valued maximal monotone mapping, then the single valued resolvent operator is defined as:

JηN(v)=[I+ηN]1(v),  vV.

Here ηR+ and I is the identity mapping.

Remark 2.2. The resolvent operator JηN has the following properties:

• (i) it is single valued and non-expansive, i.e., JηN(v)JηN(w)vw,  v,wV and for ηR+;

• (ii) it is 1-inverse strongly monotone, i.e., JηN(v)JηN(w)2vw,JηN(v)JηN(w),  v,wV.

Definition 2.6. Let N:V2V be a multi-valued maximal monotone mapping and JηN be the resolvent operator associated with it, then the Cayley operator CηN is defined as:

CηN(v)=[2JηN(v)I],  vV.

Remark 2.3. Using Remark 2.2, it can be easily seen that the Cayley operator CηN is 3-Lipschitz continuous.

In short, we denote it by C, i.e., C(v)=CηN(v). Let N:V2V be a multi-valued maximal monotone mapping, JηN be the resolvent operator associated with it and CηN be the Cayley operator, then Cayley inclusion problem is to find vV such that

0CηN(v)+N(v).

Or in short it can be written as

0C(v)+N(v).

Lemma 2.3.([4]) {\it Let N:V2V be a maximal monotone mapping and B:VV be a Lipschitz continuous mapping. Then a mapping B+N:V2V is a maximal monotone mapping.

In view of Remark 2.3 and Lemma 2.3, we can see that C+N:V2V, where C is a Cayley operator given by (2.1) is a maximal monotone. So a new resolvent operator can be defined as follows.

Definition 2.7. Let N:V2V be a maximal monotone mapping and C:VV be a cayley operator given by equation (2.1) which is Lipschitz continuous, so that C+N:V2V is also a maximal monotone mapping. A new resolvent operator associated with C+N is defined as:

JηC+N(v)=[I+η(C+N)]1(v)  vV.

Remark 2.4. The resolvent operator JηC+N is also non-expansive and 1-inverse strongly monotone.

### 3. Main Result

In this section, we will discuss an algorithm for obtaining common solutions to the problems (1.2) and (2.2). Before going to the main result, we first state the Lemma which is used in the main result.

Lemma 3.1. vV is a solution of variational inclusion problem (2.2) iff v=JηC+N(v),    ηR+.

Proof. If vV is a solution of problem (2.2), then for ηR+,

0(C+N)v0η(C+N)v      v[1+η(C+N)v]      v=[1+η(C+N)]1(v)=JηC+N(v).

Using Lemma 3.1, we develop the following iterative algorithm for obtaining common solutions to problems (1.2) and (2.2).

Algorithm: Let N:V2V be a multi-valued maximal monotone mapping, JηN be the resolvent operator associated with it, C be the Cayley operator and S:VV be a non-expansive mapping, then let

vm+1=βmv+(1βm)S(wm),wm=JηC+N(vm),  m=0,1,2,.

Now we state and prove our main result in which we show that the sequence {vm} generated by (3.1) under certain conditions converges strongly to common solutions to the problems (1.2) and (2.2).

Theorem 3.1. Let V be a Hilbert space, N:V2V be a multi-valued maximal monotone mapping, C be the Cayley operator given by (2.1), which is Lipschitz continuous so that C+N:V2V is a maximal monotone mapping by Lemma 2.3 and S:VV be a non-expansive mapping. Let F(S)VI(I,C+N)ϕ. Suppose v0V and {vm} be the sequence given by (3.1) with the following conditions:

• (i) limmβm=0; m=0βm=;

• (ii) m=0|βm+1βm|<.

Then {vm} converges strongly to F(S)VI(I,C+N).

Proof. We prove the theorem in six steps.

Step 1: First we show that the sequences {vm} and {wm} are bounded.

For zF(S)VI(I,C+N) and from Lemma 3.1, we have

z=JηC+N(z).

So, we calculate

wmz=JηC+N(vm)JηC+N(z)    vmz  m0.

Using (3.1) and (3.2), we can write

vm+1z=βm(vz)+(1βm)(Swmz)    βmvz+(1βm)wmz    βmvz+(1βm)vmz    maxvz,vmz        maxvz,v0z    =vz.

From above we conclude that the sequences {vm} and {wm} are bounded. Since S is non-expansive and C is Lipschitz continuous so {Svm} and {Cwm} are also bounded in V.

Step 2: Here we prove that

vm+1vm0  and  wm+1wm0 as m.

Since resolvent operator given by (2.3) is non-expansive, we calculate

wm+1wm=JηC+N(vm+1)JηC+N(vm)      vm+1vm.

Hence from (3.1) and (3.5), we obtain

vm+1vm=βmv+(1βm)Swm(βm1v+(1βm1)Swm1)    =(βmβm1)(vSwm1)+(1βm)(SwmSwm1)    |βmβm1|vSwm1+(1βm)SwmSwm1    |βmβm1|M+(1βm)wmwm1    |βmβm1|M+(1βm)vmvm1.

Here M=supm1vSwm1. We see that all the conditions of Lemma 2.1 are satisfied by taking em=vmvm1, fm=0 and gm=|βmβm1|M and so vm+1vm0 as m. From (3.5) wm+1wm0 as m.

Step 3: Here we prove that for zF(S)VI(I,C+N),

vmSwm0  as  m. vmSwmvmSwm1+Swm1Swm    βm1vSwm1+wm1wm.

Since βm0 and wm1wm0, therefore vmSwm0.

Step 4: Here we prove that

vmwm0;  Swmwm0.

For zF(S)VI(I,C+N) and using Remark 2.4, we obtain

wmz2=JηC+N(vm)JηC+N(z)2    vmz,wmz    =12vmz2+wmz2vmz(wmz)2    12vmz2+vmz2vmwm2.

So, we get

wmz2vmz212vmwm2.

So, using (3.1) and (3.10), we have

vm+1z2=βm(vz)(1βm)(Swmz)2    βmvz2+(1βm)Swmz2    βmvz2+(1βm)wmz2    βmvz2+(1βm)vmz212vmwm2.

This implies that

(1βm)2vmwm2βmvz2+(vmz2vm+1z2).

Since βm0 and

|vmz2vm+1z2|vm+1vm(vm+vm+1)0.

So, from (3.11), vmwm0. Also from (3.7), we obtain

SwmwmSwmvm+vmwm0.

Step 5: Here we prove that

limsupmvq,Swmq0,

here q=PF(S)VI(I,C+N)v.

Since {wm} is a bounded sequence in V, so there exists a subsequence {wmi}{wm} such that wmiwV and

limsupmvq,Swmq=limmivq,Swmiq.

Since Swmwm0, Swmiwmi0, S is non-expansive hence IS:VV is semi-closed, so S(w)=w, i.e., wF(S).

Now we prove that

wVI(I,C+N).

Since Cayley operator C is Lipschitz continuous and N is maximal monotone, therefore by Lemma 2.3 C+N is maximal monotone. Let (a,b)Graph(C+N), i.e., b(C+N)(a). Since wmi=JηC+N(vmi), we have vmi[I+(C+N)](wmi), i.e.,

1η(vmiwmi)(C+N)(wmi).

So, by maximal monotonicity (C+N), we have

awmi,b1η(vmiwmi)0.

So

awmi,bawmi,1η(vmiwmi).

Since vmiwmi0 and wmiw, we get

limmiawmi,b=aw,b0.

Because C+N is maximal monotone, this implies that 0(C+N)(w), i.e., wVI(I,C+N). So wF(S)VI(I,C+N).

Since Swmwm0 and wmiwF(S)VI(I,C+N), so from (3.13) and Remark 2.1, we get

limsupmvq,Swmq=limmivq,Swmiq        =limmivq,Swmiwmi+wmiq        =limmivq,wq0.

Hence (3.12) is proved.

Step 6: Finally we prove that

vmq=PF(S)VI(I,C+N)(v0).

Using (3.1), (3.2) and Lemma 2.2, we obtain

vm+1q2=βm(vq)+(1βm)(Swmq)2    (1βm)2(Swmq)2+2βmvq,vm+1q    (1βm)2wmq)2+2βmvq,vm+1q    (1βm)2vmq)2+2βmvq,vm+1q.

Let

γm=max0,vq,vm+1q.

Then γm0.

Now we prove that γm0.

From (3.12), it follows that for given δ>0, there exists m0 such that

vq,vm+1q<δ.

So, we have

0γm<δ,  mm0.

By the arbitrariness of δ>0, we get γm0. So we can write (3.17) as follows;

vm+1q2(1βm)2vmq)2+2βmγm.

By taking em=vm+1q2, fm=2βmγm and gm=0, all the conditions of the Lemma 2.1 are satisfied. Hence vmq as m. This proves our theorem.

### 4. Numerical Example

Example 4.1. Let V=, the set of reals and let N:2, be defined as N(v)={15(v)}  v, then we calculate resolvent operator JηN(v), Cayley operator CηN(v) and new resolvent operator JηC+N(v) for η=1 as

JηN(v)=[I+ηN]1(v)=56v.CηN(v)=[2JηN(v)I]=46v.JηC+N(v)=[I+η(C+N)]1(v)=1528v.

Let S: be defined as S(v)=v and S(v)=vβm=1m. Then all the conditions of Theorem 3.1 are satisfied and we can calculate

vm+1=1mv0+(m1)m1528v.

All codes are written in MATLAB 2012. We have taken different initial values V0=1,3.5,5.0, which show that the sequence {vm} converges to the solution of the problem. The convergence graph is shown Figure 1.

Figure 1. Convergence of vm by using Algorithm 3.1

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