Everywhere in the paper V is assumed to be a real Hilbert space with inner product and norm ⟨⋅,⋅⟩ and ‖⋅‖, respectively and F(S)={v∈V:Sv=v} to be the fixed point set of the mapping S. If A:V→V and N:V→2V are single and multi-valued mappings, respectively then the variational inclusion problem consists of obtaining v∈V such that

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

Here S:V→V. 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.

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

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

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

Definition 2.3.([14]) Let N:V→2V be a multi-valued mapping, then it is said to be

• (i) monotone if for all v,w∈V, x∈N(v), y∈N(w) such that 0≤v−w,x−y;

• (ii) strongly monotone if for all v,w∈V, x∈N(v), y∈N(w) there exists &#_120579; ∈ R⁺ such that θ‖v−w‖2≤v−w,x−y;

• (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 {g_m} be three non-negative real sequences satisfying the following condition:

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

Lemma 2.2.([5]) Let E be a real Banach space, J:E→2E* be the normalized duality mapping, then for any x,y∈E, the following conclusion holds:

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

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

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:V→K is non-expansive, i.e., ‖PK(v)−PK(w)‖≤‖v−w‖,  ∀v,w∈V;

• (ii) PK is firmly non-expansive, i.e., ‖PK(v)−PK(w)‖2≤⟨PK(v)−PK(w),v−w⟩  ∀v,w∈V;

• (iii) for each v∈V, u=PK(v)⇔⟨v−u,u−w⟩≥0,  ∀w∈K.

Definition 2.5.([14]) Let N:V→2V be a multi-valued maximal monotone mapping, then the single valued resolvent operator is defined as:

Here η∈R+ 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)‖≤‖v−w‖,  ∀v,w∈V and for η∈R+;

• (ii) it is 1-inverse strongly monotone, i.e., ‖JηN(v)−JηN(w)‖2≤⟨v−w,JηN(v)−JηN(w)⟩,  ∀v,w∈V.

Definition 2.6. Let N:V→2V be a multi-valued maximal monotone mapping and JηN be the resolvent operator associated with it, then the Cayley operator CηN is defined as:

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:V→2V 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 v∈V such that

Or in short it can be written as

Lemma 2.3.([4]) {\it Let N:V→2V be a maximal monotone mapping and B:V→V be a Lipschitz continuous mapping. Then a mapping B+N:V→2V is a maximal monotone mapping.

In view of Remark 2.3 and Lemma 2.3, we can see that C+N:V→2V, 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:V→2V be a maximal monotone mapping and C:V→V be a cayley operator given by equation (2.1) which is Lipschitz continuous, so that C+N:V→2V is also a maximal monotone mapping. A new resolvent operator associated with C+N is defined as:

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

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. v∈V is a solution of variational inclusion problem (2.2) iff v=JηC+N(v),  ∀  η∈R+.

Proof. If v∈V is a solution of problem (2.2), then for η∈R+,

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

Algorithm: Let N:V→2V be a multi-valued maximal monotone mapping, JηN be the resolvent operator associated with it, C be the Cayley operator and S:V→V be a non-expansive mapping, then let

Now we state and prove our main result in which we show that the sequence {v_m} 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:V→2V be a multi-valued maximal monotone mapping, C be the Cayley operator given by (2.1), which is Lipschitz continuous so that C+N:V→2V is a maximal monotone mapping by Lemma 2.3 and S:V→V be a non-expansive mapping. Let F(S)∩VI(I,C+N)≠ϕ. Suppose v0∈V and {v_m} be the sequence given by (3.1) with the following conditions:

• (i) limm→∞βm=0; ∑ m=0∞βm=∞;

• (ii) ∑ m=0∞|βm+1−βm|<∞.

Then {v_m} 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 {v_m} and {w_m} are bounded.

For z∈F(S)∩VI(I,C+N) and from Lemma 3.1, we have

So, we calculate

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

From above we conclude that the sequences {v_m} and {w_m} 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

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

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

Here M=supm≥1‖v−Swm−1‖. We see that all the conditions of Lemma 2.1 are satisfied by taking em=‖vm−vm−1‖, f_m=0 and gm=|βm−βm−1|M and so ‖vm+1−vm‖→0 as m→∞. From (3.5) ‖wm+1−wm‖→0 as m→∞.

Step 3: Here we prove that for z∈F(S)∩VI(I,C+N),

Since βm→0 and ‖wm−1−wm‖→0, therefore ‖vm−Swm‖→0.

Step 4: Here we prove that

For z∈F(S)∩VI(I,C+N) and using Remark 2.4, we obtain

So, we get

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

This implies that

Since βm→0 and

So, from (3.11), ‖vm−wm‖→0. Also from (3.7), we obtain

Step 5: Here we prove that

here q=PF(S)∩VI(I,C+N)v.

Since {w_m} is a bounded sequence in V, so there exists a subsequence {wmi}⊂{wm} such that wmi⇀w∈V and

Since ‖Swm−wm‖→0, ‖Swmi−wmi‖→0, S is non-expansive hence I−S:V→V is semi-closed, so S(w)=w, i.e., w∈F(S).

Now we prove that

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.,

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

So

Since ‖vmi−wmi‖→0 and wmi⇀w, we get

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

Since ‖Swm−wm‖→0 and wmi⇀w∈F(S)∩VI(I,C+N), so from (3.13) and Remark 2.1, we get

Hence (3.12) is proved.

Step 6: Finally we prove that

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

Let

Then γm≥0.

Now we prove that γm→0.

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

So, we have

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

By taking em=‖vm+1−q‖2, fm=2βmγm and g_m=0, all the conditions of the Lemma 2.1 are satisfied. Hence vm→q as m→∞. This proves our theorem.

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

Let S:ℝ→ℝ 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

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

Figure 1. Convergence of v_m by using Algorithm 3.1