Algorithmic Approach to the Split Problems

Problem 1. Find a point u^§ ∈ ℂ such that

This problem is called split feasibility problem when Ψ is a bounded linear operator. In this case, Problem 1 can be applied to many practical problems such as signal processing and image reconstruction. Specifically, we can find the prototype of Problem 1 in intensity-modulated radiation therapy; see, for example, [1–3]. Based on this relation, many mathematicians were devoted to study the split feasibility problem and develop its iterative algorithms. Related works can be found in [4–8] and the references therein.

Let 𝔸, Ψ : ℂ → ℍ be two mappings. Consider the variational inequality of finding u^† ∈ ℂ, Ψ(u^†) ∈ ℂ such that

for all Ψ(u) ∈ ℂ. We use VI (𝔸, Ψ) to denote the set of solutions of (2). Variational inequality problems have important applications in many fields such as elasticity, optimization, economics, transportation, and structural analysis, and various numerical methods have been studied by many researchers; see, for instance, [9–17].

Let ϱ : ℂ × ℂ → ℝ be an equilibrium bifunction; that is, ϱ(u, u) = 0 for each u ∈ ℂ. Consider the equilibrium problem which is to find u^* ∈ ℂ such that

Denote the set of solutions of (3) by EP (ϱ, ℂ). The equilibrium problems include fixed point problems, optimization problems, and variational inequality problems as special cases. Some algorithms have been proposed to solve the equilibrium problems; see, for example, [18–22]. Thus it is an interesting topic associated with algorithmic approach to the variational inequality and equilibrium problems. In this paper, our main purpose is to study the following split problem involved in the variational inequality and equilibrium problems. Find a point x^♮ such that

We are devoted to study (4) with operator Ψ being a nonlinear mapping. For this purpose, we develop an iterative algorithm for solving the split problem (4). We can compute x^♮ iteratively by using our algorithm. Convergence analysis is given under some mild assumptions.

Lemma 2 (see [19].)Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Let ϱ : ℂ × ℂ → ℝ be a bifunction. Assume that ϱ satisfies the following conditions:

•  

  (𝔉1)  ϱ(u, u) = 0 for all u ∈ ℂ;

•  

  (𝔉2)  ϱ is monotone, that is, ϱ(u, v) + ϱ(v, u) ≤ 0 for all u, v ∈ ℂ;

•  

  (𝔉3)  for each u, v, w ∈ ℂ, lim _t↓0ϱ(tw + (1 − t)u, v) ≤ ϱ(u, v);

•  

  (𝔉4)  for each u ∈ ℂ, v ↦ ϱ(u, v) is convex and lower semicontinuous.

Let ϖ > 0 and u ∈ ℂ. Then there exists w ∈ ℂ such that

Set Ϝ_ϖ(u) = {w ∈ ℂ : ϱ(w, v) + (1/ϖ)〈v − w, w − u〉≥0 for all v ∈ ℂ}. Then one have the following:

• (i)

  Ϝ_ϖ is single valued and Ϝ_ϖ is firmly nonexpansive,

• (ii)

  EP (ϱ, ℂ) is closed and convex and EP (ϱ, ℂ) = Fix (Ϝ_ϖ).

Lemma 3 (see [23].)Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. For x ∈ ℍ, let the mapping Ϝ_ϖ be the same as in Lemma 2. Then for μ, ν > 0 and x ∈ ℍ, one has

Lemma 4 (see [24].)Let {u_n} and {v_n} be two bounded sequences in a Banach space 𝔼, and let {κ_n} be a sequence in [0,1] satisfying 0 < liminf _n→∞κ_n ≤ limsup _n→∞κ_n < 1. Suppose u_n+1 = (1 − κ_n)v_n + κ_nu_n for all n ≥ 0 and limsup _n→∞(∥v_n+1 − v_n∥  −  ∥u_n+1 − u_n∥) ≤ 0. Then, lim _n→∞∥u_n − v_n∥ = 0.

Lemma 5 (see [25].)Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Let 𝕊 : ℂ → ℂ be a nonexpansive mapping with Fix (𝕊) ≠ ∅. Then 𝕊 is demiclosed on ℂ.

Lemma 6 (see [26].)Let {a_n}⊂[0, ∞) be a sequence. Assume that a_n+1 ≤ (1 − γ_n)a_n + δ_nγ_n, where {γ_n} is a sequence in (0,1), and {δ_n} is a sequence satisfying $$ and limsup _n→∞δ_n ≤ 0 (or $$). Then lim _n→∞a_n = 0.

Problem 7. Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Assume that

• (1)

  Ψ : ℂ → ℂ is a weakly continuous and ζ-strongly monotone mapping such that R(Ψ) = ℂ;

• (2)

  𝔸 : ℂ → ℍ is an ς-inverse strongly Ψ-monotone mapping;

• (3)

  ϱ : ℂ × ℂ → ℝ is a bifunction satisfying conditions (𝔉1)–(𝔉4) in Lemma 2.

Our objective is to

We use Υ to denote the set of solutions of (8). In the following, we assume that Υ is nonempty. For solving Problem 7, we introduce the following algorithm.

Algorithm 8.   

Step  0 (initialization). Let

Step  1. For given {u_n}, let the sequence {v_n} be generated iteratively by

where proj_ℂ is the metric projection and {μ_n} is a real number sequence.

Step  2. For given {v_n}, find {z_n} such that

where {ϖ_n}⊂(0, ∞) and {α_n}⊂[0,1] are two real number sequences.

Step  3. For the previous sequences {u_n} and {z_n}, let the (n + 1)th sequence {u_n+1} be generated by

where {κ_n}⊂[0,1] is a real number sequence.

Theorem 9. Assume that the following conditions are satisfied:

•  

  (C1)  lim _n→∞α_n = 0 and ∑_n α_n = ∞;

•  

  (C2)  0 < liminf _n→∞κ_n ≤ limsup _n→∞κ_n < 1;

•  

  (C3)  ϖ_n ∈ (η₁, η₂)⊂(0, ∞), μ_n ∈ (ξ₁, ξ₂) ⊂ (0,2ς), and ζ ∈ (0,2ς);

•  

  (C4)  lim _n→∞(μ_n+1 − μ_n) = 0 and lim _n→∞(ϖ_n+1 − ϖ_n) = 0.

Then the sequence {u_n} generated by Algorithm 8 converges strongly to x^* ∈ Υ.

Proof. Let $$. Hence $$ and $$, noting that $$ implies $$ for all ν > 0. Hence $$ for all n ≥ 0. Thus, from (10), we have

Condition (C3) and (13) imply that From Lemma 2 and (11), we get $$ for all n ≥ 0. Since $$, from Lemma 2 we deduce that $$ for all n ≥ 0. So, It follows that By induction Hence, {Ψ(u_n)} is bounded. Since Ψ is ζ-strongly monotone, we can get $$. So, $$. This implies that {u_n} is bounded. Next, we show ∥u_n+1 − u_n∥ → 0. From $$, we have Using Lemma 3, we obtain Then By condition (C3), we have ϖ_n > η₁ > 0. So, From (10), we have Therefore, It follows that Since lim _n→∞α_n = 0, lim _n→∞(μ_n+1 − μ_n) = 0, and lim _n→∞(ϖ_n+1 − ϖ_n) = 0 and the sequences {Ψ(u_n)}, {z_n}, {v_n}, and {𝔸u_n} are bounded, we deduce that Applying Lemma 4, we obtain Thus, This together with the ζ-strong monotonicity of Ψ implies that From (13) and (16), we derive Hence, Since α_n → 0, ∥Ψ(u_n+1) − Ψ(u_n)∥ → 0, and liminf _n→∞(1 − κ_n)(1 − α_n)μ_n(2ς − μ_n) > 0, we obtain Set $$ for all n. By using the firm nonexpansivity of projection, we get It follows that From (29) and (32), we have Then, we obtain Since lim _n→∞α_n = 0, lim _n→∞∥Ψ(u_n+1) − Ψ(u_n)∥ = 0, and $$, we deduce that Next, we prove limsup _n→∞〈Ψ(x^*), v_n − Ψ(x^*)〉≥0, where x^* satisfies (GVI): 〈Ψ(x^*), Ψ(x) − Ψ(x^*)〉≥0, for all x ∈ Υ (note that Ψ is ζ-strongly monotone; we can easily deduce that the solution of (GVI) is unique). We take a subsequence $$ of {v_n} such that

By the boundedness of $$, we can choose a subsequence $$ of $$ such that $$ weakly. For the convenience, we may assume that $$. This implies that $$ due to the weak continuity of Ψ. Now, we show z ∈ Υ. We firstly show Ψ(z) ∈ EP (ϱ, ℂ).

Note that ϖ_n ∈ (η₁, η₂). Then we choose a subsequence $$ of {ϖ_n} such that $$. From (26) and (36), we deduce that $$. Thus, $$. From Lemma 2, we know that Ϝ_ϖ is nonexpansive. By demiclosed principle (Lemma 5), we get immediately that Ψ(z) ∈ Fix (Ϝ_ϖ) = EP (ϱ, ℂ).

Next we prove z ∈ VI (𝔸, Ψ). Set

By [27], we know that R is maximal Ψ-monotone. Let (v, w) ∈ G(R). Since w − 𝔸v ∈ N_C(v) and u_n ∈ C, we have 〈Ψ(v) − Ψ(u_n), w − 𝔸v〉≥0. Noting that v_n = proj_ℂ(Ψ(u_n) − μ_n𝔸u_n), we get

It follows that Then, Since $$ and $$, we deduce that 〈Ψ(v) − Ψ(z), w〉≥0 by taking i → ∞ in (41). Thus, z ∈ R⁻¹0 by the maximal Ψ-monotonicity of R. Hence, z ∈ VI (𝔸, Ψ). Therefore, z ∈ Υ. From (37), we obtain From (12), we have Using Lemma 6, we conclude that Ψ(u_n) → Ψ(x^*), and hence u_n → x^*. This completes the proof.