A Unified Iterative Treatment for Solutions of Problems of Split Feasibility and Equilibrium in Hilbert Spaces

Lemma 1 (see [4].)If S is a self-mapping on ℋ, then the following assertions hold.

• (a)

  S is nonexpansive if and only if the complement I − S is 1/2-averaged.

• (b)

  If S is ν-ism and γ > 0, then γS is (ν/γ)-ism.

• (c)

  S is α-averaged if and only if I − S is (1/2α)-ism.

• (d)

  If S is α₁-averaged and T is α₂-averaged, then the composite ST is α₁ + α₂ − α₁α₂-averaged.

• (e)

  If S and U are averaged on ℋ so that Fix (S)∩Fix (U) ≠ ∅, then

  $$ ()

Lemma 2. Let A be a maximal monotone operator on ℋ. Then,

• (a)

  $$ is single-valued and firmly nonexpansive;

• (b)

  $$ and $$;

• (c)

  (the resolvent identity) for μ, λ > 0, the following identity holds:

  $$ ()

Lemma 3. Let x, y, z ∈ ℋ. Then,

• (a)

  ∥x+y∥² ≤ ∥x∥² + 2〈y, x + y〉;

• (b)

  for any λ ∈ ℝ,

  $$ ()

• (c)

  for a, b, c ∈ [0,1] with a + b + c = 1,

  $$ ()

Lemma 4 (see [11] demiclosedness principle.)Let S be a nonexpansive self-mapping on ℋ and suppose that {x_n} is a sequence in ℋ such that {x_n} converges weakly to some z ∈ ℋ and lim _n→∞∥x_n − Sx_n∥ = 0. Then, Sz = z.

Lemma 5 (see [23].)Let {s_n} be a sequence of nonnegative real numbers satisfying

where {α_n}, {μ_n}, and {ν_n} verify the following conditions:

• (i)

  {α_n}⊆[0,1], $$;

• (ii)

  limsup _n→∞μ_n ≤ 0;

• (iii)

  {ν_n}⊆[0, ∞) and $$.

Then, lim _n→∞s_n = 0.

Lemma 6 (see [25].)Let {s_n} be a sequence in ℝ that does not decrease at infinity in the sense that there exists a subsequence $$ such that

For any k ∈ ℕ, define m_k = max {j ≤ k : s_j < s_j+1}. Then, m_k → ∞ as k → ∞ and $$, for all k ∈ ℕ.

Lemma 7. Let S be a firmly nonexpansive self-mapping on ℋ with Fix (S) ≠ ∅. Then, for any x ∈ ℋ, one has

Proof. Since v ∈ Fix (S) and S is firmly nonexpansive, we have

and hence,

Proposition 8. For any γ ∈ (0, 2/∥A∥²), the operator I − γA^*(I − T)A is γ∥A∥²/2-averaged and S[I − γA^*(I − T)A] is (2 + γ∥A∥²)/4-averaged.

Proof. Using the fact that T is firmly nonexpansive, it is routine to show that A^*(I − T)A is 1/∥A∥²-ism, and so γA^*(I − T)A is 1/γ∥A∥²-ism by Lemma 1(b). Thus, Lemma 1(c) shows that I − γA^*(I − T)A is γ∥A∥²/2-averaged. As S is firmly nonexpansive, it is 1/2-averaged. Therefore, S[I − γA^*(I − T)A] is (1/2) + (γ∥A∥²/2) − (γ∥A∥²/4) = ((2 + γ∥A∥²)/4)-averaged by Lemma 1(d).

Proposition 9. Let Ω be the solution set of SFFP(5); that is, Ω = Fix (S)∩A⁻¹(Fix (T)). For any γ ∈ (0, 2/∥A∥²), let U = I − γA^*(I − T)A. Suppose that Ω ≠ ∅. Then, Fix (SU) = Fix (S)∩Fix (U) = Ω.

Proof. If x^* solves SFFP(5), we have x^* ∈ Fix (S) and Ax^* ∈ Fix (T). Now, note that Ax^* ∈ Fix (T) implies

and so which means that x^* ∈ Fix (U). Consequently, x^* ∈ Fix (S)∩Fix (U). This shows that Ω⊆Fix (S)∩Fix (U).

For the inverse inclusion, let x^* be any member of Fix (S)∩Fix (U) and pick v ∈ Ω. It is readily seen from x^* ∈ Fix (U) that

Since Av ∈ Fix (T), an application of Lemma 7 yields which together with (36) implies that This comes to conclude that Ax^* ∈ Fix (T), and hence x^* ∈ Ω once we note that x^* ∈ Fix (S). Finally, since Ω ≠ ∅ by assumption, we have Fix (S)∩Fix (U) = Ω ≠ ∅. Thus, Fix (SU) = Fix (S)∩Fix (U) follows from Lemma 1(e).

Proposition 10 (see [2], [4].)If G is an averaged self-mapping on ℋ with Fix (G) ≠ ∅, then for any x ∈ ℋ, the sequence {Gⁿx} converges weakly to a fixed point of G.

Theorem 11. Assume that SFFP(5) has a solution. Then, for any γ ∈ (0, 2/∥A∥²) and starting with any point x₁ ∈ ℋ, the sequence {x_n} generated by

converges weakly to a solution of SFFP(5).

Proposition 12 (see [2].)Let G be a α-averaged self-mapping on ℋ with Fix (G) ≠ ∅ and assume that {α_n} is a sequence in [0, 1/α] such that

Then, for any x₁ ∈ ℋ, the sequence {x_n} generated by the Mann′s algorithm converges weakly to a fixed point of G.

Theorem 13. Assume that SFFP(5) has a solution and γ ∈ (0, 2/∥A∥²). Let {α_n} be a sequence in [0, 4/(2 + γ∥A∥²)] with

Then, for any x₁ ∈ ℋ, the sequence {x_n} generated by the Mann′s algorithm converges weakly to a solution of SFFP(5).

Lemma 14. For any γ ∈ (0, 2/∥A∥²) and all x, y ∈ ℋ₁, one has

Proof. Since T is firmly nonexpansive, so is I − T. Hence, for all x, y ∈ ℋ₁, we have

Consequently,

Theorem 15. Let {a_n}, {b_n}, {c_n}, and {d_n} be sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {γ_n} be a sequence in (0, 2/∥A∥²) and let {e_n} be a bounded sequence in ℋ₁. Suppose that the solution set Ω of SFFP(16) is nonempty. For any u ∈ ℋ₁, start with any x₁ ∈ ℋ₁ and define a sequence {x_n} by

Then the sequence {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0,  $$,  $$;

• (ii)

  $$,  lim _n→∞inf b_nc_n > 0;

• (iii)

  there are two nonnegative real-valued functions κ₁ and κ₂ on ℕ with

  $$ ()

Proof. Put p = P_Ωu. For simplicity, put G_n = S_n(I − γ_nA^*(I − T_n)A). In view of Proposition 9, G_n is nonexpansive, so

from which follows that {x_n} is a bounded sequence. Taking into account Lemma 3, we get Meanwhile, we have by Lemma 14 that Therefore, we deduce that We now carry on with the proof by considering the following two cases: (I) {∥x_n − p∥} is eventually decreasing and (II) {∥x_n − p∥} is not eventually decreasing.

Case I. Suppose that {∥x_n − p∥} is eventually decreasing; that is, there is n₀ ∈ ℕ such that $$ is decreasing. In this case, lim _n→∞∥x_n − p∥ exists in ℝ. From inequality (52) we have

which together with the boundedness of {x_n} and conditions (i) and (ii) implies Then, an application of condition (iii) follows that for all m ∈ ℕ, Since {x_n} is bounded, it has a subsequence $$ such that $$ converges weakly to some z ∈ ℋ and where the last inequality follows from (24) since z ∈ Ω by Lemma 4. Choose M > 0 so that $$. From (52) we have Accordingly, because of (56) and condition (i), we can apply Lemma 5 to inequality (57) with $$, α_n = a_n + b_n, μ_n = 2〈u − p, x_n+1 − p〉, and ν_n = d_nM to conclude that

Case II. Suppose that {∥x_n − p∥} is not eventually decreasing. In this case, by Lemma 6, there exists a nondecreasing sequence {m_k} in ℕ such that m_k → ∞ and

Then, it follows from (52) and (59) that Therefore, and then proceeding just as in the proof in Case I, we obtain which in conjunction with condition (iii) shows that for all m_j and then follows that From (60) we have and thus, Letting k → ∞ and using (64) and condition (i), we obtain Also, since which together with (62) and condition (i) implies that $$, and so by virtue of (67). Consequently, we conclude that lim _k→∞∥x_k − p∥ = 0 via (59) and (69). This completes the proof.

Corollary 16. Let {a_n}, {b_n}, and {c_n} be sequences in [0,1] with a_n + b_n + c_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {γ_n} be a sequence in (0, 2/∥A∥²) and let {e_n} be a bounded sequence in ℋ₁. Suppose that the solution set Ω of SFFP(16) is nonempty. For any u ∈ ℋ₁, start with any x₁ ∈ ℋ₁ and define a sequence {x_n} by

Then, the sequence {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0,  $$,  $$;

• (ii)

  $$,  lim _n→∞inf b_nc_n > 0;

• (iii)

  there are two nonnegative real-valued functions κ₁ and κ₂ on ℕ with

  $$ ()
  $$ ()

• (iv)

  either lim _n→∞(∥e_n∥/a_n) = 0 or $$.

Proof. Put p = P_Ωu and G_n = S_n[I − γ_nA^*(I − T_n)A]. Let z₁ = x₁ and define a sequence {z_n} iteratively by

We have lim _n→∞z_n = p by Theorem 15. Since the limit lim _n→∞∥x_n − z_n∥ = 0 follows by applying Lemma 5 to (74), and thus,

Corollary 17. Let {a_n}, {b_n}, {c_n}, and {d_n} be sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {γ_n} be a sequence in (0, 2/∥A∥²) and let {e_n} be a bounded sequences in ℋ₁. Assume that the solution set Ω of SFFP(5) is nonempty. For any u ∈ ℋ₁, start with an arbitrary x₁ ∈ ℋ₁ and define the sequence {x_n} by

Then, {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0,  $$,  $$;

• (ii)

  $$,  lim _n→∞inf b_nc_n > 0.

Theorem 18. Let {a_n}, {b_n}, {c_n}, and {d_n} be sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Suppose that γ ∈ (0, 2/∥A∥²) and suppose that {e_n} is a bounded sequence in ℋ₁. Assume that the solution set Ω of SFFP(5) is nonempty. For any u ∈ ℋ₁, start with an arbitrary x₁ ∈ ℋ₁ and define the sequence {x_n} by

Then, {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0,  $$,  $$;

• (ii)

  lim _n→∞inf b_n > 0,  lim _n→∞inf c_n > 0.

Theorem 19. Let {a_n} be a sequence in (0,1). Suppose that γ ∈ (0, 2/∥A∥²) and assume that the solution set Ω of SFFP(5) is nonempty. Start with any x₁ ∈ ℋ₁ and define a sequence {x_n} by

Then, the sequence {x_n} converges strongly to the minimum norm solution of SFFP(5) provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = 0;

• (ii)

  $$;

• (iii)

  either  $$ or  lim _n→∞(|a_n+1 − a_n|/a_n) = 0.

Proof. Put  U = I − γA^*(I − T)A, G = SU, and G_n = S[(1 − a_n)]U for all n ∈ ℕ. Then,

Take p ∈ Ω. From Proposition 9, we have Hence, from which follows that {x_n} is bounded and so is {Ux_n}. Choose M > 0 so that ∥Ux_n∥ ≤ M for all n ∈ ℕ. We have In view of conditions (i), (ii), and (iii), we can apply Lemma 5 to (82) to get and then from we see that Consequently, the demiclosedness principle ensures that each weak limit point of {x_n} is a fixed point of the averaged mapping SU. And then we conclude from Proposition 9 that each weak limit point of {x_n} lies in Ω.

Let $$ be the minimum norm element of Ω; that is, $$. We shall show that {x_n} converges strongly to $$. To see this, we compute $$ as follows:

If $$, then an application of Lemma 5 to (86) yields that $$. Hence, to complete the proof, it suffices to show that $$. For this, taking into account Proposition 8, we can write U = (1 − β)I + βV for some β ∈ (0,1) and some nonexpansive mapping V. Then, from we obtain which ensures that lim _n→∞∥x_n − Vx_n∥ = 0, and hence, once we note that I − U = β(I − V). Since {x_n} is bounded, we can take a subsequence $$ so that it is weakly convergent to x^* ∈ Ω and where the last inequality comes from the characterization inequality of a metric projection. Now, applying (89) and (90) to the equality we obtain and thus, This completes the proof.

Theorem 20. Suppose that M and N are two maximal monotone operators on ℋ₁ and ℋ₂, respectively, and suppose that A : ℋ₁ → ℋ₂ is a bounded linear operator with adjoint A^*. Suppose further that {a_n}, {b_n}, {c_n}, and {d_n} are sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {α_n} and {β_n} be sequences in (0, ∞), {γ_n} a sequence in (0, 2/∥A∥²), and {e_n} a bounded sequence in ℋ₁. Assume that the solution set Ω of the problem

is nonempty. For any u ∈ ℋ₁, start with an arbitrary x₁ ∈ ℋ and define a sequence {x_n} by Then, the sequence {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0, $$, $$;

• (ii)

  $$, lim _n→∞inf b_nc_n > 0;

• (iii)

  lim _n→∞inf α_n > 0, lim _n→∞inf β_n > 0.

Proof. For any n ∈ ℕ, letting $$ and $$, then we have by Lemma 2(b) that M⁻¹0 = Fix (S_n) and N⁻¹0 = Fix (T_n) for all n ∈ ℕ, and hence, as shown in Section 1, the problem (94) becomes SFFP(16). Since all the requirements of Theorem 15 are satisfied except condition (iii), we have to check this condition. Because liminf _n→∞α_n = 0, liminf _n→∞β_n = 0 by assumption, we may assume that there is τ ∈ (0,1) so that τ < α_n and τ < β_n for all n ∈ ℕ. Let κ₁(n) = 2 + (α_n/τ) and κ₂(n) = 2 + (β_n/τ). Then, by virtue of the resolvent identity and the nonexpansiveness of $$, one has for all m ∈ ℕ that

and thus, The same argument shows for all m ∈ ℕ that Therefore, condition (iii) of Theorem 15 is true for $$ and $$, and then the desired conclusion follows from Theorem 15.

Theorem 21. Suppose that M and N are two maximal monotone operators on ℋ₁ and ℋ₂, respectively, and suppose that A : ℋ₁ → ℋ₂ is a bounded linear operator with adjoint A^*. Let {a_n} be a sequence in (0,1), γ ∈ (0, 2/∥A∥²), and α, β ∈ (0, ∞) and assume that the solution set Ω of problem (94) is nonempty. Start with any x₁ ∈ ℋ₁ and define a sequence {x_n} by

Then, the sequence {x_n} converges strongly to the minimum norm solution of problem (94) provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = 0;

• (ii)

  $$;

• (iii)

  either $$ or lim _n→∞(|a_n+1 − a_n|/a_n) = 0.

Theorem 22. Suppose that C and Q are two nonempty closed convex subsets of ℋ₁ and ℋ₂, respectively, and suppose that A : ℋ₁ → ℋ₂ is a bounded linear operator with adjoint A^*. Let f : C  ×  C → ℝ and g : Q  ×  Q → ℝ be functions satisfying conditions (A1)–(A4). Suppose further that {a_n}, {b_n}, {c_n}, and {d_n} are sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {α_n} and {β_n} be sequences in (0, ∞), {γ_n} a sequence in (0, 2/∥A∥²), and {e_n} a bounded sequence in ℋ₁. Assume that the solution set Ω of the problem

is nonempty. For any u ∈ ℋ₁, start with an arbitrary x₁ ∈ ℋ and define a sequence {x_n} by Then, the sequence {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0, $$, $$;

• (ii)

  $$, lim _n→∞inf b_nc_n > 0;

• (iii)

  lim _n→∞inf α_n > 0, lim _n→∞inf β_n > 0.

Proof. Define two set-valued mappings M_f and N_g on ℋ₁ and ℋ₂, respectively, by

As shown in Takahashi et al. [21], the set-valued mapping M_f (resp., N_g) is a maximal monotone operator with 𝒟(M_f)⊆C (resp., 𝒟(N_g)⊆Q), and $$ for any α > 0 (resp., $$ for any β > 0). With M = M_f and N = N_g in Theorem 20, the desired conclusion follows.

Corollary 23. Suppose that C is a nonempty closed convex subsets of ℋ and f : C × C → ℝ is a function satisfying conditions (A1)–(A4). Let {a_n}, {b_n}, {c_n}, and {d_n} be sequences in [0,1] with a_n + b_n + c_n + d_n = 1 and a_n ∈ (0,1) for all n ∈ ℕ. Let {α_n} and {β_n} be sequences in (0, ∞), and let {γ_n} be a sequence in (0, 2/∥A∥²) and suppose that {e_n} is a bounded sequence in ℋ₁. Assume that the solution set Ω of the problem

is nonempty. For any u ∈ ℋ₁, start with an arbitrary x₁ ∈ ℋ and define a sequence {x_n} by Then, the sequence {x_n} converges strongly to P_Ωu provided that the following conditions are satisfied:

• (i)

  lim _n→∞a_n = lim _n→∞(d_n/a_n) = 0, $$, $$;

• (ii)

  $$, lim _n→∞inf b_nc_n > 0;

• (iii)

  lim _n→∞inf α_n > 0, lim _n→∞inf β_n > 0.