Viscosity Approximation Methods for Two Accretive Operators in Banach Spaces

Lemma 1 (see [10].)In a Banach space E, the following inequality holds:

where j(x + y) ∈ J(x + y).

Lemma 2 (see [10], [13].)Let {α_n} be a sequence of nonnegative real numbers satisfying the condition

where {γ_n}⊂(0,1) and {σ_n} such that

• (i)

  lim _n→∞γ_n = 0 and $$;

• (ii)

  either limsup _n→∞σ_n ≤ 0 or $$.

Then lim _n→∞α_n = 0.

Lemma 3 (the resolvent identity [10]). For λ > 0, μ > 0 and x ∈ E,

Lemma 4 (see [3], Theorem 4.1, page 287.)Let E be a uniformly smooth Banach space, C be a closed convex subset of E, T : C → C a nonexpansive mapping with F(T) ≠ Ø, and f ∈ Σ_C. Then {z_t} defined by the following

converges strongly to a point in Fix (T). If, moreover, one defines Q : Σ_C → F(T) by then Q(f) solves the variational inequality

Lemma 5 (see [17], Theorem 2.)Let (X, d) be a complete metric space and g a weakly contractive mapping on X. Then g has a unique fixed point p in X.

Lemma 6. Let {s_n} and {γ_n} be two sequences of nonnegative real numbers and {λ_n} a sequence of positive numbers satisfying the conditions

• (i)

  $$,

• (ii)

  lim _n→∞(γ_n/λ_n) = 0.

Let the recursive inequality

be given where ψ(t) is a continuous and strict increasing function on [0, +∞) with ψ(0) = 0. Then lim _n→∞s_n = 0.

Theorem 7. Let E be a strictly convex Banach space which has a uniformly Gâteaux differentiable norm, A, B two m-accretive maps in E such that $$ is convex and A⁻¹0∩B⁻¹0 ≠ Ø, and f : C → C a fixed contraction mapping with contract constant α. Suppose that α_n ⊂ (0,1), β_n ⊂ (0,1), and r_n > 0 satisfy the following conditions:

• (i)

  $$, α_n → 0, as n → ∞;

• (ii)

  β_n → 0, and r_n → r > ε > 0 as n → ∞;

• (iii)

  $$, $$, and $$.

Let {x_n} be the composite viscosity process defined by

Then {x_n} converges strongly to p ∈ A⁻¹0∩B⁻¹0, where p is the unique solution of the following variational inequality:

Proof. First, by using Lemma 4, we know that there exists the unique solution p of a variational inequality

where p = lim _t→0 z_t and z_t is defined by z_t = tf(z_t)+(1 − t)S_r(z_t) for each r > 0 and 0 < t < 1.

Next, we will divide our discussion into the following steps.

Step  1. We will show that {x_n} is bounded.

In fact, take p ∈ A⁻¹0∩B⁻¹0. Then

Therefore, Using the induction method, we have which implies that {x_n}, {f(x_n)}, {y_n}, and {f(y_n)} are all bounded. Since $$, then $$ is bounded. Following the conditions of (i) and (ii), we obtain that

Step  2. We show that ∥x_n+1 − x_n∥ → 0.

For this, we estimate y_n+1 − y_n first. From (16), we know that

Then simple calculations show that It follows from (25) that In view of Lemma 3, we have If r_n−1 ≤ r_n, then Similarly, Thus, let $$; we have Substituting (30) into (26) we get where M₁ is a constant such that On the other hand, we have Simple calculations show that It follows that Substituting (31) into (35) we get where M₂ is a constant such that From conditions (i)–(iii), we have that Hence, noticing (36) and applying Lemma 2, we obtain ∥x_n+1 − x_n∥ → 0. Then by (22) we obtain

Step  3. We prove that ∥x_n − S_rx_n∥ → 0, ∥y_n − S_ry_n∥ → 0.

In fact, since

from (22) and (23), we have $$. Then Hence, we have

Step  4. We show that limsup _n→∞〈(I − f)p, J(y_n − p)〉≤0, limsup _n→∞〈(I − f)p, J(x_n+1 − p)〉≤0.

To prove this, let $$ be a subsequence of {y_n} such that

By Lemma 4, $$, where z_t = tf(z_t)+(1 − t)S_r(z_t). Then

For each integer n ≥ 0, let t_n ∈ (0,1) such that Using Lemma 1, we get and hence Since {x_n}, $$, and {S_ry_n} are bounded, then $$. Therefore, We also know that Notice that $$, p ∈ F(S_r), n → ∞, and J is norm to weak* uniformly continuous on bounded subset of E; then we obtain From (48), (49), and the two results mentioned above, we have Using (22) and the property of J, we obtain the result that

Step  5. lim _n→∞∥x_n − p∥ = 0.

Using (16), we have

Applying Lemma 1, we obtain where L = sup _n≥0{∥x_n − p∥, ∥y_n − p∥, ∥f(p) − p∥}. Put From the conditions (i) and (ii), the result of Step 4, and the facts that α_nβ_n/(α_n + β_n) → 0 and 〈f(p) − p, J(y_n − p) − J(x_n+1 − p)〉→0, we know that γ_n → 0, $$ and limsup _n→∞σ_n ≤ 0. In view of Lemma 2, (54) reduces to then we know that This completes the proof.

Remark 8. If we modify (16) as follows:

or Then, imitating the proof of Theorem 7, we can also get the result of Theorem 7. Therefore, from the compare of iterative scheme, the conclusions of [10, 11] are special cases of Theorem 7.

Example 9. Next we study the following optimization problem:

where C is an interior nonempty closed convex subset of a Hilbert space and h, k : C → R are two proper convex and lower semicontinuous functionals. To solve optimization problem (61), we will list the following well known results.

Proposition 10 (see [18].)Let φ : C → R be a proper convex and lower semicontinuous functional. Then

• (i)

  ∂φ : C → E^* ( ∂ denotes the subdifferential in the sense of convex analysis) is a maximal monotone mapping;

• (ii)

  ∂φ(x₀) = min _x∈Cφ(x) if and only if 0 ∈ ∂φ(x₀).

Theorem 11. Let A, B be two accretive maps in E with A⁻¹0∩B⁻¹0 ≠ Ø and satisfying the following range conditions: $$, $$ which are convex. Let f, {α_n}, {β_n}, {r_n}, and {x_n}, {y_n} be the same as those in Theorem 7. Let {x_n} be a sequence generated by (16). Then {x_n} converges strongly to p ∈ A⁻¹0∩B⁻¹0, where p is the unique solution of the following variational inequality:

Theorem 12. Let A, B be two accretive maps in E with F = A⁻¹0∩B⁻¹0 ≠ Ø and satisfying the following range conditions: $$, $$ which are convex. Let f, {α_n}, {β_n}, and {r_n} be the same as those in Theorem 7. Let g : C → C be a weakly contractive mapping with the function ψ. Let {x_n} be a sequence generated by

Then {x_n} converges strongly to p = Q(g(p)) ∈ A⁻¹0∩B⁻¹0, where Q is a sunny nonexpansive retraction from C onto F.

Proof. Since E is a uniformly smooth Banach space, then there is a sunny nonexpansive retraction Q from C onto F. Then Q∘g is a weakly contractive mapping of C into itself. Indeed, for all x, y ∈ C,

Lemma 5 assures that there exists a unique element p ∈ C such that p = Q(g(p)). Such p ∈ C is an element of A⁻¹0∩B⁻¹0. Now we define a iterative scheme as follows: Let w_n be the sequence generated by (66). Then Remark 8 (59) assures that w_n converges strongly to p = Q(g(p)) as n → ∞. For any n, we have Thus, we obtain for s_n = ∥x_n − w_n∥ the following recursive inequality: Since ∥w_n − p∥ → 0, it follows from Lemma 6 that lim _n→∞∥x_n − w_n∥ = 0. Hence This completes the proof.

Corollary 13. Let A, B be two accretive maps in E with F = A⁻¹0∩B⁻¹0 ≠ Ø and satisfying the following range conditions: $$, $$ which are convex. Let f, g ∈ Σ_C, {α_n}, {β_n}, and {r_n} be the same as those in Theorem 7. Let {x_n} be a sequence generated by

Then {x_n} converges strongly to p = Q(g(p)) ∈ A⁻¹0∩B⁻¹0, where Q is a sunny nonexpansive retraction from C onto F.

Example 14. Next we give an essential example.

Let Ω be an bounded domain in a Euclidean space R^N with Lipschitz boundary Γ. Let ϕ : Γ × R → R be a given function shch that for each x ∈ Γ

• (i)

  ϕ_x = ϕ(x, ·) : R → R is a proper, convex, lower semicontinuous function with ϕ_x(0) = 0.

• (ii)

  β_x = ∂ϕ_x (subdifferential of ϕ_x) is the maximal monotone mapping on R with 0 ∈ β_x(0) and for each t ∈ R, the function x ∈ Γ → (I + λβ_x) ⁻¹(t) ∈ R is measurable for λ > 0.

Let α : R^N → R^N be a continuous, monotone function such that there exist constants k₁, k₂ satisfying (i) |α(ξ)| ≤ k₁|ξ| and (ii) 〈α(ξ), ξ〉≥k₂|ξ|² for each ξ ∈ R^N.

Definition 15 (see [19].)One first defines a mapping A^α : H¹(Ω)→(H¹(Ω)) ^* (H¹(Ω) is a sobolev space) by

for u, v ∈ H¹(Ω). Clearly A^α is an everywhere defined, monotone, demicontinuous operator from H¹(Ω) into (H¹(Ω)) ^*. Second one defines an operator $$ for 1 < p < +∞ as follows.

(i) For p ≥ 2 one defines the domain of $$ by

Here Φ(u) = ∫_Γ ϕ_x(u|_Γ(x))dΓ(x) is the proper, convex, l.s.c. function (see [19, Lemma 3.1]). For $$ we set

(ii) For 1 < p < 2, one defines $$ as the L^p-closure of $$ defined in (i) above.

Lemma 16 (see [19], Lemma 3.4.)$$ is m-accretive operator (1 < p < +∞).

Lemma 17 (see [19], Proposition 3.2.)Let f ∈ L^p(Ω), u ∈ L^p(Ω) such that $$. Then

• (i) 

  div (α(grad u)) = f, a.e. on Ω and

• (ii) 

  〈n, α(grad u)〉∈β_x(u(x)) for a.e. x ∈ Γ.

Lemma 18 (see [19], Proposition 3.3.)Let β_x ≡ 0 for x ∈ Γ. Then $$.

Clearly for different α₁, α₂, $$ are two m-accretive operators. The above results show that

For the sake of finding a common zero, Theorems 7, 11, and 12 provided three different iterative algorithms. Therefore the study of a common zero of two accretive operators makes sense.

Remark 19. The results presented in this paper substantially improve and extend the results of Ceng et al. [10] from the following aspects.

• (1)

  Theorems 7 and 12 extend the result on the iterative construction of the zero for a single accretive operator to the case of that for common zeros of two accretive operators. If we modify two accretive operators as finite accretive operators, then, imitating the proof of Theorem 7, we can also get the result of Theorem 7.

• (2)

  Our results include one or two different viscosity items. Remark 8 shows that the conclusions of Ceng et al. are special cases of this paper.

• (3)

  The viscosity item is changed from a contractive mapping f to weakly contractive mapping g in Theorem 12.