Viscosity Approximation Methods for Equilibrium Problems, Variational Inequality Problems of Infinitely Strict Pseudocontractions in Hilbert Spaces

Lemma 2.1. Let H be a real Hilbert space. There hold the following identities:

• (i)

  ∥x − y∥² = ∥x∥² − ∥y∥² − 2〈x − y, y〉,   ∀ x, y ∈ H,

• (ii)

  ∥tx + (1 − t)y∥² = t∥x∥² + (1 − t)∥y∥² − t(1 − t)∥x − y∥²,   ∀ t ∈ [0,1],   ∀ x, y ∈ H.

Lemma 2.2 (see [18].)Assume that {α_n} is a sequence of nonnegative real numbers such that

where {ρ_n} is a sequence in (0,1) and {σ_n} is a sequence such that

• (i)

  $$,

• (ii)

  limsup _n→∞(σ_n/ρ_n) ≤ 0 or $$.

Then lim _n→∞a_n = 0.

Lemma 2.3 (see [13].)Let A : H → H be an L-Lipschitzian and η-strongly monotone operator on a Hilbert space H with L > 0,   η > 0,   0 < μ < 2η/L², and 0 < t < 1. Then S = (I − tμA) : H → H is a contraction with contractive coefficient 1 − tτ and τ = (1/2)μ(2η − μL²).

Lemma 2.4 (see [1].)Let S : C → C be a κ-strict pseudocontraction. Define T : C → C by Tx = λx + (1 − λ)Sx for each x ∈ C. Then, as λ ∈ [κ, 1), T is a nonexpansive mapping such that F(T) = F(S).

Lemma 2.5 (see [10].)Let H be a Hilbert space and f : H → H a contraction with coefficient 0 < α < 1, and A : H → H an L-Lipschitzian continuous operator and η-strongly monotone with L > 0,   η > 0. Then for 0 < γ < μη/α,

That is, μA − γf is strongly monotone with coefficient μη − γα.

Lemma 2.6 (see [8].)Let C be a nonempty closed convex subset of a strictly convex Banach space E, let $$ be nonexpansive mappings of C into itself such that $$, and let t₁, t₂, … be real numbers such that 0 < t_i ≤ b < 1, for every i = 1,2, …. Then, for any x ∈ C and k ∈ N, the limit lim _n→∞U_n,kx exists.

Lemma 2.7 (see [8].)Let C be a nonempty closed convex subset of a strictly convex Banach space E. Let $$ be nonexpansive mappings of C into itself such that $$, and let t₁, t₂, … be real numbers such that 0 < t_i ≤ b < 1, for all i ≥ 1. If K is any bounded subset of C, then

Lemma 2.8 (see [3].)Let C be a nonempty closed convex subset of a Hilbert space $$ be a family of infinite nonexpansive mappings with $$, and let t₁, t₂, … be real numbers such that 0 < t_i ≤ b < 1, for every i = 1,2, …. Then $$.

Lemma 2.9 (see [2].)Let C be a nonempty closed convex subset of H, let Φ be bifunction from C × C to ℝ satisfying (A1)–(A4), and let r > 0 and x ∈ H. Then there exists z ∈ C such that

Lemma 2.10 (see [4].)Let ϕ be a bifunction from C × C into ℝ satisfying (A1)–(A4). Then, for any r > 0 and x ∈ H, there exists z ∈ C such that

Further, if T_rx = {z ∈ C; Φ(z, y)+(1/r)〈y − z, z − x〉≥0, ∀y ∈ C}, then the following hold:

• (1)

  T_r is single-valued;

• (2)

  T_r is firmly nonexpansive;

• (3)

  F(T_r) = EP(ϕ);

• (4)

  EP(ϕ) is closed and convex.

Lemma 2.11 (see [9].)Let {x_n} and {z_n} be bounded sequences in a Banach space, and let {β_n} be a sequence of real numbers such that 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1 for all n = 0,1, 2, …. Suppose that x_n+1 = (1 − β_n)z_n   +   β_nx_n for all n = 0,1, 2, …. and lim  sup _n→∞∥z_n+1 − z_n∥−∥x_n+1 − x_n∥≤0. Then lim _n→∞∥z_n − x_n∥ = 0.

Lemma 2.12 (see [11].)Let C, H, F, and T_rx be as in Lemma 2.9. Then the following holds:

for all s, t > 0, and x ∈ H.

Lemma 2.13 (see [13].)Let H be a Hilbert space, and let C be a nonempty closed convex subset of H, and T : C → C a nonexpansive mapping with F(T) ≠ ∅. If {x_n} is a sequence in C weakly converging to x and if {(I − T)x_n} converges strongly to y, then (I − T)x = y.

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let ϕ be a bifunction from C × C → ℝ satisfying (A1)–(A4). Let S_i : C → C be a family κ_i-strict pseudocontractions for some 0 ≤ κ_i < 1. Assume the set $$. Let f be a contraction of H into itself with α ∈ (0,1), and let A be a strongly positive linear bounded operator on H with coefficient γ > 0 and $$. Let B : C → H be an ξ-inverse strongly monotone mapping. Let W_n be the mapping generated by $$ and t_i as in (2.5). Let {x_n} be a sequence generated by the following algorithm:

where {ε_n}, {β_n}, {α_n}, and {λ_n} are sequences in (0,1). Assume that the control sequences satisfy the following restrictions:

• (i)

  lim _n→∞ε_n = 0 and $$;

• (ii)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1;

• (iii)

  0 < lim _n→∞(λ_n/λ_n+1) = 1;

• (iv)

  lim _n→∞ | α_n+1 − α_n | = 0;

• (v)

  0 < μ_n ≤ 2ξ;

• (vi)

  lim _n→∞α_n = a.

Then {x_n} converges strongly to q ∈ Ω which is the unique solution of the variational inequality

or equivalent q = P_Ω(I − A + γf)(q), where P is a metric projection mapping form H onto Ω.

Proof. Since ε_n → 0, as n → ∞, we may assume, without loss of generality, that ε_n ≤ (1 − β_n)∥A∥⁻¹ for all n ∈ ℕ. By Lemma 2.3, we know that if 0 ≤ ρ ≤ ∥A∥⁻¹, then $$. We will assume that $$. Since A is a strongly positive bounded linear operator on H, we have

Observe that So this shows that (1 − β_n)I − ε_nA is positive. It follows that

Step 1. We claim that the mapping P_Ω(I − A + γf) where $$ has a unique fixed point. Let f be a contraction of H into itself with α ∈ (0,1). Then, we have

for all x, y ∈ H. Since $$, it follows that P_Ω(I − A + γf) is a contraction of H into itself. Therefore the Banach contraction mapping principle implies that there exists a unique element q ∈ H such that q = P_Ω(I − A + γf)(q).

Step 2. We shall show that (I − μ_nB) is nonexpansive. Let x, y ∈ C. Since B is ξ-inverse strongly monotone and λ_n < 2α for all n ∈ ℕ, we obtain

where μ_n ≤ 2ξ, for all n ∈ N. So we have that the mapping (I − λ_nA) is nonexpansive.

Step 3. We claim that {x_n} is bounded.

Let p ∈ Ω; from Lemma 2.10, we have

Note that It follows that By simple induction, we have Hence {x_n} is bounded. This implies that {K_n}, {f(x_n)} are also bounded.

Step 4. Show that lim sup _n→∞∥x_n+1 − x_n∥ = 0.

Observing that $$ and $$, we get

By Lemma 2.10, we obtain In particular, we have Summing up (3.14) and using (A₂), we obtain for all y ∈ C. It follows that This implies It follows that Hence, we obtain where $$. By (3.12) and (3.19), we obtain From (2.5), we have where M₁ = sup _n{∥U_n+1,n+1y_n − U_n,n+1y_n∥}.

Note that

where $$.

Suppose x_n+1 = β_nx_n + (1 − β_n)l_n, then l_n = (x_n+1 − β_nx_n)/(1 − β_n)  = (ε_nγf(x_n) + ((1 − β_n)I − ε_nF)K_n)/(1 − β_n).

Hence, we have

Then Combining with (i), (iii), and (iv), we have Hence, by Lemma 2.11, we obtain ∥l_n − x_n∥→0 as n → ∞. It follows that We also know that So

Step 5. We claim that ∥x_n − Wx_n∥→0.

Observe that

From (A1), (A3), and (3.28), using step 2, we have This implies that Next we want to show lim _n→∞∥x_n − y_n∥ = 0.

Let $$; we have

Therefore Note that From (3.34), we have It follows that From conditions (i), (vi) and (3.26), we have We also compute So, from (3.39), we get It follows that So On the other hand, we also know that and hence So Hence From (i), (3.42), and (3.26), we know that From (3.37) and (3.42), we can get On the other hand, we have By (3.30), (3.49), and using Lemma 2.7, we have

Step 6. We claim that limsup _n→∞〈(A − γf)q, q − x_n〉≤0, where q = P_Ω(I − A + γf)(q) is the unique solution of 〈(A − γf)q, x − q〉 ≥ 0, for all x ∈ Ω.

Indeed, take a subsequence $$ of {x_n} such that

Since $$ is bounded, there exists a subsequence $$ of $$, which converges weakly to p; without loss of generality, we can assume $$ and $$, we arrive at

Step 7. We show that x_n → q.

Since

so Note that which implies that Let then we have Applying Lemma 2.2, we can conclude that {x_n} converges strongly to q in norm. This completes the proof.

Corollary 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from C × C → ℝ satisfying (A1)–(A4). Let S_i : C → C be a family κ_i-strict pseudocontractions for some 0 ≤ κ_i < 1. Assume the set $$. Let f be a contraction of H into itself with α ∈ (0,1) and let A be an α-inverse strongly monotone mapping. Let F be a strongly positive linear-bounded operator on H with coefficient γ > 0 and $$ and τ < 1. Let W_n be the mapping generated by $$ and t_i, where S_i : C → C is a nonexpansive mapping with a fixed point. Let {x_n} and {u_n} be sequences generated by the following algorithm:

where{ε_n}, {β_n}, {α_n}, and {λ_n} are sequences in (0,1). Assume that the control sequences satisfy the following restrictions:

• (i)

  lim _n→∞ε_n = 0 and $$;

• (ii)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1;

• (iii)

  0 < liminf _n→∞λ_n ≤ limsup _n→∞λ_n < 1;

• (iv)

  lim _n→∞ | λ_n+1 − λ_n | = lim _n→∞ | α_n+1 − α_n | = 0;

• (v)

  0 < t_n ≤ b < 1;

• (vi)

  λ_n < 2α.

Then {x_n} converges strongly to w ∈ Ω where w = P_Ω(I − A + γf)w.