Implicit Relaxed and Hybrid Methods with Regularization for Minimization Problems and Asymptotically Strict Pseudocontractive Mappings in the Intermediate Sense

Definition 1. Let C be a nonempty subset of a normed space X. A mapping S : C → C is said to be asymptotically nonexpansive in the intermediate sense provided S is uniformly continuous and

Definition 2. Let C be a nonempty subset of a Hilbert space H. A mapping S : C → C is said to be an asymptotically κ-strict pseudocontractive mapping with sequence {γ_n} if there exists a constant κ ∈ [0,1) and a sequence {γ_n} in [0, ∞) with lim _n→∞ γ_n = 0 such that

Definition 3. Let C be a nonempty subset of a Hilbert space H. A mapping S : C → C is said to be an asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n} if there exist a constant κ ∈ [0,1) and a sequence {γ_n} in [0, ∞) with lim _n→∞ γ_n = 0 such that

Proposition 4. For given x ∈ H and z ∈ C,

• (i)

  z = P_Cx⇔〈x − z, y − z〉 ≤ 0,  for all  y ∈ C;

• (ii)

  z = P_Cx⇔ ∥x−z∥² ≤ ∥x−y∥² − ∥y−z∥²,  for all  y ∈ C;

• (iii)

  $$,  for all  y ∈ H.

Definition 5. A mapping S : C → H is said to be

• (a)

  monotone if

  $$ ()

• (b)

  η-strongly monotone if there exists a constant η > 0 such that

  $$ ()

• (c)

  α-inverse-strongly monotone (α-ism) if there exists a constant α > 0 such that

  $$ ()

Lemma 6. Let X be a real inner product space. Then,

Lemma 7 (see [7], Proposition 2.4.)Let {x_n} be a bounded sequence in a reflexive Banach space X. If ω_w({x_n}) = {x}, then x_n⇀x.

Lemma 8. Let A : C → H be a monotone mapping. Then,

Lemma 9. The following assertions hold:

• (a)

  ∥x−y∥² = ∥x∥² − ∥y∥² − 2〈x − y, y〉  for all x, y ∈ H;

• (b)

   ∥λx+μy+νz∥² = λ∥x∥² + μ∥y∥² + ν∥z∥² − λμ∥x−y∥² − μν∥y−z∥² − λ∥x−z∥²  for all x, y, z ∈ H and λ, μ, ν ∈ [0,1] with λ + μ + ν = 1 [14];

• (c)

  if {x_n} is a sequence in H such that x_n⇀x, then

  $$ ()

Lemma 10 (see [4], Lemma 2.5.)For given points x, y, z ∈ H and given also a real number a ∈ R, the set

is convex (and closed).

Lemma 11 (see [4], Lemma 2.6.)Let C be a nonempty subset of a Hilbert space H and S : C → C an asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n}. Then,

for all x, y ∈ C and n ≥ 1.

Lemma 12 (see [4], Lemma 2.7.)Let C be a nonempty subset of a Hilbert space H and S : C → C a uniformly continuous asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n}. Let {x_n} be a sequence in C such that $$ and $$ as n → ∞. Then, $$ as n → ∞.

Lemma 13 (Demiclosedness Principle, [see [4, Proposition 3.1]]). Let C be a nonempty closed convex subset of a Hilbert space H and S : C → C be a continuous asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n}. Then I − S is demiclosed at zero in the sense that if {x_n} is a sequence in C such that x_n⇀x ∈ C and $$, then (I − S)x = 0.

Lemma 14 (see [4], Proposition 3.2.)Let C be a nonempty closed convex subset of a Hilbert space H and S : C → C a continuous asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n} such that Fix(S) ≠ ∅. Then, Fix(S) is closed and convex.

Lemma 15 (see [15], page 80.)Let $$, $$, and $$ be sequences of nonnegative real numbers satisfying the inequality

If $$ and $$, then lim _n→∞ a_n exists. If, in addition, $$ has a subsequence which converges to zero, then lim _n→∞ a_n = 0.

Corollary 16 (see [16], page 303.)Let $$ and $$ be two sequences of nonnegative real numbers satisfying the inequality

If $$ converges, then lim _n→∞ a_n exists.

Lemma 17 (see [17].)Every Hilbert space H has the Kadec-Klee property; that is, given a sequence {x_n} ⊂ H and a point x ∈ H, we have

Theorem 18. Let C be a nonempty bounded closed convex subset of a real Hilbert space H. Let f : C → R be a convex functional with L-Lipschitz continuous gradient ∇f, Q : C → C a ρ-contraction with coefficient ρ ∈ [0,1), and S : C → C a uniformly continuous asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n} such that Fix(S)∩Γ is nonempty. Let {γ_n} and {c_n} be defined as in Definition 3. Let {x_n} and {y_n} be the sequences generated by

where {α_n} is a sequence (0, ∞), {λ_n}, {μ_n} are sequences in (0,1], and {σ_n}, {β_n} are sequences in [0,1] satisfying the following conditions:

• (i)

  δ ≤ β_n and σ_n + β_n ≤ 1 − κ − τ, for all n ≥ 1 for some δ, τ ∈ (0,1);

• (ii)

  $$, $$, $$ and $$;

• (iii)

  lim _n→∞ μ_n = 1;

• (iv)

  λ_n(α_n + L) < 1, for all n ≥ 1 and {λ_n}⊂[a, b] for some a, b ∈ (0, (1/L)).

Then, the sequences {x_n}, {y_n} converge weakly to some point u ∈ Fix (S)∩Γ.

Remark 19. Observe that for all x, y ∈ C and all n ≥ 1

By Banach contraction principle, we know that for each n ≥ 1, there exists a unique y_n ∈ C such that

Also, observe that for all x, y ∈ C and all n ≥ 1

Utilizing Banach contraction principle, for each n ≥ 1, there exists a unique t_n ∈ C such that

Proof of Theorem 18. Note that the L-Lipschitz continuity of the gradient ∇f implies that ∇f is (1/L)-ism [20], that is,

Thus, ∇f is monotone and L-Lipschitz continuous.

We divide the rest of the proof into several steps.

Step 1. lim _n→∞∥x_n − u∥ exists for each u ∈ Fix (S)∩Γ.

Indeed, note that $$ for all n ≥ 1. Take u ∈ Fix (S)∩Γ arbitrarily. From Proposition 4(ii), monotonicity of ∇f, and u ∈ Γ, we have

Since $$ and $$ is (α_n + L)-Lipschitz continuous, by Proposition 4(i), we have So, we have Therefore, from (33), x_n+1 = (1 − σ_n − β_n)t_n + σ_nf(t_n) + β_nSⁿt_n and u = Su. By Lemma 9(b), we have Since $$, $$, $$, and $$, from the boundedness of C, it follows that By Lemma 15, we have

Step 2. lim _n→∞∥x_n − y_n∥ = 0 and lim _n→∞∥x_n − t_n∥ = 0.

Indeed, substituting (33) in (34), we get

which hence implies that Since lim _n→∞∥x_n − u∥ exists, α_n → 0, γ_n → 0, σ_n → 0, and c_n → 0, from the boundedness of C, it follows that Meantime, utilizing the arguments similar to those in (33), we have Substituting (40) in (34), we get which hence implies that Since lim _n→∞∥x_n − u∥ exists, α_n → 0, γ_n → 0, σ_n → 0, and c_n → 0, from the boundedness of C it follows that This together with ∥x_n − y_n∥ → 0 implies that

Step 3.  lim _n→∞∥x_n+1 − x_n∥  = 0 and lim _n→∞∥x_n − Sx_n∥  = 0.

Indeed, from (33) and (34), we conclude that

which together with condition (i) yields Since lim _n→∞∥x_n − u∥ exists, α_n → 0, γ_n → 0, σ_n → 0, and c_n → 0, from the boundedness of C it follows that Note that From the boundedness of C, σ_n → 0, ∥t_n − x_n∥ → 0, and ∥Sⁿt_n − t_n∥ → 0, we deduce that Furthermore, observe that Utilizing Lemma 11, we have for every n = 1,2, …. Hence, it follows from ∥x_n − t_n∥ → 0 that ∥Sⁿt_n − Sⁿx_n∥ → 0. Thus, from (50) and ∥t_n − Sⁿt_n∥ → 0, we get ∥x_n − Sⁿx_n∥ → 0. Since ∥x_n+1 − x_n∥ → 0, ∥x_n − Sⁿx_n∥ → 0 as n → ∞, and S is uniformly continuous, we obtain from Lemma 12 that ∥x_n − Sx_n∥ → 0 as n → ∞.

Step 4. ω_w(x_n) ⊂ Fix (S)∩Γ.

Indeed, from the boundedness of {x_n}, we know that ω_w({x_n}) ≠ ∅. Take $$ arbitrarily. Then, there exists a subsequence $$ of {x_n} such that $$ converges weakly to $$. Note that S is uniformly continuous and ∥x_n − Sx_n∥ → 0. Hence, it is easy to see that ∥x_n − S^mx_n∥ → 0 for all m ≥ 1. By Lemma 13, we obtain $$. Furthermore, we show $$. Since x_n − t_n → 0 and x_n − y_n → 0, we have $$ and $$. Let

where N_Cv is the normal cone to C at v ∈ C. We have already mentioned that in this case the mapping T is maximal monotone, and 0 ∈ Tv if and only if v ∈ VI (C, ∇f); see [19] for more details. Let G(T) be the graph of T and let (v, w) ∈ G(T). Then, we have w ∈ Tv = ∇f(v) + N_Cv and hence w − ∇f(v) ∈ N_Cv. So, we have 〈v − t, w − ∇f(v)〉≥0 for all t ∈ C. On the other hand, from $$ and v ∈ C, we have and hence Therefore, from 〈v − t, w − ∇f(v)〉≥0 for all t ∈ C and $$, we have Since ∥∇f(t_n)−∇f(y_n)∥ → 0 (due to the Lipschitz continuity of ∇f), (t_n − x_n)/(λ_n) → 0 (due to {λ_n}⊂[a, b]), α_n → 0, and μ_n → 1, we obtain $$ as i → ∞. Since T is maximal monotone, we have $$ and hence $$. Clearly, $$. Consequently, $$. This implies that ω_w({x_n}) ⊂ Fix (S)∩Γ.

Step 5. {x_n} and {y_n} converge weakly to the same point u ∈ Fix (S)∩Γ.

Indeed, it is sufficient to show that ω_w(x_n) is a single-point set because x_n − y_n → 0 as n → ∞. Since ω_w(x_n) ≠ ∅, let us take two points $$ arbitrarily. Then, there exist two subsequences $$ and $$ of {x_n} such that $$ and $$, respectively. In terms of Step 4, we know that $$. Meantime, according to Step 1, we also know that there exist both lim _n→∞∥x_n − u∥ and $$. Let us show that $$. Assume that $$. From the Opial condition [18], it follows that

This leads to a contradiction. Thus, we must have $$. This implies that ω_w(x_n) is a single-point set. Without loss of generality, we may write ω_w(x_n) = {u}. Consequently, by Lemma 7, we obtain that x_n⇀u ∈ Fix (S)∩Γ. Since x_n − y_n → 0 as n → ∞, we also have This completes the proof.

Theorem 20. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → R be a convex functional with L-Lipschitz continuous gradient ∇f, Q : C → C a ρ-contraction with coefficient ρ ∈ [0,1), and S : C → C a uniformly continuous asymptotically κ-strict pseudocontractive mapping in the intermediate sense with sequence {γ_n} such that Fix(S)∩Γ is nonempty and bounded. Let {γ_n} and {c_n} be defined as in Definition 3. Let {x_n}, {y_n}, and {z_n} be the sequences generated by

where {α_n}⊂(0, ∞), {λ_n}, {μ_n}⊂(0,1], {σ_n}, {β_n}⊂[0,1], θ_n = c_n + (α_n + γ_n + σ_n)Δ_n and

Assume that the following conditions hold:

• (i)

  δ ≤ β_n and σ_n + β_n ≤ 1 − κ, for all n ≥ 1 for some δ ∈ (0,1);

• (ii)

  lim _n→∞ α_n = 0 and lim _n→∞ σ_n = 0;

• (iii)

  lim _n→∞ μ_n = 1;

• (iv)

  λ_n(α_n + L) < 1, for all n ≥ 1 and {λ_n}⊂[a, b] for some a, b ∈ (0, (1/L)).

Then, the sequences {x_n}, {y_n}, and {z_n} converge strongly to the same point P_{Fix (S)∩Γ}x.

Proof. Utilizing the condition λ_n(α_n + L) < 1 for all n ≥ 1 and repeating the same arguments as in Remark 19, we can see that {y_n} and {z_n} are defined well. Note that the L-Lipschitz continuity of the gradient ∇f implies that ∇f is (1/L)-ism [20], that is,

Observe that Hence, it follows that ∇f_α = αI + ∇f is (1/(α + L))-ism. It is clear that C_n is closed and Q_n is closed and convex for every n = 1,2, …. As the defining inequality in C_n is equivalent to the inequality By Lemma 10, we also have that C_n is convex for every n = 1,2, …. As Q_n = {z ∈ C : 〈x_n − z, x − x_n〉≥0}, we have 〈x_n − z, x − x_n〉≥0 for all z ∈ Q_n, and by Proposition 4(i), we get $$.

We divide the rest of the proof into several steps.

Step 1. {x_n}, {y_n}, and {z_n} are bounded.

Indeed, take u ∈ Fix (S)∩Γ arbitrarily. Taking into account λ_n(α_n + L) < 1, for all n ≥ 1, we deduce that

which implies that Meantime, we also have which hence implies that Thus, from (64) and (66), it follows that which hence implies that Repeating the same arguments as in (33) and (34), we can obtain that for every n = 1,2, … and hence u ∈ C_n. So, Fix (S)∩Γ ⊂ C_n for every n = 1,2, …. Next, let us show by mathematical induction that {x_n} is well-defined and Fix (S)∩Γ ⊂ C_n∩Q_n for every n = 1,2, …. For n = 1, we have Q₁ = C. Hence, we obtain Fix (S)∩Γ ⊂ C₁∩Q₁. Suppose that x_k is given and Fix (S)∩Γ ⊂ C_k∩Q_k for some integer k ≥ 1. Since Fix (S)∩Γ is nonempty, C_k∩Q_k is a nonempty closed convex subset of C. So, there exists a unique element x_k+1 ∈ C_k∩Q_k such that $$. It is also obvious that there holds 〈x_k+1 −  z, x  − x_k+1〉≥0 for every z ∈ C_k∩Q_k. Since Fix (S)∩Γ ⊂ C_k∩Q_k, we have 〈x_k+1 − z, x − x_k+1〉≥0 for every z ∈ Fix (S)∩Γ and hence Fix (S)∩Γ ⊂ Q_k+1. Therefore, we obtain Fix (S)∩Γ ⊂ C_k+1∩Q_k+1.

Step 2. lim _n→∞∥x_n − x_n+1∥ = 0 and lim _n→∞∥x_n − z_n∥ = 0.

Indeed, let q = P_{Fix (S)∩Γ}x. From $$ and q ∈ Fix (S)∩Γ ⊂ C_n∩Q_n, we have

for every n = 1,2, …. Therefore, {x_n} is bounded. From (64), (66), and (70), we also obtain that {y_n}, {t_n}, and {z_n} are bounded. Since x_n+1 ∈ C_n∩Q_n ⊂ Q_n and $$, we have for every n = 1,2, …. Therefore, there exists lim _n→∞∥x_n − x∥. Since $$ and x_n+1 ∈ Q_n, using Proposition 4(ii), we have for every n = 1,2, …. This implies that Since x_n+1 ∈ C_n, we have which implies that Hence, we get for every n = 1,2, …. From ∥x_n+1 − x_n∥ → 0 and θ_n → 0, we conclude that ∥x_n − z_n∥ → 0 as n → ∞.

Step 3. lim _n→∞∥x_n − y_n∥ = 0, lim _n→∞∥x_n − t_n∥ = 0 and lim _n→∞∥x_n − Sx_n∥ = 0.

Indeed, substituting (69) in (70), we get

which hence implies that Since ∥x_n − z_n∥ → 0, α_n → 0, γ_n → 0, σ_n → 0, and c_n → 0, from the boundedness of {x_n}, {y_n}, {t_n}, and {z_n}, it follows that Meantime, utilizing the arguments similar to those in (69), we have Substituting (81) in (70), we get which hence implies that Since ∥x_n − z_n∥ → 0, α_n → 0, γ_n → 0, σ_n → 0, and c_n → 0, from the boundedness of {x_n}, {y_n}, {t_n}, and {z_n}, it follows that This together with ∥x_n − y_n∥ → 0 implies that

Since z_n = (1 − σ_n − β_n)t_n + σ_nQt_n + β_nSⁿt_n, we have β_n(Sⁿt_n − t_n) = (z_n − t_n) − σ_n(Qt_n − t_n). Then,

and hence ∥t_n − Sⁿt_n∥ → 0. Furthermore, observe that Utilizing Lemma 11, we have for every n = 1,2, …. Hence, it follows from ∥x_n − t_n∥ → 0 that ∥Sⁿt_n − Sⁿx_n∥ → 0. Thus, from (87) and ∥t_n − Sⁿt_n∥ → 0, we get ∥x_n − Sⁿx_n∥ → 0. Since ∥x_n+1 − x_n∥ → 0, ∥x_n − Sⁿx_n∥ → 0 as n → ∞, and S is uniformly continuous, we obtain from Lemma 12 that ∥x_n − Sx_n∥ → 0 as n → ∞.

Step 4. ω_w(x_n) ⊂ Fix (S)∩Γ.

Indeed, repeating the same arguments as in Step 4 of the proof of Theorem 18, we can derive the desired conclusion.

Step 5. {x_n}, {y_n}, and {z_n} converge strongly to q = P_{Fix (S)∩Γ}x.

Indeed, take $$ arbitrarily. Then, $$ according to Step 4. Moreover, there exists a subsequence $$ of {x_n} such that $$. Hence, from q = P_{Fix (S)∩Γ}x, $$, and (71), we have

So, we obtain From $$, we have $$ due to the Kadec-Klee property of Hilbert spaces [17]. So, it is clear that $$. Since $$ and q ∈ Fix (S)∩Γ ⊂ C_n∩Q_n ⊂ Q_n, we have As i → ∞, we obtain $$ by q = P_{Fix (S)∩Γ}x and $$. Hence, we have $$. This implies that x_n → q. It is easy to see that y_n → q and z_n → q. This completes the proof.