An extragradient inertial algorithm for solving split fixed-point problems of demicontractive mappings, with equilibrium and variational inequality problems

Example 1.1

Let H be the Hilbert space L 2 ( [ 0 , 1 ] ) of square-integrable, measurable functions with the inner product ⟨ x , y ⟩ = ∫ 0 1 x ( t ) y ( t ) d t and the induced norm as follows:

Take two positive real numbers R and r such that R > r > k k + 1 R for some k > 1 . Consider the operator F : H → H defined by F ( x ) = ( R − ‖ x ‖ ) x for each x ∈ H , and set Γ = { x ∈ H : ‖ x ‖ ≤ r } . Since R > r > k k + 1 R and k > 1 , we obtain immediately that R k + 1 < r k < r . Thus, there exists x 0 ∈ Γ such that R k + 1 < ‖ x 0 ‖ < r k . Take y 0 = k x 0 . Then ‖ y 0 ‖ = k ‖ x 0 ‖ < k . r k = r , which implies that y 0 also belongs to Γ . By simple computation, we obtain

Thus, by R k + 1 < ‖ x 0 ‖ , we obtain ⟨ F ( x 0 ) − F ( y 0 ) , x 0 − y 0 ⟩ < 0 . This implies that the operator F is not monotone on Γ . Indeed, ⟨ F ( x ) , y − x ⟩ ≥ 0 for some x , y ∈ Γ , i.e. ⟨ ( R − ‖ x ‖ ) x , y − x ⟩ ≥ 0 . Then ⟨ x , y − x ⟩ ≥ 0 because R − ‖ x ‖ > r − ‖ x ‖ ≥ 0 . Thus,

because y ∈ Γ and R − ‖ y ‖ > r − ‖ y ‖ ≥ 0 . Hence, the operator F is pseudomonotone on Γ .

Theorem 1.2

[29] Let H 1 and H 2 be real Hilbert spaces, C be a closed convex subset H 1 , A : H 1 → H 2 be a bounded linear operator with its adjoint operator A ∗ . Let S : H 1 → H 1 and T : H 2 → H 2 k 2 be k 1 -demicontractive mappings such that S − I and T − I are demiclosed at zero, g be a bifunction of C × C into R satisfying the conditions ( A 1 ) – ( A 4 ) in Lemma 2.3. Suppose F ( S ) ∩ A − 1 F ( T ) ∩ E P ( g ) ≠ ∅ . Let { x n } be a sequence generated by

1. Choose x 0 , x 1 ∈ H 1 and n = 1 .

2. Given x n compute u n and y n using

   y n = ( 1 − β n ) ( x n + θ n ( x n − x n − 1 ) ) , u n = T σ n g ( y n + γ n A ∗ ( T − I ) A y n ) ,

   where { γ n } is bounded and γ n is chosen in such a way that { γ n } is bounded and

   γ n ∈ ϵ , ( 1 − k ) ‖ ( T − I ) A w n ‖ 2 ‖ A ∗ ( T − I ) A w n ‖ 2 − ϵ i f n ∈ Γ = { κ : ( T − I ) A w k ≠ 0 } , ∃ ϵ > 0 ,

   otherwise γ n = γ ( γ is any nonnegative real number).

3. Compute

   x n + 1 = ( 1 − α n ) u n + α n S u n .

Set n ≔ n + 1 and go to Step 2.

The control sequences { θ n } , { β n } , { α n } ⊂ ( 0 , 1 ) , and { σ n } are real sequences in [ σ , ∞ ) ⊂ ( 0 , ∞ ) satisfying the following conditions:

1. 0 < a ≤ α n b < 1 − k 2 , for all n ∈ N ;

2. lim n → ∞ β n = 0 and ∑ n = 1 ∞ β n = ∞ ;

3. lim n → ∞ θ n β n ‖ x n − x n − 1 ‖ = 0 .

Definition 2.1

Let C be a nonempty closed convex subset of of a real Hilbert space H . A mapping T : C → C is said to be

1. L -Lipschitz-continuous with L > 0 if

   ‖ T x − T y ‖ ≤ L ‖ x − y ‖ , ∀ x , y ∈ C ;

   if L = 1 , then the operator T is called nonexpansive, and if L ∈ ( 0 , 1 ) , then T is called contraction.

2. firmly nonexpansive if

   ‖ T x − T y ‖ 2 ≤ ‖ x − y ‖ 2 − ‖ ( I − T ) x − ( I − T ) y ‖ 2 , ∀ x , y ∈ C ;

3. k -demicontractive if F ( T ) ≠ ∅ and there exists k ∈ [ 0 , 1 ) such that

   ‖ T x − y ‖ 2 ≤ ‖ x − y ‖ 2 + k ‖ x − T x ‖ 2 , ∀ x ∈ C , y ∈ F ( T ) .

4. monotone if

   ⟨ T x − T y , x − y ⟩ ≥ 0 ∀ x , y ∈ C .

5. pseudomonotone if

   ⟨ T x , y − x ⟩ ≥ 0 ⇒ ⟨ T y , y − x ⟩ ≥ 0 ∀ x , y ∈ C .

6. α -strongly monotone if there exists a constant α > 0 such that

   ⟨ T x − T y , x − y ⟩ ≥ α ‖ x − y ‖ 2 ∀ x , y ∈ C .

Lemma 2.2

Let C be a nonempty, closed, and convex set in H and x ∈ H . Then

1. ⟨ P C x − x , y − P C x ⟩ ≥ 0 , ∀ y ∈ C .

2. ‖ y − P C x ‖ 2 ≤ ‖ x − y ‖ 2 − ‖ P C x − x ‖ 2 , ∀ y ∈ C .

Lemma 2.3

[25] Let C be a nonempty, closed, and convex subset of H and g be a bifunction of C × C into R satisfying the following:

1. g ( x , x ) = 0 for all x ∈ C ;

2. g is monotone, i.e. g ( x , y ) + g ( y , x ) ≤ 0 for all x , y ∈ C ;

3. for each x , y , z ∈ C ,

   limsup t ↓ 0 g ( t z + ( 1 − t ) x , y ) ≤ g ( x , y ) ;

4. g ( x , ⋅ ) is a convex and lower semicontinuous for all x ∈ C .

Lemma 2.4

[31] Let C be a nonempty, closed, and convex subset of H and g be a bifunction of C × C into R satisfying conditions ( A 1 ) – ( A 4 ) . For r > 0 , define a mapping T r g : X → C of g by

Then, the following hold:

1. T r g is single-valued;

2. T r g is firmly nonexpansive;

3. Fix ( T r g ) = E P ( g ) ;

4. E P ( g ) is closed and convex.

Lemma 2.5

[32] Let λ ∈ ( 0 , 1 ] , T : C → H be a nonexpansive mapping, and the mapping T l : C → H defined by T l x ≔ T x − l μ G ( T x ) ∀ x ∈ C , where G : H → H is κ -Lipschitzian and η -strongly monotone. Then T λ is a contraction provided 0 < μ < 2 η / κ 2 , i.e. ‖ T l x − T l y ‖ ≤ ( 1 − l τ ) ‖ x − y ‖ for all x , y ∈ C , where τ = 1 − 1 − μ ( 2 η − μ κ 2 ) ∈ ( 0 , 1 ] .

Lemma 2.6

[33] Let H 1 and H 2 be real Hilbert spaces and A : H 1 → H 2 be a bounded linear operator with adjoint operator A ∗ . If T : H 2 → H 2 is a κ -demicontractive mapping with κ < 1 and A − 1 Fix ( T ) ≠ ∅ , then

Lemma 2.7

[34] Let C be a nonempty, closed, and convex subset of a real Hilbert space H, and let F : C → H be pseudomonotone and continuous. Then, x ∗ is a solution of V I ( C , F ) if and only if

Lemma 2.8

[35] Assume { a n } is a sequence of nonnegative real numbers such that

where { γ n } is a sequence in ( 0 , 1 ) and { δ n } is a sequence in R such that

1. ∑ n = 0 ∞ γ n = ∞ ,

2. limsup n → ∞ δ n ≤ 0 .

3. Then lim n → ∞ a n = 0 .

Lemma 2.9

[36] Let { a n } be a sequence of real numbers such that there exists a subsequence { n j } of { n } such that a n j < a n j + 1 for all j ∈ N . Then, there exists a nondecreasing sequence { m k } ⊂ N such that m k → ∞ and the following properties are satisfied by all (sufficiently large) number k ∈ N :

In fact, m k = max { j ≤ k : a j < a j + 1 } .

Lemma 3.1

[29] Let H 1 and H 2 be real Hilbert spaces and C be nonempty, closed, and convex subset of H 1 . Let A : H 1 → H 2 be a bounded linear operator with its adjoint operator A ∗ , T : H 2 → H 2 be a k-demicontractive mapping, and g be a bifunction of C × C into R satisfying conditions ( A 1 ) – ( A 4 ) . Let { w n } be a sequence in H 1 and let

where { γ n } is a real sequence in ( 0 , ∞ ) and { σ n } is a real sequence in [ σ , ∞ ) for some σ > 0 . Then, we have

for all n ∈ N and all p ∈ E P ( g ) such that A p ∈ Fix ( T ) .

Proof

The proof is included for completeness. Let p ∈ E P ( g ) be such that A p ∈ Fix ( T ) . Since T σ n g ( p ) = p and T σ n g is firmly nonexpansive, we have

Since A is a bounded linear operator with its adjoint A ∗ and T is a k -demicontractive mapping, by Lemma 4(ii), we deduce that

From (12) and (13), we obtain

Assumption 3.2

Let H 1 and H 2 be real Hilbert spaces. Suppose the following conditions hold:

1. The set C is a nonempty, closed and convex subset of H 1 .

2. F : H 1 → H 1 is pseudomonotone, Lipschitz continuous on H 1 , and satisfy the following property:

   whenever { u n } ⊂ C , u n ⇀ z one has ‖ F z ‖ ≤ liminf ‖ F ( u n ) ‖ .

3. The solution set ϒ ≔ { z ∈ V I ( C , F ) ∩ E P ( g ) : A z ∈ F ( T ) } is nonempty.

4. T : H 2 → H 2 is k -demicontractive mapping such that T − I is demiclosed at zero and A : H 1 → H 2 be a bounded linear operator with its adjoint operator A ∗ .

5. We assume that f : C → C is a contractive mapping with coefficient δ ∈ [ 0 , 1 ) and g : C × C → R is a bifunction satisfying conditions ( A 1 ) – ( A 4 ) in Lemma 2.3.

6. ν δ < τ ≔ 1 − 1 − ρ ( 2 η − ρ κ 2 ) for ν ≥ 0 and ρ ∈ 0 , 2 η κ 2 .

Assumption 3.3

Suppose that { θ n } { l n } , { ϵ n } { α n } , and { σ n } are positive sequences satisfying the following conditions:

1. { α n } ∈ ( 0 , 1 ) with lim n → ∞ α n = 0 , ∑ n = 1 ∞ α n = ∞ .

2. lim n → ∞ lim ϵ n α n = 0 and sup n ≥ 1 ( θ n / α n ) < ∞ .

3. 0 < liminf n → ∞ l n ≤ liminf n → ∞ l n < 1 and α n + l n ≤ 1 ∀ ≥ 1 .

4. { σ n } is a real sequence in [ σ , ∞ ) ⊂ ( 0 , ∞ ) .

Algorithm 3.4

Remark 3.5

1. The stepsize given by (18) are generated at each iteration by some simple computations, which allows it to be easily implemented without the prior knowledge of the Lipschitz contact of the operator F .

2. Step 1 Algorithm 3.4 is also easily implemented since the value of ‖ x n − x n − 1 ‖ is a priori known before choosing α n .

3. It is obvious that { λ n } is a monotonically decreasing sequence since F is Lipschitz-continuous mapping with constant L > 0 , and the case of ⟨ F ( u n ) − F ( y n ) , z n − y n ⟩ ≥ 0 , we have

   μ ( ‖ u n − y n ‖ 2 + ‖ z n − y n ‖ 2 ) 2 ⟨ F ( u n ) − F ( y n ) , z n − y n ⟩ ≥ 2 μ ‖ u n − y n ‖ ‖ z n − y n ‖ 2 ‖ F ( u n ) − F ( y n ) ‖ ‖ z n − y n ‖ ≥ μ ‖ u n − y n ‖ L ‖ u n − y n ‖ = μ L .

   Clearly, the sequence { λ n } has the lower bound min { μ / L , λ n } .

4. The limit of { λ n } exists and we denote λ = lim n → ∞ λ n as it is obvious that λ > 0 .