New iterative schemes for solving variational inequality and fixed points problems involving demicontractive and quasi-nonexpansive mappings in Banach spaces

From the definition of A, we note that strongly positive bounded linear operator A is a ∥A∥-Lipschitzian and k-strongly monotone operator.

Let H = ℝ and K = [–1, 1]. Define T : K → K by

Tx=23xsin⁡(1x),x≠00x=0. $$\begin{array}{} \displaystyle Tx= \left \{ \begin{array}{ll} \dfrac{2}{3}x\sin(\frac{1}{x}),\,\,\, x\neq 0\\\\ 0\,\,\,\,\,\,\, x=0. \end{array} \right. \end{array}$$(3)

Clearly Fix(T) = {0}. For x ∈ K, we have

|Tx−0|2=|23xsin⁡(1x)|2≤|23x|2≤|x|2≤|x−0|2+k|x−Tx|2∀k∈[0,1). $$\begin{array}{} \begin{split} \displaystyle \vert Tx-0 \vert^2 &=& \vert \dfrac{2}{3}x\sin(\frac{1}{x}) \vert^2\\ &\leq & \vert \dfrac{2}{3}x \vert^2\\ &\leq & \vert x \vert^2\\ &\leq & \vert x-0 \vert^2+ k \vert x-Tx\vert^2\,\,\, \forall k\in [0, 1). \end{split} \end{array}$$

Thus T is k demi-contratcive for k ∈ [0, 1). To see that T is not k strictly pseudo-contractive, choose x = 2π $\begin{array}{} \dfrac{2}{\pi} \end{array}$ and y = 23π $\begin{array}{} \dfrac{2}{3\pi} \end{array}$, then

|Tx−Ty|2>|x−y|2+k|x−y−(Tx−Ty)|2. $$\begin{array}{} \displaystyle \vert Tx-Ty \vert^2 \gt \vert x-y \vert^2+k \vert x-y-(Tx-Ty)\vert^2. \end{array}$$

Hence, T is not k strictly pseudo-contractive mapping for k ∈ [0, 1).

(Example of a Demicontractive Function which is not Quasi-nonexpansive and is not Pseudocontractive). Let f be a real function defined by f(x) = –x² – x; it can be seen that f : [–2, 1] → [–2, 1]. This function is demicontractive on [–2, 1] and continuous. It is not quasi-nonexpansive and is not pseudocontractive on [–2, 1] (check for instance the condition of pseudocontractivity for x = –1 .5 and y = –0 . 6).

Can we construct an iterative method with a strongly accretive and Lipschitzian operator for solving a variational inequality problem with quasi-nonexpansive and demicontractive mappings in real Banach spaces?

Note also that a duality mapping exists in each Banach space. We recall from [1] some of the examples of this mapping in l_p, L_p, W^m,p-spaces, 1 < p < ∞.

lp:Jx=∥x∥lp2−py∈lq, $\begin{array}{} \displaystyle l_p: \,\,Jx=\|x\|^{2-p}_{l_p}y\in l_q, \end{array}$ x = (x₁, x₂, ⋯, x_n, ⋯), y = (x₁|x₁|^p–2, x₂|x₂|^p–2, ⋯, x_n|x_n|^p–2, ⋯),

[17] Let C and D be nonempty subsets of a real Banach space E with D ⊂ C and Q_D : C → D a retraction from C into D. Then Q_D is sunny and nonexpansive if and only if

〈z−QDz,j(y−QDz)〉≤0 $$\begin{array}{} \displaystyle \langle z-Q_D z, j(y-Q_D z)\rangle \leq 0 \end{array}$$

for all z ∈ C and y ∈ D.

If K is a nonempty closed convex subset of a Hilbert space H, then the nearest point projection P_K from H to K is the sunny nonexpansive retraction.

If φ(t) = t^q–1, q > 1, inequality (9) becomes

〈Ax−Ay,Jq(x−y)〉≥k∥x−y∥q. $$\begin{array}{} \displaystyle \langle Ax-Ay,J_q(x-y)\rangle \geq k\|x-y\|^q. \end{array}$$

[7] Let E be a Banach space satisfying Opial’s property, K be a closed convex subset of E, and T : K → K be a nonexpansive mapping such that F(T) ≠ ∅. Then I – T is demiclosed; that is,

{xn}⊂K,xn⇀x∈Kand(I−T)xn→yimpliesthat(I−T)x=y. $$\begin{array}{} \displaystyle \{x_n\}\subset K,\,\,x_n\rightharpoonup x\in K\,\,\, {and} \,\,\,(I-T)x_n\to y\,\,\,{implies ~~that}\,\, (I-T)x=y. \end{array}$$

([9]). Assume that a Banach space E has a weakly continous duality mapping J_φ with jauge φ.

Φ(∥x+y∥)≤Φ(∥x∥)+〈y,Jφ(x+y)〉 $$\begin{array}{} \displaystyle \varPhi( \lVert x+ y\rVert) \leq \varPhi(\lVert x\rVert)+ \langle y, J_\varphi(x+y)\rangle \end{array}$$

for all x, y ∈ E.

In particular, for all x, y ∈ E,

∥x+y∥2≤∥x∥2+2〈y,J(x+y)〉. $$\begin{array}{} \displaystyle \lVert x+ y\rVert^2 \leq \lVert x\rVert^2+2\langle y, J(x+y)\rangle . \end{array}$$

[5] Let q > 1 be a fixed real number and E be a smooth Banach space. Then the following statements are equivalent:

E is q-uniformly smooth.

[20] Let E be a uniformly convex real Banach space. For arbitrary r > 0, let B(0)_r := {x ∈ E : ||x|| ≤ r} and λ ∈ [0, 1]. Then there exists a continuous, strictly increasing and convex function

g:[0,2r]→R+,g(0)=0, $$\begin{array}{} \displaystyle g: [0,2r] \rightarrow \mathbb{R}^{+},\; g(0) = 0, \end{array}$$

such that for all x, y ∈ B(0)_r,

∥λx+(1−λ)y∥2≤λ∥x∥2+(1−λ)∥y∥2−(1−λ)λg(∥x−y∥). $$\begin{array}{} \displaystyle \| \lambda x + (1 - \lambda )y\|^{2} \leq \lambda \|x\|^{2} + (1 - \lambda) \|y\|^{2} - (1-\lambda)\lambda g(\|x - y\|). \end{array}$$

[21] Assume that {a_n} is a sequence of nonnegative real numbers such that a_n+1 ≤ (1 – α_n)a_n + α_nσ_n + β_n, n ≥ 0, where {α_n}, {β_n} and {σ_n} satisfy the conditions:

α_n ⊂ (0, 1), ∑n=0∞αn=∞, $\begin{array}{} \displaystyle \sum_{n=0}^\infty\alpha_n = \infty, \end{array}$

[13] Let t_n be a sequence of real numbers that does not decrease at infinity in a sense that there exists a subsequence t_ni of t_n such that t_ni such that t_ni ≤ t_ni+1 for all i ≥ 0. For sufficiently large numbers n ∈ ℕ, an integer sequence {τ(n)} is defined as follows:

τ(n)=max{k≤n:tk≤tk+1}. $$\begin{array}{} \displaystyle \tau(n) = \max\lbrace k\leq n:\,\,t_k \leq t_{k+1} \rbrace. \end{array}$$

Then, τ(n) → ∞ as n → ∞ and

max{tτ(n),tn}≤tτ(n)+1. $$\begin{array}{} \displaystyle \max\lbrace t_{\tau(n)},\,\,\, t_{n} \rbrace\leq t_{\tau(n)+1} . \end{array}$$

([12], Proposition 2.1). Assume K is a closed convex subset of a Hilbert space H. Let T : K → K be a self-mapping of C. If T is a k-demicontractive mapping, then the fixed point set Fix(T) is closed and convex.

[12] Let K be a nonempty closed convex subset of a real Hilbert space H and T : K → K be a mapping.

If T is a k-strictly pseudo-contractive mapping, then T satisfies the Lipschitzian condition

∥Tx−Ty∥≤1+k1−k∥x−y∥. $$\begin{array}{} \displaystyle \Vert Tx-Ty\Vert \leq \dfrac{1+k}{1-k}\Vert x-y\Vert. \end{array}$$

[18] Let q > 1 be a fixed real number and E be a q-uniformly smooth real Banach space with constant d_q. Let K be a nonempty, closed convex subset of E and A : K → E be a k-strongly accretive and L-Lipschitzian operator with k > 0, L > 0. Assume that 0<η<(kqdqLq)1q−1andτ=η(k−dqLqηq−1q). $\begin{array}{} \displaystyle 0 \lt \eta \lt \Big(\dfrac{kq}{ d_qL{^q}}\Big)^{\frac{1}{q-1}}~~ and ~~\tau=\eta\Big(k-\dfrac{d_q L^q \eta^{q-1}}{q}\Big). \end{array}$ Then for each t ∈ (0,min{1,1τ}), $\begin{array}{} \displaystyle \Big(0, min\{1,\,\, \dfrac{1}{\tau}\}\Big), \end{array}$ we have

∥(I−tηA)x−(I−tηA)y∥≤(1−tτ)∥x−y∥,∀x,y∈K. $$\begin{array}{} \displaystyle \| (I-t\eta A)x-(I-t\eta A)y\| \leq (1- t\tau) \| x-y\|,\,\, \forall\, x,y\in K. \end{array}$$

Let q > 1 be a fixed real number and E be a q-uniformly smooth and uniformly convex real Banach space having a weakly continuous duality map J_φ and f : E → E be an b-Lipschitzian mapping with a constant b ≥ 0. Let A : E → E be an μ-strongly accretive and L-Lipschitzian operator with 0<η<(μqdqLq)1q−1 $\begin{array}{} \displaystyle 0 \lt \eta \lt \Big(\dfrac{\mu q}{d_qL^q}\Big)^{\frac{1}{q-1}} \end{array}$ and 0 ≤ γb < τ, where τ=η(μ−dqLqηq−1q). $\begin{array}{} \displaystyle \tau=\eta\Big(\mu-\dfrac{d_q L^q \eta^{q-1}}{q}\Big). \end{array}$ Let T₁ : E → E be a k-demicontractive mapping and T₂ : E → E be a quasi-nonexpansive mapping such that Γ := Fix(T₁) ∩ Fix(T₂) ≠ ∅. Let {x_n} be a sequence defined as follows:

x0∈E,zn=(1−θn)xn+θnT1xn,yn=βnzn+(1−βn)T2zn,xn+1=αnγf(xn)+(I−ηαnA)yn, $$\begin{array}{} \displaystyle \left \{ \begin{array}{lll} x_0\in E,\\ z_n= (1-\theta_n) x_n + \theta_n T_1x_n, \\\\ y_n= \beta_{n} z_n+ (1-\beta_n) T_2 z_n,\\\\ x_{n+1}= \alpha_n \gamma f(x_n) + (I- \eta \alpha_n A)y_n, \end{array} \right. \end{array}$$(10)

with the conditions θ_n ∈ [a, b] ⊂ (0, π) where

π:=min{1,(qωq−1dq)1q−1},withω=1−k2, $$\begin{array}{} \displaystyle \pi: = {\min}\Big \{1,\Big (\frac{q \omega^{q-1}}{d_q}\Big )^{\frac{1}{q-1}}\Big\} ,\,\,\, \textit{with}\,\,\, \omega= \dfrac{ 1-k}{2}, \end{array}$$

{α_n}, {θ_n} and {β_n} are the sequences such that:

limn→∞αn=0,∑n=0∞αn=∞, $\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\alpha_n=0,\,\, \, \displaystyle\sum_{n=0}^\infty \alpha_n= \infty, \end{array}$

From the choice of η and γ, properties of Q_Γ, (ηA – γf) is strongly accretive, then the variational inequality (11) has a unique solution in Γ. Without loss of generality, we can assume αn∈(0,min{1,1τ}). $\begin{array}{} \displaystyle \alpha_n\in\Big(0, min\{1\,\, , \dfrac{1}{\tau}\}\Big). \end{array}$ In what follows, we denote x^∗ to be the unique solution of (11). Fixing p ∈ Γ. We prove that the sequence {x_n} is bounded. Using (10), inequality (ii) of Theorem 7 and inequality (8), we can compute

∥zn−p∥q=∥(1−θn)(xn−p)+θn(T1xn−p)∥q=∥(1−θn)(xn−p)+θn(T1xn−xn)+θn(xn−p)∥q=∥xn−p+θn(T1xn−xn)∥q≤∥xn−p∥q−qθn〈xn−T1xn,Jq(xn−p)〉+dq∥θn(T1xn−xn)∥q≤∥xn−p∥q−qθnωq−1∥xn−T1xn∥q+dq∥θn(T1xn−xn)∥q. $$\begin{array}{} \begin{split} \displaystyle \| z_{n}- p\|^q &=&\Big\| (1-\theta_n) (x_n-p) + \theta_n (T_1 x_n-p)\Big\|^q\\ \\ &=&\Big\| (1-\theta_n) (x_n-p) + \theta_n (T_1x_n-x_n) + \theta_n (x_n-p)\Big\|^q \\ \\\ &=&\Big\|x_n-p + \theta_n (T_1 x_n-x_n)\Big\|^q \\ \\ &\leq&\|x_n-p\|^q - q \theta_n \langle x_n-T_1x_n, J_q(x_n-p)\rangle + d_q\Big\| \theta_n (T_1x_n-x_n)\Big\|^q \\ \\ &\leq&\|x_n-p\|^q - q \theta_n \omega^{q-1}\|x_n-T_1x_n\|^q + d_q\Big\|\theta_n (T_1x_n-x_n)\Big\|^q. \end{split} \end{array}$$(12)

By inequality (12), it then follows that:

∥zn−p∥q≤∥xn−p∥q−qθnωq−1∥xn−T1xn∥q+dqθnq∥T1xn−xn∥q.=∥xn−p∥q−θn[qωq−1−dqθnq−1]∥xn−T1xn∥q. $$\begin{array}{} \begin{split} \displaystyle \Big \|z_{n}-p\Big \|^q &\leq&\Big \|x_n-p\Big \|^q - q \theta_n \omega^{q-1}\Big \|x_n-T_1x_n\Big \|^q + d_q \theta_n^q\Big \|T_1x_n-x_n\Big \|^q. \\ &=& \Big \|x_n-p\Big \|^q -\theta_n \Big [q \omega^{q-1}-d_q \theta_n^{q-1}\Big ]\Big \|x_n-T_1x_n\Big \|^q. \end{split} \end{array}$$(13)

Since qωq−1−dqθnq−1>0, $\begin{array}{} \displaystyle q\omega^{q-1}-d_q \theta_n^{q-1} \gt 0, \end{array}$ we obtain

∥zn−p∥≤∥xn−p∥. $$\begin{array}{} \displaystyle \|z_n-p\Big \|\leq \|x_{n}-p\Big \|. \end{array}$$(14)

From (10) and T₂ is quasi-nonexpansive, it follows that

∥yn−p∥=∥βnzn+(1−βn)T2zn−p∥≤βn∥zn−p∥+(1−βn)∥T2zn−p∥≤∥zn−p∥. $$\begin{array}{} \begin{split} \displaystyle \Vert y_n-p \Vert &=&\Vert \beta_{n} z_n+ (1-\beta_{n}) T_2z_n -p\Vert\\ & \leq & \beta_{n} \Vert z_n-p\Vert+ (1-\beta_{n}) \Vert T_2z_n -p \Vert\\ & \leq & \Vert z_n-p\Vert. \end{split} \end{array}$$

Therefore, we have

∥yn−p∥≤∥zn−p∥≤∥xn−p∥. $$\begin{array}{} \displaystyle \Vert y_n-p \Vert \leq \Vert z_n-p \Vert \leq \Vert x_n-p \Vert. \end{array}$$(15)

By Lemma 13, inequalities (15) and (14), we have

∥xn+1−p∥=∥αnγf(xn)+(I−ηαnA)yn−p∥≤αnγ∥f(xn)−f(p)∥+(1−ταn)∥yn−p∥+αn∥γf(p)−ηAp∥≤(1−αn(τ−bγ))∥xn−p∥+αn∥γf(p)−ηAp∥≤max{∥xn−p∥,∥γf(p)−ηAp∥τ−bγ}. $$\begin{array}{} \begin{split} \displaystyle \lVert x_{n+1}-p\rVert &=& \lVert \alpha_n \gamma f(x_n)+ (I-\eta \alpha_n A) y_n -p\rVert\\ &\leq& \alpha_n \gamma \lVert f(x_n) -f(p)\rVert + (1-\tau\alpha_n) \lVert y_n -p\rVert + \alpha_n\lVert \gamma f(p)- \eta Ap\rVert \\ &\leq&(1-\alpha_n(\tau- b\gamma ))\lVert x_n-p\rVert+\alpha_n\lVert \gamma f(p)- \eta Ap\rVert\\ &\leq& \max{\{\lVert x_n - p\|,\dfrac{\lVert \gamma f(p)- \eta A p \|}{\tau- b\gamma}\}}. \end{split} \end{array}$$

By induction, it is easy to see that

∥xn−p∥≤max{∥x0−p∥,∥γf(p)−ηAp∥τ−bγ},n≥1. $$\begin{array}{} \displaystyle \lVert x_{n}-p\rVert \leq \max{\{\lVert x_0 - p\|,\dfrac{\lVert \gamma f(p)- \eta A p \|}{\tau- b\gamma}\}}, \,\,\,\, n \geq 1. \end{array}$$

Hence, {x_n} is bounded also are {f(x_n)}, and {Ax_n}.

Thus we have

∥xn+1−p∥q=∥αnγf(xn)+(I−ηαnA)yn−p∥q≤∥yn−p+αnγf(xn)−αnηAyn∥q≤∥yn−p∥q−qαn〈ηAyn−γf(xn),Jq(yn−p)〉+dq∥αnγf(xn)+αnηAyn∥q≤∥yn−p∥q+qαn∥ηAyn−γf(xn)∥∥yn−p∥q−1+dq∥αnγf(xn)−αnηAyn∥q≤∥xn−p∥q−θn[qωq−1−dqθnq−1]∥xn−T1xn∥q+qαn∥ηAyn−γf(xn)∥∥yn−p∥q−1+dq∥αnγf(xn)−αnηAyn∥q. $$\begin{array}{} \begin{split} \displaystyle \lVert x_{n+1}-p\rVert^q &= & \lVert \alpha_n \gamma f(x_n)+ ( I-\eta \alpha_n A)y_n -p\rVert^q\\ &\leq& \lVert y_n -p + \alpha_n \gamma f(x_n)- \alpha_n\eta Ay_n \rVert^q\\ &\leq& \| y_n -p \|^q - q \alpha_n\langle \eta Ay_n-\gamma f(x_n), J_q( y_n -p) \rangle + d_q\Big\|\alpha_n \gamma f(x_n)+\alpha_n\eta Ay_n \Big\|^q\\ &\leq & \lVert y_n-p\rVert^q + q \alpha_n \lVert \eta Ay_n -\gamma f(x_n)\rVert\lVert y_n-p\rVert^{q-1}+ d_q\Big\|\alpha_n \gamma f(x_n)-\alpha_n\eta Ay_n \Big\|^q\\ \\ &\leq&\Big \|x_n-p\Big \|^q -\theta_n \Big [q \omega^{q-1}-d_q \theta_n^{q-1}\Big ]\Big \|x_n-T_1 x_n\Big \|^q + q \alpha_n\lVert \eta Ay_n -\gamma f(x_n)\rVert\lVert y_n-p\rVert^{q-1}\\&&+ d_q\Big\|\alpha_n \gamma f(x_n)-\alpha_n\eta Ay_n \Big\|^q. \end{split} \end{array}$$

Since {x_n} and {y_n} are bounded, then there exists a constant C > 0 such that

θn[qωq−1−dqθnq−1]∥xn−T1xn∥q≤∥xn−p∥q−∥xn+1−p∥q+αnC. $$\begin{array}{} \displaystyle \theta_n \Big [q \omega^{q-1}-d_q\theta_n^{q-1}\Big ] \|x_n-T_1 x_n \|^q \leq \Big \|x_n-p\Big \|^q -\lVert x_{n+1}-p\rVert^q + \alpha_n C. \end{array}$$(16)

We show that the sequence {x_n} converges strongly to a point x^∗. Now we divide the rest of the proof into two cases. Case 1. Assume that the sequence {∥x_n – p∥} is monotonically decreasing. Then {∥x_n – p∥} is convergent. Clearly, we have

limn→∞[∥xn−p∥−∥xn+1−p∥]=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty} \Big[ \lVert x_{n}-p \rVert-\lVert x_{n+1}-p\rVert\Big]= 0. \end{array}$$(17)

It then implies from (16) that

limn→∞θn[qωq−1−qdqθnq−1]∥xn−T1xn∥q=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\theta_n \Big [q \omega^{q-1}- qd_q\theta_n^{q-1}\Big ]\Big \|x_n- T_1x_n\Big \|^q =0. \end{array}$$(18)

Since θ_n ∈ [a, b] ⊂ (0, π) and qωq−1−dqθnq−1>0, $\begin{array}{} \displaystyle q \omega^{q-1}-d_q\theta_n^{q-1} \gt 0, \end{array}$ we have

limn→∞∥xn−T1xn∥=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty} \|x_n- T_1x_n \|=0. \end{array}$$(19)

Now, we observe that,

∥zn−xn∥=∥(1−θn)xn+θnT1xn−xn∥=∥(1−θn)xn+θnT1xn−θnxn−(1−θn)xn∥=θn∥T1xn−xn∥. $$\begin{array}{} \begin{split} \displaystyle \Vert z_n-x_n\Vert &= & \Vert (1-\theta_n) x_n + \theta_n T_1x_n-x_n\Vert\\ &= & \Vert (1-\theta_n) x_n + \theta_n T_1 x_n- \theta_n x_n- (1-\theta_n) x_n\Vert\\ & =& \theta_n \Vert T_1 x_n- x_n\Vert. \end{split} \end{array}$$

Therefore, from (19) we have

limn→∞∥zn−xn∥=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty} \| z_n-x_n \|=0. \end{array}$$(20)

We show that lim supn→+∞ $\begin{array}{} \displaystyle \limsup_{n\to +\infty} \end{array}$ 〈ηAx^∗ – γf(x^∗), J_φ(x^∗ – x_n)〉 ≤ 0. First, we note that there exists a subsequence {x_nj} of {x_n} such that x_nj converges weakly to a in E and

lim supn→+∞〈ηAx∗−γf(x∗),Jφ(x∗−xn)〉=limj→+∞〈ηAx∗−γf(x∗),Jφ(x∗−xnj)〉. $$\begin{array}{} \displaystyle \limsup_{n\to +\infty}\langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_n)\rangle=\displaystyle\lim_{j\to +\infty}\langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n_{j}})\rangle. \end{array}$$

Since {x_nj} is bounded, there exists a subsequence {x_nji} of {x_nj} which converges weakly to a. Without loss of generality, we can assume that {x_nj} converges weakly to the point a. From (19) and I – T₁ is demiclosed, we obtain a ∈ Fix(T₁). From Lemma 8, the fact that T₂ is quasi-nonexpansive and (15), we have

∥yn−p∥2=∥βnzn+(1−βn)T2zn−p∥2=βn∥zn−p∥2+(1−βn)∥T2zn−p∥2−(1−βn)βng(∥T2zn−zn∥)≤∥xn−p∥2−(1−βn)βng(∥T2zn−zn∥). $$\begin{array}{} \begin{split} \displaystyle \Vert y_{n}-p\Vert^2&=&\Vert \beta_{n} z_n+ (1-\beta_{n}) T_2z_n-p \Vert^2\\ &=&\beta_{n}\Vert z_n-p \Vert^2+(1-\beta_{n}) \Vert T_2z_n-p \Vert^2 - (1-\beta_{n})\beta_{n} g(\Vert T_2z_n - z_n\Vert)\\ &\leq& \Vert x_n-p \Vert^2-(1-\beta_{n})\beta_{n} g(\Vert T_2z_n - z_n\Vert) . \end{split} \end{array}$$

Hence,

∥xn+1−p∥2≤∥αnγf(xn)+(I−ηαnA)yn−p∥2≤∥(I−αnηA)(yn−p)−αn(ηAp−γf(xn))∥2≤αn2∥ηAp−γf(xn)∥2+(1−αnτ)2∥yn−p∥2+2αn(1−αnτ)∥ηAp−γf(xn)∥∥yn−p∥≤αn2∥ηAp−γf(xn)∥2+(1−αnτ)2∥xn−p∥2−(1−αnτ)2(1−βn)βng(∥T2zn−zn∥)+2αn(1−αnτ)∥ηAp−γf(xn)∥∥xn−p∥. $$\begin{array}{} \begin{split} \displaystyle \Vert x_{n+1}-p\Vert^2 &\leq & \Vert \alpha_n \gamma f(x_n) + (I- \eta \alpha_n A)y_n -p\Vert^2\\ &\leq& \Vert (I-\alpha_n \eta A )( y_n -p)-\alpha_n (\eta Ap-\gamma f(x_n))\Vert^2\\ &\leq& \alpha_n^2 \Vert \eta Ap-\gamma f(x_n) \Vert^2 + (1-\alpha_n\tau)^2 \Vert y_n -p\Vert^2+2 \alpha_n(1-\alpha_n\tau)\Vert \eta Ap-\gamma f(x_n) \Vert\Vert y_n -p\Vert\\ &\leq& \alpha_n^2 \Vert \eta Ap-\gamma f(x_n) \Vert^2 + (1-\alpha_n\tau)^2\Vert x_n-p \Vert^2-(1-\alpha_n\tau)^2 (1-\beta_{n})\beta_{n} g(\Vert T_2z_n - z_n\Vert) \\ &&+2 \alpha_n(1-\alpha_n\tau )\Vert \eta Ap-\gamma f(x_n) \Vert\Vert x_n -p\Vert. \end{split} \end{array}$$

Thus, we get

(1−αnτ)2βn(1−βn)g(∥T2zn−zn∥)≤∥xn−p∥2−∥xn+1−p∥2+αn2∥ηAp−γf(xn)∥2+2αn(1−αnτ)∥ηAp−γf(xn)∥∥xn−p∥. $$\begin{array}{} \begin{split} \displaystyle (1-\alpha_n\tau)^2\beta_{n} (1-\beta_{n}) g(\Vert T_2z_n - z_n\Vert) \leq \Vert x_{n}-p \Vert^2-\Vert x_{n+1}-p\Vert^2+\alpha_n^2 \Vert \eta Ap-\gamma f(x_n) \Vert^2 \\ +2 \alpha_n(1-\alpha_n\tau)\Vert \eta Ap-\gamma f(x_n) \Vert\Vert x_n -p\Vert. \end{split} \end{array}$$(21)

Since {x_n} is bounded, then there exists a constant B > 0 sucht that

(1−αnτ)2βn(1−βn)g(∥T2zn−zn∥)≤∥xn−p∥2−∥xn+1−p∥2+αnB. $$\begin{array}{} \displaystyle (1-\alpha_n\tau)^2\beta_{n} (1-\beta_{n}) g(\Vert T_2z_n - z_n\Vert) \leq \Vert x_{n}-p \Vert^2-\Vert x_{n+1}-p\Vert^2+\alpha_nB. \end{array}$$(22)

Thus we have

limn→∞βn(1−βn)g(∥T2zn−zn∥)=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\beta_{n} (1-\beta_{n}) g(\Vert T_2z_n - z_n\Vert) =0. \end{array}$$(23)

Since lim_n→∞ inf(1 – β_n)β_n > 0 and property of g, we have

limn→∞∥T2zn−zn∥=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\Vert T_2z_n - z_n\Vert =0. \end{array}$$(24)

Since z_nj ⇀ a, it follows from (24) and I – T₂ is demiclosed that a ∈ Fix(T₂). Therefore, a ∈ Γ. On the other hand, by using x^∗ solves (11) and the assumption that the duality mapping J_φ is weakly continuous, we have,

lim supn→+∞〈ηAx∗−γf(x∗),Jφ(x∗−xn)〉=limj→+∞〈ηAx∗−γf(x∗),Jφ(x∗−xnj)〉=〈ηAx∗−γf(x∗),Jφ(x∗−a)〉≤0. $$\begin{array}{} \begin{split} \displaystyle \limsup_{n\to +\infty}\langle \eta Ax^{*}-\gamma f(x^*) , J_{\varphi}(x^*-x_n)\rangle &=&\displaystyle\lim_{j\to +\infty}\langle \eta Ax^{*}-\gamma f(x^*), J_{\varphi}(x^*-x_{n_{j}})\rangle\\ &=&\langle \eta Ax^{*}-\gamma f(x^*), J_{\varphi}(x^*-a)\rangle\leq 0. \end{split} \end{array}$$

Finally, we show that x_n → x^∗. In fact, since Φ(t) = ∫0t $\begin{array}{} \int_ {0}^{t} \end{array}$ φ(σ)dσ, ∀t ≥ 0, and φ is a gauge function, then for 1 ≥ k ≥ 0, Φ(kt) ≤ kΦ(t). From (10), Lemmas 6 and 13, observe that

Φ(∥xn+1−x∗∥)=Φ(∥αnγf(xn)+(I−ηαnA)yn−x∗∥)≤Φ(∥αnγf(xn)+(I−ηαnA)yn−x∗∥)≤Φ(∥αn(γf(xn)−γf(x∗)+(I−αnηA)(yn−x∗)∥)+αn〈ηAx∗−γf(x∗),Jφ(x∗−xn+1)〉≤Φ(αnγ∥f(xn)−f(x∗)∥+∥(I−αnηA)(yn−x∗)∥)+αn〈ηAx∗−γf(x∗),Jφ(x∗−xn+1)〉≤Φ(αnbγ∥xn−x∗∥+(1−αnτ)∥yn−x∗∥)+αn〈ηAx∗−γf(x∗),Jφ(x∗−xn+1)〉≤Φ((1−αn(τ−bγ))∥xn−x∗∥)+αn〈ηAx∗−γf(x∗),Jφ(x∗−xn+1)〉≤(1−αn(τ−bγ))Φ(∥xn−x∗∥)+αn〈ηAx∗−γf(x∗),Jφ(x∗−xn+1)〉. $$\begin{array}{} \begin{split} \displaystyle \Phi(\| x_{n+1}-x^*\|)&=&\Phi(\| \alpha_n \gamma f(x_n)+ (I-\eta \alpha_n A) y_n -x^* \|)\\ &\leq& \Phi(\| \alpha_n \gamma f(x_n)+ (I-\eta \alpha_n A) y_n -x^* \|)\\ & \leq& \Phi(\| \alpha_n (\gamma f( x_n)-\gamma f(x^*) + (I-\alpha_n \eta A)( y_{n}-x^*)\|) + \alpha_n \langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n+1})\rangle\\ & \leq& \Phi( \alpha_n \gamma \| f( x_n)-f(x^*)\| + \| (I-\alpha_n \eta A)( y_{n}-x^*)\|)+ \alpha_n \langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n+1})\rangle\\ & \leq& \Phi( \alpha_n b \gamma \| x_n-x^*\| + (1-\alpha_n \tau) \| y_{n}-x^*\|) + \alpha_n \langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n+1})\rangle\\ & \leq& \Phi( (1-\alpha_n(\tau- b\gamma)) \| x_n-x^*\|)+ \alpha_n \langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n+1})\rangle\\ & \leq& ( 1-\alpha_n(\tau- b\gamma)) \Phi( \| x_n-x^*\|)+ \alpha_n \langle \eta Ax^*-\gamma f(x^*), J_{\varphi}(x^*-x_{n+1})\rangle. \end{split} \end{array}$$

Hence, by Lemma 9, we conclude that the sequence {x_n} converge strongly to the point x^∗ ∈ Γ.

Case 2. Assume that the sequence {∥x_n – x^∗∥} is not monotonically decreasing. Set B_n = ∥ x_n – x^∗∥ and τ : ℕ → ℕ be a mapping for all n ≥ n₀ (for some n₀ large enough) by τ(n) = max{k ∈ ℕ : k ≤ n, B_k ≤ B_k+1}. Obviously, {τ(n)} is a non-decreasing sequence such that τ(n) → ∞ as n → ∞ and B_τ(n) ≤ B_τ(n)+1 for n ≥ n₀. From (16), we have

θτ(n)[qωq−1−dqθτ(n)q−1]∥xτ(n)−T1xτ(n)∥q≤ατ(n)C. $$\begin{array}{} \displaystyle \theta_{\tau(n)} \Big [q \omega^{q-1}-d_q \theta_{\tau(n)}^{q-1}\Big ]\Big \|x_ {\tau(n)}- T_1 x_{\tau(n)} \Big \|^q \leq \alpha_{\tau(n)} C. \end{array}$$

Furthermore, we have

limn→∞θτ(n)[qωq−1−dqθτ(n)q−1]∥xτ(n)−T1xτ(n)∥q=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\theta_{\tau(n)} \Big [q \omega^{q-1}-d_q \theta_{\tau(n)}^{q-1}\Big ]\Big \|x_{\tau(n)}- T_1 x_{\tau(n)} \Big \|^q =0. \end{array}$$

Since qωq−1−dqθτ(n)q−1>0 $\begin{array}{} \displaystyle q\omega^{q-1}-d_q \theta_{\tau(n)}^{q-1} \gt 0 \end{array}$ and property of π, we have

limn→∞∥xτ(n)−T1xτ(n)∥=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\Big \|x_{\tau(n)}- T_1 x_{\tau(n)} \Big \|=0. \end{array}$$(25)

By same argument as in case 1, we can show that x_τ(n) and y_τ(n) are bounded in E and lim supτ(n)→+∞ $\begin{array}{} \displaystyle \limsup_{\tau(n)\to +\infty} \end{array}$ 〈ηAx^∗ – γf(x^∗), J_φ(x^∗ – x_τ(n))〉 ≤ 0. We have for all n ≥ n₀,

0≤Φ(∥xτ(n)+1−x∗∥)−Φ(∥xτ(n)−x∗∥)≤ατ(n)[−(τ−bγ)Φ(∥xτ(n)−x∗∥)+〈ηAx∗−γf(x∗),Jφ(x∗−xτ(n)+1)〉], $$\begin{array}{} \displaystyle 0\leq \Phi(\lVert x_{\tau(n)+1}-x^* \rVert)-\Phi(\lVert x_{\tau(n)}-x^* \rVert)\leq \alpha_{\tau(n)}[- (\tau- b\gamma)\Phi(\lVert x_{\tau(n)}-x^* \rVert) +\langle \eta Ax^*-\gamma f(x^*) ,J_{\varphi}(x^*-x_{\tau(n)+1})\rangle], \end{array}$$

which implies that

Φ(∥xτ(n)−x∗∥)≤1τ−bγ〈ηAx∗−γf(x∗),Jφ(x∗−xτ(n)+1)〉. $$\begin{array}{} \displaystyle \Phi(\lVert x_{\tau(n)}-x^* \rVert) \leq \dfrac{ 1}{\tau- b\gamma } \langle \eta Ax^*-\gamma f(x^*) ,J_{\varphi}(x^*- x_{\tau(n)+1})\rangle. \end{array}$$

Then, we have

limn→∞Φ(∥xτ(n)−x∗∥)=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\Phi(\lVert x_{\tau(n)}-x^* \rVert) =0. \end{array}$$

Therefore,

limn→∞Bτ(n)=limn→∞Bτ(n)+1=0. $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty} B_{\tau(n)}=\displaystyle\lim_{n\rightarrow \infty} B_{\tau(n)+1}=0. \end{array}$$

Thus, by Lemma 10, we conclude that

0≤Bn≤max{Bτ(n),Bτ(n)+1}=Bτ(n)+1. $$\begin{array}{} \displaystyle 0\leq B_n\leq \max\lbrace B_{\tau(n)},\,\,B_{\tau(n)+1}\rbrace=B_{\tau(n)+1}. \end{array}$$

Hence, limn→∞ $\begin{array}{} \displaystyle \lim_{n\rightarrow \infty} \end{array}$ B_n = 0, that is {x_n} converges strongly to x^∗. This completes the proof.□

Let q > 1 be a fixed real number and E be a q-uniformly smooth and uniformly convex real Banach space having a weakly continuous duality map and f : E → E be an b-Lipschitzian mapping with a constant b ≥ 0. Let A : E → E be an μ-strongly accretive and L-Lipschitzian operator with 0<η<(μqdqLq)1q−1 $\begin{array}{} \displaystyle 0 \lt \eta \lt \Big(\dfrac{\mu q}{d_qL^q}\Big)^{\frac{1}{q-1}} \end{array}$ and 0 ≤ γb < τ, where τ=η(μ−dqLqηq−1q). $\begin{array}{} \displaystyle \tau=\eta\Big(\mu-\dfrac{d_q L^q \eta^{q-1}}{q}\Big). \end{array}$ Let T₁ : E → E and T₂ : E → E two nonexpansive mappings such that Γ := Fix(T₁) ∩ Fix(T₂) ≠ ∅. Let {x_n} be a sequence defined as follows:

x0∈E,zn=(1−θn)xn+θnT1xn,yn=βnzn+(1−βn)T2zn,xn+1=αnγf(xn)+(I−ηαnA)yn, $$\begin{array}{} \displaystyle \left \{ \begin{array}{lll} x_0\in E,\\ z_n= (1-\theta_n) x_n + \theta_n T_1x_n, \\\\ y_n= \beta_{n} z_n+ (1-\beta_n) T_2 z_n,\\\\ x_{n+1}= \alpha_n \gamma f(x_n) + (I- \eta \alpha_n A)y_n, \end{array} \right. \end{array}$$(26)

with the conditions θ_n ∈ [a, b] ⊂ (0, π) where

π:=min{1,(q2q−1dq)1q−1}, $$\begin{array}{} \displaystyle \pi: = {\min}\Big \{1,\Big (\frac{q }{2^{q-1} d_q}\Big )^{\frac{1}{q-1}}\Big\}, \end{array}$$

{α_n}, {θ_n} and {β_n} are the sequences such that:

limn→∞αn=0,∑n=0∞αn=∞, $\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}\alpha_n=0,\,\, \, \displaystyle\sum_{n=0}^\infty \alpha_n= \infty, \end{array}$

Since every nonexpansive mapping is quasi-nonexpansive and 0-demicontractive. The proof follows Lemma 5 and Theorem 14.□

Assume that E = l_q, 1 < q < ∞ and f : E → E be an b-Lipschitzian mapping with a constant b ≥ 0. Let A : E → E be an μ-strongly accretive and L-Lipschitzian operator with 0<η<(μqdqLq)1q−1 $\begin{array}{} \displaystyle 0 \lt \eta \lt \Big(\dfrac{\mu q}{d_qL^q}\Big)^{\frac{1}{q-1}} \end{array}$ and 0 ≤ γb < τ, where τ=η(μ−dqLqηq−1q). $\begin{array}{} \displaystyle \tau=\eta\Big(\mu-\dfrac{d_q L^q \eta^{q-1}}{q}\Big). \end{array}$ Let T₁ : E → E be a k-demicontractive mapping and T₂ : E → E be a quasi-nonexpansive mapping such that Γ := Fix(T₁) ∩ Fix(T₂) ≠ ∅. Let {x_n} be a sequence defined as follows:

x0∈E,zn=(1−θn)xn+θnT1xn,yn=βnzn+(1−βn)T2zn,xn+1=αnγf(xn)+(I−ηαnA)yn, $$\begin{array}{} \displaystyle \left \{ \begin{array}{lll} x_0\in E,\\ z_n= (1-\theta_n) x_n + \theta_n T_1x_n, \\\\ y_n= \beta_{n} z_n+ (1-\beta_n) T_2 z_n,\\\\ x_{n+1}= \alpha_n \gamma f(x_n) + (I- \eta \alpha_n A)y_n, \end{array} \right. \end{array}$$(28)