Convergence of the modified Mann's iterative method for asymptotically κ-strictly pseudocontractive mappings

Abstract

Let E be a real uniformly convex Banach space which has the Fréchet differentiable norm, and K a nonempty, closed, and convex subset of E. Let T : K → K be an asymptotically κ-strictly pseudocontractive mapping with a nonempty fixed point set. We prove that (I - T) is demiclosed at 0 and obtain a weak convergence theorem of the modified Mann's algorithm for T under suitable control conditions. Moreover, we also elicit a necessary and sufficient condition that guarantees strong convergence of the modified Mann's iterative sequence to a fixed point of T in a real Banach spaces with the Fréchet differentiable norm.

2000 AMS Subject Classification: 47H09; 47H10.

1 Introduction

Let E and E* be a real Banach space and the dual space of E, respectively. Let K be a nonempty subset of E. Let J denote the normalized duality mapping from E into 2^E* given by J(x) = {f ∈ E* : 〈x, f〉 = ||x||² = ||f||²}, for all x ∈ E, where 〈·,·〉 denotes the duality pairing between E and E*. In the sequel, we will denote the set of fixed points of a mapping T : K → K by F (T) = {x ∈ K : Tx = x}.

A mapping T : K → K is called asymptotically κ-strictly pseudocontractive with sequence ${\left\{{\kappa }_n\right\}}_{n=1}^{\infty }\subseteq \left[1,\infty \right)$ such that lim _n→∞ κ _n = 1 (see, e.g., [1–3]) if for all x, y ∈ K, there exist a constant κ ∈ [0, 1) and j(x - y) ∈ J(x - y) such that

$$⟨T^{n}x-T^{n}y,j\left(x-y\right)⟩\le {\kappa }_n\parallel x-y{\parallel }^2-\kappa \parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2,\forall n\ge 1.$$

(1)

If I denotes the identity operator, then (1) can be written as

$$⟨\left(I-T^n\right)x-\left(I-T^n\right)y,j\left(x-y\right)⟩\ge \kappa \parallel \left(I-T^n\right)x-\left(I-T^n\right)y{\parallel }^2-\left({\kappa }_n-1\right)\parallel x-y{\parallel }^2,\forall n\ge 1.$$

(2)

The class of asymptotically κ-strictly pseudocontractive mappings was first introduced in Hilbert spaces by Qihou [3]. In Hilbert spaces, j is the identity and it is shown by Osilike et al. [2] that (1) (and hence (2)) is equivalent to the inequality

$$\parallel T^{n}x-T^{n}y{\parallel }^2\le {\lambda }_n\parallel x-y{\parallel }^2+\lambda \parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2,$$

where lim _n→∞ λ _n = lim _n→∞ [1 + 2(κ _n - 1)] = 1, λ = (1 - 2κ) ∈ [0, 1).

A mapping T with domain D(T) and range R(T) in E is called strictly pseudocontractive of Browder-Petryshyn type [4], if for all x, y ∈ D(T), there exists κ ∈ [0, 1) and j(x - y) ∈ J(x - y) such that

$$⟨Tx-Ty,j\left(x-y\right)⟩\le \parallel x-y{\parallel }^2-\kappa \parallel x-y-\left(Tx-Ty\right){\parallel }^2.$$

(3)

If I denotes the identity operator, then (3) can be written as

$$⟨\left(I-T\right)x-\left(I-T\right)y,j\left(x-y\right)⟩\ge \kappa \parallel \left(I-T\right)x-\left(I-T\right)y{\parallel }^2.$$

(4)

In Hilbert spaces, (3) (and hence (4)) is equivalent to the inequality

$$\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+k\parallel x-y-\left(Tx-Ty\right){\parallel }^2,k=\left(1-2\kappa \right)<1,$$

It is shown in [5] that the class of asymptotically κ-strictly pseudocontractive mappings and the class of κ-strictly pseudocontractive mappings are independent.

A mapping T is said to be uniformly L-Lipschitzian if there exists a constant L > 0 such that

$$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,n\ge 1$$

for all x, y ∈ K and is said to be demiclosed at a point p if whenever {x _n } ⊂ D(T) such that {x _n } converges weakly to x ∈ D(T ) and {Tx _n } converges strongly to p, then Tx = p.

Kim and Xu [6] studied weak and strong convergence theorems for the class of asymptotically κ-strictly pseudocontractive mappings in Hilbert space. They obtained a weak convergence theorem of modified Mann iterative processes for this class of mappings. Moreover, a strong convergence theorem was also established in a real Hilbert space by hybrid projection method. They proved the following.

Theorem KX [6] Let K be a closed and convex subset of a Hilbert space H. Let T : K → K be an asymptotically κ-strictly pseudocontractive mapping for some 0 ≤ κ < 1 with sequence {κ _n } ⊂ [1, ∞) such that${\sum }_{n=1}^{\infty }\left({\kappa }_n-1\right)<\infty$and F(T ) ≠ ∅. Let${\left\{x_n\right\}}_{n=1}^{\infty }$be a sequence generated by the modified Mann's iteration method:

$$x_{n+1}={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T^n{x}_n,n\ge 1,$$

Assume that the control sequence ${\left\{{\alpha }_n\right\}}_{n=1}^{\infty }$ is chosen in such a way that κ + λ ≤ α _n ≤ 1 - λ for all n, where λ ∈ (0, 1) is a small enough constant. Then, {x _n } converges weakly to a fixed point of T.

The modified Mann's iteration scheme was introduced by Schu [7, 8] and has been used by several authors (see, for example, [1–3, 9–11]). One question is raised naturally: is the result in Theorem KX true in the framework of the much general Banach space?

Osilike et al. [5] proved the convergence theorems of modified Mann iteration method in the framework of q-uniformly smooth Banach spaces which are also uniformly convex. They also obtained that a modified Mann iterative process {x _n } converges weakly to a fixed point of T under suitable control conditions. However, the control sequence {α _n } ⊂ [0,1] depended on the Lipschizian constant L and excluded the natural choice ${\alpha }_n=\frac{1}{n},n\ge 1.$ These are motivations for us to improve the results. We prove the demiclosedness principle and weak convergence theorem of the modified Mann's algorithm for T in the framework of uniformly convex Banach spaces which have the Fréchet differentiable norm. Moreover, we also elicit a necessary and sufficient condition that guarantees strong convergence of the modified Mann's iterative sequence to a fixed point of T in a real Banach spaces with the Fréchet differentiable norm.

We will use the notation:

• 1. ⇀ for weak convergence.

• 2. ${\omega }_{\mathcal{W}}\left(x_n\right)=\left\{x:\exists x_{n_j}\rightharpoonup x\right\}$ denotes the weak ω-limit set of {x _n }.

2 Preliminaries

Let E be a real Banach space. The space E is called uniformly convex if for each ε > 0, there exists a δ > 0 such that for x, y ∈ E with ||x|| ≤ 1, ||y|| ≤ 1, ||x - y|| ≥ ε, we have $\parallel \frac{1}{2}\left(x+y\right)\parallel \le 1-\delta$. The modulus of convexity of E is defined by

$${\delta }_E\left(\epsilon \right)=inf\left\{1-\parallel \frac{1}{2}\left(x+y\right)\parallel :\parallel x\parallel \le 1,\parallel y\parallel \le 1,\parallel x-y\parallel \ge \epsilon ,\right\}\forall x,y\in E$$

for all ε ∈ [0,2]. E is uniformly convex if δ _E (0) = 0 and δ _E (ε) > 0 for all ε ∈ (0, 2]. The modulus of smoothness of E is the function ρ _E : [0, ∞) ∈ [0, ∞) defined by

$${\rho }_E\left(\tau \right)=sup\left\{\frac{1}{2}\left(\parallel x+y\parallel +\parallel x-y\parallel \right)-1:\parallel x\parallel \le 1,\parallel y\parallel \le \tau \right\},\forall x,y\in E.$$

E is uniformly smooth if and only if $\underset{\tau \to 0}{lim}\frac{{\rho }_E\left(\tau \right)}{\tau }=0.$

E is said to have a Fréchet differentiable norm if for all x ∈ U = {x ∈ E : ||x|| = 1}

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists and is attained uniformly in y ∈ U. In this case, there exists an increasing function b : [0, ∞) → [0, ∞) with $\underset{t\to 0}{lim}\left[b\left(t\right)∕t\right]=0$ such that for all x, h ∈ E

$$\frac{1}{2}\parallel x{\parallel }^2+⟨h,j\left(x\right)⟩\le \frac{1}{2}\parallel x+h{\parallel }^2\le \frac{1}{2}\parallel x{\parallel }^2+⟨h,j\left(x\right)⟩+b\left(\parallel h\parallel \right).$$

(5)

It is well known (see, for example, [[12], p. 107]) that uniformly smooth Banach space has a Fréchet differentiable norm.

Lemma 2.1 [2, p. 80] Let ${\left\{a_n\right\}}_{n=1}^{\infty }$, ${\left\{b_n\right\}}_{n=1}^{\infty }$, ${\left\{{\delta }_n\right\}}_{n=1}^{\infty }$ be nonnegative sequences of real numbers satisfying the following inequality

$$a_{n+1}\le \left(1+{\delta }_n\right)a_n+b_n,\forall n\ge 1.$$

If ${\sum }_{n=1}^{\infty }{\delta }_n<\infty$ and ${\sum }_{n=1}^{\infty }{b}_n<\infty$, then lim _{n →∞} a _n exists. If in addition ${\left\{a_n\right\}}_{n=1}^{\infty }$ has a subsequence which converges strongly to zero, then lim _{n →∞} a _n = 0.

Lemma 2.2 [2, p. 78] Let E be a real Banach space, K a nonempty subset of E, and T : K → K an asymptotically κ-strictly pseudocontractive mapping. Then, T is uniformly L-Lipschitzian.

Lemma 2.3 [[13], p. 29] Let K be a nonempty, closed, convex, and bounded subset of a uniformly convex Banach space E, and let T : K → E be a nonexpansive mappings. Let {x _n } be a sequence in K such that {x _n } converges weakly to some point x ∈ K. Then, there exists an increasing continuous function h : [0, ∞) → [0, ∞) with h(0) = 0 depending on the diameter of K such that

$$h\left(\parallel x-Tx\parallel \right)\le \underset{n\to \infty }{liminf}\parallel x_n-T{x}_n\parallel .$$

Lemma 2.4 [[14], p. 9] Let E be a real Banach space with the Fréchet differentiable norm.

For x ∈ E, let β*(t) be defined for 0 < t < ∞ by

$${\beta }^{*}\left(t\right)=\underset{y\in U}{sup}\left|\frac{\parallel x+ty{\parallel }^2-\parallel x{\parallel }^2}{t}-2⟨y,j\left(x\right)⟩\right|.$$

Then, lim _{t →0 +} β*(t) = 0 and

$$\parallel x+h{\parallel }^2\le \parallel x{\parallel }^2+2⟨h,j\left(x\right)⟩+\parallel h\parallel {\beta }^{*}\left(\parallel h\parallel \right),\forall h\in E\backslash \left\{0\right\}.$$

(6)

Remark 2.5 In a real Hilbert space, we can see that β*(t) = t for t > 0. In our more general setting, throughout this article we will still assume that

$${\beta }^{*}\left(t\right)\le 2t,$$

where β* is a function appearing in (6).

Then, we prove the demiclosedness principle of T in the uniformly convex Banach space which has the Fréchet differentiable norm.

Lemma 2.6 Let E be a real uniformly convex Banach space which has the Fréchet differentiable norm. Let K be a nonempty, closed, and convex subset of E and T : K → K an asymptotically κ-strictly pseudocontractive mapping with F(T) ≠ ∅. Then, (I - T) is demiclosed at 0.

Proof. Let {x _n } be a sequence in K which converges weakly to p ∈ K and {x _n - Tx _n } converges strongly to 0. We prove that (I - T)(p) = 0. Let x* ∈ F(T). Then, there exists a constant r > 0 such that ||x _n - x*|| ≤ r, ∀n ≥ 1. Let ${\bar{B}}_r=\left\{x\in E:\parallel x-x^{*}\parallel \le r\right\}$, and let $C=K\cap {\bar{B}}_r$. Then, C is nonempty, closed, convex, and bounded, and {x _n } ⊆ C. Choose any α ∈ (0, κ) and let T _α,n : K → K be defined for all x ∈ K by

$$T_{\alpha ,n}x=\left(1-\alpha \right)x+\alpha T^{n}x,n\ge 1,$$

Then for all x, y ∈ K,

$$\begin{array}{ll}\hfill \parallel T_{\alpha ,n}x-T_{\alpha ,n}y{\parallel }^2& =\parallel \left(x-y\right)-\alpha \left[\left(I-T^n\right)x-\left(I-T^n\right)y\right]{\parallel }^2\\ \le \parallel x-y{\parallel }^2-2\alpha ⟨\left(I-T^n\right)x-\left(I-T^n\right)y,j\left(x-y\right)⟩\\ +\alpha \parallel x-y-\left(T^{n}x-T^{n}y\right)\parallel {\beta }^{*}\left[\alpha \parallel x-y-\left(T^{n}x-T^{n}y\right)\parallel \right]\\ \parallel x-y{\parallel }^2-2\alpha \left[\kappa \parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2-\left({\kappa }_n-1\right)\parallel x-y{\parallel }^2\right]\\ +2{\alpha }^2\parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2\\ =\left[1+2\alpha \left({\kappa }_n-1\right)\right]\parallel x-y{\parallel }^2-2\alpha \left(\kappa -\alpha \right)\parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2\\ \le {\tau }_n^2\parallel x-y{\parallel }^2,\end{array}$$

(7)

where ${\tau }_n={\left[1+2\alpha \left({\kappa }_n-1\right)\right]}^{\frac{1}{2}}$. (In fact, in (7) the domain of β*(·) requires ||x - y - (Tⁿx-Tⁿy)|| ≠ 0. But when ||x - y - (Tⁿx-Tⁿy)|| = 0, we have ||T _α,n x-T _α,n y||² = ||x - y||², which still satisfies the inequality $\parallel T_{\alpha ,n}x-T_{\alpha ,n}y{\parallel }^2\le {\tau }_n^2\parallel x-y{\parallel }^2$. So we do not specially emphasize the situation that the argument of β*(·) equals 0 in this inequality and the following proof of Theorem 3.1.) Define G _α,m : K → E by

$$G_{\alpha ,m}x=\frac{1}{{\tau }_m}{T}_{\alpha ,m}x,m\ge 1.$$

Then, G _α,m is nonexpansive and it follows from Lemma 2.3 that there exists an increasing continuous function h : [0, ∞) → [0, ∞) with h(0) = 0 depending on the diameter of K such that

$$h\left(\parallel p-G_{\alpha ,m}p\parallel \right)\le \underset{n\to \infty }{liminf}\parallel x_n-G_{\alpha ,m}{x}_n\parallel .$$

(8)

Observe that

$$\begin{array}{ll}\hfill \parallel x_n-G_{\alpha ,m}{x}_n\parallel & =\parallel x_n-\frac{1}{{\tau }_m}{T}_{\alpha ,m}{x}_n\parallel \\ \le \parallel x_n-T_{\alpha ,m}{x}_n\parallel +\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_m\parallel x_n-x^{*}\parallel +\parallel x^{*}\parallel \right)\\ \le \parallel x_n-T_{\alpha ,m}{x}_n\parallel +\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right),\end{array}$$

(9)

and as n → ∞

$$\parallel x_n-T_{\alpha ,m}{x}_n\parallel =\alpha \parallel x_n-T^m{x}_n\parallel \le \sum _{j=1}^m\parallel T^{j-1}{x}_n-T^j{x}_n\parallel \le \left[1+L\left(m-1\right)\right]\parallel x_n-T{x}_n\parallel \to 0.$$

(10)

Thus, it follows from (9) and (10) that

$$\underset{n\to \infty }{limsup}\parallel x_n-G_{\alpha ,m}{x}_n\parallel \le \left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right),$$

so that (8) implies that

$$h\left(\parallel p-G_{\alpha ,m}p\parallel \right)\le \left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right).$$

Observe that

$$\begin{array}{ll}\hfill \parallel p-G_{\alpha ,m}p\parallel & \ge \parallel p-T_{\alpha ,m}p\parallel -\left(1-\frac{1}{{\tau }_m}\right)\parallel T_{\alpha ,m}p\parallel \\ \ge \parallel p-T_{\alpha ,m}p\parallel -\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right),\end{array}$$

so that

$$\begin{array}{ll}\hfill \parallel p-T_{\alpha ,m}p\parallel & \le \parallel p-G_{\alpha ,m}p\parallel +\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right)\\ \le h^{-1}\left[\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right)\right]+\left(1-\frac{1}{{\tau }_m}\right)\left({\tau }_{m}r+\parallel x^{*}\parallel \right)\to 0,asm\to \infty .\end{array}$$

Since T is continuous, we have (I - T)(p) = 0, completing the proof of Lemma 2.6.    □

Lemma 2.7 Let E be a real uniformly convex Banach space which has the Fréchet differentiable norm, and let K be a nonempty, closed, and convex subset of E. Let T : K → K be an asymptotically κ-strictly pseudocontractive mapping with F(T) ≠ ∅. Let ${\left\{x_n\right\}}_{n=1}^{\infty }$ be the sequence satisfying the following conditions:

• (a) $\underset{n\to \infty }{lim}\parallel x_n-p\parallel$ exists for every p ∈ F(T );

• (b) $\underset{n\to \infty }{lim}\parallel x_n-T{x}_n\parallel =0$;

• (c) $\underset{n\to \infty }{lim}\parallel t{x}_n+\left(1-t\right)p_1-p_2\parallel$ exists for all t ∈ [0, 1] and for all p₁, p₂ ∈ F (T ).

Then, the sequence {x _n } converges weakly to a fixed point of T.

Proof. Since lim _{n →∞} ||x _n - p|| exists, then {x _n } is bounded. By (b) and Lemma 2.6, we have ${\omega }_{\mathcal{W}}\left(x_n\right)\subset F\left(T\right)$. Assume that $p_1,p_2\in {\omega }_{\mathcal{W}}\left(x_n\right)$ and that $\left\{x_{n_i}\right\}$ and $\left\{x_{m_j}\right\}$ are subsequences of {x _n } such that $x_{n_i}\rightharpoonup p_1$ and $x_{m_j}\rightharpoonup p_2$, respectively. Since E has the Fréchet differentiable norm, by setting x = p₁ - p₂, h = t(x _n - p₁) in (5) we obtain

$$\begin{array}{ll}\hfill \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t⟨x_n-p_1,j\left(p_1-p_2\right)⟩& \le \frac{1}{2}\parallel t{x}_n+\left(1-t\right)p_1-p_2{\parallel }^2\\ \le \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t⟨x_n-p_1,j\left(p_1-p_2\right)⟩+b\left(t\parallel x_n-p_1\parallel \right),\end{array}$$

where b is an increasing function. Since ||x _n - p₁|| ≤ M, ∀n ≥ 1, for some M > 0, then

$$\begin{array}{ll}\hfill \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t⟨x_n-p_1,j\left(p_1-p_2\right)⟩& \le \frac{1}{2}\parallel t{x}_n+\left(1-t\right)p_1-p_2{\parallel }^2\\ \le \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t⟨x_n-p_1,j\left(p_1-p_2\right)⟩+b\left(tM\right).\end{array}$$

Therefore,

$$\begin{array}{ll}\hfill \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t\underset{n\to \infty }{limsup}⟨x_n-p_1,j\left(p_1-p_2\right)⟩& \le \frac{1}{2}\underset{n\to \infty }{lim}\parallel t{x}_n+\left(1-t\right)p_1-p_2{\parallel }^2\\ \le \frac{1}{2}\parallel p_1-p_2{\parallel }^2+t\underset{n\to \infty }{liminf}⟨x_n-p_1,j\left(p_1-p_2\right)⟩+b\left(tM\right).\end{array}$$

Hence, $lim\underset{n\to \infty }{sup}⟨x_n-p_1,j\left(p_1-p_2\right)⟩\le lim\underset{n\to \infty }{inf}⟨x_n-p_1,j\left(p_1-p_2\right)⟩+\frac{b\left(tM\right)}{t}$. Since $\underset{t\to 0^{+}}{lim}\frac{b\left(tM\right)}{t}=0$, then lim _{n →∞} 〈x _n - p₁, j(p₁ - p₂)〉 exists. Since lim _{n →∞} 〈x _n - p₁, j(p₁ - p₂)〉 = 〈p - p₁, j(p₁ - p₂)〉, for all $p\in {\omega }_{\mathcal{W}}\left(x_n\right)$. Set p = p₂. We have 〈p₂ - p₁, j(p₁ - p₂)〉 = 0, that is, p₂ = p₁. Hence, ${\omega }_{\mathcal{W}}\left(x_n\right)$ is singleton, so that {x _n } converges weakly to a fixed point of T.    □

3 Main results

Theorem 3.1 Let E be a real uniformly convex Banach space which has the Fréchet differentiable norm, and let K be a nonempty, closed, and convex subset of E. Let T : K → K be an asymptotically κ-strictly pseudocontractive mapping for some 0 ≤ κ < 1 with sequence ${\left\{{\kappa }_n\right\}}_{n=1}^{\infty }\subset \left[1,\infty \right)$, such that ${\sum }_{n=1}^{\infty }\left({\kappa }_n-1\right)<\infty ,$ and let F(T) ≠ ∅. Assume that the control sequence ${\left\{{\alpha }_n\right\}}_{n=1}^{\infty }$ is chosen so that

• (i*) 0 < α _n < κ, n ≥ 1;

• (ii*) $\sum _{n=1}^{\infty }{\alpha }_n\left(\kappa -{\alpha }_n\right)=\infty$. (11)

Given x₁ ∈ K, then the sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ is generated by the modified Mann's algorithm:

$$x_{n+1}=\left(1-{\alpha }_n\right)x_n+{\alpha }_n{T}^n{x}_n,$$

(12)

converges weakly to a fixed point of T.

Proof. Pick a p ∈ F(T). We firstly show that lim _{n →∞} ||x _n - p|| exists. To see this, using (2) and (6), we obtain

$$\begin{array}{ll}\hfill \parallel x_{n+1}-p{\parallel }^2& =\parallel \left(x_n-p\right)-{\alpha }_n\left(x_n-T^n{x}_n\right){\parallel }^2\\ \le \parallel x_n-p{\parallel }^2-2{\alpha }_n⟨x_n-T^n{x}_n,j\left(x_n-p\right)⟩+{\alpha }_n\parallel x_n-T^n{x}_n\parallel {\beta }^{*}\left({\alpha }_n\parallel x_n-T^n{x}_n\parallel \right)\\ \le \parallel x_n-p{\parallel }^2-2{\alpha }_n\left[\kappa \parallel x_n-T^n{x}_n{\parallel }^2-\left({\kappa }_n-1\right)\parallel x_n-p{\parallel }^2\right]+2{\alpha }_n^2\parallel x_n-T^n{x}_n{\parallel }^2\\ =\left[1+2{\alpha }_n\left({\kappa }_n-1\right)\right]\parallel x_n-p{\parallel }^2-2{\alpha }_n\left(\kappa -{\alpha }_n\right)\parallel x_n-T^n{x}_n{\parallel }^2.\end{array}$$

(13)

Obviously,

$$\parallel x_{n+1}-p{\parallel }^2\le \left[1+2{\alpha }_n\left({\kappa }_n-1\right)\right]\parallel x_n-p{\parallel }^2.$$

(14)

Let δ _n = 1 + 2α _n (κ _n - 1). Since ${\sum }_{n=1}^{\infty }\left({\kappa }_n-1\right)<\infty$, we have

$$\sum _{n=1}^{\infty }\left({\delta }_n-1\right)\le 2\sum _{n=1}^{\infty }\left({\kappa }_n-1\right)<\infty ,$$

then (14) implies lim _{n →∞} ||x _n - p|| exists by Lemma 2.1 (and hence the sequence {||x _n - p||} is bounded, that is, there exists a constant M > 0 such that ||x _n - p|| < M ).

Then, we prove lim _{n →∞} ||x _n - Tx _n || = 0. In fact, it follows from (13) that

$$\begin{array}{ll}\hfill \sum _{n=1}^{j}2{\alpha }_n\left(\kappa -{\alpha }_n\right)\parallel x_n-T^n{x}_n{\parallel }^2& \le \sum _{n=1}^j\left(\parallel x_n-p{\parallel }^2-\parallel x_{n+1}-p{\parallel }^2\right)+\sum _{n=1}^j\left[2{\alpha }_n\left({\kappa }_n-1\right)\right]\parallel x_n-p{\parallel }^2\\ \le \sum _{n=1}^j\left(\parallel x_n-p{\parallel }^2-\parallel x_{n+1}-p{\parallel }^2\right)+\sum _{n=1}^j\left({\delta }_n-1\right)M^2.\end{array}$$

Then,

$$\sum _{n=1}^{\infty }2{\alpha }_n\left(\kappa -{\alpha }_n\right)\parallel x_n-T^n{x}_n{\parallel }^2<\parallel x_1-p{\parallel }^2+M^2\sum _{n=1}^{\infty }\left({\delta }_n-1\right)<\infty .$$

(15)

Since ${\sum }_{n=1}^{\infty }{\alpha }_n\left(\kappa -{\alpha }_n\right)=\infty$, then (15) implies that lim inf _{n →∞} ||x _n - Tⁿx _n || = 0. Thus lim _{n →∞} ||x _n - Tⁿx _n || = 0.

By Lemma 2.2 we know that T is uniformly L-Lipschitzian, then there exists a constant L > 0, such that

$$\begin{array}{ll}\hfill \parallel x_n-T{x}_n\parallel & \le \parallel x_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T{x}_n\parallel \le \parallel x_n-T^n{x}_n\parallel +L\parallel T^{n-1}{x}_n-x_n\parallel \\ \le \parallel x_n-T^n{x}_n\parallel +L\parallel T^{n-1}{x}_n-T^{n-1}{x}_{n-1}\parallel +L\parallel T^{n-1}{x}_{n-1}-x_n\parallel \\ \le \parallel x_n-T^n{x}_n\parallel +L^2\parallel x_n-x_{n-1}\parallel +L\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel +L\parallel x_n-x_{n-1}\parallel \\ \le \parallel x_n-T^n{x}_n\parallel +L\left(2+L\right)\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel \end{array}$$

Hence, lim _{n →∞} ||x _n - Tx _n || = 0.

Now we prove that for all p₁, p₂ ∈ F(T), lim _{n →∞} ||tx _n + (1 - t)p₁ - p₂|| exists for all t ∈ [0, 1]. Let σ _n (t) = ||tx _n + (1 - t)p₁ - p₂||. It is obvious that lim _{n →∞} σ _n (0) = ||p₁ - p₂|| and lim _{n →∞} σ _n (1) = lim _{n →∞} ||x _n - p₂|| exist. So, we only need to consider the case of t ∈ (0, 1).

Define T _n : K → K by

$$T_{n}x=\left(1-{\alpha }_n\right)x+{\alpha }_n{T}^{n}x,x\in K.$$

Then for all x, y ∈ K,

$$\begin{array}{ll}\hfill \parallel T_{n}x-T_{n}y{\parallel }^2& =\parallel \left(x-y\right)-{\alpha }_n\left[\left(I-T^n\right)x-\left(I-T^n\right)y\right]{\parallel }^2\\ \le \parallel x-y{\parallel }^2-2{\alpha }_n⟨\left(I-T^n\right)x-\left(I-T^n\right)y,j\left(x-y\right)⟩\\ +{\alpha }_n\parallel x-y-\left(T^{n}x-T^{n}y\right)\parallel {\beta }^{*}\left[{\alpha }_n\parallel x-y-\left(T^{n}x-T^{n}y\right)\parallel \right]\\ \le \parallel x-y{\parallel }^2-2{\alpha }_n\left[\kappa \parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2-\left({\kappa }_n-1\right)\parallel x-y{\parallel }^2\right]\\ +2{\alpha }_n^2\parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2\\ =\left[1+2{\alpha }_n\left({\kappa }_n-1\right)\right]\parallel x-y{\parallel }^2-2{\alpha }_n\left(\kappa -{\alpha }_n\right)\parallel x-y-\left(T^{n}x-T^{n}y\right){\parallel }^2.\end{array}$$

By the choice of α _n , we have ||T _n x - T _n y||² ≤ [1 + 2α _n (κ _n - 1)]||x - y||². For the convenience of the following discussing, set ${\lambda }_n={\left[1+2{\alpha }_n\left({\kappa }_n-1\right)\right]}^{\frac{1}{2}}$, then ||T _n x - T _n y|| ≤ λ _n ||x - y||.

Set S _n,m = T _{n + m -1}T _{n + m -2} ··· T _n , m ≥ 1. We have

$$\parallel S_{n,m}x-S_{n,m}y\parallel \le \left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x-y\parallel forallx,y\in K,$$

and

$$S_{n,m}{x}_n=x_{n+m},S_{n,m}p=pforallp\in F\left(T\right).$$

Set b _n,m = ||S _n,m (tx _n + (1 - t)p₁) - tS _n,m x _n - (1 - t)S _n,m p₁||. If ||x _n - p₁|| = 0 for some n₀, then x _n = p₁ for any n ≥ n₀ so that lim _{n →∞} ||x _n - p₁|| = 0, in fact {x _n } converges strongly to p₁ ∈ F(T). Thus, we may assume ||x _n - p₁|| > 0 for any n ≥ 1. Let δ denote the modulus of convexity of E. It is well known (see, for example, [[15], p. 108]) that

$$\begin{array}{ll}\hfill \parallel tx+\left(1-t\right)y\parallel & \le 1-2min\left\{t,\left(1-t\right)\right\}\delta \left(\parallel x-y\parallel \right)\\ \le 1-2t\left(1-t\right)\delta \left(\parallel x-y\parallel \right)\end{array}$$

(16)

for all t ∈ [0, 1] and for all x, y ∈ E such that ||x|| ≤ 1, ||y|| ≤ 1. Set

$$\begin{array}{c}{w}_{n,m}=\frac{S_{n,m}{p}_1-S_{n,m}\left(t{x}_n+\left(1-t\right)p_1\right)}{t\left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel }\\ z_{n,m}=\frac{S_{n,m}\left(t{x}_n+\left(1-t\right)p_1\right)-S_{n,m}{x}_n}{\left(1-t\right)\left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel }\end{array}$$

Then, ||w _n , _m || ≤ 1 and ||z _n , _m || ≤ 1 so that it follows from (16) that

$$2t\left(1-t\right)\delta \left(\parallel w_{n,m}-z_{n,m}\parallel \right)\le 1-\parallel t{w}_{n,m}+\left(1-t\right)z_{n,m}\parallel .$$

(17)

Observe that

$$\parallel w_{n,m}-z_{n,m}\parallel =\frac{b_{n,m}}{t\left(1-t\right)\left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel }$$

and

$$\parallel t{w}_{n,m}+\left(1-t\right)z_{n,m}\parallel =\frac{\parallel S_{n,m}{x}_n-S_{n,m}{p}_1\parallel }{\left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel },$$

it follows from (17) that

$$\begin{array}{c}2t\left(1-t\right)\left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel \delta \left(\frac{b_{n,m}}{t\left(1-t\right)\left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel }\right)\\ \le \left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel -\parallel S_{n,m}{x}_n-S_{n,m}{p}_1\parallel =\left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel -\parallel x_{n+m}-p_1\parallel .\end{array}$$

(18)

Since E is uniformly convex, then $\frac{\delta \left(s\right)}{s}$ is nondecreasing, and since $\left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel \le \left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right){\lambda }_{n-1}\parallel x_{n-1}-p_1\parallel \le \cdot \cdot \cdot \le \left({\prod }_{j=n}^{n+m-1}{\lambda }_j\right)\left({\prod }_{j=1}^{n-1}{\lambda }_j\right)\parallel x_1-p_1\parallel =\left({\prod }_{j=1}^{n+m-1}{\lambda }_j\right)\parallel x_1-p_1\parallel$, hence it follows from (18) that

$$\begin{array}{c}\frac{\left(\prod _{j=1}^{n+m-1}{\lambda }_j\right)\parallel x_1-p_1\parallel }{2}\delta \left(\frac{4}{\left(\prod _{j=1}^{n+m-1}{\lambda }_j\right)\parallel x_1-p_1\parallel }{b}_{n,m}\right)\le \left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel x_n-p_1\parallel -\parallel x_{n+m}-p_1\parallel \\ \left(\mathsf{\text{since}}t\left(1-t\right)\le \frac{1}{4}\mathsf{\text{for all}}t\in \left[0,1\right]\right).\end{array}$$

Since $\underset{n\to \infty }{lim}{\prod }_{j=1}^{n+m-1}{\lambda }_j=1$ and since δ(0) = 0 and lim _{n →∞} ||x _n - p₁|| exists, then the continuity of δ yields lim _{n →∞} b _n,m = 0 uniformly for all m ≥ 1. Observe that

$$\begin{array}{ll}\hfill {\sigma }_{n+m}\left(t\right)& \le \parallel t{x}_{n+m}+\left(1-t\right)p_1-p_2+\left(S_{n,m}\left(t{x}_n+\left(1-t\right)p_1\right)-t{S}_{n,m}{x}_n-\left(1-t\right)S_{n,m}{p}_1\right)\parallel \\ +\parallel S_{n,m}\left(t{x}_n+\left(1-t\right)p_1\right)-t{S}_{n,m}{x}_n-\left(1-t\right)S_{n,m}{p}_1\parallel \\ =\parallel S_{n,m}\left(t{x}_n+\left(1-t\right)p_1\right)-S_{n,m}{p}_2\parallel +b_{n,m}\\ \le \left(\prod _{j=n}^{n+m-1}{\lambda }_j\right)\parallel t{x}_n+\left(1-t\right)p_1-p_2\parallel +b_{n,m}\\ =\left(\prod _{j=n}^{n+m-1}{\lambda }_j\right){\sigma }_n\left(t\right)+b_{n,m}.\\ \hfill \text{(6) }\end{array}$$

Hence, lim sup _{n →∞} σ _n (t) ≤ lim inf _{n →∞} σ _n (t), this ensures that lim _{n →∞} σ _n (t) exists for all t ∈ (0, 1).

Now, apply Lemma 2.7 to conclude that {x _n } converges weakly to a fixed point of T.    □

Theorem 3.2 Let E be a real Banach space with the Fréchet differentiable norm, and let K be a nonempty, closed, and convex subset of E. Let T : K → K be an asymptotically κ-strictly pseudocontractive mapping for some 0 ≤ κ < 1 with sequence {κ _n } ⊂ [1, ∞) such that ${\sum }_{n=1}^{\infty }\left({\kappa }_n-1\right)<\infty$, let F(T) ≠ ∅. Let {α _n } be a real sequence satisfying the condition (11). Given x₁ ∈ K, let ${\left\{x_n\right\}}_{n=1}^{\infty }$ be the sequence generated by the modified Mann's algorithm (12). Then, the sequence {x _n } converges strongly to a fixed point of T if and only if

$$\underset{n\to \infty }{liminf}d\left(x_n,F\left(T\right)\right)=0,$$

where d(x _n , F(T)) = inf _{p ∈ F} ( _T )||x _n - p||.

Proof. In the real Banach space E with the Fréchet differentiable norm, we still have

$$\parallel x_{n+1}-p{\parallel }^2\le {\delta }_n\parallel x_n-p{\parallel }^2.$$

(19)

as we have already proved in Theorem 3.1. Thus, [d(x _{n +1} - p)]² ≤ δ _n [d(x _n - p)]² and it follows from Lemma 2.1 that lim _{n -∞} d(x _n , F(T )) exists.

Now if {x _n } converges strongly to a fixed point p of T, then lim _{n →∞} ||x _n - p|| = 0. Since

$$0\le d\left(x_n,F\left(T\right)\right)\le \parallel x_n-p\parallel ,$$

we have lim inf _{n →∞} d(x _n , F(T )) = 0.

Conversely, suppose lim inf _{n →∞} d(x _n , F(T)) = 0, then the existence of lim _{n →∞} d(x _n , F (T)) implies that lim _{n →∞} d(x _n , F(T)) = 0. Thus, for arbitrary ε > 0 there exists a positive integer n₀ such that $d\left(x_n,F\left(T\right)\right)<\frac{\epsilon }{2}$ for any n ≥ n₀.

From (19), we have

$$\parallel x_{n+1}-p{\parallel }^2\le \parallel x_n-p{\parallel }^2+M^2\left({\delta }_n-1\right),n\ge 1,$$

and for some M > 0, ||x _n - p|| < M. Now, an induction yields

$$\begin{array}{ll}\hfill \parallel x_n-p{\parallel }^2& \le \parallel x_{n-1}-p{\parallel }^2+M^2\left({\delta }_{n-1}-1\right)\\ \le \parallel x_{n-2}-p{\parallel }^2+M^2\left({\delta }_{n-2}-1\right)+M^2\left({\delta }_{n-1}-1\right)\\ \le \cdot \cdot \cdot \le \parallel x_l-p{\parallel }^2+M^2\sum _{j=l}^{n-1}\left({\delta }_j-1\right),n-1\ge l\ge 1,\end{array}$$

Since ${\sum }_{n=1}^{\infty }\left({\delta }_n-1\right)<\infty$, then there exists a positive integer n₁ such that ${\sum }_{j=n}^{\infty }\left({\delta }_j-1\right)<{\left(\frac{\epsilon }{2M}\right)}^2$, ∀n ≥ n₁. Choose N = max{n₀, n₁}, then for all n, m ≥ N + 1 and for all p ∈ F (T ) we have

$$\begin{array}{ll}\hfill \parallel x_n-x_m\parallel & \le \parallel x_n-p\parallel +\parallel x_m-p\parallel \\ \le {\left[\parallel x_N-p{\parallel }^2+M^2\sum _{j=N}^{n-1}\left({\delta }_j-1\right)\right]}^{\frac{1}{2}}+{\left[\parallel x_N-p{\parallel }^2+M^2\sum _{j=N}^{m-1}\left({\delta }_j-1\right)\right]}^{\frac{1}{2}}\\ \le {\left[\parallel x_N-p{\parallel }^2+M^2\sum _{j=N}^{\infty }\left({\delta }_j-1\right)\right]}^{\frac{1}{2}}+{\left[\parallel x_N-p{\parallel }^2+M^2\sum _{j=N}^{\infty }\left({\delta }_j-1\right)\right]}^{\frac{1}{2}}.\end{array}$$

Taking infimum over all p ∈ F(T), we obtain

$$\begin{array}{l}\left|\right|x_n-x_m\left|\right|\le {\{{[d(x_N,F(T))]}^2+M^2{\displaystyle \sum _{j=N}^{\infty }}({\delta }_j-1)\}}^{\frac{1}{2}}+{\{{[d(x_N,F(T))]}^2+M^2{\displaystyle \sum _{j=N}^{\infty }({\delta }_j-1)}\}}^{\frac{1}{2}}\\ <2{[{(\frac{\epsilon }{2})}^2+M^2{(\frac{\epsilon }{2M})}^2]}^{\frac{1}{2}}<2\epsilon .\end{array}$$

Thus, ${\left\{x_n\right\}}_{n=0}^{\infty }$ is Cauchy. We can also prove lim _{n →∞} ||x _n - Tx _n || = 0 as we have done in Theorem 3.1. Suppose lim _{n →∞} x _n = u. Then,

$$0\le \parallel u-Tu\parallel \le \parallel u-x_n\parallel +\parallel x_n-T{x}_n\parallel +L\parallel x_n-u\parallel \to 0,asn\to \infty .$$

Thus, u ∈ F(T).    □