Fixed point and weak convergence theorems for point-dependent λ-hybrid mappings in Banach spaces

Abstract

The purpose of this article is to study the fixed point and weak convergence problem for the new defined class of point-dependent λ-hybrid mappings relative to a Bregman distance D _f in a Banach space. We at first extend the Aoyama-Iemoto-Kohsaka-Takahashi fixed point theorem for λ-hybrid mappings in Hilbert spaces in 2010 to this much wider class of nonlinear mappings in Banach spaces. Secondly, we derive an Opial-like inequality for the Bregman distance and apply it to establish a weak convergence theorem for this new class of nonlinear mappings. Some concrete examples in a Hilbert space showing that our extension is proper are also given.

2010 MSC: 47H09; 47H10.

1 Introduction

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

(1.1) nonexpansive if ||Tx - Ty|| ≤ ||x - y||, ∀x, y ∈ C, cf. [1, 2];

(1.2) nonspreading if ||Tx - Ty||² ≤ ||x - y||² + 2 〈x - Tx, y - Ty〉, ∀x, y ∈ C, cf. [3–5];

(1.3) hybrid if ||Tx - Ty||² ≤ ||x - y||² + 〈x - Tx, y - Ty〉, ∀x, y ∈ C, cf. [3, 5–7].

As shown in [3], (1.2) is equivalent to

$$2\left|\right|Tx-Ty|{|}^2\le ||Tx-y|{|}^2+\left|\right|x-Ty|{|}^2$$

for all x, y ∈ C.

In 1965, Browder [1] established the following

Browder fixed point Theorem. Let C be a nonempty closed convex subset of a Hilbert space H, and let T : C → C be a nonexpansive mapping. Then, the following are equivalent:

• (a) There exists x ∈ C such that {Tⁿx} _n∈ℕ is bounded;

• (b) T has a fixed point.

The above result is still true for nonspreading mappings which was shown in Kohsaka and Takahashi [4]. (We call it the Kohsaka-Takahashi fixed point theorem.)

Recently, Aoyama et al. [8] introduced a new class of nonlinear mappings in a Hilbert space containing the classes of nonexpansive mappings, nonspreading mappings and hybrid mappings. For λ ∈ ℝ, they call a mapping T : C → H

(1.4) λ-hybrid if ||Tx - Ty||² ≤ ||x - y||² + λ 〈x - Tx, y - Ty〉, ∀x, y ∈ C.

And, among other things, they establish the following

Aoyama-Iemoto-Kohsaka-Takahashi fixed point Theorem. [8]Let C be a nonempty closed convex subset of a Hilbert space H, and let T : C → C be a λ-hybrid mapping. Then, the following are equivalent:

• (a) There exists x ∈ C such that {Tⁿx} _n∈ℕ is bounded;

• (b) T has a fixed point.

Obviously, T is nonexpansive if and only if it is 0-hybrid; T is nonspreading if and only if it is 2-hybrid; T is hybrid if and only if it is 1-hybrid.

Motivated by the above works, we extend the concept of λ-hybrid from Hilbert spaces to Banach spaces in the following way:

Definition 1.1. For a nonempty subset C of a Banach space X, a Gâteaux differentiable convex function f : X → (-∞,∞] and a function λ : C → ℝ, a mapping T : C → X is said to be point-dependent λ-hybrid relative to D _f if

(1.5) D _f (Tx, Ty) ≤ D _f (x, y) + λ(y) 〈x - Tx, f'(y) - f(Ty)〉, ∀x, y ∈ C,

where D _f is the Bregman distance associated with f and f'(x) denotes the Gâteaux derivative of f at x.

In this article, we study the fixed point and weak convergence problem for mappings satisfying (1.5). This article is organized in the following way: Section 2 provides preliminaries. We investigate the fixed point problem for point-dependent λ-hybrid mappings in Section 3, and we give some concrete examples showing that even in the setting of a Hilbert space, our fixed point theorem generalizes the Aoyama-Iemoto-Kohsaka-Takahashi fixed point theorem properly in Section 4. Section 5 is devoting to studying the weak convergence problem for this new class of nonlinear mappings.

2 Preliminaries

In what follows, X will be a real Banach space with topological dual X* and f : X → (-∞,∞] will be a convex function. $\mathcal{D}$ denotes the domain of f, that is,

$$\mathcal{D}=\left\{x\in X:f\left(x\right)<\infty \right\},$$

and ${\mathcal{D}}^{\circ }$ denotes the algebraic interior of $\mathcal{D}$, i.e., the subset of $\mathcal{D}$ consisting of all those points $x\in \mathcal{D}$ such that, for any y ∈ X \ {x}, there is z in the open segment (x, y) with $\left[x,z\right]\subseteq \mathcal{D}$. The topological interior of $\mathcal{D}$, denoted by $\mathsf{\text{Int}}\left(\mathcal{D}\right)$, is contained in ${\mathcal{D}}^{\circ }$. f is said to be proper provided that $\mathcal{D}\ne \varnothing$. f is called lower semicontinuous (l.s.c.) at x ∈ X if f(x) ≤ lim inf _y→x f (y). f is strictly convex if

$$f\left(\alpha x+\left(1-\alpha \right)y\right)<\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)$$

for all x, y ∈ X and α ∈ (0, 1).

The function f : X → (-∞, ∞] is said to be Gâteaux differentiable at x ∈ X if there is f'(x) ∈ X* such that

$$\underset{t\to 0}{lim}\frac{f\left(x+ty\right)-f\left(x\right)}{t}=〈y,f^{\prime }\left(x\right)〉$$

for all y ∈ X.

The Bregman distance D _f associated with a proper convex function f is the function $D_f:\mathcal{D}\times \mathcal{D}\to \left[0,\infty \right]$ defined by

$$D_f\left(y,x\right)=f\left(y\right)-f\left(x\right)+f^{\circ }\left(x,x-y\right),$$

(1)

where $f^{\circ }(x,x-y)=\underset{t\to 0^{+}}{\lim }\frac{f(x+t(x-y))-f(x)}{t}$. D _f (y, x) is finite valued if and only if $x\in {\mathcal{D}}^{\mathsf{\text{o}}}$, cf. Proposition 1.1.2 (iv) of [9]. When f is Gâteaux differentiable on D, (1) becomes

$$D_f\left(y,x\right)=f\left(y\right)-f\left(x\right)-〈y-x,f^{\prime }\left(x\right)〉,$$

(2)

and then the modulus of total convexity is the function ${\nu }_f:{\mathcal{D}}^{\circ }\times \left[0,\infty \right)\to \left[0,\infty \right]$ defined by

$${\nu }_f\left(x,t\right)=inf\left\{D_f\left(y,x\right):y\in \mathcal{D},\left|\right|y-x\left|\right|=t\right\}.$$

It is known that

$${\nu }_f\left(x,ct\right)\ge c{\nu }_f\left(x,t\right)$$

(3)

for all t ≥ 0 and c ≥ 1, cf. Proposition 1.2.2 (ii) of [9]. By definition it follows that

$$D_f\left(y,x\right)\ge {\nu }_f\left(x,\left|\right|y-x\left|\right|\right).$$

(4)

The modulus of uniform convexity of f is the function δ _f : [0, ∞) → [0, ∞] defined by

$${\delta }_f\left(t\right)=inf\left\{f\left(x\right)+f\left(y\right)-2f\left(\frac{x+y}{2}\right):x,y\in \mathcal{D},\left|\right|x-y\left|\right|\ge t\right\}.$$

The function f is called uniformly convex if δ _f (t) > 0 for all t > 0. If f is uniformly convex then for any ε > 0 there is δ > 0 such that

$$f\left(\frac{x+y}{2}\right)\le \frac{f\left(x\right)}{2}+\frac{f\left(y\right)}{2}-\delta$$

(5)

for all $x,y\in \mathcal{D}$ with ||x - y|| ≥ ε.

Note that for $y\in \mathcal{D}$ and $x\in {\mathcal{D}}^{\circ }$, we have

$$\begin{array}{c}f\left(x\right)+f\left(y\right)-2f\left(\frac{x+y}{2}\right)\\ =f\left(y\right)-f\left(x\right)-\frac{f\left(x+\frac{y-x}{2}\right)-f\left(x\right)}{\frac{1}{2}}\\ \le f\left(y\right)-f\left(x\right)-f^{\circ }\left(x,y-x\right)\le D_f\left(y,x\right),\end{array}$$

where the first inequality follows from the fact that the function t → f(x + tz) - f(x)/t is nondecreasing on (0, ∞). Therefore,

$${\nu }_f\left(x,t\right)\ge {\delta }_f\left(t\right)$$

(6)

whenever $x\in {\mathcal{D}}^{\circ }$ and t ≥ 0. For other properties of the Bregman distance D _f , we refer readers to [9].

The normalized duality mapping J from X to 2 ^X* is defined by

$$Jx=\left\{x^{*}\in X^{*}:〈x,x^{*}〉=\left|\right|x|{|}^2=|\left|x^{*}\right|{|}^2\right\}$$

for all x ∈ X.

When f(x) = ||x||² in a smooth Banach space X, it is known that f'(x) = 2J(x) for x ∈ X, cf. Corollaries 1.2.7 and 1.4.5 of [10]. Hence, we have

$$\begin{array}{ll}\hfill D_f\left(y,x\right)& =\left|\right|y|{|}^2-|\left|x\right|{|}^2-〈y-x,f^{\prime }\left(x\right)〉\\ =\left|\right|y|{|}^2-|\left|x\right|{|}^2-2〈y-x,Jx〉\\ =\left|\right|y|{|}^2+|\left|x\right|{|}^2-2〈y,Jx〉.\end{array}$$

Moreover, as the normalized duality mapping J in a Hilbert space H is the identity operator, we have

$$D_f\left(y,x\right)=\left|\right|y|{|}^2+|\left|x\right|{|}^2-2〈y,x〉=\left|\right|y-x|{|}^2.$$

Thus, in case λ is a constant function and f(x) = ||x||² in a Hilbert space, (1.5) coincides with (1.4). However, in general, they are different.

A function g : X → (-∞,∞] is said to be subdifferentiable at a point x ∈ X if there exists a linear functional x* ∈ X* such that

$$g\left(y\right)-g\left(x\right)\ge 〈y-x,x^{*}〉,\forall y\in X.$$

We call such x* the subgradient of g at x. The set of all subgradients of g at x is denoted by ∂g(x) and the mapping ∂g : X → 2 ^X* is called the subdifferential of g. For a l.s.c. convex function f, ∂f is bounded on bounded subsets of $\mathsf{\text{Int}}\left(\mathcal{D}\right)$ if and only if f is bounded on bounded subsets there, cf. Proposition 1.1.11 of [9]. A proper convex l.s.c. function f is Gâteaux differentiable at $x\in \mathsf{\text{Int}}\left(\mathcal{D}\right)$ if and only if it has a unique subgradient at x; in such case ∂f(x) = f'(x), cf. Corollary 1.2.7 of [10].

The following lemma will be quoted in the sequel.

Lemma 2.1. (Proposition 1.1.9 of [9]) If a proper convex function f : X → (-∞, ∞] is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ in a Banach space X, then the following statements are equivalent:

• (a) The function f is strictly convex on $Int\left(\mathcal{D}\right)$.

• (b) For any two distinct points$x,y\in Int\left(\mathcal{D}\right)$, one has D _f (y, x) > 0.

• (c) For any two distinct points $x,y\in Int\left(\mathcal{D}\right)$, one has

  $$〈x-y,f^{\prime }\left(x\right)-f^{\prime }\left(y\right)〉>0.$$

Throughout this article, F(T) will denote the set of all fixed points of a mapping T.

3 Fixed point theorems

In this section, we apply Lemma 2.1 to study the fixed point problem for mappings satisfying (1.5).

Theorem 3.1. Let X be a reflexive Banach space and let f : X → (-∞,∞] be a l.s.c. strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of $Int\left(\mathcal{D}\right)$. Suppose $C\subseteq Int\left(\mathcal{D}\right)$ is a nonempty closed convex subset of X and T: C → C is point-dependent λ-hybrid relative to D _f for some function λ : C → ℝ. For x ∈ C and any n ∈ ℕ define

$$S_{n}x=\frac{1}{n}\sum _{k=0}^{n-1}{T}^{k}x,$$

where T⁰is the identity mapping on C. If {Tⁿx}_n∈ℕis bounded, then every weak cluster point of {S _n x}_n∈ℕis a fixed point of T.

Proof. Since T is point-dependent λ-hybrid relative to D _f , we have, for any y ∈ C and k ∈ ℕ ∪ {0},

$$\begin{array}{c}0\le D_f\left(T^{k}x,y\right)-D_f\left(T^{k+1}x,Ty\right)+\lambda \left(y\right)〈T^{k}x-T^{k+1}x,f^{\prime }\left(y\right)-f^{\prime }\left(Ty\right)〉\\ =f\left(T^{k}x\right)-f\left(y\right)-〈T^{k}x-y,f^{\prime }\left(y\right)〉-f\left(T^{k+1}x\right)+f\left(Ty\right)+〈T^{k+1}x-Ty,f^{\prime }\left(Ty\right)〉\\ +\lambda \left(y\right)〈T^{k}x-T^{k+1}x,f^{\prime }\left(y\right)-f^{\prime }\left(Ty\right)〉\\ =\left[f\left(T^{k}x\right)-f\left(T^{k+1}x\right)\right]+\left[f\left(Ty\right)-f\left(y\right)\right]+〈\lambda \left(y\right)\left(T^{k}x-T^{k+1}x\right)-T^{k}x+y,f^{\prime }\left(y\right)〉\\ +〈T^{k+1}x-Ty-\lambda \left(y\right)\left(T^{k}x-T^{k+1}x\right),f^{\prime }\left(Ty\right)〉.\end{array}$$

Summing up these inequalities with respect to k = 0, 1,..., n - 1, we get

$$\begin{array}{c}0\le \left[f\left(x\right)-f\left(T^{n}x\right)\right]+n\left[f\left(Ty\right)-f\left(y\right)\right]+〈\lambda \left(y\right)\left(x-T^{n}x\right)+ny-n{S}_{n}x,f^{\prime }\left(y\right)〉\\ +〈\left(n+1\right)S_{n+1}x-x-nTy-\lambda \left(y\right)\left(x-T^{n}x\right),f^{\prime }\left(Ty\right)〉.\end{array}$$

Dividing the above inequality by n, we have

$$\begin{array}{c}0\le \frac{f\left(x\right)-f\left(T^{n}x\right)}{n}+\left[f\left(Ty\right)-f\left(y\right)\right]+〈\frac{\lambda \left(y\right)\left(x-T^{n}x\right)}{n}+y-S_{n}x,f^{\prime }\left(y\right)〉\\ +〈\frac{n+1}{n}{S}_{n+1}x-\frac{x}{n}-Ty-\frac{\lambda \left(y\right)\left(x-T^{n}x\right)}{n},f^{\prime }\left(Ty\right)〉.\end{array}$$

(7)

Since {Tⁿx}_n∈ℕis bounded, {S _n x}_n∈ℕis bounded, and so, in view of X being reflexive, it has a subsequence ${\left\{S_{n_i}x\right\}}_{i\in \mathbb{N}}$ so that $S_{n_i}x$ converges weakly to some v ∈ C as n _i → ∞. Replacing n by n _i in (7), and letting n _i → ∞, we obtain from the fact that {Tⁿx}_n∈ℕand {f(Tⁿx)}_n∈ℕare bounded that

$$0\le f\left(Ty\right)-f\left(y\right)+〈y-v,f^{\prime }\left(y\right)〉+〈v-Ty,f^{\prime }\left(Ty\right)〉.$$

(8)

Putting y = v in (8), we get

$$0\le f\left(Tv\right)-f\left(v\right)+〈v-Tv,f^{\prime }\left(Tv\right)〉,$$

that is,

$$0\le -D_f\left(v,Tv\right),$$

from which follows that D _f (v, Tv) = 0. Therefore Tv = v by Lemma 2.1. □

The following theorem comes from Theorem 3.1 immediately.

Theorem 3.2. Let X be a reflexive Banach space and let f : X → (-∞,∞] be a l.s.c. strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of $Int\left(\mathcal{D}\right)$. Suppose $C\subseteq Int\left(\mathcal{D}\right)$ is a nonempty closed convex subset of X and T: C → C is point-dependent λ-hybrid relative to D _f for some function λ : C → ℝ. Then, the following two statements are equivalent:

• (a) There is a point x ∈ C such that {Tⁿx}_n∈ℕ is bounded.

• (b) F(T) ≠ ∅.

Taking λ(x) = λ, a constant real number, for all x ∈ C and noting the function f(x) = ||x||² in a Hilbert space H satisfies all the requirements of Theorem 3.2, the corollary below follows immediately.

Corollary 3.3. [8]Let C be a nonempty closed convex subset of Hilbert space H and suppose T : C → C is λ-hybrid. Then, the following two statements are equivalent:

• (a) There exists x ∈ C such that {Tⁿ (x)}_n∈ℕ is bounded.

• (b) T has a fixed point.

We now show that the fixed point set F(T) is closed and convex under the assumptions of Theorem 3.2.

A mapping T : C → X is said to be quasi-nonexpansive with respect to D _f if F(T) ≠ ∅ and D _f (v, Tx) ≤ D _f (v, x) for all x ∈ C and all v ∈ F(T).

Lemma 3.4. Let f : X → (-∞,∞] be a proper strictly convex function on a Banach space X so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$, and let $C\subseteq Int\left(\mathcal{D}\right)$ be a nonempty closed convex subset of X. If T: C → C is quasi-nonexpansive with respect to D _f , then F(T) is a closed convex subset.

Proof. Let $x\in \overline{F(T)}$ and choose {x _n }_n∈ℕ⊆ F(T) such that x _n → x as n → ∞. By the continuity of D _f (·, Tx) and D _f (x _n , T _x ) ≤ D _f (x _n , x), we have

$$D_f\left(x,Tx\right)=\underset{n\to \infty }{lim}{D}_f\left(x_n,Tx\right)\le \underset{n\to \infty }{lim}{D}_f\left(x_n,x\right)=D_f\left(x,x\right)=0.$$

Thus, due to the strict convexity of f, it follows from Lemma 2.2 that Tx = x. This shows F(T) is closed. Next, let x, y ∈ F(T) and α ∈ [0, 1]. Put z = αx + (1 - α)y. We show that Tz = z to conclude F(T) is convex. Indeed,

$$\begin{array}{c}{D}_f\left(z,Tz\right)\\ =f\left(z\right)-f\left(Tz\right)-〈z-Tz,f^{\prime }\left(Tz\right)〉\\ =f\left(z\right)+\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]-f\left(Tz\right)-〈z-Tz,f^{\prime }\left(Tz\right)〉-\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ =f\left(z\right)+\alpha \left[f\left(x\right)-f\left(Tz\right)-〈x-Tz,f^{\prime }\left(Tz\right)〉\right]\\ +\left(1-\alpha \right)\left[f\left(y\right)-f\left(Tz\right)-〈y-Tz,f^{\prime }\left(Tz\right)〉\right]-\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ =f\left(z\right)+\alpha D_f\left(x,Tz\right)+\left(1-\alpha \right)D_f\left(y,Tz\right)-\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ \le f\left(z\right)+\alpha D_f\left(x,z\right)+\left(1-\alpha \right)D_f\left(y,z\right)-\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ =f\left(z\right)+\alpha \left[f\left(x\right)-f\left(z\right)-〈x-z,f^{\prime }\left(z\right)〉\right]+\left(1-\alpha \right)\left[f\left(y\right)-f\left(z\right)-〈y-z,f^{\prime }\left(z\right)〉\right]\\ -\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ =f\left(z\right)+\alpha f\left(x\right)-\alpha f\left(z\right)-〈\alpha x-\alpha z,f^{\prime }\left(z\right)〉+\left(1-\alpha \right)f\left(y\right)-\left(1-\alpha \right)f\left(z\right)\\ -〈\left(1-\alpha \right)y-\left(1-\alpha \right)z,f^{\prime }\left(z\right)〉-\left[\alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right)\right]\\ =-〈\alpha x+\left(1-\alpha \right)y-\left(\alpha z+\left(1-\alpha \right)z\right),f^{\prime }\left(z\right)〉\\ =-〈0,f^{\prime }\left(z\right)〉=0.\end{array}$$

Therefore, Tz = z by the strictly convex of f. This completes the proof. □

Proposition 3.5. Let f : X → (-∞,∞] be a proper strictly convex function on a reflexive Banach space X so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of Int(D), and let $C\subseteq Int\left(\mathcal{D}\right)$ be a nonempty closed convex subset of X. Suppose T: C → C is point-dependent λ-hybrid relative to D _f for some function λ : C → ℝ and has a point x₀ ∈ C such that {Tⁿ (x₀)}_n∈ℕis bounded. Then, T is quasi-nonexpansive with respect to D _f , and therefore, F(T) is a nonempty closed convex subset of C.

Proof. In view of Theorem 3.2, F(T) ≠ ∅. Now, for any v ∈ F(T) and any y ∈ C, as T is point-dependent λ-hybrid relative to D _f , we have

$$\begin{array}{ll}\hfill D_f\left(v,Ty\right)& =D_f\left(Tv,Ty\right)\\ \le D_f\left(v,y\right)+\lambda \left(y\right)〈v-Tv,f^{\prime }\left(y\right)-f^{\prime }\left(Ty\right)〉\\ =D_f\left(v,y\right)\end{array}$$

for all y ∈ C, so T is quasi-nonexpansive with respect to D _f , and hence, F(T) is a nonempty closed convex subset of C by Lemma 3.4. □

For the remainder of this section, we establish a common fixed point theorem for a commutative family of point-dependent λ-hybrid mappings relative to D _f .

Lemma 3.6. Let X be a reflexive Banach space and let f : X → (-∞,∞] be a l.s.c. strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of $Int\left(\mathcal{D}\right)$. Suppose $C\subseteq Int\left(\mathcal{D}\right)$ is a nonempty bounded closed convex subset of X and{T₁, T₂,..., T _N } is a commutative finite family of point-dependent λ-hybrid mappings relative to D _f for some function λ : C → ℝ from C into itself. Then {T₁, T₂,..., T _N } has a common fixed point.

Proof. We prove this lemma by induction with respect to N. To begin with, we deal with the case that N = 2. By Proposition 3.5, we see that F(T₁) and F(T₂) are nonempty bounded closed convex subsets of X. Moreover, F(T₁) is T₂-invariant. Indeed, for any v ∈ F(T₁), it follows from T₁T₂ = T₂T₁ that T₁T₂v = T₂T₁v = T₂v, which shows that T₂v ∈ F(T₁). Consequently, the restriction of T₂ to F(T₁) is point-dependent λ-hybrid relative to D _f , and hence by Theorem 3.2, T₂ has a fixed point u ∈ F(T₁), that is, u ∈ F(T₁) ∩ F(T₂).

By induction hypothesis, assume that for some n ≥ 2, $E={\cap }_{k=1}^{n}F\left(T_k\right)$ is nonempty. Then, E is a nonempty closed convex subset of X and the restriction of T _n+1 to E is a point-dependent λ-hybrid mapping relative to D _f from E into itself. By Theorem 3.2, T _n+1 has a fixed point in X. This shows that E ∩ F(T _n+1 ) ≠ ∅, that is, ${\cap }_{k=1}^{n+1}F\left(T_k\right)\ne \varnothing$, completing the proof. □.

Theorem 3.7. Let X be a reflexive Banach space and let f : X → (-∞,∞] be a l.s.c. strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$. Suppose $C\subseteq Int\left(\mathcal{D}\right)$ is a nonempty bounded closed convex subset of X and{T _i } _i∈I is a commutative family of point-dependent λ-hybrid mappings relative to D _f for some function λ : C → ℝ from C into itself. Then, {T _i } _i∈I has a common fixed point.

Proof. Since C is a nonempty bounded closed convex subset of the reflexive Banach space X, it is weakly compact. By Proposition 3.5, each F(T _i ) is a nonempty weakly compact subset of C. Therefore, the conclusion follows once we note that {F(T _i )} _i∈I has the finite intersection property by Lemma 3.6. □.

4 Examples

In this section, we give some concrete examples for our fixed point theorem. At first, we need a lemma.

Lemma 4.1. Let h and k be two real numbers in [0, 1]. Then, the following two statements are true.

• (a) (h ² - k ²)² - (h - k)² ≥ 0, if $\frac{h+k}{2}>0.5$.

• (b) (h ² - k ²)² - (h - k)² ≤ 0, if $\frac{h+k}{2}\le 0.5$.

Proof. First, we represent h and k by

$$h=0.5+a,\mathsf{\text{where}}-0.5\le a\le 0.5,$$

and

$$k=0.5+b,\mathsf{\text{where}}-0.5\le b\le 0.5.$$

Then, we have

$${\left(h^2-k^2\right)}^2-{\left(h-k\right)}^2={\left(a-b\right)}^2\left(a+b\right)\left(a+b+2\right).$$

If $\frac{h+k}{2}>0.5$, then a + b > 0, and so through the above equation, we obtain that (h² - k²)² - (h - k)² ≥ 0. On the other hand, $\frac{h+k}{2}\le 0.5$ implies a + b ≤ 0, and hence, (h² - k²)² - (h - k)² ≤ 0.

Example 4.2. Let $C=\left\{x\in l^2\left(\mathbb{N}\right):x=\left(x_1,x_2,\dots ,x_n,\dots \right),0\le x_i\le 1-\frac{1}{i+1}\right\}$ and δ be a positive number so small that $\sqrt{\delta }<0.5$. Define a mapping T: C → C by

$$Tx=\left(T{x}_1,T{x}_2,\dots ,T{x}_n,\dots \right):T{x}_i=\left\{\begin{array}{cc}\hfill x_i^2,\hfill & \hfill if\sqrt{\delta }<x_i\le 1-\frac{1}{i+1};\hfill \\ \hfill \delta ,\hfill & \hfill if\delta <x_i\le \sqrt{\delta };\hfill \\ \hfill x_i,\hfill & \hfill if0\le x_i\le \delta .\hfill \end{array}\right.$$

Then for any λ ∈ ℝ, T is not λ-hybrid. However, for each x ∈ C, if we let $n_x=min\left\{n:\sum _{i=n+1}^{\infty }{x}_i^2\le {\delta }^2\right\}$ and define λ: C → ℝ by

$$\lambda \left(x\right)=\frac{1}{{\left(\frac{1}{n_x+1}-\frac{1}{{\left(n_x+1\right)}^2}\right)}^2},$$

then T is point-dependent λ-hybrid, that is,

$$\left|\right|Tx-Ty|{|}^2\le ||x-y|{|}^2+\lambda \left(y\right)〈x-Tx,y-Ty〉$$

(9)

for all x, y ∈ C. Therefore, we can apply Theorem 3.2 to conclude that T has a fixed point, while the Aoyama-Iemoto-Kohsaka-Takahashi fixed point theorem fails to give us the desired conclusion.

Proof. Let x and y be two elements from C so that the m^th coordinate of x is $1-\frac{1}{m+1}$ the m^th coordinate of y is 0.5 and the rest coordinates of x and y are zero. We have

$$\begin{array}{c}\left|\right|Tx-Ty|{|}^2-||x-y|{|}^2-m〈x-Tx,y-Ty〉\\ ={\left[{\left(1-\frac{1}{m+1}\right)}^2-{\left(0.5\right)}^2\right]}^2-{\left[\left(1-\frac{1}{m+1}\right)-0.5\right]}^2\\ -m\left[\left(1-\frac{1}{m+1}\right)-{\left(1-\frac{1}{m+1}\right)}^2\right]\left[0.5-{\left(0.5\right)}^2\right]\\ =\frac{9}{16}-\frac{2}{m+1}+\frac{9}{2{\left(m+1\right)}^2}-\frac{4}{{\left(m+1\right)}^3}+\frac{1}{{\left(m+1\right)}^4}-\frac{m^2}{4{\left(m+1\right)}^2}\\ \to \frac{5}{16}\mathsf{\text{ as }}m\to \infty .\end{array}$$

Since the value of above equality is always positive as m is large enough, we conclude that there is no constant λ to satisfy the inequality:

$$\left|\right|Tx-Ty|{|}^2\le ||x-y|{|}^2+\lambda 〈x-Tx,y-Ty〉$$

for all x, y ∈ C.

It remains to show that T satisfies the inequality (9). We can rewrite the inequality as

$$\sum _{i=1}^{\infty }{\left(T{x}_i-T{y}_i\right)}^2\le \sum _{i=1}^{\infty }{\left(x_i-y_i\right)}^2+\sum _{i=1}^{\infty }\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).$$

Thus, if we can show that for all i ∈ ℕ,

$${\left(T{x}_i-T{y}_i\right)}^2\le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right),$$

(10)

then the assertion follows. We prove inequality (10) holds for all i ∈ ℕ by considering the following two cases: (I) i > min{n _x , n _y } and (II) i ≤ min{n _x , n _y }.

● Case (I). i > min{n _x , n _y }.

In this case, at least one of x _i and y _i is less than or equal to δ. Suppose that 0 ≤ x _i ≤ δ. There are three subcases to discuss.

(I-1): If $\sqrt{\delta }<y_i\le 1-\frac{1}{i+1}$, then we have

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(x_i-y_i^2\right)}^2& \le {\left(x_i-y_i\right)}^2\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

(I-2): $\delta <y_i\le \sqrt{\delta }$, then we have

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(x_i-\delta \right)}^2& \le {\left(x_i-y_i\right)}^2\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

(I-3): If 0 ≤ y _i ≤ δ, then we have

$${\left(T{x}_i-T{y}_i\right)}^2={\left(x_i-y_i\right)}^2\le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).$$

The case that 0 ≤ y _i ≤ δ can be proved in the same manner.

● Case (II). i ≤ min{n _x , n _y }.

In this case, there are 9 subcases to discuss.

(II-1): $\sqrt{\delta }<x_i\le 1-\frac{1}{i+1}$ and $\sqrt{\delta }<y_i\le 1-\frac{1}{i+1}$.

If $\frac{x_i+y_i}{2}\le 0.5$, it follows from Lemma 4.1 that

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(x_i^2-y_i^2\right)}^2& \le {\left(x_i-y_i\right)}^2\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

If $\frac{x_i+y_i}{2}>0.5$, then both x _i and y _i are greater than $\frac{1}{i+1}$, and so by considering the graph of the function g(z) = z - z² in ℝ, which is symmetric to the line L : x = 0.5, we have

$$\left(x_i-x_i^2\right)\ge \left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2\ge \left(\frac{1}{n_y+1}\right)-{\left(\frac{1}{n_y+1}\right)}^2$$

and

$$\left(y_i-y_i^2\right)\ge \left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2\ge \left(\frac{1}{n_y+1}\right)-{\left(\frac{1}{n_y+1}\right)}^2.$$

Consequently, we obtain

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(x_i^2-y_i^2\right)}^2\le 1& \le \frac{1}{{\left(\frac{1}{n_y+1}-\frac{1}{{\left(n_y+1\right)}^2}\right)}^2}\left(x_i-x_i^2\right)\left(y_i-y_i^2\right)\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

(II-2): $\delta <x_i\le \sqrt{\delta }$ and $\sqrt{\delta }<y_i\le 1-\frac{1}{i+1}$.

If y _i ≤ 0.5, then $\frac{x_i+y_i}{2}<0.5$. Thus, from Lemma 4.1, we have

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(\delta -y_i^2\right)}^2& \le {\left(x_i^2-y_i^2\right)}^2\\ \le {\left(x_i-y_i\right)}^2\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

If y _i > 0.5, we have either

$$\delta <x_i\le \delta +\left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2$$

or

$$\delta +\left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2<x_i\le \sqrt{\delta }.$$

When $\delta <x_i\le \delta +\left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2$, by considering the graph of the function g(z) = z - z² in ℝ, we have

$$y_i-y_i^2\ge \left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2\ge x_i-\delta .$$

and thus, we obtain

$$y_i-x_i\ge y_i^2-\delta >0.$$

Therefore,

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2& ={\left(\delta -y_i^2\right)}^2\\ \le {\left(x_i-y_i\right)}^2\le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

When $\delta +\left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2<x_i\le \sqrt{\delta }$, both of x _i -δ and $y_i-y_i^2$ are greater than $\left(\frac{1}{i+1}\right)-{\left(\frac{1}{i+1}\right)}^2$ and thus also greater than $\left(\frac{1}{n_y+1}\right)-{\left(\frac{1}{n_y+1}\right)}^2$.

Therefore,

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(\delta -y_i^2\right)}^2\le 1& \le \frac{1}{{\left(\frac{1}{n_y+1}-\frac{1}{{\left(n_y+1\right)}^2}\right)}^2}\left(x_i-\delta \right)\left(y_i-y_i^2\right)\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

Likely, we can prove the case:

(II-3): $\sqrt{\delta }<x_i\le 1-\frac{1}{i+1}$ and $\delta <y_i\le \sqrt{\delta }$.

(II-4): 0 ≤·x _i ≤ δ and $\sqrt{\delta }<y_i\le 1-\frac{1}{i+1}$.

Then, we have

$$\begin{array}{ll}\hfill {\left(T{x}_i-T{y}_i\right)}^2={\left(x_i-y_i^2\right)}^2& \le {\left(x_i-y_i\right)}^2\\ \le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).\end{array}$$

Similarly, we can prove the case:

(II-5): $\sqrt{\delta }<x_i\le 1-\frac{1}{i+1}$ and 0 ≤ y _i ≤ δ.

(II-6): $\delta <x_i\le \sqrt{\delta }$ and $\delta <y_i\le \sqrt{\delta }$.

In this case, we have

$${\left(T{x}_i-T{y}_i\right)}^2={\left(\delta -\delta \right)}^2=0\le {\left(x_i-y_i\right)}^2+\lambda \left(y\right)\left(x_i-T{x}_i\right)\left(y_i-T{y}_i\right).$$

(II-7): 0 ≤ x _i ≤ δ and $\delta <y_i\le \sqrt{\delta }$.

This case can be treated as (I-2).

(II-8): 0 ≤ x _i ≤ δ and 0 ≤ y _i ≤ δ.

This case can be treated as (I-3).

(II-9): $\delta <x_i\le \sqrt{\delta }$ and 0 ≤ y _i ≤ δ.

This case can be treated as (I-2). □

To end this section, we give another example which shows that the concept of a nonspreading mapping in the sense of (1.2) is generally different from that of a 2-hybrid mapping relative to some D _f in Hilbert spaces.

Example 4.3. Define f : ℝ → ℝ by f(x) = x¹⁰for all x ∈ ℝ, and define T : [0, 0.85] → [0, 0.85] by Tx = x²for all x ∈ [0, 0.85]. Then, T is neither nonexpansive nor nonspreading, but it is λ-hybrid relative to D _f for any λ ≥ 0. Thus, we can apply Theorem 3.2 to conclude T has a fixed point, while both of the Browder Fixed Point Theorem and the Kohsaka-Takahashi fixed point theorem fail.

Proof. It is easy to check that T is not nonexpansive. As for not nonspreading, taking x = 0.85 and y = 0.5, we have

$$\left|\right|Tx-Ty|{|}^2={\left(x^2-y^2\right)}^2={\left[{\left(0.85\right)}^2-{\left(0.5\right)}^2\right]}^2=0.22325625$$

while

$$\begin{array}{c}\left|\right|x-y|{|}^2+2〈x-Tx,y-Ty〉\\ ={\left(x-y\right)}^2+2\left(x-x^2\right)\left(y-y^2\right)\\ ={\left(0.85-0.5\right)}^2+2\left[0.85-{\left(0.85\right)}^2\right]\left[0.5-{\left(0.5\right)}^2\right]=0.18625.\end{array}$$

Hence, T is not nonspreading in the sense of (1.2). It remains to show that for any λ ≥ 0, T is λ-hybrid relative to D _f . Note at first that, for all λ ≥ 0 and for all x, y ∈ [0, 0.85],

$$\begin{array}{c}\lambda 〈x-Tx,f^{\prime }\left(y\right)-f^{\prime }\left(Ty\right)〉\\ =\lambda \left(x-x^2\right)\left(10y^9-10y^{18}\right)\ge 0.\end{array}$$

Hence, it suffices to prove that T is 0-hybrid relative to D _f , that is, to show that

$$D_f\left(Tx,Ty\right)-D_f\left(x,y\right)\le 0,\forall x,y\in \left[0,0.85\right].$$

Fixed any x ∈ [0, 0.85], let h(y) = D _f (T _x , T _y ) - D _f (x, y). Then

$$\begin{array}{ll}\hfill h\left(y\right)& =f\left(Tx\right)-f\left(Ty\right)-〈Tx-Ty,f^{\prime }\left(Ty\right)〉-\left[f\left(x\right)-f\left(y\right)-〈x-y,f^{\prime }\left(y\right)〉\right]\\ =x^{20}+9y^{20}-10x^2{y}^{18}-x^{10}-9y^{10}+10x{y}^9.\end{array}$$

We have

$$\begin{array}{ll}\hfill h^{\prime }\left(y\right)& =180y^{19}-180x^2{y}^{17}-90y^9+90x{y}^8\\ =90y^8\left(2y^{11}-2x^2{y}^9-y+x\right)\\ =90y^8\left[2y^9\left(y^2-x^2\right)-\left(y-x\right)\right]\\ =90y^8\left[2y^9\left(y+x\right)\left(y-x\right)-\left(y-x\right)\right]\\ =90y^8\left(y-x\right)\left[2y^9\left(y+x\right)-1\right].\end{array}$$

Since y and x are in [0, 0.85], one has

$$2y^9\left(y+x\right)-1<2{\left(0.85\right)}^9\left(0.85+0.85\right)-1<0,$$

and hence

$$h^{\prime }\left(y\right)\left\{\begin{array}{cc}\hfill \ge 0,\hfill & \hfill ify\le x;\hfill \\ \hfill \le 0,\hfill & \hfill ify>x.\hfill \end{array}\right.$$

Moreover, we know h(y) = 0 if x = y. Therefore, h(y) is always less than or equal to zero and we have proved that D _f (Tx, Ty) - D _f (x, y) ≤ 0 for all x, y ∈ [0, 0.85]. □

5 Weak convergence theorems

In this section, we discuss the demiclosedness and the weak convergence problem of point-dependent λ-hybrid relative to D _f . We denote the weak convergence and strong convergence of a sequence {x _n } to v in a Banach space by x _n ⇀ v and x _n → v, respectively. For a nonempty closed convex subset C of a Banach space X, a mapping T : C → X is demiclosed if for any sequence {x _n } in C with x _n ⇀ v and x _n - Tx _n → 0, one has Tv = v.

We first derive an Opial-like inequality for the Bregman distance. For the Opial's inequality, we refer readers to Lemma 1 of [11].

Lemma 5.1. Suppose f : X → (-∞,∞] is a proper strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ in a Banach space X and{x _n }_n∈ℕis a sequence in$\mathcal{D}$such that x _n ⇀ v for some $v\in Int\left(\mathcal{D}\right)$. Then

$$\underset{n\to \infty }{liminf}{D}_f\left(x_n,v\right)<\underset{n\to \infty }{liminf}{D}_f\left(x_n,y\right),\forall y\in \mathsf{\text{Int}}\left(\mathcal{D}\right)withy\ne v.$$

Proof. Since

$$\begin{array}{l}{D}_f(x_n,v)-D_f(x_n,y)\\ =f(x_n)-f(v)-〈x_n-v,f^{\prime }(v)〉-[f(x_n)-f(y)-〈x_n-y,f^{\prime }(y)〉]\\ =f(x_n)-f(v)-〈x_n-v,f^{\prime }(v)〉-f(x_n)+f(y)+〈x_n-y,f^{\prime }(y)〉]\\ +〈x_n-v,f^{\prime }(y)〉-〈x_n-v,f^{\prime }(y)〉\\ =-[f(v)-f(y)-〈v-y,f^{\prime }(y)〉]+〈x_n-v,f^{\prime }(y)-f^{\prime }(v)〉\\ =-D_f(v,y)+〈x_n-v,f^{\prime }(y)-f^{\prime }(v)〉\end{array}$$

and x _n ⇀ v, we have

$$\underset{n\to \infty }{lim}\left[D_f\left(x_n,v\right)-D_f\left(x_n,y\right)\right]=-D_f\left(v,y\right).$$

Consequently,

$$\begin{array}{ll}\hfill \underset{n\to \infty }{liminf}{D}_f\left(x_n,v\right)& =\underset{n\to \infty }{liminf}\left[\left(D_f\left(x_n,v\right)-D_f\left(x_n,y\right)\right)+D_f\left(x_n,y\right)\right]\\ =\underset{n\to \infty }{lim}\left(D_f\left(x_n,v\right)-D_f\left(x_n,y\right)\right)+\underset{n\to \infty }{liminf}{D}_f\left(x_n,y\right)\\ =-D_f\left(v,y\right)+\underset{n\to \infty }{liminf}{D}_f\left(x_n,y\right),\end{array}$$

and hence in view of D _f (v, y) > 0 for y ≠ v we obtain

$$\underset{n\to \infty }{liminf}{D}_f\left(x_n,v\right)<\underset{n\to \infty }{liminf}{D}_f\left(x_n,y\right).$$

□

Proposition 5.2. Let f : X → (-∞,∞] be a strictly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of $Int\left(\mathcal{D}\right)$. Suppose C is a closed convex subset of $Int\left(\mathcal{D}\right)$ and T: C → C is point-dependent λ-hybrid relative to D _f for some λ : C → ℝ. Then T is demiclosed.

Proof. Let {x _n } be any sequence in C with x _n ⇀ v and x _n - Tx _n → 0. We have to show that Tv = v. Since f is bounded on bounded subsets, by Proposition 1.1.11 of [9] there exists a constant M > 0 such that

$$max\left\{sup\left\{\left|\right|f^{\prime }\left(x_n\right)\left|\right|:n\in \mathbb{N}\right\},\left|\right|\lambda \left(v\right)\left|\right|,\left|\right|f^{\prime }\left(Tv\right)\left|\right|,\left|\right|f^{\prime }\left(v\right)\left|\right|\right\}\le M.$$

Rewrite D _f (x _n , Tv) as

$$\begin{array}{ll}\hfill D_f\left(x_n,Tv\right)& =f\left(x_n\right)-f\left(Tv\right)-〈x_n-Tv,f^{\prime }\left(Tv\right)〉\\ =f\left(x_n\right)+f\left(T{x}_n\right)-f\left(T{x}_n\right)-f\left(Tv\right)-〈x_n-Tv,f^{\prime }\left(Tv\right)〉\\ +〈T{x}_n-Tv,f^{\prime }\left(Tv\right)〉-〈T{x}_n-Tv,f^{\prime }\left(Tv\right)〉\\ =\left[f\left(T{x}_n\right)-f\left(Tv\right)-〈T{x}_n-Tv,f^{\prime }\left(Tv\right)〉\right]+f\left(x_n\right)-f\left(T{x}_n\right)\\ +〈T{x}_n-x_n,f^{\prime }\left(Tv\right)〉\\ =D_f\left(T{x}_n,Tv\right)+f\left(x_n\right)-f\left(T{x}_n\right)+〈T{x}_n-x_n,f^{\prime }\left(Tv\right)〉.\end{array}$$

(11)

Noting f(x _n ) - f(Tx _n ) ≤ 〈x _n - Tx _n , f'(x _n )〉 and T is point-dependent λ-hybrid relative to D _f , we have from (11) that

$$\begin{array}{c}{D}_f\left(x_n,Tv\right)\\ \le D_f\left(T{x}_n,Tv\right)+〈x_n-T{x}_n,f^{\prime }\left(x_n\right)〉-〈x_n-T{x}_n,f^{\prime }\left(Tv\right)〉\\ \le D_f\left(x_n,v\right)+\lambda \left(v\right)〈x_n-T{x}_n,f^{\prime }\left(v\right)-f^{\prime }\left(Tv\right)〉+〈x_n-T{x}_n,f^{\prime }\left(x_n\right)-f^{\prime }\left(Tv\right)〉\\ \le D_f\left(x_n,v\right)+\left[\left|\lambda \left(v\right)\right|\left(\left|\right|f^{\prime }\left(v\right)\left|\right|+\left|\right|f^{\prime }\left(Tv\right)\left|\right|\right)+\left(\left|\right|f^{\prime }\left(x_n\right)\left|\right|+\left|\right|f^{\prime }\left(Tv\right)\left|\right|\right)\right]\left|\right|x_n-T{x}_n\left|\right|\\ \le D_f\left(x_n,v\right)+2M\left(M+1\right)\left|\right|x_n-T{x}_n\left|\right|.\end{array}$$

(12)

If Tv ≠ v, then Lemma 5.1 and (12) imply that

$$\begin{array}{c}\underset{n\to \infty }{liminf}{D}_f\left(x_n,v\right)\\ <\underset{n\to \infty }{liminf}{D}_f\left(x_n,Tv\right)\\ \le \underset{n\to \infty }{liminf}\left[D_f\left(x_n,v\right)+2M\left(M+1\right)\left|\right|x_n-T{x}_n\left|\right|\right]=\underset{n\to \infty }{liminf}{D}_f\left(x_n,v\right),\end{array}$$

a contradiction. This completes the proof. □

A mapping T : C → C is said to be asymptotically regular if, for any x ∈ C, the sequence {Tⁿ⁺¹x - Tⁿx} tends to zero as n → ∞.

Theorem 5.3. Suppose the following conditions hold:

(5.3.1) f : X → (-∞,∞] is l.s.c. uniformly convex function so that it is Gâteaux differentiable on $Int\left(\mathcal{D}\right)$ and is bounded on bounded subsets of $Int\left(\mathcal{D}\right)$ in a reflexive Banach space X.

(5.3.2) $C\subseteq Int\left(\mathcal{D}\right)$is a closed convex subset of X.

(5.3.3) T : C → C is point-dependent λ-hybrid relative to D _f for some λ : C → ℝ and is asymptotically regular with a bounded sequence {Tⁿx₀}_n∈ℕfor some x₀ ∈ C.

(5.3.4) The mapping x → f'(x) for x ∈ X is weak-to-weak* continuous.

Then for any x ∈ C, {Tⁿx}_n∈ℕis weakly convergent to an element v ∈ F(T).

Proof. Let v ∈ F(T) and x ∈ C. If {Tⁿx}_n∈ℕis not bounded, then there is a subsequence ${\left\{T^{n_i}x\right\}}_{i\in \mathbb{N}}$ such that $\left|\right|v-T^{n_i}x\left|\right|\ge 1$ for all i ∈ ℕ and $\left|\right|v-T^{n_i}x\left|\right|\to \infty$ as i → ∞. From (5.3.3), for any n ∈ ℕ, we have

$$\begin{array}{ll}\hfill D_f\left(v,T^{n+1}x\right)& =D_f\left(Tv,T^{n+1}x\right)\\ \le D_f\left(v,T^{n}x\right)+\lambda \left(T^{n}x\right)\left\{v-Tv,f^{\prime }\left(T^{n}x\right)-f^{\prime }\left(T^{n+1}x\right)\right\}=D_f\left(v,T^{n}x\right)\\ \le D_f\left(v,x\right),\end{array}$$

which in conjunction with (3), (4), and (6) implies that

$$\begin{array}{ll}\hfill D_f\left(v,x\right)& \ge D_f\left(v,T^{n_i}x\right)\ge {\nu }_f\left(T^{n_i}x,\left|\right|v-T^{n_i}x\left|\right|\right)\\ \ge \left|\right|v-T^{n_i}x\left|\right|{\nu }_f\left(T^{n_i}x,1\right)\\ \ge \left|\right|v-T^{n_i}x\left|\right|{\delta }_f\left(1\right)\to \infty ,\mathsf{\text{as }}i\to \infty ,\end{array}$$

a contradiction. Therefore, for any x ∈ X, {Tⁿx}_n∈ℕis bounded, and so it has a subsequence ${\left\{T^{n_j}x\right\}}_{j\in \mathbb{N}}$ which is weakly convergent to w for some w ∈ C. As $T^{n_j}x-T^{n_j+1}x\to 0$, it follows from the demiclosedness of T that w ∈ F(T). It remains to show that Tⁿx ⇀ w as n → ∞. Let ${\left\{T^{n_k}x\right\}}_{n\in \mathbb{N}}$ be any subsequence of {Tⁿx}_n∈ℕso that $T^{n_k}x\rightharpoonup u$ for some u ∈ C. Then u ∈ F(T). Since both of {D _f (w, Tⁿx)}_n∈ℕand {D _f (u, Tⁿx)}_n∈ℕare decreasing, we have

$$\underset{n\to \infty }{lim}\left[D_f\left(w,T^{n}x\right)-D_f\left(u,T^{n}x\right)\right]=\underset{n\to \infty }{lim}\left[f\left(w\right)-f\left(u\right)-〈w-u,f^{\prime }\left(T^{n}x\right)〉\right]=a$$

for some a ∈ ℝ. Particularly, from (5.3.4) we obtain

$$a=\underset{n_j\to \infty }{lim}\left[f\left(w\right)-f\left(u\right)-〈w-u,f^{\prime }\left(T^{n_j}x\right)〉\right]=f\left(w\right)-f\left(u\right)-〈w-u,f^{\prime }\left(w\right)〉$$

and

$$a=\underset{n_k\to \infty }{lim}\left[f\left(w\right)-f\left(u\right)-〈w-u,f^{\prime }\left(T^{n_k}x\right)〉\right]=f\left(w\right)-f\left(u\right)-〈w-u,f^{\prime }\left(u\right)〉.$$

Consequently, 〈w - u, f'(w) - f'(u)〉 = 0, and hence w = u by the strict convexity of f. This shows that Tⁿx ⇀ w for some w ∈ F(T).□

Adopting the technique of [8], we have the following ergodic theorem for point-dependent λ-hybrid mappings in Hilbert spaces.

Theorem 5.4. Suppose

(5.4.1) C is nonempty closed convex subset of a Hilbert space H.

(5.4.2) T : C → C is a point-dependent λ-hybrid mapping for some function λ : C → ℝ, that is,

$$\left|\right|Tx-Ty|{|}^2\le ||x-y|{|}^2+\lambda \left(y\right)〈x-Tx,y-Ty〉,\forall x,y\in C.$$

(5.4.3) F(T) ≠ ∅.

Then for any x ∈ C, the sequence {S _n (x)}_n∈ℕdefined by

$$S_n\left(x\right)=\frac{1}{n}\sum _{k=0}^{n-1}{T}^{k}x$$

converges weakly to some point v ∈ F(T).