Convergence analysis of Agarwal et al. iterative scheme for Lipschitzian hemicontractive mappings

Abstract

In this paper, we establish strong convergence for the Agarwal et al. iterative scheme associated with Lipschitzian hemicontractive mappings in Hilbert spaces.

MSC:47H10, 47J25.

1 Introduction and preliminaries

Let K be a nonempty subset of a Hilbert space H and $T:K\to K$ be a mapping.

The mapping T is called Lipschitzian if there exists $L>0$ such that

$$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in K.$$

If $L=1$, then T is called nonexpansive and if $0\le L<1$, then T is called contractive.

The mapping $T:K\to K$ is said to be pseudocontractive (see, for example, [1, 2]) if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in K,$$

and it is said to be strongly pseudocontractive if there exists $k\in (0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in K.$$

Let $F(T):=\{x\in H:Tx=x\}$, and the mapping $T:K\to K$ is called hemicontractive if $F(T)\ne \mathrm{\varnothing }$ and

$${\parallel Tx-x^{\ast }\parallel }^2\le {\parallel x-x^{\ast }\parallel }^2+{\parallel x-Tx\parallel }^2,\mathrm{\forall }x\in K,x^{\ast }\in F(T).$$

It is easy to see that the class of pseudocontractive mappings with fixed points is a subclass of the class of hemicontractions. For the importance of fixed points of pseudocontractions, the reader may consult [1].

In 1974, Ishikawa [3] proved the following result.

Theorem 1.1 Let K be a compact convex subset of a Hilbert space H, and let $T:K\to K$ be a Lipschitzian pseudocontractive mapping. For arbitrary $x_1\in K$, let $\{x_n\}$ be a sequence defined iteratively by

$$\{\begin{array}{c}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{y}_n,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences satisfying

• (i) $0\le {\alpha }_n\le {\beta }_n\le 1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to a fixed point of T.

Another iteration scheme has been studied extensively in connection with fixed points of pseudocontractive mappings.

In 2007, Agarwal et al. [4] introduced the new iterative scheme as in the following.

The sequence $\{x_n\}$ defined by, for arbitrary $x_1\in K$,

$$\{\begin{array}{c}{x}_{n+1}=(1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $[0,1]$, is known as the Agarwal et al. iterative scheme.

In this paper, we establish the strong convergence for the Agarwal et al. iterative scheme associated with Lipschitzian hemicontractive mappings in Hilbert spaces.

2 Main results

We need the following lemma.

Lemma 2.1 [5]

For all $x,y\in H$ and $\lambda \in [0,1]$, we have

$${\parallel (1-\lambda )x+\lambda y\parallel }^2=(1-\lambda ){\parallel x\parallel }^2+\lambda {\parallel y\parallel }^2-\lambda (1-\lambda ){\parallel x-y\parallel }^2.$$

Now we prove our main results.

Theorem 2.2 Let K be a compact convex subset of a real Hilbert space H, and let $T:K\to K$ be a Lipschitzian hemicontractive mapping satisfying

$$\parallel x-Ty\parallel \le \parallel Tx-Ty\parallel ,\mathrm{\forall }x,y\in K.$$

(C)

Let $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ be sequences in $[0,1]$ satisfying

• (ii) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=\mathrm{\infty }$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$.

For arbitrary $x_1\in K$, let $\{x_n\}$ be a sequence iteratively defined by

$$\{\begin{array}{c}{x}_{n+1}=(1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,n\ge 1.\hfill \end{array}$$

(2.1)

Then the sequence $\{x_n\}$ converges strongly to the fixed point $x^{\ast }$ of T.

Proof From Schauder’s fixed point theorem, $F(T)$ is nonempty since K is a convex compact set and T is continuous, let $x^{\ast }\in F(T)$.

By using condition (C), we have

$$\begin{array}{rl}\parallel x-Tx\parallel & \le \parallel x-Ty\parallel +\parallel Tx-Ty\parallel \\ \le 2\parallel Tx-Ty\parallel .\end{array}$$

(2.2)

Using the fact that T is hemicontractive, we obtain

$${\parallel T{x}_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2+{\parallel x_n-T{x}_n\parallel }^2$$

(2.3)

and

$${\parallel T{y}_n-x^{\ast }\parallel }^2\le {\parallel y_n-x^{\ast }\parallel }^2+{\parallel y_n-T{y}_n\parallel }^2.$$

(2.4)

With the help of (2.1), (2.3) and Lemma 2.1, we obtain

$$\begin{array}{rl}{\parallel y_n-x^{\ast }\parallel }^2=& {\parallel (1-{\beta }_n)x_n+{\beta }_{n}T{x}_n-x^{\ast }\parallel }^2\\ =& {\parallel (1-{\beta }_n)(x_n-x^{\ast })+{\beta }_n(T{x}_n-x^{\ast })\parallel }^2\\ =& (1-{\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n{\parallel T{x}_n-x^{\ast }\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T{x}_n\parallel }^2\\ \le & (1-{\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n({\parallel x_n-x^{\ast }\parallel }^2+{\parallel x_n-T{x}_n\parallel }^2)\\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T{x}_n\parallel }^2\\ =& {\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n^2{\parallel x_n-T{x}_n\parallel }^2\end{array}$$

(2.5)

and

$$\begin{array}{rl}{\parallel y_n-T{y}_n\parallel }^2=& {\parallel (1-{\beta }_n)x_n+{\beta }_{n}T{x}_n-T{y}_n\parallel }^2\\ =& {\parallel (1-{\beta }_n)(x_n-T{y}_n)+{\beta }_n(T{x}_n-T{y}_n)\parallel }^2\\ =& (1-{\beta }_n){\parallel x_n-T{y}_n\parallel }^2+{\beta }_n{\parallel T{x}_n-T{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T{x}_n\parallel }^2.\end{array}$$

(2.6)

Substituting (2.5) and (2.6) in (2.4), we obtain

$$\begin{array}{rl}{\parallel T{y}_n-x^{\ast }\parallel }^2\le & {\parallel x_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel x_n-T{y}_n\parallel }^2\\ +{\beta }_n{\parallel T{x}_n-T{y}_n\parallel }^2-{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2.\end{array}$$

(2.7)

Also, with the help of conditions (2.2) and (2.7), we have

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ ={\parallel (1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n-x^{\ast }\parallel }^2\\ ={\parallel (1-{\alpha }_n)(T{x}_n-x^{\ast })+{\alpha }_n(T{y}_n-x^{\ast })\parallel }^2\\ =(1-{\alpha }_n){\parallel T{x}_n-x^{\ast }\parallel }^2+{\alpha }_n{\parallel T{y}_n-x^{\ast }\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel T{x}_n-T{y}_n\parallel }^2\\ \le (1-{\alpha }_n)({\parallel x_n-x^{\ast }\parallel }^2+{\parallel x_n-T{x}_n\parallel }^2)+{\alpha }_n({\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\beta }_n){\parallel x_n-T{y}_n\parallel }^2+{\beta }_n{\parallel T{x}_n-T{y}_n\parallel }^2\\ -{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2)\\ ={\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\parallel x_n-T{x}_n\parallel }^2+{\alpha }_n{\beta }_n{\parallel T{x}_n-T{y}_n\parallel }^2\\ -{\alpha }_n{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel x_n-T{y}_n\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+(4(1-{\alpha }_n)+{\alpha }_n{\beta }_n+{\alpha }_n(1-{\beta }_n)){\parallel T{x}_n-T{y}_n\parallel }^2\\ -{\alpha }_n{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+\theta {\alpha }_n{\parallel T{x}_n-T{y}_n\parallel }^2-{\alpha }_n{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2,\end{array}$$

(2.8)

because by (iv), there exists $n_0\in \mathbb{N}$ such that for all $n\ge n_0$,

$$1-{\alpha }_n\le \frac{\theta -1}{\theta +3},$$

(2.9)

where $\theta >1$, which implies that

$$4(1-{\alpha }_n)+{\alpha }_n{\beta }_n+{\alpha }_n(1-{\beta }_n)\le \theta {\alpha }_n.$$

(2.10)

Hence (2.8) yields

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+\theta {\alpha }_n{L}^2{\parallel x_n-y_n\parallel }^2-{\alpha }_n{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2\\ ={\parallel x_n-x^{\ast }\parallel }^2+\theta {\alpha }_n{\beta }_n^2{L}^2{\parallel x_n-T{x}_n\parallel }^2-{\alpha }_n{\beta }_n(1-2{\beta }_n){\parallel x_n-T{x}_n\parallel }^2\\ ={\parallel x_n-x^{\ast }\parallel }^2-{\alpha }_n{\beta }_n(1-(2+\theta L^2){\beta }_n){\parallel x_n-T{x}_n\parallel }^2.\end{array}$$

(2.11)

Now, by (ii), since ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, there exists $n_0\in \mathbb{N}$ such that for all $n\ge n_0$,

$${\beta }_n\le \frac{1}{2(2+\theta L^2)}.$$

(2.12)

With the help of (iii) and (2.12), (2.11) yields

$${\parallel x_{n+1}-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-\frac{1}{2}{\alpha }_n{\beta }_n{\parallel x_n-T{x}_n\parallel }^2,$$

which implies that

$$\frac{1}{2}{\alpha }_n{\beta }_n{\parallel x_n-T{x}_n\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2,$$

so that

$$\frac{1}{2}\sum _{j=N}^n{\alpha }_j{\beta }_j{\parallel x_j-T{x}_j\parallel }^2\le {\parallel x_N-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2.$$

The rest of the argument follows exactly as in the proof of theorem of [3]. This completes the proof. □

Theorem 2.3 Let K be a compact convex subset of a real Hilbert space H; let $T:K\to K$ be a Lipschitzian hemicontractive mapping satisfying condition (C). Let $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ be sequence in $[0,1]$ satisfying conditions (ii)-(iv).

Let $P_K:H\to K$ be the projection operator of H onto K. Let $\{x_n\}$ be a sequence defined iteratively by

$$\{\begin{array}{c}{x}_{n+1}=P_K((1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n),\hfill \\ y_n=P_K((1-{\beta }_n)x_n+{\beta }_{n}T{x}_n),n\ge 1.\hfill \end{array}$$

Then the sequence$\{x_n\}$ converges strongly to a fixed point of T.

Proof The operator $P_K$ is nonexpansive (see, e.g., [2]). K is a Chebyshev subset of H so that $P_K$ is a single-valued mapping. Hence, we have

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& ={\parallel P_K((1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n)-P_K{x}^{\ast }\parallel }^2\\ \le {\parallel (1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n-x^{\ast }\parallel }^2\\ ={\parallel (1-{\alpha }_n)(x_n-x^{\ast })+{\alpha }_n(T{y}_n-x^{\ast })\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\alpha }_n{\beta }_n(1-(2+\theta L^2){\beta }_n){\parallel x_n-T{x}_n\parallel }^2.\end{array}$$

The set $K=K\cup T(K)$ is compact and so the sequence $\{\parallel x_n-T{x}_n\parallel \}$ is bounded. The rest of the argument follows exactly as in the proof of Theorem 2.2. This completes the proof. □

Example 2.4 The choice for the control parameters is ${\alpha }_n=\frac{n}{n+1}$ and ${\beta }_n=\frac{1}{n}$.

Remark 2.5 (1) We remove the condition ${\alpha }_n\le {\beta }_n$ as introduced in [3].

(2) The condition (C) is not new and it is due to [6].