Fixed point theorems for nonlinear non-self mappings in Hilbert spaces and applications

Abstract

Recently, Kawasaki and Takahashi (J. Nonlinear Convex Anal. 14:71-87, 2013) defined a broad class of nonlinear mappings, called widely more generalized hybrid, in a Hilbert space which contains generalized hybrid mappings (Kocourek et al. in Taiwan. J. Math. 14:2497-2511, 2010) and strict pseudo-contractive mappings (Browder and Petryshyn in J. Math. Anal. Appl. 20:197-228, 1967). They proved fixed point theorems for such mappings. In this paper, we prove fixed point theorems for widely more generalized hybrid non-self mappings in a Hilbert space by using the idea of Hojo et al. (Fixed Point Theory 12:113-126, 2011) and Kawasaki and Takahashi fixed point theorems (J. Nonlinear Convex Anal. 14:71-87, 2013). Using these fixed point theorems for non-self mappings, we proved the Browder and Petryshyn fixed point theorem (J. Math. Anal. Appl. 20:197-228, 1967) for strict pseudo-contractive non-self mappings and the Kocourek et al. fixed point theorem (Taiwan. J. Math. 14:2497-2511, 2010) for super hybrid non-self mappings. In particular, we solve a fixed point problem.

MSC:Primary 47H10; secondary 47H05.

1 Introduction

Let ℝ be the real line and let $[0,\frac{\pi }{2}]$ be a bounded, closed and convex subset of ℝ. Consider a mapping $T:[0,\frac{\pi }{2}]\to \mathbb{R}$ defined by

$$Tx=(1+\frac{1}{2}x)cosx-\frac{1}{2}{x}^2$$

for all $x\in [0,\frac{\pi }{2}]$. Such a mapping T has a unique fixed point $z\in [0,\frac{\pi }{2}]$ such that $cosz=z$. What kind of fixed point theorems can we use to find such a unique fixed point z of T?

Let H be a real Hilbert space and let C be a non-empty subset of H. Kocourek, Takahashi and Yao [1] introduced a class of nonlinear mappings in a Hilbert space which covers nonexpansive mappings, nonspreading mappings [2] and hybrid mappings [3]. A mapping $T:C\to H$ is said to be generalized hybrid if there exist $\alpha ,\beta \in \mathbb{R}$ such that

$$\alpha {\parallel Tx-Ty\parallel }^2+(1-\alpha ){\parallel x-Ty\parallel }^2\le \beta {\parallel Tx-y\parallel }^2+(1-\beta ){\parallel x-y\parallel }^2$$

(1.1)

for all $x,y\in C$. We call such a mapping an $(\alpha ,\beta )$-generalized hybrid mapping. An $(\alpha ,\beta )$-generalized hybrid mapping is nonexpansive for $\alpha =1$ and $\beta =0$, i.e.,

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$$

for all $x,y\in C$. It is nonspreading for $\alpha =2$ and $\beta =1$, i.e.,

$$2{\parallel Tx-Ty\parallel }^2\le {\parallel x-Ty\parallel }^2+{\parallel y-Tx\parallel }^2$$

for all $x,y\in C$. Furthermore, it is hybrid for $\alpha =\frac{3}{2}$ and $\beta =\frac{1}{2}$, i.e.,

$$3{\parallel Tx-Ty\parallel }^2\le {\parallel x-Ty\parallel }^2+{\parallel y-Tx\parallel }^2+{\parallel y-x\parallel }^2$$

for all $x,y\in C$. They proved fixed point theorems and nonlinear ergodic theorems of Baillon type [4] for generalized hybrid mappings; see also Kohsaka and Takahashi [5] and Iemoto and Takahashi [6]. Very recently, Kawasaki and Takahashi [7] introduced a broader class of nonlinear mappings than the class of generalized hybrid mappings in a Hilbert space. A mapping T from C into H is called widely more generalized hybrid if there exist $\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta \in \mathbb{R}$ such that

$$\begin{array}{r}\alpha {\parallel Tx-Ty\parallel }^2+\beta {\parallel x-Ty\parallel }^2+\gamma {\parallel Tx-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {\parallel x-Tx\parallel }^2+\zeta {\parallel y-Ty\parallel }^2+\eta {\parallel (x-Tx)-(y-Ty)\parallel }^2\le 0\end{array}$$

(1.2)

for all $x,y\in C$. Such a mapping T is called an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping. In particular, an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping is generalized hybrid in the sense of Kocourek, Takahashi and Yao [1] if $\alpha +\beta =-\gamma -\delta =1$ and $\epsilon =\zeta =\eta =0$. An $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping is strict pseudo-contractive in the sense of Browder and Petryshyn [8] if $\alpha =1$, $\beta =\gamma =0$, $\delta =-1$, $\epsilon =\zeta =0$, $\eta =-k$, where $0\le k<1$. A generalized hybrid mapping with a fixed point is quasi-nonexpansive. However, a widely more generalized hybrid mapping is not quasi-nonexpansive in general even if it has a fixed point. In [7], Kawasaki and Takahashi proved fixed point theorems and nonlinear ergodic theorems of Baillon type [4] for such widely more generalized hybrid mappings in a Hilbert space. In particular, they proved directly the Browder and Petryshyn fixed point theorem [8] for strict pseudo-contractive mappings and the Kocourek, Takahashi and Yao fixed point theorem [1] for super hybrid mappings by using their fixed point theorems. However, we cannot use Kawasaki and Takahashi fixed point theorems to solve the above problem. For a nice synthesis on metric fixed point theory, see Kirk [9].

In this paper, motivated by such a problem, we prove fixed point theorems for widely more generalized hybrid non-self mappings in a Hilbert space by using the idea of Hojo, Takahashi and Yao [10] and Kawasaki and Takahashi fixed point theorems [7]. Using these fixed point theorems for non-self mappings, we prove the Browder and Petryshyn fixed point theorem [8] for strict pseudo-contractive non-self mappings and the Kocourek, Takahashi and Yao fixed point theorem [1] for super hybrid non-self mappings. In particular, we solve the above problem by using one of our fixed point theorems.

2 Preliminaries

Throughout this paper, we denote by ℕ the set of positive integers. Let H be a (real) Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm , respectively. From [11], we know the following basic equality: For $x,y\in H$ and $\lambda \in \mathbb{R}$, we have

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

(2.1)

Furthermore, we know that for $x,y,u,v\in H$,

$$2〈x-y,u-v〉={\parallel x-v\parallel }^2+{\parallel y-u\parallel }^2-{\parallel x-u\parallel }^2-{\parallel y-v\parallel }^2.$$

(2.2)

Let C be a non-empty, closed and convex subset of H and let T be a mapping from C into H. Then we denote by $F(T)$ the set of fixed points of T. A mapping $S:C\to H$ is called super hybrid [1, 12] if there exist $\alpha ,\beta ,\gamma \in \mathbb{R}$ such that

$$\begin{array}{r}\alpha {\parallel Sx-Sy\parallel }^2+(1-\alpha +\gamma ){\parallel x-Sy\parallel }^2\\ \le (\beta +(\beta -\alpha )\gamma ){\parallel Sx-y\parallel }^2+(1-\beta -(\beta -\alpha -1)\gamma ){\parallel x-y\parallel }^2\\ +(\alpha -\beta )\gamma {\parallel x-Sx\parallel }^2+\gamma {\parallel y-Sy\parallel }^2\end{array}$$

(2.3)

for all $x,y\in C$. We call such a mapping an $(\alpha ,\beta ,\gamma )$-super hybrid mapping. An $(\alpha ,\beta ,0)$-super hybrid mapping is $(\alpha ,\beta )$-generalized hybrid. Thus the class of super hybrid mappings contains generalized hybrid mappings. The following theorem was proved in [12]; see also [1].

Theorem 2.1 ([12])

Let C be a non-empty subset of a Hilbert space H and let α, β and γ be real numbers with $\gamma \ne -1$. Let S and T be mappings of C into H such that $T=\frac{1}{1+\gamma }S+\frac{\gamma }{1+\gamma }I$. Then S is $(\alpha ,\beta ,\gamma )$-super hybrid if and only if T is $(\alpha ,\beta )$-generalized hybrid. In this case, $F(S)=F(T)$. In particular, let C be a nonempty, closed and convex subset of H and let α, β and γ be real numbers with $\gamma \ge 0$. If a mapping $S:C\to C$ is $(\alpha ,\beta ,\gamma )$-super hybrid, then the mapping $T=\frac{1}{1+\gamma }S+\frac{\gamma }{1+\gamma }I$ is an $(\alpha ,\beta )$-generalized hybrid mapping of C into itself.

In [1], Kocourek, Takahashi and Yao also proved the following fixed point theorem for super hybrid mappings in a Hilbert space.

Theorem 2.2 ([1])

Let C be a non-empty, bounded, closed and convex subset of a Hilbert space H and let α, β and γ be real numbers with $\gamma \ge 0$. Let $S:C\to C$ be an $(\alpha ,\beta ,\gamma )$-super hybrid mapping. Then S has a fixed point in C. In particular, if $S:C\to C$ is an $(\alpha ,\beta )$-generalized hybrid mapping, then S has a fixed point in C.

A super hybrid mapping is not quasi-nonexpansive in general even if it has a fixed point. There exists a class of nonlinear mappings in a Hilbert space defined by Kawasaki and Takahashi [13] which covers contractive mappings and generalized hybrid mappings. A mapping T from C into H is said to be widely generalized hybrid if there exist $\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta \in \mathbb{R}$ such that

$$\begin{array}{r}\alpha {\parallel Tx-Ty\parallel }^2+\beta {\parallel x-Ty\parallel }^2+\gamma {\parallel Tx-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +max\{\epsilon {\parallel x-Tx\parallel }^2,\zeta {\parallel y-Ty\parallel }^2\}\le 0\end{array}$$

for any $x,y\in C$. Such a mapping T is called $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta )$-widely generalized hybrid. Kawasaki and Takahashi [13] proved the following fixed point theorem.

Theorem 2.3 ([13])

Let H be a Hilbert space, let C be a non-empty, closed and convex subset of H and let T be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta )$-widely generalized hybrid mapping from C into itself which satisfies the following conditions (1) and (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$;

• (2) $\epsilon +\alpha +\gamma >0$, or $\zeta +\alpha +\beta >0$.

Then T has a fixed point if and only if there exists $z\in C$ such that $\{T^{n}z\mid n=0,1,\dots \}$ is bounded. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the condition (1).

Very recently, Kawasaki and Takahashi [7] also proved the following fixed point theorem which will be used in the proofs of our main theorems in this paper.

Theorem 2.4 ([7])

Let H be a Hilbert space, let C be a non-empty, closed and convex subset of H and let T be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping from C into itself, i.e., there exist $\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta \in \mathbb{R}$ such that

$$\begin{array}{r}\alpha {\parallel Tx-Ty\parallel }^2+\beta {\parallel x-Ty\parallel }^2+\gamma {\parallel Tx-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {\parallel x-Tx\parallel }^2+\zeta {\parallel y-Ty\parallel }^2+\eta {\parallel (x-Tx)-(y-Ty)\parallel }^2\le 0\end{array}$$

for all $x,y\in C$. Suppose that it satisfies the following condition (1) or (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +\eta >0$ and $\zeta +\eta \ge 0$;

• (2) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$ and $\epsilon +\eta \ge 0$.

Then T has a fixed point if and only if there exists $z\in C$ such that $\{T^{n}z\mid n=0,1,\dots \}$ is bounded. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the conditions (1) and (2).

In particular, we have the following theorem from Theorem 2.4.

Theorem 2.5 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping from C into itself which satisfies the following condition (1) or (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +\eta >0$ and $\zeta +\eta \ge 0$;

• (2) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$ and $\epsilon +\eta \ge 0$.

Then T has a fixed point. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the conditions (1) and (2).

3 Fixed point theorems for non-self mappings

In this section, using the fixed point theorem (Theorem 2.5), we first prove the following fixed point theorem for widely more generalized hybrid non-self mappings in a Hilbert space.

Theorem 3.1 Let C be a non-empty, bounded, closed and convex subset of a Hilbert space H and let $\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta \in \mathbb{R}$. Let $T:C\to H$ be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping. Suppose that it satisfies the following condition (1) or (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +\eta >0$, $\alpha +\beta +\zeta +\eta \ge 0$ and $\zeta +\eta \ge 0$;

• (2) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$, $\alpha +\gamma +\epsilon +\eta \ge 0$ and $\epsilon +\eta \ge 0$.

Assume that there exists a positive number $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$. Then T has a fixed point in C. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the conditions (1) and (2).

Proof We give the proof for the case of (1). By the assumption, we have that for any $x\in C$, there exist $y\in C$ and t with $0<t\le m$ such that $Tx=x+t(y-x)$. From this, we have $Tx=ty+(1-t)x$ and hence

$$y=\frac{1}{t}Tx+\frac{t-1}{t}x.$$

Define $Ux\in C$ as follows:

$$Ux=(1-\frac{t}{m})x+\frac{t}{m}y=(1-\frac{t}{m})x+\frac{t}{m}(\frac{1}{t}Tx+\frac{t-1}{t}x)=\frac{1}{m}Tx+\frac{m-1}{m}x.$$

Taking $\lambda >0$ with $m=1+\lambda$, we have that

$$Ux=\frac{1}{1+\lambda }Tx+\frac{\lambda }{1+\lambda }x$$

and hence

$$T=(1+\lambda )U-\lambda I.$$

(3.1)

Since $T:C\to H$ is an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping, we have from (3.1) and (2.1) that for any $x,y\in C$,

$$\begin{array}{r}\alpha {\parallel (1+\lambda )Ux-\lambda x-((1+\lambda )Uy-\lambda y)\parallel }^2\\ +\beta {\parallel x-((1+\lambda )Uy-\lambda y)\parallel }^2+\gamma {\parallel (1+\lambda )Ux-\lambda x-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {\parallel x-((1+\lambda )Ux-\lambda x)\parallel }^2+\zeta {\parallel (1+\lambda )Uy-\lambda y-y\parallel }^2\\ +\eta {\parallel x-((1+\lambda )Ux-\lambda x)-(y-((1+\lambda )Uy-\lambda y))\parallel }^2\\ =\alpha {\parallel (1+\lambda )(Ux-Uy)-\lambda (x-y)\parallel }^2\\ +\beta {\parallel (1+\lambda )(x-Uy)-\lambda (x-y)\parallel }^2+\gamma {\parallel (1+\lambda )(Ux-y)-\lambda (x-y)\parallel }^2\\ +\delta {\parallel x-y\parallel }^2+\epsilon {\parallel (1+\lambda )(x-Ux)\parallel }^2+\zeta {\parallel (1+\lambda )(y-Uy)\parallel }^2\\ +\eta {\parallel (1+\lambda )(x-Ux)-(1+\lambda )(y-Uy)\parallel }^2\\ =\alpha (1+\lambda ){\parallel Ux-Uy\parallel }^2-\alpha \lambda {\parallel x-y\parallel }^2+\alpha \lambda (1+\lambda ){\parallel x-y-(Ux-Uy)\parallel }^2\\ +\beta (1+\lambda ){\parallel x-Uy\parallel }^2-\beta \lambda {\parallel x-y\parallel }^2+\beta \lambda (1+\lambda ){\parallel y-Uy\parallel }^2\\ +\gamma (1+\lambda ){\parallel Ux-y\parallel }^2-\gamma \lambda {\parallel x-y\parallel }^2+\gamma \lambda (1+\lambda ){\parallel x-Ux\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {(1+\lambda )}^2{\parallel x-Ux\parallel }^2+\zeta {(1+\lambda )}^2{\parallel y-Uy\parallel }^2\\ +\eta {(1+\lambda )}^2{\parallel x-Ux-(y-Uy)\parallel }^2\\ =\alpha (1+\lambda ){\parallel Ux-Uy\parallel }^2+\beta (1+\lambda ){\parallel x-Uy\parallel }^2+\gamma (1+\lambda ){\parallel Ux-y\parallel }^2\\ +(-\alpha \lambda -\beta \lambda -\gamma \lambda +\delta ){\parallel x-y\parallel }^2\\ +(\gamma \lambda +\epsilon \lambda +\epsilon )(1+\lambda ){\parallel x-Ux\parallel }^2+(\beta \lambda +\zeta \lambda +\zeta )(1+\lambda ){\parallel y-Uy\parallel }^2\\ +(\alpha \lambda +\eta \lambda +\eta )(1+\lambda ){\parallel x-y-(Ux-Uy)\parallel }^2\le 0.\end{array}$$

This implies that U is widely more generalized hybrid. Since $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +\eta >0$, $\alpha +\beta +\zeta +\eta \ge 0$ and $\zeta +\eta \ge 0$, we obtain that

$$\begin{array}{r}\begin{array}{r}\alpha (1+\lambda )+\beta (1+\lambda )+\gamma (1+\lambda )-\alpha \lambda -\beta \lambda -\gamma \lambda +\delta =\alpha +\beta +\gamma +\delta \ge 0,\end{array}\\ \begin{array}{r}\alpha (1+\lambda )+\gamma (1+\lambda )+(\gamma \lambda +\epsilon \lambda +\epsilon )(1+\lambda )+(\alpha \lambda +\eta \lambda +\eta )(1+\lambda )\\ =(1+\lambda )(\alpha +\gamma +\epsilon +\eta +\lambda (\gamma +\epsilon +\alpha +\eta ))\\ ={(1+\lambda )}^2(\alpha +\gamma +\epsilon +\eta )>0,\end{array}\\ \begin{array}{r}(\beta \lambda +\zeta \lambda +\zeta )(1+\lambda )+(\alpha \lambda +\eta \lambda +\eta )(1+\lambda )\\ =((\alpha +\beta +\zeta +\eta )\lambda +\zeta +\eta )(1+\lambda )\ge 0.\end{array}\end{array}$$

By Theorem 2.5, we obtain that $F(U)\ne \mathrm{\varnothing }$. Therefore, we have from $F(U)=F(T)$ that $F(T)\ne \mathrm{\varnothing }$. Suppose that $\alpha +\beta +\gamma +\delta >0$. Let $p_1$ and $p_2$ be fixed points of T. We have that

$$\begin{array}{r}\alpha {\parallel T{p}_1-T{p}_2\parallel }^2+\beta {\parallel p_1-T{p}_2\parallel }^2+\gamma {\parallel T{p}_1-p_2\parallel }^2+\delta {\parallel p_1-p_2\parallel }^2\\ +\epsilon {\parallel p_1-T{p}_1\parallel }^2+\zeta {\parallel p_2-T{p}_2\parallel }^2+\eta {\parallel (p_1-T{p}_1)-(p_2-T{p}_2)\parallel }^2\\ =(\alpha +\beta +\gamma +\delta ){\parallel p_1-p_2\parallel }^2\le 0\end{array}$$

and hence $p_1=p_2$. Therefore, a fixed point of T is unique.

Similarly, we can obtain the desired result for the case when $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$, $\alpha +\gamma +\epsilon +\eta \ge 0$ and $\epsilon +\eta \ge 0$. This completes the proof. □

The following theorem is a useful extension of Theorem 3.1.

Theorem 3.2 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping from C into H which satisfies the following condition (1) or (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +\eta >0$, $\alpha +\beta +\zeta +\eta \ge 0$ and $[0,1)\cap \{\lambda \mid (\alpha +\beta )\lambda +\zeta +\eta \ge 0\}\ne \mathrm{\varnothing }$;

• (2) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$, $\alpha +\gamma +\epsilon +\eta \ge 0$ and $[0,1)\cap \{\lambda \mid (\alpha +\gamma )\lambda +\epsilon +\eta \ge 0\}\ne \mathrm{\varnothing }$.

Assume that there exists $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$. Then T has a fixed point. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the conditions (1) and (2).

Proof Let $\lambda \in [0,1)\cap \{\lambda \mid (\alpha +\beta )\lambda +\zeta +\eta \ge 0\}$ and define $S=(1-\lambda )T+\lambda I$. Then S is a mapping from C into H. Since $\lambda \ne 1$, we obtain that $F(S)=F(T)$. Moreover, from $T=\frac{1}{1-\lambda }S-\frac{\lambda }{1-\lambda }I$ and (2.1), we have that

$$\begin{array}{r}\alpha {\parallel (\frac{1}{1-\lambda }Sx-\frac{\lambda }{1-\lambda }x)-(\frac{1}{1-\lambda }Sy-\frac{\lambda }{1-\lambda }y)\parallel }^2\\ +\beta {\parallel x-(\frac{1}{1-\lambda }Sy-\frac{\lambda }{1-\lambda }y)\parallel }^2+\gamma {\parallel (\frac{1}{1-\lambda }Sx-\frac{\lambda }{1-\lambda }x)-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {\parallel x-(\frac{1}{1-\lambda }Sx-\frac{\lambda }{1-\lambda }x)\parallel }^2+\zeta {\parallel y-(\frac{1}{1-\lambda }Sy-\frac{\lambda }{1-\lambda }y)\parallel }^2\\ +\eta {\parallel (x-(\frac{1}{1-\lambda }Sx-\frac{\lambda }{1-\lambda }x))-(y-(\frac{1}{1-\lambda }Sy-\frac{\lambda }{1-\lambda }y))\parallel }^2\\ =\alpha {\parallel \frac{1}{1-\lambda }(Sx-Sy)-\frac{\lambda }{1-\lambda }(x-y)\parallel }^2\\ +\beta {\parallel \frac{1}{1-\lambda }(x-Sy)-\frac{\lambda }{1-\lambda }(x-y)\parallel }^2\\ +\gamma {\parallel \frac{1}{1-\lambda }(Sx-y)-\frac{\lambda }{1-\lambda }(x-y)\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +\epsilon {\parallel \frac{1}{1-\lambda }(x-Sx)\parallel }^2+\zeta {\parallel \frac{1}{1-\lambda }(y-Sy)\parallel }^2\\ +\eta {\parallel \frac{1}{1-\lambda }(x-Sx)-\frac{1}{1-\lambda }(y-Sy)\parallel }^2\\ =\frac{\alpha }{1-\lambda }{\parallel Sx-Sy\parallel }^2+\frac{\beta }{1-\lambda }{\parallel x-Sy\parallel }^2\\ +\frac{\gamma }{1-\lambda }{\parallel Sx-y\parallel }^2+(-\frac{\lambda }{1-\lambda }(\alpha +\beta +\gamma )+\delta ){\parallel x-y\parallel }^2\\ +\frac{\epsilon +\gamma \lambda }{{(1-\lambda )}^2}{\parallel x-Sx\parallel }^2+\frac{\zeta +\beta \lambda }{{(1-\lambda )}^2}{\parallel y-Sy\parallel }^2\\ +\frac{\eta +\alpha \lambda }{{(1-\lambda )}^2}{\parallel (x-Sx)-(y-Sy)\parallel }^2\le 0.\end{array}$$

Therefore S is an $(\frac{\alpha }{1-\lambda },\frac{\beta }{1-\lambda },\frac{\gamma }{1-\lambda },-\frac{\lambda }{1-\lambda }(\alpha +\beta +\gamma )+\delta ,\frac{\epsilon +\gamma \lambda }{{(1-\lambda )}^2},\frac{\zeta +\beta \lambda }{{(1-\lambda )}^2},\frac{\eta +\alpha \lambda }{{(1-\lambda )}^2})$-widely more generalized hybrid mapping. Furthermore, we obtain that

$$\begin{array}{c}\frac{\alpha }{1-\lambda }+\frac{\beta }{1-\lambda }+\frac{\gamma }{1-\lambda }-\frac{\lambda }{1-\lambda }(\alpha +\beta +\gamma )+\delta =\alpha +\beta +\gamma +\delta \ge 0,\hfill \\ \frac{\alpha }{1-\lambda }+\frac{\gamma }{1-\lambda }+\frac{\epsilon +\gamma \lambda }{{(1-\lambda )}^2}+\frac{\eta +\alpha \lambda }{{(1-\lambda )}^2}=\frac{\alpha +\gamma +\epsilon +\eta }{{(1-\lambda )}^2}>0,\hfill \\ \frac{\alpha }{1-\lambda }+\frac{\beta }{1-\lambda }+\frac{\zeta +\beta \lambda }{{(1-\lambda )}^2}+\frac{\eta +\alpha \lambda }{{(1-\lambda )}^2}=\frac{\alpha +\beta +\zeta +\eta }{{(1-\lambda )}^2}\ge 0,\hfill \\ \frac{\zeta +\beta \lambda }{{(1-\lambda )}^2}+\frac{\eta +\alpha \lambda }{{(1-\lambda )}^2}=\frac{(\alpha +\beta )\lambda +\zeta +\eta }{{(1-\lambda )}^2}\ge 0.\hfill \end{array}$$

Furthermore, from the assumption, there exists $m>1$ such that for any $x\in C$,

$$Sx=(1-\lambda )Tx+\lambda x=(1-\lambda )(x+t(y-x))+\lambda x=t(1-\lambda )(y-x)+x,$$

where $y\in C$ and $0<t\le m$. From $0\le \lambda <1$, we have $0<t(1-\lambda )\le m$. Putting $s=t(1-\lambda )$, we have that there exists $m>1$ such that for any $x\in C$,

$$Sx=x+s(y-x)$$

for some $y\in C$ and s with $0<s\le m$. Therefore, we obtain from Theorem 3.1 that $F(S)\ne \mathrm{\varnothing }$. Since $F(S)=F(T)$, we obtain that $F(T)\ne \mathrm{\varnothing }$.

Next, suppose that $\alpha +\beta +\gamma +\delta >0$. Let $p_1$ and $p_2$ be fixed points of T. As in the proof of Theorem 3.1, we have $p_1=p_2$. Therefore a fixed point of T is unique.

In the case of $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$, $\alpha +\gamma +\epsilon +\eta \ge 0$ and $[0,1)\cap \{\lambda \mid (\alpha +\gamma )\lambda +\epsilon +\eta \ge 0\}\ne \mathrm{\varnothing }$, we can obtain the desired result by replacing the variables x and y. □

Remark 1 We can also prove Theorems 3.1 and 3.2 by using the condition

$$-\beta -\delta +\epsilon +\eta >0,\text{or}-\gamma -\delta +\epsilon +\eta >0$$

instead of the condition

$$\alpha +\gamma +\epsilon +\eta >0,\text{or}\alpha +\beta +\zeta +\eta >0,$$

respectively. In fact, in the case of the condition $-\beta -\delta +\epsilon +\eta >0$, we obtain from $\alpha +\beta +\gamma +\delta \ge 0$ that

$$0<-\beta -\delta +\epsilon +\eta \le \alpha +\gamma +\epsilon +\eta .$$

Thus we obtain the desired results by Theorems 3.1 and 3.2. Similarly, in the case of $-\gamma -\delta +\epsilon +\eta >0$, we can obtain the results by using the case of $\alpha +\beta +\zeta +\eta >0$.

4 Fixed point theorems for well-known mappings

Using Theorem 3.1, we first show the following fixed point theorem for generalized hybrid non-self mappings in a Hilbert space; see also Kocourek, Takahashi and Yao [1].

Theorem 4.1 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be a generalized hybrid mapping from C into H, i.e., there exist $\alpha ,\beta \in \mathbb{R}$ such that

$$\alpha {\parallel Tx-Ty\parallel }^2+(1-\alpha ){\parallel x-Ty\parallel }^2\le \beta {\parallel Tx-y\parallel }^2+(1-\beta ){\parallel x-y\parallel }^2$$

for any $x,y\in C$. Suppose $\alpha -\beta \ge 0$ and assume that there exists $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$. Then T has a fixed point.

Proof An $(\alpha ,\beta )$-generalized hybrid mapping T from C into H is an $(\alpha ,1-\alpha ,-\beta ,-(1-\beta ),0,0,0)$-widely more generalized hybrid mapping. Furthermore, $\alpha +(1-\alpha )-\beta -(1-\beta )=0$, $\alpha +(1-\alpha )+0+0=1>0$, $\alpha -\beta +0+0=\alpha -\beta \ge 0$ and $0+0=0$, that is, it satisfies the condition (2) in Theorem 3.1. Furthermore, since there exists $m\ge 1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$, we obtain the desired result from Theorem 3.1. □

Using Theorem 3.1, we can also show the following fixed point theorem for widely generalized hybrid non-self mappings in a Hilbert space; see Kawasaki and Takahashi [13].

Theorem 4.2 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta )$-widely generalized hybrid mapping from C into H which satisfies the following condition (1) or (2):

• (1) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon >0$ and $\alpha +\beta \ge 0$;

• (2) $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta >0$ and $\alpha +\gamma \ge 0$.

Assume that there exists $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and $t\in \mathbb{R}$ with $0<t\le m$. Then T has a fixed point. In particular, a fixed point of T is unique in the case of $\alpha +\beta +\gamma +\delta >0$ under the conditions (1) and (2).

Proof Since T is $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta )$-widely generalized hybrid, we obtain that

$$\begin{array}{r}\alpha {\parallel Tx-Ty\parallel }^2+\beta {\parallel x-Ty\parallel }^2+\gamma {\parallel Tx-y\parallel }^2+\delta {\parallel x-y\parallel }^2\\ +max\{\epsilon {\parallel x-Tx\parallel }^2,\zeta {\parallel y-Ty\parallel }^2\}\le 0\end{array}$$

for any $x,y\in C$. In the case of $\alpha +\gamma +\epsilon >0$, from

$$\epsilon {\parallel x-Tx\parallel }^2\le max\{\epsilon {\parallel x-Tx\parallel }^2,\zeta {\parallel y-Ty\parallel }^2\},$$

we obtain that

$$\alpha {\parallel Tx-Ty\parallel }^2+\beta {\parallel x-Ty\parallel }^2+\gamma {\parallel Tx-y\parallel }^2+\delta {\parallel x-y\parallel }^2+\epsilon {\parallel x-Tx\parallel }^2\le 0,$$

that is, it is an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,0,0)$-widely more generalized hybrid mapping. Furthermore, we have that $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\gamma +\epsilon +0=\alpha +\gamma +\epsilon >0$, $\alpha +\beta +0+0=\alpha +\beta \ge 0$ and $0+0=0$, that is, it satisfies the condition (1) in Theorem 3.1. Furthermore, since there exists $m\ge 1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$, we obtain the desired result from Theorem 3.1. In the case of $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta >0$ and $\alpha +\gamma \ge 0$, we can obtain the desired result by replacing the variables x and y. □

We know that an $(\alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta ,\eta )$-widely more generalized hybrid mapping with $\alpha =1$, $\beta =\gamma =\epsilon =\zeta =0$, $\delta =-1$ and $\eta =-k\in (-1,0]$ is a strict pseudo-contractive mapping in the sense of Browder and Petryshyn [8]. We also define the following mapping: $T:C\to H$ is called a generalized strict pseudo-contractive mapping if there exist $r,k\in \mathbb{R}$ with $0\le r\le 1$ and $0\le k<1$ such that

$${\parallel Tx-Ty\parallel }^2\le r{\parallel x-y\parallel }^2+k{\parallel (x-Tx)-(y-Ty)\parallel }^2$$

for any $x,y\in C$. Using Theorem 3.2, we can show the following fixed point theorem for generalized strict pseudo-contractive non-self mappings in a Hilbert space.

Theorem 4.3 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be a generalized strict pseudo-contractive mapping from C into H, that is, there exist $r,k\in \mathbb{R}$ with $0\le r\le 1$ and $0\le k<1$ such that

$${\parallel Tx-Ty\parallel }^2\le r{\parallel x-y\parallel }^2+k{\parallel (x-Tx)-(y-Ty)\parallel }^2$$

for all $x,y\in C$. Assume that there exists $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and $t\in \mathbb{R}$ with $0<t\le m$. Then T has a fixed point. In particular, if $0\le r<1$, then T has a unique fixed point.

Proof A generalized strict pseudo-contractive mapping T from C into H is a $(1,0,0,-r,0,0,-k)$-widely more generalized hybrid mapping. Furthermore, $1+0+0+(-r)\ge 0$, $1+0+0+(-k)=1-k>0$, $1+0+0+(-k)=1-k>0$ and $[0,1)\cap \{\lambda \mid (1+0)\lambda +0-k\ge 0\}=[k,1)\ne \mathrm{\varnothing }$, that is, it satisfies the condition (1) in Theorem 3.2. Furthermore, since there exists $m\ge 1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$, we obtain the desired result from Theorem 3.2. In particular, if $0\le r<1$, then $1+0+0+(-r)>0$. We have from Theorem 3.2 that T has a unique fixed point. □

Let us consider the problem in the Introduction. A mapping $T:[0,\frac{\pi }{2}]\to \mathbb{R}$ was defined as follows:

$$Tx=(1+\frac{1}{2}x)cosx-\frac{1}{2}{x}^2$$

(4.1)

for all $x\in [0,\frac{\pi }{2}]$. We have that

$$\begin{array}{r}Tx=(1+\frac{1}{2}x)cosx-\frac{1}{2}{x}^2\\ ⟺\frac{1}{1+\frac{1}{2}x}Tx+\frac{\frac{1}{2}x}{1+\frac{1}{2}x}x=cosx.\end{array}$$

Thus we have that for any $x\in [0,\frac{\pi }{2}]$,

$$\begin{array}{r}\frac{1+\frac{1}{2}x}{1+\pi }(\frac{1}{1+\frac{1}{2}x}Tx+\frac{\frac{1}{2}x}{1+\frac{1}{2}x})+(1-\frac{1+\frac{1}{2}x}{1+\pi })x\\ =\frac{1+\frac{1}{2}x}{1+\pi }cosx+(1-\frac{1+\frac{1}{2}x}{1+\pi })x,\end{array}$$

and hence

$$\frac{1}{1+\pi }Tx+\frac{\pi }{1+\pi }x=\frac{1+\frac{1}{2}x}{1+\pi }cosx+\frac{\pi -\frac{1}{2}x}{1+\pi }x.$$

Using this, we also have from (2.1) that for any $x,y\in [0,\frac{\pi }{2}]$,

$$\begin{array}{r}|\frac{1}{1+\pi }Tx+\frac{\pi }{1+\pi }x-(\frac{1}{1+\pi }Ty+\frac{\pi }{1+\pi }y){|}^2\\ =|\frac{1+\frac{1}{2}x}{1+\pi }cosx+\frac{\pi -\frac{1}{2}x}{1+\pi }x-(\frac{1+\frac{1}{2}y}{1+\pi }cosy+\frac{\pi -\frac{1}{2}y}{1+\pi }y){|}^2\end{array}$$

and hence

$$\begin{array}{r}\frac{1}{1+\pi }{|Tx-Ty|}^2+\frac{\pi }{1+\pi }{|x-y|}^2-\frac{\pi }{{(1+\pi )}^2}{|x-y-(Tx-Ty)|}^2\\ =|\frac{1+\frac{1}{2}x}{1+\pi }cosx+\frac{\pi -\frac{1}{2}x}{1+\pi }x-(\frac{1+\frac{1}{2}y}{1+\pi }cosy+\frac{\pi -\frac{1}{2}y}{1+\pi }y){|}^2.\end{array}$$

(4.2)

Define a function $f:[0,\frac{\pi }{2}]\to \mathbb{R}$ as follows:

$$f(x)=\frac{1+\frac{1}{2}x}{1+\pi }cosx+\frac{\pi -\frac{1}{2}x}{1+\pi }x$$

for all $x\in [0,\frac{\pi }{2}]$. Then we have

$$f^{\mathrm{\prime }}(x)=\frac{\frac{1}{2}}{1+\pi }cosx-\frac{1+\frac{1}{2}x}{1+\pi }sinx+\frac{\pi }{1+\pi }-\frac{x}{1+\pi }$$

and

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=-\frac{1}{1+\pi }sinx-\frac{1+\frac{1}{2}x}{1+\pi }cosx-\frac{1}{1+\pi }.$$

Since

$$f^{\mathrm{\prime }}(0)=\frac{\frac{1}{2}+\pi }{1+\pi },f^{\mathrm{\prime }}\left(\frac{\pi }{2}\right)=\frac{-1+\frac{1}{4}\pi }{1+\pi }$$

and $f^{\mathrm{\prime }\mathrm{\prime }}(x)<0$ for all $x\in [0,\frac{\pi }{2}]$, we have from the mean value theorem that there exists a positive number r with $0<r<1$ such that

$$|\frac{1+\frac{1}{2}x}{1+\pi }cosx+\frac{\pi -\frac{1}{2}x}{1+\pi }x-(\frac{1+\frac{1}{2}y}{1+\pi }cosy+\frac{\pi -\frac{1}{2}y}{1+\pi }y){|}^2\le r{|x-y|}^2$$

for all $x,y\in [0,\frac{\pi }{2}]$. Therefore, we have from (4.2) that

$$\frac{1}{1+\pi }{|Tx-Ty|}^2+\frac{\pi }{1+\pi }{|x-y|}^2\le r{|x-y|}^2+\frac{\pi }{{(1+\pi )}^2}{|x-y-(Tx-Ty)|}^2$$

for all $x,y\in [0,\frac{\pi }{2}]$. Furthermore, we have from (4.1) that

$$Tx=(1+\frac{1}{2}x)(cosx-x)+x$$

for all $x\in [0,\frac{\pi }{2}]$. Take $m=1+\pi$ and let $t=1+\frac{1}{2}x$ and $y=cosx$ for all $x\in [0,\frac{\pi }{2}]$. Then we have that

$$Tx=t(y-x)+x,y=cosx\in [0,\frac{\pi }{2}]\text{and}0<t=1+\frac{1}{2}x\le 1+\pi .$$

Using Theorem 3.2, we have that T has a unique fixed point $z\in [0,\frac{\pi }{2}]$. We also know that $z=Tz$ is equivalent to $cosz=z$. In fact,

$$\begin{array}{rl}z=Tz& ⟺z=(1+\frac{1}{2}z)(cosz-z)+z\\ ⟺0=(1+\frac{1}{2}z)(cosz-z)\\ ⟺0=cosz-z.\end{array}$$

Using Theorem 3.2, we can also show the following fixed point theorem for super hybrid non-self mappings in a Hilbert space; see [1].

Theorem 4.4 Let H be a Hilbert space, let C be a non-empty, bounded, closed and convex subset of H and let T be a super hybrid mapping from C into H, that is, there exist $\alpha ,\beta ,\gamma \in \mathbb{R}$ such that

$$\begin{array}{r}\alpha {\parallel Tx-Ty\parallel }^2+(1-\alpha +\gamma ){\parallel x-Ty\parallel }^2\\ \le (\beta +(\beta -\alpha )\gamma ){\parallel Tx-y\parallel }^2+(1-\beta -(\beta -\alpha -1)\gamma ){\parallel x-y\parallel }^2\\ +(\alpha -\beta )\gamma {\parallel x-Tx\parallel }^2+\gamma {\parallel y-Ty\parallel }^2\end{array}$$

for all $x,y\in C$. Assume that there exists $m>1$ such that for any $x\in C$,

$$Tx=x+t(y-x)$$

for some $y\in C$ and t with $0<t\le m$. Suppose that $\alpha -\beta \ge 0$ or $\gamma \ge 0$. Then T has a fixed point.

Proof An $(\alpha ,\beta ,\gamma )$-super hybrid mapping T from C into H is an $(\alpha ,1-\alpha +\gamma ,-\beta -(\beta -\alpha )\gamma ,-1+\beta +(\beta -\alpha -1)\gamma ,-(\alpha -\beta )\gamma ,-\gamma ,0)$-widely more generalized hybrid mapping. Furthermore, $\alpha +(1-\alpha +\gamma )+(-\beta -(\beta -\alpha )\gamma )+(-1+\beta +(\beta -\alpha -1)\gamma )=0$, $\alpha +(1-\alpha +\gamma )+(-\gamma )+0=1>0$ and $\alpha -\beta -(\beta -\alpha )\gamma -(\alpha -\beta )\gamma +0=\alpha -\beta \ge 0$, that is, it satisfies the conditions $\alpha +\beta +\gamma +\delta \ge 0$, $\alpha +\beta +\zeta +\eta >0$ and $\alpha +\gamma +\epsilon +\eta \ge 0$ in (2) of Theorem 3.2. Moreover, we have that

$$\begin{array}{r}[0,1)\cap \{\lambda \mid (\alpha +(-\beta -(\beta -\alpha )\gamma ))\lambda +(-(\alpha -\beta )\gamma )+0\ge 0\}\\ =[0,1)\cap \{\lambda \mid (\alpha -\beta )((1+\gamma )\lambda -\gamma )\ge 0\}.\end{array}$$

If $\alpha -\beta >0$, then

$$\begin{array}{rcl}[0,1)\cap \{\lambda \mid (\alpha -\beta )((1+\gamma )\lambda -\gamma )\ge 0\}& =& [0,1)\cap \{\lambda \mid (1+\gamma )\lambda -\gamma \ge 0\}\\ =& \{\begin{array}{cc}[0,1)\hfill & \text{if }\gamma <0,\hfill \\ [\frac{\gamma }{1+\gamma },1)\hfill & \text{if }\gamma \ge 0\hfill \end{array}\\ \ne & \mathrm{\varnothing },\end{array}$$

that is, it satisfies the condition $[0,1)\cap \{\lambda \mid (\alpha +\gamma )\lambda +\epsilon +\eta \ge 0\}\ne \mathrm{\varnothing }$ in (2) of Theorem 3.2. If $\alpha -\beta =0$, then

$$[0,1)\cap \{\lambda \mid (\alpha -\beta )((1+\gamma )\lambda -\gamma )\ge 0\}=[0,1)\ne \mathrm{\varnothing },$$

that is, it satisfies the condition $[0,1)\cap \{\lambda \mid (\alpha +\gamma )\lambda +\epsilon +\eta \ge 0\}\ne \mathrm{\varnothing }$ in (2) of Theorem 3.2. If $\alpha -\beta <0$ and $\gamma \ge 0$, then

$$\begin{array}{r}[0,1)\cap \{\lambda \mid (\alpha -\beta )((1+\gamma )\lambda -\gamma )\ge 0\}\\ =[0,1)\cap \{\lambda \mid (1+\gamma )\lambda -\gamma \le 0\}\\ =[0,\frac{\gamma }{1+\gamma }]\ne \mathrm{\varnothing },\end{array}$$

that is, it again satisfies the condition $[0,1)\cap \{\lambda \mid (\alpha +\gamma )\lambda +\epsilon +\eta \ge 0\}\ne \mathrm{\varnothing }$ in (2) of Theorem 3.2. Then we obtain the desired result from Theorem 3.2. Similarly, we obtain the desired result from Theorem 3.2 in the case of (1). □

We remark that some recent results related to this paper have been obtained in [14–17].