Using Theorem 3.1, we first show the following fixed point theorem for generalized hybrid nonself 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 nonempty, 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 TxTy\parallel}^{2}+(1\alpha ){\parallel xTy\parallel}^{2}\le \beta {\parallel Txy\parallel}^{2}+(1\beta ){\parallel xy\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,
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,
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 nonself mappings in a Hilbert space; see Kawasaki and Takahashi [13].
Theorem 4.2 Let H be a Hilbert space, let C be a nonempty, 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,
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 TxTy\parallel}^{2}+\beta {\parallel xTy\parallel}^{2}+\gamma {\parallel Txy\parallel}^{2}+\delta {\parallel xy\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}+max\{\epsilon {\parallel xTx\parallel}^{2},\zeta {\parallel yTy\parallel}^{2}\}\le 0\end{array}
for any x,y\in C. In the case of \alpha +\gamma +\epsilon >0, from
\epsilon {\parallel xTx\parallel}^{2}\le max\{\epsilon {\parallel xTx\parallel}^{2},\zeta {\parallel yTy\parallel}^{2}\},
we obtain that
\alpha {\parallel TxTy\parallel}^{2}+\beta {\parallel xTy\parallel}^{2}+\gamma {\parallel Txy\parallel}^{2}+\delta {\parallel xy\parallel}^{2}+\epsilon {\parallel xTx\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,
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 pseudocontractive mapping in the sense of Browder and Petryshyn [8]. We also define the following mapping: T:C\to H is called a generalized strict pseudocontractive mapping if there exist r,k\in \mathbb{R} with 0\le r\le 1 and 0\le k<1 such that
{\parallel TxTy\parallel}^{2}\le r{\parallel xy\parallel}^{2}+k{\parallel (xTx)(yTy)\parallel}^{2}
for any x,y\in C. Using Theorem 3.2, we can show the following fixed point theorem for generalized strict pseudocontractive nonself mappings in a Hilbert space.
Theorem 4.3 Let H be a Hilbert space, let C be a nonempty, bounded, closed and convex subset of H and let T be a generalized strict pseudocontractive 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 TxTy\parallel}^{2}\le r{\parallel xy\parallel}^{2}+k{\parallel (xTx)(yTy)\parallel}^{2}
for all x,y\in C. Assume that there exists m>1 such that for any x\in C,
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 pseudocontractive 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)=1k>0, 1+0+0+(k)=1k>0 and [0,1)\cap \{\lambda \mid (1+0)\lambda +0k\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,
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}\\ \phantom{\rule{1em}{0ex}}\u27fa\phantom{\rule{1em}{0ex}}\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\\ \phantom{\rule{1em}{0ex}}=\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}\\ \phantom{\rule{1em}{0ex}}=\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}{TxTy}^{2}+\frac{\pi}{1+\pi}{xy}^{2}\frac{\pi}{{(1+\pi )}^{2}}{xy(TxTy)}^{2}\\ \phantom{\rule{1em}{0ex}}=\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},\phantom{\rule{2em}{0ex}}{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{xy}^{2}
for all x,y\in [0,\frac{\pi}{2}]. Therefore, we have from (4.2) that
\frac{1}{1+\pi}{TxTy}^{2}+\frac{\pi}{1+\pi}{xy}^{2}\le r{xy}^{2}+\frac{\pi}{{(1+\pi )}^{2}}{xy(TxTy)}^{2}
for all x,y\in [0,\frac{\pi}{2}]. Furthermore, we have from (4.1) that
Tx=(1+\frac{1}{2}x)(cosxx)+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(yx)+x,y=cosx\in [0,\frac{\pi}{2}]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}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\phantom{\rule{1em}{0ex}}& \u27fa\phantom{\rule{1em}{0ex}}z=(1+\frac{1}{2}z)(coszz)+z\\ \u27fa\phantom{\rule{1em}{0ex}}0=(1+\frac{1}{2}z)(coszz)\\ \u27fa\phantom{\rule{1em}{0ex}}0=coszz.\end{array}
Using Theorem 3.2, we can also show the following fixed point theorem for super hybrid nonself mappings in a Hilbert space; see [1].
Theorem 4.4 Let H be a Hilbert space, let C be a nonempty, 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 TxTy\parallel}^{2}+(1\alpha +\gamma ){\parallel xTy\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}\le (\beta +(\beta \alpha )\gamma ){\parallel Txy\parallel}^{2}+(1\beta (\beta \alpha 1)\gamma ){\parallel xy\parallel}^{2}\\ \phantom{\rule{2em}{0ex}}+(\alpha \beta )\gamma {\parallel xTx\parallel}^{2}+\gamma {\parallel yTy\parallel}^{2}\end{array}
for all x,y\in C. Assume that there exists m>1 such that for any x\in C,
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\}\\ \phantom{\rule{1em}{0ex}}=[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\}\\ \phantom{\rule{1em}{0ex}}=[0,1)\cap \{\lambda \mid (1+\gamma )\lambda \gamma \le 0\}\\ \phantom{\rule{1em}{0ex}}=[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].