Around averaged mappings

Abstract

This paper is intended for a general mathematical audience. The examples show how the study of existence of fixed points of averaged mappings $T_{\lambda }= (1-\lambda )I+ \lambda T$, where $0<\lambda <1$ and I is the identity operator, can help in the study of existence of fixed points of mappings T.

1 Known results

Let X be a Banach space with norm $\Vert \cdot \Vert$, and C be a nonempty closed convex subset of X. Let $T:C\rightarrow C$ be a map. We denote the set of fixed points of T by ${\text {Fix}}(T)$.

Definition 1.1

[1] Let T be a self-mapping on a nonempty convex subset C of a Banach space. By an averaged mapping we mean one of the form $T_{\lambda }= (1-\lambda )I+ \lambda T:C\rightarrow C$, where $0<\lambda <1$, I is the identity operator.

Generally,

• (1) T and $T_{\lambda }$, both have the same fixed point set (${\text {Fix}}(T_{\lambda })={\text {Fix}}(T)$).

  Indeed, if $p\in {\text {Fix}}(T)$, then $T_{\lambda }p=(1-\lambda )p+\lambda Tp=p$. If $p\in {\text {Fix}}(T_{\lambda })$, then $p=T_{\lambda }p=(1-\lambda )p+\lambda Tp$, so $Tp=p$.

• (2) An averaged mapping of the averaged mapping is an averaged mapping.

  Let $\lambda _1,\lambda _2\in (0,1)$. If $T_{\lambda _1}=(1-\lambda _1)I+\lambda _1T$, then $(T_{\lambda _1})_{\lambda _2}=(1-\lambda _2)I+\lambda _2T_{\lambda _1}=(1-\lambda _2)I+ \lambda _2[(1-\lambda _1)I+\lambda _1T)]=(1-\lambda _1\lambda _2)I+\lambda _1\lambda _2T$.

• (3) In the class of nonexpansive mappings $F:C\rightarrow C$ (satisfying $\Vert Fx-Fy\Vert \leqslant \Vert x-y\Vert$ for all $x,y\in C$), if $S=(1-\lambda )T+\lambda V$ for some $\lambda \in (0,1)$ and if T is averaged and V is nonexpansive, then S is averaged.

  Indeed, if T is averaged, we have a number $\alpha \in (0,1)$ and a nonexpansive mapping $U:C\rightarrow C$ such that $T=(1-\alpha )I+\alpha U$. Put $\gamma =\lambda + \alpha (1-\lambda )\in (0,1)$ and $W=\frac{(1-\lambda )\alpha }{\gamma }U+\frac{\lambda }{\gamma }V$. Since W is a convex combination of two nonexpansive mappings, W remains nonexpansive. Now, we can express S as

  $$\begin{aligned} S= & {} (1-\lambda )T+\lambda V= (1-\lambda )((1-\alpha )I+\alpha U)+\lambda V=\\= & {} (1-\lambda )(1-\alpha )I+(1-\lambda )U+\lambda V=(1-\gamma )I+\gamma W. \end{aligned}$$

  So S is averaged.

• (4) In the class of nonexpansive mappings $F:C \rightarrow C$, if S and T are both averaged, then the composite ST is averaged.

  If S is averaged, we have a number $\alpha \in (0,1)$ and a nonexpansive mapping $U:C\rightarrow C$ such that $S=(1-\alpha )I+\alpha U$. Then $ST=(1-\alpha )T+\alpha UT$. Since UT is again nonexpansive, by (2), ST is averaged.

• (5) Let H be a real Hilbert space with inner product $\langle \cdot ,\cdot \rangle$ and C given a closed convex subset of H. Projection $P_C:H\rightarrow C$ is the unique point in C such that $\Vert x - P_Cx\Vert =\inf \{\Vert x-y\Vert :y\in C\}$. In particular, $P_C$ is nonexpansive and can be written as the mean average of a nonexpansive mapping and the identity: $P_C=\frac{1}{2}(I+S)$.

  Mapping $T:H\rightarrow H$ such that $\langle x-y,Tx -Ty\rangle \geqslant \Vert Tx - Ty\Vert ^2$ for all $x,y\in H$ is firmly nonexpansive (a projection is firmly nonexpansive). It is easily seen that T is firmly nonexpansive if and only if T can be expressed as $\frac{1}{2}(I+V)$, where V is nonexpansive, see [14].

  Thus, a firmly nonexpansive mapping (in particular, a projection) is an averaged map with $\lambda = \frac{1}{2}$, see (2).

• (5) Let H be a real Hilbert space. If $S,T:H\rightarrow H$ are both averaged (in the class of nonexpansive mappings) and if S and T have a common fixed point, then ${\text {Fix}}(S)\cap {\text {Fix}}(T)={\text {Fix}}(ST)$, details in [24].

In Banach spaces when T is nonexpansive mapping, $T_{\lambda }$ may have much more felicitous asymptotic behavior than the original mapping.

Krasnosielskij [19] was the first to notice this regularizing effect (for $\lambda =\frac{1}{2}$). He proved

Theorem 1.2

Let C be a nonempty compact convex subset of a uniformly convex Banach space X and $T:C\rightarrow C$ a nonexpansive mapping. Then the iterates $\{T^n_{\frac{1}{2}}\}$ (i.e. the sequence $x_{n+1}=\frac{1}{2}(x_n+Tx_n)$, $x_1\in C$ be an arbitrary, $n=1,2,3,\ldots$ ) converges strongly to a fixed point of T.

Since the existence of a fixed point is guaranteed by Schauder’s theorem, the novelty of this result lays in this asymptotic statement.

Shortly thereafter, Schaefer [23] proved that $\{T^n_{\lambda }\}$ converges strongly for all $0<\lambda <1$, and in 1966, Edelstein [13] noted that the uniform convexity of X can be relaxed to strict convexity.

In 1966, Browder and Petryshyn [9] separated out the notion of asymptotic regularity: T is said to be asymptotically regular on C if for each $x\in C$, $T^{n+1}x-T^nx \rightarrow 0$ as $n\rightarrow \infty$. Obviously, every Banach contraction or Kannan mapping is an asymptotically regular mapping having exactly one fixed point [7, 15]. For more information, see [1, 10, 12, 21, 22].

In 1976, Ishikawa [17] proved the truly astonishing result with no restrictions on the geometry of the Banach space!

Theorem 1.3

If C is nonempty bounded closed convex subset of a Banach space X and $T:C\rightarrow C$ is nonexpansive, then the mapping $T_{\lambda }$ is asymptotically regular for each $0<\lambda <1$.

Kirk [18] has extended Ishikawa’s method to metric spaces with convexity structure. For the rate of asymptotic regularity, see Bruck [11] and reference therein.

2 Examples

The search for fixed points for mappings $T: C\rightarrow C$ that do not fit into the “known patterns” is difficult.

Instead of the mapping T, we can, for example, look for the fixed points of mapping $T^n$, $n=2,3,\ldots$, nth iterate of T. Sometimes it is easier, see the following example.

Example 2.1

Let $\Vert \cdot \Vert$ be the usual norm on $\mathbb {R}$. Define $Tx=0$ if $x\leqslant 1$ and $Tx=1$ if $x>1$. Then T is discontinuous, T does not satisfy the Kannan type condition, i.e. there is no constant $K\in [0,\infty )$ such that for all $x,y\in X$ satisfying $\Vert Tx-Ty\Vert \leqslant K\cdot \{\Vert x-Tx\Vert +\Vert y-Ty\Vert \}$, take $x=0$ and $y=1+\varepsilon$, $\varepsilon >0$.

On the other hand, $T^2x\equiv 0$, so $T^2$ is a contraction and $p=0$ is the unique fixed point of T.

When the examination of successive iterations fails, we can look for fixed points of the averaged mapping instead of the T mapping. As we already know, the averaged mappings may have more favorable properties. In particular, averaged mapping $T_{\lambda }$ (for $\lambda \in (0,1)$) has a fixed point if and only if mapping T has it, see (1).

Example 2.2

Let $X=\mathbb {C}$ with the norm $\Vert z\Vert =\vert z\vert$ for complex numbers, $C=\{z\in \mathbb {C}:\vert z\vert \leqslant 1\}$, and $T:C\rightarrow C$ be a mapping defined by $Tz=iz$ (T is an anticlockwise rotation of $\frac{\pi }{4}$ about the origin of coordinates). Then T is nonexpansive ($\vert T((0,0))-T((1,0))\vert =1 =\vert (0,0)-(1,0)\vert$) with the origin as the only fixed point and T is not asymptotically regular. Indeed, let $z\in \mathbb {C}\setminus \{0\}$. Then,

$$\begin{aligned} \vert T^{n+1}z-T^nz\vert =\vert i^{n+1}z -i^nz\vert =\vert i^n\vert \vert 1-i\vert \vert z\vert =\sqrt{2}\vert z\vert \nrightarrow 0,~\hbox {as}~n\rightarrow \infty . \end{aligned}$$

However, $T_{\frac{1}{2}}=\frac{1}{2}(z+iz)$, $z\in \mathbb {C}$, is a contraction, for $z_1,z_2\in C$ we have,

$$\begin{aligned} \vert T_{\frac{1}{2}}z_1- T_{\frac{1}{2}}z_2\vert =\vert \frac{1}{2}(1+i)z_1- \frac{1}{2}(1+i)z_2\vert =\Big \vert \frac{1+i}{2}\Big \vert \vert z_1-z_2\vert =\frac{\sqrt{2}}{2}\vert z_1-z_2\vert . \end{aligned}$$

Thus, ${\text {Fix}}(T_{\frac{1}{2}})=\{0\}={\text {Fix}}(T)$ and $T^n_{\frac{1}{2}}z\rightarrow 0$ for all $z\in C$ (so $T_{\frac{1}{2}}$ is asymptotically regular).

Example 2.3

Let $X=\mathbb {R}$ with the usual norm $\Vert x\Vert =\vert x\vert$. Define a mapping $T:\mathbb {R}\rightarrow \mathbb {R}$ by $Tx=-3x+2$. For $x=0$ and $y=1$, we have $\vert T0-T1\vert =3>1= \vert 0-1\vert$, so T is not a nonexpansive mapping. But then $T_{\frac{1}{2}}x=\frac{1}{2}(x+Tx) = 1-x$, $x\in \mathbb {R}$, is isometry and $T_{\frac{1}{2}}$ does not satisfy Kannan’s condition ($\vert T_{\frac{1}{2}}0-T_{\frac{1}{2}}1\vert =1$ and $\{\vert 0-T_{\frac{1}{2}}0\vert +\vert 1-T_{\frac{1}{2}}1\vert \}=2$). But averaged mapping $(T_{\frac{1}{2}})_{\frac{1}{2}}=T_{\frac{1}{4}}x=(1-\frac{1}{4})x+ \frac{1}{4}Tx=\frac{1}{2}$ for all $x\in \mathbb {R}$, and $T_{\frac{1}{4}}$ is a contraction (see [4, Example 1 (2)]). Thus, ${\text {Fix}}(T_{\frac{1}{4}})={\text {Fix}}(T)$. Unfortunately, this procedure is not the rule. We will show this in the next example.

Example 2.4

Let $X=[1,\infty )$ with the usual norm $\Vert x\Vert =\vert x\vert$. Define a mapping $T:X\rightarrow X$ by $Tx=2x-1$. For $x=1$ and $y=2$, we have $\vert T1-T2\vert =2>1= \vert 1-2\vert$, so T is not a nonexpansive mapping. Similarly for mappings $T_{\lambda }=(1-\lambda )I+\lambda T$, $0< \lambda <1$, we have $\vert T_{\lambda }1-T_{\lambda }2\vert =1+\lambda >1 =\vert 1-2\vert$.

Example 2.5

Let $X=l_{\infty }$ be a normed space of all bounded sequences $x=(x_1,x_2,\ldots )$ of real numbers with the norm $\Vert x\Vert =\sup \limits _{i}\vert x_i\vert$. Let $C=\{x\in l_{\infty }:\Vert x \Vert \leqslant 1\}$. Define $T:C\rightarrow C$ by $Tx=(0,x_1^2,x_2^2,x_3^2,\ldots )$ for $x=(x_1,x_2,x_3,\ldots )$ in C. Obviously, ${\text {Fix}}(T)=\{0\}$, where $0=(0,0,0,\ldots )$.

If $a=(\frac{3}{4},\frac{3}{4},\frac{3}{4},\ldots )$ and $b=(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots )$, then $\Vert a-b\Vert =\Vert (\frac{1}{4},\frac{1}{4},\frac{1}{4},\ldots )\Vert =\frac{1}{4}$, and

$$\begin{aligned} \Vert Ta-Tb\Vert =\Big \Vert (0,\frac{5}{16},\frac{5}{16},\ldots )\Big \Vert =\frac{5}{16}>\frac{1}{4}=\Vert a-b\Vert . \end{aligned}$$

Let $T_{\lambda }x=(1-\lambda )x+\lambda Tx$, $0<\lambda <1$. Then

$$\begin{aligned}&\Vert T_{\lambda }a -T_{\lambda }b\Vert = \Big \Vert \Big ( (1-\lambda )(\frac{3}{4}-\frac{1}{2}), (1-\lambda )(\frac{3}{4}-\frac{1}{2})+\lambda ((\frac{3}{4})^2-(\frac{1}{2})^2),\\&\qquad (1-\lambda )(\frac{3}{4}-\frac{1}{2})+\lambda ((\frac{3}{4})^2-(\frac{1}{2})^2),\ldots \Big )\Big \Vert =\\&\quad =\Big \Vert \Big ((1-\lambda )\frac{1}{4},\frac{1}{4}+\lambda \cdot \frac{1}{16},\frac{1}{4}+\lambda \cdot \frac{1}{16},\ldots \Big )\Big \Vert =\frac{1}{4}+\lambda \cdot \frac{1}{16}> \frac{1}{4}=\Vert a-b \Vert . \end{aligned}$$

Hence, T and averaged mappings $T_{\lambda }$ for all $0<\lambda <1$ are not nonexpansive.

We will illustrate the procedure described above with another examples.

Example 2.6

[15, Example 2.12] Let $X=[0,\frac{\pi }{2}]$ with the usual norm $\Vert x\Vert =\vert x\vert$ and $T:X\rightarrow X$ be given by $Tx=\cos x$. Then T is not a contraction. Suppose there exists $M<1$ such that $\vert \frac{\cos x - \cos y}{x - y}\vert \leqslant M$ for all $x\ne y$. Letting $y\rightarrow x$, we get $\vert \sin x\vert \leqslant M$ for all $x,y\in [0,\frac{\pi }{2}]$, which is false. Of course, T as a continuous mapping, by Schauder’s Theorem has a fixed point. This can be obtained by weaker means. Consider the mapping $T_{\frac{1}{2}}x=\frac{1}{2}(x+\cos x)$, $x\in X$. Let $x,y\in X$ and $x<y$. By Mean Value Theorem,

$$\begin{aligned} \vert T_{\frac{1}{2}}x-T_{\frac{1}{2}}y\vert =\vert \frac{1}{2}- \frac{1}{2}\sin \xi \vert \cdot \vert x-y\vert \leqslant \frac{1}{2}\vert x -y\vert , \end{aligned}$$

for some $x<\xi <y$, so $T_{\frac{1}{2}}$ is a contraction and ${\text {Fix}}(T)={\text {Fix}}(T_{\frac{1}{2}})=\{p\}$, $p\approx 0.739$.

Example 2.7

Let $X=[1,\infty )$ with the usual norm $\Vert x\Vert =\vert x\vert$ and $T:X\rightarrow X$ be given by $Tx=\frac{1}{x^2}$. Since $\vert T1-T(1+\frac{1}{10})\vert = \frac{21}{121}\approx 0.17 > 0.1 =\vert 1-(1+\frac{1}{10})\vert$, T is not a nonexpansive mapping. Consider the mapping $T_{\frac{1}{2}}x=\frac{1}{2}(x+\frac{1}{x^2})$, $x\geqslant 1$. By Mean Value Theorem,

$$\begin{aligned} \vert T_{\frac{1}{2}}x-T_{\frac{1}{2}}y\vert =\vert \frac{1}{2}- \frac{1}{\xi ^3}\vert \cdot \vert x-y\vert \leqslant \frac{1}{2}\vert x -y\vert ,~~x,y\geqslant 1 \end{aligned}$$

for some $x<\xi <y$, so $T_{\frac{1}{2}}$ is a contraction and ${\text {Fix}}(T)={\text {Fix}}(T_{\frac{1}{2}})=\{1\}$.

Remark 2.8

Let F be a self-mapping on a nonempty convex subset C of a Banach space X. By a backwards mapping for F we mean the mapping $T_F:D\rightarrow X$ ($D\subset X$) such that $F=(1-\lambda )I+\lambda T_F$ for some $0<\lambda <1$, i.e., $T_F=(1-\frac{1}{\lambda })I+\frac{1}{\lambda }F$.

Generally,

• (6) F and $T_F$($=$the backwards mapping for F) both have the same fixed point set, ${\text {Fix}}(F)={\text {Fix}}(T_F)$.

  Indeed, if $p\in {\text {Fix}}(F)$, then $T_Fp=(1-\frac{1}{\lambda })p+\frac{1}{\lambda }Fp=p$. If $p\in {\text {Fix}}(T_F)$, then $T_Fp=(1-\frac{1}{\lambda })p+\frac{1}{\lambda }Fp=p$, so $Fp=p$.

• (7) A backwards mapping of the backwards mapping is a backwards mapping.

  Let $\lambda _1,\lambda _2\in (0,1)$ and $F:C\rightarrow C$. If $T_F=(1-\frac{1}{\lambda _1})I+\frac{1}{\lambda _1}F$, then $G_{T_F}=(1-\frac{1}{\lambda _2})I+\frac{1}{\lambda _2}T_F=(1-\frac{1}{\lambda _2})I+\frac{1}{\lambda _2}[(1-\frac{1}{\lambda _1})I+\frac{1}{\lambda _1}F]=(1-\frac{1}{\lambda _1\lambda _2})I+\frac{1}{\lambda _1\lambda _2}F$.

3 New results

The following two problems are significant in the study of fixed points:

• (1) What conditions on the structure of the ambient space X and / or on the properties of T must be added to assure that mapping T has at least one fixed point?

• (2) How one can locate and approximate such a fixed point?

The literature on the subject is very extensive and multi-threaded. In recently published papers by Berinde [2, 3], Berinde and Pǎcurar [5, 6] drew attention to the so-called enriched nonexpansive (contractive, Kannan, Chatterjea) mappings. In the next theorems, we will give some extension of the Berinde and Pǎcurar result for enriched contractions [4] and some earlier results from paper [16].

Let $b\in [0,\infty )$ and let $\mathcal {S}$ denote the class of those functions $\alpha :[0,\infty )\rightarrow [0,b+1)$ which satisfy the simple condition: $\alpha (t_n)\rightarrow b+1 \Rightarrow t_n\rightarrow 0$. (We do not assume that $\alpha$ is continuous in any sense.) For example, $\alpha _1 (t)=(b+1)e^{-t}$, $\alpha _2(t)=\frac{b+1}{t+1}$, $t>0$.

Theorem 3.1

Let $(X,\Vert \cdot \Vert )$ be a Banach space. Suppose $T:X\rightarrow X$ is a mapping for which exists $0\leqslant b,K<\infty$, $\alpha \in \mathcal {S}$, such that for each $x,y\in X$,

$$\begin{aligned} \Vert b(x-y)+Tx-Ty \Vert \leqslant \alpha (\Vert x-y \Vert )\cdot \Vert x-y \Vert +K\cdot \{\Vert x-Tx \Vert + \Vert y-Ty \Vert \}\, \,(8) \end{aligned}$$

and $T_{\frac{1}{b+1}}$ is an asymptotically regular mapping. If T is continuous, then T has a unique fixed point $z\in X$ and for each $x_0\in X$ a sequence given by $x_{n+1}=T_{\frac{1}{b+1}}x_n$, $n=0,1,\ldots$, converges to z as $n\rightarrow \infty$.

Proof

Let us consider two cases.

Case 1.:

When $b=0$, then we can apply [16, Theorem 2.1].

Case 2.:

Let $b>0$ and let us denote $\lambda = \frac{1}{b+1}\in (0,1)$. By (8), we get

$$\begin{aligned} \Vert \left( \frac{1-\lambda }{\lambda }\right) (x-y)+Tx-Ty \Vert \leqslant \alpha (\Vert x-y \Vert )\cdot \Vert x-y \Vert +K\cdot \{\Vert x-Tx \Vert + \Vert y-Ty \Vert \}, \end{aligned}$$

which can be written in an equivalent form as

$$\begin{aligned} \Vert [(1-\lambda )x+\lambda Tx]-[(1-\lambda )y+\lambda Ty] \Vert\leqslant & {} \lambda \cdot \alpha (\Vert x-y \Vert )\cdot \Vert x-y \Vert \\&+K\cdot \{\Vert x-[(1-\lambda )x+\lambda Tx] \Vert \\&+ \Vert y-[(1-\lambda )y+\lambda Ty] \Vert , \end{aligned}$$

and

$$\begin{aligned} \Vert T_{\lambda }x-T_{\lambda }y \Vert \leqslant \lambda \cdot \alpha (\Vert x-y \Vert )\cdot \Vert x-y \Vert +K\cdot \{\Vert x-T_{\lambda }x \Vert + \Vert y-T_{\lambda }y \Vert \}. \,\,(9) \end{aligned}$$

Fix $x_0\in X$ and let $x_{n+1}=T_{\lambda }x_n$, $n=0,1,2,\ldots$ If $x_{n+1}=x_n$ for some $n\in \mathbb {N}\cup \{0\}$, then $(1-\lambda )x_n+\lambda Tx_n=x_n$, so $Tx_n=x_n$ is the fixed point of T. Suppose $x_{n+1}\ne x_n$ for all $n\geqslant 0$. Assume that $\{x_n\}$ is not a Cauchy sequence. Then $\limsup \limits _{n,m\rightarrow \infty }\Vert x_n-x_m\Vert >0$. In view of triangle inequality and (9), we get

$$\begin{aligned} \Vert x_n- x_m\Vert\leqslant & {} \Vert x_n-x_{n+1}\Vert +\Vert x_{n+1}-x_{m+1}\Vert +\Vert x_{m+1}-x_m\Vert \\\leqslant & {} \lambda \cdot \alpha (\Vert x_n-x_m\Vert )\Vert x_n-x_m\Vert +(K+1)\{\Vert x_n-x_{n+1}\Vert +\Vert x_m-x_{m+1}\Vert \}. \end{aligned}$$

Under the assumption $\limsup \limits _{n,m\rightarrow \infty }\Vert x_n-x_m\Vert >0$, asymptotic regularity of $\{x_n\}$, the inequality

$$\begin{aligned} \dfrac{\Vert x_n-x_m\Vert }{\Vert x_n-x_{n+1}\Vert +\Vert x_m-x_{m+1}\Vert }\leqslant \dfrac{K+1}{1-\lambda \cdot \alpha (\Vert x_n-x_m\Vert )} \end{aligned}$$

now implies

$$\begin{aligned} \limsup _{n,m\rightarrow \infty }\dfrac{K+1}{1-\lambda \cdot \alpha (\Vert x_n-x_m\Vert )}=+\infty , \end{aligned}$$

from which

$$\begin{aligned} \limsup _{n,m\rightarrow \infty }\alpha (\Vert x_n-x_m\Vert )=\frac{1}{\lambda }=b+1. \end{aligned}$$

But since $\alpha \in \mathcal {S}$ this implies $\limsup \limits _{n,m\rightarrow \infty }\Vert x_n-x_m\Vert =0$, which is a contradiction. Thus, $\{x_n\}$ is a Cauchy sequence. Because X is a Banach space $\lim \limits _{n\rightarrow \infty }x_n=z\in X$.

Using the continuity of $T_{\lambda }$, we immediately obtain $z=T_{\lambda }z$, so by (1), $Tz=z$.

Assume that $u\ne z$ is another fixed point of $T_{\lambda }$. Then by (9),

$$\begin{aligned} 0<\Vert z-u\Vert \leqslant \lambda \cdot \alpha (\Vert z-u\Vert )\Vert z-u\Vert < \Vert z-u\Vert , \end{aligned}$$

a contradiction. Hence, z is the unique fixed point of T.

It is clear that for each $x_0\in X$, the sequence $\{x_{n+1}=T_{\lambda }x_n\}$ converges to z. $\square$

The next result is inspired by theorem of Boyd and Wong [8]. A mapping T satisfying $\Vert Tx-Ty\Vert \leqslant \varphi (\Vert x-y\Vert )$, $\varphi (t)<t$ for each $t>0$, may not posses a fixed point unless some additional condition is assumed on $\varphi$. Boyd and Wong [8] assumed $\varphi$ to be upper semi-continuous from the right.

Let $b\in [0,\infty )$ and let $\mathcal {U}$ denote the class of all mappings $\varphi :[0,\infty )\rightarrow [0,\infty )$ satisfying

• (a) $\varphi (t)<t(b+1)$ for all $t>0$,

• (b) $\varphi$ is upper semi-continuous, that is $t_n\rightarrow t\geqslant 0 \Rightarrow \limsup \limits _{n\rightarrow \infty }\varphi (t_n)\leqslant \varphi (t)$.

Theorem 3.2

Let $(X,\Vert \cdot \Vert )$ be a Banach space. Suppose $T:X\rightarrow X$ is a mapping for which exists $0\leqslant b,K<\infty$, $\varphi \in \mathcal {U}$ such that for each $x,y\in X$,

$$\begin{aligned} \Vert b(x-y)+ Tx -Ty\Vert \leqslant \varphi (\Vert x-y\Vert )+K\cdot \{\Vert x-Tx\Vert +\Vert y-Ty\Vert \}\, \,(10) \end{aligned}$$

and $T_{\frac{1}{b+1}}$ is an asymptotically regular mapping. If T is continuous, then T has a unique fixed point $z\in X$ and for each $x_0\in X$ a sequence given by $x_{n+1}=T_{\frac{1}{b+1}}x_n$, $n=0,1,2,\ldots$, converges to z as $n\rightarrow \infty$.

Proof

Let us consider two cases.

Case 1.:

When $b=0$, then we can apply [16, Theorem 2.2].

Case 2.:

Let $b>0$ and let us denote $\lambda = \frac{1}{b+1}\in (0,1)$. Analogically as in the proof of theorem 2.6, we get

$$\begin{aligned} \Vert T_{\lambda }x-T_{\lambda }y \Vert \leqslant \lambda \cdot \varphi (\Vert x-y \Vert )+K\cdot \{\Vert x-T_{\lambda }x \Vert + \Vert y-T_{\lambda }y \Vert \}.\, \,(11) \end{aligned}$$

Let $x_0\in X$ and let $x_{n+1}=T_{\lambda }x_n$, $n=0,1,2,\ldots$ If $x_{n+1}=x_n$ for some $n\in \mathbb {N}\cup \{0\}$, then $(1-\lambda )x_n+\lambda Tx_n=x_n$, so $Tx_n=x_n$ is the fixed point of T. Suppose $x_{n+1}\ne x_n$ for all $n\geqslant 0$. Assume that $\{x_n\}$ is not a Cauchy sequence. Then there exists $\varepsilon >0$ and integers $n_i,m_i\in \mathbb {N}$ such that $m_i>n_i\geqslant i$ and

$$\begin{aligned} \Vert x_{n_i}-x_{m_i}\Vert \geqslant \varepsilon ~~\hbox {for}~~i=1,2,\ldots \end{aligned}$$

Also, choosing $m_i$ small as possible, if may be assumed that

$$\begin{aligned} \Vert x_{n_i}-x_{m_i-1}\Vert <\varepsilon . \end{aligned}$$

Hence, for each $i\in \mathbb {N}$, we have

$$\begin{aligned} \varepsilon \leqslant \Vert x_{n_i}-x_{m_i}\Vert \leqslant&~\Vert x_{n_i}-x_{m_i-1}\Vert +\Vert x_{m_i-1}-x_{m_i}\Vert<\\ <&~\varepsilon + \Vert x_{m_i-1}-x_{m_i}\Vert , \end{aligned}$$

and it follows from asymptotic regularity that

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert x_{n_i}-x_{m_i}\Vert =\varepsilon . \end{aligned}$$

Next observe, by (11), that

$$\begin{aligned} \Vert x_{n_i}-x_{m_i}\Vert \leqslant&~ \Vert x_{n_i}-x_{n_i+1}\Vert +\Vert x_{n_i+1}-x_{m_i+1}\Vert +\Vert x_{m_i+1}-x_{m_i}\Vert \leqslant \\ \leqslant&~\lambda \cdot \varphi (\Vert x_{n_i}-x_{m_i}\Vert )+\\&~+(K+1)\{\Vert x_{n_i}-x_{n_i+1}\Vert +\Vert x_{m_i}-x_{m_i+1}\Vert \}. \end{aligned}$$

Letting $i\rightarrow \infty$, using asymptotic regularity and upper semi-continuity of $\varphi$, we obtain

$$\begin{aligned} 0<\varepsilon =\lim _{i\rightarrow \infty }\Vert x_{n_i}-x_{m_i}\Vert \leqslant \limsup _{i\rightarrow \infty }\lambda \cdot \varphi (\Vert x_{n_i}-x_{m_i}\Vert )\leqslant \lambda \cdot \varphi (\varepsilon )<\varepsilon , \end{aligned}$$

which is a contradiction. Hence, $\{x_n\}$ is a Cauchy sequence. The conclusion of the proof follows as in the proof of Theorem 3.1. $\square$

Example 3.3

[16, Example 2.3] If $b=0$ and $Tx=\dfrac{x}{x+1}$ if $x\geqslant 0$, then T does not satisfy the Banach theorem, take $x=0$ and $y=\varepsilon$, where $\varepsilon > 0$ and $\varepsilon \downarrow 0$. On the other hand, T satisfy $\vert Tx -Ty\vert \leqslant \varphi (\vert x-y\vert ))$ with $\varphi (t)=\dfrac{t}{t+1}$. Obviously, the mapping T is asymptotically regular and has a unique fixed point $p=0$.

Now we note that T does not satisfy the Kannan type condition, i.e. there is no constant $K\in [0,\infty )$ such that for all $x,y\geqslant 0$ satisfying $\vert Tx-Ty\vert \leqslant K\cdot \{\vert x-Tx\vert +\vert y-Ty\vert \}$, take $x=0$ and $y=\varepsilon$, where $\varepsilon$ is sufficiently small. Therefore we cannot apply [15, Theorem 2.6] or [7, Theorem 2.1].

In the next result, we show that the assumption of the existence of constant $K\in [0,\infty )$ can be relaxed to weaker concept using certain functions, see [20].

Let $\mathcal {B}$ denote the class of functions $\beta : [0,\infty ) \rightarrow [0,\infty )$ satisfying the following condition, $\limsup \limits _{t\rightarrow 0}\beta (t) <\infty$.

Using the approach of Theorem 3.2, we prove the following result:

Theorem 3.4

Let $(X,\Vert \cdot \Vert )$ be a Banach space. Suppose $T:X\rightarrow X$ is a mapping for which exists $0\leqslant b<\infty$, $\varphi \in \mathcal {U}$, $\beta _1,\beta _2\in \mathcal {B}$ such that for each $x,y\in X$,

$$\begin{aligned} \Vert b(x-y)+ Tx -Ty\Vert\leqslant & {} \\\leqslant & {} \varphi (\Vert x-y\Vert )+ \beta _1\left( \frac{1}{b+1}\Vert x-Tx \Vert \right) \cdot \Vert x-Tx \Vert \\&+ \beta _2\left( \frac{1}{b+1}\Vert y-Ty \Vert \right) \cdot \Vert y-Ty \Vert \, \qquad \qquad \qquad \,(12) \end{aligned}$$

and $T_{\frac{1}{b+1}}$ is an asymptotically regular mapping. If T is continuous, then T has a unique fixed point $z\in X$ and for each $x_0\in X$ a sequence given by $x_{n+1}=T_{\frac{1}{b+1}}x_n$, $n=0,1,2,\ldots$, converges to z as $n\rightarrow \infty$.

Proof

Let us consider two cases.

Case 1.:

When $b=0$, then we can apply [20, Theorem 3, for ${{\varvec{a}}}={{\varvec{b}}}=0$ and $\ell _1=\ell _2=1$].

Case 2.:

Let $b>0$ and let us denote $\lambda = \frac{1}{b+1}\in (0,1)$. Analogically as in the proof of Theorem 3.1, we get

$$\begin{aligned} \Vert T_{\lambda }x-T_{\lambda }y \Vert\leqslant & {} \\\leqslant & {} \lambda \cdot \varphi (\Vert x-y \Vert )+ \beta _1(\Vert x-T_{\lambda }x \Vert )\cdot \Vert x-T_{\lambda }x \Vert \\&+ \beta _2(\Vert y-T_{\lambda }y \Vert )\cdot \Vert y-T_{\lambda }y \Vert . \, \qquad \qquad \qquad \qquad \qquad \qquad \,(13) \end{aligned}$$

Let $x_0\in X$ and let $x_{n+1}=T_{\lambda }x_n$, $n=0,1,2,\ldots$ If $x_{n+1}=x_n$ for some $n\in \mathbb {N}\cup \{0\}$, then $(1-\lambda )x_n+\lambda Tx_n=x_n$, so $Tx_n=x_n$ is the fixed point of T. Suppose $x_{n+1}\ne x_n$ for all $n\geqslant 0$. Assume that $\{x_n\}$ is not a Cauchy sequence. Then there exists $\varepsilon >0$ and integers $n_i,m_i\in \mathbb {N}$ such that $m_i>n_i\geqslant i$ and

$$\begin{aligned} \Vert x_{n_i}-x_{m_i}\Vert \geqslant \varepsilon ~~\hbox {for}~~i=1,2,\ldots \end{aligned}$$

We may choose $m_i$ small as possible such that

$$\begin{aligned} \Vert x_{n_i}-x_{m_i-1}\Vert <\varepsilon . \end{aligned}$$

In view of triangle inequality, we have for each $i\in \mathbb {N}$ that

$$\begin{aligned} \varepsilon \leqslant \Vert x_{n_i}-x_{m_i}\Vert \leqslant \Vert x_{n_i}-x_{m_i-1}\Vert +\Vert x_{m_i-1}-x_{m_i}\Vert < \varepsilon + \Vert x_{m_i-1}-x_{m_i}\Vert . \end{aligned}$$

It follows from asymptotic regularity that

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert x_{n_i}-x_{m_i}\Vert =\varepsilon . \end{aligned}$$

Next observe from (13), we have for each i that

$$\begin{aligned} \Vert x_{n_i}- x_{m_i}\Vert \leqslant&\Vert x_{n_i}- x_{n_i+1}\Vert + \Vert x_{m_i}- x_{m_i+1}\Vert + \Vert x_{n_i+1}- x_{m_i+1}\Vert \leqslant \\ \leqslant&\Vert x_{n_i}- x_{n_i+1}\Vert + \Vert x_{m_i}- x_{m_i+1}\Vert + \\&+\lambda \cdot \varphi (\Vert x_{n_i}- x_{m_i}\Vert ) + \\&+ \beta _1(\Vert x_{n_i}-x_{n_i+1}\Vert )\cdot \Vert x_{n_i}- x_{n_i+1}\Vert ) + \\&+ \beta _2(\Vert x_{m_i}- x_{m_i+1}\Vert )\cdot \Vert x_{m_i}- x_{m_i+1}\Vert . \end{aligned}$$

It follows from $\mathcal {B}$ that there exist $i_1\ge i_0 \in \mathbb {N}$ and $L>0$ such that

$$\begin{aligned} \beta _1(\Vert x_{n_i}-x_{n_i+1}\Vert )\leqslant L,\quad \text{ and }\quad \beta _2(\Vert x_{m_i}-x_{m_i+1}\Vert )\leqslant L \end{aligned}$$

for all $i\ge i_1$. Thus, by above inequality, we have for all $i\ge i_1$,

$$\begin{aligned} \Vert x_{n_i}-x_{m_i}\Vert \leqslant&\Vert x_{n_i}-x_{n_i+1}\Vert + \Vert x_{m_i}-x_{m_i+1}\Vert + \\&+ \lambda \cdot \varphi (\Vert x_{n_i}-x_{m_i}\Vert ) + \\&+ L\{\Vert x_{n_i}-x_{n_i+1}\Vert + \Vert x_{m_i}-x_{m_i+1}\Vert \}. \end{aligned}$$

Making $i\rightarrow \infty$, using asymptotic regularity and upper semi-continuity of $\varphi$, we get

$$\begin{aligned} 0<\varepsilon =\lim _{i\rightarrow \infty }\Vert x_{n_i}-x_{m_i}\Vert \leqslant \limsup _{i\rightarrow \infty }\lambda \cdot \varphi (\Vert x_{n_i}-x_{m_i}\Vert )\leqslant \lambda \cdot \varphi (\varepsilon )<\varepsilon , \end{aligned}$$

which is a contradiction. Hence, $\{x_n\}$ is a Cauchy sequence. The conclusion of the proof follows as in the proof of Theorem 3.1. $\Box$ Now notice that mappings satisfying the condition

$$\begin{aligned} \vert Tx-Ty\vert <\vert x-y\vert +K\cdot \{\vert x-Tx\vert + \vert y-Ty\vert \}~\hbox {for all}~x,y\in X, x\ne y, \end{aligned}$$

where $0\leqslant K<\infty$, may not have a fixed point.

Example 3.5

[16, Example 2.5] Let $X=\mathbb {R}$ be with the usual norm and let $Tx=\ln (1+e^x)$, $x\in \mathbb {R}$. Then T is asymptotically regular ($T^nx=\ln (n+e^x)$, $n=1,2,\ldots$) and fixed point free ($Tx>x$ for all $x\in \mathbb {R}$). It is easy to see that T satisfies

$$\begin{aligned} \vert Tx-Ty\vert <\vert x-y\vert ~~\hbox {for}~~x\ne y. \end{aligned}$$

Now we note, that T does not satisfy the Kannan type condition, i.e. there is no constant $K\in [0,\infty )$ such that for all $x,y\geqslant 0$ satisfying $\vert Tx-Ty\vert \leqslant K\cdot \{\vert x-Tx\vert +\vert y-Ty\vert \}$: for sufficiently large values $n\in \mathbb {N}$, we have

$$\begin{aligned} \vert T(2n)-T(n)\vert \sim n~~\hbox {and}~~\max \{\vert n-T(n)\vert , \vert 2n- T(2n)\vert \}\sim 0. \end{aligned}$$

Added in proof. Recently published paper Popescu, O., A new class of contractive mappings, Acta Math. Hungar. https://doi.org/10.1007/s10474-021-01154-6, contains additional information about enriched contractions, enriched Kannan mappings and enriched Chatterjea mappings.