In what follows, we always assume that \((Y,d,\eta )\) is a complete b-metric space. Moreover, E denotes a nonempty set and, given \(\nu ,\mu \in {Y}^E\), we define a mapping \(d(\nu ,\mu ):E\rightarrow \mathbb {R}_+\) by the formula:
$$\begin{aligned} d(\nu ,\mu )(t):=d(\nu (t),\mu (t)),\qquad \nu ,\mu \in {Y}^E,t\in E. \end{aligned}$$
Similarly, as in the usual metric spaces, if \((\chi _n)_{n\in \mathbb {N}}\) is a sequence in \(Y^E\), then we say that \(\chi \in {Y}^E\) is a pointwise limit of \((\chi _n)_{n\in \mathbb {N}}\) when
$$\begin{aligned} \lim _{n\rightarrow \infty } d(\chi ,\chi _n)(t)=0,\qquad t\in E; \end{aligned}$$
\(\chi \in {Y}^E\) is a uniform limit of \((\chi _n)_{n\in \mathbb {N}}\) if
$$\begin{aligned} \lim _{n\rightarrow \infty }\;\sup _{t\in E} \,d(\chi ,\chi _n)(t)=0. \end{aligned}$$
We say that a subset \(\mathcal {F}\ne \emptyset \) of \({Y}^E\) is p-closed (u-closed, respectively) if each \(\chi \in {Y}^E\) that is a pointwise (uniform, resp.) limit of a sequence of elements of \(\mathcal {F}\), is an element of \(\mathcal {F}\).
Further, if \(f,g\in \mathbb {R}^E\), then we write \(f\le g\) provided \(f(t)\le g(t)\) for \(t\in E\).
First note that from Theorem 1 we can easily deduce the following partial generalization of Theorem 2 (i.e., of [13, Theorem 2.1]), in which we do not need to presume any particular form of \(\varLambda \), but has to replace (4) with a somewhat different estimate (see the definition of \(\varepsilon ^*\) in (2)) and formulate the statement on uniqueness similarly as in Theorem 1.
Theorem 7
Let \(\varLambda :{\mathbb {R}_+}^E\rightarrow {\mathbb {R}_+}^E\) fulfil \((\mathcal {C}_0)\), \(\mathcal {T}:Y^E\rightarrow Y^E\) be \(\varLambda \)-contractive, and \(\varepsilon :E\rightarrow \mathbb {R}_+\) and \(\varphi :E\rightarrow Y\) be such that
$$\begin{aligned} d\big ((\mathcal {T}\varphi )(t),\varphi (t)\big )\le \varepsilon (t),\qquad \sum _{k\in \mathbb {N}_0} ({\varLambda _{\xi }} ^{k}\varepsilon ^{\,\xi })(t)<\infty , \qquad t\in E, \end{aligned}$$
(7)
where \(\xi := log_{\,2\eta } \,2\) and \(\varepsilon ^{\,\xi }:E\rightarrow \mathbb {R}_+\), \({\varLambda _{\xi }}:{\mathbb {R}_+}^E\rightarrow {\mathbb {R}_+}^E\) are given by
$$\begin{aligned} \varepsilon ^{\,\xi }(t):=(\varepsilon (t))^{\xi },\qquad ({\varLambda _{\xi }}\delta )(t)=\big (({\varLambda \delta )(t)\big )^{\xi }},\qquad \delta \in {\mathbb {R}_+}^E, t\in E. \end{aligned}$$
Then the limit
$$\begin{aligned} \psi (t):=\lim _{n\rightarrow \infty }(\mathcal {T}^n\varphi )(t) \end{aligned}$$
exists for each \(t\in E\) and the function \(\psi \in Y^E\), defined in this way, is a unique fixed point of \(\mathcal {T}\) with
$$\begin{aligned} d\big (\mathcal {T}^n\varphi (t),\psi (t)\big )^{\xi }\le 4\sum _{k=n}^{\infty } \big (({\varLambda _{\xi }}^{k}\varepsilon ^{\,\xi }\,)(t)\big )^{\xi },\qquad t\in E,n\in \mathbb {N}_0. \end{aligned}$$
Proof
According to Theorem 5, (7) and the \(\varLambda \)-contractivity of \(\mathcal {T}\),
$$\begin{aligned} D_d\big ((\mathcal {T}\varphi )(t),\varphi (t)\big )&\le d\big ((\mathcal {T}\varphi )(t),\varphi (t)\big )^{\xi }\le \varepsilon ^{\,\xi }(t), \\ D_d\big ((\mathcal {T}\nu )(t),(\mathcal {T}\mu )(t)\big )&\le d\big ((\mathcal {T}\nu )(t),(\mathcal {T}\mu )(t)\big )^{\xi } \\&\le \big (\big (\varLambda d(\nu ,\mu )\big )(t)\big )^{\xi }=\big (\varLambda _{\xi } d(\nu ,\mu )\big )(t) \end{aligned}$$
for every \(\nu ,\mu \in Y^E \) and \(t\in E.\) Moreover, note that the completeness of \((Y,d,\eta )\) implies the completeness of the metric space \((Y,D_d)\) (see (6)). So, by Theorem 5 and Theorem 1 (with \(\rho =D_d\) and \(\varLambda \) and \(\varepsilon \) replaced by \(\varLambda _{\xi }\) and \(\varepsilon ^{\xi }\)), we obtain the statement. \(\square \)
The next fixed point theorem provides a further, a bit different, generalization of Theorem 2; it is the main result of this paper. To simplify some formulas, we denote by \(\varLambda _0\) the identity operator on \({\mathbb {R_+}}^{E}\), i.e., \(\varLambda _0 \delta =\delta \) for all \(\delta \in {\mathbb {R_+}}^{E}\).
Theorem 8
Let \(\mathcal {C} \subset {Y}^E\) be nonempty, \(\mathcal {T}:\mathcal {C}\rightarrow \mathcal {C}\), \(\varLambda _n:{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}\) for \(n\in \mathbb {N}\), and \(\mathcal {T}^n\) be \(\varLambda _{n}\) – contractive for \(n\in \mathbb {N}\). Suppose there are \(\varepsilon \in {\mathbb {R_+}}^{E}\) and \(\varphi \in \mathcal {C}\) with
$$\begin{aligned} d(\mathcal {T} \varphi ,\varphi )\le \varepsilon ,\qquad \varepsilon ^*(t):=\sum _{i=0}^{\infty }\big ((\varLambda _i\varepsilon )(t)\big )^{\xi }<\infty ,\qquad t\in E, \end{aligned}$$
(8)
and one of the following two hypotheses is valid.
-
(i)
\(\mathcal {C}\) is p-closed.
-
(ii)
\(\mathcal {C}\) is u-closed and the sequence \(\Big (\sum _{i=0}^{n} (\varLambda _i\varepsilon )^{\xi }\Big )_{n\in \mathbb {N}}\) tends uniformly to \(\varepsilon ^*\).
Then, for each \(t\in E\), the limit
$$\begin{aligned} \psi (t):=\lim _{n\rightarrow \infty }(\mathcal {T}^n\varphi )(t) \end{aligned}$$
(9)
exists and the function \(\psi \in {Y}^E\), so defined, belongs to \(\mathcal {C}\) and fulfils the inequalities
$$\begin{aligned} d(\mathcal {T}^{n}\varphi ,\psi )\le \Big (4\sum _{i=n}^{\infty } (\varLambda _{i}\varepsilon )^{\xi }\Big )^{1/\xi }=: \varepsilon ^*_n,\qquad n\in \mathbb {N}_0. \end{aligned}$$
(10)
Moreover, the following three statements are valid.
-
(a)
If \(l \in \mathbb {N}\) and
$$\begin{aligned} \liminf _{n\rightarrow \infty }\big (\varLambda _l\varepsilon ^*_n\big )(t)=0,\qquad t\in E, \end{aligned}$$
(11)
then \(\psi \) is a fixed point of operator \(\mathcal {T}^l\).
-
(b)
If \(l\in \mathbb {N}\), \(t\in E\) and \(\gamma \in {\mathbb {R}_+}^E\) are such that
$$\begin{aligned} \liminf _{n\rightarrow \infty }(\varLambda _{ln}\gamma )(t)=0, \end{aligned}$$
(12)
then \(\psi (t)=\omega (t)\) for every fixed point \(\omega \in \mathcal {C}\) of \(\mathcal {T}^l\), satisfying the inequality
$$\begin{aligned} d(\varphi ,\omega ) \le \gamma . \end{aligned}$$
(13)
-
(c)
If \(t\in E\) and \(\varepsilon ^*_n(t)\ne 0\) for \(n\in \mathbb {N}\), then \(\tau (t)=\psi (t)\) for every \(\tau \in \mathcal {C}\) with
$$\begin{aligned} \liminf _{n\rightarrow \infty } \frac{d\big ((\mathcal {T}^{n}\varphi )(t),\tau (t)\big )}{\varepsilon ^*_{n}(t)}< \infty . \end{aligned}$$
(14)
Proof
Let \(D_d\) be the metric defined as in Theorem 5. Since \(\mathcal {T}^n\) is \(\varLambda _n\)-contractive, (8) implies that
$$\begin{aligned} d(\mathcal {T}^{n+1}\varphi ,\mathcal {T}^n\varphi )\le \varLambda _{n}\varepsilon ,\qquad n\in \mathbb {N}, \end{aligned}$$
whence
$$\begin{aligned} D_d \left( \mathcal {T}^{n+k}\varphi ,\mathcal {T}^n\varphi \right) \le&\; \sum _{i=0}^{k-1} D_d \left( \mathcal {T}^{n+i+1}\varphi ,\mathcal {T}^{n+i}\varphi \right) \nonumber \\ \le&\; \sum _{i=0}^{k-1} d \left( \mathcal {T}^{n+i+1}\varphi ,\mathcal {T}^{n+i}\varphi \right) ^{\xi }\nonumber \\ \le&\; \sum _{i=n}^{n+k-1} \left( \varLambda _{i}\varepsilon \right) ^{\xi }\le \sum _{i=n}^{\infty } \left( \varLambda _{i}\varepsilon \right) ^{\xi }, \end{aligned}$$
(15)
and consequently
$$\begin{aligned} d(\mathcal {T}^{n+k}\varphi ,\mathcal {T}^{n+r}\varphi )^{\xi }\le&\; 4D_d \left( \mathcal {T}^{n+k}\varphi ,\mathcal {T}^{n+r}\varphi \right) \\ \le&\; 4D_d \left( \mathcal {T}^{n+k}\varphi ,\mathcal {T}^n\varphi \right) +4D_d \left( \mathcal {T}^n\varphi ,\mathcal {T}^{n+r}\varphi \right) \\ \le&\; 4\sum _{i=n}^{\infty } (\varLambda _{i}\varepsilon )^{\xi } +4\sum _{i=n}^{\infty } (\varLambda _{i}\varepsilon )^{\xi } \end{aligned}$$
for any \(k,r\in \mathbb {N}\) and \(n\in \mathbb {N}_0\), where \((\varLambda _{i}\varepsilon )^{\xi }(t):=\big ((\varLambda _{i}\varepsilon )(t)\big )^{\xi }\) for \(t\in E\). So, by (8), \((\mathcal {T}^n\varphi (t))_{n\in \mathbb {N}}\) is a Cauchy sequence in \((Y,d,\eta )\) for each \(t\in E\), which means that it is convergent. Consequently, (9) defines a function \(\psi \in Y^E\). Clearly, if (i) holds, then \(\psi \in \mathcal {C}\).
Note that (9) holds also in the metric space \((Y,D_d)\), in view of (6). So, letting \(k\rightarrow \infty \) in (15), for each \(t\in E\) we get
$$\begin{aligned} d(\psi (t),(\mathcal {T}^n\varphi )(t))^{\xi }\le&\;4D_d(\psi (t),(\mathcal {T}^n\varphi )(t)) \nonumber \\ \le&\;4\sum _{i=n}^{\infty } ((\varLambda _{i}\varepsilon )(t))^{\xi },\qquad n\in \mathbb {N}_0, \end{aligned}$$
(16)
which is (10). Since, in the case of (ii), we have
$$\begin{aligned} \lim _{n\rightarrow \infty }\;\sup _{t\in E}\Big (\sum _{i=n}^{\infty } \big ((\varLambda _i\varepsilon )(t)\big )^{\xi }\Big )=0, \end{aligned}$$
it is easily seen that (10) implies that \(\psi \in \mathcal {C}\) also in this case.
Next, observe that (16) implies
$$\begin{aligned} d \left( \mathcal {T}^{n+l}\varphi ,\mathcal {T}^l\psi \right) \le \varLambda _l\Big (\Big (4\sum _{i=n}^{\infty } \big (\varLambda _{i}\varepsilon \big )^{\xi }\Big )^{1/\xi }\Big )=\varLambda _l \varepsilon ^*_n \end{aligned}$$
for \(l,n\in \mathbb {N}_0\). So, for each \(l,n\in \mathbb {N}_0\),
$$\begin{aligned} d \left( \psi ,\mathcal {T}^l\psi \right)&\;\le \eta d \left( \psi ,\mathcal {T}^{n+l}\varphi \right) +\eta d \left( \mathcal {T}^{n+l}\varphi ,\mathcal {T}^l\psi \right) \\&\;\le \eta d \left( \psi ,\mathcal {T}^{n+l}\varphi \right) +\eta \varLambda _l \varepsilon ^*_n. \end{aligned}$$
Hence, if (11) is valid for a fixed \(l\in \mathbb {N}\), then with \(n\rightarrow \infty \) we get
$$\begin{aligned} d \left( \psi (t),(\mathcal {T}^l\psi )(t)\right) = 0,\qquad t\in E, \end{aligned}$$
and consequently \(\mathcal {T}^l \psi =\psi \).
Now, let \(l\in \mathbb {N}\), \(\gamma :E\rightarrow \mathbb {R}_+\) and \(\omega \in \mathcal {C}\) be a fixed point of \(\mathcal {T}^l\) satisfying (13). Then, for each \(n\in \mathbb {N}\), we get
$$\begin{aligned} d(\psi ,\omega )&=d \left( \psi ,\mathcal {T}^{ln}\omega \right) \\&\le \eta d(\psi ,\mathcal {T}^{ln}\varphi )+\eta d \left( \mathcal {T}^{ln}\varphi ,\mathcal {T}^{ln}\omega \right) \end{aligned}$$
whence
$$\begin{aligned} d(\psi (t),\omega (t)) \le \eta d \left( \psi (t),\left( \mathcal {T}^{ln}\varphi \right) (t)\right) +\eta \left( \varLambda _{ln}\gamma \right) (t),\qquad t\in E. \end{aligned}$$
Consequently, letting \(n\rightarrow \infty \), by (9), we can easily see that \(\omega (t)=\psi (t)\) for each \(t\in E\) such that (12) holds.
Finally, let \(t\in E\) and \(\tau \in \mathcal {C}\) be such that \(\varepsilon ^*_n(t)\ne 0\) for \(n\in \mathbb {N}\) and (14) is valid. Write
$$\begin{aligned} M_t:= 1+\liminf _{n\rightarrow \infty }\;\frac{d\big ((\mathcal {T}^{n}\varphi )(t),\tau (t)\big )}{\varepsilon ^*_{n}(t)}. \end{aligned}$$
Then there is a strictly increasing sequence \((t_n)_{n\in \mathbb {N}}\) in \(\mathbb {N}\) such that
$$\begin{aligned} d \left( \left( \mathcal {T}^{t_n}\varphi \right) (t),\tau (t)\right) ^{\xi }< \left( M_t\varepsilon ^*_{t_n}(t)\right) ^{\xi }=M_t^{\xi }\sum _{i=t_n}^{\infty } \big (\left( \varLambda _{i}\varepsilon \right) (t)\big )^{\xi },\qquad n\in \mathbb {N}, \end{aligned}$$
and consequently, by (16),
$$\begin{aligned} d \left( \tau (t),\psi (t)\right) ^{\xi }\le&\; 4D_d \left( \tau (t),(\mathcal {T}^{t_n}\varphi )(t)\right) + 4D_d \left( \left( \mathcal {T}^{t_n}\varphi \right) (t),\psi (t)\right) \\ \le&\; 4d \left( \tau (t),(\mathcal {T}^{t_n}\varphi )(t)\right) ^{\xi }+ 4d \left( \left( \mathcal {T}^{t_n}\varphi \right) (t),\psi (t)\right) ^{\xi }\\ <&\;4(M_t^{\xi }+1)\sum _{i=t_n}^{\infty } \big (\left( \varLambda _{i}\varepsilon \right) (t)\big )^{\xi },\qquad n\in \mathbb {N}, \end{aligned}$$
whence we get \(\tau (t)=\psi (t)\) with \(n\rightarrow \infty \) (in view of (8)). \(\square \)
Remark 9
From statement (b) of Theorem 8 we can deduce, in particular, the following: If \(\psi \) is not a fixed point of \(\mathcal {T}^l\) for some \(l\in \mathbb {N}\), then the same is true for every function \(\omega \in \mathcal {C}\), which satisfies (13) with some \(\gamma \in \Gamma _l\), where \(\Gamma _l\) denotes the family of all functions \(\gamma :E\rightarrow \mathbb {R}_+\) such that (12) holds for each \(t\in E\).
But, if \(\psi \) is a fixed point of \(\mathcal {T}^l\) for some \(l\in \mathbb {N}\), then
$$\begin{aligned} \psi (t)= \lim _{n \rightarrow \infty } (\mathcal {T}^{lk_n} \chi ) (t) \end{aligned}$$
for every \(\chi \in \mathcal {C}\), \(t\in E\), and every sequence \((k_n)_{n\in \mathbb {N}}\) in \(\mathbb {N}\) such that there is \(\gamma \in {\mathbb {R}_+}^E\) with \(d(\chi ,\psi )\le \gamma \) and
$$\begin{aligned} \lim _{n\rightarrow \infty }(\varLambda _{lk_n}\gamma )(t)=0. \end{aligned}$$
(17)
In fact, if \(\gamma :E\rightarrow \mathbb {R}_+\) and \(\chi \in \mathcal {C}\) satisfies the inequality \(d(\chi ,\psi ) \le \gamma \), then for each \(t\in E\) and each sequence \((k_n)_{n\in \mathbb {N}}\) in \(\mathbb {N}\) such that (17) holds, we have
$$\begin{aligned} d((\mathcal {T}^{l k_n} \chi )(t), \psi (t))=&\;d((\mathcal {T}^{l k_n} \chi )(t), (\mathcal {T}^{l k_n} \psi )(t)) \\ \le&\;(\varLambda _{l k_n} \gamma )(t),\qquad n \in \mathbb {N}, \end{aligned}$$
which implies that
$$\begin{aligned} \psi (t)= \lim _{n \rightarrow \infty } (\mathcal {T}^{lk_n} \chi ) (t). \end{aligned}$$
To avoid any ambiguity in the sequel, we need one more definition.
Definition 10
Let \((D,\rho ,\eta )\) be a b-metric space, T be a nonempty subset of D, \(S:T\rightarrow D\) and \(\lambda :\mathbb {R_+}\rightarrow \mathbb {R_+}\). We say that S is a \(\lambda \)-contraction provided
$$\begin{aligned} \rho (S(t),S(s))\le \lambda (\delta ) \end{aligned}$$
for every \(t,s\in T\) and every \(\delta \in \mathbb {R}_+\) such that \(\rho (t,s)\le \delta \).
It is clear that, for nondecreasing \(\lambda \), S is a \(\lambda \)-contraction if and only if
$$\begin{aligned} \rho (S(t),S(s))\le \lambda (\rho (t,s)),\qquad t,s\in T. \end{aligned}$$
Note yet that, if E has only one element, i.e., \(E=\{t_0\}\), then \({Y}^E\) can be identified with Y. Therefore, Theorem 8 yields at once the following generalization of the Banach Contraction Principle.
Corollary 11
Let \(T:Y\rightarrow Y\) and \(\lambda _n:\mathbb {R_+}\rightarrow \mathbb {R_+}\) for \(n\in \mathbb {N}\). Suppose that \(T^n\) is a \(\lambda _n\)-contraction for \(n\in \mathbb {N}\) and there exist \(e \in {\mathbb {R_+}}\) and \(\phi \in Y\) such that
$$\begin{aligned} \sum _{i=1}^{\infty } \lambda _i (e)^{\xi }<\infty ,\qquad d(T (\phi ),\phi )\le e. \end{aligned}$$
Then there exists the limit
$$\begin{aligned} \psi _0 :=\lim _{n\rightarrow \infty }T^n(\phi ) \end{aligned}$$
and
$$\begin{aligned} d(T^n(\phi ),\psi _0)\le \Big (4\sum _{i=n}^{\infty } \lambda _{i}(e)^{\xi }\Big )^{1/\xi }=:e_n^*, \qquad n\in \mathbb {N}_0, \end{aligned}$$
where \(\lambda _0(d)=d\) for \(d\in \mathbb {R_+}\). Moreover, the following statements are valid.
- (\(\alpha \)):
-
If \(l\in \mathbb {N}\) and
$$\begin{aligned} \liminf _{n\rightarrow \infty }\lambda _l (e_n^*)=0, \end{aligned}$$
then \(\psi _0\) is a fixed point of \(T^l\).
- (\(\beta \)):
-
If \(e_0\in \mathbb {R}_+\), \(l\in \mathbb {N}\) and
$$\begin{aligned} \liminf _{n\rightarrow \infty }\lambda _{ln} (e_0)=0, \end{aligned}$$
then every fixed point \(\phi _0\in Y\) of \(T^l\), satisfying the inequality
$$\begin{aligned} d(\phi ,\phi _0)\le e_0, \end{aligned}$$
is equal to \(\psi _0\).
- (\(\gamma \)):
-
If \(e_n^*\ne 0\) for \(n\in \mathbb {N}\), then \(\psi _0\) is the unique element of the set
$$\begin{aligned} \Big \{z\in Y:\;\liminf _{n\rightarrow \infty }\;\frac{d\big ((T^{n}(\phi ),z\big )}{e^*_{n}}< \infty \Big \}. \end{aligned}$$
Remark 12
Analogously as in Remark 9, from statement (\(\beta \)) of Corollary 11 we can deduce, in particular, the following: If \(\psi _0\) is not a fixed point of \(T^l\) for some \(l\in \mathbb {N}\), then the same is true for each \(z\in E\) with \(d(\psi _0,z)<\sup D\), where D is the set of all \(c\in \mathbb {R}_+\) such that there is a sequence \((k_n)_{n\in \mathbb {N}}\) in \(\mathbb {N}\) with
$$\begin{aligned} \lim _{n\rightarrow \infty } k_n=\infty ,\qquad \liminf _{n\rightarrow \infty }\lambda _{lk_n} (c)=0. \end{aligned}$$
However, if \(\psi _0\) is a fixed point of \(T^l\) for some \(l\in \mathbb {N}\), then (again analogously as in Remark 9) it is easy to show that
$$\begin{aligned} \psi _0= \lim _{n \rightarrow \infty } T^{lk_n} (v) \end{aligned}$$
for every \(v\in E\) and every sequence \((k_n)_{n\in \mathbb {N}}\) in \(\mathbb {N}\) such that there is \(d\in {\mathbb {R}_+}^E\) with \(d(v,\psi _0)\le d\) and
$$\begin{aligned} \lim _{n\rightarrow \infty } \lambda _{lk_n}(d)=0. \end{aligned}$$
Theorem 8 also yields the subsequent.
Corollary 13
Let \(\mathcal {C} \subset {Y}^E\) be nonempty, \(\varLambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}\), \(\mathcal {T}:\mathcal {C}\rightarrow \mathcal {C}\) be \(\varLambda \)-contractive, and functions \(\varepsilon \in {\mathbb {R_+}}^{E}\) and \(\varphi \in \mathcal {C}\) be such that
$$\begin{aligned} d(\mathcal {T} \varphi ,\varphi )\le \varepsilon ,\qquad \varepsilon ^{\star }(t):=\sum _{i=0}^{\infty }\big ((\varLambda ^i\varepsilon )(t)\big )^{\xi }<\infty ,\qquad t\in E. \end{aligned}$$
(18)
Next, assume one of the following two hypotheses.
-
(i)
\(\mathcal {C}\) is p-closed.
-
(ii)
\(\mathcal {C}\) is u-closed and the sequence \(\Big (\sum _{i=0}^{n} (\varLambda ^i\varepsilon )^{\xi }\Big )_{n\in \mathbb {N}}\) tends uniformly to \(\varepsilon ^{\star }\).
Then, for each \(t\in E\), there exists the limit
$$\begin{aligned} \psi (t):=\lim _{n\rightarrow \infty }(\mathcal {T}^n\varphi )(t) \end{aligned}$$
(19)
and the function \(\psi \in {Y}^E\), so defined, belongs to \(\mathcal {C}\) and fulfils the inequalities
$$\begin{aligned} d(\mathcal {T}^n\varphi ,\psi )\le \Big (4\sum _{i=n}^{\infty } \big (\varLambda ^{i}\varepsilon \big )^{\xi }\Big )^{1/\xi }=:\varepsilon ^{\star }_n,\qquad n\in \mathbb {N}_0. \end{aligned}$$
(20)
Moreover, the following statements are valid.
-
(A)
If
$$\begin{aligned} \liminf _{n\rightarrow \infty } \big (\varLambda \varepsilon ^{\star }_n\big )(t) =0,\qquad t\in E, \end{aligned}$$
(21)
then \(\psi \) is a fixed point of \(\mathcal {T}\).
-
(B)
If \(t\in E\) and \(\gamma \in {\mathbb {R}_+}^E\) are such that
$$\begin{aligned} \liminf _{n\rightarrow \infty }(\varLambda ^{n}\gamma )(t)=0, \end{aligned}$$
(22)
then \(\psi (t)=\omega (t)\) for every fixed point \(\omega \in \mathcal {C}\) of \(\mathcal {T}\), satisfying the inequality \(d(\varphi ,\omega ) \le \gamma \).
-
(C)
If \(t\in E\) and \(\varepsilon ^{\star }_n(t)\ne 0\) for \(n\in \mathbb {N}\), then \(\tau (t)=\psi (t)\) for every \(\tau \in \mathcal {C}\) with
$$\begin{aligned} \liminf _{n\rightarrow \infty }\;\frac{d\big ((\mathcal {T}^{n}\varphi )(t),\tau (t)\big )}{\varepsilon ^{\star }_{n}(t)}<\infty . \end{aligned}$$
Proof
It is easy to see that \(\mathcal {T}^n\) is \(\varLambda ^{n}\) – contractive for each \(n\in \mathbb {N}\). So, it is enough to use Theorem 8 with \(l=1\) and \(\varLambda _{n}:=\varLambda ^{n}\) for \(n\in \mathbb {N}\). \(\square \)
Remark 14
Fix \(j\in \mathbb {N}\). Let \(f_{i}:E\rightarrow E\) and \(L_i :E\rightarrow \mathbb {R}_+\) for \(i=1,\ldots ,j\). Define \(\varLambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}\) by the formula
$$\begin{aligned} \varLambda \delta (t):=&\;\sum _{i=1}^{j}L_i(t)\delta (f_i(t)),\qquad \delta \in {\mathbb {R}_+}^{E}, t\in E. \end{aligned}$$
Then \((\mathcal {C}_0)\) holds for such \(\varLambda \) (see Remark 3). This means that Theorem 1 is a simple consequence of Corollary 13. We show that this is also the case for Theorem 2.
First note that \((\mathcal {C}_0)\) and (18) imply (21). Therefore the statement of Theorem 2 that \(\psi \) is a fixed point of \(\mathcal {T}\) can be deduced from (A). We are yet to show that the statement on uniqueness of \(\psi \) can be derived from (B).
To this end fix \(\gamma \in {\mathbb {R}_+}^E\) and \(M>0\) such that
$$\begin{aligned} \gamma ^*_0(t):= \Big (\sum _{k=0}^{\infty } \big ((\varLambda ^{k}\gamma )(t)\big )^{\xi }\Big )^{1/\xi }<\infty ,\qquad t\in E, \end{aligned}$$
$$\begin{aligned} \gamma ^*_0(t)\le M \sum _{i=0}^{\infty }(\varLambda ^i\gamma )(t),\qquad t\in E. \end{aligned}$$
(23)
We prove that, for each \(n\in \mathbb {N}_0\),
$$\begin{aligned} (\varLambda ^n \gamma ^*_0)(t)\le M \sum _{i=n}^{\infty }(\varLambda ^i\gamma )(t),\qquad t\in E. \end{aligned}$$
(24)
Clearly, the case \(n=0\) is trivial in view of (23). Next, fix \(n\in \mathbb {N}_0\) and suppose that (24) is valid. Then
$$\begin{aligned} (\varLambda ^{n+1}\gamma ^*_0)(t)=&\;\sum _{i=1}^{j} L_i(t) (\varLambda ^{n}\gamma ^*_0)(f_i(t))\le \sum _{i=1}^{j} L_i(t) M \sum _{k=n}^{\infty }(\varLambda ^k\gamma )(f_i(t))\\ \le&\;M\sum _{k=n}^{\infty } \sum _{i=1}^{j} L_i(t) (\varLambda ^{k}\gamma )(f_i(t))\\ =&\;M\sum _{k=n}^{\infty }(\varLambda ^{k+1}\gamma )(t)=M\sum _{k=n+1}^{\infty }(\varLambda ^{k}\gamma )(t),\qquad t\in E. \end{aligned}$$
Thus, by induction, we have proved (24). Therefore, in this case, (24) yields (22). Consequently, (22) results also from the condition
$$\begin{aligned} \gamma ^*_0(t)\le M\sum _{k=1}^{\infty } (\varLambda ^{k}\gamma )(t),\qquad t\in E. \end{aligned}$$
This means that (5) (cf. [13, (2.6)]) implies that
$$\begin{aligned} \liminf _{n\rightarrow \infty }(\varLambda ^{n}\varepsilon ^*_0)(t)=0,\qquad t\in E, \end{aligned}$$
and therefore from (B) we can infer that \(\psi \) is the unique fixed point of \(\mathcal {T}\), satisfying the inequality \(d(\varphi ,\psi ) \le \varepsilon ^*_0\).
Thus we have shown that Theorem 2 (i.e., [13, Theorem 2.1]) is a simple consequence of Corollary 13.
The next example shows that, in some natural situations, assumption (8) can be significantly weaker than (18) with \(\varLambda :=\varLambda _{1}\). It is a bit trivial, but this allows to expose clearly the main differences between (8) and (18).
Example 3
Assume that \((X,\Vert \cdot \Vert ,\eta )\) is a real quasi–normed space. Let \(Y=X^2\) and define \(d:Y^2\rightarrow \mathbb {R}_+\) by
$$\begin{aligned} d((x_1,x_2),(y_1,y_2))=\sum _{i=1}^2\Vert x_i-y_i\Vert , \qquad (x_1,x_2),(y_1,y_2)\in Y. \end{aligned}$$
Clearly, \((Y,d,\eta )\) is a b–metric space.
Fix \(\tau _1,\tau _2\in X^E\) and define the operator \(\mathcal {T}:Y^{E}\rightarrow Y^{E}\) by
$$\begin{aligned} (\mathcal {T}\phi )(t)=(\tau _1(t),\phi _1(t)+\tau _2(t)),\qquad t\in E, \end{aligned}$$
for all \(\phi =(\phi _1,\phi _2)\in Y^{E}\). Then
$$\begin{aligned} d((\mathcal {T}\phi )(t),(\mathcal {T}\mu )(x))=&\;d((\tau _1(t),\phi _1(t)+\tau _2(t)),(\tau _1(t),\mu _1(t)+\tau _2(t)))\\ \le&\; d(\phi (t),\mu (t)),\qquad t\in E, \end{aligned}$$
for every \(\phi =(\phi _1,\phi _2),\mu =(\mu _1,\mu _2)\in Y^{E}\). Therefore, with any \(\varLambda _1:{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}\) such that \(\varLambda _1\delta \ge \delta \) for \(\delta \in {\mathbb {R}_+}^{E}\), we have \(\varLambda _1\)-contractivity of \(\mathcal {T}\). Next,
$$\begin{aligned} (\mathcal {T}^n\phi )(t)=(\tau _1(t),\tau _1(t)+\tau _2(t)),\qquad t\in E,n\in \mathbb {N}, n\ge 2, \end{aligned}$$
for each \(\phi =(\phi _1,\phi _2)\in Y^{E}\) and we can take \(\varLambda _n\delta (t)=0\) for \(\delta \in {\mathbb {R}_+}^{E}\), \(t\in E\) and \(n\in \mathbb {N}\), \(n\ge 2\). This means that the second inequality in (8) is fulfilled for every \(\varepsilon \in {\mathbb {R}_+}^{E}\).
Fix \(\delta \in {\mathbb {R}_+}^{E}\) and \(u\in X\) with \(\Vert u\Vert =1\). Let functions \(\phi ,\psi \in Y^{E}\) be defined by:
$$\begin{aligned} \phi (t)=(\delta (t)u,0),\qquad \psi (t)=(0,0),\qquad t\in E. \end{aligned}$$
Clearly,
$$\begin{aligned} (\mathcal {T}\phi )(t)= & {} (\tau _1(t),\delta (t)u+\tau _2(t)), \\ (\mathcal {T}\psi )(t)= & {} (\tau _1(t),\tau _2(t)),\quad t\in E, \\ d(\phi (t),\psi (t))= & {} \Vert \delta (t)u\Vert =\delta (t),\quad t\in E. \end{aligned}$$
Assume now that \(\mathcal {T}\) is \(\varLambda _1\)-contractive with some \(\varLambda _1:{\mathbb {R}_+}^{E}\rightarrow {\mathbb {R}_+}^{E}\). Then
$$\begin{aligned} \delta (t)=\Vert \delta (t)u\Vert =d((\mathcal {T}\phi )(t),(\mathcal {T}\psi )(t))\le \varLambda _1 \delta (t),\qquad t\in E. \end{aligned}$$
Thus we have shown that
$$\begin{aligned} \varLambda _1\delta \ge \delta ,\qquad \delta \in {\mathbb {R}_+}^{E}. \end{aligned}$$
Consequently, by an easy induction, for each \(n\in \mathbb {N}\), we obtain
$$\begin{aligned} {\varLambda _1}^n\delta \ge \delta ,\qquad \delta \in {\mathbb {R}_+}^{E}, \end{aligned}$$
whence
$$\begin{aligned} \sum _{i=0}^{\infty }({\varLambda _1}^i\delta )(t)=\infty ,\qquad t\in E,\delta \in {\mathbb {R}_+}^{E},\delta (t)\ne 0. \end{aligned}$$