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Fixed point theorems of orderLipschitz mappings in Banach algebras
Fixed Point Theory and Applications volume 2016, Article number: 30 (2016)
Abstract
In this paper, by introducing the concept of Picardcompleteness and using the sandwich theorem in the sense of wconvergence, we first prove some fixed point theorems of orderLipschitz mappings in Banach algebras with nonnormal cones which improve the result of Sun’s since the normality of the cone was removed. Moreover, we reconsider the case with normal cones and obtain a fixed point theorem under the assumption relating to the spectral radius, which partially improves the results of Krasnoselskii and Zabreiko’s. In addition, we present some suitable examples which show the usability of our theorems.
Introduction
This work is mainly concerned with fixed point theory of orderLipschitz mappings in Banach algebras relating to the improvements of the Banach contraction principle which states that each Banach contraction on a complete metric space has a unique fixed point. Let \((X, d)\) be a metric space. A Banach contraction in a metric space (resp. a cone metric space) is also called a Lipschitz mapping with respect to the metric (resp. the cone metric); see [1]. Let P be a cone of a Banach algebra \((E,\Vert \cdot \Vert )\) and ⪯ the partial order in E introduced by P. A mapping \(T:E\rightarrow E\) is called an orderLipschitz mapping if there exist \(l, k\in P\) such that the following Lipschitz condition with respect to the partial order is satisfied:
In particular when k, l are nonnegative real numbers, Sun [2] obtained the following fixed point theorem of orderLipschitz mappings in Banach spaces by using the sandwich theorem in the sense of normconvergence.
Theorem 1
(see [2])
Let P be a normal cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(u_{0}, v_{0}\in E\) with \(u_{0}\preceq v_{0}\). Assume that \(T:[u_{0}, v_{0}]\rightarrow E\) is an orderLipschitz mapping with \(l\in[0,+\infty)\) and \(k\in[0, 1)\) such that
Then T has a unique fixed point \(x^{*}\in[u_{0}, v_{0}]\). And for each \(x_{0}\in[u_{0}, v_{0}]\), let \(\{x_{n}\}\) be the Picard iterative sequence (i.e., \(x_{n}=T^{n}x_{0}\) for each n), then we must have \(x_{n}\overset{\Vert \cdot \Vert }{\rightarrow} x^{*}\).
Remark 1
The normality of the cone is essential for ensuring that the sandwich theorem holds in the sense of normconvergence which plays an important role in the proof of Theorem 1. However, if P is nonnormal then the sandwich theorem does not hold in the sense of normconvergence, and consequently, the method used in the proof of Theorem 1 may become invalid.
Krasnoselskii and Zabreiko [3] considered orderLipschitz mappings in Banach spaces restricted with linear bounded mappings (i.e., k, l are linear bounded mappings), and proved the following fixed point theorem by using the Banach contraction principle.
Theorem 2
(see [3])
Let P be a normal solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(T:E\rightarrow E\) an orderLipschitz mapping with \(k=l\), where \(k:P\rightarrow P\) is a linear bounded mapping. If \(\Vert k\Vert <1\), then T has a unique fixed point \(x^{*}\in E\). And for each \(x_{0}\in E\), let \(\{x_{n}\}\) be the Picard iterative sequence, then we must have \(x_{n}\overset{\Vert \cdot \Vert }{\rightarrow} x^{*}\).
To our knowledge, in all the works concerned with fixed points of orderLipschitz mappings, the involving cone is necessarily assumed to be normal. In this paper, we shall remove the normality of the cone in Theorem 1 and extend Theorems 1 and 2 to Banach algebras. From Remark 1 we know that the method in [2] is not applicable for the case with nonnormal cones, and so we need to find a new way to solve it. By introducing the concept of Picardcomplete and using the sandwich theorem in the sense of wconvergence established in [4], we first prove some fixed point theorems of orderLipschitz mappings in Banach algebras with nonnormal cones. Motivated by [3], we reconsider the case with normal cones, and we obtain a fixed point theorem of orderLipschitz mappings in Banach algebras under the assumption that \(r(k)<1\) by showing that there exists some \(n_{0}\) such that \(T^{n_{0}}\) is a Banach contraction in \((E,\Vert \cdot \Vert _{0})\), where \(\Vert \cdot \Vert _{0}\) is a newly introduced norm which is equivalent to \(\Vert \cdot \Vert \); see Lemma 5. In addition, some suitable examples are presented to show the usability of our theorems.
Preliminaries and lemmas
A Banach space \((E,\Vert \cdot \Vert )\) is called a Banach algebra [5] if there exists a multiplication in E such that, for each \(x, y, z \in E\) and \(a\in \mathbb{R}\), the following conditions are satisfied: (I) \((xy)z = x(yz)\); (II) \(x(y + z) = xy + xz\) and \((x + y)z = xz + yz\); (III) \(a(xy) = (ax)y = x(ay)\); (IV) \(\Vert xy\Vert \leq \Vert x\Vert \Vert y\Vert \). If there exists some \(e\in E\) such that \(ex =xe=x\) for each \(x\in E\) then e is called a unit (i.e., a multiplicative identity) of E. A nonempty closed subset P of a Banach space \((E,\Vert \cdot \Vert )\) is a cone [5, 6] if it is such that the following conditions are satisfied: (V) \(ax+by\in P\) for each \(x,y\in P\) and each \(a,b\geq0\); (VI) \(P\cap(P)=\{\theta\}\), where θ is the zero element of E. A nonempty closed subset P of a Banach algebra \((E,\Vert \cdot \Vert )\) is a cone [1, 5] if it is such that (V) and (VI) are satisfied and (VII) \(\{e\}\subset P\) and \(P^{2}=PP\subset P\).
Each cone P of a Banach space E determines a partial order ⪯ on E by \(x\preceq y\Leftrightarrow yx\in P \) for each \(x,y\in X\). For each \(u_{0},v_{0}\in E\) with \(u_{0}\preceq v_{0}\), we set \([u_{0},v_{0}]=\{u\in E:u_{0}\preceq u\preceq v_{0}\}\), \([u_{0},+\infty)=\{x\in E:u_{0}\preceq x\}\) and \((\infty,v_{0}]=\{x\in E:x\preceq v_{0}\}\). A cone P is solid [5, 6] if \(\operatorname {\mathsf {int}}P\neq\O\), where int P denotes the interior of P. For each \(x,y\in E\) with \(yx\in \operatorname {\mathsf {int}}P\), we write \(x\ll y\).
Definition 1
Let P be a solid cone of a Banach space E, \(\{x_{n}\}\subset E\) and \(D\subset E\).

(i)
the sequence \(\{x_{n}\}\) is wconvergent [4, 7] if for each \(\epsilon\in \operatorname {\mathsf {int}}P\), there exist some positive integer \(n_{0}\) and \(x\in E\) such that \(x\epsilon\ll x_{n}\ll x+\epsilon\) for each \(n\geq n_{0}\) (denote \(x_{n}\overset{w}{\rightarrow} x\) and x is called a wlimit of \(\{ x_{n}\}\));

(ii)
the sequence \(\{x_{n}\}\) is wCauchy if for each \(\epsilon\in \operatorname {\mathsf {int}}P\), there exists some positive integer \(n_{0}\) such that \(\epsilon\ll x_{n}x_{m}\ll\epsilon\) for each \(m,n\geq n_{0}\), i.e., \(x_{n}x_{m}\overset{w}{\rightarrow} \theta\) (\(m,n\rightarrow\infty\));

(iii)
the subset D is wclosed if for each \(\{x_{n}\}\subset D\), \(x_{n}\overset{w}{\rightarrow}x\) implies \(x\in D\).
Lemma 1
Let P be a solid cone of a Banach space E, \(\{x_{n}\}\) a wconvergent sequence of E and \(u_{0},v_{0}\in E\) with \(u_{0}\preceq v_{0}\). Then \(\{x_{n}\}\) has a unique wlimit, and the partial order intervals \([u_{0},v_{0}]\), \([u_{0},+\infty)\) and \((\infty,v_{0}]\) are wclosed.
Proof
Suppose that there exists \(x,y\in E\) such that \(x_{n}\overset{w}{\rightarrow} x\) and \(x_{n}\overset{w}{\rightarrow} y\). From Definition 1 we find that, for each \(\epsilon\in \operatorname {\mathsf {int}}P\), there exists a positive integer \(n_{0}\) such that \(x\epsilon\ll x_{n}\ll x+\epsilon\) and \(y\epsilon\ll x_{n}\ll y+\epsilon \) for each \(n\geq n_{0}\). This forces that \(xy2\epsilon\ll x_{n}x_{n}=\theta\ll xy+2\epsilon\) for each \(n\geq n_{0}\). So we have \(2\epsilon\ll xy\ll2\epsilon\), which together with the arbitrary property of ϵ implies that \(x=y\). This shows that \(\{x_{n}\}\) has a unique wlimit.
Let \(\{x_{n}\}\) be a sequence of \([u_{0},v_{0}]\) such that \(x_{n}\overset {w}{\rightarrow} x\). For each \(\epsilon\in \operatorname {\mathsf {int}}P\), there exists a positive integer \(n_{0}\) such that \(x\epsilon\ll x_{n}\ll x+\epsilon\) for each \(n\geq n_{0}\). Thus we get
which together with the arbitrary property of ϵ implies that \(u_{0}\preceq x\) and \(x\preceq v_{0}\), i.e., \(x\in[u_{0},v_{0}]\). This shows that \([u_{0},v_{0}]\) is wclosed. Similarly, we can show \([u_{0},+\infty)\) and \((\infty,v_{0}]\) are wclosed. The proof is complete. □
A cone P of a Banach space E is normal if there is some positive number N such that \(x, y \in E\) and \(\theta\preceq x\preceq y\) implies that \(\Vert x\Vert \leq N\Vert y\Vert \), and the minimal N is called a normal constant of P. Note that an equivalent condition of a normal cone is that \(\inf\{\Vert x+y\Vert :x,y\in P \text{ and } \Vert x\Vert =\Vert y\Vert =1\}>0\), then it is not hard to conclude that a cone P is nonnormal if and only if there exist \(\{u_{n}\},\{v_{n}\}\subset P\) such that \(u_{n}+v_{n}\overset{\Vert \cdot \Vert }{\rightarrow} \theta\nRightarrow u_{n} \overset{\Vert \cdot \Vert }{\rightarrow }\theta \). This implies that the sandwich theorem does not hold in the sense of normconvergence. Recently, without using the normality of P Li and Jiang [4] proved the following sandwich theorem in the sense of wconvergence, which is very important for our further discussions.
Lemma 2
(see [4])
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(\{x_{n}\},\{y_{n}\},\{z_{n}\}\subset E\) with \(x_{n}\preceq y_{n}\preceq z_{n}\) for each n. If \(x_{n}\overset{w}{\rightarrow}z\) and \(z_{n}\overset{w}{\rightarrow}z\), then \(y_{n}\overset{w}{\rightarrow }z\).
Lemma 3
(see [7])
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(x_{n}\subset E\). Then \(x_{n}\overset{\Vert \cdot \Vert }{\rightarrow} x\) implies \(x_{n}\overset {w}{\rightarrow} x\). Moreover, if P is normal then \(x_{n}\overset{w}{\rightarrow} x \Leftrightarrow x_{n}\overset{\Vert \cdot \Vert }{\rightarrow} x\).
Lemma 4
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\). Then there is \(\tau>0\) such that, for each \(x\in E\), there exist \(y,z\in P\) with \(\Vert y\Vert \leq\tau \Vert x\Vert \) and \(\Vert z\Vert \leq\tau \Vert x\Vert \) such that \(x=yz\).
Lemma 5
Let P be a normal solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and define \(\Vert \cdot \Vert _{0}\) in E by
Then \(\Vert \cdot \Vert _{0}\) is equivalent to \(\Vert \cdot \Vert \) and \((E,\Vert \cdot \Vert _{0})\) is a Banach space.
Proof
It follows from Lemma 4 that there is a \(\tau>0\) such that, for each \(x\in E\), there exist \(y,z\in P\) with \(\Vert y\Vert \leq\tau \Vert x\Vert \) and \(\Vert z\Vert \leq\tau \Vert x\Vert \) such that \(x=yz\), and so we have
Then it is clear that, for each \(x\in E\), there exists \(u\in P\) such that
and hence the definition of \(\Vert \cdot \Vert _{0}\) is meaningful. It is easy to check that \(\Vert \cdot \Vert _{0}\) is a norm in E. For each \(x\in E\), by (6) and the normality of P, we get \(\Vert x\Vert \leq \Vert x+u\Vert +\Vert u\Vert \leq(2N+1)\Vert u\Vert \), and hence \(\Vert x\Vert \leq (2N+1)\inf_{u\in P}\Vert u\Vert =(2N+1)\Vert x\Vert _{0}\) by (4). On the other hand, by (5) we get \(\Vert x\Vert _{0}\leq \Vert y+z\Vert \leq2\tau \Vert x\Vert \) for each \(x\in E\). Thus we have \(\frac{\Vert x\Vert }{2N+1}\leq \Vert x\Vert _{0}\leq2\tau \Vert x\Vert \) for each \(x\in E\). This shows that \(\Vert \cdot \Vert \) is equivalent to \(\Vert \cdot \Vert _{0}\) and hence \((E,\Vert \cdot \Vert _{0})\) is a Banach space. The proof is complete. □
Let P be a cone of a Banach space E and \(T:E\rightarrow E\). For each \(x_{0}\in E\), set \(O(T,x_{0})=\{x_{n}\}\), where \(\{x_{n}\}\) is the Picard iterative sequence (i.e., \(x_{n}=T^{n}x_{0}\) for each n).
Definition 2
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\), \(x_{0}\in E\) and \(T:E\rightarrow E\). If the Picard iterative sequence \(O(T,x_{0})\) is wconvergent provided that it is wCauchy, then T is said to be Picardcomplete at \(x_{0}\). Moreover, if T is Picardcomplete at each \(x\in E\), then T is said to be Picardcomplete on E.
Remark 2

(i)
If \(O(T,x_{0})\) is wconvergent then T is certainly Picardcomplete at \(x_{0}\).

(ii)
If P is normal then each mapping \(T:E\rightarrow E\) is Picardcomplete on E by Lemma 3.
Fixed point theorems
We first state and prove a fixed point result of orderLipschitz mappings in Banach algebras with nonnormal cones as follows.
Theorem 3
Let P be a solid cone of a Banach algebra \((E,\Vert \cdot \Vert )\) and \(u_{0},v_{0}\in E\) with \(u_{0}\preceq v_{0}\). Assume that \(T:[u_{0},v_{0}]\rightarrow E\) is a nondecreasing orderLipschitz mapping with \(k,l\in P\) such that (2) is satisfied. If \(r(k)<1\) and T is Picardcomplete at \(u_{0}\) and \(v_{0}\), then T has a unique fixed point \(x^{*}\in[u_{0},v_{0}]\). And for each \(x_{0}\in[u_{0},v_{0}]\), we have \(x_{n}\overset{w}{\rightarrow} x^{*}\), where \(\{x_{n}\}=O(T,x_{0})\).
Proof
Let \(u_{n}=T^{n}u_{0}\) and \(v_{n}=T^{n}v_{0}\) for each n. From (2) and the nondecreasing property of T on \([u_{0},v_{0}]\) it follows that
By (1) and \(P^{2}\subset P\), we get
Thus for each \(m>n\), by (4) and (5) we have
It follows from Gelfand’s formula and \(r(k)<1\) that there exist \(n_{0}\) and \(\alpha\in[0,1)\) such that
which implies that \(k^{n}\overset{\Vert \cdot \Vert }{\rightarrow} \theta\). Note that \(\Vert k^{n}(v_{0}u_{0})\Vert \leq \Vert k^{n}\Vert \Vert v_{0}u_{0}\Vert \) by (IV), then we obtain \(k^{n}(v_{0}u_{0})\overset{\Vert \cdot \Vert }{\rightarrow}\theta\). Moreover, by Lemma 3, we get
Thus it follows from (9), (11), and Lemma 2 that \(\{u_{n}\}\) and \(\{v_{n}\}\) are wCauchy sequences. Since T is Picardcomplete at \(u_{0}\) and \(v_{0}\), there exist \(u^{*},v^{*}\in E\) such that
Letting \(n\rightarrow\infty\) in (8), by (11) and Lemma 2 we get \(u^{*}=v^{*}\). Set \(x^{*}=u^{*}=v^{*}\), then \(x^{*}\in[u_{0},v_{0}]\) by (7) and Lemma 1. For each \(m>n\), by (7) we get
Letting \(m\rightarrow\infty\) in (13), by (12) and Lemma 1 we have
with together with the nondecreasing property of T on \([u_{0},v_{0}]\) implies that
Letting \(n\rightarrow\infty\) in (15), we get \(x^{*}=Tx^{*}\) by (12), Lemma 1 and Lemma 2. Hence \(x^{*}\) is a fixed point of T.
For each \(x_{0}\in[u_{0},v_{0}]\), let \(x_{n}=T^{n}x_{0}\). It is clear that \(u_{1}=Tu_{0}\preceq Tx_{0}=x_{1}\preceq Tv_{0}=v_{1}\) since T is nondecreasing on \([u_{0},v_{0}]\). Then by induction, we obtain
which together with (12) and Lemma 2 implies that \(x_{n}\overset{w}{\rightarrow} x^{*}\). Let \(x\in[u_{0},v_{0}]\) be another fixed point of T and \(y_{n}=T^{n}x\). Similarly, we can show that \(y_{n}\overset{w}{\rightarrow} x^{*}\). Note that \(y_{n}=T^{n}x\equiv x\) implies that \(y_{n}\overset{w}{\rightarrow} x\), then \(x=x^{*}\) by Lemma 1. Hence T has a unique fixed point \(x^{*}\in [u_{0},v_{0}]\). The proof is complete. □
Example 1
Let \(E=C_{\mathbb{R}}^{1}[0,1]\) be endowed with the norm \(\Vert u\Vert =\Vert u\Vert _{\infty}+\Vert u'\Vert _{\infty}\) and \(P=\{u\in E:u(t)\geq0, \forall t\in[0,1]\}\), where \(\Vert u\Vert _{\infty}=\max_{t\in[0,1]}u(t)\) for each \(u\in C_{\mathbb{R}}[0,1]\). Define a multiplication in E by \((xy)(t)=x(t)y(t)\) for each \(x,y\in E\) and \(t\in[0,1]\). Clearly, \((E,\Vert \cdot \Vert )\) is a Banach algebra with a unit \(e(t)\equiv1\) and P is a nonnormal solid cone.
Let \(Tx=x^{2}\), \(u_{0}=\theta\) and \(v_{0}(t)\equiv a\), where \(a\in[0,\frac {1}{2})\). Clearly, \(Tu_{0}\preceq u_{0}\) and \(Tv_{0}\preceq v_{0}\). For each \(x,y\in[u_{0},v_{0}]\) with \(y\preceq x\), we have \(0\leq(Tx)(t)(Ty)(t)=x^{2}(t)y^{2}(t)=(x(t)+y(t))(x(t)y(t))\leq k(x(t)y(t))\) for each \(t\in[0,1]\), where \(k=2ae\). This shows that \(T:[u_{0},v_{0}]\rightarrow E\) is a nondecreasing orderLipschitz mapping with \(r(k)=2a<1\). Let \(\{u_{n}\}\) and \(\{v_{n}\}\) be the Picard iterative sequences of \(u_{0}\) and \(v_{0}\), then \(u_{n}=\theta\) and \(v_{n}(t)\equiv a^{2^{n}}\) for each n, and so \(\Vert u_{n}\Vert \equiv0\) and \(\Vert v_{n}\Vert =a^{2^{n}}\) for each n, which forces that \(u_{n}\overset{\Vert \cdot \Vert }{\rightarrow}\theta\) and \(v_{n}\overset{\Vert \cdot \Vert }{\rightarrow}\theta\). This together with (i) of Remark 2 and Lemma 3 implies that T is Picardcomplete at \(u_{0}\) and \(v_{0}\). Hence by Theorem 3, T has a unique fixed point in \([u_{0},v_{0}]\).
However, Theorems 1 and 2 are not applicable here since P is nonnormal.
In analogy to the proof of Theorem 3, we can prove the following fixed point theorem of orderLipschitz mappings in Banach space.
Theorem 4
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(u_{0},v_{0}\in E\) with \(u_{0}\preceq v_{0}\). Assume that \(T:[u_{0},v_{0}]\rightarrow E\) is a nondecreasing orderLipschitz mapping with \(k\in[0,1)\) such that (2) is satisfied. If T is Picardcomplete at \(u_{0}\) and \(v_{0}\), then T has a unique fixed point \(x^{*}\in[u_{0},v_{0}]\). And for each \(x_{0}\in[u_{0},v_{0}]\), we have \(x_{n}\overset{w}{\rightarrow} x^{*}\), where \(\{x_{n}\}=O(T,x_{0})\).
Corollary 1
Let P be a solid cone of a Banach space \((E,\Vert \cdot \Vert )\) and \(u_{0},v_{0}\in E\) with \(u_{0}\preceq v_{0}\). Assume that \(T:[u_{0},v_{0}]\rightarrow E\) is an orderLipschitz mapping with \(l\in[0,+\infty)\) and \(k\in[0,1)\) such that (2) is satisfied. If A is Picardcomplete at \(u_{0}\) and \(v_{0}\), where \(Ax=\frac{Tx+lx}{1+l}\) for each \(x\in E\), then T has a unique fixed point \(x^{*}\in[u_{0},v_{0}]\).
Proof
Set \(k_{1}=\frac{l+k}{1+l}\). By (1) and (2) we get
which indicates that \(A:[u_{0},v_{0}]\rightarrow E\) is a nondecreasing orderLipschitz mapping. Note that \(k_{1}\in[0,1)\) and A is Picardcomplete at \(u_{0}\) and \(v_{0}\), then A has a unique fixed point \(x^{*}\in[u_{0},v_{0}]\) by Theorem 4. Thus we have \(Tx^{*}+lx^{*}=x^{*}+lx^{*}\) and so \(Tx^{*}=x^{*}\). Let \(x\in[u_{0},v_{0}]\) be another fixed point of T, then \(Tx=x\) and hence \(Ax=x\). Moreover, by the uniqueness of fixed point of A in \([u_{0},v_{0}]\), we get \(x=x^{*}\). Hence \(x^{*}\) is the unique fixed point of T in \([u_{0},v_{0}]\). The proof is complete. □
Remark 3
By (ii) of Remark 2, Theorem 1 immediately follows from Corollary 1, which indeed improves Theorem 1 since the normality of P has been removed.
Motivated by [3], we reconsider the case with normal cones, and we obtain the following fixed point result.
Theorem 5
Let P be a normal solid cone of a Banach algebra \((E,\Vert \cdot \Vert )\) and \(T:E\rightarrow E\) an orderLipschitz mapping with \(l=k\in P\). If \(r(k)<1\), then T has a unique fixed point \(x^{*}\in E\). And for each \(x_{0}\in E\), we have \(x_{n}\overset{\Vert \cdot \Vert }{\rightarrow} x^{*}\), where \(\{x_{n}\} =O(T,x_{0})\).
Proof
Since P is solid, it follows from (6) that, for each \(x,y\in E\), there exists \(u\in P\) such that
and so
We observe that
where \(u\in P\) satisfies (16). By (1) and (17), for each \(x,y\in E\) we have
and
which can be rewritten as
Adding (19) and (20), by (II), we get
which implies that (10) holds for \(n=1\). Suppose that (18) holds for n, then for each \(x,y\in E\), we get
Moreover, by (1), for each \(x,y\in E\) we get
and
which can be rewritten as
Adding (21) and (22), by (II) we get \(k^{n+1}u\preceq T^{n+1}xT^{n+1}y\preceq k^{n+1}u\) for each \(x,y\in E\), and hence (18) holds for \(n+1\). Therefore (18) holds true by induction. For each \(x,y\in E\) and arbitrary \(u\in P\) such that (16) is satisfied, by (18) and (IV) we get \(\Vert T^{n}xT^{n}y\Vert _{0}\leq \Vert k^{n}u\Vert \leq \Vert k^{n}\Vert \Vert u\Vert \) for each n. Then by (4), (10), and the arbitrariness property of u, we have
which together with \(\alpha^{n_{0}}<1\) implies that \(T^{n_{0}}:E\rightarrow E\) is a Banach contraction. Note that \((E,\Vert \cdot \Vert _{0})\) is a Banach space by Lemma 5, then by the Banach contraction principle, \(T^{n_{0}}\) has a unique fixed point \(x^{*}\in E\) (i.e., \(T^{n_{0}}x^{*}=x^{*}\)). It is clear that \(T^{n_{0}}(Tx^{*})=T^{n_{0}+1}x^{*}=T(T^{n_{0}}x)=Tx^{*}\), i.e., \(Tx^{*}\) is a fixed point of \(T^{n_{0}}\), then \(x^{*}=Tx^{*}\) by the uniqueness of fixed point of \(T^{n_{0}}\). Let \(x\in E\) be another fixed point of T. Then x is also a fixed point of \(T^{n_{0}}\), and so \(x=x^{*}\) by the unique existence of fixed point of \(T^{n_{0}}\). Hence \(x^{*}\) is the unique fixed point of T.
For each \(x_{0}\in E\), let \(x_{n}=T^{n}x_{0}\) and \(y_{n}^{i}=T^{nn_{0}}x_{i}\) for each n and each \(0\leq i\leq n_{0}1\). It is clear that \(y_{n}^{i}\overset{\Vert \cdot \Vert _{0}}{\rightarrow}x^{*}\) for each i, and so for each \(\varepsilon>0\), there exists a positive integer \(m_{0}^{i}\) such that
Set \(m_{0}=\max_{1\leq i\leq n_{0}1} m_{0}^{i}\), then for each i, we get
which together with \(\{x_{n}\}=\bigcup_{i=0}^{n_{0}1} \{y_{n}^{i}\}\) implies that
This shows that \(x_{n}\overset{\Vert \cdot \Vert _{0}}{\rightarrow }x^{*}\) and hence \(x_{n}\overset{\Vert \cdot \Vert }{\rightarrow}x^{*}\) since \(\Vert \cdot \Vert \) and \(\Vert \cdot \Vert _{0}\) are equivalent by Lemma 5. The proof is complete. □
Remark 4
Theorem 5 partially improves Theorem 2 since the norm condition \(\Vert k\Vert <1\) is replaced by the spectral radius condition \(r(k)<1\).
The following example will show Theorem 5 is more applicable than many other fixed point results.
Example 2
Let \(E=P=\mathbb{R}_{+}^{2}=\{ x=(x_{1},x_{2}):x_{1},x_{2}\geq0\}\) with the norm \(\Vert x\Vert =\vert x_{1}\vert +\vert x_{2}\vert \). Clearly, P is a normal solid cone. Define a multiplication in E by
then E is a Banach algebra with \(e=(1,0)\). Define a mapping \(T:E\rightarrow E\) by
where \(a>1\), \(b\geq\sqrt{a1}\) and \(c>0\), then for each \(x,y\in E\) with \(y\preceq x\), by the Lagrange mean value theorem, we have
where \(k=(\frac{1}{a},c)\in P\). This implies that T is an orderLipschitz mapping with \(k=l=(\frac{1}{a},c)\). Note that \(k^{2}=(\frac{1}{a^{2}},\frac{2c}{a})\) and \(k^{3}=(\frac{1}{a^{3}},\frac{3c}{a^{2}})\), then by induction we obtain \(k^{n}=(\frac{1}{a^{n}},\frac{nc}{a^{n1}})\) for each n. Moreover, by Gelfand’s formula, we get
Hence by Theorem 5, T has a unique fixed point.
In the case that \(c>1\frac{1}{a}\), we get \(\Vert k\Vert =\frac {1}{a}+c>1\), and hence Theorem 2 is not applicable even taking k as a linear bounded mapping.
In the case that \(c>2\), we get \(\Vert TxTy\Vert \geq c\vert x_{1}y_{1}\vert >2\vert x_{1}y_{1}\vert \geq \vert x_{1}y_{1}\vert +\vert x_{2}y_{2}\vert =\Vert xy\Vert \) for each \(x,y\in E\) with \(\vert x_{1}y_{1}\vert \geq \vert x_{2}y_{2}\vert \), and hence the Banach contraction principle is not applicable.
In the case that \(c>1\), we get \(\arctan(b+x_{2})\arctan(b+y_{2})+c(x_{1}y_{1})\geq c(x_{2}y_{2})>x_{2}y_{2}\) for each \(x,y\in E\) with \(y\preceq x\) and \(x_{1}y_{1}\geq x_{2}y_{2}\). This implies that there does not exist \(l\in[0,1)\) such that \(TxTy\preceq l(xy)\). Consequently, Theorem 1 is not applicable.
Remark 5
The normality of P is essential for the completeness of \((E,\Vert \cdot \Vert _{0})\) (see Lemma 5), which leads to that the Banach contraction principle is applicable in Theorem 5. Naturally, one may wonder whether the normality of P in Theorem 5 could be removed by the method used in Theorem 1 or other methods.
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Acknowledgements
The work was supported by the National Natural Science Foundation of China (11161022, 11561026, 71462015), the Natural Science Foundation of Jiangxi Province (20142BCB23013, 20143ACB21012, 20151BAB201003, 20151BAB201023), the Natural Science Foundation of Jiangxi Provincial Education Department (KJLD14034, GJJ150479).
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MSC
 06A07
 47H10
Keywords
 Fixed point theorem
 Banach algebra
 orderLipschitz mapping
 Picardcomplete