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Nonlinear \((m,\infty )\)-isometries and \((m,\infty )\)-expansive (contractive) mappings on normed spaces
Journal of Inequalities and Applications volume 2021, Article number: 87 (2021)
Abstract
Let S be a self-mapping on a normed space \({\mathcal{X}}\). In this paper, we introduce three new classes of mappings satisfying the following conditions:
for all \(x,y\in {\mathcal{X}}\), where m is a positive integer. We prove some properties of these classes of mappings.
1 Introduction
The notion of an m-isometry in the setting of Hilbert spaces was introduced by Agler [1]: a bounded linear operator T on a Hilbert space \({\mathcal{H}}\) is an m-isometry (integer \(m \geq 1\)) if
where \(T^{*}\) denotes the adjoint operator of T. It is clear that (1.1) is equivalent to
A 1-isometry is an isometry and vice versa. We refer the reader to the trilogy [1–3] by Agler and Stankus for the fundamentals of the theory of m-isometries.
In the last years, a generalization of m-isometries to operators on general Banach spaces has been presented by several authors. Bayart [5] introduced the notion of \((m, p)\)-isometries on general (real or complex) Banach spaces. An operator T on a Banach space \({\mathcal{X}}\) is called an \((m, p)\)-isometry if there exist an integer \(m \geq 1 \) and \(p \in [1, \infty )\) with
In [11] the authors took off the restriction \(p \geq 1\). They considered equation (1.3) for \(p \in ( 0, \infty )\) and studied the role of the second parameter p and also discussed the case \(p = \infty \).
Let \(m \in \mathbb{{N}} \). An operator T acting on a Banach space \({\mathcal{X}}\) is called an \((m,\infty )\)-isometry (or \((m,\infty )\)-isometric operator) if
Let X and Y be metric spaces. A mapping \(S: X \longrightarrow Y\) is called an isometry if it satisfies \(d_{Y} (Sx, Sy) = d_{X}(x, y) \) for all \(x, y \in X\), where \(d_{X}(\cdot, \cdot )\) and \(d_{Y} (\cdot, \cdot )\) denote the metrics in the spaces X and Y, respectively.
In [6] the authors introduced the concept of \((m, q)\)-isometry for maps on a metric space \((X,d_{X})\) as follows: a mapping \(S: X \rightarrow X\) is called an \((m, q)\)-isometry for integer \(m \geq 1 \) and real \(q> 0 \) if it satisfies
Very recently, in [4] the present author studied a class of mappings, called \((m,\infty )\)-isometries, acting on a metric space. A mapping S acting on a metric space \((X,d_{X})\) is called an \((m, \infty )\)-isometry for some positive integer m if for all \(x,y \in X\),
In [9] the author considers \(A(m, p)\)-isometries, where for an operator \(A \in {\mathcal{B} }({\mathcal{X}})\), \(T \in {\mathcal{B}}({\mathcal{X}})\) (the algebra of bounded linear operators) is \(A(m, p)\)-isometric if
Evidently, an \(I(m, p)\)-isometry is an \((m, p)\)-isometry; if \({\mathcal{X}} = {\mathcal{H}}\) is a Hilbert space, then
If \(\beta _{m}^{(p)}(T,A,x) \leq 0 \) (resp., \(\beta _{m}^{(p)}(T,A,x)\geq 0\)) for \(x\in {\mathcal{X}}\), then T is said to be \((A,m,p)\)-expansive (resp., \((A,m,p)\)-contractive). We refer the interested reader to [10, 13] for details.
A mapping S (not necessarily linear) on a normed space \({\mathcal{X}}\) [12] is an \((m, p)\)-isometry for integer \(m \geq 1\) and real \(p > 0 \) if for all \(x,y \in {\mathcal{X}}\),
When \(m=1\), (1.5) is equivalent to \(\|Sx-Sy\|=\|x-y\|\) for \(x,y \in {\mathcal{X}}\), and when \(m=2\), (1.5) is equivalent to
After a short introduction and some connections with known results in this context, we present the main results of the paper as follows. In Sect. 2, we introduce and study some properties of \((m, \infty )\)-isometric mappings. Exactly, we give conditions under which a self-mapping S is an \((m, \infty )\)-isometry (Proposition 2.5, Corollary 2.7, Proposition 2.16). An \((m, \infty )\)-isometry becomes isometry (Theorems 2.10 and 2.20). An \((m, \infty )\)-isometric mapping becomes an \((m + 1, \infty )\)-isometric mapping. The product of two \((m, \infty )\)-isometries is an \((m, \infty )\)-isometry (Theorem 2.17), and a power of a \((2, \infty )\)-isometry is again a \((2, \infty )\)-isometry (Theorem 2.18). In Sect. 3, we present a parallel study of the classes of nonlinear \((m,\infty )\)-expansive and \((m,\infty )\)-contractive mappings.
2 Nonlinear \((m,\infty )\)-isometric mappings
This section is devoted to the study of some basic properties of the class of \((m,\infty )\)-isometric mappings (not necessary linear) on a normed space \({\mathcal{X}}\). Our inspiration comes from the papers [7, 11], and [14].
Let \(S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\) be an \((m,p)\)-isometric mapping. It obvious that
By taking the limit as \(p \to \infty \) we arrive at the following definition of an \((m,\infty )\)-isometric nonlinear mapping.
Definition 2.1
A nonlinear mapping \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\) is said to be an \(( m, \infty )\)-isometric mapping for some positive integer m if for all \(x,y\in {\mathcal{X}}\),
Remark 2.2
(1) A self-mapping S on \({\mathcal{X}}\) is an \((1,\infty )\)-isometry if for all \(x,y\in {\mathcal{X}}\),
(2) A self-mapping S on \({\mathcal{X}}\) is a \((2,\infty )\)-isometry if for all \(x,y\in {\mathcal{X}}\),
(3) A self-mapping S on \({\mathcal{X}}\) is a \((3,\infty )\)-isometry if for all \(x,y \in {\mathcal{X}}\),
Remark 2.3
The following remarks are obvious consequences of Definition 2.1.
(1) Every \((1,\infty )\)-isometry is an isometry and vice versa.
(2) Every isometric mapping is an \(( m,\infty )\)-isometric mapping for all \(m\geq 1\).
Indeed, the classes of \(( m,\infty )\)-isometries is a generalization of the class of
isometries.
(3) If S is an \((m,\infty )\)-isometry that satisfies \(S^{2}=I\) (the identity map), then S
is an isometry.
In the next example, we show that \((m, \infty )\)-isometries are in general neither continuous nor linear.
Example 2.4
Let \({\mathcal{X}} = \mathbb{R}\) with the usual norm \(\|x\| = |x|\). Consider the map \(S: \mathbb{R}\longrightarrow \mathbb{R}\) defined by
It is easy verify that S is a \((2,\infty )\)-isometry, but S is neither continuous nor linear.
Proposition 2.5
An mapping \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\) is an \(( m, \infty )\)-isometric if and only if
for all \(x,y \in {\mathcal{X}}\) and \(j\in \mathbb{N}_{0}: =\mathbb{N}\cup \{0\} \), where \(\mathbb{N}\) is the set of positive integers.
Proof
The proof follows by substituting x by \(S^{j}x\) and y by \(S^{j}y\) into (2.1) for \(j\in \mathbb{N}_{0}\), □
Proposition 2.6
([11, Lemma 5.3])
For all \(k \in \mathbb{N}_{0}\), let \(\pi (k) = k \textit{ mod } 2 \) denote the parity of k. Let further \(m \in \mathbb{N}\) with \(m \geq 1\), and let \((a_{k})_{k\in \mathbb{N}_{0}} \subset \mathbb{R}\). The following are equivalent.
(1) \((a_{k})_{k\in \mathbb{N}_{0}}\) satisfies
(2) \((a_{k})_{k\in \mathbb{N}_{0}}\) attains a maximum, and
Corollary 2.7
Let \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\), and let \(m\in \mathbb{N}\). Then S is an \(( m,\infty )\)-isometric mapping if and only if
for all \(x,y \in {\mathcal{X}}\) and \(j\in \mathbb{N}_{0}\).
Proof
The proof is essentially an application of Proposition 2.6. It suffices to consider \((a_{k})_{k}:= (\|S^{k}x-S^{k}y\| )_{k}\) for all \(x,y\in {\mathcal{X}}\). □
Definition 2.8
A self-mapping S on a normed space \({\mathcal{X}}\) is called power bounded if
Corollary 2.9
Let \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\) be an \((m,\infty )\)-isometry. Then for all \(n\in \mathbb{N}\) and \(x \in {\mathcal{X}}\),
In particular, S is power bounded.
Proof
From Corollary 2.7 we have
This gives that \(\max_{k \in \mathbb{N}_{0}}\|S^{k}x-S^{k}y\| < \infty \). Further, we see that for all \(n\in \mathbb{N}_{0}\),
In particular,
Therefore S is a power bounded mapping. □
In the following theorem, we show that if S is a self-mapping on a normed space \({\mathcal{X}}\) that is an \((m, \infty )\)-isometry, then there exists a metric \(d_{\infty }\) on \({\mathcal{X}}\) such that S is a \((1,\infty )\)-isometry on \(({\mathcal{X}}, d_{\infty })\).
Theorem 2.10
Let \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\) be an \(( m,\infty )\)-isometry. Then there exists a metric \(d_{\infty }\) on \({\mathcal{X}}\) such that S is an isometry on \(({\mathcal{X}}, d_{\infty })\). Moreover, \(d_{\infty }\) is given by
Proof
Since S is an \(( m,\infty )\)-isometric mapping, we have by Corollary 2.7 that
Define the map \(d_{\infty }:{\mathcal{X}} \times {\mathcal{X}}\to \mathbb{R}_{+}\) by
It is easy to show that the map \(d_{\infty }\) define a metric on \({\mathcal{X}}\). On the other hand, since S is an \((m, \infty )\)-isometry, it follows that
Consequently, \(d_{\infty }(x,y)=d_{\infty }(Sx,Sy)\). So, S is an isometry on \(({\mathcal{X}},d_{\infty })\), and the proof is complete. □
Proposition 2.11
Let \({\mathcal{X}}\) be a normed space, and let \(S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\) be a mapping (not necessarily linear). If S is an \(( m,\infty )\)-isometry, then S is an \(( m+1,\infty )\)-isometry.
Proof
Since S is an \((m,\infty )\)-isometry, it follows that
for all \(x,y \in {\mathcal{X}}\) and \(j \in \mathbb{N}_{0}\). Hence, for for all \(x,y\in {\mathcal{X}}\) and \(j \in \mathbb{N}_{0}\), we have
Consequently,
So, S is an \(( m+1,\infty )\)-isometry. □
Remark 2.12
In general, an \(( m,\infty )\)-isometry is not necessary an \(( m-1,\infty )\)-isometry as shown in the following example.
Example 2.13
Let \({\mathcal{X}}=\mathbb{R}^{2}\) be equipped with the norm \(\|(x,y)\|=|x|+|y|\). Define the map \(S:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \) by \(S(x,y)=(y+1,-x+y)\). A simple calculation shows that,
From the above calculation we easily see that
and
Consequently, S is a \(( 5,\infty )\)-isometry but not a \(( 4,\infty )\)-isometry.
Proposition 2.14
Let \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\). If \(S^{n}\) is an isometry for odd integer n, then S is an \(( m,\infty )\)-isometry for \(m\geq 2n-1\).
Proof
In view of Proposition 2.11, it suffices to show that S is a \(( 2n-1,\infty )\)-isometry.
Assume that \(S^{n}\) is an isometry. Then we have
Since n is an odd integer, for \(k \in \mathbb{N}_{0}\), we have that k is even if and only if \(n+k\) is odd. Since \(S^{n}\) is an isometry, it follows that
from which we deduce that S is a \(( 2n-1,\infty )\)-isometry. □
Corollary 2.15
Let \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\) be a mapping such that \(S^{n}\) is an isometry for an odd integer n. Then \(S^{k}\) is a \((2n-1,\infty )\)-isometry for any integer \(k\in \mathbb{N}\).
Proof
If \(k=1\), then the result follows from Proposition 2.14.
For \(k> 1\), if \(S^{n}\) is an isometry, then \((S^{k})^{n}\) is also an isometry, so by Proposition 2.14 we get that \(S^{k}\) is a \((2n-1,\infty )\)-isometry. □
The following proposition generalizes [11, Proposition 5.8].
Proposition 2.16
Let \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\), and let \(m\in \mathbb{N}, m\geq 2\). Then the following properties hold.
-
(1)
If \(m\geq 3\) and S satisfy the conditions
-
(i)
\(\|S^{m}x-S^{m}y\|=\|S^{m-1}x-S^{m-1}y\|\) and
-
(ii)
\(\|S^{m}x-S^{m}y\|\geq \|S^{k}x-S^{k}y\|\) for \(k=0, \ldots,m-2\) and all \(x,y\in {\mathcal{X}}\),
then S is an \(( m,\infty )\)-isometry.
-
(i)
-
(2)
If \(m=2\), then S is an \((2,\infty )\)-isometry if and only if
$$ \bigl\Vert S^{2}x-S^{2}y \bigr\Vert = \Vert Sx-Sy \Vert \quad \textit{and}\quad \bigl\Vert S^{2}x-S^{2}y \bigr\Vert \geq \Vert x-y \Vert ,\quad \forall x,y\in {\mathcal{X}}. $$
Proof
(1) In view of conditions (i) and (ii), it is clear that
so that S is an \(( m,\infty )\)-isometry.
(2) Assume that S is an \(( 2,\infty )\)-isometry. Then we have
and it follows that
Replacing x by Sx and y by Sy, we get
and then
So we have
The converse follows from statement (1). □
The authors in [6] proved that if \(T,S:{\mathcal{X}}\longrightarrow \mathcal{ X}\) are two linear maps such that \(TS=ST\), T is an \((m,p)\)-isometry, and S is an \((n,p)\)-isometry, then TS is an \((m+n-1,p)\)-isometry. A similar result was proved in [4, Theorem 2.4]. In the following theorem, we show if T is an \((m,\infty )\)-isometry and S is a \((2,\infty )\)-isometry for which \(TS=ST\), then TS is an \(( m,\infty )\)-isometry.
Theorem 2.17
Let \(T,S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\) be two nonlinear mappings such that \(ST=TS\). If T is an \(( m,\infty )\)-isometry and S is a \((2,\infty )\)-isometry, then TS is an \((m, \infty )\)-isometry.
Proof
Since S is a \(( 2,\infty )\)-isometry, by statement (2) of Proposition 2.16 we have
Now assume that T is a \((2,\infty )\)-isometry. Then it follows that for all \(x,y \in {\mathcal{X}}\),
Consequently,
This implies that TS is a \(( 2,\infty )\)-isometry by statement (2) of Proposition 2.16.
We further suppose that \(m> 2\). By the inequality
for all \(k=1,2,\ldots \) , we have
Thus for each \(x,y \in {\mathcal{X}}\), we have
Using this inequality, for all \(x,y\in {\mathcal{X}}\), we have
We obtain
On the other hand, it is obvious that for all \(x,y \in {\mathcal{X}}\),
Then we have
Using this inequality, we get
In same way, we also have
Since T is an \(( m,\infty )\)-isometry, we deduce that
So, the desired conclusion is an immediate consequence of Definition 2.1. □
Patel [14] showed that if S is a 2-isometric operator on a Hilbert space, then \(S^{2}\) is a 2-isometric operator. We now generalize this result to a \((2,\infty )\)-isometric mappings.
Theorem 2.18
A power of a \((2,\infty )\)-isometric nonlinear mapping is again a \((2, \infty )\)-isometric mapping.
Proof
Let \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}}\) be a \(( 2,\infty )\)-isometric mapping. We need to prove that \(S^{k} \) is a \(( 2,\infty )\)-isometric mapping for all positive integers k.
By statement (2) of Proposition 2.16 it suffices to show that
Using mathematical induction on k we will show that
For \(k=1\), it is true since S is an \((2,\infty )\)-isometry. Assume that this equality is true for k and prove it for \(k+1\). Indeed, we have
Thus by induction we have proved that \(\|S^{2k}x-S^{2k}y\|=\| S^{k}x-S^{k}y\|\) for all \(x,y\in {\mathcal{X}}\), for all \(k =1,2,\dots \).
It remains to show that for all \(x,y \in {\mathcal{X}}\), \(\|S^{k}x-S^{k}y\|\geq \|x-y\|\) for all \(k=1,2,\ldots \) .
Indeed, since \(\| Sx-Sy\|\geq \|x-y\|\) for all \(x,y \in {\mathcal{X}}\), by using the same inequality we have that for all \(x,y \in {\mathcal{X}}\),
By induction on k it follows that
Therefore \(S^{k}\) also is a \(( 2,\infty )\)-isometry. This completes the proof. □
Theorem 2.19
Let \(S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\) be an invertible \(( m,\infty )\)-isometry. Then the following statements hold.
-
(i)
\(S^{-1}\) is an \((m,\infty )\)-isometry.
-
(ii)
If m is even, then S is an \(( m-1,\infty )\)-isometry.
Proof
(i) Since S is an \(( m,\infty )\)-isometry, from Definition 2.1 it follows that
Replacing x by \(S^{-m}x\) and \(S^{-m}y\), we obtain
or, equivalently,
which implies
Consequently, \(S^{-1}\) is an \(( m,\infty )\)-isometry.
(ii) Since S is an \(( m,\infty )\)-isometry, it follows that
and, moreover,
Since S is invertible and \(\pi (j-1)\neq\pi (m-2+j)\) for even m, we get that
This shows that
Hence the proof of the statement (ii) is complete. □
Theorem 2.20
Let \(S:{\mathcal{X}} \longrightarrow {\mathcal{X}} \) be a mapping such that \(S^{2}\) is an isometry. Then the following conditions are equivalent.
(1) S is an isometry,
(2) S is an \((m,\infty )\)-isometry.
Proof
Since \(S^{2}\) is an isometry, it follows that
and
This shows that \((1)\Longleftrightarrow (2)\). □
Similarly to the \((m,q)\)-isometry (see [6, Proposition 2.18], we obtain the following theorem.
Theorem 2.21
For \(i=1,2,\ldots,n\), let \(({\mathcal{X}}_{i},\|\cdot \|_{i})\) be a normed space, and let \(S_{i}:{\mathcal{X}_{i}}\longrightarrow \mathcal{X}_{i}\), \(m_{i}\geq 1\). Denote by \({\mathcal{X}}={\mathcal{X}}_{i}\times {\mathcal{X}}_{2}\times \cdots \times {\mathcal{X}}_{n}\) the product space endowed with the product norm \(\|(x_{1},x_{2},\ldots,x_{n})\|:=\max_{1\leq i\leq n} (\|x_{i}\|_{i} )\). Let \(S:=S_{1}\times S_{2}\times \cdots \times S_{n}: {\mathcal{X}} \rightarrow {\mathcal{X}}\) be the mapping defined by
If each \(S_{i}\) is an \(( m_{i},\infty )\)-isometry for \(i=1,2,\ldots,n\), then S is an \(( m,\infty )\)-isometry, where \(m=\max (m_{1},\ldots,m_{n})\).
Proof
Let \(m=\max (m_{1},\ldots,m_{n})\) and consider, for all \(x,y \in X\),
Since each \(S_{i}\) is an \(( m_{i},\infty )\)-isometric operator for \(i=1,2,\ldots,n\), it follows that \(S_{i}\) is an \(( m,\infty )\)-isometry for \(i=1,2,\ldots,n\) by Proposition 2.11, from the above equality we have
Thus we have
for all \(x,y\in {\mathcal{X}}\). Therefore S is an \(( m,\infty )\)-isometric operator. □
3 Nonlinear \((m,\infty )\)-expansive and \((m,\infty )\)-contractive mappings
In this section, we introduce and study \(( m,\infty )\)-expansive and \((m,\infty )\)-contractive nonlinear mappings on a normed space. We observe that
Taking the limit as \(p \to \infty \), we arrive at the following definition of an \((m,\infty )\)-expansive mapping.
Definition 3.1
Let \(m \in \mathbb{N}\). A mapping \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\) is said to be
(1) \((m, \infty )\)-expansive if
(2) \((m, \infty )\)-hyperexpansive if S is \(( k,\infty )\)-expansive for \(k=1,\ldots,m\);
(3) completely ∞-hyperexpansive if S is \((k,\infty )\)-expansive for all \(k \in \mathbb{N}\).
Similarly,
Taking the limit as \(p \to \infty \), we arrive at the following definition of an \((m,\infty )\)-contractive mapping.
Definition 3.2
Let \(m \in \mathbb{N}\). A mapping \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\) is said to be
(1) \(( m, \infty )\)-contractive if
(2) \((m, \infty )\)-hypercontractive if S is \((k,\infty )\)-contractive for \(k=1,\ldots,m\);
(3) completely ∞-hypercontractive if S is \((k,\infty )\)-contractive for all \(k \in \mathbb{N}\).
Remark 3.3
We make the following observations.
(1) Every \((1,\infty )\)-expansive mapping is expansive, that is, \(\|Sx-Sy\| \geq \|x-y\|\) for all \(x,y \in {\mathcal{X}}\).
(2) Every \((1,\infty )\)-contractive mapping is contractive, that is, \(\|Sx-Sy\| \leq \|x-y\|\) for all \(x,y \in {\mathcal{X}}\).
(3) S is \((2,\infty )\)-expansive if for all \(x,y\in {\mathcal{X}}\),
(4) S is \((2,\infty )\)-contractive if for all \(x,y\in {\mathcal{X}}\),
Remark 3.4
Observe that every \((m,\infty )\)-isometry is an \(( m,\infty )\)-expansive and \(( m,\infty )\)-contractive mapping.
Theorem 3.5
Let \(S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\). The we have the following properties:
(1) T is \(( m, \infty )\)-expansive if and only if
(2) S is \(( m, \infty )\)-contractive if and only if
Proof
Let \(j\in \mathbb{N}_{0}\). The desired characterizations follow by substituting \(S^{j}x\) for x and \(S^{j}y\) for y in statement (1) of Definition 3.1 and statement (2) of Definition 3.2. □
Proposition 3.6
Let \(S: {\mathcal{X}}\longrightarrow {\mathcal{X}}\) be a mapping such that \(S^{2}\) is an isometry. Then the following are equivalent.
(1) S is \((m,\infty )\)-expansive,
(2) S is expansive,
(3) S is an isometry,
(4) S is contractive,
(5) S is \((m,\infty )\)-contractive.
Proof
Since \(S^{2}\) is an isometry, it follows that
and
and this shows that \((1)\Longleftrightarrow (2)\) and \((4) \Longleftrightarrow (5)\). The equivalence of (2), (3), and (4) follows on replacing x by Sx and y by Sy. □
Theorem 3.7
Let \(S:{\mathcal{X}}\longrightarrow {\mathcal{X}}\) be a is invertible mapping. Then we have:
-
(1)
If S is \((m,\infty )\)-expansive, then
-
(i)
\(S^{-1}\) is \((m,\infty )\)-expansive for even m, and
-
(ii)
\(S^{-1}\) is \(( m,\infty )\)-contractive for odd m.
-
(i)
-
(2)
If S is \((m,\infty )\)-contractive, then
-
(i)
\(S^{-1}\) is \(( m,\infty )\)-contractive for even m, and
-
(ii)
\(S^{-1}\) is \((m,\infty )\)-expansive for odd m.
-
(i)
Proof
(1) Since S is an invertible \(( m,\infty )\)-expansive mapping, we have
for all \(x,y\in {\mathcal{X}}\). Replacing x by \(S^{-m}x\) and y by \(S^{-m}y\) in this inequality, we get
for all \(x,y\in {\mathcal{X}}\). From this it immediately follows that
proving the first statement.
(2) This statement is proved in the same way as statement (1). □
Corollary 3.8
Let \(S: {\mathcal{X}} \longrightarrow {\mathcal{X}}\) be an invertible mapping. We have:
(1) If S is \((2,\infty )\)-expansive, then S is a \((1,\infty )\)-isometry.
(2) If S is \((2,\infty )\)-contractive, then S is a \((1,\infty )\)-isometry.
Proof
(1) If S is \((2,\infty )\)-expansive, then we have
By Theorem 3.7, \(S^{-1}\) is \((2,\infty )\)-expansive, so
This means that \(\|x-y\|\geq \|Sx-Sy\|\) for all \(x,y \in {\mathcal{X}}\). Therefore
(2) This statement is proved in the same way as statement (1). □
Proposition 3.9
Let \(S: {\mathcal{X}} \longrightarrow {\mathcal{X}}\) be a \((2,\infty )\)-expansive mapping and an \(( m, \infty )\)-isometry, then S is a \((2,\infty )\)-isometry.
Proof
Since S is a \((2,\infty )\)-expansive mapping and an \(( m, \infty )\)-isometry, it follows that
and
Combining (3.2) and (3.3), we obtain
So, S is a \((2,\infty )\)-isometry. □
Theorem 3.10
For \(i=1,2,\ldots,n\), let \(({\mathcal{X}}_{i},\|\cdot \|_{i})\) be a normed space, and let \(S_{i}:{\mathcal{X}}_{i}\rightarrow {\mathcal{X}}_{i}\), \(m_{i}\geq 1\). Denote by \({\mathcal{X}}={\mathcal{X}}_{i}\times {\mathcal{X}}_{2}\times \cdots \times {\mathcal{X}}_{n}\) the product space endowed with the product norm \(\|(x_{1},x_{2},\ldots,x_{n})\|:=\max_{1\leq i\leq n} (\|x_{i}\| )\). Let \(S:=S_{1}\times S_{2}\times \cdots \times S_{n}: {\mathcal{X}} \rightarrow {\mathcal{X}}\) be the mapping defined by
Then we have:
(1) If each \(S_{i}\) is \((m_{i},\infty )\)-hyperexpansive for \(i=1,2,\ldots,n\), then S is \((m,\infty )\)-expansive, where \(m=\min (m_{1},\ldots,m_{n})\).
(2) If each \(S_{i}\) is \((m_{i},\infty )\)-hypercontractive for \(i=1,2,\ldots,n\), then S is \((m,\infty )\)-contractive, where \(m=\min (m_{1},\ldots,m_{n})\).
(3) If each \(S_{i}\) is completely ∞-hyperexpansive for \(i=1,2,\ldots,n\), then so is S.
(4) If each \(S_{i}\) is completely ∞-hypercontractive for \(i=1,2,\ldots,n\), then so is S.
Proof
(1) Let \(m=\min (m_{1},\ldots,m_{n})\) and consider, for all \(x,y \in X\),
Since \(S_{i}\) is \((m_{i},\infty )\)-hyperexpansive for \(i=1,2,\ldots,n\), it follows that \(S_{i}\) is \((m,\infty )\)-expansive for \(i=1,2,\ldots,n\), and hence
Thus we have
Consequently, S is an \((m,\infty )\)-expansive mapping.
(2) This statement follows from statement (1) by reversing the above inequality.
(3) Suppose that \(S_{i}\) is completely ∞-hyperexpansive for each \(i=1,2,\ldots,n\), and hence each \(S_{i}\) is \((k,\infty )\)-expansive for any \(k \in \mathbb{ N}\). As a consequence of this observation, we deduce the following inequality for all \(x,y \in X\):
from which statement (3) follows.
(4) This statement is proved in the same way as statement (3). □
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Ayed Al-Ahmadi, A.M. Nonlinear \((m,\infty )\)-isometries and \((m,\infty )\)-expansive (contractive) mappings on normed spaces. J Inequal Appl 2021, 87 (2021). https://doi.org/10.1186/s13660-021-02607-w
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DOI: https://doi.org/10.1186/s13660-021-02607-w