In this section, the proofs of results and theorems will be given.

*Proof of Lemma 1* Since {lim}_{\parallel x\parallel \to \mathrm{\infty}}sup\parallel {T}^{\prime}(x)\parallel =\lambda, for any \epsilon >0, there is b>0 such that \parallel {T}^{\prime}(x)\parallel \le \lambda +\epsilon as \parallel x-{x}^{\ast}\parallel \ge b. Let F(b)=\{x\in {R}^{n}:\parallel x-{x}^{\ast}\parallel \le b\}. Since *T* is continuously differentiable in {R}^{n}, \parallel {T}^{\prime}(x)\parallel is a continuous function in the compact set F(b). Thus, we have {sup}_{z\in F(b)}\parallel {T}^{\prime}(z)\parallel <\mathrm{\infty}. This implies that M={sup}_{z\in {R}^{n}}\parallel {T}^{\prime}(z)\parallel <\mathrm{\infty} holds. From the mid-value theorem, for any x,y\in {R}^{n}, one obtains

\parallel Tx-Ty\parallel \le \underset{z\in {R}^{n}}{sup}\parallel {T}^{\prime}(z)\parallel \cdot \parallel x-y\parallel \le M\cdot \parallel x-y\parallel .

This implies that {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})\le M<\mathrm{\infty} holds. Since {T}^{j} is continuously differentiable, with the definition of the Gateaux differential, for any x\in {R}^{n} and any h\in {R}^{n} with \parallel h\parallel =1,

\parallel {\left({T}^{j}\right)}^{\prime}(x)\cdot h\parallel =\underset{t\to {0}^{+}}{lim}\frac{\parallel {T}^{j}(x+t\cdot h)-{T}^{j}x\parallel}{t}=\underset{t\to {0}^{+}}{lim}\frac{\parallel {T}^{j}(x+t\cdot h)-{T}^{j}x\parallel}{\parallel t\cdot h\parallel}\le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{j},{R}^{n}).

This implies that

\parallel {\left({T}^{j}\right)}^{\prime}(x)\parallel =\underset{\parallel h\parallel =1}{sup}\parallel {\left({T}^{j}\right)}^{\prime}(x)\cdot h\parallel \le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{j},{R}^{n})\le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}{(T,{R}^{n})}^{j}.

The proof is done. □

*Proof of Theorem 1* The equivalent relationships between (C), (D) and (E) can be easily gotten from Wang and Xu [11]. And it is also easy to deduce (B) from (A). Next, we mainly prove (C) ⇒ (A) and (B) ⇒ (C).

Firstly, prove (C) ⇒ (A).

If Lip(T,{R}^{n})<1, then system (2) is of global exponential stability in {R}^{n}. With equation (6), for any sufficiently small \epsilon \in (0,1-Lip(T,{R}^{n})), there is a strongly equivalent metric {d}_{\epsilon}(\cdot ,\cdot )\in \mathrm{\Omega} such that {L}_{{d}_{\epsilon}}(T,{R}^{n})\le Lip(T,{R}^{n})+\epsilon <1. Assume that {d}_{\epsilon}(\cdot ,\cdot ) and \parallel \cdot \parallel have the following strongly equivalent relationships:

C{(\epsilon )}_{1}\cdot {d}_{\epsilon}(x,y)\le \parallel x-y\parallel \le C{(\epsilon )}_{2}\cdot {d}_{\epsilon}(x,y)\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y\in {R}^{n},

(7)

where C{(\epsilon )}_{2}>C{(\epsilon )}_{1}>0 are two constants. For any positive integer *k* and any x\in {R}^{n}, we have

\begin{array}{rcl}\parallel {T}^{k}x-{x}^{\ast}\parallel & =& \parallel {T}^{k}x-{T}^{k}{x}^{\ast}\parallel \le C{(\epsilon )}_{2}\cdot {d}_{\epsilon}({T}^{k}x,{T}^{k}{x}^{\ast})\\ \le & C{(\epsilon )}_{2}\cdot {L}_{{d}_{\epsilon}}{(T,{R}^{n})}^{k}\cdot {d}_{\epsilon}(x,{x}^{\ast})\\ \le & C{(\epsilon )}_{2}\cdot {(Lip(T,{R}^{n})+\epsilon )}^{k}\cdot {d}_{\epsilon}(x,{x}^{\ast})\\ \le & \frac{C{(\epsilon )}_{2}}{C{(\epsilon )}_{1}}\cdot {(Lip(T,{R}^{n})+\epsilon )}^{k}\cdot \parallel x-{x}^{\ast}\parallel .\end{array}

(8)

Thus, system (2) is of global exponential stability in {R}^{n} with the exponential bound Lip(T,{R}^{n})+\epsilon.

Secondly, prove (B) ⇒ (C); namely, if system (2) is of global exponential stability in {R}^{n} and \rho ({T}^{\prime}({x}^{\ast}))<1, then Lip(T,{R}^{n})<1.

Since {lim}_{\parallel x\parallel \to \mathrm{\infty}}sup\parallel {T}^{\prime}(x)\parallel =\lambda \le \alpha <1, for any \epsilon \in (0,1-\alpha ), there is a constant b>0 such that \parallel {T}^{\prime}(x)\parallel \le \alpha +\epsilon <1 as \parallel x-{x}^{\ast}\parallel \ge b. Let E(b)=\{x\in {R}^{n}:\parallel x-{x}^{\ast}\parallel >b\}, F(b)=\{x\in {R}^{n}:\parallel x-{x}^{\ast}\parallel \le b\}, B(b)=\{x\in {R}^{n}:\parallel x-{x}^{\ast}\parallel \le 2b\}. For any z\in E(b) and any given positive integer *m*, if {T}^{i}z\in E(b) holds for any positive integer i=0,1,\dots ,m-1, then by the chain-rule of derivative, one gets

\parallel {\left({T}^{m}\right)}^{\prime}(z)\parallel =\parallel \prod _{i=1}^{m}{T}^{\prime}\left({T}^{m-i}z\right)\parallel \le \prod _{i=1}^{m}\parallel {T}^{\prime}\left({T}^{m-i}z\right)\parallel \le {(\alpha +\epsilon )}^{m}.

(9)

For the fixed \epsilon \in (0,1-\alpha ), there exists an equivalent norm {\parallel \cdot \parallel}_{\epsilon} such that the subordinated matrix norm {\parallel {T}^{\prime}({x}^{\ast})\parallel}_{\epsilon}\le \rho ({T}^{\prime}({x}^{\ast}))+\epsilon /2. Since *T* is continuously differentiable, {\parallel {T}^{\prime}(x)\parallel}_{\epsilon} is a continuous function of x\in {R}^{n}. It implies that there exists a spherical neighborhood U(r)=\{x\in {R}^{n}:{\parallel x-{x}^{\ast}\parallel}_{\epsilon}\le r\} such that for any x\in U(r),

{\parallel {T}^{\prime}(x)\parallel}_{\epsilon}\le {\parallel {T}^{\prime}\left({x}^{\ast}\right)\parallel}_{\epsilon}+\epsilon /2\le \rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon .

By the mid-value theorem, for any x,y\in U(r), we have

{\parallel Tx-Ty\parallel}_{\epsilon}\le \underset{z\in U(r)}{sup}{\parallel {T}^{\prime}(z)\parallel}_{\epsilon}\cdot {\parallel x-y\parallel}_{\epsilon}\le (\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )\cdot {\parallel x-y\parallel}_{\epsilon}.

(10)

If system (2) is of global exponential stability in {R}^{n}, it is globally uniformly asymptotically stable in {R}^{n} as well (Elaydi [16]). Therefore, there exists a positive integer *N* such that for any positive integer k\ge N and any x\in B(b), {T}^{k}x\in U(r) holds. With equation (10), for any x,y\in B(b), if k>N, we have

\begin{array}{rcl}{\parallel {T}^{k}x-{T}^{k}y\parallel}_{\epsilon}& \le & (\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )\cdot {\parallel {T}^{k-1}x-{T}^{k-1}y\parallel}_{\epsilon}\le \cdots \\ \le & {(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{k-N}\cdot {\parallel {T}^{N}x-{T}^{N}y\parallel}_{\epsilon}\\ \le & {(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{k-N}\cdot {L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}{(T,{R}^{n})}^{N}\cdot {\parallel x-y\parallel}_{\epsilon}.\end{array}

It implies that for any positive integer k>N, one obtains

{L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}({T}^{k},B(b))\le {(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{k-N}\cdot {L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}{(T,{R}^{n})}^{N}.

Since {L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}(T,{R}^{n})<\mathrm{\infty}, then

\begin{array}{rcl}Lip(T,B(b))& =& \underset{k\to \mathrm{\infty}}{lim}{L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}{({T}^{k},B(b))}^{1/k}\\ \le & \underset{k\to \mathrm{\infty}}{lim}{(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{(k-N)/k}\cdot \underset{k\to \mathrm{\infty}}{lim}{L}_{{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}_{\epsilon}}{(T,{R}^{n})}^{N/k}\\ \le & (\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon ).\end{array}

Lip(T,B(b))\le \rho ({T}^{\prime}({x}^{\ast})) holds for *ε* is an arbitrary positive number. Thus, there exists a positive integer {N}_{1} such that {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{k},B(b))\le {(\rho ({T}^{\prime}({x}^{\ast}))+\epsilon )}^{k} for the fixed \epsilon >0 as a positive integer k\ge {N}_{1}. For any positive integer k\ge {N}_{1}, {T}^{k} is continuously differentiable, and for any x\in F(b)\subset B(b), *x* is an inner point of B(b). By the definition of the Gateaux differential, for any x\in F(b), k\ge {N}_{1} and h\in {R}^{n} with \parallel h\parallel =1, one can obtain

\begin{array}{rcl}\parallel {\left({T}^{k}\right)}^{\prime}(x)\cdot h\parallel & =& \underset{t\to {0}^{+}}{lim}\frac{\parallel {T}^{k}(x+t\cdot h)-{T}^{k}x\parallel}{t}=\underset{t\to {0}^{+}}{lim}\frac{\parallel {T}^{k}(x+t\cdot h)-{T}^{k}x\parallel}{\parallel t\cdot h\parallel}\\ \le & {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{k},B(b)).\end{array}

It implies that for any x\in F(b) and k\ge {N}_{1},

\begin{array}{rcl}\parallel {\left({T}^{k}\right)}^{\prime}(x)\parallel & =& \underset{\parallel h\parallel =1}{sup}\parallel {\left({T}^{k}\right)}^{\prime}(x)\cdot h\parallel \le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{k},B(b))\\ \le & {(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{k}\le {(\alpha +\epsilon )}^{k}\end{array}

(11)

holds. For any z\in {R}^{n}, either z\in F(b) or z\in E(b), the result that if z\in F(b), then \parallel {({T}^{k})}^{\prime}(z)\parallel \le \beta holds for any positive integer k\ge {N}_{1} is shown above. Next, we will prove that a similar result exists for any z\in E(b). For any given z\in E(b) and any given positive integer k\ge {N}_{1}, there are the following two different cases.

Case 1: {T}^{i}z\in E(b) for any positive integer i<k. From equation (9), one may have \parallel {({T}^{k})}^{\prime}(z)\parallel \le {(\alpha +\epsilon )}^{k}.

Case 2: There exists a positive integer j<k such that {T}^{i}z\in E(b) for any positive integer i<j and {T}^{j}z\in F(b). If k-j\ge {N}_{1}, then by the chain-rule of derivative and equations (9) and (11), we have

\begin{array}{rcl}\parallel {\left({T}^{k}\right)}^{\prime}(z)\parallel & =& \parallel {\left({T}^{k-j}\right)}^{\prime}\left({T}^{j}z\right)\cdot {\left({T}^{j}\right)}^{\prime}(z)\parallel \\ \le & \parallel {\left({T}^{k-j}\right)}^{\prime}\left({T}^{j}z\right)\parallel \cdot \parallel {\left({T}^{j}\right)}^{\prime}(z)\parallel \\ \le & {(\alpha +\epsilon )}^{k-j}\cdot {(\alpha +\epsilon )}^{j}\le {(\alpha +\epsilon )}^{k}.\end{array}

If k-j<{N}_{1}, then from equation (9) and Lemma 1, one gets

\begin{array}{rcl}\parallel {\left({T}^{k}\right)}^{\prime}(z)\parallel & =& \parallel {\left({T}^{k-j}\right)}^{\prime}\left({T}^{j}z\right)\cdot {\left({T}^{j}\right)}^{\prime}(z)\parallel \le \parallel {\left({T}^{k-j}\right)}^{\prime}\left({T}^{j}z\right)\parallel \cdot \parallel {\left({T}^{j}\right)}^{\prime}(z)\parallel \\ \le & {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}{(T,{R}^{n})}^{k-j}\cdot {(\alpha +\epsilon )}^{j}\le {(\alpha +\epsilon )}^{k}\cdot {\left(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha +\epsilon}\right)}^{k-j}.\end{array}

(12)

Since \epsilon >0 is an arbitrary positive number, with equation (12), we have \parallel {({T}^{k})}^{\prime}(z)\parallel \le {\alpha}^{k}\cdot {(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha})}^{k-j}. And since \parallel {T}^{\prime}(x)\parallel \le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n}) holds for any x\in {R}^{n} (Wang *et al.* [9]), we have \lambda {\le}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}L(T,{R}^{n}) and \rho ({T}^{\prime}({x}^{\ast}))\le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n}). It implies that \frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha}\ge 1 and \parallel {({T}^{k})}^{\prime}(z)\parallel \le {\alpha}^{k}\cdot {(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha})}^{{N}_{1}}.

From the above two cases, for any k\ge {N}_{1} and z\in E(b),

\parallel {\left({T}^{k}\right)}^{\prime}(z)\parallel \le {(\alpha +\epsilon )}^{k}\cdot {\left(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha}\right)}^{{N}_{1}}.

(13)

By combining equations (11) and (13), one can conclude that for any positive integer k\ge {N}_{1} and any z\in {R}^{n},

\parallel {\left({T}^{k}\right)}^{\prime}(z)\parallel \le {(\alpha +\epsilon )}^{k}\cdot {\left(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha}\right)}^{{N}_{1}}.

For any x,y\in {R}^{n} and k\ge {N}_{1},

{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{k},{R}^{n})=\underset{x,y\in {R}^{n},x\ne y}{sup}\frac{\parallel {T}^{k}x-{T}^{k}y\parallel}{\parallel x-y\parallel}\le \underset{z\in {R}^{n}}{sup}\parallel {\left({T}^{k}\right)}^{\prime}(z)\parallel \le {(\alpha +\epsilon )}^{k}\cdot {\left(\frac{{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}(T,{R}^{n})}{\alpha}\right)}^{{N}_{1}}.

It implies that

Lip(T,{R}^{n})=\underset{k\to \mathrm{\infty}}{lim}{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}{({T}^{k},{R}^{n})}^{1/k}\le (\alpha +\epsilon )\cdot \underset{k\to \mathrm{\infty}}{lim}{\left(\frac{L(T,{R}^{n})}{\alpha}\right)}^{\frac{{N}_{1}}{k}}=\alpha +\epsilon <1.

Since *ε* is an arbitrary positive number, Lip(T,{R}^{n})\le \alpha holds. □

*Proof of Theorem 2* Since system (2) is of global exponential stability in {R}^{n}, there are constants M>0 and 0<\alpha <1 such that for any {x}_{0}\in {R}^{n} and any positive integer *k*, we have \parallel {x}_{k}-{x}^{\ast}\parallel \le M\cdot {\alpha}^{k}\cdot \parallel {x}_{0}-{x}^{\ast}\parallel. With Lemma 1, for any positive integer *k* and x\in {R}^{n}, then \parallel {({T}^{k})}^{\prime}({x}^{\ast})\parallel \le {L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}({T}^{k},{R}^{n}). By the chain-rule of derivative, for any given positive integer *k*, one obtains {({T}^{k})}^{\prime}({x}^{\ast})={\prod}_{i=1}^{k}{T}^{\prime}({T}^{k-i}{x}^{\ast})={T}^{\prime}{({x}^{\ast})}^{k}. This implies that

\begin{array}{rcl}\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)& =& \underset{k\to \mathrm{\infty}}{lim}{\parallel {T}^{\prime}{\left({x}^{\ast}\right)}^{k}\parallel}^{1/k}=\underset{k\to \mathrm{\infty}}{lim}{\parallel {\left({T}^{k}\right)}^{\prime}\left({x}^{\ast}\right)\parallel}^{1/k}\\ \le & \underset{k\to \mathrm{\infty}}{lim}{L}_{\parallel \phantom{\rule{0.25em}{0ex}}\parallel}{({T}^{k},{R}^{n})}^{1/k}=Lip(T,{R}^{n}).\end{array}

Lip(T,{R}^{n})\le \alpha =\rho ({T}^{\prime}({x}^{\ast})) holds from Theorem 1 and Lip(T,{R}^{n})=\rho ({T}^{\prime}({x}^{\ast})) if system (2) is globally exponentially stable in {R}^{n}. And from equation (8), for any \epsilon >0 and *k*, we have

\parallel {T}^{k}x-{x}^{\ast}\parallel \le \frac{C{(\epsilon )}_{2}}{C{(\epsilon )}_{1}}\cdot {(Lip(T,{R}^{n})+\epsilon )}^{k}\cdot \parallel x-{x}^{\ast}\parallel =\frac{C{(\epsilon )}_{2}}{C{(\epsilon )}_{1}}\cdot {(\rho \left({T}^{\prime}\left({x}^{\ast}\right)\right)+\epsilon )}^{k}\cdot \parallel x-{x}^{\ast}\parallel ,

where C{(\epsilon )}_{2}\ge C{(\epsilon )}_{1}>0 are the same constants as in equation (7). It implies that \rho ({T}^{\prime}({x}^{\ast})) is not less than the infimum of exponential bounds of convergent trajectories. Moreover, from Wang *et al.* [9], Wang and Xu [11] if system (2) is of global exponential stability in {R}^{n} with an exponential bound *α*, then \alpha \ge \rho ({T}^{\prime}({x}^{\ast})). Therefore, \rho ({T}^{\prime}({x}^{\ast})) is not larger than the infimum of exponential bounds of convergent trajectories. Finally, \rho ({T}^{\prime}({x}^{\ast})) is the infimum of exponential bounds of convergent trajectories. □