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I was wondering how could I prove such a property stated in [Angeli, 2004]. For instance, consider the system $\dot{x}=f(x)$, where $f:\mathbb{R}^n\to\mathbb{R}^n$ is Lipschitz continuous.

Claim. Suppose that the system is locally stable and globally attractive. Then, it is globally asymptotically stable.

Proof. I was trying to prove it as follows. From local stability definition, there exists neighborhood $N$ of the origin for which, $$\forall\varepsilon>0, \exists\delta>0:|\xi|<\delta\Rightarrow|\phi(t,\xi)|<\varepsilon,\forall t\geq0,\forall \xi\in N,$$ and \begin{equation*} \lim_{t\to\infty}\phi(t,\xi)=0,\quad\forall \xi\in N. \end{equation*} where $\phi$ is a solution for the system.

Let $\tilde{\xi}\notin N$ be an initial condition for a solution $\phi$ of the system. From the global attracvity, $$\lim_{t\to\infty}\phi(t,\tilde{\xi})=0.$$ From the local asymptotic stability of the origin, there exists $T<\infty$ such that, $\forall t>T$, $\phi(T,\tilde{\xi})\in N$. Moreover, \begin{equation*} \forall\varepsilon>0, \exists\delta>0:|\phi(T,\tilde{\xi})|<\delta\Rightarrow|\phi(t,\phi(T,\tilde{\xi}))|<\varepsilon,\forall t\geq0. \end{equation*} From the time-invariance property and unicity of solutions, $$\forall t\geq0,\quad\phi(t,\phi(T,\tilde{\xi}))=\phi(t+T,\tilde{\xi}).$$

Thus, $\forall\xi\in\mathbb{R}^n$, \begin{equation*} \forall\varepsilon>0,\exists\delta>0:|\xi|<\delta\Rightarrow|\phi(t,\xi)|<\varepsilon \end{equation*} and \begin{equation*} \lim_{t\to\infty}\phi(t,\xi)=0. \end{equation*} Hence, the origin of the system is globally asymptotically stable.$\square$

Is this proof correct?

References:

[Angeli, 2004] Angeli, D. (2004). An Almost Global Notion of Input-to-State Stability. IEEE Trans. Aut. Control, 49(6):866–874.

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I think you misunderstood the paper. The property "global asymptotic stability" is defined to be the combination of local stability and global attractivity. There is nothing to prove here; it's a definition. Here is the relevant part:

definition

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No, the proof is not correct. Because $\phi(t+T,\tilde \xi) < \epsilon \not \Rightarrow \phi(t,\xi) < \epsilon$

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