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Being $V(x(t))$ a Lyapunov function, why is $\lim_{t \to \infty} V(x(t))=0 \Rightarrow \lim_{t \to \infty} x(t)=0$ ???

I don't know why is true that implication. I don't now from where to start. The only thing I think is that $\lim_{t\to \infty}V(x(t))=0 \Leftrightarrow \lim_{t\to \infty}||x(t)-0||=0$, true?. But I don't know if it serves for something... Could anyone help me?

User160
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    A Lyapunov function $V$ is positive definite (in the dynamical system context), that is, $V(x)>0$ for all $x\neq0$. The unique exception is $V(0)=0$. From this point, a proof with $\epsilon-\delta$ should be straightforward. – Julian Feb 04 '23 at 17:47
  • I understand what you say about $V$ being positive definite. But what do I have to do with $\epsilon - \delta$ ? Could you write it with more details, please? Thanks!! @Julian – User160 Feb 04 '23 at 17:54

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Suppose that $x(t)$ does not converge to $0$ when $t\to\infty$. Therefore, there must exists some $\epsilon>0$ for which $\epsilon<V(x(t))<V(x(0))$ for all $t>0$, given that $$\frac{d}{dt}V(x(t))<0 \text{ for all }x(t)\neq0.$$ Now, recall that the level sets of a Lyapunov function are bounded, or equivalently, $V(x)\to\infty$ when $||x||\to\infty.$ With this, the set $$\mathcal{C}=\{y\ |\ \epsilon<V(y)\leq V(x(0))\}$$ is closed and bounded. As $0\not \in \mathcal{C}$, $$-\gamma=\sup_{y\in\mathcal{C}} \dot{V}(y)<0,$$ and $$V(x(t))=V(x(0))+\int_0^t \frac{d}{dt}V(x(t))dt\leq V(x(0)) - \gamma t.$$ From the last equation we get a contradiction, because taking $t$ sufficiently large would imply that $V(x(t))$ is negative.

Julian
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