Lyapunov's stability theorem proof

Question

Lyapunov's Stability Theorem states that if $E \subset \mathbb{R}^n$ is open, $f\in C(E)$ and there exists $V \in C^1(E)$ such that $V(y) > 0$ for all $y \neq y_0$ where $y_0$ is an equilibrium point os the non linear system $f(y) = y'$, then: $$\begin{align} 1)\; & \text{If}\; V'(y) \leq 0\; \forall y\in E, \; \text{$y_0$ is an stable point of the system}\\ 2)\; & \text{If}\; V'(y) < 0\; \forall y\in E\setminus\{y_0\}, \; \text{$y_0$ is an asymptotically stable point of the system}\\ 3)\; & \text{If}\; V'(y) > 0\; \forall y\in E\setminus \{y_0\}, \; \text{$y_0$ is an unstable point of the system} \end{align}$$

Now the proof I know of this theorem goes by proving $1)$ and $2)$ first and then deduce $3)$ from $b)$ by letting $t = -t$, and I'm not sure I understood correctly how this works, so if anyone could help me with that, I'd be grateful. Thank you in advance!

Answer 1

Yes, you can deduce $3)$ from $2)$.

However, you can give a much simpler proof of $3)$: Note that at a point $y$ where $\dot V(y)>0$, the function $V$ grows in an neighborhood of that point along trajctories (the notation $V'(y)$ is bad since we are not taking the derivative of $V$). Since $V(y_0)=0$ (surely you have this condition), no trajectory can converge to $y_0$.

With the same proof you can show that $y_0$ is unstable provided that there is a sequence $y_n\to y_0$ with $\dot V(y_n)>0$ for each $n$. No need to require that $\dot V(y)>0$ in a whole neighborhood outside $y_0$.