Proof of Lyapunov Stability Theorem

Question

I am following the proof presented on page 8 here. I think I follow most of the argument, except for one part. The crux of the proof seems to rest on the fact that when we assume a non-zero $c$, $\dot{V}$ has an upper bound $-\gamma$ on the interval $\alpha \leq \|x\| \leq \epsilon$, which the trajectory is localised to. This seems to be linked to the existence of the ball $B_{\alpha} = \{x \in{S} | \|x\| < \alpha\}$, because this is the only extra condition allowed by a non-zero $c$. I'm not sure why this is needed though - if $B_{\alpha}$ is empty, can't we just obtain an upper bound on $0 \leq \|x\| \leq \epsilon$ instead and then apply the same logic to reach the same contradiction?

Answer 1

Ah ok, I've got it. The crucial part I was missing was that $\dot{V}(x) < 0 \ \forall \ x \in{\mathbb{R^n}}$ except at $x = 0$.

$\dot{V}(0) = 0$, because $\dot{V}(0) = \dot{V}(x(t))|_{x=0} = \nabla{V}\cdot{\dot{x}(t)|_{x=0} = \nabla{V}\cdot{f(0)}} = 0$

The upper bound on $\dot{V}$ over the interval $0 \leq \|x\| \leq \epsilon$ would therefore be $0$, so $$V(x(t)) = V(x(0)) + \int_0^t{\dot{V}(x(\tau))d{\tau}}$$ $$\leq V(x(0)) + \int_0^t{0d{\tau}} = V(x(0))$$

$V(x(t)) \leq V(x(0))$ is true, so there is no contradiction for the case where c = 0.