 Solution

The First Lyapunov theorem states that for a locally Lipschitz autonomous system $\dot{x}=f(x)$ with equilibrium zero, if a $C^1$ function $V$ is positive definite near zero and $\dot{V}=\nabla V\cdot f\le 0$, then the equilibrium is Lyapunov stable. To prove it, choose a small sphere $|x|=\epsilon$ inside that neighborhood. Compactness and positivity give $m={\min }_{|x|=\epsilon }V(x)>0$. Continuity at zero gives a smaller ball $|x|<\delta$ where $V<m$. Since $V$ cannot increase along a trajectory, a trajectory starting there cannot first cross the sphere where $V\ge m$. It stays within the sphere for all future time, with continuation justified by confinement in a compact subset of the vector field's domain. This proves stability for every sufficiently small $\epsilon$.

The LaSalle invariance principle states that on a compact positively invariant set on which $\dot{V}\le 0$, every trajectory approaches the largest invariant subset of $\{\dot{V}=0\}$. If that subset is just the equilibrium, the trapping and stability argument proves asymptotic stability.

For example take $\dot{x}=y$, $\dot{y}=-x-y$ and $V=(x^2+y^2)/2$. Then $\dot{V}=-y^2$, which is not strictly negative on the punctured plane. Nevertheless its sublevel discs are compact and invariant, and a trajectory confined to $y=0$ must have $\dot{y}=-x=0$, hence $x=0$. LaSalle gives asymptotic stability despite the non-strict orbital derivative of this particular energy function.

The printed phrase about nonexistence of a strict Lyapunov function must be understood as a limitation of the chosen candidate. In this example a strict function actually exists:More generally the converse Lyapunov theorem supplies local strict functions for asymptotically stable sufficiently smooth systems. LaSalle avoids having to find such a function; it does not show that none exists.