= 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 $\dot V=\nabla V\cdot f\le0$, then the equilibrium is <Lyapunov stable>. To prove it, choose a small sphere $|x|=\varepsilon$ inside that neighborhood. Compactness and positivity give $m=\min_{|x|=\varepsilon}V(x)>0$. 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 $\varepsilon$.
The <LaSalle invariance principle> states that on a compact positively invariant set on which $\dot V\le0$, 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.
\b[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:
$$
W=\tfrac34x^2+\tfrac12xy+\tfrac12y^2,\qquad \dot W=-\tfrac12(x^2+y^2).
$$
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.
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