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  "body": "\n## Proof\n\nWe first show that a solution which stays in a compact set exists for all $t \\ge 0$. Then we use that $V$ does not increase along solutions. Here $\\norm{\\cdot}$ is the Euclidean norm, and $\\gamma_{x_0} : (\\alpha, \\omega) \\to U$ is the [[maximal solution|019VSXSFMQ]] with $\\gamma_{x_0}(0) = x_0$, as in the definition of the [[flow|02903SSA48]].\n\nSolutions in a compact set exist for all $t \\ge 0$.\n\n1. Let $C \\subseteq U$ be compact, and let $x_0 \\in U$ with $\\gamma_{x_0}(t) \\in C$ for all $t \\in [0, \\omega)$. Write $\\gamma = \\gamma_{x_0}$. We show $\\omega = \\infty$. Suppose instead that $\\omega \u003c \\infty$.\n2. A locally Lipschitz map is continuous. By the [[extreme value theorem|0V4DET59NW]], there is $M \\ge 0$ with $\\norm{f(x)} \\le M$ for $x \\in C$.\n3. For $0 \\le s \u003c t \u003c \\omega$, the derivative of $\\gamma$ on $[s, t]$ is $f(\\gamma(r))$, of norm at most $M$. By the [[mean value inequality|02CYB184RC]], $\\norm{\\gamma(t) - \\gamma(s)} \\le M (t - s)$.\n4. So for every sequence $t_m \\to \\omega$ in $[0, \\omega)$, the sequence $\\gamma(t_m)$ is Cauchy. Since [[$\\R^n$ is complete|0ABP2F2ZP7]], it converges. Merging two such sequences shows that the limit $p$ does not depend on the sequence.\n5. The set $C$ is [[closed|0MYFJVMN7M]], so $p \\in C \\subseteq U$. Since $f$ is locally Lipschitz, there are $r \u003e 0$ and $K \\ge 0$ with $\\bar B(p, 2r) \\subseteq U$. Moreover $f$ is $K$-[[Lipschitz|055VVSN00V]] on this closed ball.\n6. The ball $\\bar B(p, 2r)$ is compact by the [[Heine–Borel theorem|0PHT6NBYPB]]. So $\\norm{f} \\le L$ on it for some $L \u003e 0$. Let $h = r / L$.\n7. By step 4, choose $t_1 \\in [0, \\omega)$ with $\\omega - t_1 \u003c h$ and $\\norm{\\gamma(t_1) - p} \\le r$. Let $y_1 = \\gamma(t_1)$. The closed ball $B_1 = \\bar B(y_1, r)$ lies in $\\bar B(p, 2r)$.\n8. Let $g(t, x) = f(x)$ for $x \\in B_1$ and $g(t, x) = 0$ otherwise, for $t \\in [t_1 - h, t_1 + h]$. Then $g$ satisfies the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]] at $(t_1, y_1)$ with constants $r, L, K$, since $L h = r$.\n9. By the [[Picard–Lindelöf theorem|0P6KRD5B7A]], there is $\\beta : [t_1 - h, t_1 + h] \\to B_1$ with $\\beta(t_1) = y_1$ and $\\beta' = g(t, \\beta) = f(\\beta)$. So $\\beta$ solves $x' = f(x)$, with values in $U$.\n10. On the open interval $(\\max(\\alpha, t_1 - h), \\omega)$, the curves $\\gamma$ and $\\beta$ are solutions that agree at $t_1$. By [[uniqueness of solutions|025PKW8AMR]], they agree on this interval.\n11. Let $\\tilde\\gamma = \\gamma$ on $(\\alpha, \\omega)$ and $\\tilde\\gamma = \\beta$ on $[\\omega, t_1 + h)$. By step 10, $\\tilde\\gamma = \\beta$ on $(\\max(\\alpha, t_1 - h), t_1 + h)$. So $\\tilde\\gamma$ is a solution on $(\\alpha, t_1 + h)$ that extends $\\gamma$.\n12. Since $t_1 + h \u003e \\omega$, this contradicts the maximality of $\\gamma$. Hence $\\omega = \\infty$.\n\n$V$ does not increase along solutions.\n\n13. Let $\\gamma$ be a solution on an open interval $I$, and let $[c, d] \\subseteq I$ with $\\gamma(t) \\in W$ for $t \\in [c, d]$. Let $v(t) = V(\\gamma(t))$.\n14. By the [[chain rule|0A8TJ0VHC7]], at each $t \\in [c, d]$ we have $v'(t) = \\nabla V(\\gamma(t)) \\cdot f(\\gamma(t)) = \\dot V(\\gamma(t)) \\le 0$, as $V$ is a [[Lyapunov function|04P7B5XC22]].\n15. So $-v$ has a non-negative derivative on $[c, d]$, and $-v$ is [[increasing|026BJBQNGJ]]. Thus $v$ is non-increasing on $[c, d]$.\n\nPart (1).\n\n16. Let $\\varepsilon \u003e 0$. Choose $r \\in (0, \\varepsilon]$ with $\\bar B(x^*, r) \\subseteq W$. The sphere $S = \\set{x : \\norm{x - x^*} = r}$ is compact by Heine–Borel and non-empty.\n17. By the extreme value theorem, $V$ has a minimum $m$ on $S$. Since $V \u003e 0$ on $W \\setminus \\set{x^*}$, we have $m \u003e 0$.\n18. Since $V$ is continuous and $V(x^*) = 0$, there is $\\delta \\in (0, r)$ with $V(x) \u003c m$ whenever $\\norm{x - x^*} \u003c \\delta$.\n19. Let $x_0 \\in U$ with $\\norm{x_0 - x^*} \u003c \\delta$, and let $\\gamma = \\gamma_{x_0}$. We claim $\\norm{\\gamma(t) - x^*} \u003c r$ for all $t \\in [0, \\omega)$.\n20. If not, let $T$ be the infimum of the times $t \\in [0, \\omega)$ with $\\norm{\\gamma(t) - x^*} \\ge r$. By continuity, $T \u003e 0$ and $\\norm{\\gamma(T) - x^*} = r$. Then $\\gamma(t) \\in \\bar B(x^*, r) \\subseteq W$ for $t \\in [0, T]$.\n21. By step 15, $V(\\gamma(T)) \\le V(x_0) \u003c m$. But $\\gamma(T) \\in S$, so $V(\\gamma(T)) \\ge m$. This contradiction proves the claim.\n22. So $\\gamma(t)$ lies in the compact set $\\bar B(x^*, r) \\subseteq U$ for $t \\in [0, \\omega)$. By steps 1 to 12, $\\omega = \\infty$.\n23. Hence $\\varphi_t(x_0)$ exists and $\\norm{\\varphi_t(x_0) - x^*} \u003c r \\le \\varepsilon$ for all $t \\ge 0$. So $x^*$ is [[Lyapunov stable|092S6VBECQ]].\n\nA convergence lemma for strict $V$.\n\n24. Let $V$ be strict, let $C \\subseteq W$ be compact, and let $\\gamma$ be a solution on an interval containing $[0, \\infty)$ with $\\gamma(t) \\in C$ for $t \\ge 0$. We show $\\gamma(t) \\to x^*$.\n25. Let $v = V \\circ \\gamma$. By step 15, $v$ is non-increasing on $[0, \\infty)$. Also $v \\ge 0$. So $v(t)$ tends to a limit $c \\ge 0$, and $v(t) \\ge c$ for all $t \\ge 0$.\n26. Suppose $c \u003e 0$. By continuity of $V$ at $x^*$, there is $\\rho \u003e 0$ with $V(x) \u003c c$ for $x \\in W$ and $\\norm{x - x^*} \u003c \\rho$.\n27. So $\\gamma(t) \\in A$ for $t \\ge 0$, where $A = \\set{x \\in C : \\norm{x - x^*} \\ge \\rho}$. The set $A$ is a [[closed subset of a compact set|066HW6D9A7]], hence compact, and it is non-empty.\n28. The function $\\dot V = \\nabla V \\cdot f$ is continuous on $W$, and it is negative on $A$ because $x^* \\notin A$. By the extreme value theorem, $\\dot V \\le -\\mu$ on $A$ for some $\\mu \u003e 0$.\n29. By step 14, $v' \\le -\\mu$ on $[0, \\infty)$. By the [[mean value theorem|0BARQY52N5]], $v(t) \\le v(0) - \\mu t$, which is negative for large $t$. This contradicts $v \\ge 0$. So $c = 0$.\n30. Now let $\\rho \u003e 0$ and $A_\\rho = \\set{x \\in C : \\norm{x - x^*} \\ge \\rho}$. If $A_\\rho$ is non-empty, it is compact and $V$ has a minimum $m_\\rho \u003e 0$ on it.\n31. Since $v(t) \\to 0$, there is $T$ with $v(t) \u003c m_\\rho$ for $t \\ge T$. Then $\\gamma(t) \\notin A_\\rho$, so $\\norm{\\gamma(t) - x^*} \u003c \\rho$ for $t \\ge T$. Hence $\\gamma(t) \\to x^*$.\n\nPart (2).\n\n32. Choose $r_0 \u003e 0$ with $\\bar B(x^*, r_0) \\subseteq W$. Steps 16 to 23 with $\\varepsilon = r_0$ give $\\eta \u003e 0$ as follows. If $\\norm{x_0 - x^*} \u003c \\eta$, then $\\varphi_t(x_0)$ exists for $t \\ge 0$ and lies in $\\bar B(x^*, r_0)$.\n33. By the lemma with $C = \\bar B(x^*, r_0)$, $\\varphi_t(x_0) \\to x^*$. With part (1), $x^*$ is [[asymptotically stable|05Y1JAM3Y5]].\n\nPart (3).\n\n34. Let $x_0 \\in \\R^n$ and $\\Omega = \\set{x : V(x) \\le V(x_0)}$. Since $V(x) \\to \\infty$ as $\\norm{x} \\to \\infty$, there is $R$ with $V(x) \u003e V(x_0)$ for $\\norm{x} \u003e R$. So $\\Omega$ is bounded.\n35. It is closed because $V$ is continuous, so it is compact by Heine–Borel.\n36. Let $\\gamma = \\gamma_{x_0}$. Since $W = \\R^n$, step 15 on $[0, t]$ gives $V(\\gamma(t)) \\le V(x_0)$, so $\\gamma(t) \\in \\Omega$ for $t \\in [0, \\omega)$. By steps 1 to 12, $\\omega = \\infty$.\n37. By the lemma with $C = \\Omega$, $\\varphi_t(x_0) \\to x^*$. This holds for each $x_0 \\in \\R^n$, and $x^*$ is Lyapunov stable by part (1). So $x^*$ is globally asymptotically stable.\n\n## Idea\n\nAlong a solution, $V$ can only decrease, so a solution that starts where $V$ is small can never reach a sphere on which $V$ is larger. Being trapped in a compact set, the solution cannot blow up and exists for all time. If $V$ strictly decreases away from $x^*$, it must decrease to $0$, and $V$ near $0$ forces the solution near $x^*$.\n",
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