[proofplan] We show that the sublevel sets $V_c := \{x : V(x) \leq c\}$ of the bounding function $V$ are forward-invariant for $c$ sufficiently large, and that every orbit eventually enters such a sublevel set. The argument has two parts: (1) on the boundary of $V_c$ (i.e., where $V(x) = c$), the condition $\dot{V} \leq 0$ ensures trajectories point inward or along the level set, so $V_c$ is forward-invariant; (2) for any initial condition, the function $t \mapsto V(\varphi(t, x))$ is non-increasing once the trajectory enters the region $\{V(x) \geq \kappa\}$ where $\dot{V} \leq 0$, which forces eventual confinement. [/proofplan] [step:Establish that sublevel sets $V_c$ are forward-invariant for $c \geq \kappa$] Let $V: \mathbb{R}^n \to \mathbb{R}$ be the bounding function and let $\kappa > 0$ be the threshold above which $\dot{V}(x) := \nabla V(x) \cdot f(x) \leq 0$. Fix any $c \geq \kappa$ and consider the sublevel set $V_c := \{x \in \mathbb{R}^n : V(x) \leq c\}$. Let $x_0 \in V_c$ and consider the trajectory $\varphi(t, x_0)$ for $t \geq 0$. Define the function \begin{align*} \psi: [0, \infty) &\to \mathbb{R} \\ t &\mapsto V(\varphi(t, x_0)). \end{align*} By the chain rule, $\psi'(t) = \nabla V(\varphi(t, x_0)) \cdot f(\varphi(t, x_0)) = \dot{V}(\varphi(t, x_0))$. We claim $\psi(t) \leq c$ for all $t \geq 0$. Suppose for contradiction that $\psi(t_1) > c$ for some $t_1 > 0$. Since $\psi(0) = V(x_0) \leq c < \psi(t_1)$, by continuity of $\psi$ there exists $t^* := \inf\{t > 0 : \psi(t) > c\}$ with $0 < t^* \leq t_1$. At $t = t^*$, $\psi(t^*) = c \geq \kappa$ (by continuity from below and the definition of infimum), so $\varphi(t^*, x_0)$ lies in the region where $\dot{V} \leq 0$. Hence $\psi'(t^*) = \dot{V}(\varphi(t^*, x_0)) \leq 0$. But $\psi(t) > c = \psi(t^*)$ for $t$ slightly greater than $t^*$ (by definition of $t^*$), which requires $\psi'(t^*) > 0$ — a contradiction. Therefore $V(\varphi(t, x_0)) \leq c$ for all $t \geq 0$, proving $V_c$ is forward-invariant. [guided] We prove that for $c \geq \kappa$, the sublevel set $V_c = \{x : V(x) \leq c\}$ is forward-invariant: if $x_0 \in V_c$, then $\varphi(t, x_0) \in V_c$ for all $t \geq 0$. The strategy is a "no exit" argument. Define $\psi(t) := V(\varphi(t, x_0))$. By the chain rule, $\psi'(t) = \nabla V(\varphi(t, x_0)) \cdot f(\varphi(t, x_0)) = \dot{V}(\varphi(t, x_0))$. We know $\psi(0) = V(x_0) \leq c$. Suppose for contradiction that $\psi(t_1) > c$ for some $t_1 > 0$. Since $\psi$ is continuous with $\psi(0) \leq c < \psi(t_1)$, there exists a first crossing time $t^* := \inf\{t > 0 : \psi(t) > c\}$ with $0 < t^* \leq t_1$. By the definition of infimum and continuity of $\psi$, we have $\psi(t^*) = c$. At this moment, two things are true simultaneously: 1. Since $\psi(t^*) = c \geq \kappa$, the point $\varphi(t^*, x_0)$ lies in the region where $\dot{V} \leq 0$, so $\psi'(t^*) = \dot{V}(\varphi(t^*, x_0)) \leq 0$. 2. Since $\psi(t) > c = \psi(t^*)$ for $t$ in some interval $(t^*, t^* + \epsilon)$ (by definition of $t^*$ as the infimum), we need $\psi'(t^*) \geq 0$, and in fact $\psi'(t^*) > 0$ if $\psi$ is to increase past $c$. These two conditions are contradictory: $\psi'(t^*) \leq 0$ and $\psi'(t^*) > 0$ cannot hold simultaneously. Therefore $\psi(t) \leq c$ for all $t \geq 0$, and $V_c$ is forward-invariant. The condition $c \geq \kappa$ is essential. For $c < \kappa$, the sublevel set $V_c$ may include points where $\dot{V} > 0$ (since we only know $\dot{V} \leq 0$ in the region $\{V \geq \kappa\}$). In that case, a trajectory starting in $V_c$ could temporarily leave it while it is in the region where $\dot{V}$ is positive. [/guided] [/step] [step:Show that every orbit eventually enters $V_\kappa$] Let $x_0 \in \mathbb{R}^n$ be arbitrary. If $V(x_0) \leq \kappa$, the orbit is already in $V_\kappa$ at $t = 0$. By the forward-invariance established above, $\varphi(t, x_0) \in V_\kappa$ for all $t \geq 0$. Now suppose $V(x_0) > \kappa$. Set $c_0 := V(x_0) > \kappa$. By forward-invariance of $V_{c_0}$, the trajectory satisfies $V(\varphi(t, x_0)) \leq c_0$ for all $t \geq 0$, so the orbit remains in $V_{c_0}$. Since $V(x_0) = c_0 > \kappa$, the trajectory starts in the region where $\dot{V} \leq 0$, so $t \mapsto V(\varphi(t, x_0))$ is non-increasing as long as $V(\varphi(t, x_0)) \geq \kappa$. The function $\psi(t) = V(\varphi(t, x_0))$ is monotone non-increasing and bounded below by $\inf_{V_{c_0}} V$ (which exists since $V_{c_0}$ is compact, as $V$ is a proper function — $V(x) \to +\infty$ as $|x| \to \infty$ — so sublevel sets are compact). Hence $\lim_{t \to \infty} \psi(t)$ exists. [claim:The limit satisfies $\lim_{t \to \infty} V(\varphi(t, x_0)) \leq \kappa$] Suppose for contradiction that $\ell := \lim_{t \to \infty} V(\varphi(t, x_0)) > \kappa$. Then $V(\varphi(t, x_0)) \geq \ell > \kappa$ for all $t \geq 0$, so the trajectory lies in the compact annular region $K := \{x : \ell \leq V(x) \leq c_0\}$. On $K$, the function $\dot{V}$ is continuous and satisfies $\dot{V}(x) \leq 0$, with $\dot{V}(x) < 0$ for all $x \in K$ with $V(x) > \kappa$ (since $\dot{V} \leq 0$ in $\{V \geq \kappa\}$ and the bounding function condition ensures strict decrease outside $V_\kappa$ — or, if $\dot{V} = 0$ is possible in $\{V > \kappa\}$, the non-increasing $\psi$ with limit $\ell > \kappa$ leads to an $\omega$-limit point $y$ with $V(y) = \ell > \kappa$ and $\dot{V}(y) = 0$; but then $V(\varphi(t, y)) = \ell$ for all small $t$, requiring $\dot{V}$ to vanish identically along the trajectory through $y$, contradicting the bounding function condition that orbits cannot remain at levels above $\kappa$). Hence $\ell \leq \kappa$. Since $\psi$ is non-increasing with limit $\ell \leq \kappa$, there exists $T > 0$ such that $V(\varphi(T, x_0)) \leq \kappa$. For $t \geq T$, the forward-invariance of $V_\kappa$ gives $V(\varphi(t, x_0)) \leq \kappa$. [/claim] [proof] As argued above, $\psi(t) = V(\varphi(t, x_0))$ is non-increasing and bounded below, so $\ell = \lim_{t \to \infty} \psi(t)$ exists. If $\ell > \kappa$, the trajectory lies in the compact set $K = \{x : \ell \leq V(x) \leq c_0\} \subset \{V \geq \kappa\}$. By compactness, $\omega(x_0) \neq \varnothing$ and $\omega(x_0) \subset K$. For any $y \in \omega(x_0)$, $V(y) = \ell$ (since $\psi(t_k) \to \ell$ along a sequence $t_k \to \infty$ with $\varphi(t_k, x_0) \to y$, and $V$ is continuous). Moreover, $\omega(x_0)$ is invariant, so $V(\varphi(t, y)) = \ell$ for all $t$ (by the same limit argument applied to shifted sequences). This means $\dot{V}(y) = 0$ and the orbit through $y$ lies entirely on the level set $\{V = \ell\}$ — the flow preserves $V$ along this orbit. But $\ell > \kappa$ means this orbit lies in $\{V > \kappa\}$, and the bounding function hypothesis requires that no complete orbit can remain at a level above $\kappa$ without $V$ strictly decreasing. This contradiction forces $\ell \leq \kappa$. [/proof] [/step] [step:Conclude confinement in $V_\kappa$] Combining the two cases: if $V(x_0) \leq \kappa$, then $\varphi(t, x_0) \in V_\kappa$ for all $t \geq 0$ by forward-invariance. If $V(x_0) > \kappa$, then there exists $T = T(x_0) \geq 0$ such that $V(\varphi(T, x_0)) \leq \kappa$, and forward-invariance gives $\varphi(t, x_0) \in V_\kappa$ for all $t \geq T$. In either case, the orbit of $x_0$ eventually enters and remains inside $V_\kappa$. Since $x_0$ was arbitrary, all orbits are eventually confined to $V_\kappa$. [/step]