[proofplan] The proof is the standard shifted-horizon MPC argument followed by a direct discrete-time Lyapunov argument. Starting from an optimizer at $x$, we drop the first input, append the terminal controller at the old terminal state, and use terminal invariance of $X_f$ to obtain a feasible candidate at the successor state. Comparing this candidate cost with the old optimal cost gives the Lyapunov decrease for $V_N$. The local stability claim is then proved directly on the compact invariant sublevel set $\Omega_c$: monotonicity of $V_N$ gives invariance and Lyapunov stability, while the lower bound $\ell(x,u)\ge \alpha(|x|)$ forces closed-loop trajectories to converge to the origin. [/proofplan]

[step:Shift an optimal trajectory and append the terminal controller]Fix $x \in \mathcal X_N$. Let \begin{align*} ((x_0^*,\dots,x_N^*),(u_0^*,\dots,u_{N-1}^*)) \in \mathcal A_N(x) \end{align*} be the optimizer selected in the definition of $\kappa_N(x)$. Thus $x_0^*=x$, $u_0^*=\kappa_N(x)$, $x_{j+1}^*=F(x_j^*,u_j^*)$ for every $j \in \{0,\dots,N-1\}$, and $x_N^* \in X_f$. Define the closed-loop successor \begin{align*} x^+ := F(x,\kappa_N(x)). \end{align*} Since $u_0^*=\kappa_N(x)$ and $x_1^*=F(x_0^*,u_0^*)$, we have $x^+=x_1^*$. Define the shifted input sequence \begin{align*} \tilde u := (\tilde u_0,\dots,\tilde u_{N-1}) \in U^N \end{align*} by $\tilde u_j:=u_{j+1}^*$ for $j \in \{0,\dots,N-2\}$ and $\tilde u_{N-1}:=\kappa_f(x_N^*)$. This is an element of $U^N$ because $u_j^* \in U$ for all $j$ and $\kappa_f:X_f\to U$ with $x_N^*\in X_f$. Define the shifted state sequence \begin{align*} \tilde x := (\tilde x_0,\dots,\tilde x_N) \in X^{N+1} \end{align*} by $\tilde x_j:=x_{j+1}^*$ for $j \in \{0,\dots,N-1\}$ and $\tilde x_N:=F(x_N^*,\kappa_f(x_N^*))$. Since $F:X\times U\to X$, all entries of $\tilde x$ lie in $X$. Also $\tilde x_0=x_1^*=x^+$. For $j \in \{0,\dots,N-2\}$, \begin{align*} \tilde x_{j+1}=x_{j+2}^*=F(x_{j+1}^*,u_{j+1}^*)=F(\tilde x_j,\tilde u_j). \end{align*} For the last transition, \begin{align*} \tilde x_N=F(x_N^*,\kappa_f(x_N^*))=F(\tilde x_{N-1},\tilde u_{N-1}). \end{align*} Finally, because $x_N^*\in X_f$ and the terminal controller satisfies $F(y,\kappa_f(y))\in X_f$ for every $y\in X_f$, we have $\tilde x_N\in X_f$. Therefore \begin{align*} ((\tilde x_0,\dots,\tilde x_N),(\tilde u_0,\dots,\tilde u_{N-1})) \in \mathcal A_N(x^+). \end{align*} Thus $x^+ \in \mathcal X_N$.[/step]

[guided]The point of recursive feasibility is to construct, from a feasible optimal plan at $x$, at least one feasible plan at the next state $x^+$. We begin with the selected optimizer \begin{align*} ((x_0^*,\dots,x_N^*),(u_0^*,\dots,u_{N-1}^*)) \in \mathcal A_N(x). \end{align*} By definition of $\mathcal A_N(x)$, this means $x_0^*=x$, $x_{j+1}^*=F(x_j^*,u_j^*)$ for each $j \in \{0,\dots,N-1\}$, every $u_j^*$ lies in $U$, every $x_j^*$ lies in $X$, and the terminal state $x_N^*$ lies in $X_f$. The MPC feedback is the first input of this selected optimizer, so $u_0^*=\kappa_N(x)$. Therefore the closed-loop successor is \begin{align*} x^+=F(x,\kappa_N(x))=F(x_0^*,u_0^*)=x_1^*. \end{align*} This identity is the reason the shifted plan starts at the correct state. We now define the candidate input sequence at $x^+$ by dropping the already-used input $u_0^*$ and appending the terminal feedback at the old terminal state. Explicitly, define \begin{align*} \tilde u := (\tilde u_0,\dots,\tilde u_{N-1}) \in U^N \end{align*} by $\tilde u_j:=u_{j+1}^*$ for $j \in \{0,\dots,N-2\}$ and $\tilde u_{N-1}:=\kappa_f(x_N^*)$. The last input belongs to $U$ because $x_N^*\in X_f$ and $\kappa_f:X_f\to U$. The matching state sequence is \begin{align*} \tilde x := (\tilde x_0,\dots,\tilde x_N) \in X^{N+1}, \end{align*} where $\tilde x_j:=x_{j+1}^*$ for $j \in \{0,\dots,N-1\}$ and $\tilde x_N:=F(x_N^*,\kappa_f(x_N^*))$. Since $F:X\times U\to X$, the final state $\tilde x_N$ belongs to $X$. The initial state condition is also correct: \begin{align*} \tilde x_0=x_1^*=x^+. \end{align*} It remains to verify the dynamics and terminal constraint. For every $j \in \{0,\dots,N-2\}$, the old optimal dynamics give \begin{align*} \tilde x_{j+1}=x_{j+2}^*=F(x_{j+1}^*,u_{j+1}^*)=F(\tilde x_j,\tilde u_j). \end{align*} For the last transition, the definition of the appended state gives \begin{align*} \tilde x_N=F(x_N^*,\kappa_f(x_N^*))=F(\tilde x_{N-1},\tilde u_{N-1}). \end{align*} Finally, terminal invariance says that $F(y,\kappa_f(y))\in X_f$ for every $y\in X_f$. Applying this with $y=x_N^*$ gives $\tilde x_N\in X_f$. Hence the shifted pair belongs to $\mathcal A_N(x^+)$, so $\mathcal A_N(x^+)\neq \varnothing$ and $x^+\in \mathcal X_N$.[/guided]

[step:Compare the shifted candidate cost with the old optimal cost] Since the shifted pair constructed above is feasible for $x^+$, the optimal value at $x^+$ is no larger than its cost: \begin{align*} V_N(x^+) \le \sum_{j=0}^{N-1}\ell(\tilde x_j,\tilde u_j)+V_f(\tilde x_N). \end{align*} Using the definitions of $\tilde x$ and $\tilde u$, the right-hand side is \begin{align*} \sum_{j=1}^{N-1}\ell(x_j^*,u_j^*)+\ell(x_N^*,\kappa_f(x_N^*))+V_f(F(x_N^*,\kappa_f(x_N^*))). \end{align*} The terminal decrease condition applied at $x_N^*\in X_f$ gives \begin{align*} \ell(x_N^*,\kappa_f(x_N^*))+V_f(F(x_N^*,\kappa_f(x_N^*))) \le V_f(x_N^*). \end{align*} Therefore \begin{align*} V_N(x^+) \le \sum_{j=1}^{N-1}\ell(x_j^*,u_j^*)+V_f(x_N^*). \end{align*} Because the selected trajectory is optimal at $x$, \begin{align*} V_N(x)=\sum_{j=0}^{N-1}\ell(x_j^*,u_j^*)+V_f(x_N^*). \end{align*} Subtracting the nonnegative first-stage term in this identity yields \begin{align*} V_N(x^+) \le V_N(x)-\ell(x_0^*,u_0^*). \end{align*} Since $x_0^*=x$ and $u_0^*=\kappa_N(x)$, this is \begin{align*} V_N(x^+)-V_N(x) \le -\ell(x,\kappa_N(x)). \end{align*} [/step]

[step:Iterate recursive feasibility to obtain constraint admissibility] Let $x_0\in \mathcal X_N$, and define the closed-loop sequence inductively by \begin{align*} x_{k+1}:=F(x_k,\kappa_N(x_k)) \end{align*} whenever $x_k\in \mathcal X_N$. The previous step proves that $x_k\in \mathcal X_N$ implies $x_{k+1}\in \mathcal X_N$. By induction over $k\in \mathbb N\cup\{0\}$, the entire trajectory is well-defined and satisfies $x_k\in \mathcal X_N\subset X$ for every $k\ge 0$. Since $\kappa_N:\mathcal X_N\to U$, we also have $\kappa_N(x_k)\in U$ for every $k\ge 0$. This proves constraint admissibility. [/step]

[step:Use the Lyapunov decrease to prove positive invariance of $\Omega_c$] Assume now that $c>0$ is such that \begin{align*} \Omega_c=\{x\in \mathcal X_N:V_N(x)\le c\} \end{align*} is a neighbourhood of $0$ relative to $\mathcal X_N$ and $\Omega_c\subset D$. Let $x\in \Omega_c$, and set $x^+:=F(x,\kappa_N(x))$. Recursive feasibility gives $x^+\in\mathcal X_N$. The Lyapunov decrease and nonnegativity of $\ell$ give \begin{align*} V_N(x^+) \le V_N(x)-\ell(x,\kappa_N(x)) \le V_N(x)\le c. \end{align*} Hence $x^+\in\Omega_c$. Therefore $\Omega_c$ is positively invariant for the closed-loop map. [/step]

[step:Show that the origin is an equilibrium on the invariant sublevel set] Since $0\in\Omega_c$, define \begin{align*} x_1^0:=F(0,\kappa_N(0)). \end{align*} Positive invariance gives $x_1^0\in\Omega_c\subset D$. The Lyapunov decrease at $x=0$ gives \begin{align*} V_N(x_1^0)-V_N(0)\le -\ell(0,\kappa_N(0)). \end{align*} Because $V_N(0)=0$ and $\ell\ge 0$, this implies $V_N(x_1^0)\le 0$. Since $V_N$ takes values in $[0,\infty)$, we get $V_N(x_1^0)=0$. The point $x_1^0$ lies in $D$, and $V_N|_D$ is positive definite, so $x_1^0=0$. Thus \begin{align*} F(0,\kappa_N(0))=0. \end{align*} [/step]

[step:Prove Lyapunov stability relative to $\Omega_c$]We prove stability in the metric inherited from $\mathbb R^n$, with initial conditions restricted to $\Omega_c$. Let $\varepsilon>0$ be given. Define \begin{align*} A_\varepsilon:=\{x\in\Omega_c: |x|\ge \varepsilon\}. \end{align*} The set $\Omega_c$ is compact by the assumed properness, because $\Omega_c\subset D$. The set $A_\varepsilon$ is a closed subset of $\Omega_c$, hence compact. If $A_\varepsilon=\varnothing$, then every trajectory in $\Omega_c$ already satisfies $|x_k|<\varepsilon$ for all $k$ by positive invariance. Assume $A_\varepsilon\neq\varnothing$. Since $V_N|_D$ is continuous and $A_\varepsilon\subset D$, the minimum \begin{align*} \eta_\varepsilon:=\min_{x\in A_\varepsilon}V_N(x) \end{align*} exists. Since $A_\varepsilon\subset D\setminus\{0\}$ and $V_N|_D$ is positive definite, $\eta_\varepsilon>0$. Continuity of $V_N|_D$ at $0$, together with $V_N(0)=0$, gives a number $\delta>0$ such that $x\in D$ and $|x|<\delta$ imply \begin{align*} V_N(x)<\eta_\varepsilon. \end{align*} Shrinking $\delta$ if necessary, we may also ensure that every $x\in\mathcal X_N$ with $|x|<\delta$ lies in $\Omega_c$, because $\Omega_c$ is a neighbourhood of $0$ relative to $\mathcal X_N$. Let $x_0\in\Omega_c$ satisfy $|x_0|<\delta$, and let $(x_k)_{k\ge 0}$ be the closed-loop trajectory. Positive invariance gives $x_k\in\Omega_c\subset D$ for every $k\ge 0$. The Lyapunov decrease gives $V_N(x_{k+1})\le V_N(x_k)$ for every $k$, so \begin{align*} V_N(x_k)\le V_N(x_0)<\eta_\varepsilon \end{align*} for every $k\ge 0$. Hence $x_k\notin A_\varepsilon$ for every $k$, and therefore $|x_k|<\varepsilon$ for every $k\ge 0$. This proves Lyapunov stability of the origin relative to $\Omega_c$.[/step]

[guided]We need a stability estimate that does not use continuity of the feedback $\kappa_N$. The only object we know is continuous is $V_N$ on $D$, so we use sublevel sets of $V_N$. Fix $\varepsilon>0$. The set of points in the invariant region that are at least $\varepsilon$ away from the origin is \begin{align*} A_\varepsilon:=\{x\in\Omega_c: |x|\ge \varepsilon\}. \end{align*} Because $\Omega_c=\{x\in\mathcal X_N:V_N(x)\le c\}$ is contained in $D$, the properness assumption says that $\Omega_c$ is compact. The condition $|x|\ge\varepsilon$ defines a closed condition in $\mathbb R^n$, so $A_\varepsilon$ is a closed subset of the compact set $\Omega_c$. Hence $A_\varepsilon$ is compact. If $A_\varepsilon$ is empty, then no point of $\Omega_c$ has norm at least $\varepsilon$, and positive invariance immediately gives $|x_k|<\varepsilon$ for every trajectory starting in $\Omega_c$. Suppose now that $A_\varepsilon\neq\varnothing$. Since $V_N|_D$ is continuous and $A_\varepsilon\subset D$, $V_N$ attains a minimum on $A_\varepsilon$: \begin{align*} \eta_\varepsilon:=\min_{x\in A_\varepsilon}V_N(x). \end{align*} Every point of $A_\varepsilon$ is nonzero, and $A_\varepsilon\subset D$. Positive definiteness of $V_N|_D$ therefore gives $V_N(x)>0$ on $A_\varepsilon$, so the compact minimum satisfies $\eta_\varepsilon>0$. The role of continuity at the origin is to make the initial value of $V_N$ smaller than this separating level. Since $V_N(0)=0$ and $V_N|_D$ is continuous at $0$, there is $\delta>0$ such that $x\in D$ and $|x|<\delta$ imply \begin{align*} V_N(x)<\eta_\varepsilon. \end{align*} Because $\Omega_c$ is a neighbourhood of $0$ relative to $\mathcal X_N$, we shrink $\delta$ if necessary so that every $x\in\mathcal X_N$ with $|x|<\delta$ belongs to $\Omega_c$. Now take any initial state $x_0\in\Omega_c$ with $|x_0|<\delta$. Positive invariance gives $x_k\in\Omega_c\subset D$ for all $k\ge 0$. The Lyapunov decrease implies monotonicity: \begin{align*} V_N(x_{k+1})\le V_N(x_k) \end{align*} for every $k\ge 0$. Hence, by induction, \begin{align*} V_N(x_k)\le V_N(x_0)<\eta_\varepsilon \end{align*} for every $k\ge 0$. If some $x_k$ had $|x_k|\ge\varepsilon$, then $x_k\in A_\varepsilon$, which would force $V_N(x_k)\ge\eta_\varepsilon$, contradicting the displayed inequality. Therefore $|x_k|<\varepsilon$ for every $k\ge 0$. This is Lyapunov stability relative to $\Omega_c$.[/guided]

[step:Use the stage-cost lower bound to prove convergence to the origin] Let $x_0\in\Omega_c$, and let $(x_k)_{k\ge 0}$ be the closed-loop trajectory. By positive invariance, $x_k\in\Omega_c\subset\mathcal X_N$ for every $k\ge 0$. Applying the Lyapunov decrease at $x_k$ gives \begin{align*} V_N(x_{k+1})-V_N(x_k)\le -\ell(x_k,\kappa_N(x_k)). \end{align*} Using $\ell(x,u)\ge\alpha(|x|)$ with $u=\kappa_N(x_k)$, we obtain \begin{align*} V_N(x_{k+1})-V_N(x_k)\le -\alpha(|x_k|). \end{align*} Summing this inequality from $k=0$ to $M$ gives \begin{align*} \sum_{k=0}^{M}\alpha(|x_k|)\le V_N(x_0)-V_N(x_{M+1})\le V_N(x_0) \end{align*} for every $M\ge 0$, because $V_N\ge 0$. Therefore the nonnegative series \begin{align*} \sum_{k=0}^{\infty}\alpha(|x_k|) \end{align*} converges, and in particular $\alpha(|x_k|)\to 0$ as $k\to\infty$. Since $\alpha$ is class-$\mathcal K$, it is strictly increasing, continuous, and satisfies $\alpha(0)=0$. Hence $\alpha(r)>0$ for every $r>0$. If $|x_k|$ did not converge to $0$, there would exist $\varepsilon>0$ and a subsequence $(x_{k_i})_{i\ge 1}$ such that $|x_{k_i}|\ge\varepsilon$ for every $i$. Then \begin{align*} \alpha(|x_{k_i}|)\ge \alpha(\varepsilon)>0 \end{align*} for every $i$, contradicting $\alpha(|x_k|)\to 0$. Thus $|x_k|\to 0$, so $x_k\to 0$ in $\mathbb R^n$. Combining equilibrium of the origin, Lyapunov stability relative to $\Omega_c$, and convergence of every trajectory starting in $\Omega_c$ proves that the origin is asymptotically stable for the closed-loop system restricted to $\Omega_c$. [/step]