[proofplan] The proof first extracts the immediate consequence of the control Lyapunov condition: at any nonzero point where $b$ vanishes, the uncontrolled derivative $a$ must already be negative. For points where $b\ne0$, direct substitution of Sontag's feedback gives the exact identity $a+bk=-\sqrt{a^2+b^4}$. The small-control property then controls the quotient in the feedback formula near the origin by comparing it with arbitrarily small admissible controls. Finally, a direct Lyapunov argument on compact sublevel sets gives relative stability and attractivity for the stated class of forward closed-loop solutions. [/proofplan]

[step:Show that vanishing of $b$ forces $a$ to be negative away from the origin] Fix $x \in D \setminus \{0\}$ and suppose $b(x)=0$. By the control Lyapunov hypothesis, there exists $u \in \mathbb{R}$ such that \begin{align*} a(x)+b(x)u<0. \end{align*} Since $b(x)=0$, this inequality reduces to \begin{align*} a(x)<0. \end{align*} Thus every nonzero point with $b(x)=0$ already has strictly negative derivative along the drift field $f$. [/step]

[step:Substitute the universal feedback where $b$ is nonzero] Let $x \in D$ satisfy $b(x)\ne0$. By the definition of $k$, \begin{align*} b(x)k(x)=-a(x)-\sqrt{a(x)^2+b(x)^4}. \end{align*} Adding $a(x)$ to both sides gives \begin{align*} a(x)+b(x)k(x)=-\sqrt{a(x)^2+b(x)^4}. \end{align*} Since \begin{align*} \nabla V(x)\cdot\bigl(f(x)+g(x)k(x)\bigr)=\nabla V(x)\cdot f(x)+k(x)\nabla V(x)\cdot g(x)=a(x)+b(x)k(x), \end{align*} we obtain \begin{align*} \nabla V(x)\cdot\bigl(f(x)+g(x)k(x)\bigr)=-\sqrt{a(x)^2+b(x)^4}. \end{align*} If $x\ne0$, this quantity is strictly negative. Indeed, if it were zero, then $a(x)=0$ and $b(x)=0$, contradicting the assumption $b(x)\ne0$. [/step]

[step:Verify strict decrease at points where $b$ vanishes]Let $x \in D \setminus \{0\}$ satisfy $b(x)=0$. By definition, $k(x)=0$, so \begin{align*} \nabla V(x)\cdot\bigl(f(x)+g(x)k(x)\bigr)=\nabla V(x)\cdot f(x)=a(x). \end{align*} The first step gives $a(x)<0$. Therefore \begin{align*} \nabla V(x)\cdot\bigl(f(x)+g(x)k(x)\bigr)<0 \end{align*} at every nonzero point with $b(x)=0$.[/step]

[guided]At a point where $b(x)=0$, the control vector field is invisible to the derivative of $V$, because \begin{align*} \nabla V(x)\cdot g(x)=b(x)=0. \end{align*} Thus no scalar control $u$ can change the value of $a(x)+b(x)u$ at that point. The control Lyapunov hypothesis says that for this same nonzero $x$, there is some $u \in \mathbb{R}$ with \begin{align*} a(x)+b(x)u<0. \end{align*} Substituting $b(x)=0$ gives \begin{align*} a(x)<0. \end{align*} The feedback definition also gives $k(x)=0$ when $b(x)=0$. Therefore the closed-loop derivative of $V$ at this point is \begin{align*} \nabla V(x)\cdot\bigl(f(x)+g(x)k(x)\bigr)=\nabla V(x)\cdot f(x)+k(x)\nabla V(x)\cdot g(x)=a(x)+0=a(x). \end{align*} Since $a(x)<0$, the closed-loop derivative is strictly negative. This verifies the only case not covered by the quotient formula, namely the points where the denominator in Sontag's expression vanishes.[/guided]

[step:Use the small-control property to prove continuity of $k$ at the origin]Since $V \in C^1(D;\mathbb{R})$, $V(0)=0$, and $V(x)>0$ for $x \in D\setminus\{0\}$, the point $0$ is a local minimizer of $V$. Hence \begin{align*} \nabla V(0)=0. \end{align*} Because $f$, $g$, and $\nabla V$ are continuous on $D$, and $f(0)=0$, it follows that \begin{align*} a(x)\to0 \end{align*} and \begin{align*} b(x)\to0 \end{align*} as $x\to0$. Let $\eta>0$ be arbitrary. Choose $\varepsilon>0$ so small that \begin{align*} \varepsilon+\sqrt{2}\varepsilon<\eta. \end{align*} By the small-control property, there exists $\delta_1>0$ such that, whenever $x \in D$ and $0<|x|<\delta_1$, there exists $u_x \in \mathbb{R}$ with $|u_x|<\varepsilon$ and \begin{align*} a(x)+b(x)u_x<0. \end{align*} Since $b(x)\to0$, there exists $\delta_2>0$ such that $|b(x)|<\varepsilon$ whenever $x \in D$ and $|x|<\delta_2$. Let $\delta:=\min\{\delta_1,\delta_2\}$. If $x \in D$, $0<|x|<\delta$, and $b(x)=0$, then $k(x)=0$, so $|k(x)|<\eta$. If instead $b(x)\ne0$, define \begin{align*} z(x):=\frac{a(x)}{|b(x)|}. \end{align*} From $a(x)+b(x)u_x<0$ and $|u_x|<\varepsilon$, we get \begin{align*} a(x)<-b(x)u_x\le |b(x)|\,|u_x|<\varepsilon |b(x)|. \end{align*} Thus $z(x)<\varepsilon$. Also, \begin{align*} |k(x)|=\frac{a(x)+\sqrt{a(x)^2+b(x)^4}}{|b(x)|}=z(x)+\sqrt{z(x)^2+b(x)^2}. \end{align*} The function $z \mapsto z+\sqrt{z^2+b(x)^2}$ is increasing on $\mathbb{R}$, so using $z(x)<\varepsilon$ and $|b(x)|<\varepsilon$ gives \begin{align*} |k(x)|\le \varepsilon+\sqrt{\varepsilon^2+\varepsilon^2}=\varepsilon+\sqrt{2}\varepsilon<\eta. \end{align*} Since $k(0)=0$, this proves \begin{align*} \lim_{x\to0}k(x)=0. \end{align*} Therefore $k$ is continuous at $0$.[/step]

[guided]The only delicate point in the theorem is the behaviour of the quotient defining $k$ when $x$ approaches the origin and $b(x)$ may approach $0$. The small-control property is exactly the hypothesis that prevents this quotient from becoming large. First note that $0$ is a local minimizer of $V$. Indeed, $V(0)=0$ and $V(x)>0$ for every nonzero $x \in D$. Since $V$ is $C^1$, the first-order necessary condition for a local minimum gives \begin{align*} \nabla V(0)=0. \end{align*} The maps $f: D \to \mathbb{R}^n$, $g: D \to \mathbb{R}^n$, and $\nabla V: D \to \mathbb{R}^n$ are continuous, and $f(0)=0$. Therefore \begin{align*} a(x)=\nabla V(x)\cdot f(x)\to0 \end{align*} and \begin{align*} b(x)=\nabla V(x)\cdot g(x)\to0 \end{align*} as $x\to0$. Fix an error tolerance $\eta>0$. We will prove that $|k(x)|<\eta$ for all sufficiently small nonzero $x$. Choose $\varepsilon>0$ so small that \begin{align*} \varepsilon+\sqrt{2}\varepsilon<\eta. \end{align*} The small-control property gives a radius $\delta_1>0$ such that, for every $x \in D$ with $0<|x|<\delta_1$, there exists a control $u_x \in \mathbb{R}$ satisfying $|u_x|<\varepsilon$ and \begin{align*} a(x)+b(x)u_x<0. \end{align*} Because $b(x)\to0$, there is another radius $\delta_2>0$ such that $|b(x)|<\varepsilon$ whenever $|x|<\delta_2$. Set $\delta:=\min\{\delta_1,\delta_2\}$. Now fix $x \in D$ with $0<|x|<\delta$. If $b(x)=0$, then the definition of the feedback gives $k(x)=0$, so $|k(x)|<\eta$. Suppose instead that $b(x)\ne0$. Define the normalized scalar \begin{align*} z(x):=\frac{a(x)}{|b(x)|}. \end{align*} The small admissible control $u_x$ gives \begin{align*} a(x)<-b(x)u_x\le |b(x)|\,|u_x|<\varepsilon |b(x)|. \end{align*} Dividing by $|b(x)|>0$ gives \begin{align*} z(x)<\varepsilon. \end{align*} We rewrite the feedback magnitude in terms of $z(x)$. Since $a(x)+\sqrt{a(x)^2+b(x)^4}\ge0$, \begin{align*} |k(x)|=\frac{a(x)+\sqrt{a(x)^2+b(x)^4}}{|b(x)|}=z(x)+\sqrt{z(x)^2+b(x)^2}. \end{align*} For the fixed value of $b(x)$, the scalar function $z \mapsto z+\sqrt{z^2+b(x)^2}$ is increasing because its derivative is \begin{align*} 1+\frac{z}{\sqrt{z^2+b(x)^2}}>0. \end{align*} Thus $z(x)<\varepsilon$ implies \begin{align*} |k(x)|\le \varepsilon+\sqrt{\varepsilon^2+b(x)^2}. \end{align*} Since $|b(x)|<\varepsilon$, we obtain \begin{align*} |k(x)|\le \varepsilon+\sqrt{\varepsilon^2+\varepsilon^2}=\varepsilon+\sqrt{2}\varepsilon<\eta. \end{align*} This proves that $k(x)\to0$ as $x\to0$. Since $k(0)=0$ by definition, $k$ is continuous at the origin.[/guided]

[step:Derive relative stability from sublevel sets] Fix $c>0$ such that \begin{align*} \Omega_c:=\{x \in D: V(x)\le c\} \end{align*} is compactly contained in $D$. Let \begin{align*} F: D \to \mathbb{R}^n \end{align*} denote the closed-loop vector field defined by \begin{align*} F(x):=f(x)+g(x)k(x). \end{align*} Let \begin{align*} \gamma:[0,\infty)\to\Omega_c \end{align*} be a forward classical solution of $\dot{x}=F(x)$. Since $\gamma$ is classical and $V \in C^1(D;\mathbb{R})$, the chain rule gives, for every $t\ge0$, \begin{align*} \frac{d}{dt}V(\gamma(t))=\nabla V(\gamma(t))\cdot F(\gamma(t)). \end{align*} If $\gamma(t)\ne0$, the strict decrease already proved gives \begin{align*} \frac{d}{dt}V(\gamma(t))<0. \end{align*} If $\gamma(t)=0$, then $f(0)=0$, $b(0)=0$, and $k(0)=0$ give $F(0)=0$, so \begin{align*} \frac{d}{dt}V(\gamma(t))=\nabla V(0)\cdot F(0)=0. \end{align*} Thus the derivative of $t\mapsto V(\gamma(t))$ is everywhere nonpositive on $[0,\infty)$. By the one-variable [mean value theorem](/theorems/186) on each compact interval $[s,t]\subset[0,\infty)$, $V(\gamma(t))\le V(\gamma(s))$ whenever $0\le s\le t$. This establishes global monotonicity without invoking uniqueness of the closed-loop ODE. Let $U$ be any neighbourhood of $0$ relative to $\Omega_c$. Since $V$ is continuous, positive definite on $D$, and $\Omega_c \setminus U$ is compact, the number \begin{align*} m_U:=\min_{x\in \Omega_c\setminus U}V(x) \end{align*} exists and satisfies $m_U>0$ whenever $\Omega_c\setminus U$ is nonempty. Choose $\rho_U>0$ such that \begin{align*} \{x\in \Omega_c: |x|<\rho_U\}\subset \{x\in \Omega_c: V(x)<m_U\}. \end{align*} If $\gamma(0)\in\Omega_c$ and $|\gamma(0)|<\rho_U$, then the monotonicity of $V(\gamma(t))$ gives \begin{align*} V(\gamma(t))\le V(\gamma(0))<m_U \end{align*} for all $t\ge0$. Therefore $\gamma(t)\notin \Omega_c\setminus U$ for all $t\ge0$, so $\gamma(t)\in U$ for all $t\ge0$. This is Lyapunov stability relative to $\Omega_c$. [/step]

[step:Prove convergence to the origin for trajectories remaining in $\Omega_c$] Let $\gamma:[0,\infty)\to\Omega_c$ be a forward classical solution of the closed-loop system. The function $t\mapsto V(\gamma(t))$ is nonincreasing and bounded below by $0$, hence has a limit \begin{align*} \ell:=\lim_{t\to\infty}V(\gamma(t))\ge0. \end{align*} We prove $\ell=0$. Suppose instead that $\ell>0$. Define the compact annulus \begin{align*} K_\ell:=\{x\in\Omega_c: V(x)\ge \ell/2\}. \end{align*} For all sufficiently large $t$, $\gamma(t)\in K_\ell$. Define $W:K_\ell\to(0,\infty)$ by \begin{align*} W(x):=-\nabla V(x)\cdot F(x). \end{align*} The strict negativity already proved gives $W(x)>0$ on $K_\ell$. On points of $K_\ell$ with $b(x)\ne0$, \begin{align*} W(x)=\sqrt{a(x)^2+b(x)^4}. \end{align*} On points of $K_\ell$ with $b(x)=0$, \begin{align*} W(x)=-a(x). \end{align*} These formulas agree continuously across points where $b=0$ in $K_\ell$, because such points have $a<0$, and therefore $\sqrt{a^2+b^4}\to |a|=-a$. Hence $W$ is continuous on the compact set $K_\ell$. Therefore \begin{align*} \mu_\ell:=\min_{x\in K_\ell}W(x)>0. \end{align*} Choose $T_0\ge 0$ such that $\gamma(t)\in K_\ell$ for every $t\ge T_0$. For all $t\ge T_0$, \begin{align*} \frac{d}{dt}V(\gamma(t))=-W(\gamma(t))\le -\mu_\ell. \end{align*} Integrating over any interval $[T_0,t]$ with $t\ge T_0$ gives \begin{align*} V(\gamma(t))\le V(\gamma(T_0))-\mu_\ell(t-T_0). \end{align*} For large enough $t$, the right-hand side is negative, contradicting $V\ge0$. Hence $\ell=0$. Since $V$ is positive definite and continuous on the compact set $\Omega_c$, convergence $V(\gamma(t))\to0$ implies $\gamma(t)\to0$ in the relative topology of $\Omega_c$. Together with relative Lyapunov stability, this proves asymptotic stability of the origin relative to $\Omega_c$ for every forward classical solution that exists on $[0,\infty)$ and remains in $\Omega_c$. [/step]