[proofplan] We prove the result through the linearization of the closed-loop system in the coordinates $(x,e)$. Exactness implies that the error vector field vanishes whenever $e=0$, so the derivative of the error equation with respect to $x$ at the origin is zero. The full Jacobian is therefore block upper triangular. The nominal local exponential stability gives a Hurwitz $x$-block, while the Lyapunov inequality for $W$ gives local exponential stability of the frozen error subsystem and hence a Hurwitz $e$-block. The [Lyapunov indirect method](/theorems/6852) for autonomous $C^1$ systems then yields local exponential, and therefore local asymptotic, stability of the full closed-loop equilibrium. [/proofplan] [step:Define the closed-loop vector field and record the exact error invariance] Define the nominal vector field $\phi: X_0 \to \mathbb{R}^n$ by \begin{align*} \phi(x)=f(x,k(x)). \end{align*} Define the closed-loop state component $\psi: X_0 \times E_0 \to \mathbb{R}^n$ by \begin{align*} \psi(x,e)=f(x,k(x+e)). \end{align*} Define the error vector field $g: X_0 \times E_0 \to \mathbb{R}^n$ by \begin{align*} g(x,e)=q(x+e,h(x),k(x+e))-f(x,k(x+e)). \end{align*} Then the closed-loop system is \begin{align*} \dot{x}=\psi(x,e), \qquad \dot{e}=g(x,e). \end{align*} Since $0\in X_0\subset X_{\mathrm{ex}}$, $h(0)=0$, and $k(0)=0$, the exactness assumption gives \begin{align*} g(0,0)=q(0,h(0),k(0))-f(0,k(0))=q(0,0,0)-f(0,0)=0. \end{align*} Also $\psi(0,0)=f(0,k(0))=f(0,0)=0$. Hence $(0,0)$ is an equilibrium. For every $x\in X_0$, the exactness assumption gives \begin{align*} g(x,0)=q(x,h(x),k(x))-f(x,k(x))=0. \end{align*} Thus the map $x\mapsto g(x,0)$ is identically zero on $X_0$. [/step] [step:Compute the block upper triangular linearization] Because $f$, $h$, $k$, and $q$ are $C^1$, the maps $\psi$ and $g$ are $C^1$ on $X_0\times E_0$. Let $J$ denote the Jacobian matrix of the map $(x,e)\mapsto (\psi(x,e),g(x,e))$ at $(0,0)$ with respect to the standard coordinates on $\mathbb{R}^n\times \mathbb{R}^n$. Since $x\mapsto g(x,0)$ is identically zero on a neighbourhood of $0$, its derivative at $0$ is the zero [linear map](/page/Linear%20Map). Therefore \begin{align*} D_x g_{(0,0)}=0. \end{align*} Define the linear maps $A: \mathbb{R}^n\to \mathbb{R}^n$, $B: \mathbb{R}^n\to \mathbb{R}^n$, and $C: \mathbb{R}^n\to \mathbb{R}^n$ by $A=D_x\psi_{(0,0)}$, $B=D_e\psi_{(0,0)}$, and $C=D_e g_{(0,0)}$. With respect to the decomposition $\mathbb{R}^n\times\mathbb{R}^n$, the linearization is \begin{align*} J(\xi,\eta)=(A\xi+B\eta,C\eta). \end{align*} Thus $J$ is block upper triangular, with diagonal blocks $A$ and $C$. Moreover, since $\psi(x,0)=f(x,k(x))=\phi(x)$ for $x\in X_0$, differentiating with respect to $x$ at $0$ gives \begin{align*} D_x\psi_{(0,0)}=D\phi_0. \end{align*} [guided] The central point is that the error equation has no linear dependence on $x$ at $e=0$. We first isolate the three vector fields involved. The nominal closed-loop vector field is $\phi: X_0\to \mathbb{R}^n$, defined by $\phi(x)=f(x,k(x))$. The actual state component in error coordinates is $\psi: X_0\times E_0\to \mathbb{R}^n$, defined by $\psi(x,e)=f(x,k(x+e))$. The error component is $g: X_0\times E_0\to \mathbb{R}^n$, defined by \begin{align*} g(x,e)=q(x+e,h(x),k(x+e))-f(x,k(x+e)). \end{align*} The exactness assumption says precisely that, when the observer state equals the plant state, the observer reproduces the plant dynamics. In error coordinates, equality of observer and plant state means $e=0$. Therefore, for every $x\in X_0\subset X_{\mathrm{ex}}$, \begin{align*} g(x,0)=q(x,h(x),k(x))-f(x,k(x))=0. \end{align*} This identity holds on an open neighbourhood of $0$, not merely at the point $0$. Since $g$ is $C^1$, the derivative of the map $x\mapsto g(x,0)$ at $0$ exists, and because that map is identically zero, its derivative is the zero linear map: \begin{align*} D_x g_{(0,0)}=0. \end{align*} Define $A=D_x\psi_{(0,0)}$, $B=D_e\psi_{(0,0)}$, and $C=D_e g_{(0,0)}$ as linear maps from $\mathbb{R}^n$ to $\mathbb{R}^n$. Thus the Jacobian matrix of the full vector field $(\psi,g)$ at $(0,0)$ acts on $(\xi,\eta)\in\mathbb{R}^n\times\mathbb{R}^n$ by \begin{align*} J(\xi,\eta)=(A\xi+B\eta,C\eta). \end{align*} This is exactly the block upper triangular separation structure: the linearized error equation does not receive a first-order forcing term from $x$. Finally, when $e=0$, the state component is exactly the nominal feedback vector field: \begin{align*} \psi(x,0)=f(x,k(x))=\phi(x). \end{align*} Differentiating this identity with respect to $x$ at $0$ gives \begin{align*} D_x\psi_{(0,0)}=D\phi_0. \end{align*} [/guided] [/step] [step:Show that the nominal block is Hurwitz] The map $\phi:X_0\to\mathbb{R}^n$ is $C^1$ because $f$ and $k$ are $C^1$, and $X_0$ is a neighbourhood of $0$. Also $\phi(0)=f(0,k(0))=f(0,0)=0$, so $0$ is an equilibrium of the autonomous system $\dot{x}=\phi(x)$. By hypothesis, this equilibrium is locally exponentially stable. The standard converse linearization criterion for $C^1$ autonomous systems therefore applies: by differentiability of the local flow with respect to initial data at the equilibrium, local exponential stability of the nonlinear equilibrium implies exponential stability of the linear variational flow, so the Jacobian matrix of the vector field at that equilibrium is Hurwitz. Applying this result to $\phi$ gives that $D\phi_0=D_x\psi_{(0,0)}$ is Hurwitz. [/step] [step:Use the error Lyapunov bound to show that the error block is Hurwitz] Consider the frozen error subsystem at $x=0$: \begin{align*} \dot{e}=g(0,e). \end{align*} By hypothesis, $W: E_0\to \mathbb{R}$ is a $C^1$ function satisfying, for every $e\in E_0$, \begin{align*} c_1|e|^2 \le W(e) \le c_2|e|^2. \end{align*} Along any solution of the frozen error subsystem that remains in $E_0$, the chain rule gives \begin{align*} \frac{d}{dt}W(e(t))=\nabla W(e(t))\cdot g(0,e(t)). \end{align*} The assumed frozen Lyapunov decay inequality gives \begin{align*} \frac{d}{dt}W(e(t))\le -c_4|e(t)|^2. \end{align*} Using $W(e)\le c_2|e|^2$, equivalently $|e|^2\ge c_2^{-1}W(e)$, we obtain \begin{align*} \frac{d}{dt}W(e(t))\le -\frac{c_4}{c_2}W(e(t)). \end{align*} Choose a radius $r>0$ such that the closed Euclidean ball $\overline{B}(0,r)$ is contained in $E_0$, and define $\alpha=c_1r^2$. If $W(e_0)<\alpha$, then the sublevel set $\{e\in E_0: W(e)\le W(e_0)\}$ is contained in $B(0,r)$, because $c_1|e|^2\le W(e)\le W(e_0)<c_1r^2$ implies $|e|<r$. The differential inequality above implies that $W(e(t))$ is nonincreasing on every time interval on which the solution remains in $E_0$; therefore a solution starting with $W(e_0)<\alpha$ cannot exit this sublevel set through $\partial B(0,r)$. Hence the Lyapunov estimates remain valid for all forward time for which the local solution is continued, and the comparison inequality gives local exponential decay. Together with the quadratic bounds on $W$, this proves local exponential stability of $e=0$ for the frozen $C^1$ system $\dot{e}=g(0,e)$. Applying again the standard converse implication that local exponential stability of a $C^1$ autonomous equilibrium implies a Hurwitz linearization, by differentiability of the local flow with respect to initial data and exponential stability of the resulting linear variational flow, the matrix $D_e g_{(0,0)}$ is Hurwitz. [guided] We now extract stability of the linearized error block from the frozen Lyapunov hypothesis. The statement gives a $C^1$ Lyapunov function $W: E_0\to \mathbb{R}$ for the frozen system at $x=0$. Its quadratic bounds say that, for every $e\in E_0$, \begin{align*} c_1|e|^2 \le W(e) \le c_2|e|^2. \end{align*} Now let $e:(a,b)\to E_0$ be a solution of the frozen error equation \begin{align*} \dot{e}(t)=g(0,e(t)). \end{align*} The chain rule applies because $W$ and $g(0,\cdot)$ are $C^1$. Therefore \begin{align*} \frac{d}{dt}W(e(t))=\nabla W(e(t))\cdot g(0,e(t)). \end{align*} The frozen Lyapunov decay assumption in the theorem gives \begin{align*} \nabla W(e(t))\cdot g(0,e(t))\le -c_4|e(t)|^2. \end{align*} Combining these two formulas, \begin{align*} \frac{d}{dt}W(e(t))\le -c_4|e(t)|^2. \end{align*} To convert this into exponential decay of the Lyapunov function itself, use the upper quadratic bound $W(e)\le c_2|e|^2$. This implies $|e|^2\ge c_2^{-1}W(e)$, and hence \begin{align*} \frac{d}{dt}W(e(t))\le -\frac{c_4}{c_2}W(e(t)). \end{align*} The comparison inequality then gives exponential decay of $W(e(t))$ as long as the solution remains in the local neighbourhood. Since $W$ is also bounded below by $c_1|e|^2$, this exponential decay of $W$ gives exponential decay of $|e(t)|$. To justify that this decay is genuinely local in the ODE sense, choose $r>0$ with $\overline{B}(0,r)\subset E_0$ and set $\alpha=c_1r^2$. If $W(e(0))<\alpha$, then every point in the sublevel set $\{e\in E_0:W(e)\le W(e(0))\}$ lies in $B(0,r)$, because $c_1|e|^2\le W(e)<c_1r^2$. The inequality $\frac{d}{dt}W(e(t))\le 0$ on $E_0$ keeps the trajectory in this sublevel set for as long as the local solution exists. Therefore the Lyapunov estimates remain valid for all forward time in the local solution, and the comparison inequality yields local exponential stability of $e=0$ for the frozen $C^1$ system $\dot{e}=g(0,e)$. For a $C^1$ autonomous system, local exponential stability of an equilibrium forces the Jacobian matrix at that equilibrium to be Hurwitz: differentiability of the local flow with respect to initial data identifies the derivative of the nonlinear flow at the equilibrium with the linear variational flow, and the nonlinear exponential estimate passes to that derivative. Applying this standard result to the vector field $e\mapsto g(0,e)$ gives that its derivative at $0$, namely $D_e g_{(0,0)}$, is Hurwitz. This is the only conclusion about the error subsystem needed for the block triangular argument. [/guided] [/step] [step:Apply the indirect method to the block triangular Jacobian] The eigenvalues of a block upper triangular linear map are exactly the eigenvalues of its diagonal blocks. Since $D_x\psi_{(0,0)}=D\phi_0$ is Hurwitz and $D_e g_{(0,0)}$ is Hurwitz, every eigenvalue of the linear map $J$ has strictly negative real part. Hence $J$ is Hurwitz. The closed-loop vector field $(x,e)\mapsto(\psi(x,e),g(x,e))$ is $C^1$ on a neighbourhood of $(0,0)$ and has equilibrium $(0,0)$. By the Lyapunov indirect method for autonomous $C^1$ systems, a Hurwitz linearization implies that the equilibrium is locally exponentially stable. Local exponential stability implies local asymptotic stability. Therefore $(x,e)=(0,0)$ is a locally asymptotically stable equilibrium of the observer-based output-feedback closed loop. [guided] From the previous steps, the derivative of the full closed-loop vector field at $(0,0)$ acts by \begin{align*} J(\xi,\eta)=(A\xi+B\eta,C\eta), \end{align*} where $A=D_x\psi_{(0,0)}=D\phi_0$ and $C=D_e g_{(0,0)}$. This is a block upper triangular linear map with diagonal blocks $A$ and $C$. For such a linear map, the characteristic polynomial factors as the product of the characteristic polynomials of the diagonal blocks, so its eigenvalues are precisely the eigenvalues of $A$ together with the eigenvalues of $C$. The nominal block $A$ is Hurwitz by the nominal local exponential stability argument, and the error block $C$ is Hurwitz by the frozen-error Lyapunov argument. Hence every eigenvalue of $J$ has strictly negative real part, so $J$ is Hurwitz. We may now apply the Lyapunov indirect method for autonomous $C^1$ systems. Its hypotheses are satisfied: the closed-loop vector field $(x,e)\mapsto(\psi(x,e),g(x,e))$ is $C^1$ on a neighbourhood of $(0,0)$, the point $(0,0)$ is an equilibrium, and its Jacobian matrix $J$ is Hurwitz. The theorem therefore gives local exponential stability of $(0,0)$. Since local exponential stability includes convergence to the equilibrium for all sufficiently small initial data and stability under sufficiently small perturbations of the initial condition, it implies local asymptotic stability. This proves the stated local asymptotic stability of the observer-based output-feedback closed loop. [/guided] [/step]