[proofplan] The proof is a uniform Taylor expansion on the compact tube around the observer trajectory. We first choose a radius so that every point $\hat{x}(t)+v$ stays inside the prescribed compact subset of $V$, and we record uniform bounds for the second derivatives of $f$ in the state variable and of $h$. Subtracting the observer equation from the plant equation gives the exact error equation; Taylor's formula separates its linear part, $A(t)e(t)-K(t)C(t)e(t)$, from a remainder. The boundedness of $K$ and the uniform second-derivative bounds give the quadratic estimate for that remainder. [/proofplan]

[step:Fix compatible norms and uniform compact sets]Use the Euclidean norms on $\mathbb{R}^n$, $\mathbb{R}^m$, and $\mathbb{R}^p$, and use the induced operator norm on matrices and multilinear maps. Since $K:I\to\mathbb{R}^{n\times p}$ is bounded, define \begin{align*} K_0:=\sup_{t\in I}\|K(t)\|_{\mathrm{op}}<\infty. \end{align*} Let $U_c\subset W$ be a compact set such that $u(I)\subset U_c$. The continuity of $u$ is retained as part of the observer hypotheses, while the proof uses the stated compact containment of its image to obtain uniform bounds. Choose \begin{align*} \rho:=\rho_0. \end{align*} For this choice, \begin{align*} \mathcal{T}_{\rho}:=\{\hat{x}(t)+z:t\in I,\ |z|\le \rho\} \end{align*} is a compact subset of $V$ by hypothesis. Therefore $\mathcal{T}_{\rho}\times U_c$ is a compact subset of $V\times W$. Because $f$ is $C^2$ on $V\times W$, the second derivative of $f$ in the state variables is continuous. Define \begin{align*} L_f:=\sup_{(q,w)\in \mathcal{T}_{\rho}\times U_c}\|D_x^2 f_{(q,w)}\|_{\mathrm{op}}<\infty. \end{align*} Because $h$ is $C^2$ on $V$, define \begin{align*} L_h:=\sup_{q\in \mathcal{T}_{\rho}}\|D^2 h_q\|_{\mathrm{op}}<\infty. \end{align*}[/step]

[guided]We need constants that are uniform in time, not just pointwise Taylor estimates at each fixed $t$. The compactness assumptions are exactly what provide this uniformity. We use the Euclidean norms on $\mathbb{R}^n$, $\mathbb{R}^m$, and $\mathbb{R}^p$. For matrices and bilinear maps we use the corresponding induced operator norm. Since $K:I\to\mathbb{R}^{n\times p}$ is bounded, the number \begin{align*} K_0:=\sup_{t\in I}\|K(t)\|_{\mathrm{op}} \end{align*} is finite. The statement assumes that $u(I)$ is contained in a compact subset of $W$. Fix one such compact set and call it $U_c$. The continuity of $u$ is part of the deterministic observer setup, but the uniform Taylor bounds below use this compact image containment. We take $\rho:=\rho_0$. Then the set \begin{align*} \mathcal{T}_{\rho}:=\{\hat{x}(t)+z:t\in I,\ |z|\le \rho\} \end{align*} is compact and contained in $V$ by hypothesis. Thus, whenever $t\in I$ and $|v|\le \rho$, the point $\hat{x}(t)+v$ lies in $V$. This is the domain condition needed before Taylor expansion can even be stated. Now $\mathcal{T}_{\rho}\times U_c$ is compact in $V\times W$. Since $f$ is $C^2$, its second derivative with respect to the state variable is continuous, so its operator norm attains a finite supremum on this compact set. We define \begin{align*} L_f:=\sup_{(q,w)\in \mathcal{T}_{\rho}\times U_c}\|D_x^2 f_{(q,w)}\|_{\mathrm{op}}. \end{align*} Similarly, since $h$ is $C^2$ and $\mathcal{T}_{\rho}$ is compact, define \begin{align*} L_h:=\sup_{q\in \mathcal{T}_{\rho}}\|D^2 h_q\|_{\mathrm{op}}. \end{align*} Both constants are finite. These two constants are the source of the eventual quadratic bound.[/guided]

[step:Define the nonlinear remainder as a map on the error variable] Define the state Taylor remainder \begin{align*} R_f:B(0,\rho)\times I\to\mathbb{R}^n \end{align*} by \begin{align*} R_f(v,t):=f(\hat{x}(t)+v,u(t))-f(\hat{x}(t),u(t))-A(t)v. \end{align*} Define the output Taylor remainder \begin{align*} R_h:B(0,\rho)\times I\to\mathbb{R}^p \end{align*} by \begin{align*} R_h(v,t):=h(\hat{x}(t)+v)-h(\hat{x}(t))-C(t)v. \end{align*} Finally define \begin{align*} r:B(0,\rho)\times I\to\mathbb{R}^n \end{align*} by \begin{align*} r(v,t):=R_f(v,t)-K(t)R_h(v,t). \end{align*} This definition uses only values of $f$ and $h$ on their domains because $\hat{x}(t)+v\in\mathcal{T}_{\rho}\subset V$ for $|v|<\rho$, and $u(t)\in U_c\subset W$. [/step]

[step:Subtract the observer equation from the plant equation] Let $t\in I$ be such that $|e(t)|<\rho$. Since $e(t)=x(t)-\hat{x}(t)$, we have \begin{align*} x(t)=\hat{x}(t)+e(t). \end{align*} Using the plant and observer equations, and using $y(t)=h(x(t))$, we obtain \begin{align*} \dot{e}(t)=\dot{x}(t)-\dot{\hat{x}}(t). \end{align*} Hence \begin{align*} \dot{e}(t)=f(x(t),u(t))-f(\hat{x}(t),u(t))-K(t)\bigl(h(x(t))-h(\hat{x}(t))\bigr). \end{align*} Substituting $x(t)=\hat{x}(t)+e(t)$ gives \begin{align*} \dot{e}(t)=f(\hat{x}(t)+e(t),u(t))-f(\hat{x}(t),u(t))-K(t)\bigl(h(\hat{x}(t)+e(t))-h(\hat{x}(t))\bigr). \end{align*} By the definitions of $R_f$, $R_h$, and $r$, this becomes \begin{align*} \dot{e}(t)=A(t)e(t)-K(t)C(t)e(t)+r(e(t),t). \end{align*} Therefore \begin{align*} \dot{e}(t)=\bigl(A(t)-K(t)C(t)\bigr)e(t)+r(e(t),t). \end{align*} [/step]

[step:Bound the Taylor remainders uniformly by quadratic terms]Fix $v\in B(0,\rho)$ and $t\in I$. Define \begin{align*} \gamma_{v,t}:[0,1]\to V \end{align*} by \begin{align*} \gamma_{v,t}(s):=\hat{x}(t)+sv. \end{align*} Since $\gamma_{v,t}([0,1])\subset\mathcal{T}_{\rho}$ and $u(t)\in U_c$, Taylor's formula in the one-dimensional variable $s$ applied to the map $s\mapsto f(\gamma_{v,t}(s),u(t))$ gives \begin{align*} |R_f(v,t)|\le \frac{1}{2}L_f|v|^2. \end{align*} Similarly, Taylor's formula applied to $s\mapsto h(\gamma_{v,t}(s))$ gives \begin{align*} |R_h(v,t)|\le \frac{1}{2}L_h|v|^2. \end{align*} Using the definition $r(v,t)=R_f(v,t)-K(t)R_h(v,t)$ and the induced operator norm, \begin{align*} |r(v,t)|\le |R_f(v,t)|+\|K(t)\|_{\mathrm{op}}|R_h(v,t)|. \end{align*} Therefore \begin{align*} |r(v,t)|\le \frac{1}{2}\bigl(L_f+K_0L_h\bigr)|v|^2. \end{align*} Set \begin{align*} M:=\frac{1}{2}\bigl(L_f+K_0L_h\bigr). \end{align*} Then $M\ge 0$, and if $M=0$ we may replace it by $1$ to obtain a strictly positive constant. Thus there exists $M>0$ such that \begin{align*} |r(v,t)|\le M|v|^2 \end{align*} for every $v\in B(0,\rho)$ and every $t\in I$.[/step]

[guided]The purpose of this step is to prove that the nonlinear part is genuinely second order in the error, uniformly in time. Fix $v\in B(0,\rho)$ and $t\in I$. Introduce the line segment map \begin{align*} \gamma_{v,t}:[0,1]\to V \end{align*} by \begin{align*} \gamma_{v,t}(s):=\hat{x}(t)+sv. \end{align*} For every $s\in[0,1]$ we have $|sv|\le |v|<\rho$, so $\gamma_{v,t}(s)\in\mathcal{T}_{\rho}\subset V$. This verifies the domain condition for applying Taylor's formula along the whole segment. For the state equation, consider the function \begin{align*} \phi_{v,t}:[0,1]\to\mathbb{R}^n \end{align*} defined by \begin{align*} \phi_{v,t}(s):=f(\gamma_{v,t}(s),u(t)). \end{align*} The parameter $u(t)$ is fixed while $s$ varies. Since $f$ is $C^2$, the map $\phi_{v,t}$ is $C^2$. Its first derivative is \begin{align*} \phi_{v,t}'(0)=J_x f_{(\hat{x}(t),u(t))}v=A(t)v. \end{align*} Its second derivative is obtained by differentiating again in the state direction $v$: \begin{align*} \phi_{v,t}''(s)=D_x^2 f_{(\gamma_{v,t}(s),u(t))}[v,v]. \end{align*} By the definition of $L_f$ and by the operator norm bound for bilinear maps, \begin{align*} |\phi_{v,t}''(s)|\le L_f|v|^2 \end{align*} for all $s\in[0,1]$. Taylor's formula with quadratic remainder for $\phi_{v,t}$ at $s=0$ gives \begin{align*} R_f(v,t)=\phi_{v,t}(1)-\phi_{v,t}(0)-\phi_{v,t}'(0). \end{align*} The remainder estimate is therefore \begin{align*} |R_f(v,t)|\le \frac{1}{2}L_f|v|^2. \end{align*} The same argument applies to the output map. Define \begin{align*} \psi_{v,t}:[0,1]\to\mathbb{R}^p \end{align*} by \begin{align*} \psi_{v,t}(s):=h(\gamma_{v,t}(s)). \end{align*} Then $\psi_{v,t}$ is $C^2$, \begin{align*} \psi_{v,t}'(0)=Jh_{\hat{x}(t)}v=C(t)v, \end{align*} and \begin{align*} \psi_{v,t}''(s)=D^2h_{\gamma_{v,t}(s)}[v,v]. \end{align*} Since $\gamma_{v,t}(s)\in\mathcal{T}_{\rho}$, the definition of $L_h$ gives \begin{align*} |\psi_{v,t}''(s)|\le L_h|v|^2. \end{align*} Taylor's formula gives \begin{align*} |R_h(v,t)|\le \frac{1}{2}L_h|v|^2. \end{align*} Now combine the two bounds. Since \begin{align*} r(v,t)=R_f(v,t)-K(t)R_h(v,t), \end{align*} the triangle inequality and the induced operator norm give \begin{align*} |r(v,t)|\le |R_f(v,t)|+\|K(t)\|_{\mathrm{op}}|R_h(v,t)|. \end{align*} Using $\|K(t)\|_{\mathrm{op}}\le K_0$, we obtain \begin{align*} |r(v,t)|\le \frac{1}{2}\bigl(L_f+K_0L_h\bigr)|v|^2. \end{align*} Thus, with \begin{align*} M:=\frac{1}{2}\bigl(L_f+K_0L_h\bigr), \end{align*} or with $M:=1$ in the degenerate case where this number is $0$, the desired uniform quadratic estimate holds for every $v\in B(0,\rho)$ and every $t\in I$.[/guided]

[step:Identify the first-order deterministic EKF error model] The preceding steps show that for every $t\in I$ with $|e(t)|<\rho$, \begin{align*} \dot{e}(t)=\bigl(A(t)-K(t)C(t)\bigr)e(t)+r(e(t),t), \end{align*} where $r:B(0,\rho)\times I\to\mathbb{R}^n$ satisfies \begin{align*} |r(v,t)|\le M|v|^2 \end{align*} for all $v\in B(0,\rho)$ and all $t\in I$. Hence the linear part of the local deterministic EKF error dynamics is precisely \begin{align*} \dot{e}(t)=\bigl(A(t)-K(t)C(t)\bigr)e(t), \end{align*} and the discarded nonlinear term is uniformly quadratic for errors $v\in B(0,\rho)$ based at points of the compact tube $\mathcal{T}_{\rho}$. This proves the theorem. [/step]