[proofplan] We enclose the reference solution graph over the compact interval $J$ in a closed tube contained in $I \times E$. Compactness gives a uniform bound and a uniform state-Lipschitz constant for $f$ on that tube. For nearby initial data, the existence of a unique maximal solution theorem [citetheorem:8375] supplies maximal nearby solutions, and Gronwall's inequality controls their distance from the reference solution as long as they remain in the tube. The control is strictly smaller than the tube radius, so an exit or loss of existence before the end of $J$ is impossible; applying the argument on both sides of $t_0$ gives existence on all of $J$ and [uniform convergence](/page/Uniform%20Convergence). [/proofplan]

[step:Choose a compact tube around the reference solution] Write $J=[a,b]$, where $a,b \in \mathbb{R}$ and $a \leq t_0 \leq b$. Throughout the proof, let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Since $x:J_x \to E$ is continuous and $J$ is compact, the set \begin{align*} X_J:=x(J)=\{x(t):t \in J\} \end{align*} is a compact subset of $E$. Define the Euclidean distance-to-$X_J$ map \begin{align*} d_{X_J}:\mathbb{R}^n &\to [0,\infty) \end{align*} by $d_{X_J}(y):=\inf\{|y-z|:z \in X_J\}$. We write $\operatorname{dist}(y,X_J):=d_{X_J}(y)$. Because $E \subset \mathbb{R}^n$ is open, there exists $\rho>0$ such that \begin{align*} \{y \in \mathbb{R}^n:\operatorname{dist}(y,X_J)\leq \rho\}\subset E. \end{align*} Define the closed tube \begin{align*} \mathcal{K}:=\{(t,y)\in J \times E: |y-x(t)|\leq \rho\}. \end{align*} Then $\mathcal{K}$ is compact in $I \times E$. By continuity of $f:I \times E \to \mathbb{R}^n$ and compactness of $\mathcal{K}$, define \begin{align*} M:=\sup_{(t,y)\in \mathcal{K}} |f(t,y)|<\infty. \end{align*} We next derive a single state-Lipschitz constant on $\mathcal{K}$. For $w\in\mathbb{R}^n$ and $r>0$, write $B(w,r):=\{u\in\mathbb{R}^n:|u-w|<r\}$ for the open Euclidean ball. For each $(s,w)\in\mathcal{K}$, the local Lipschitz hypothesis gives a relative time neighbourhood $U_s\subset I$, an open state neighbourhood $V_w\subset E$, and a constant $L_{s,w}\geq 0$ such that $|f(t,u)-f(t,v)|\leq L_{s,w}|u-v|$ for $t\in U_s$ and $u,v\in V_w$. Shrinking $V_w$ if necessary, choose $r_{s,w}>0$ such that $B(w,2r_{s,w})\subset V_w$. The sets $U_s\times B(w,r_{s,w})$ cover $\mathcal{K}$, so compactness gives finitely many centres $(s_i,w_i)$, radii $r_i>0$, relative time neighbourhoods $U_i\subset I$, and constants $L_i\geq0$, for $1\leq i\leq N$, whose smaller products $U_i\times B(w_i,r_i)$ cover $\mathcal{K}$ and whose larger products $U_i\times B(w_i,2r_i)$ are Lipschitz neighbourhoods. Let $\lambda>0$ be a Lebesgue number for this finite cover of the compact [metric space](/page/Metric%20Space) $\mathcal{K}$, and define \begin{align*} L_0:=\max_{1\leq i\leq N} L_i. \end{align*} If $(t,y),(t,z)\in\mathcal{K}$ have the same time coordinate $t$, the segment $\gamma:[0,1]\to E$ defined by $\gamma(\theta):=(1-\theta)y+\theta z$ lies in the fibre ball $\{u\in\mathbb{R}^n:|u-x(t)|\leq\rho\}\subset E$, so $(t,\gamma(\theta))\in\mathcal{K}$ for every $\theta\in[0,1]$. Choose an integer $m\geq1$ with $|y-z|/m<\lambda/2$, and set $y_j:=\gamma(j/m)$ for $0\leq j\leq m$. For each $j$, the two points $(t,y_{j-1})$ and $(t,y_j)$ have distance less than $\lambda$ in $I\times E$, so they lie in one larger Lipschitz neighbourhood from the finite cover. Hence \begin{align*} |f(t,y_j)-f(t,y_{j-1})|\leq L_0|y_j-y_{j-1}|. \end{align*} Summing over $j$ gives \begin{align*} |f(t,y)-f(t,z)|\leq L_0\sum_{j=1}^{m}|y_j-y_{j-1}|=L_0|y-z|. \end{align*} Set $L:=\max\{L_0,1\}$. Then $L>0$ and, for all $(t,y),(t,z)\in\mathcal{K}$ with the same time coordinate $t$, \begin{align*} |f(t,y)-f(t,z)|\leq L|y-z|. \end{align*} [/step]

[step:Construct maximal nearby solutions from the nearby initial data] By the existence of a unique maximal solution theorem [citetheorem:8375] applied to the initial data $(t_0,x_{0,k}) \in I \times E$, for every $k \in \mathbb{N}$ there exists a unique maximal solution \begin{align*} y_k:J_k \to E \end{align*} of \begin{align*} \dot{y}_k(t)=f(t,y_k(t)), \qquad y_k(t_0)=x_{0,k}, \end{align*} where $J_k \subset I$ is an interval containing $t_0$ and open in the relative topology of $I$. Let \begin{align*} T:=\max\{b-t_0,t_0-a,0\}. \end{align*} Since $x_{0,k}\to x_0$, choose $k_0 \in \mathbb{N}$ such that for every $k\geq k_0$, \begin{align*} e^{LT}|x_{0,k}-x_0|<\frac{\rho}{2}. \end{align*} Fix $k\geq k_0$ for the rest of the proof. [/step]

[step:Control the nearby solution as long as it remains inside the tube]Let $c \in [t_0,b]$ be such that $[t_0,c]\subset J_k$ and \begin{align*} |y_k(s)-x(s)|\leq \rho \end{align*} for every $s\in [t_0,c]$. Then $(s,y_k(s))\in \mathcal{K}$ and $(s,x(s))\in \mathcal{K}$ for every $s\in [t_0,c]$. Since both $x$ and $y_k$ satisfy their integral equations on $[t_0,c]$, for every $t\in [t_0,c]$, \begin{align*} y_k(t)-x(t)=x_{0,k}-x_0+\int_{t_0}^{t}\bigl(f(s,y_k(s))-f(s,x(s))\bigr)\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms and using the Lipschitz bound on $\mathcal{K}$ gives \begin{align*} |y_k(t)-x(t)|\leq |x_{0,k}-x_0|+L\int_{t_0}^{t}|y_k(s)-x(s)|\,d\mathcal{L}^1(s). \end{align*} Applying the integral form of Gronwall's inequality yields \begin{align*} |y_k(t)-x(t)|\leq e^{L(t-t_0)}|x_{0,k}-x_0|\leq e^{LT}|x_{0,k}-x_0|<\frac{\rho}{2}. \end{align*}[/step]

[guided]The purpose of this step is to obtain an estimate that is valid before any possible exit from the tube. Fix $c\in[t_0,b]$ such that $[t_0,c]\subset J_k$ and such that $y_k$ remains within distance $\rho$ of $x$ on this interval. This assumption places both points $(s,y_k(s))$ and $(s,x(s))$ in the compact tube $\mathcal{K}$ for every $s\in[t_0,c]$, so the uniform Lipschitz constant $L$ is available. The integral form of the differential equations is \begin{align*} y_k(t)=x_{0,k}+\int_{t_0}^{t}f(s,y_k(s))\,d\mathcal{L}^1(s) \end{align*} and \begin{align*} x(t)=x_0+\int_{t_0}^{t}f(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Subtracting these two identities gives, for every $t\in[t_0,c]$, \begin{align*} y_k(t)-x(t)=x_{0,k}-x_0+\int_{t_0}^{t}\bigl(f(s,y_k(s))-f(s,x(s))\bigr)\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms and using the triangle inequality for the Bochner integral in $\mathbb{R}^n$ gives \begin{align*} |y_k(t)-x(t)|\leq |x_{0,k}-x_0|+\int_{t_0}^{t}|f(s,y_k(s))-f(s,x(s))|\,d\mathcal{L}^1(s). \end{align*} Because both arguments lie in $\mathcal{K}$ and have the same time coordinate $s$, the state-Lipschitz estimate gives \begin{align*} |f(s,y_k(s))-f(s,x(s))|\leq L|y_k(s)-x(s)|. \end{align*} Therefore \begin{align*} |y_k(t)-x(t)|\leq |x_{0,k}-x_0|+L\int_{t_0}^{t}|y_k(s)-x(s)|\,d\mathcal{L}^1(s). \end{align*} We now apply the integral form of Gronwall's inequality. Its hypotheses are satisfied because the function \begin{align*} u:[t_0,c]\to[0,\infty), \qquad u(t):=|y_k(t)-x(t)| \end{align*} is continuous, the coefficient $L$ is nonnegative and constant, and the additive term $|x_{0,k}-x_0|$ is nonnegative. Hence \begin{align*} |y_k(t)-x(t)|\leq e^{L(t-t_0)}|x_{0,k}-x_0|. \end{align*} Since $t-t_0\leq T$ for $t\in[t_0,c]$ and $k\geq k_0$, the choice of $k_0$ gives \begin{align*} |y_k(t)-x(t)|\leq e^{LT}|x_{0,k}-x_0|<\frac{\rho}{2}. \end{align*} The strict improvement from $\rho$ to $\rho/2$ is the key point: once the solution is known to stay inside the tube, the estimate proves it actually stays a positive distance away from the tube boundary.[/guided]

[step:Rule out a first exit before the right endpoint of $J$] We prove that $[t_0,b]\subset J_k$ and that the estimate \begin{align*} |y_k(t)-x(t)|\leq e^{L(t-t_0)}|x_{0,k}-x_0| \end{align*} holds for every $t\in[t_0,b]$. Let $R_k$ be the set of all $c\in[t_0,b]$ such that $[t_0,c]\subset J_k$ and \begin{align*} |y_k(s)-x(s)|<\rho \end{align*} for every $s\in[t_0,c]$. The set $R_k$ is nonempty because $t_0\in R_k$ and $|x_{0,k}-x_0|<\rho/2$. Let \begin{align*} \tau:=\sup R_k. \end{align*} If $\tau<b$, then for every $c\in R_k$ the previous step gives \begin{align*} |y_k(t)-x(t)|<\frac{\rho}{2} \end{align*} for every $t\in[t_0,c]$. We first show that $y_k(t)$ has a limit as $t\uparrow \tau$. On every interval $[t_0,c]$ with $c\in R_k$, one has $(s,y_k(s))\in\mathcal{K}$, hence $|\dot{y}_k(s)|\leq M$ for all $s\in[t_0,c]$. Therefore, for $r,s\in[t_0,c]$ with $r\leq s$, \begin{align*} |y_k(s)-y_k(r)|\leq \int_r^s |\dot{y}_k(u)|\,d\mathcal{L}^1(u)\leq M|s-r|. \end{align*} This Lipschitz estimate implies that $y_k(t)$ is Cauchy as $t\uparrow \tau$, so there exists $y_\tau\in\mathbb{R}^n$ such that \begin{align*} \lim_{t\uparrow\tau}y_k(t)=y_\tau. \end{align*} Because $x:J\to E$ is continuous and $y_k(t)\to y_\tau$ as $t\uparrow\tau$, passing to the limit in $|y_k(t)-x(t)|\leq \rho/2$ gives \begin{align*} |y_\tau-x(\tau)|\leq \frac{\rho}{2}. \end{align*} Thus $y_\tau\in E$ by the definition of the tube. Since $\tau<b$ and $J\subset \operatorname{int}_I(J_x)$, the point $\tau$ belongs to $I$ and has a right-hand neighbourhood inside $I$. We have already shown $y_\tau\in E$, so $(\tau,y_\tau)\in I\times E$. The hypotheses of [citetheorem:8375] are precisely the standing hypotheses on $I$, $E$, and $f$, so applying it to the initial condition $(\tau,y_\tau)$ gives a solution \begin{align*} z:(\tau-\delta,\tau+\delta)\cap I\to E \end{align*} for some $\delta>0$, with $z(\tau)=y_\tau$. Choose $\delta>0$ smaller, if necessary, so that $\tau+\delta\leq b$ and $[\tau,\tau+\delta)\cap I$ is nonempty to the right of $\tau$. If $\tau\notin J_k$, first extend $y_k$ continuously to the endpoint by setting $\overline{y}_k(\tau):=y_\tau$ and $\overline{y}_k(t):=y_k(t)$ for $t\in J_k$ with $t<\tau$. For every $r\in J_k$ with $r<\tau$, the integral equation for $y_k$ on $[r,t]$ and the bound $|\dot y_k|\leq M$ on the tube imply, by passing to the limit as $t\uparrow\tau$, \begin{align*} y_\tau-y_k(r)=\int_r^\tau f(s,\overline{y}_k(s))\,d\mathcal{L}^1(s). \end{align*} Indeed the integrand is bounded on a compact neighbourhood of the graph segment and converges pointwise to $f(s,\overline{y}_k(s))$. Thus $\overline{y}_k$ solves the integral equation up to the endpoint $\tau$. On the common interval $J_k\cap(\tau-\delta,\tau)$, the functions $\overline{y}_k$ and $z$ are therefore two solutions with the same value $y_\tau$ at time $\tau$. Applying uniqueness in [citetheorem:8375] to this terminal initial value, equivalently to the same equation read on the reversed time interval, gives \begin{align*} z(t)=y_k(t) \end{align*} for every $t\in J_k\cap(\tau-\delta,\tau)$. If $\tau\notin J_k$, define the interval \begin{align*} \widetilde{J}_k:=J_k\cup((\tau-\delta,\tau+\delta)\cap I) \end{align*} and define the pasted map \begin{align*} \widetilde{y}_k:\widetilde{J}_k\to E \end{align*} by $\widetilde{y}_k(t)=y_k(t)$ for $t\in J_k$ and $\widetilde{y}_k(t)=z(t)$ for $t\in((\tau-\delta,\tau+\delta)\cap I)\setminus J_k$. The agreement just proved on the overlap makes this well-defined, and the one-sided limit $y_k(t)\to y_\tau=z(\tau)$ as $t\uparrow\tau$ makes $\widetilde{y}_k$ continuous at $\tau$. On subintervals contained in $J_k$ or in $(\tau-\delta,\tau+\delta)\cap I$, the map satisfies the integral equation because $y_k$ and $z$ do. For an interval crossing $\tau$, add the integral equation for $\overline{y}_k$ from the left endpoint to $\tau$ and the integral equation for $z$ from $\tau$ to the right endpoint. Hence $\widetilde{y}_k$ is a solution on the interval $\widetilde{J}_k$, which properly extends the maximal solution $y_k$, contradicting maximality. If $\tau\in J_k$, then both $y_k$ and $z$ solve the same [initial value problem](/page/Initial%20Value%20Problem) at $(\tau,y_\tau)$ on the common relative neighbourhood $J_k\cap(\tau-\delta,\tau+\delta)\cap I$. By uniqueness in [citetheorem:8375], they agree on this overlap. Since $J_k$ is open in the relative topology of $I$, there is $\varepsilon_1>0$ such that $[t_0,\tau+\varepsilon_1]\subset J_k$. By continuity of $y_k$ and $x$, together with $|y_\tau-x(\tau)|\leq\rho/2$, there is $\varepsilon_2>0$ such that $|y_k(s)-x(s)|<\rho$ for $s\in[\tau,\tau+\varepsilon_2]$. Taking $\varepsilon:=\min\{\varepsilon_1,\varepsilon_2,b-\tau\}>0$ gives $[t_0,\tau+\varepsilon]\subset J_k$ and $|y_k(s)-x(s)|<\rho$ on that interval, contradicting the definition of $\tau$ as the supremum of $R_k$. Hence $\tau=b$, and the estimate from the previous step holds on $[t_0,b]$. [/step]

[step:Repeat the exit argument on the left endpoint of $J$] We prove that $[a,t_0]\subset J_k$. Let $c\in[a,t_0]$ be such that $[c,t_0]\subset J_k$ and \begin{align*} |y_k(s)-x(s)|\leq \rho \end{align*} for every $s\in[c,t_0]$. For every $t\in[c,t_0]$, the integral equations give \begin{align*} y_k(t)-x(t)=x_{0,k}-x_0-\int_t^{t_0}\bigl(f(s,y_k(s))-f(s,x(s))\bigr)\,d\mathcal{L}^1(s). \end{align*} Taking norms and applying the Lipschitz estimate on $\mathcal{K}$ gives \begin{align*} |y_k(t)-x(t)|\leq |x_{0,k}-x_0|+L\int_t^{t_0}|y_k(s)-x(s)|\,d\mathcal{L}^1(s). \end{align*} The backward form of Gronwall's inequality, equivalently the forward form applied to the variable $r=t_0-t$, gives \begin{align*} |y_k(t)-x(t)|\leq e^{L(t_0-t)}|x_{0,k}-x_0|\leq e^{LT}|x_{0,k}-x_0|<\frac{\rho}{2}. \end{align*} Define $L_k$ to be the set of all $c\in[a,t_0]$ such that $[c,t_0]\subset J_k$ and \begin{align*} |y_k(s)-x(s)|<\rho \end{align*} for every $s\in[c,t_0]$. This set is nonempty, and let \begin{align*} \sigma:=\inf L_k. \end{align*} If $\sigma>a$, the same bounded-derivative estimate $|\dot{y}_k|\leq M$ on the tube implies that $y_k(t)$ has a limit $y_\sigma\in\mathbb{R}^n$ as $t\downarrow\sigma$. Since $x$ is continuous and $y_k(t)\to y_\sigma$ as $t\downarrow\sigma$, passing to the limit in the strict-tube estimate gives \begin{align*} |y_\sigma-x(\sigma)|\leq \frac{\rho}{2}. \end{align*} Thus $y_\sigma\in E$ by the definition of the tube. Since $\sigma>a$ and $J\subset \operatorname{int}_I(J_x)$, the point $\sigma$ belongs to $I$ and has a left-hand neighbourhood inside $I$. Hence $(\sigma,y_\sigma)\in I\times E$. The hypotheses of [citetheorem:8375] are again the standing hypotheses on $I$, $E$, and $f$, so applying it at $(\sigma,y_\sigma)$ gives a local solution \begin{align*} w:(\sigma-\delta,\sigma+\delta)\cap I\to E \end{align*} for some $\delta>0$, with $w(\sigma)=y_\sigma$. Choose $\delta>0$ smaller, if necessary, so that $\sigma-\delta\geq a$ and $(\sigma-\delta,\sigma]\cap I$ is nonempty to the left of $\sigma$. If $\sigma\notin J_k$, first extend $y_k$ continuously to the endpoint by setting $\overline{y}_k(\sigma):=y_\sigma$ and $\overline{y}_k(t):=y_k(t)$ for $t\in J_k$ with $t>\sigma$. For every $r\in J_k$ with $r>\sigma$, the integral equation for $y_k$ on $[t,r]$ and the boundedness of $f$ on the tube imply, by passing to the limit as $t\downarrow\sigma$, \begin{align*} y_k(r)-y_\sigma=\int_\sigma^r f(s,\overline{y}_k(s))\,d\mathcal{L}^1(s). \end{align*} Thus $\overline{y}_k$ solves the integral equation up to the endpoint $\sigma$. On the common interval $J_k\cap(\sigma,\sigma+\delta)$, the functions $\overline{y}_k$ and $w$ are two solutions with the same value $y_\sigma$ at time $\sigma$, so uniqueness in [citetheorem:8375] gives \begin{align*} w(t)=y_k(t) \end{align*} for every $t\in J_k\cap(\sigma,\sigma+\delta)$. If $\sigma\notin J_k$, define \begin{align*} \widehat{J}_k:=((\sigma-\delta,\sigma+\delta)\cap I)\cup J_k \end{align*} and define \begin{align*} \widehat{y}_k:\widehat{J}_k\to E \end{align*} by $\widehat{y}_k(t)=w(t)$ for $t\in((\sigma-\delta,\sigma+\delta)\cap I)\setminus J_k$ and $\widehat{y}_k(t)=y_k(t)$ for $t\in J_k$. The agreement just proved on the overlap makes this well-defined. The one-sided limit $y_k(t)\to y_\sigma=w(\sigma)$ as $t\downarrow\sigma$ makes $\widehat{y}_k$ continuous at $\sigma$, and $\widehat{J}_k$ is an interval open in the relative topology of $I$. On the two pieces of $\widehat{J}_k$, the integral equation holds by construction. If a subinterval crosses $\sigma$, split the integral at $\sigma$, use the integral equation for $w$ on the left and for $\overline{y}_k$ on the right. Hence $\widehat{y}_k$ is a solution on $\widehat{J}_k$, so it strictly extends $y_k$, contradicting maximality. If $\sigma\in J_k$, then $y_k$ and $w$ solve the same initial value problem at $(\sigma,y_\sigma)$ on the common relative neighbourhood $J_k\cap(\sigma-\delta,\sigma+\delta)\cap I$. By uniqueness in [citetheorem:8375], they agree on this overlap. Since $J_k$ is open in the relative topology of $I$, there is $\varepsilon_1>0$ such that $[\sigma-\varepsilon_1,t_0]\subset J_k$. By continuity of $y_k$ and $x$, together with $|y_\sigma-x(\sigma)|\leq\rho/2$, there is $\varepsilon_2>0$ such that $|y_k(s)-x(s)|<\rho$ for $s\in[\sigma-\varepsilon_2,\sigma]$. Taking $\varepsilon:=\min\{\varepsilon_1,\varepsilon_2,\sigma-a\}>0$ gives $[\sigma-\varepsilon,t_0]\subset J_k$ and $|y_k(s)-x(s)|<\rho$ on that interval, contradicting the definition of $\sigma$ as the infimum of $L_k$. Hence $\sigma=a$, so $[a,t_0]\subset J_k$ and the backward estimate holds on $[a,t_0]$. [/step]

[step:Restrict the maximal solutions to $J$ and pass to the uniform limit] The previous two steps show that $J=[a,b]\subset J_k$ for every $k\geq k_0$. Define \begin{align*} x_k:J\to E, \qquad x_k(t):=y_k(t). \end{align*} Then $x_k$ is a solution on $J$ and satisfies $x_k(t_0)=x_{0,k}$. Combining the forward and backward estimates gives, for every $t\in J$, \begin{align*} |x_k(t)-x(t)|\leq e^{L|t-t_0|}|x_{0,k}-x_0|\leq e^{LT}|x_{0,k}-x_0|. \end{align*} Taking the supremum over $t\in J$ yields \begin{align*} \sup_{t\in J}|x_k(t)-x(t)|\leq e^{LT}|x_{0,k}-x_0|. \end{align*} Since $x_{0,k}\to x_0$ in $\mathbb{R}^n$, the right-hand side tends to $0$. Therefore \begin{align*} \lim_{k\to\infty}\sup_{t\in J}|x_k(t)-x(t)|=0. \end{align*} This proves continuous dependence on the initial data. [/step]