[proofplan] We prove existence and uniqueness of the solution to the Cauchy problem $\dot{X}(t) = h(t, X(t))$, $X(t_0) = X_0$ by reformulating it as a fixed-point problem for an integral operator on a complete metric space. We define the operator $T$ on the space of continuous functions remaining within a ball of radius $r$ around $X_0$, verify that $T$ maps this space into itself using the boundedness hypothesis, and then show by induction that the $n$-th iterate $T^n$ has Lipschitz constant $K^n/n!$ (where $K = \|k\|_{L^1}$). Since $K^n/n! \to 0$, some iterate $T^N$ is a strict contraction, and a generalisation of the [Contraction Mapping Principle](/page/Contraction%20Mapping%20Principle) delivers a unique fixed point. [/proofplan] [step:Reformulate the ODE as a fixed-point problem for an integral operator] A [continuous](/page/Continuity) [function](/page/Function) $X: I_\tau \to E$ solves the Cauchy problem if and only if it satisfies the integral equation: \begin{align*} X(t) = X_0 + \int_{t_0}^t h(s, X(s)) \, d\mathcal{L}^1(s) \quad \text{for all } t \in I_\tau. \end{align*} Define $\tau := r \min\{1, 1/m\}$ and the [complete metric space](/page/Complete%20Metric%20Space): \begin{align*} Z := \left\{\phi \in C(I_\tau; E) : \sup_{t \in I_\tau} \|\phi(t) - X_0\|_E \le r\right\}, \end{align*} equipped with the uniform metric $d(\phi, \psi) = \sup_{t \in I_\tau} \|\phi(t) - \psi(t)\|_E$. Since $Z$ is a closed subset of the [Banach space](/page/Banach%20Space) $C(I_\tau; E)$, it is complete. Define the [integral](/page/Integral) operator: \begin{align*} T: Z &\to C(I_\tau; E) \\ [T(\phi)](t) &:= X_0 + \int_{t_0}^t h(s, \phi(s)) \, d\mathcal{L}^1(s). \end{align*} A fixed point of $T$ in $Z$ is a solution to the Cauchy problem on $I_\tau$. [guided] The equivalence between the differential equation and the integral equation is standard: if $X$ is $C^1$ and satisfies $\dot{X}(t) = h(t, X(t))$ with $X(t_0) = X_0$, then integrating both sides over $[t_0, t]$ gives the integral equation. Conversely, if $X$ is continuous and satisfies the integral equation, then $t \mapsto \int_{t_0}^t h(s, X(s)) \, d\mathcal{L}^1(s)$ has a continuous integrand, so by the [Fundamental Theorem of Calculus](/theorems/632), $X$ is $C^1$ and $\dot{X}(t) = h(t, X(t))$. We work on the space $Z$ rather than all of $C(I_\tau; E)$ because we need the graphs $(t, \phi(t))$ to remain inside the ball $B_r$ where the Lipschitz and boundedness hypotheses hold. The choice $\tau = r \min\{1, 1/m\}$ ensures two things: $|t - t_0| \le \tau \le r$ (so the time component stays in the ball), and $m\tau \le r$ (so the image under $T$ stays within distance $r$ of $X_0$). Define: \begin{align*} T: Z &\to C(I_\tau; E) \\ [T(\phi)](t) &:= X_0 + \int_{t_0}^t h(s, \phi(s)) \, d\mathcal{L}^1(s). \end{align*} A fixed point $X = T(X)$ satisfies the integral equation and hence the ODE. [/guided] [/step] [step:Verify that $T$ maps $Z$ into itself] Let $\phi \in Z$. For each $t \in I_\tau$, we have $\|\phi(t) - X_0\|_E \le r$ and $|t - t_0| \le \tau \le r$, so $(t, \phi(t)) \in B_r \subseteq U$. By the boundedness hypothesis $\|h(t, X)\|_E \le m$ on $B_r$: \begin{align*} \|[T(\phi)](t) - X_0\|_E &= \left\|\int_{t_0}^t h(s, \phi(s)) \, d\mathcal{L}^1(s)\right\|_E \\ &\le \int_{t_0}^t \|h(s, \phi(s))\|_E \, d\mathcal{L}^1(s) \\ &\le m\,|t - t_0| \le m\tau \le r. \end{align*} The last inequality holds by the definition $\tau = r\min\{1, 1/m\}$, which gives $m\tau \le r$. Therefore $T(\phi) \in Z$, confirming $T: Z \to Z$. [/step] [step:Prove by induction that $T^n$ has Lipschitz constant $K^n/n!$] Let $K := \int_{I_\tau} k(s) \, d\mathcal{L}^1(s) = \int_{t_0 - \tau}^{t_0 + \tau} k(s) \, d\mathcal{L}^1(s)$. [claim:Iterate contraction estimate] For all $n \ge 1$, all $\phi_1, \phi_2 \in Z$, and all $t \in I_\tau$: \begin{align*} \|[T^n(\phi_1)](t) - [T^n(\phi_2)](t)\|_E \le \frac{1}{n!}\left(\left|\int_{t_0}^t k(s) \, d\mathcal{L}^1(s)\right|\right)^n \|\phi_1 - \phi_2\|_{C(I_\tau; E)}. \end{align*} [/claim] [proof] **Base case ($n = 1$).** By the Lipschitz hypothesis on $h$: \begin{align*} \|[T(\phi_1)](t) - [T(\phi_2)](t)\|_E &= \left\|\int_{t_0}^t \bigl(h(s, \phi_1(s)) - h(s, \phi_2(s))\bigr) \, d\mathcal{L}^1(s)\right\|_E \\ &\le \int_{t_0}^t k(s)\,\|\phi_1(s) - \phi_2(s)\|_E \, d\mathcal{L}^1(s) \\ &\le \left(\int_{t_0}^t k(s) \, d\mathcal{L}^1(s)\right) \|\phi_1 - \phi_2\|_{C(I_\tau; E)}. \end{align*} **Inductive step.** Assume the estimate holds for $n - 1$. For $t \ge t_0$ (the case $t \le t_0$ is analogous): \begin{align*} \|[T^n(\phi_1)](t) - [T^n(\phi_2)](t)\|_E &\le \int_{t_0}^t k(s)\,\|[T^{n-1}(\phi_1)](s) - [T^{n-1}(\phi_2)](s)\|_E \, d\mathcal{L}^1(s) \\ &\le \frac{\|\phi_1 - \phi_2\|_{C(I_\tau; E)}}{(n-1)!} \int_{t_0}^t k(s)\left(\int_{t_0}^s k(\sigma) \, d\mathcal{L}^1(\sigma)\right)^{n-1} d\mathcal{L}^1(s). \end{align*} Recognise the integrand as a derivative. Let $\Psi(s) := \int_{t_0}^s k(\sigma) \, d\mathcal{L}^1(\sigma)$. Then $\Psi'(s) = k(s)$ a.e., and: \begin{align*} \int_{t_0}^t k(s)\,\Psi(s)^{n-1} \, d\mathcal{L}^1(s) = \int_{t_0}^t \Psi'(s)\,\Psi(s)^{n-1} \, d\mathcal{L}^1(s) = \frac{\Psi(t)^n}{n}. \end{align*} Substituting: \begin{align*} \|[T^n(\phi_1)](t) - [T^n(\phi_2)](t)\|_E \le \frac{1}{n!}\left(\int_{t_0}^t k(s) \, d\mathcal{L}^1(s)\right)^n \|\phi_1 - \phi_2\|_{C(I_\tau; E)}. \end{align*} [/proof] [/step] [step:Apply the generalised contraction mapping principle to obtain the unique fixed point] Taking the supremum over $t \in I_\tau$ in the iterate estimate: \begin{align*} \|T^n(\phi_1) - T^n(\phi_2)\|_{C(I_\tau; E)} \le \frac{K^n}{n!}\,\|\phi_1 - \phi_2\|_{C(I_\tau; E)}. \end{align*} Since $K^n/n! \to 0$ as $n \to \infty$, there exists $N \in \mathbb{N}$ such that $K^N/N! < 1$. The iterate $T^N: Z \to Z$ is then a strict contraction on the [complete metric space](/page/Complete%20Metric%20Space) $Z$. By a generalisation of the [Contraction Mapping Principle](/page/Contraction%20Mapping%20Principle) (if $T^N$ has a unique fixed point $X^*$, then $X^* = T^N(X^*) = T^N(T(X^*))$, so $T(X^*)$ is also a fixed point of $T^N$, hence $T(X^*) = X^*$ by uniqueness), the operator $T$ itself has a unique fixed point $X \in Z$. This fixed point satisfies: \begin{align*} X(t) = X_0 + \int_{t_0}^t h(s, X(s)) \, d\mathcal{L}^1(s) \quad \text{for all } t \in I_\tau, \end{align*} and hence $X \in C^1(I_\tau; E)$ is the unique solution to the Cauchy problem on $I_\tau$. [guided] Why does a contraction of an iterate imply a unique fixed point for $T$ itself? Suppose $T^N$ is a contraction with unique fixed point $X^*$. Then $T^N(X^*) = X^*$. Applying $T$ to both sides: $T^{N+1}(X^*) = T(X^*)$, which means $T^N(T(X^*)) = T(X^*)$. So $T(X^*)$ is also a fixed point of $T^N$. By uniqueness of the fixed point of $T^N$, we conclude $T(X^*) = X^*$. For uniqueness of the fixed point of $T$: if $\tilde{X}$ were another fixed point of $T$ in $Z$, then $T^N(\tilde{X}) = \tilde{X}$ as well, contradicting the uniqueness of the fixed point of $T^N$. Taking the supremum over $t \in I_\tau$ in the iterate estimate gives: \begin{align*} \|T^n(\phi_1) - T^n(\phi_2)\|_{C(I_\tau; E)} \le \frac{K^n}{n!}\,\|\phi_1 - \phi_2\|_{C(I_\tau; E)}. \end{align*} The ratio $K^n/n!$ is the $n$-th term of the Taylor series of $e^K$, which converges to $0$. So for some $N$, $K^N/N! < 1$, and $T^N$ is a strict contraction. The [Contraction Mapping Principle](/page/Contraction%20Mapping%20Principle) applied to $T^N$ on the complete metric space $Z$ gives a unique fixed point $X^* \in Z$, and the argument above shows $X^*$ is the unique fixed point of $T$ as well. The fixed point satisfies the integral equation, and since $h(s, X^*(s))$ is continuous in $s$, the [Fundamental Theorem of Calculus](/theorems/632) yields $X^* \in C^1(I_\tau; E)$ with $\dot{X}^*(t) = h(t, X^*(t))$ and $X^*(t_0) = X_0$. [/guided] [/step]