Most of Cauchy existence theorems for a differential equation
$$\begin{equation} \textbf{x}^\prime = \textbf{f}(t,\textbf{x}) \end{equation}$$ where $t$ is a real variable and $\textbf{x}$ a vector on a real vectorial space $E$ are valid when $E$ is of finite dimension or a Banach space. This is however not true for the Peano existence theorem.

Take for $E$ the space of real sequences converging to $0$ for the norm $\Vert \textbf{x} \Vert = \sup_n \vert x_n \vert$. For all $\textbf{x} \in E$ let’s note $\textbf{y}=\textbf{f}(\textbf{x})$ the sequence $(y_n)$ defined by:
$$\begin{equation*} y_n=\sqrt{\vert x_n \vert} + \frac{1}{n+1} \end{equation*}$$ As $\lim\limits_{n \to +\infty} y_n=0$, $\textbf{y}$ belongs to $E$. As the real function $\sqrt{\vert x \vert}$ is uniformly continuous on $\mathbb{R}$, the function $\textbf{y}=\textbf{f}(\textbf{x})$ is continuous on $E$.

However, we’ll prove that the differential equation
$$\begin{equation} \textbf{x}^\prime = \textbf{f}(\textbf{x}) \label{eq:eqglobal} \end{equation}$$ has no solution in $E$ with initial condition $\textbf{x}(0)=\textbf{0}$. If $\textbf{u}$ was a solution, there would exist a sequence $\textbf{u}(t)=[u_n(t)]$ of differentiable functions in a neighborhood of $0$ satisfying:
$$\begin{equation} u_n^\prime(t)=\sqrt{\vert u_n(t) \vert} + \frac{1}{n+1} \label{eq:eqwithn} \end{equation}$$ and $u_n(0)=0$ for all $n \in \mathbb{N}$.

By direct integration it can be proved that each equation $\eqref{eq:eqwithn}$ has a unique solution satisfying $u_n(0)=0$. For all $n \in \mathbb{N}$, $u_n$ is an odd function defined on all $\mathbb{R}$ and verifying for $t \ge 0$: $u_n(t) \ge \frac{t^2}{4}$. This last inequality, valid for all $n \in \mathbb{N}$ contradicts the ability for $\textbf{u}$ to belongs to $E$. Therefore, the equation $\eqref{eq:eqglobal}$ has no solution.

This counterexample is from the French mathematician Jean Dieudonné.