In this post, I mention that Peano existence theorem is valid for finite dimensional vector spaces, but not for Banach spaces of infinite dimension. I highlight here a second property of ordinary differential equations which is valid for finite dimensional vector spaces but not for infinite dimensional Banach spaces.

We consider a Banach space $E$, a real interval $I$ with $t_0$ as right endpoint and a function $\textbf{f}$ defined on $I \times E$ with codomain $E$ having following properties:

• $\textbf{f}$ is continuous.
• $\textbf{f}$ is locally Lipschitz continuous which means that for every point $(t,\textbf{x}) \in I \times E$ we can find a neighborhood $V$ of $t$, a neighborhood $W$ of $\textbf{x}$ and a number $k>0$ such that $\Vert \textbf{f}(s,\textbf{x}_1)-\textbf{f}(s,\textbf{x}_2) \Vert \le k \Vert \textbf{x}_1 – \textbf{x}_2 \Vert$ for all $s \in V$ and all $\textbf{x}_1,\textbf{x}_2 \in W$.

With above conditions met and according to the Picard–Lindelöf theorem, the differential equation $\textbf{x}^\prime = \textbf{f}(t,\textbf{x})$ is having a unique maximal solution $\textbf{u}$ to the initial value problem $\textbf{x}(t)=\textbf{x}_0$ defined on an interval $J \subseteq I$.

If $J \neq I$ and $E$ of finite dimension, one can prove that $\lim\limits_{t \rightarrow a^+} \textbf{u}(t) = + \infty$ where $a$ is $J$ lower endpoint. This might not be the case when $E$ if of infinite dimension. Following counterexample is from the French mathematician Jean Dieudonné.

We pick up for $E$ the Banach space of real sequences converging to $0$ for the norm $\Vert \textbf{x} \Vert = \sup_n \vert x_n \vert$ and define as $\textbf{e}_n$ the sequence whose terms are all vanishing except the one of index $n$ whose value is equal to $1$. For all $n \in \mathbb{N}$, we define the map from $E$ to $E$:
$$\textbf{f}_n(\textbf{x})=[2(x_n+x_{n+1})-1]^+(\textbf{e}_{n+1}-\textbf{e}_{n})$$ with $\textbf{x}=(x_m) \in E$.
$\textbf{f}_n$ is lipschitz continuous, vanishes on the closed ball $\Vert \textbf{x} \Vert \le \frac{1}{4}$ and $\textbf{f}_n(\textbf{x})=\textbf{e}_{n+1}-\textbf{e}_n$ for $\textbf{x}=\lambda \textbf{e}_n + (1-\lambda) \textbf{e}_{n+1}$ where $\lambda$ is any real. For all integers $n > 0$, we pick up a continuous positive map $\varphi_n$ defined on the interval $[\frac{1}{n+1},\frac{1}{n}]$, vanishing at the endpoints of this interval and such that $\displaystyle \int_{\frac{1}{n+1}}^{\frac{1}{n}} \varphi_n(t) dt=1$.

Let $I=(-\infty,1]$ and define the map:
$$\begin{array}{l|rcl} \textbf{f} : & I \times E & \longrightarrow & E\\ & (t,\textbf{x}) & \longmapsto & \textbf{0} \text{ for } t \le 0\\ & (t,\textbf{x}) & \longmapsto & \varphi_n(t)\textbf{f}_n(\textbf{x}) \text{ for } t \in [\frac{1}{n+1},\frac{1}{n}]\end{array}$$
It is clear that $\textbf{f}$ is continuous and locally Lipschitz continuous at all points $(t_0,\textbf{x}_0) \in I \times E$ for $t_0 \neq 0$. We prove that the assertion is also true at all points $(0,\textbf{a})$ with $\textbf{a}=(a_n)$. Because $\textbf{a}$ converges to zero, one can find $m \in \mathbb{N}$ such that $\vert a_n \vert \le \frac{1}{8}$ for $n \ge m$. Hence for $\textbf{x} \in E$ and $\Vert \textbf{x} – \textbf{a} \Vert \le \frac{1}{8}$ we get $\vert x_n \vert \le \frac{1}{4}$ for $n \ge m$ and therefore $\textbf{f}_n(\textbf{x})=\textbf{0}$ for all $n \ge m$. Finally, $\textbf{f}(t,\textbf{x})$ vanishes for $t \in [0,\frac{1}{m}]$ and $\Vert \textbf{x} – \textbf{a} \Vert \le \frac{1}{8}$ leading to the expected conclusion.

We now define a sequence (for $n \ge 1$) of maps $\textbf{v}_n: I\ \mapsto E$:
$$\textbf{v}_1(t) = \begin{cases} 0 & \text{ for } t < \frac{1}{2}\\ \textbf{e}_1+(\textbf{e}_2-\textbf{e}_1) \int_t^1 \varphi_1(s) ds & \text{ for } \frac{1}{2} \le t \le 1 \end{cases}$$ and for $n \ge 2$: $$\textbf{v}_n(t) = \begin{cases} 0 & \text{ for } t < \frac{1}{n+1} \text{ or } t > \frac{1}{n}\\ (\textbf{e}_{n+1}-\textbf{e}_n) \int_t^{\frac{1}{n}} \varphi_n(s) ds & \text{ for } \frac{1}{n+1} \le t \le \frac{1}{n} \end{cases}$$ For $0 < t \le 1$, the series $\textbf{u}(t) = \displaystyle \sum_{n=1}^\infty \textbf{v}_n(t)$ is eventually vanishing and therefore well defined. For $\frac{1}{n+1} \le t \le \frac{1}{n}$ we have $\textbf{u}(t)=\textbf{e}_n+(\textbf{e}_{n+1}-\textbf{e}_n) \int_t^{\frac{1}{n}} \varphi_n(s) ds$ which implies that $\Vert \textbf{u}(t) \Vert \le 1$ for $0 < t \le 1$, and is differentiable in this interval. Considering the definition of $\textbf{f}$ we get $\textbf{u}^\prime(t)=-\textbf{f}(t,\textbf{u}(t))$. We also notice that $\textbf{u}(1/n)=\textbf{e}_n$ for all positive integers. Hence, $\textbf{u}(t)$ has no limit when $t$ tends to $0$. Consequently $\textbf{u}$ is a solution of the initial value problem $\textbf{x}^\prime(t)=-\textbf{f}(t,\textbf{x}(t))$ and $\textbf{x}(1)=\textbf{e}_1$. $J=(0,1]$ is the maximal interval of the solution $\textbf{u}$. We have proven that $J \neq I$ while $\textbf{u}(t)$ remains bounded on $J$ as desired.