Counter-example to Cauchy-Peano-Arzela theorem I was looking for a counter-example to Cauchy-Peano-Arzela theorem. I found this paper (in french) from Dieudonné.
Link to Dieudonné's paper
Take $E = c_0$ to be the space of real sequences converging to $0$, endowed with the sup norm (which makes it a Banach space).
Let $f_n(x) = \sqrt{\vert x \vert} + \frac{1}{n+1}$, and $f((x_n)_{n \in \mathbb{N}}) = (f_n(x_n))_{n \in \mathbb{N}}$.
$f$ is continuous from $E$ to itself, and consider the equation $(E) : x' = f(x)$, $x(0) = (0)_{n \in \mathbb{N}}$. Then for all $n$, the $n$-th component of a hypothetical solution $ (x_n)_{n\in \mathbb{N}}$ of $ (E)$ verifies $(E_n) : x_n'(t) = f_n(x_n(t))$, $x_n(0) = 0$. Dieudonné then says that the solutions of $(E_n)$ are unique, and verify $x_n(t) \geq \frac{t^2}{4}$ for $t\geq 0$.
I don't know how to prove those two last assertions. In fact, I find the first disturbing, since I know that without the $\frac{1}{n+1}$, the solutions would not be unique... And I think the $f_n$ don't turn locally Lipschitz when I add the $\frac{1}{n+1}$... And then, I don't know how to prove the minoration.
Could you help me with this ?
Or perhaps do you know a simpler counter-example ?
 A: You don't need to prove uniqueness, just the lower bound estimate.  But I'll prove uniqueness at the end.
Let $x$ be a solution to $(E_n)$.  Since $x$ is differentiable, using the definition of derivative at $t = 0$, we see $x(\epsilon) = \frac{1}{n+1}\epsilon + o(\epsilon)$ as $\epsilon\to 0$.  Hence for sufficiently small $\epsilon$, we know that $x(\epsilon)$ is greater than zero.  Now $\sqrt x$ is Lipschitz on any domain of the form $[\delta,\infty)$, so this means that the $x(t)$ is uniquely determined and increasing for $t \ge \epsilon$.  And on this interval
$$ \frac{d}{dt} (2\sqrt{x}) = \frac{x'}{\sqrt{x}} \ge 1 ,$$
and hence for $t \ge \epsilon$
$$ 2\sqrt{x(t)} - 2\sqrt{x(\epsilon)} \ge t-\epsilon.$$
Letting $\epsilon \to 0$, it follows that
$$ x(t) \ge \frac{t^2}4 \quad (1).$$
Now let's prove uniqueness.  Suppose that $x$ and $y$ are two solutions.  Then
$$ x' - y' = \sqrt{x} - \sqrt{y} = \frac{x-y}{\sqrt x + \sqrt y} .$$
Applying Gronwall's inequality, we obtain
$$ |x(t) - y(t)| \le |x(\epsilon) - y(\epsilon)| \exp\left(\int_{\epsilon}^t \frac{d\tau}{\sqrt{x(\tau)}+\sqrt{y(\tau)}} \right).$$
Since estimate (1) holds for both $x$ and $y$, we see that
$$ \int_{\epsilon}^t \frac{d\tau}{\sqrt{x(\tau)}+\sqrt{y(\tau)}} \le \log\left(\frac t\epsilon\right) $$
Hence
$$ |x(t) - y(t)| \le \frac{|x(\epsilon) - y(\epsilon)|t}\epsilon = \frac {o(\epsilon)t}\epsilon .$$
Letting $\epsilon \to 0$, we obtain $x(t) = y(t)$.
