Demonstration of Peano's Existence theorem by delayed argument I am studying the proof of the Peano Theorem presented in the book A Short Course in Ordinary Differential Equations by Qingkai Kong.
The theorem is the Lemma 1.3.2 in the page 13 of the book. I have a specific doubt when "uniformly bounded" is demonstrated.
Consider the IVP, $x'=f(t,x)$, $x(t_0)=x_0$, where $f\in C(G,\mathbb{R}^n)$ with $G=\{ (t,x):~|t-t_0|\leq a,~|x-x_0|\leq b \}$. Let $x_r(t)$ be defined by
$$x_r(t) = \begin{cases}
    x_0 & \quad t_0-r<t\leq t_0 \\
    x_0 + \int_{t_0}^{t} f(s,x_r(s-r))~ds & \quad t_0\leq t\leq t_0+\gamma        
\end{cases}
$$
where $0<r<\gamma=\min\{a,b/M\}$ with $M=\max_{(t,x)\in G}|f(t,x)|$.
Assume $|x_r(t)-x_0|\leq b$ holds for $t \in [t_0+(k−1)r, t_0+kr]\cap[t_0, t_0+\gamma]$. Then $(t, x_r(t−r)) \in G$ and thus $|f(t, x_r(t−r))| \leq M$ for $t \in [t_0 + (k − 1)r, t_0 + kr] \cap [t_0, t_0 + \gamma]$.
When $t \in [t_0 + kr, t_0 + (k + 1)r] \cap [t_0, t_0 + \gamma]$, we have
$$|x_r(t) − x_0| = \left|\int_{t_0}^{t} f(s,x_r(s-r))~ds\right| \leq \int_{t_0}^{t} |f(s,x_r(s-r))|~ds \leq M(t − t_0) \leq M\gamma \leq b$$
It was this step that I didn't understand. How was it possible to conclude that in the interval $[t_0 + kr, t_0 + (k + 1)r] \cap [t_0, t_0 + \gamma]$, $|f(s,x_r(s-r))|\leq M$ ?
This demonstration can be found here too, but without the details.
 A: It seems like your issues will be solved once we show that $x_r$ is actually well-defined and always satisfies $|x_r-x_0|\le b$.  We will do this inductively in $t$-intervals of size $r$.
The base case i.e. the first interval we need to check is $[t_0-r,t_0]$,  but here there is no issue with the definition $x_r(t):=x_0$, so lets check the next one i.e. $$t \in [t_0,t_0+r] \cap [t_0, t_0 + \gamma].$$ Then for $t_0< s< t$, we have  $x_r(s-r) = x_0$ so $|x_0 -  x_r(s-r)|=0\le b$. Consequently $f(s,x_r(s-r))=f(s,x_0)$ is well defined, so the definition of $x_r$ makes sense for $t\in[t_0,t_0+r]\cap[t_0,t_0+\gamma]$, and
$$ |x_r(t) - x_0| \le  \int_{t_0}^t|f(s,x_0)| ds  \le M|t-t_0| \le M\gamma \le b.$$
Induction step: assume that $x_r$ has been defined for  $$ t \in [t_0 + (k-1)r, t_0 + k r] \cap [t_0, t_0 + \gamma], \quad k=0,1,2,\dots,k_0,$$
and that $|x_r(t)-x_0|\le b$ there.
We need to define $x_r(t)$  for
$$ t \in [t_0 + k_0r, t_0 + (k_0 +1)r] \cap [t_0, t_0 + \gamma].$$
If this set is empty, we're done. Else, for $t_0<s<t$,
$$ s \in (t_0 , t_0 + (k_0 + 1)r] \cap (t_0, t_0 + \gamma],$$
so we have
$$ s-r\in (t_0 -r, t_0 + k_0r] \cap [t_0-r, t_0 -r + \gamma] .$$
For each $s$ there is some $0\le k\le k_0$ such that
$$ s-r \in (t_0+(k-1)r , t_0 + kr] \cap [t_0, t_0  + \gamma]$$
Induction hypothesis gives that each $x_r(s-r)$ is already well-defined, and that $|x_r(s-r)-x_0|\le b$. By the choice of $\gamma$, $|s-t_0| \le a$. So in fact $(s,x_r(s-r))\in G$, and $f(s,x_r(s-r))$ makes sense, and by the same reasoning as the base case,
$$|x_r(t) -x_0| \le b.$$
This defines $x_r(t)$ for all $t\in[t_0-r,t_0+\gamma]$ as an element of $\{ x: |x-x_0|\le b\}$.
