Let $y(x,\xi)$ be the value at point $x$ of the maximal solution that satisfies the initial condition $y(0)=\xi$.
- Prove that the domain of $y(\cdot, y(s, \xi))$ is $I-s$, where $I$ is the domain of $y(\cdot ,\xi)$.
- Prove that for all $s, t$ such that $y(s, \xi)$ and $y(t+s, \xi)$ exist, then $y(t,y(s,\xi))$ also exists and $y(t,y(s,\xi))=y(t+s, \xi)$.
- If $y$ is a maximal solution and there exists $T>0$ such that $y(0)=y(t)$ and $f(y(0))\neq 0$, then $y$ is a periodic solution and not constant.
In the above problem, I'm missing only that in 3., $y$ is defined on $\mathbb R$. I asked about this problem here. The user Artem has provided some help in the comments regarding this, but I can't transform it into an answer.
I have a theorem that says that if for all $a,b\in I$ such that $y$ is defined on $(a,b)$, there exists $K$ such that $|y(x)|\leq K$ for all $x\in (a,b)$, then $y$ is defined on $I$.
How can I use the above to solve the problem? Why is $y$ bounded?
Otherwise, how I can use the 1. and 2. to prove that $y$ is defined on $\Bbb R$?
Some detail is appreciated, not because I'm lazy but because I feel like I need it.
