ODE Teschl, global solution fails for $f(t,x) \leq M(T) +L(T) |x|^{\alpha}, \alpha >1$ 
Theorem: Suppose $U=\mathbb{R}\times \mathbb{R}^n$ and for every $T> 0 $ there are constants $M(T),L(T)$ such that $$ |f(t,x)| \leq M(T) +L(T)|x|, \text{ for } (t,x) \in[-T,T] \times \mathbb{R^n}$$ Then all solutions of the IVP for 
  $$\begin{cases} x(t) = f(t,x) \\ x(t_0) = x_0 \end{cases}$$ are defined for all $t\geq 0$
However, if it is replaced by $$ |f(t,x)| \leq M(T) +L(T)|x|^{\alpha}, \text{ for } \alpha>1$$ then above result fails.

The theorem is easily proven by Gronwall inequality. But I have no idea how to prove the problem. I want to show that there is $T>0$ such that $x(t)$ is not bounded on $[-T,T]$, using $$x(t) = x(t_0) + \int_{t_0}^t f(t,x(s))ds$$
But I cannot go further. Can anyone give some hints? (Thank you : ))

There are lots of counter examples. For example, consider the following IVP
$$ x'=x^2, x(0)=1. \tag{1}$$
For every $T>0$, if there are constants $M(t), L(t)$ such that
$$ |f(x)| \le M(T)+L(T)|x|. $$
Then one must have, for every $x\in\mathrm{R}$,
$$ |x|^2\le M(T)+L(T)|x|. \tag{2}$$
But when $|x|$ is large, (2) is not true. In fact, (1) has the solution $x(t)=\frac{1}{1-t}$ which is not a global solution.
