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Let $f:mathbb{R}timesmathbb{R}^nto mathbb{R}^n$ be a continuously differentiable function. Suppose that exists $v:mathbb{R}to[0,infty)$ continuous such that $||f(t,x)||leq v(t)||x||; forall (t,x)in mathbb{R}timesmathbb{R}^n$.
Prove that all maximal solutions for $dot{x}=f(t,x)$ are defined in $mathbb{R}$; what happens if $f$ is bounded?

I believe that the Picard–Lindelöf theorem is the way to go, but I don't know how to prove that $f$ satisfies the Lipschitz condition. In order to use it, if I understand correctly, I should prove that $v$ is bounded for every $t$ in a neighborhood of $(t_0,x_0)$ being the last a solution for the differential equation.

If $dot{x}=f(t,x)$ then $x(t)=displaystyleint f(t,x)dt$. Considering the hypothesis, $$displaystyle{||x||leqint||f(t,x)||dtleqint v(t) ||x||dt}\1leqint||f(t,x)||dtleqint v(t)dt$$

which doesn't seem useful at all....

First update

Alright, let's try it this way: The goal is to show that we always have a is defined in $mathbb{R}$ for any IVP. Suppose $(t_0,x_0)$ is a maximal solution for $dot{x}=f(t,x)$ with $x(t_0)=a$, then I should prove that exists an interval $[t_0-c,t_0+c]$ in which the solution exists.

$v$ is continuos, then for any $t_0$ we can pick $c$ and $[t_0-c,t_0+c]$ will be be bounded. Define $K=operatorname{max}{v(t):tin [t_0-c,t_0+c]}$. Then $f$ is Lipschitz for every $(t,x)in [t_0-c,t_0+c]times mathbb{R}^n$, this is, $||f(t,x)||leq v(t)||x||leq K ||x||;forall (t,x)in [t_0-c,t_0+c]timesmathbb{R}^n$

By the Picard-Lindelöf theorem, the IVP has an unique maximal solution. And $t_0,c$ were arbitrary, can I conclude that all maximal solutions are defined?

Second update

Looking about extensibility I found these notes about differential equation. The extensibility theorem and the corollary seem to be enough to prove that the solutions exists for all $mathbb{R}$.

I'll quote both here:

• 
  

  Suppose that $f$ is $C^1$ on $mathbb{R}^n$. Denote the unique solution by $x(t)$ and suppose $J:=(a,b)$ is the maximal interval of existence.

  
  
  

  Theorem 1.20 (Extensibility Theorem): For each compact set $Ksubset mathbb{R}^n$ there is a $tin J$ such that $x(t)notin K$; thus, $$lim_{tto b^-}|x(t)|=lim_{tto a^+}|x(t)|=+infty .$$

  
  
  

  Corollary 1.21: Without loss of generality, if $f:mathbb{R}^nto mathbb{R}^n$ is continuous, then the solutions to $dot{x}=f(x)$, $x(0)=x_0$ exist $forall tin mathbb{R}$.

  
  

With the corollary and the continuity of $f:mathbb{R}timesmathbb{R}^ntomathbb{R}^n$, can I affirm that all maximal solutions can be defined in $mathbb{R}$ for the equation $dot{x}=f(t,x)$?

By the Picard-Lindelöf theorem, the IVP has an unique maximal solution. And $t_0,,c$ were arbitrary, can I conclude that all maximal solutions are defined?

No, you cannot conclude this, because, as I mentioned, Picard's theorem is local.

To finish your prove you need to have
$$
x(t)=x_0+int_{t_0}^t f(tau,x)dtau,
$$
which implies that
$$
|x(t)|leq |x_0|+K|x|(t-t_0),
$$
using your notations. After this you will need a Grownwall's inequality, to find that
$$
|x(t)|leqldots,
$$
which, together with my previous remarks, finished the proof. I think you should be able to fill in the necessary details.