Let $X$ be a vector field on $\mathbb R^n$, and suppose that $\|X\|$ is bounded, where the norm is taken with respect to the Euclidean inner product. I am trying to show that $X$ has globally defined flow.
To prove this, it will suffice to find some $\epsilon > 0$, independent of $x \in \mathbb R^n$, such that the flow of $X$ can be defined on $(-\epsilon , \epsilon)$ for each $x\in \mathbb R^n$. But, I can't seem to get such an $\epsilon$ just from the boundedness hypothesis. Is there a simple proof of this fact?
I am assuming that $X$ is locally Lipschitz. Fix $x_0\in\mathbb{R}^N$ and $t_0\in\mathbb{R}$ and suppose that $x(t_0)=x_0$, where $x:(\alpha,\beta)\to\mathbb{R}^N$ satisfies $$x'(t)=X(x(t))$$
and $(\alpha,\beta)$ is the maximal interval of definition of $x$. Suppose for example that $\beta<\infty$ and remember that $$x(t)=x_0+\int_{t_0}^t X(x(\tau))d\tau$$
By using the boundedness of $X$, we can conclude that $$\|x(t)-x(s)\|\leq \|X\||t-s|$$
The last inequality implies that $x$ is uniformly continuous on $(\alpha,\beta)$. But then it is possible to extend $x$ to $\beta$, because $\beta<\infty$, which is an absurd.