Suppose $f(0) = 0$ and $f \in C^1(E)$ with $ E$ an open subset of $\mathbb{R}^n$ . If $x(t)$ is a solution of $x' = f(x)$ and $x(0) = x_0$ on $[0, T]$, $0 < T <\infty$. Show that there exists $L > 0$ such that $$|x(t)| ≤ |x_0|e^{Lt},\:\: on \:\:[0, T].$$ Hint: use the property that all partial derivatives of f are uniformly continuous in a compact subset of $E$, and consider the inequality that $|x(t)|$ satisfies.
I'm stuck on this problem, especially the existence of $L$. I'm not also sure about the other parts of my proof. I appreciate any help.
As, $x(t)$ is a solution of $x'=f(x)$, we have $\int_0^t{x'(s)ds}=\int_0^tf(x(s))ds$, which implies that $x(t)-x(0)=\int_0^{t}f(x(s))ds$, and so $x(t)=x_0+\int_0^{t}f(x(s))ds$ on $[0,T]$. Hence,
$$|x(t)|\leq |x_0|+\int_0^{t}|f(x(s))|ds.$$
If I can show there exists $L$ satisfying $|f(x(s))|\leq L|x(s)|$. then I can get: $$|x(t)|\leq |x_0|+L\int_0^{t}|x(s)|ds.$$
Now, $|x(t)|$ satisfies in gronwall inequality (as $|x(t)|>0$, and continuous for all $t\in[0,T]$, and so
$$|x(t)|\leq |x_0|e^{Lt}.$$