Does the asymptotic stability of a differential equation imply uniform boundedness? Consider the system of differential equations $$\dot{x}(t) = f(t,x)$$ where $x \in \mathbb{R}^n$ and $f$ is a $C^{\infty}$ function. Suppose that every trajectory which begins in the closed unit ball tends to zero as $t \rightarrow \infty$. Does it follow that there exists a ball around the origin $\mathcal{B}$ such that every trajectory that begins in the unit ball stays in $\mathcal{B}$?
My natural inclination is to think the answer ought to be yes and one should be able to show this by picking a convergent subsequence somehow. I'm having trouble making this work, however. 
Here is where I am stuck. Supposing $x_i(t)$ is a trajectory that begins from $x_i(0)$ at $t=0$ and has distance at least $i$ away from the origin at some later time; and supposing $\lim_{i \rightarrow \infty} x_i(0) = x$; then I can't see how to derive a contradiction between the fact that the trajectory beginning from $x$ approaches zero, and is consequently bounded, and the the fact that $||x_i(t_i)||_2 \geq i$ for some $t_i$. If each $t_i$ was below some $T$ a contradiction is easy to obtain, but how to deal with $t_i$'s which blow up?
Finally, this is not homework. 
 A: In fact, the forward image $x(t,{\mathcal{B}})$ need not be uniformly bounded in ${\mathbb R}^n$ for $n>1$.  We can modify any given stable vector field, say $x'=-x$, by messing with a small slice of phase space for a fixed interval of time, forcing trajectories in this slice to take large excursions. During these excursions, other trajectories can be frozen by using a smooth cutoff in time. And no trajectory ever gets messed with more than once, so ultimately every trajectory approaches the origin.
The details: Fix a smooth bump function $u:{\mathbb R}\to[0,2]$ such that 
$u(t)>0$ for $0<t<1$ and $u(t)=0$ otherwise,  and $\int_0^1 u(t)\,dt = 1$. 
Let $c(t)=\sum_{k\in\mathbb Z} u(t-2k-1)$. Then $c$ is $2$-periodic and 
 $c(t)=0$ for $t\in[2k,2k+1]$, and every solution of the system $$x'=-c(t)x$$ approaches 0 as $t\to\infty$, with $x(t+2)=e^{-1} x(t)$ for all $t$. 
Now consider a modified system of the form
$$ x_1' = -c(t)x_1+ \sum_{k=1}^\infty k\, u'(t-2k)v_k(x_2),\quad  x_j'= -c(t)x_j \quad (j>1).$$
 We choose $v_k$ to be a non-negative cutoff function supported in $[a_k,b_k]\subset (0,1)$, of the form
$$
v_k(y)= u\left(\frac{y-a_k}{b_k-a_k}\right).
$$
For $t\in[2k,2k+1]$, $x_2(t)$ is constant, and $x_1(t)-x_1(2k)=k\,u(t-2k)v_k(x_2).$
Thus, for $x_2(t)\in (a_k,b_k)$ at time $t=2k$,  $x_1(t)$ makes an excursion of order $k$, and returns to its starting value at time $t=2k+1$.  
Finally, we choose $a_k, b_k\to0$ as $k\to \infty$ with $b_k<e^{-2}a_k$.
Then for any solution of the system, $x_2(2k)\in(a_k,b_k)$ for at most one $k$, hence the solution $x(t)\to0$ as $t\to\infty$. On the other hand, solutions with $x_2(2k)\in(a_k,b_k)$ correspond to initial data with $x_2(0)\in (e^ka_k,e^kb_k)$.
Since $e^kb_k\to0$, the forward image $x(t,{\mathcal B})$ is not bounded uniformly in $t$.
