Is the state norm of an asymptotically stable linear system always bounded?

Question

Suppose I have the dynamical system $$x_{i+1} = A_{i+1} x_{i}$$ with state vector $x \in \mathbb{R}^{n}$. If its given that $\lim_{i \to \infty} \Vert x_{i} \Vert = 0$, does this imply $\Vert x_{i} \Vert \leq M$? How do I prove this?

Answer 1

A convergent sequence is automatically bounded, no matter how it has been generated.

Of course the bound $M$ depends on the sequence, i.e. in this case on $x_1$.

Answer 2

No. Counterexample: $A_i = i, i \leq p$, and 0 afterwards. The state increases beyond exponentially until time p, the crashes to zero. There exist systems that are convergent but not stable.

Answer 3

It seems that there may be confusion here. We have the following definitions:

• The equilibrium point $x^*$ is globally attractive if $\lim_{t\to\infty}||x(t)||=x^*$ for all initial conditions $x(t_0)$.

• The equilibrium point $x^*$ is stable if for all $\epsilon>0$, there exists an $\eta(t_0)>0$ such that $||x(t_0)-x^*||\le\eta$ implies $||x(t)-x^*||\le\epsilon$ for all $t\ge t_0$.

• The equilibrium point $x^*$ is globally asymptotically stable it is both stable and attractive.

If those properties are independent of the initial time $t_0$, they are said to be uniform; e.g. uniform (asymptotic) stability.

As shown in another answer, an equilibrium point can be attractive but not stable.

It seems, however, that you are looking for a bound on the state. In that case, this will follow from the attractivity of the process since we have

$$\lim_{t\to\infty}\prod_{\tau=t_0}^t||A(\tau)||=0$$

which means that $$\prod_{\tau=t_0}^t||A(\tau)||$$ is bounded for all $t$ by a constant $M(t_0)$.

Note, however, that there may be no bound $M$ that is independent of $t_0$ and valid for all $t_0\ge0$.