Is this statement correct?

Randy Yates wrote:

> Bernhard,
> 
> I'm not sure I follow your analogy completely. It is my belief
> that the filter can be completely stable if the proper ROC is
> chosen.
> 
> --Randy
> 

In an introduction to the explanation of Lyapunov stability, 
I found this fundamental thought (transl. from German):

If the total enery of a mechanical (or electrical) system is 
continuously dissipating, the system (linear or non-linear) 
approaches an equilibrium condition.

Based on that, I'd expect that a completely stable filter will
end with zero output, when let alone with no excitation.
Despite any earlier excitation.

If the filter requires, that the ROC is never overridden (maybe by 
any spurious/transient signal which had occurred in the past),
I would not call it stable.

In contrary, I use such a filter to create an oscillator. 
Just by relying on spurious effects which cause the filter to leave 
the ROC, it is started.
It then oscillates at a certain frequency which is defined by 
carefully selecting the position of the poles.  

If your filter becomes stable, under the condition that a certain 
ROC is kept, then the filter contains such a ROC-keeper.
Your transfer function currently doesn't.
If you would include it, this would certainly modify your transfer 
function so that the poles after that modification would be inside 
the unit circle.

 Bernhard