Solution of infinite dimension linear system

Suppose that ${a_n}$ and $b_n$ is decreasing sequence such that $a_0=A$, $lim_{n->\infty}a_n=0$ and $b_0=B$, $lim_{n->\infty}b_n=0$.

For fix n, we can construct n dimension linear equation system such that

\begin{eqnarray} a_n x_n+a_{n-1}x_{n-1}+\cdots+a_0x_0&=&b_0\\ a_{n}x_{n-1}+\cdots+a_1x_0&=&b_1\\ \vdots\\ a_{n}x_0&=&b_n \end{eqnarray}

then solutions are uniquely determined. If n increases to $\infty$, i want know whether the solutions are bounded. If so, how can it be proved? If not, which condition is added to satisfy the bounded solution?

• Your question does not really make sense as stated, because when $n \rightarrow \infty$, there is an infinite sum in each row. Therefore, you have to at least state in which sense your equations should hold. The only way I can see your system of equations make sense is that you construct an operator from $\ell^1$ to $\ell^\infty$. But I do not see why this operator should be invertible. Jul 13 '14 at 15:38
• @Kofi: I don't understand your objection. For each $n$, we have a perfectly well defined vector $x^{(n)} =(x^{(n)}_0,\ldots,x^{(n)}_n)$, and we can ask about the properties of this sequence as $n\to\infty$. For example, do we have a uniform bound (obviously we don't in general: at the very least, $b_n/a_n$ needs to be bounded)? Jul 13 '14 at 20:01
It seems, you want to bound $$X_n = A_n^{-1} B_n$$. Let's say we simplify, that $A_n$ and $B_n$ are independent, and then start by assuming $B_n$ is bounded and then find conditions for bounded inverse of a triangular toeplitz matrix. (I did this because I thought that bounded inverses is a nice problem that many people would have worked on already. ) Then we can google "bounded inverse of triangular toeplitz matrix". I found Inversion of Toeplitz matrices II which gives a necessary and sufficient condition for bounded inverse. There is also Uniform bounds on the 1-norm of the inverse of lower triangular Toeplitz matrices. Let us know if that was helpful.