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A matrix $A$ is known to converge such that $\lim_{k\rightarrow \infty} A^k = \bar{A} \neq 0$. We have an iteration defined as $$x(k+1) = A x(k) + B u(k), \ \ k\in \mathbb{Z}_+.$$

$\{u(k), k=0,1,..\}$ is an infinite sequence. Under what kind of sequence $\{u(k)\}$, will $x(k)$ converge as $k \rightarrow \infty$?

I suspect the sequence converges if $\{u(k)\}$ is absolutely summable and, in that case, converges to $\bar{A}x_0 + \bar{A} B \sum_{k=0}^\infty u(k)$ where $x_0 = x(0)$. However, I am not sure how to prove it. Would really appreciate any help on this.

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A sufficient condition is summability of $u$.

The solution to the difference equation is $x_k = A^k x_0 + \sum_{i=0}^k A^{k-i} B u_i$.

Since $A^k \to \overline{A}$, we see that $\|A^k\| \le M$ for some $M$ and for all $k \ge 0$. (It follows that $\|\overline{A}\| \le M$.)

We have$\sum_{i=0}^k \|A^{k-i}\| \|B\| \|u_i\| \le M \|B\| \sum_{i=0}^k \|u_i\| \le M \|B\| \sum_{i=0}^\infty \|u_i\|$, hence $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i$ exists. It follows that $x_k \to \overline{A} x_0 + x_1$.

If $u$ is summable, then $\|u_k \| \le M'$ for some $M'$ and for all $k \ge 0$. Let $ \epsilon >0$ and choose $N$ such that $2 M \|B\| \sum_{i>N} \|u_i\| < \frac{1}{2} \epsilon$. Now choose $N' \ge N$ such that for $k \ge N'$, $\|B\| M' \max(\|A^k-\overline{A}\|, \|A^{k-1}-\overline{A}\|,..., \|A^{k-N}-\overline{A}\|) < \frac{1}{2(N+1)} \epsilon$.

If $k \ge N'$, then \begin{eqnarray} \| \sum_{i=0}^k (A^{k-i} - \overline{A}) B u_i \| &\le& \| \sum_{i=0}^N (A^{k-i} - \overline{A}) B u_i \| + \| \sum_{i>N} (A^{k-i} - \overline{A}) B u_i \| \\ &\le& \|B\|M' \sum_{i=0}^N \| A^{k-i} - \overline{A} \| + 2 M \|B\| \sum_{i>N} \|u_i\| \\ &<& (N+1)\frac{1}{2(N+1)} \epsilon + \frac{1}{2} \epsilon \\ &=& \epsilon \end{eqnarray} It follows that $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i = \lim_{k \to \infty} \sum_{i=0}^k \overline{A} B u_i = \overline{A} B \sum_{i=0}^\infty u_i$, and so $\lim_{k \to \infty} x_k = \overline{A} x_0 + \overline{A} B \sum_{i=0}^\infty u_i$.

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Thanks for the proof. So, the proof shows that for arbitrarily small positive $\epsilon$, there exists an $N'$ such that the sum $\| \sum_{i=0}^k(A^{k-i} - \bar{A})B u_i\| is bounded by $\epsilon$. –  Gemmi May 31 '13 at 15:42
    
Just a query - is the sufficient condition the absolute summability or just summability of u? Its seems we assume $\sum_{i=0}^\infty\|u(i)\|$ to be bounded. On a different note, any idea about some book that can be referred to for this kind of results? –  Gemmi May 31 '13 at 15:58
    
The sufficient condition is absolute summability. Without that I'm not sure how to show $\sum_{i=0}^k A^{k-i} B u_i$ converges (I imagine it is true, since $A^k \to \overline{A}$, but would need to rethink the proof). –  copper.hat May 31 '13 at 17:14
    
The latter part of the proof is just showing that $\lim_{k \to \infty} \| \sum_{i=0}^k (A^{k-i} - \overline{A}) B u_i \| = 0$. –  copper.hat May 31 '13 at 17:16
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