Given a linear operation $f:\mathbb{R}^m\to \mathbb{R}^m$, such that for any initiation $x_0$, iteration $f^{(n)}(x_0)$ converges to $0$ as $n\to\infty$. For simplicity, we may assume $m=1$. Consider an updating rule $x_{n+1}=f(x_n)+c_n$, where $c_n\to 0$ as $n\to\infty$.
How can I conclude that $x_{n}$ converges to zero? Or what conditions should I add to make this conclusion?
I think in this setting $x_n \to 0$ is true. I consider $m=1$ (the case $m > 1$ works similar but is a bit more technical). First note that $f(x)=\alpha x$ for some $\alpha \in (-1,1)$. Fixing $x_0 \in \mathbb{R}$ we have $$ x_{n+1} = \alpha^{n+1}x_0 + \sum_{k=0}^n \alpha^kc_{n-k}. $$ Triangle inequality gives $$ |x_{n+1}| \le |\alpha|^{n+1}|x_0| + \sum_{k=0}^n |\alpha|^k |c_{n-k}|. $$ Thus, it is sufficient to prove $$ (\ast) \quad \sum_{k=0}^n |\alpha|^k |c_{n-k}| \to 0 \quad (n \to \infty). $$ As $(c_n)$ tends to $0$ it is bounded: $|c_n| \le b$ for all $n$. Let $\varepsilon > 0$. Choose $N \in \mathbb{N}$ with $$ \frac{b|\alpha|^{N+1}}{1-|\alpha|} < \varepsilon. $$ Then, choose $n_0> N$ such that $|c_{n-N}| < \varepsilon (1-|\alpha|)$ $(n \ge n_0)$. Now, for $n \ge n_0$: $$ \sum_{k=0}^n |\alpha|^k |c_{n-k}| = \sum_{k=0}^N |\alpha|^k |c_{n-k}| + \sum_{k=N+1}^n |\alpha|^k |c_{n-k}| $$ $$ \le \frac{\varepsilon(1-|\alpha|)}{1-|\alpha|} + \frac{b|\alpha|^{N+1}}{1-|\alpha|} < 2 \varepsilon. $$ As $\varepsilon$ was arbitrary we have $(\ast)$. Hope everything is correct (please check).
In case $m > 1$ the function $f$ is of the form $f(x)=Ax$ with a real $m \times m$ matrix $A$, and all eigenvalues of $A$ have absolut value $< 1$. Thus there should be some $\lambda \in [0,1)$ and some $M \ge 0$ such that $\|A^n\| \le M \lambda^n$ $(n \ge 0)$. With this inequalities a similar reasoning as in the case $m=1$ should be possible.
