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I am trying to proof the following, of which I am sure it is true. I would be pleased if you could give me hints, how to approach such a problem.

Given a finite set of matrices $\mathcal{A}=\{A_j\in\mathbb{R}^{s\times s},j=1,\ldots,J\}$. Define the set $$ \mathcal{M}=\{ (A_{i_n}\cdots A_{i_1})_n\in\mathcal{A}^\mathbb{N} \ : \ %i_j\in\{1,\ldots,J\},\ \rho(A_{i_n}\cdots A_{i_1})<1 \text{ and }\ A_{i_{n-1}}\cdots A_{i_1}\in\mathcal{M} \}. $$

Thus, $\mathcal{M}$ consists of sequences of products, all of whose entries have spectral radius then less then one.

Show that there exists $C>0$ such that for all $x\in\mathbb{R}^s$ and all sequences $(A_{i_n}\cdots A_{i_1})_{n\in\mathbb{N}}\in\mathcal{M}$

$$ \sup_{n\in\mathbb{N}} \|A_{i_n}\cdots A_{i_1}x\|<C. $$


EDIT: After more thoughts about the problem, and making some numerical experiments, the above would follow from the following generalization of Gelfands formula, which I suppose is true.

Let $(i_n)_n\in\{1,\ldots,J\}$. Then for $(B_n)_n=(A_{i_n}\cdots A_{i_1})_n$, $$ \lim_{n\rightarrow\infty} \left(\frac{\rho(B_n)}{\|B_n\|}\right)^{1/n}=1 $$

Clearly $\rho(M_n)\leq\|M_n\|$, but the other direction I could not proof yet.

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