$P$ a $n \times n$ stochastic matrix, that is each row is a discrete probability distribution. Let $0 \leq \alpha < 1$ and define $A = \sum_{i=0}^\infty \alpha^i P^i$. Is there an upper bound on the operator norm $\|A\|_{1,\beta}$ that is independent of $n$?
I tried writing it as
$$ \|A\|_{1,\beta} \leq \sum_{i=0}^\infty \alpha^i \|P\|_{1,\beta}^{i} $$
But what can we say about $\|P\|_{1,\beta}$ of a stochastic matrix, particularly for $\beta = 1, 2$?
For a $n\times n$ stochastic matrix $P$ there is no bound of $\|P\|_{1,\beta}$ independent on $n.$ Indeed let $$p_{ij}=\begin{cases} 1 & j=1\\ 0 & 2\le j\le n \end{cases}$$ Then $$\|P\|_{1,\beta}\ge \|Pe_1\|_\beta =n^{1/\beta}$$ Actually $\|P\|_{1,\beta}=n^{1/\beta},$ as $$\displaylines{\|Px\|_\beta \le \sum_{k=1}^n|x_k|\|Pe_k\|_\beta\le \|x\|_1\max_k\|Pe_k\|_\beta \\ =\|x\|_1\|Pe_1\|_\beta=n^{1/\beta}\|x\|_1}$$ Moreover $P^k=P$ for any $k\ge 1.$ Thus $$A=I+\sum_{i=1}^\infty a^iP=I +{a\over 1-a}P$$ and $$\|A-I\|_{1,\beta}={a\over 1-a}n^{1/\beta}$$
Remark If $P$ is doubly stochastic then $$\|P\|_{1,\beta}\le \|P\|_{1,1}=1$$ and $$\|A\|_{1,\beta} \le {1\over 1-a}$$