Let $(π,||_{\infty})$ be a metric space and $π : πβπ$ be a function mapping S into itself. $S$ is a space of bounded and Lipschitz continuous function.
For each $π\inβ$, $\tau_{n}\in T$ satisfies the Blackwell sufficient conditions, which implies the existence of unique fixed point. Let $s_{n}$ be the fixed point function of each $\tau_{n}$.
Suppose $\{\tau_{n}\}_{n=1}^{\infty}$ converges to $\tau_{0}$, with fixed point $s_{0}$.
Unfortunately, here I am not sure what type of convergence is required, but what I want to show is the following :
For any $u \in S$,
$$\lim_{k\rightarrow \infty} \tau_{k} \circ \tau_{k-1} \circ …\circ \tau_{1} (u)=s_{0}$$
The only references that I could find was about $\{s_{n}\}_{n=1}^{\infty} \rightarrow s_{0}$. I thought that my conjecture might hold because of the convergence of mappings and the continuity of output functions. There must be some $N \in β$ which guarantees $\tau_{N}$ to be a sufficiently similar contraction mapping to $\tau_{0}$, and as we keep applying $\tau_{n>N}$, the input function converges to the fixed point of $s_{0}$.
Ali left an answer, which I believe correct at the moment. Regarding $\{s_{n}\}_{n=1}^{\infty} \rightarrow s_{0}$, I leave the following theorem just in case some of you find needed; the proof is by F.F Bonsall "Lectures on Some Fixed Point Theorems of Functional Analysis".
[Theorem]
Let $S$ be a complete metric space, and let $\tau_{0}$ and $\tau_{k}$ be contraction mappings of S into itself with the same Lipschitz
constant $\beta < 1$, and with fixed points $s_{0}$ and $s_{k}$ respectively. Suppose that $lim_{k \rightarrow \infty} \tau_{k}(u) = \tau(u) $ for every $u \in S$. Then $lim_{k \rightarrow \infty} s_{k}= s_{0} $
At the end, the uniform Lipschitz constant for every $\tau_{k}$ seems to be an essential condition.
Best Answer
Define $u_{n}$ according to $u_{k+1}=\tau_{k+1}(u_{k})$. Assume, as suggested in the comments, \begin{equation*} \lim_{k\to\infty}\lVert s_{k}-s_{0}\rVert=0\qquad\text{and} \end{equation*} \begin{equation*} \exists \beta\in(0,1)\quad\text{s.t.}\quad\lVert\tau_{k}(u)-\tau_{k}(v)\rVert \le \beta \lVert u - v \rVert \quad\text{for all }k,u,v. \end{equation*} Let $\epsilon>0$ be given. Choose $\delta$ such that $\delta/(1-\beta)<\epsilon/2$ and $n$ such that $m\ge n$ implies $(1+\beta)\,\lVert s_{m}-s_{0}\rVert<\delta$. Then for $m \ge n$ we have \begin{eqnarray*} \lVert u_{m+1}- s_{0}\rVert &\le& \lVert \tau_{m+1}(u_{m})-\tau_{m+1}(s_{m+1})\rVert +\lVert s_{m+1} - s_{0}\rVert \\ &\le& \beta \lVert u_{m}-s_{m+1}\rVert +\lVert s_{m+1} - s_{0}\rVert\\ &\le& \beta \lVert u_{m}-s_{0}\rVert +(1+\beta)\lVert s_{m+1} - s_{0}\rVert. \end{eqnarray*} With $x_{n}=\lVert u_{n}-s_{0}\rVert$ and $x_{m+1}=\beta x_{m}+\delta$ for $m\ge n$, we have $x_{m}$ an upper bound for $\lVert u_{m}-s_{0}\rVert$. In fact \begin{equation*} x_{n+k}=\beta^{k}x_{n}+\frac{1-\beta^{k}}{1-\beta}\delta < \beta^{k}x_{n}+\frac{1}{2}\epsilon. \end{equation*} So just choose $k$ such that $\beta^{k}x_{n}<\epsilon/2$ and for all $m\ge n+k$ we will have $x_{m}<\epsilon$.