Convergence of discrete function satisfying $|f_{k+1}(X_k)|\leq \delta\Vert f\Vert_\infty$ for $X_k$ Markov chain Let $N\subset \mathbb N$ be a finite set. Define the sequence of functions $f_{k}: N\to\mathbb R$. Let  $X_k$ be a homogeneous irreducible aperiodic Markov chain taking values in $N$ with no absorbing states. Assume that $f_{k+1}(X_k)$ satisfies the following:
\begin{align}\tag{1}
| f_{k+1}(X_k) |\leq \delta \Vert f_k\Vert_\infty
\ \ \ \text{a. s.} 
\end{align}
with $\delta\in(0,1)$.
Question: Can we know something about the convergence of $f_k$? If nothing, what kind of conditions do we need to get some kind of convergence?

Attempts. I think that $f_k\to 0$ at least pointwise. But I cannot use usual tools, because there is some randomness going on here. My stochastic process knowledge is weak, but I tried the following. Define $\tau_n:=\sum_{j=1}^n Y_j$ where $Y_j$ is the time it takes to get to see each state of $X_k$ at least once starting from where we were last in $Y_{j-1}$. Then from $(1)$ we immediately see the following:
$$\Vert f_{\tau_{n+1}}\Vert_\infty \leq \delta \Vert f_{\tau_n}\Vert_\infty$$ Clearly $f_{\tau_n}\to 0$ as $n\to\infty$ (I don't know what kind of convergence do we have here?). It seems that now I have something like: given $\varepsilon>0$, then there is some random time $\tau$ such that for all $k>\tau$ one has:
\begin{align}
\Vert f_k \Vert_\infty \leq \Vert f_{\tau}\Vert_\infty <\varepsilon
\end{align}
Can we conclude $f_k\to 0$ then? What bothers me is the random time $\tau$. Also I think that I need some assumptions on $\tau_n$ like $\tau_n<\infty$ a.s., is that true? 
I appreciate any help. Even hints like  "You need to study that part of that book.". Thanks in advance!
(This is actually part of a larger problem where I'm trying to show that the error made in an algorithm will be zero eventuallly)
 A: For simplicity, assume that the initial condition of the Markov chain is fixed, say, $X_0=0$ surely, where $0$ is a particular state in $N$. Your equality (1) is equivalent to: For all $k \in \{0, 1, 2, ...\}$ we have
$$ |f_{k+1}(X_k)|\leq \delta \max_{x \in N} |f_k(x)| \quad \mbox{(with prob 1)} \quad (1)$$
where $\delta \in (0,1)$ is a given constant. 
If the chain is irreducible and aperiodic
Since the chain is finite state, irreducible, and aperiodic, there is a (deterministic) time index $K$ such that for every $y \in N$, we have
$$ P[X_k = y] >0 \quad \forall k \geq K$$
(This can be proven from the fact that aperiodic means there is a $\tilde{K}$ such that it is possible to go from state $0$ to state $0$ in $k$ hops whenever $k \geq \tilde{K}$.) Fix $y \in N, k \geq K$.  If $f_{k+1}(y)>\delta \max_{x \in N} |f_k(x)|$ then 
$$P\left[|f_{k+1}(X_k)|> \delta \max_{x \in N}|f_k(x)|\right] \geq P[X_k=y]>0$$
which contradicts (1).  Thus, 
$$ |f_{k+1}(y) |\leq \delta \max_{x \in N} |f_k(x)| \quad \forall y \in N, \forall k \geq K$$
Taking a max of both sides over all $y \in N$ gives
$$\max_{y \in N} |f_{k+1}(y)|\leq \delta \max_{x\in N} |f_k(x)| \quad \forall k \geq K$$
and so $\lim_{k\rightarrow\infty}\max_{x \in N} |f_k(x)|= 0$.
If the chain is irreducible but not aperiodic
Here is a counter-example:  Let $N=\{0,1\}$ and consider the Markov chain that deterministically alternates between states 0 and 1 over time $k \in \{0, 1, 2, ...\}$: 
$$ X_k =  \left\{ \begin{array}{ll}
0 &\mbox{ if $k$ is even} \\
1  & \mbox{ if $k$ is odd} 
\end{array}
\right.$$
It is easy to construct functions $f_k:\{0,1\}\rightarrow\mathbb{R}$ that satisfy (1) with $\delta = 1/2$ such that 
$$ \lim_{k\rightarrow\infty}\max_{x \in \{0,1\}} |f_k(x)| = \infty$$
