a linear operator on the space of continuous maps from $C[0,1]$ to $\mathbb R$ $X_{n+1}=f_{\omega_n}(X_n)$.
$f_k$ be finite sets of continuous mapping from $S\to S$, $S$ is a compact metric space, $\omega_0,\dots$ are iid sequence of discrete RV. $p_k=\mathbb P(\omega_i=k)$, define the operator on $C[0,1]$, $f: S\to \mathbb R$,
$T(f)(x)=\mathbb E(f(X_{n+1})| X_n=x)= \sum_{k=1}^{n} p_k (f\circ f_k)(x)$
(1)Could anyone help me to understand what is happening in the equation?
(2) I don't understand how the expectation part is equal to the summation part and how the expectation part is equal to $T(f)(x)$.
(3)It would be great if you could give me an explanation. I understand that the summation part is equal to $T(f)(x)$.
so according to that
$T(f+g)= \sum p_k(f+g)\circ f_k= \sum p_k f\circ f_k+\sum p_k g\circ f_k $.
(4)It would be also great what is the expression for  $T^n(f)(x)$ for some natural number $n$ as per as his definition of the operator?
Thanks a lot.
 A: First of all the operator $T$ acts on the function $f$ not on the value
$f(x)$. So it is appropriate to write
$Tf(x)$ or even $(Tf)(x)$ but not $T(f(x))$.
If $X_{0}=x$ then there is probability $p_{1}$ that $X_{1}=f_{1}(x)$,
probability $p_{2}$ that $X_{1}=f_{2}(x)$, probability $p_{3}$ that
$X_{1}=f_{3}(x)$, etc.
Therefore the sum is an expectation
\begin{equation*}
  \sum_{k}p_{k} f_{k}(x)=\mathbb{E}[X_{1}\mid X_{0}=x]
\end{equation*}
\begin{equation*}
  \sum_{k}p_{k} f(f_{k}(x))=\mathbb{E}[f(X_{1})\mid X_{0}=x]=Tf(x).
\end{equation*}
Now that expression is linear in $f$, as I think you have worked out
yourself. So we can make the linear operator $T$ acting on $f$.
\begin{align*}
  Tf(x)=\mathbb{E}\left(f(X_{n})\mid X_{n-1}=x\right)&=
  \sum_{k}p_{k}(f\circ f_{k})(x)\\
  T^{2}f(x)=\mathbb{E}\left(f(X_{n})\mid X_{n-2}=x\right)&=
  \sum_{k_{1}k_{2}}p_{k_{1}}p_{k_{2}}(f\circ f_{k_{1}}\circ f_{k_{2}})(x)\\
  T^{3}f(x)=\mathbb{E}\left(f(X_{n})\mid X_{n-3}=x\right)&=
  \sum_{k_{1}k_{2}k_{3}}p_{k_{1}}p_{k_{2}}p_{k_{3}}(f\circ f_{k_{1}}\circ f_{k_{2}}\circ f_{k_{3}})(x)\\
\end{align*}
That means $(Tf)(x)$, $(T^2 f)(x)$ and $(T^3 f)(x)$ respectively.
