This afternoon, I read the proof of Banach fixed-point theorem in
Wikipedia. It’s said that
\begin{equation}
d(x^*,x_n) \le \frac{q^n}{1 - q} d(x_1,x_0).
\label{eq:inf_err}
\end{equation}
In the proofs for the lemmas, I could only find something like $x_k$
inside the brackets, but not $x^$. Thus, I *couldn’t figure out
how one can derive inequality \eqref{eq:inf_err} from an inequality
derived in the proof of Lemma 2.
\begin{equation}
d(x_m,x_n) \le \frac{q^n}{1 - q} d(x_1,x_0),
\text{ where } m > n.
\label{eq:finite_err}
\end{equation}
I googled for some notes, and found one which told me to take the
limit of the L.H.S. of inequality \eqref{eq:finite_err} as $m \to
\infty$. After looking at Corollary 2.4 in the PDF file in
footnote #2 for a while, I know what I’ve missed.
If $\left\{ p_k \right\}$
converges to $p$,
\begin{equation}
\lim_{k \to \infty} d(p_k,q) = d(p,q)
= d\left(\lim_{k \to \infty} p_k,q \right)
\label{eq:dist_limit}
\end{equation}
That’s why I wrote the previous post.
With equation \eqref{eq:dist_limit}, I can now derive
\eqref{eq:inf_err} from \eqref{eq:finite_err}.
$\because \lim\limits_{k \to \infty}
x_k = x_*$
\[
\begin{aligned}
d(x^*,x_n) =& d\left( \lim_{k \to \infty} x_k,x_n\right) \\
=& \lim_{k \to \infty} d(x_k,x_n) \\
\le& \lim_{k \to \infty} \frac{q^n}{1 - q} d(x_1,x_0) \qquad
\text{(by \eqref{eq:finite_err})} \\
=& \frac{q^n}{1 - q} d(x_1,x_0)
\end{aligned}
\]