Complete metric space and Banach fixed point theorem(Complete space와 바나흐 고정점 정리)

This article referenced the lecture notes below.
https://nptel.ac.in/courses/108106024

  In this article, we are going to talk about the definition of complete metric space and Banach fixed point theorem, also prove the Banach fixed point theorem. So first, we need to talk about the definition of Cauchy sequence and complete metric space.

Definition of Cauchy sequence) Let $\mathcal{X}$ be a metric space and $\{p_n\}$ be a sequence in $\mathcal{X}$; Then $\{p_n\}$ is a Cauchy sequence if and only if $\forall$ $\epsilon>0$, $\exist$ $N \in \mathbb{N}$ such that

Definition of complete metric space) A metric space $\mathcal{X}$ is complete if and only if every Cauchy sequence converges to a point of $\mathcal{X}$;

  And we can show that closed subset of a complete metric space is also complete. So, we are going to prove Banach fixed point theorem. Before proving the theorem, we need to define contractive mapping.

Definition of contractive mapping) A map $P:\mathbb{X} \rightarrow \mathbb{X}$ is said to be contractive if and only if there exists a $0\le\rho<1$ such that

  Thus we say that the sequence from $x_2$ to $x_1$ gets contracted under the action of P. Now we are ready to prove Banach fixed point theorem.

Theorem) Let $\mathbb{X}$ be a complete metric space and let $S$ be a closed subset of $\mathbb{X}$; Suppose $T:S \rightarrow S$ is contractive. Then :

1. There exists a unique fixed point $x^*$ (i.e., $T(x^*)=x^*$) in $S$.

2. $x^*$ can be found by successively operating $T$ on any $x_0 \in S$.

Proof of the theorem)

  Because $S$ is a closed subset of the complete metric space $\mathbb{X}$, we can say that $S$ is also complete. For a sequence ${x_n}$ in $S$, let $x_{n+1} = T(x_{n})$ for any $n\ge0$. We first need to show that $\{x_n\}$ is a Cauchy hence $x^*$ exists. See that

  Then, for any given $r \in \mathbb{N}$, let $m=n+r;$ by triangular inequality:

Thus

  Note that as $n \rightarrow \infty$ then $d_{\mathcal{S}}(x_m, x_n) \rightarrow 0$ for arbitrary $n$ and $m$. Thus we show that $\{x_n\}$ is a Cauchy. Because $S$ is complete, $\{x_n\}$ converges to a limit $x^*\in S$.

  Now we need to show that $x^*$ is a fixed point $: T(x^*) = x^*$. See that

hence we can say that $x^*$ is fixed.

  And we show that $x^*$ is the unique fixed point. Let $\bar{x}$ be another fixed point:

  So we can conclude that $x^* = \bar{x}$.

  Hence we have proved that if there is a contraction mapping on $S$, then there exists a unique fixed point $x^*$ where we can find it by successively operating $T$ on any $x_0 \in S$; $\blacksquare$

  Finally we can define Banach space.

Definintion of Banach space) A complete normed vector space is called Banach space;