## Fixed point theorem in complete metric spaces

Definition. Let $(X,d)$ be a metric space. $X$ is said to be complete if every Cauchy sequence in $X$ is convergent.

Definition. A mapping $f: X \rightarrow X$ is called contractive if there exists $0\leq k < 1$ such that

$d(f(x), f(y)) \leq kd(x,y)$ for all $x, y\in X$.

Remark. If $f: X \rightarrow X$ is a contractive, then $f$ is continuous.

Definition. A point $x^* \in X$ is called a fixed point of $f: X\rightarrow X$ if $f(x^*) = x^*$.

Theorem. Assume that $(X,d)$ is a complete metric space, and $f: X\rightarrow X$ is contractive. Then $f$ has a unique fixed point.

Proof. Take any $x_0\in X$. For any $n\geq 1$, we let $x_n = f(x_{n-1})$. Use the assumption of $f$, we have

$d(x_2, x_1) = d(f(x_1), f(x_0)) \leq kd(x_1,x_0)$

$d(x_3,x_2) = d(f(x_2),f(x_1)) \leq kd(x_2,x_1)\leq k^2d(x_1,x_0)$

………. (similarly)

$d(x_{n+1},x_n)\leq k^n d(x_1,x_0)$.

Let $m. Using triangle inequality, we get

$d(x_m,x_n) \leq d(x_m,x_{m+1}) + d(x_{m+1}, x_{m+2}) + \ldots + d(x_{n-1},x_n)$

$\leq (k^{m} + k^{m+1}+\ldots + k^{n-1})d(x_1,x_0)$

$\leq k^m(1+k+k^2+\ldots)d(x_1,x_0)$

$= k^m\dfrac{1}{1-k}d(x_1,x_0) \longrightarrow 0$

as $m\rightarrow +\infty$ since $k<1$.

This implies that $\{x_n\}_{n\geq 0}$ is a Cauchy sequence. Because $X$ is a complete metric space, we conclude that there exists $x^*$ such that $lim_{n\rightarrow \infty}x_n = x^*$.

We claim that $f(x^*) = x^*$. By the continuity of $f$ (see Remark), we find that $\lim_{n\rightarrow \infty}f(x_n) = f(x^*)$. Take any $\epsilon >0$. We choose $N$ such that

$d(x_n, x^*) < \epsilon, d(f(x_n),f(x^*)) <\epsilon$ for all $n>N$.

Now we have, keep in mind that $f(x_n) = x_{n+1}$,

$d(f(x^*), x^*) \leq d(f(x^*), f(x_n)) + d(x_{n+1}, x^*) < 2\epsilon$.

Since $\epsilon$ is arbitrary, we get $d(f(x^*), x^*) =$, thus $f(x^*) = x^*$.

For the uniqueness, we assume that $f(y) = y$ for some $y\in X$. Then

$d(x^*, y) = d(f(x^*), f(y)) \leq kd(x^*,y)$

Because $0\leq k<1$, we obtain $x^* = y$.